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Courses/Purdue/Purdue_Chem_26100%3A_Organic_Chemistry_I_(Wenthold)/Chapter_04%3A_Stereochemistry/4.5_Racemic_Mixtures
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What do you notice about these three pictures? Count the number of left gloves and right gloves. 6 left and 6 right gloves, correct? What about this one: I count 8 right gloves, 4 left gloves. So there’s a slight excess of right gloves here. Finally, this figure: ONLY right hand gloves here. 12 right gloves, zero left gloves. Application to organic chemistry? Gloves are chiral objects. That is, they lack an internal plane of symmetry. Left gloves and right gloves are mirror images of each other, but they can’t be superimposed. In chemistry, there’s a word we have to describe a pair of non-superimposable mirror images – they’re called enantiomers. Tying it back to the drawings, we can have three types of situations. Racemic Mixture : In the first drawing, we have an equal number of left and right gloves (i.e. enantiomers). This is called a racemic mixture of enantiomers. Enantiomeric excess : In the second drawing, we have an excess of right gloves compared to left gloves. In a situtation like this we can say we have an “enantiomeric excess” of gloves, or alternatively, the mixture is “enantioenriched” in the right-hand glove. [We can also calculate the "excess" here: the mixture is 66% right and 33% left - so we have a 33% "excess" of the right-hand enantiomer]. Enantiomeric pure: In the third drawing, we have only right-hand gloves. This is said to be an “enantiomerically pure” mixture of gloves, since we have only one enantiomer present. To tie it back to chemistry, let’s say we have a solution of a chiral molecule, like 2-butanol, which can exist as either the (R)-enantiomer or the (S)-enantiomer. A solution containing equal amounts of (R)-2-butanol and (S)-2-butanol is a racemic mixture. A solution containing an excess of either the (R)-enantiomer or the (S)-enantiomer would be enantioenriched. A solution containing only the (R)-enantiomer or the (S)-enantiomer will be enantiomerically pure . Contributors James Ashenhurst ( MasterOrganicChemistry.com ) A big thanks to Agnieszka at IlluScientia for the glove drawings. 4.5.1 Enantiomeric Excess and Optical Purity Enantiomeric Excess For non-racemic mixtures of enantiomers, one enantiomer is more abundant than the other. The composition of these mixtures is described by the enantiomeric excess , which is the difference between the relative abundance of the two enantiomers. Therefore, if a mixture contains 75% of the R enantiomer and 25% S, the enantiomeric excess if 50%. Similarly, a mixture that is 95% of one enantiomer, the enantiomeric excess is 90%, etc. Enantiomeric excess is useful because it reflects the optical activity of the mixture. The standard optical rotation by the mixture (\([\alpha]_{mix}\)) is equal to the product of the standard optical rotation of the major isomer (\([\alpha]_{major}\)) and the enantiomer ic excess (\(EE\)): \[[\alpha]_{mix} = EE \times [\alpha]_{major}\] In the same way, the enantiomeric excess in a mixture can be measured if the optical rotation of the pure enantiomer is known. Diastereomeric Excess A similar approach can be used to describe mixtures of diastereomers , resulting in the diastereomeric excess . 4.5.2 Resolution of Enantiomers As noted earlier, chiral compounds synthesized from achiral starting materials and reagents are generally racemic (i.e. a 50:50 mixture of enantiomers). Separation of racemates into their component enantiomers is a process called resolution . Since enantiomers have identical physical properties, such as solubility and melting point, resolution is extremely difficult. Diastereomers, on the other hand, have different physical properties, and this fact is used to achieve resolution of racemates. Reaction of a racemate with an enantiomerically pure chiral reagent gives a mixture of diastereomers, which can be separated. Reversing the first reaction then leads to the separated enantiomers plus the recovered reagent. Many kinds of chemical and physical reactions, including salt formation, may be used to achieve the diastereomeric intermediates needed for separation. The following diagram illustrates this general principle by showing how a nut having a right-handed thread (R) could serve as a "reagent" to discriminate and separate a mixture of right- and left-handed bolts of identical size and weight. Only the two right-handed partners can interact to give a fully-threaded intermediate, so separation is fairly simple. The resolving moiety, i.e. the nut, is then removed, leaving the bolts separated into their right and left-handed forms. Chemical reactions of enantiomers are normally not so dramatically different, but a practical distinction is nevertheless possible. To learn more about chemical procedures for achieving resolution Click Here . Contributors William Reusch, Professor Emeritus ( Michigan State U. ), Virtual Textbook of Organic Chemistry Further Reading on Racemic Mixtures Cliffs Notes Racemic Mixtures: Resolving Enantiomers Web Pages Racemic Mixtures Racemic Mixtures and Enantiomeric Excess Further Reading on Enantiomeric Excess and Optical Purity Carey 4 th Edition On-Line Activity Determining Optical Purity
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Courses/Tennessee_State_University/Inorganic_Chemistry_(CHEM_5000_4200)/01%3A_Map-_Inorganic_Chemistry-I_(LibreTexts)/08%3A_Molecular_Orbitals/8.04%3A_Larger_(Polyatomic)_Molecules/8.4.03%3A_HO
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Construct SALCs and the molecular orbital diagram for H\(_2\)O. This is the first example so far that is not a linear molecule. Water is a bent molecule, and so it is important to remember that interactions of pendant ligands are dependent on their positions in space. You should consider the positions of the three atoms in water to be essentially fixed in relation to each other. The process for constructing the molecular orbital diagram for a non-linear molecule , like water, is similar to the process for linear molecules. We will walk through the steps below to construct the molecular orbital diagram of water. Preliminary Steps Step 1. Find the point group of the molecule and assign Cartesian coordinates so that z is the principal axis. The H\(_2\)O molecule is bent and its point group is \(C_{2v}\). The \(z\) axis is collinear with the principal axis, the \(C_2\) axis. There is no need to simplify this problem, as we had done for previous examples. The \(C_{2v}\) point group is simple enough. Step 2. Identify and count the pendant atoms' valence orbitals. Each of the two pendant hydrogen atoms has one valence orbital, the \(1s\). Thus, we can expect a total of two SALCs from these two atoms. Generate SALCs The SALCs for H\(_2\)O are quite simple, yet we will systematically derive the SALCs here to demonstrate the process. Step 3. Generate the \(\Gamma\)'s Use the \(C_{2v}\) character table to generate one reducible representation (\(\Gamma\)); in this case we need only one \(\Gamma\) because there is only one type of valence orbital (the \(1s\)). For each \(s\) orbital, assign a value of 1 if it remains in place during the operation or zero if it moves out of its original place. The \(\Gamma\) is given below: \[\begin{array}{|c|cccc|} \hline \bf{C_{2v}} & E & C_2 &\sigma_v (xz) & \sigma_v' (yz) \\ \hline \bf{\Gamma_{1s}} & 2 & 0 & 2 & 0 \\ \hline \end{array} \nonumber \] Step 4. Break \(\Gamma\)'s into irreducible representations for individual SALCs Reduce each \(\Gamma\) into its component irreducible representations. Using either of the processes described previously, we find that the \(\Gamma\) reduces to the two irreducible representations \(A_1\) and \(B_1\) under the \(C_{2v}\) point group. \[\begin{array}{|c|cccc|} \hline \bf{C_{2v}} & E & C_2 &\sigma_v (xz) & \sigma_v' (yz) \\ \hline \bf{\Gamma_{1s}} & \bf 2 & \bf 0 & \bf 2 & \bf 0 \\ A_{1} & 1 & 1 & 1 & 1 \\ B_{1} & 1 & -1 & 1 & -1 \\ \hline \end{array} \nonumber \] Step 5. Sketch the SALCs From the systematic process above, you have found the symmetries (the irreducible representations) of both SALCs under the \(C_{2v}\) point group. To sketch the SALC that corresponds to each irreducible representation, again we use the \(C_{2v}\) character table , and specifically the functions listed on the right side columns of the table. One \(A_1\) SALC: The \(A_1\) SALC is singly degenerate and symmetric with respect to both the principle axis (\(z\)) and the inversion center (\(i\)) (based on its Mulliken Label ). We can look at the functions in the \(C_{2v}\) character table that correspond to \(A_1\) and see that it is completely symmetric under the group (because the combination of \(x^2,y^2,z^2\) shows that it is totally symmetric). This would be the same symmetry as an \(s\) orbital on the central oxygen atom. From this information, we know that this SALC must have symmetry compatible with an \(s\) orbital on the central atom. We can also see from the character table that the \(z\) axis, and thus a \(p_z\) orbital on oxygen, also possesses \(A_1\) symmetry. This tells us that in addition to being compatible with the oxygen \(s\) orbital, it should also be compatible with the oxygen \(p_z\) orbital. From this information we can draw the \(A_1\) SALC shown in Figure \(\PageIndex{2}\). If it is not obvious how the \(A_1\) sketch in Fig. \(\PageIndex{2}\) is compatible with the \(s\) and \(p_z\) orbitals of oxygen, inspect the drawings corresponding to molecular orbitals \(2a_1\) and \(3a_1\) in Fig. \(\PageIndex{3}\). One \(B_{1}\) SALC: The Mulliken Label tells us that the \(B_{1}\) SALC is singly degenerate and antisymmetric with respect to both the principal axis and the inversion center. The function, \(x\), appearing with \(B_{1}\) in the character table tells us that this SALC has the same symmetry as the \(x\) axis, or a \(p_x\) orbital on the central oxygen atom. From this information, we know that this SALC should be compatible with a \(p_x\) orbital on the central atom, and we can draw the \(B_{1}\) SALCs shown in Figure \(\PageIndex{2}\). If it is not obvious how the \(B_1\) sketch in Fig. \(\PageIndex{2}\) is compatible with the \(p_x\) orbital of oxygen, inspect the drawing corresponding to molecular orbital \(1b_1\) in Fig. \(\PageIndex{3}\) . Not convinced about the sketches of these SALCs? You can convince yourself by putting them to the test. Try the exercise below. Exercise \(\PageIndex{1}\) Perform all operations of the \(C_{2v}\) point group on the two sketches of H SALCs shown in Figure \(\PageIndex{2}\), and convince yourself that each sketch does possess the \(A_1\) and \(B_1\) symmetries assigned to them, respectively, under the \(C_{2v}\) point group. Answer Add texts here. Do not delete this text first. Draw the MO diagram for \(H_2O\) Step 6. Combine SALCs with AO’s of like symmetry. First we must identify the valence orbitals on the central oxygen: there are four including \(2s\), \(2p_x\), \(2p_y\), and \(2p_z\). Now we identify the symmetry of each using the \(C_{2v}\) character table. The symmetry of a central \(2s\) orbital corresponds to the combination of functions \(x^2\), \(y^2\), and \(z^2\) in the character table; this is \(A_1\). The \(p_z\) orbital also corresponds to \(A_1\). And so on... The symmetries of oxygen valence orbitals are listed below. \[2s =A_1 \\ 2p_x = B_1 \\ 2p_y = B_2 \\ 2p_z = A_1 \nonumber \] Now that we have identified the symmetries of the two hydrogen SALCs and the four valence orbitals on oxygen, we know which atomic orbitals and SALCs may combine based on compatible symmetries. We also need to know the relative orbital energy levels so that we can predict the relative strength of orbital interactions. The orbital ionization energies are listed in Section 5.3 .1. With knowledge of both orbital symmetries and energies, we can construct the molecular orbital diagram. The valence orbitals of oxygen go on one side of the diagram while the hydrogen group orbitals are drawn on the opposite side. Molecular orbitals are drawn in the center column of the diagram, as shown in Figure \(\PageIndex{3}\): Example \(\PageIndex{1}\) In the previous examples shown for the molecular orbital diagrams of the bifluoride anion and carbon dioxide, we discussed differences in the understanding of those molecules from molecular orbital theory compared to Lewis structures. Water contains two lone pairs in its Lewis structure. Compare the predictions about water's lone pairs of electrons and its reactivity based on (1) the combination of elementary models (Lewis, Valence Bond, and Hybridized Orbital theories) and (2) Molecular Orbital theory. Specifically address and explain how the elementary models differ from molecular orbital theory in the following respects: Where are the lone pairs in water? Are the two lone pairs equivalent or are they different? Where are sites on the molecule that will undergo reaction with electrophiles and nucleophiles? Solution 1) Elementary models: The Lewis structure predicts that two lone pairs are (a) localized on the oxygen atom of water and that (b) both lone pairs are equivalent. The Lewis structure, combined with Valence Bond Theory, would predict that lone pairs occupy two equivalent hybridized \(sp^3\) atomic orbitals on oxygen. (c) Lewis theory would predict that the oxygen lone pairs are nucleophiles; thus an electrophile would react with the oxygen atom lone pairs. The polarized O-H bond leaves the H atoms as the most electrophilic locations on the water molecule; thus nucleophiles would react at the H atoms of water. 2) Molecular orbital theory: The molecular orbitals predict that (a) the two lone pairs of water are not equivalent, and that (b) each is distributed over the entire molecule. One lone pair is in the truly non-bonding \(1b_2\) orbital (Figure \(\PageIndex{3}\)), which is also the HOMO. There is not another truly non-bonding orbital in the molecule, but the lowest-energy \(2a_1\) orbital could be considered mostly non-bonding due to the large energy difference between it and other valence orbitals. (c) Although the \(2a_1\) orbital can be considered mostly non-bonding, we cannot expect the electrons in \(2a_1\) to react readily, as they are in the lowest-energy molecular orbital. Molecular orbital theory predicts that reactions occur at the HOMO and LUMO orbitals. The HOMO reacts with electrophiles, and in this case, the HOMO is distributed over the top and bottom faces of the molecule, and is centered on the oxygen atom. The LUMO would react with nucleophiles. The LUMO is the orbital labeled \(3a_1^*\) in Figure \(\PageIndex{3}\), and has a major lobe that is distributed more heavily over the H atoms, and from this we should predict H atoms to be the preferred site of reaction for nucleophiles. Expressing molecular orbitals in terms of \(\Psi\) The general expression for a molecular orbital, or the linear combination of atomic orbitals (LCAO), was given previously as \(\Psi=c_{a} \psi_{a}+c_{b} \psi_{b}\). In this expression, the wavefunction of two atoms (\(\psi_a\) and \(\psi_b\)) is combined to form the wavefunction of the molecular orbital. The coefficients \(c_a\) and \(c_b\) quantify the contribution of each atomic \(\psi\) to the molecular \(\Psi\). \[\Psi=c_{a} \psi_{a}+c_{b} \psi_{b} \nonumber \] In the case of polyatomic orbitals, each LCAO is constructed of atomic orbitals from a central atom and group orbitals (SALCs) from the pendent atoms. In the case of water, the expression above could be modified to give the expression below. \[\Psi = c_{oxygen}\left(\psi_{oxygen}\right) + c_{SALCs}\left(N (\psi_{H_a} \pm \psi_{H_b})\right) \nonumber \] N represents the normalizing requirement that was mentioned previously in reference to the requirements for electron wavefunctions and probability functions. The normalizing requirement simply stated is the requirement that the probability of finding the electron in any orbital, including group orbitals, is 1. In general, the value of N for a group orbital is... \[N=\frac{1}{\sum{c_i^2} } \nonumber \] ...where \(c_i\) is the coefficient of each unique atomic orbital that contributes to the group. For the hydrogen SALCs of water, there are two atomic orbitals that contribute to equally to each SALC, and so the coefficient on each of the two orbitals is 1. This gives \(N=\left(\frac{1}{\sqrt{1^2 + 1^2}}\right) = \frac{1}{\sqrt{2}}\) for the H SALCs of water. There are only two hydrogen SALCs: the one in which the hydrogen wavefunctions are added (\(N(\psi_{H_a} + \psi_{H_b})\)) has \(A_1\) symmetry, and the one in which the hydrogen wavefunctions are subtracted (\(N(\psi_{H_a} - \psi_{H_b})\)) has \(B_1\) symmetry. The molecular orbitals from Figure \(\PageIndex{3}\) are expressed below in terms of their LCAO of individual wavefunctions. Yet, note that these expressions are simplifications that ignore orbital mixing. For example, as expressed below, they ignore contributions of oxygen \(2s\) to the higher energy moelcular orbitals with \(A_1\) symmetry (the \(3a_1\) and \(3a_1^*\) in Figure \(\PageIndex{3}\)). \[\begin {array}{|rcccc|l|} \hline MO & & Oxygen AO & & Hydrogen SALC & Description \\ \hline \Psi_{2a_1} & = & c_{(ox1)}\psi_{(2s)} & + & c_{(hy1)} [N(\psi_{H_a} + \psi_{H_b})] & c_{(hy1)} \text{ is positive; bonding, slightly nonbonding} \\ \Psi_{1b_1} & = & c_{(ox2)}\psi_{(2p_x)} & + & c_{(hy2)} [N(\psi_{H_a} - \psi_{H_b})] & c_{(hy2)} \text{ is positive; bonding} \\ \Psi_{3a_1} & = & c_{(ox3)}\psi_{(2p_z)} & + & c_{(hy3)} [N(\psi_{H_a} + \psi_{H_b})] & c_{(hy3)} \text{ is positive; bonding} \\ \Psi_{1b_2} & = & \psi_{(2p_y)} & & & \text{nonbonding} \\ \Psi_{3a_1}^* & = & c_{(ox4)}\psi_{(2p_z)} & + & c_{(hy4)} [N(\psi_{H_a} + \psi_{H_b})] & c_{(hy4)} \text{ is negative; antibonding} \\ \Psi_{1b_1}^* & = & c_{(ox5)}\psi_{(2p_x)} & + & c_{(hy5)} [N(\psi_{H_a} - \psi_{H_b})] & c_{(hy5)} \text{ is negative; antibonding} \\ \hline \end {array} \nonumber \]
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Courses/Mendocino_College/Introduction_to_Chemistry_(CHM_200)/09%3A_Solutions_and_Aqueous_Mixtures/9.08%3A_Properties_of_Solutions
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Learning Objectives To describe how the properties of solutions differ from those of pure solvents. Solutions are likely to have properties similar to those of their major component—usually the solvent. However, some solution properties differ significantly from those of the solvent. Here, we will focus on liquid solutions that have a solid solute, but many of the effects we will discuss in this section are applicable to all solutions. Colligative Properties Solutes affect some properties of solutions that depend only on the concentration of the dissolved particles. These properties are called colligative properties . Four important colligative properties that we will examine here are vapor pressure depression, boiling point elevation, freezing point depression, and osmotic pressure. Molecular compounds separate into individual molecules when they are dissolved, so for every 1 mol of molecules dissolved, we get 1 mol of particles. In contrast, ionic compounds separate into their constituent ions when they dissolve, so 1 mol of an ionic compound will produce more than 1 mol of dissolved particles. For example, every mole of NaCl that dissolves yields 1 mol of Na + ions and 1 mol of Cl − ions, for a total of 2 mol of particles in solution. Thus, the effect on a solution’s properties by dissolving NaCl may be twice as large as the effect of dissolving the same amount of moles of glucose (C 6 H 12 O 6 ). Vapor Pressure Depression All liquids evaporate. In fact, given enough volume, a liquid will turn completely into a vapor. If enough volume is not present, a liquid will evaporate only to the point where the rate of evaporation equals the rate of vapor condensing back into a liquid. The pressure of the vapor at this point is called the vapor pressure of the liquid. The presence of a dissolved solid lowers the characteristic vapor pressure of a liquid so that it evaporates more slowly. (The exceptions to this statement are if the solute itself is a liquid or a gas, in which case the solute will also contribute something to the evaporation process. We will not discuss such solutions here.) This property is called vapor pressure depression and is depicted in Figure \(\PageIndex{1}\). Boiling Point Elevation A related property of solutions is that their boiling points are higher than the boiling point of the pure solvent. Because the presence of solute particles decreases the vapor pressure of the liquid solvent, a higher temperature is needed to reach the boiling point. This phenomenon is called boiling point elevation. For every mole of particles dissolved in a liter of water, the boiling point of water increases by about 0.51°C. The addition of one mole of sucrose (molecular compound) in one liter of water will raise the boiling point from 100°C to 100.51°C but the addition of one mole of NaCl in one liter of water will raise the boiling point by 2 x 0.51°C = 1.02°C. Furthermore, the addition of one mole of \(\ce{CaCl2}\) in one liter of water will raise the boiling point by 3 x 0.51°C = 1.53°C. Some people argue that putting a pinch or two of salt in water used to cook spaghetti or other pasta makes a solution that has a higher boiling point, so the pasta cooks faster. In actuality, the amount of solute is so small that the boiling point of the water is practically unchanged. Freezing Point Depression The presence of solute particles has the opposite effect on the freezing point of a solution. When a solution freezes, only the solvent particles come together to form a solid phase, and the presence of solute particles interferes with that process. Therefore, for the liquid solvent to freeze, more energy must be removed from the solution, which lowers the temperature. Thus, solutions have lower freezing points than pure solvents do. This phenomenon is called freezing point depression. For every mole of particles in a liter of water, the freezing point decreases by about 1.86°C. Both boiling point elevation and freezing point depression have practical uses. For example, solutions of water and ethylene glycol (C 2 H 6 O 2 ) are used as coolants in automobile engines because the boiling point of such a solution is greater than 100°C, the normal boiling point of water. In winter, salts like NaCl and \(\ce{CaCl_2}\) are sprinkled on the ground to melt ice or keep ice from forming on roads and sidewalks (Figure \(\PageIndex{3}\)). This is because the solution made by dissolving sodium chloride or calcium chloride in water has a lower freezing point than pure water, so the formation of ice is inhibited. Comparing the Freezing and Boiling Point of Solutions Recall that covalent and ionic compounds do not dissolve in the same way. Ionic compounds break up into cations and anions when they dissolve. Covalent compounds typically do not break up. For example a sugar/water solution stays as sugar and water, with the sugar molecules staying as molecules. Remember that colligative properties are due to the number of solute particles in the solution. Adding 10 molecules of sugar to a solvent will produce 10 solute particles in the solution. When the solute is ionic, such as \(\ce{NaCl}\) however, adding 10 formulas of solute to the solution will produce 20 ions (solute particles) in the solution. Therefore, adding enough \(\ce{NaCl}\) solute to a solvent to produce a \(0.20 \: \text{m}\) solution will have twice the effect of adding enough sugar to a solvent to produce a \(0.20 \: \text{m}\) solution. Colligative properties depend on the number of solute particles in the solution. "\(i\)" is the number of particles that the solute will dissociate into upon mixing with the solvent. For example, sodium chloride, \(\ce{NaCl}\), will dissociate into two ions so for \(\ce{NaCl}\), \(i = 2\); for lithium nitrate, \(\ce{LiNO_3}\), \(i = 2\); and for calcium chloride, \(\ce{CaCl_2}\), \(i = 3\). For covalent compounds, \(i\) is always equal to 1. By knowing the molality of a solution and the number of particles a compound will dissolve to form, it is possible to predict which solution in a group will have the lowest freezing point. To compare the boiling or freezing points of solutions, follow these general steps: Label each solute as ionic or covalent. If the solute is ionic, determine the number of ions in the formula. Be careful to look for polyatomic ions. Multiply the original molality (\(\text{m}\)) of the solution by the number of particles formed when the solution dissolves. This will give you the total concentration of particles dissolved. Compare these values. The higher total concentration will result in a higher boiling point and a lower freezing point. Example \(\PageIndex{1}\) Rank the following solutions in water in order of increasing (lowest to highest) freezing point: \(0.1 \: \text{m} \: \ce{NaCl}\) \(0.1 \: \text{m} \: \ce{C_6H_{12}O_6}\) \(0.1 \: \text{m} \: \ce{CaI_2}\) Solution To compare freezing points, we need to know the total concentration of all particles when the solute has been dissolved. \(0.1 \: \text{m} \: \ce{NaCl}\): This compound is ionic (metal with nonmetal), and will dissolve into 2 parts. The total final concentration is: \(\left( 0.1 \: \text{m} \right) \left( 2 \right) = 0.2 \: \text{m}\) \(0.1 \: \text{m} \: \ce{C_6H_{12}O_6}\): This compound is covalent (nonmetal with nonmetal), and will stay as 1 part. The total final concentration is: \(\left( 0.1 \: \text{m} \right) \left(1 \right) = 0.1 \: \text{m}\) \(0.1 \: \text{m} \: \ce{CaI_2}\): This compound is ionic (metal with nonmetal), and will dissolve into 3 parts. The total final concentration is: \(\left( 0.1 \: \text{m} \right) \left( 3 \right) = 0.3 \: \text{m}\) Remember, the greater the concentration of particles, the lower the freezing point will be. \(0.1 \: \text{m} \: \ce{CaI_2}\) will have the lowest freezing point, followed by \(0.1 \: \text{m} \: \ce{NaCl}\), and the highest of the three solutions will be \(0.1 \: \text{m} \: \ce{C_6H_{12}O_6}\), but all three of them will have a lower freezing point than pure water. Example \(\PageIndex{1}\) Which solution’s freezing point deviates more from that of pure water—a 1 M solution of NaCl or a 1 M solution of \(\ce{CaCl_2}\)? Solution Colligative properties depend on the number of dissolved particles, so the solution with the greater number of particles in solution will show the greatest deviation. When NaCl dissolves, it separates into two ions, Na + and Cl − . But when \(\ce{CaCl2}\) dissolves, it separates into three ions—one Ca 2 + ion and two Cl − ions. Thus, mole for mole, \(\ce{CaCl2}\) will have 50% more impact on freezing point depression than NaCl. Exercise \(\PageIndex{1}\) Which solution’s boiling point deviates more from that of pure water—a 1 M solution of \(\ce{CaCl2}\) or a 1 M solution of MgSO 4 ? Answer \(\ce{CaCl2}\)
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Courses/Indiana_Tech/EWC%3A_CHEM_1000_-_Introductory_Chemistry_(Budhi)/14%3A_Acids_and_Bases/14.08%3A_Water_-_Acid_and_Base_in_One
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Learning Objectives Describe the autoionization of water. Calculate the concentrations of \(\ce{H3O^{+}}\) and \(\ce{OH^{−}}\) in aqueous solutions, knowing the other concentration. We have already seen that \(\ce{H2O}\) can act as an acid or a base: \[\color{blue}{\underbrace{\ce{NH3}}_{\text{base}}} + \color{red}{\underbrace{\ce{H2O}}_{\text{acid}}} \color{black} \ce{<=> NH4^{+} + OH^{−}} \nonumber \] where \(\ce{H2O}\) acts as an \(\color{red}{\text{acid}}\) (in red). \[\color{red}{\underbrace{\ce{HCl}}_{\text{acid}}} + \color{blue}{\underbrace{\ce{H2O}}_{\text{base}}} \color{black} \ce{-> H3O^{+} + Cl^{−}} \nonumber \] where \(\ce{H2O}\) acts as an \(\color{blue}{\text{base}}\) (in blue). It may not surprise you to learn, then, that within any given sample of water, some \(\ce{H2O}\) molecules are acting as acids, and other \(\ce{H2O}\) molecules are acting as bases. The chemical equation is as follows: \[\color{red}{\underbrace{\ce{H2O}}_{\text{acid}}} + \color{blue}{\underbrace{\ce{H2O}}_{\text{base}}} \color{black} \ce{<=> H3O^{+} + OH^{−}} \label{Auto} \] This occurs only to a very small degree: only about 6 in 10 8 \(\ce{H2O}\) molecules are participating in this process, which is called the autoionization of water . At this level, the concentration of both \(\ce{H3O^{+}(aq)}\) and \(\ce{OH^{−}(aq)}\) in a sample of pure \(\ce{H2O}\) is about \(1.0 \times 10^{−7}\, M\) (at room temperature). If we use square brackets—[ ]—around a dissolved species to imply the molar concentration of that species, we have \[\color{red}{\ce{[H3O^{+}]}} \color{black}{ = } \color{blue}{\ce{[OH^{-}]}} \color{black} = 1.0 \times 10^{-7} \label{eq5} \] for any sample of pure water because H 2 O can act as both an acid and a base. The product of these two concentrations is \(1.0\times 10^{−14}\): \[\color{red}{\ce{[H3O^{+}]}} \color{black}{\times} \color{blue}{\ce{[OH^{-}]}} \color{black} = (1.0 \times 10^{-7})( 1.0 \times 10^{-7}) = 1.0 \times 10^{-14} \nonumber \] For acids, the concentration of \(\ce{H3O^{+}(aq)}\) (i.e., \(\ce{[H3O^{+}]}\)) is greater than \(1.0 \times 10^{−7}\, M\). For bases the concentration of \(\ce{OH^{−}(aq)}\) (i.e., \(\ce{[OH^{−}]}\)) is greater than \(1.0 \times 10^{−7}\, M\). However, the product of the two concentrations—\(\ce{[H3O^{+}][OH^{−}]}\)—is always equal to \(1.0 \times 10^{−14}\), no matter whether the aqueous solution is an acid, a base, or neutral: \[\color{red}{\ce{[H_3O^+]}} \color{blue}{\ce{[OH^{-}]}} \color{black} = 1.0 \times 10^{-14} \nonumber \] This value of the product of concentrations is so important for aqueous solutions that it is called the autoionization constant of water and is denoted \(K_w\): \[K_w = \color{red}{\ce{[H_3O^+]}} \color{blue}{\ce{[OH^{-}]}} \color{black} = 1.0 \times 10^{-14} \label{eq10} \] This means that if you know \(\ce{[H3O^{+}]}\) for a solution, you can calculate what \(\ce{[OH^{−}]}\)) has to be for the product to equal \(1.0 \times 10^{−14}\); or if you know \(\ce{[OH^{−}]}\)), you can calculate \(\ce{[H3O^{+}]}\). This also implies that as one concentration goes up, the other must go down to compensate so that their product always equals the value of \(K_w\). Warning: Temperature Matters The degree of autoionization of water (Equation \ref{Auto})—and hence the value of \(K_w\)—changes with temperature, so Equations \ref{eq5} - \ref{eq10} are accurate only at room temperature. Example \(\PageIndex{1}\): Hydroxide Concentration What is \(\ce{[OH^{−}]}\)) of an aqueous solution if \(\ce{[H3O^{+}]}\) is \(1.0 \times 10^{−4} M\)? Solution Steps for Problem Solving Unnamed: 1 Identify the "given" information and what the problem is asking you to "find." Given: \(\ce{[H3O^{+}]} =1.0 \times 10^{−4}\, M\) Find: [OH−] = ? M List other known quantities. none Plan the problem. Using the expression for \(K_w\), (Equation \ref{eq10}), rearrange the equation algebraically to solve for [OH−]. \[\left [ \ce{OH^{-}} \right ]=\dfrac{1.0\times 10^{-14}}{\left [ H_3O^+ \right ]} \nonumber \] Calculate. Now substitute the known quantities into the equation and solve. \[\left [\ce{ OH^{-}} \right ]=\dfrac{1.0\times 10^{-14}}{1.0\times 10^{-4}}=1.0\times 10^{-10}M\nonumber \] It is assumed that the concentration unit is molarity, so \(\ce{[OH^{−}]}\) is 1.0 × 10−10 M. Think about your result. The concentration of the acid is high (> 1 x 10-7 M), so \(\ce{[OH^{−}]}\) should be low. Exercise \(\PageIndex{1}\) What is \(\ce{[OH^{−}]}\) in a 0.00032 M solution of H 2 SO 4 ? Hint Assume both protons ionize from the molecule...although this is not the case. Answer \(3.1 \times 10^{−11}\, M\) When you have a solution of a particular acid or base, you need to look at the formula of the acid or base to determine the number of H 3 O + or OH − ions in the formula unit because \(\ce{[H_3O^{+}]}\) or \(\ce{[OH^{−}]}\)) may not be the same as the concentration of the acid or base itself. Example \(\PageIndex{2}\): Hydronium Concentration What is \(\ce{[H_3O^{+}]}\) in a 0.0044 M solution of \(\ce{Ca(OH)_2}\)? Solution Steps for Problem Solving Unnamed: 1 Identify the "given" information and what the problem is asking you to "find." Given: \([\ce{Ca(OH)_2}] =0.0044 \,M\) Find: \(\ce{[H_3O^{+}]}\) = ? M List other known quantities. We begin by determining \(\ce{[OH^{−}]}\). The concentration of the solute is 0.0044 M, but because \(\ce{Ca(OH)_2}\) is a strong base, there are two OH− ions in solution for every formula unit dissolved, so the actual \(\ce{[OH^{−}]}\) is two times this: \[\ce{[OH^{−}] = 2 \times 0.0044\, M = 0.0088 \,M.} \nonumber \] Plan the problem. Use the expression for \(K_w\) (Equation \ref{eq10}) and rearrange the equation algebraically to solve for \(\ce{[H_3O^{+}]}\). \[\left [ H_3O^{+} \right ]=\dfrac{1.0\times 10^{-14}}{\left [ OH^{-} \right ]} \nonumber \] Calculate. Now substitute the known quantities into the equation and solve. \[\left [ H_3O^{+} \right ]=\dfrac{1.0\times 10^{-14}}{(0.0088)}=1.1\times 10^{-12}M \nonumber \] \(\ce{[H_3O^{+}]}\) has decreased significantly in this basic solution. Think about your result. The concentration of the base is high (> 1 x 10-7 M) so \(\ce{[H_3O^+}]}\) should be low. Exercise \(\PageIndex{2}\) What is \(\ce{[H_3O^{+}]}\) of an aqueous solution if \(\ce{[OH^{−}]}\) is \(1.0 \times 10^{−9}\, M\)? Answer 1.0 × 10 −5 M In any aqueous solution, the product of \(\ce{[H_3O^{+}]}\) and \(\ce{[OH^{−}]}\) equals \(1.0 \times 10^{−14}\) (at room temperature). Contributions & Attributions
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Courses/Sacramento_City_College/SCC%3A_CHEM_300_-_Beginning_Chemistry/SCC%3A_CHEM_300_-_Beginning_Chemistry_(Faculty)
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This is a lecture and laboratory course that covers the fundamental concepts of chemistry. This course assumes no previous knowledge of chemistry, presenting both chemical problem solving and laboratory skills. This course is intended primarily to prepare students for CHEM 400 . Front Matter 1: The Chemical World 2: Measurement and Problem Solving 3: Matter and Energy 4: Atoms and Elements 5: Molecules and Compounds 6: Chemical Composition 7: Chemical Reactions 8: Quantities in Chemical Reactions 9: Electrons in Atoms and the Periodic Table 10: Chemical Bonding 11: Gases 12: Liquids, Solids, and Intermolecular Forces 13: Solutions 14: Acids and Bases Back Matter
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Courses/Cornell_College/CC_CHM_411%3A_Advanced_Analytical_Chemistry/03%3A_Mass_Spectrometry/3.07%3A_Mass_Spectra/3.7.03%3A_Mass_Spectroscopy-_Quizzes
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0 1 2 3 1.0 Identify the molecular ion in this spectrum. Identify the molecular ion in this spectrum. NaN NaN NaN NaN NaN NaN NaN Click on image to enlarge NaN NaN NaN NaN NaN NaN NaN a) 45 NaN NaN NaN b) 46 NaN NaN NaN c) 47 NaN NaN NaN NaN NaN NaN NaN NaN NaN 2.0 For this same spectrum, choose the compound that the spectrum represents. For this same spectrum, choose the compound that the spectrum represents. NaN NaN NaN Web Reference NaN NaN NaN Click on image to enlarge NaN NaN NaN NaN NaN NaN a) formic acid NaN NaN b) 1-propanol NaN NaN c) ethanol NaN NaN d) methanol NaN NaN e) isopropyl alcohol NaN NaN NaN NaN NaN NaN NaN NaN NaN 3.0 Find the alkyl ion series in the spectrum below. (Check the hint!) Find the alkyl ion series in the spectrum below. (Check the hint!) NaN NaN NaN Web Reference NaN NaN NaN Click on image to enlarge NaN NaN NaN NaN NaN NaN a) 15, 29, 43, 57 NaN NaN b) 31, 45 NaN NaN c) none NaN NaN NaN NaN NaN NaN NaN NaN NaN 4.0 Find the alkyl LOSS ion series in the same spectra shown below. (Check the hint!) Find the alkyl LOSS ion series in the same spectra shown below. (Check the hint!) NaN NaN NaN Web Reference NaN NaN NaN Click on image to enlarge NaN NaN NaN NaN NaN NaN a) 15, 29, 43, 57 NaN NaN b) 31, 45 NaN NaN c) none NaN NaN NaN NaN NaN NaN NaN NaN NaN 5.0 For the same spectrum shown in the previous two questions, choose the compound that the spectrum represents. For the same spectrum shown in the previous two questions, choose the compound that the spectrum represents. NaN NaN NaN Web Reference NaN NaN NaN Click on image to enlarge NaN NaN NaN NaN NaN NaN a) 2-methyl-2-propanol NaN NaN b) 1-butanol NaN NaN c) 2-butanol NaN NaN d) 1-pentanol NaN NaN e) 2-methyl-1-propanol NaN NaN NaN NaN NaN NaN NaN NaN NaN 6.0 Identify the molecular ion in this spectrum. Identify the molecular ion in this spectrum. NaN NaN NaN Web Reference NaN NaN NaN Click on image to enlarge NaN NaN NaN NaN NaN NaN a) 70 NaN NaN b) 71 NaN NaN c) 87 NaN NaN d) 88 NaN NaN NaN NaN NaN NaN NaN NaN NaN 7.0 For the same spectrum, choose the compound that the spectrum represents. For the same spectrum, choose the compound that the spectrum represents. NaN NaN NaN Web Reference NaN NaN NaN Click on image to enlarge NaN NaN NaN NaN NaN NaN a) 1-hexanol NaN NaN b) 1-pentanol NaN NaN c) 2-methyl-2-butanol NaN NaN d) 2-pentanol NaN NaN e) 1-butanol NaN NaN NaN NaN NaN NaN NaN NaN NaN 8.0 Choose the compound that this spectrum represents. Choose the compound that this spectrum represents. NaN NaN NaN Web Reference NaN NaN NaN Click on image to enlarge NaN NaN NaN NaN NaN NaN a) 2-methyl-2-butanol NaN NaN b) 2-methyl-2-propanol NaN NaN c) 2-butanol NaN NaN d) 1-butanol NaN NaN e) 2-methyl-1-propanol NaN NaN NaN NaN NaN NaN NaN NaN NaN 9.0 Choose the compound that this spectrum represents. Choose the compound that this spectrum represents. NaN NaN NaN Web Reference NaN NaN NaN Click on image to enlarge NaN NaN NaN NaN NaN NaN a) 2-pentanol NaN NaN b) 2-methyl-2-propanol NaN NaN c) 1-butanol NaN NaN d) 2-methyl-1-propanol NaN NaN e) 2-butanol NaN NaN NaN NaN NaN NaN NaN NaN NaN Contributors
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Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/2_p-Block_Elements/Group_17%3A_The_Halogens/1Group_17%3A_General_Reactions/Testing_for_Halide_Ions
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This page discusses the tests for halide ions (fluoride, chloride, bromide and iodide) using silver nitrate and ammonia. Using silver nitrate solution This test is carried out in a solution of halide ions. The solution is acidified by adding dilute nitric acid. The nitric acid reacts with, and removes, other ions that might also form precipitates with silver nitrate. Silver nitrate solution is then added, and the halide can be identified from the following products: ion present observation F- no precipitate Cl- white precipitate Br- very pale cream precipitate I- very pale yellow precipitate The chloride, bromide and iodide precipitates are shown in the photograph: The chloride precipitate is easily identified, but the other two are quite similar to each other. They can only be differentiated in a side-by-side comparison. All the precipitates change color if they are exposed to light, taking on gray or purple tints. The absence of a precipitate with fluoride ions is unhelpful unless it is known that a halogen is present; otherwise, it indicates that there is no chloride, bromide, or iodide. The chemistry of the test The precipitates are insoluble silver halides: silver chloride, silver bromide or silver iodide. The formation of these is illustrated in the following equations: \[ Ag^+_{aq} + Cl^-_{(aq)} \rightarrow AgCl_{(s)} \nonumber \] \[ Ag^+_{aq} + Br^-_{(aq)} \rightarrow AgBr_{(s)} \nonumber \] \[ Ag^+_{aq} + I^-_{(aq)} \rightarrow AgI_{(s)} \nonumber \] Silver fluoride is soluble, so no precipitate is formed. \[ Ag^+_{aq} + F^-_{(aq)} \rightarrow Ag^+_{aq} + F^-_{(aq)} \nonumber \] Confirming the precipitate using ammonia solution Ammonia solution is added to the precipitates. original precipitate Observation AgCl precipitate dissolves to give a colorless solution AgBr precipitate is almost unchanged using dilute ammonia solution, but dissolves in concentrated ammonia solution to give a colorless solution AgI precipitate is insoluble in ammonia solution of any concentration There are no absolutely insoluble ionic compounds. A precipitate forms if the concentrations of the ions in solution in water exceed a certain value, unique to every compound. This value is known as the solubility product . For the silver halides, the solubility product is given by the expression: \[ K_{sp} = [Ag^+][X^-] \nonumber \] The square brackets indicate molar concentrations, with units of mol L -1 . If the product of the concentrations of ions is less than the solubility product, no precipitate is formed. If the product of the concentrations exceeds this value, a precipitate is formed. Essentially, the product of the ionic concentrations is never greater than the solubility product value. Enough solid is always precipitated to lower the ionic product to the solubility product. The table below lists solubility products from silver chloride to silver iodide (a solubility product for silver fluoride cannot be reported because it is too soluble). Unnamed: 0 Ksp (mol2dm-6) AgCl 1.8 x 10-10 AgBr 7.7 x 10-13 AgI 8.3 x 10-17 The compounds are all quite insoluble, but become even less so down the group. The purpose of ammonia The ammonia combines with silver ions to produce a complex ion called the diamminesilver(I) ion, [Ag(NH 3 ) 2 ] + . This is a reversible reaction, but the complex is very stable, and the position of equilibrium lies well to the right. The equation for this reaction is given below: A solution in contact with one of the silver halide precipitates contains a very small concentration of dissolved silver ions. The effect of adding the ammonia is to lower this concentration still further. If the adjusted silver ion concentration multiplied by the halide ion concentration is less than the solubility product, some precipitate dissolves to restore equilibrium. This occurs with silver chloride, and with silver bromide if the ammonia is concentrated. The more concentrated ammonia pushes the equilibrium even further to the right, lowering the silver ion concentration even more. The silver iodide is so insoluble that ammonia cannot lower the silver ion concentration enough for the precipitate to dissolve. An alternative test using concentrated sulfuric acid Adding concentrated sulfuric acid to a solid sample of one of the halides gives the following results: ion present observation F- steamy acidic fumes (of HF) Cl- steamy acidic fumes (of HCl) Br- steamy acidic fumes (of HBr) contaminated with brown bromine vapor I- some HI fumes with large amounts of purple iodine vapor and a red compound in the reaction vessel The only possible confusion is between a fluoride and a chloride—they behave identically under these conditions. They can be distinguished by dissolving the original solid in water and then testing with silver nitrate solution. The chloride gives a white precipitate; the fluoride produces none.
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Courses/Chabot_College/Introduction_to_General_Organic_and_Biochemistry/03%3A_Matter_and_Energy/3.09%3A_Temperature_-_Random_Motion_of_Molecules_and_Atoms
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Learning Objectives Identify the different between temperature and heat. Recognize the different scales used to measure temperature The concept of temperature may seem familiar to you, but many people confuse temperature with heat. Temperature is a measure of how hot or cold an object is relative to another object (its thermal energy content), whereas heat is the flow of thermal energy between objects with different temperatures. Temperature is a measure of the average kinetic energy of the particles in matter. In everyday usage, temperature indicates a measure of how hot or cold an object is. Temperature is an important parameter in chemistry. When a substance changes from solid to liquid, it is because there was in increase in the temperature of the material. Chemical reactions usually proceed faster if the temperature is increased. Many unstable materials (such as enzymes) will be viable longer at lower temperatures. Three different scales are commonly used to measure temperature: Fahrenheit (expressed as °F), Celsius (°C), and Kelvin (K). Thermometers measure temperature by using materials that expand or contract when heated or cooled. Mercury or alcohol thermometers, for example, have a reservoir of liquid that expands when heated and contracts when cooled, so the liquid column lengthens or shortens as the temperature of the liquid changes. The Fahrenheit Scale The first thermometers were glass and contained alcohol, which expanded and contracted as the temperature changed. The German scientist, Daniel Gabriel Fahrenheit used mercury in the tube, an idea put forth by Ismael Boulliau. The Fahrenheit scale was first developed in 1724 and tinkered with for some time after that. The main problem with this scale is the arbitrary definitions of temperature. The freezing point of water was defined as \(32^\text{o} \text{F}\) and the boiling point as \(212^\text{o} \text{F}\). The Fahrenheit scale is typically not used for scientific purposes. The Celsius Scale The Celsius scale of the metric system is named after Swedish astronomer Anders Celsius (1701-1744). The Celsius scale sets the freezing point and boiling point of water at \(0^\text{o} \text{C}\) and \(100^\text{o} \text{C}\) respectively. The distance between those two points is divided into 100 equal intervals, each of which is one degree. Another term sometimes used for the Celsius scale is "centigrade" because there are 100 degrees between the freezing and boiling points of water on this scale. However, the preferred term is "Celsius". The Kelvin Scale The Kelvin temperature scale is named after Scottish physicist and mathematician Lord Kelvin (1824-1907). It is based on molecular motion, with the temperature of \(0 \: \text{K}\), also known as absolute zero, being the point where all molecular motion ceases. The freezing point of water on the Kelvin scale is \(273.15 \: \text{K}\), while the boiling point is \(373.15 \: \text{K}\). Notice that there is no "degree" used in the temperature designation. Unlike the Fahrenheit and Celsius scales where temperatures are referred to as "degrees \(\text{F}\)" or "degrees \(\text{C}\)", we simply designate temperatures in the Kelvin scale as kelvins. Converting Between Scales The Kelvin is the same size as the Celsius degree, so measurements are easily converted from one to the other. The freezing point of water is 0°C = 273.15 K; the boiling point of water is 100°C = 373.15 K. The Kelvin and Celsius scales are related as follows: \[T \,\text{(in °C)} + 273.15 = T \, \text{(in K)} \tag{3.10.1} \label{3.10.1} \] \[T \; \text{ (in K)} − 273.15 = T \; \text{(in °C)} \tag{3.10.2} \label{3.10.2} \] Degrees on the Fahrenheit scale, however, are based on an English tradition of using 12 divisions, just as 1 ft = 12 in. The relationship between degrees Fahrenheit and degrees Celsius is as follows: where the coefficient for degrees Fahrenheit is exact. (Some calculators have a function that allows you to convert directly between °F and °C.) There is only one temperature for which the numerical value is the same on both the Fahrenheit and Celsius scales: −40°C = −40°F. The relationship between the scales is as follows: \[°C = \dfrac{(°F-32)}{1.8} \tag{3.10.3} \label{3.10.3} \] \[°F = 1.8 \times (°C)+32 \tag{3.10.4} \label{3.10.4} \] Example \(\PageIndex{1}\): Temperature Conversions A student is ill with a temperature of 103.5°F. What is her temperature in °C and K? Solution Converting from Fahrenheit to Celsius requires the use of Equation \ref{3.10.3}: \[\begin{align} °C &= \dfrac{(103.5°F - 32)}{1.8} \\ &= 39.7 \,°C \end{align} \nonumber \] Converting from Celsius to Kelvin requires the use of Equation \ref{3.10.1}: \[\begin{align} K &= 39.7 \,°C + 273.15 \\ &= 312.9\,K \end{align} \nonumber \] Exercise \(\PageIndex{1}\) Convert each temperature to °C and °F. the temperature of the surface of the sun (5800 K) the boiling point of gold (3080 K) the boiling point of liquid nitrogen (77.36 K) Answer (a) 5527 K, 9980 °F Answer (b) 2807 K, 5084 °F Answer (c) -195.79 K, -320.42 °F Summary Three different scales are commonly used to measure temperature: Fahrenheit (expressed as °F), Celsius (°C), and Kelvin (K). Contributions & Attributions
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Bookshelves/Environmental_Chemistry/Geochemistry_(Lower)/03%3A_The_Atmosphere/3.02%3A_Origin_and_Evolution_of_the_Atmosphere
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Prebiotic atmosphere The atmosphere of the Earth (and also of Venus and Mars) is generally believed to have its origin in relatively volatile compounds that were incorporated into the solids from which these planets accreted. Such compounds could include nitrides (a source of N 2 ), water (which can be taken up by silica, for example), carbides, and hydrogen compounds of nitrogen and carbon. Many of these compounds (and also some noble gases) can form clathrate complexes with water and some minerals which are fairly stable at low temperatures. The high temperatures developed during the later stages of accretion as well as subsequent heating produced by decay of radioactive elements presumably released these gases the surface. Even at the present time, large amounts of CO 2 , water vapor, N 2 , HCl, SO 2 and H 2 S are emitted from volcanos. The more reactive of these gases would be selectively removed from the atmosphere by reaction with surface rocks or dissolution in the ocean, leaving an atmosphere enriched in its present major components with the exception of oxygen which is discussed in the next section. Any hydrogen present would tend to escape into space, causing the atmosphere to gradually became less reducing. However there is now some doubt that hydrogen and other volatiles (mainly the inert gases) were present in the newly accreted planets in anything like their cosmic abundance. The main evidence against this is the observation that gases such as helium, neon and argon, which are among the ten most abundant elements in the universe, are depleted on the earth by factors of 10 –7 to 10 –11 . This implies that there was a selective removal of these volatiles prior to or during the planetary accretion process. The overall oxidation state of the earth’s mantle is not consistent with what one would expect from equilibration with highly reduced volatiles, and there is no evidence to suggest that the composition of the mantle has not remained the same. If this is correct, then the primitive atmosphere may well have had about the same composition as the gases emitted by volcanos at the present time. These consist mainly of water and CO 2 , together with small amounts of N 2 , H 2 , H 2 S, SO 2 , CO, CH 4 , NH 3 , HCl and HF. If water vapor was a major component of outgassing of the accreted earth, it must have condensed quite rapidly into rain. Any significant concentration of water vapor in the atmosphere would have led to a runaway greenhouse effect, resulting in temperatures as high as 400°C. Origin of Atmospheric Oxygen Free oxygen is never more than a trace component of most planetary atmospheres. Thermodynamically, oxygen is much happier when combined with other elements as oxides; the pressure of O 2 in equilibrium with basaltic magmas is only about 10 –7 atm. Photochemical decomposition of gaseous oxides in the upper atmosphere is the major source of O 2 on most planets. On Venus, for example, CO 2 is broken down into CO and O 2 . On the earth, the major inorganic source of O 2 is the photolysis of water vapor; most of the resulting hydrogen escapes into space, allowing the O2 concentration to build up. An estimated 2 10 11 g of O 2 per year is generated in this way. Integrated over the earth’s history, this amounts to less than 3% of the present oxygen abundance. The partial pressure of O 2 in the prebiotic atmosphere is estimated to be no more than 10 –3 atm, and may have been several orders of magnitude less. The major source of atmospheric oxygen on the earth is photosynthesis carried out by green plants and certain bacteria: \[\ce{H2O + CO2 -> (CH2O)_x + O2}\] A historical view of the buildup of atmospheric oxygen concentration since the beginning of the sedimentary record (3.7 10 9 ybp) can be worked out by making use of the fact that the carbon in the product of the above reaction has a slightly lower C 13 content than does carbon of inorganic origin. Isotopic analysis of carbon-containing sediments thus provides a measure of the amounts of photosynthetic O 2 produced at various times in the past. Figure \(\PageIndex{1}\): Evolution of atmospheric oxygen content. Note carefully that the curve plots cumulative O 2 production, but that until a few hundred million years ago, most of this was taken up by Fe(II) compounds in the crust and by reduced sulfur; only after this massive "oxygen sink" became filled did free O 2 begin to accumulate in the atmosphere. Carbon Dioxide Carbon dioxide has probably always been present in the atmosphere in the relatively small absolute amounts now observed (around 54 x 10 15 mol = 54 Pmol). The reaction of CO 2 with silicate-containing rocks to form precambrian limestones suggest a possible moderating influence on its atmospheric concentration throughout geological time. \[\ce{ CaSiO_3 + CO_2 \rightarrow CaCO_3 + SiO_2 }\] About ten percent of the atmospheric \(CO_2\) is taken up each year by photosynthesis . Of this, all except 0.05 percent is returned by respiration , almost entirely due to microorganisms. The remainder leaks into the slow part of the geochemical cycle, mostly as buried carbonate sediments. Since the advent of the industrial revolution in 1860, the amount of \(CO_2\) in the atmosphere has been increasing. Isotopic analysis shows that most of this has been due to fossil-fuel combustion; in recent years, the mass of carbon released to the atmosphere by this means has been more than ten times the estimated rate of natural removal into sediments. The large-scale destruction of tropical forests, which has accelerated greatly in recent years, is believed to exacerbate this effect by removing a temporary sink for \(CO_2\). The oceans have a large absorptive capacity for \(CO_2\) owing to its reaction with carbonate: \[ \ce{CO_2 + CO_3^{2–} \rightleftharpoons 2 HCO_3^–}\] There is about 60 times as much inorganic carbon in the oceans as in the atmosphere. However, efficient transfer takes place only into the topmost (100 m) wind-mixed layer of the ocean, which contains only about one atmosphere equivalent of \(CO_2\). Further uptake is limited by the very slow transport of water into the deep ocean, which takes around 1000 years. For this reason, the buffering effect of the oceans on atmospheric \(CO_2\) is not very effective; only about ten percent of the added \(CO_2\) is taken up by the oceans. Figure. This set of plots, taken from an ETE page with data from NASA, shows the dramatic increase in atmospheric CO 2 since the beginning of the industrial age. The "squiggles" on the Mauna Loa data reflect seasonal variations; more CO 2 is taken up during the longer-daylight periods of the summer.
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Courses/Widener_University/CHEM_175_-_General_Chemistry_I_(Van_Bramer)/04%3A_Stoichiometry_of_Chemical_Reactions/4.E%3A_Stoichiometry_of_Chemical_Reactions-_Homework
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For Chapter 4 you MUST Know Solubility rules for: Always Soluble: lithium, sodium, potassium, ammonium, acetate, and nitrate Usually Soluble with exceptions for silver (I) and lead (II) chloride, bromide, iodide sulfate with additional exceptions for calcium, strontium, and barium Usually NOT soluble with exceptions for ammonium and alkali metals carbonate phosphate hydroxides added exceptions for strontium, and barium Turn in your answers for the following questions - show your work Balance the following equations. Give the names of the reactants and the products. Given 10.0 grams of each reactant. How many grams of each product could be produced? What is the total mass of reactants and products when the reaction is complete? CH 4 (g) + O 2 (g) --> CO 2 (g) + H 2 O (g) NaCl (aq) + AgNO 3 (aq) --> AgCl (s) + NaNO 3 (aq) Solution Reaction 10.00 grams of solid potassium hydroxide is added to 250.0 mL of distilled water. Write a chemical equation that describes what happens. Calculate the moles of each product. 10.00 grams of pure hydrochloric acid is added to 250.0 mL of distilled water. Write a chemical equation that describes what happens. Calculate the moles of each product. The two solutions are mixed together. Write a chemical equation that describes what happens. Calculate the moles of each product. Calculate the molarity of each product. Solution Reaction 5.00 grams of sodium chloride is diluted to 100.0 mL with distilled water. Write a chemical equation that describes what happens, calculate the concentration of each ion in solution, calculate the mass of any solid. 5.00 grams of silver (I) nitrate is diluted to 100.0 mL with distilled water. Write a chemical equation that describes what happens, calculate the concentration of each ion in solution, and calculate the mass of any solid. The two solutions above are mixed together. Write a chemical equation that describes what happens Calculate the moles of each ion in the final solution Calculate the mass of any solid product Calculate the molarity of each ion in the final solution The Following Questions are for your practice - Do Not Turn In. They include answers so you can check your work Writing and Balancing Chemical Equations Write a balanced molecular equation describing each of the following chemical reactions. Solid calcium carbonate is heated and decomposes to solid calcium oxide and carbon dioxide gas. Gaseous butane, C 4 H 10 , reacts with diatomic oxygen gas to yield gaseous carbon dioxide and water vapor. Answer a \(\ce{CaCO3}(s)\rightarrow \ce{CaO}(s)+\ce{CO2}(g)\); Answer b \(\ce{2C4H10}(g)+\ce{13O2}(g)\rightarrow \ce{8CO2}(g)+\ce{10H2O}(g)\); From the balanced molecular equations, write the complete ionic and net ionic equations for the following: \(\ce{K2C2O4}(aq)+\ce{Ba(OH)2}(aq)\rightarrow \ce{2KOH}(aq)+\ce{BaC2O2}(s)\) answer \[\ce{2K+}(aq)+\ce{C2O4^2-}(aq)+\ce{Ba^2+}(aq)+\ce{2OH-}(aq)\rightarrow \ce{2K+}(aq)+\ce{2OH-}(aq)+\ce{BaC2O4}(s)\hspace{20px}\ce{(complete)}\] \[\ce{Ba^2+}(aq)+\ce{C2O4^2-}(aq)\rightarrow \ce{BaC2O4}(s)\hspace{20px}\ce{(net)}\] Classifying Chemical Reactions Complete and balance the equations for the following acid-base neutralization reactions. If water is used as a solvent, write the reactants and products as aqueous ions. In some cases, there may be more than one correct answer, depending on the amounts of reactants used. \(\ce{Mg(OH)2}(s)+\ce{HClO4}(aq)\rightarrow \) \(\ce{SO3}(g)+\ce{H2O}(l)\rightarrow \) (assume an excess of water and that the product dissolves) answer a \(\ce{Mg(OH)2}(s)+\ce{2HClO4}(aq)\rightarrow \ce{Mg^2+}(aq)+\ce{2ClO4-}(aq)+\ce{2H2O}(l)\) ; answer b \(\ce{SO3}(g)+\ce{2H2O}(l)\rightarrow \ce{H3O+}(aq)+\ce{HSO4-}(aq)\), (a solution of H 2 SO 4 ); Reaction Stoichiometry H 2 is produced by the reaction of 118.5 mL of a 0.8775-M solution of H 3 PO 4 according to the following equation: \(\ce{2Cr + 2H3PO4 \rightarrow 3H2 + 2CrPO4}\). Outline the steps necessary to determine the number of moles and mass of H 2 . Answer Convert mL to L Multiply L by the molarity to determine moles of H 3 PO 4 Convert moles of H 3 PO 4 to moles of H 2 Multiply moles of H 2 by the molar mass of H 2 to get the answer in grams Perform the calculations outlined. Answer 1. \(118.5\: mL\times \dfrac{1\: L}{1000\: mL} = 0.1185\: L\) 2. \(0.1185\: L \times \dfrac{0.8775\: moles\: \ce{H3PO4}}{1\: L} = 0.1040\: moles\: \ce{H3PO4}\) 3. \(0.1040\: moles\: \ce{H3PO4} \times \dfrac{3\: moles\:\ce{H_2}}{2\: moles\: \ce{H3PO4}} = 0.1560\: moles\: \ce{H2}\) 4. \(0.1560\: moles\: \ce{H2} \times \dfrac{2.02 g}{1\: mole} = 0.3151g\: \ce{H2}\) Carborundum is silicon carbide, SiC, a very hard material used as an abrasive on sandpaper and in other applications. It is prepared by the reaction of pure sand, SiO 2 , with carbon at high temperature. Carbon monoxide, CO, is the other product of this reaction. Write the balanced equation for the reaction, and calculate how much SiO 2 is required to produce 3.00 kg of SiC. Answer \(\ce{SiO2 + 3C \rightarrow SiC + 2CO}\), 4.50 kg SiO 2 Urea, CO(NH 2 ) 2 , is manufactured on a large scale for use in producing urea-formaldehyde plastics and as a fertilizer. What is the maximum mass of urea that can be manufactured from the CO 2 produced by combustion of 1.00×10 3 kg of carbon followed by the reaction? \[\ce{CO2}(g)+\ce{2NH3}(g)\rightarrow \ce{CO(NH2)2}(s)+\ce{H2O}(l)\] Answer 5.00 × 10 3 kg Reaction Yields Freon-12, CCl 2 F 2 , is prepared from CCl 4 by reaction with HF. The other product of this reaction is HCl. Outline the steps needed to determine the percent yield of a reaction that produces 12.5 g of CCl 2 F 2 from 32.9 g of CCl 4 . Freon-12 has been banned and is no longer used as a refrigerant because it catalyzes the decomposition of ozone and has a very long lifetime in the atmosphere. Determine the percent yield. Answer \(\ce{g\: CCl4\rightarrow mol\: CCl4\rightarrow mol\: CCl2F2 \rightarrow g\: CCl2F2}, \mathrm{\:percent\: yield=48.3\%}\) How many molecules of the sweetener saccharin can be prepared from 30 C atoms, 25 H atoms, 12 O atoms, 8 S atoms, and 14 N atoms? Answer Only four molecules can be made. Quantitative Chemical Analysis Titration of a 20.0-mL sample of acid rain required 1.7 mL of 0.0811 M NaOH to reach the end point. If we assume that the acidity of the rain is due to the presence of sulfuric acid, what was the concentration of sulfuric acid in this sample of rain? Answer 3.4 × 10 −3 M H 2 SO 4 A 0.025-g sample of a compound composed of boron and hydrogen, with a molecular mass of ~28 amu, burns spontaneously when exposed to air, producing 0.063 g of B 2 O 3 . What are the empirical and molecular formulas of the compound. Answer The empirical formula is BH 3 . The molecular formula is B 2 H 6 .
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Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/07%3A_Alkyl_Halides-_Nucleophilic_Substitution_and_Elimination/7.22%3A__Additional_Exercises
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SN2 7-1 Predict which compound in each pair would undergo the SN2 reaction faster. a) or b) or c) or d) or 7-2 Predict the products of these nucleophilic substitution reactions, including stereochemistry when appropriate. a) b) c) d) 7-3 Show how each compound might be synthesized using S N 2 reaction. a) b) c) 7-4 Show 2 ways to synthesize the ether below using S N 2 reaction 7-5 Of the two ways of synthesizing the compound in previous question ( 7-4 ), which one would be the most efficient? Why? Show the mechanism of the reaction as part of your explanation. 7-6 Arrange the compounds below in increasing order of nucleophilicity. H 2 O; NH 2 - ; CH 3 O - ; CH 3 CH 2 O - SN1 7-7 List the following carbocations in order of increasing stability. 7-8 Give the solvolysis product expected when the compound is heated in methanol. a) b) c) d) 7-9 Predict with compound in each pair will undergo an SN1 reaction more quickly. a) or b) or c) or d) or 7-10 Show how the following carbocations would rearrange to become more stable. Draw the mechanism of the rearrangement. a) b) c) d) SN2 vs SN1 7-11 Predict whether each compound below would be more likely to undergo a S N 2 or S N 1 reaction. a) b) c) d) 7-12 Predict the product of the following reactions. a) b) c) d) 7-13 Show how each compound may be synthesized using nucleophilic substitution reactions. a) b) c) d) e) f) g) E2 vs E1 7-14 Predict the major products of the following reactions. a) b) c) 7-15 Draw the expected major product when each of the following compounds is treated with hydroxide to give an E2 reaction. a) b) c) 7-16 Predict all the elimination products of the following reactions and label the major product. a) b) c) d) Substitution vs Elimination 7-17 Identify the function of the following reagents. The reagents will be a strong/weak nucleophile and/or a strong/weak base. a) Cl - b) NaH c) t-BuO - d) OH - e) H 2 O f) HS - g) MeOH 7-18 Identify which mechanism the following reactions would undergo. a) b) c) d) e) 7-19 Identify all the products of the following reactions and specify the major product. a) b) c) d) e) 7-20 The following reaction yields five different products. Give the mechanisms for how each is formed.
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Bookshelves/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/13%3A_Organometallic_Chemistry/13.05%3A_Bonding_between_Metal_Atoms_and_Organic_Pi_Systems/13.5.01%3A_Linear__Systems
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General Properties The π bonding orbitals of alkenes, alkynes, carbonyls, and other unsaturated compounds may overlap with dσ orbitals on metal centers. This is the classic ligand HOMO → metal LUMO interaction that we’ve beaten into the ground over the last few posts. Because of this electron donation from the π system to the metal center, coordinated π systems often act electrophilic, even if the starting alkene was nucleophilic (the Wacker oxidation is a classic example; water attacks a palladium-coordinated alkene). The π → dσ orbital interaction is central to the structure and reactivity of π-system complexes. Then again, a theme of the last three posts has been the importance of orbital interactions with the opposite sense: metal HOMO → ligand LUMO. Like CO, phosphines, and NHCs, π systems are often subject to important backbonding interactions. We’ll focus on alkenes here, but these same ideas apply to carbonyls, alkynes, and other unsaturated ligands bound through their π clouds. For alkene ligands, the relative importance of “normal” bonding and backbonding is nicely captured by the relative importance of the two resonance structures in the figure below. Resonance forms of alkene ligands. Complexes of weakly backbonding metals, such as the electronegative late metals, are best represented by the traditional dative resonance structure 1 . But complexes of strong backbonders, such as electropositive Ti(II), are often best drawn in the metallacyclopropane form 2 . Organic hardliners may have a tough time believing that 1 and 2 are truly resonance forms, but several studies—e.g. of the Kulinkovich cyclopropanation—have shown that independent synthetic routes to metallacyclopropanes and alkene complexes containing the same atoms result in the same compound. Furthermore, bond lengths and angles in the alkene change substantially upon coordination to a strongly backbonding metal. We see an elongation of the C=C bond (consistent with decreased bond order) and some pyramidalization of the alkene carbons (consistent with a change in hybridization from sp2 to sp3). A complete orbital picture of “normal” bonding and backbonding in alkenes is shown in the figure below. Normal bonding and backbonding in alkene complexes. Here’s an interesting question with stereochemical implications: what is the orientation of the alkene relative to the other ligands? From what we’ve discussed so far, we can surmise that one face of the alkene must point toward the metal center. Put differently, the bonding axis must be normal to the plane of the alkene. However, this restriction says nothing about rotation about the bonding axis, which spins the alkene ligand like a pinwheel. Is a particular orientation preferred, or can we think about the alkene as a circular smudge over time? The figure below depicts two possible orientations of the alkene ligand in a trigonal planar complex. Other orientations make less sense because they would involve inefficient orbital overlap with the metal’s orthogonal d orbitals. Which one is favored? Two limiting cases for alkene orientation in a trigonal planar complex. First of all, we need to notice that these two complexes are diastereomeric . They have different energies as a result, so one must be favored over the other. Naive steric considerations suggest that complex 4 ought to be more stable (in most complexes, steric factors dictate alkene orientation). To dig a little deeper, let’s consider any electronic factors that may influence the preferred geometry. We’ve already seen that electronic factors can overcome steric considerations when it comes to complex geometry! To begin, we need to consider the crystal field orbitals of the complex as a whole. Verify on your own that in this d10, Pt(0) complex, the crystal-field HOMOs are the dxy and dx2–y2 orbitals. Where are these orbitals located in space? In the xy-plane! Only the alkene in 3 can engage in efficient backbonding with the metal center. In cases when the metal is electron rich and/or the alkene is electron poor, complexes like 3 can sometimes be favored in spite of sterics. The thought process here is very similar to the one developed in an earlier post on geometry. However, please note that this situation is fairly rare—steric considerations often either match or dominate electronics where alkenes are concerned. Dr. Michael Evans ( Georgia Tech )
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Courses/East_Tennessee_State_University/CHEM_3110%3A_Descriptive_Inorganic_Chemistry/04%3A_Molecular_Symmetry_and_Point_Groups
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Symmetry in Chemistry Symmetry is actually a concept of mathematics and not of chemistry. However, symmetry, and the underlying mathematical theory for symmetry, group theory, are of tremendous importance in chemistry because they can be applied to many chemistry problems. For example it helps us to classify the structures of molecules and crystals, understand chemical bonding, predict vibrational spectra, and determine the optical activity of compounds. We will therefore first discuss the general foundations of symmetry and group theory, and then how they can be applied to chemical problems. Let us first find a definition for symmetry. The human brain innately recognizes symmetry and patterns and we associate symmetry with beauty, but very familiar things are not necessarily easy to define scientifically. One common definition is that symmetry is the self-similarity of an object. The more similar parts it has the more symmetric it appears. For example, we would argue that the two wings of the butterfly depicted look similar. If the left wing was very different from the right wing the butterfly would look less symmetric. In other classes you have learned different ways to describe symmetry, radial and bilateral symmetry in biology; R and S enantiomers or cis and trans isomers in organic chemistry. Group theory gives us a universal language and tools to define and describe the symmetry of any object or molecule. This TED talk explains symmetry from a mathematical and artistic perspective. Learning Objectives Identify symmetry elements and operations in molecules and objects Use flowchart to assign a molecule to a point group Recognize the components of a character table Determine whether a vibrational mode is infrared allowed Thumbnail image shows all the reflection planes in ferrocene. Image from Otterbein University symmetry gallery 4.1: Symmetry Operations and Elements The symmetry of a molecule consists of symmetry operations and symmetry elements. A symmetry operation is an operation that is performed to a molecule which leaves it indistinguishable and superimposable on the original position. Symmetry operations are performed with respect to symmetry elements (points, lines, or planes). 4.2: Point Groups In group theory, molecules or other objects can be organized into point groups based on the type and number of symmetry operations they possess. Every molecule in a point group will have all of the same symmetry operations as any other molecule in that same point group. 4.3: Introduction to Character Tables Although the method of assigning a point group to a molecule depends on some knowledge of the symmetry elements the molecule has, it does not require the consideration of all elements. This is where the character table of the point group comes into play. 4.4: Applications of Symmetry in Chemistry There are many ways that symmetry and group theory can be applied to chemical problems. Examples include determining orbital overlap in molecular orbital theory and determining spectroscopic selections rules.
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Courses/Kenyon_College/Chemistry_231_and_232_-_Kenyon_College_(Getzler_Hofferberth_and_Hunsen)/18%3A_Enols_Enolates_and_the_Aldol_Condensation%3A_ab-Unsaturated_Aldehydes_and_Ketones/18.10%3A_12-_and__14-Additions_of_Organometallic_Reagents
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One of the largest and most diverse classes of reactions is composed of nucleophilic additions to a carbonyl group. Conjugation of a double bond to a carbonyl group transmits the electrophilic character of the carbonyl carbon to the beta-carbon of the double bond. These conjugated carbonyl are called enones or α, β unsaturated carbonyls. A resonance description of this transmission is shown below. From this formula it should be clear that nucleophiles may attack either at the carbonyl carbon, as for any aldehyde, ketone or carboxylic acid derivative, or at the beta-carbon. These two modes of reaction are referred to as 1,2-addition and 1,4-addition respectively. A 1,4-addition is also called a conjugate addition. Basic reaction of 1,2 addition Here the nucleophile adds to the carbon which is in the one position. The hydrogen adds to the oxygen which is in the two position. Basic reaction of 1,4 addition In 1,4 addition the Nucleophile is added to the carbon β to the carbonyl while the hydrogen is added to the carbon α to the carbonyl. Mechanism for 1,4 addition 1) Nucleophilic attack on the carbon β to the carbonyl 2) Proton Transfer Here we can see why this addition is called 1,4. The nucleophile bonds to the carbon in the one position and the hydrogen adds to the oxygen in the four position. 3) Tautomerization Going from reactant to products simplified 1,2 vs. 1,4 addition Whether 1,2 or 1,4-addition occurs depends on multiple variables but mostly it is determined by the nature of the nucleophile. During the addition of a nucleophile there is a competition between 1,2 and 1,4 addition products. If the nucleophile is a strong base, such as Grignard reagents , both the 1,2 and 1,4 reactions are irreversible and therefor are under kinetic control. Since 1,2-additions to the carbonyl group are fast, we would expect to find a predominance of 1,2-products from these reactions. If the nucleophile is a weak base, such as alcohols or amines, then the 1,2 addition is usually reversible. This means the competition between 1,2 and 1,4 addition is under thermodynamic control. In this case 1,4-addition dominates because the stable carbonyl group is retained. Nucleophiles which add 1,4 to α, β unsaturated carbonyls Water Alcohols Thiols 1 o Amines 2 o Amines HBr Cyanides Gilman Reagents Another important reaction exhibited by organometallic reagents is metal exchange. Organolithium reagents react with cuprous iodide to give a lithium dimethylcopper reagent, which is referred to as a Gilman reagent. Gilman reagents are a source of carbanion like nucleophiles similar to Grignard and Organo lithium reagents. However, the reactivity of organocuprate reagents is slightly different and this difference will be exploited in different situations. In the case of α, β unsaturated carbonyls organocuprate reagents allow for an 1,4 addition of an alkyl group. As we will see later Grignard and Organolithium reagents add alkyl groups 1,2 to α, β unsaturated carbonyls Organocuprate reagents are made from the reaction of organolithium reagents and \(CuI\) \[ 2 RLi + CuI \rightarrow R_2CuLi + LiI\] This acts as a source of R: - \[2 CH_3Li + CuI \rightarrow (CH_3)_2CuLi + LiI\] Example Nucleophiles which add 1,2 to α, β unsaturated carbonyls Metal Hydrides Grignard Reagents Organolithium Reagents Contributors Prof. Steven Farmer ( Sonoma State University ) William Reusch, Professor Emeritus ( Michigan State U. ), Virtual Textbook of Organic Chemistry
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Courses/Providence_College/CHM_331_Advanced_Analytical_Chemistry_1/04%3A__The_Vocabulary_of_Analytical_Chemistry/4.07%3A_The_Importance_of_Analytical_Methodology
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The importance of the issues raised in this chapter is evident if we examine environmental monitoring programs. The purpose of a monitoring program is to determine the present status of an environmental system, and to assess long term trends in the system’s health. These are broad and poorly defined goals. In many cases, an environmental monitoring program begins before the essential questions are known. This is not surprising since it is difficult to formulate questions in the absence of results. Without careful planning, however, a poor experimental design may result in data that has little value. These concerns are illustrated by the Chesapeake Bay Monitoring Program. This research program, designed to study nutrients and toxic pollutants in the Chesapeake Bay, was initiated in 1984 as a cooperative venture between the federal government, the state governments of Maryland, Virginia, and Pennsylvania, and the District of Columbia. A 1989 review of the program highlights the problems common to many monitoring programs [D’Elia, C. F.; Sanders, J. G.; Capone, D. G. Envrion. Sci. Technol. 1989 , 23 , 768–774]. At the beginning of the Chesapeake Bay monitoring program, little attention was given to selecting analytical methods, in large part because the eventual use of the data was not yet specified. The analytical methods initially chosen were standard methods already approved by the Environmental Protection Agency (EPA). In many cases these methods were not useful because they were designed to detect pollutants at their legally mandated maximum allowed concentrations. In unpolluted waters, however, the concentrations of these contaminants often are well below the detection limit of the EPA methods. For example, the detection limit for the EPA approved standard method for phosphate was 7.5 ppb. Since the actual phosphate concentrations in Chesapeake Bay were below the EPA method’s detection limit, it provided no useful information. On the other hand, the detection limit for a non-approved variant of the EPA method, a method routinely used by chemical oceanographers, was 0.06 ppb, a more realistic detection limit for their samples. In other cases, such as the elemental analysis for particulate forms of carbon, nitrogen and phosphorous, EPA approved procedures provided poorer reproducibility than nonapproved methods.
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Bookshelves/Inorganic_Chemistry/Organometallic_Chemistry_(Evans)/01%3A_Introduction_to_Organometallic_Chemistry/1.02%3A_What_is_Organometallic_Chemistry
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Let’s begin with a few simple questions: what is organometallic chemistry? What, after studying organometallic chemistry, will we know about the world that we didn’t know before? Why is the subject worth studying? And what kinds of problems is the subject meant to address? The purpose of this post is to give the best answers I currently know of to these questions. The goal of this otherwise content-free post is twofold: (1) to help motivate us as we move forward (that is, to constantly remind us that there is a point to all this!); and (2) to illustrate the kinds of problems we’ll be able to address using concepts from the field. You might be surprised by the spine-chilling power you feel after learning about the behavior of organometallic compounds and reactions! Put most bluntly, organometallic (OM) chemistry is the study of compounds containing, and reactions involving, metal-carbon bonds. The metal-carbon bond may be transient or temporary, but if one exists during a reaction or in a compound of interest, we’re squarely in the domain of organometallic chemistry. Despite the denotational importance of the M-C bond, bonds between metals and the other common elements of organic chemistry also appear in OM chemistry: metal-nitrogen, metal-oxygen, metal-halogen, and even metal-hydrogen bonds all play a role. Metals cover a vast swath of the periodic table and include the alkali metals (group 1), alkali earth metals (group 2), transition metals (groups 3-12), the main group metals (groups 13-15, “under the stairs”), and the lanthanides and actinides. We will focus most prominently on the behavior of the transition metals, so called because they cover the transition between the electropositive group 2 elements and the more electron-rich main group elements. Why is the subject worth studying? Well, for me, it mostly comes down to synthetic flexibility. There’s a reason the “organo” comes first in “organometallic chemistry”—our goal is usually the creation of new bonds in organic compounds. The metals tend to just be along for the ride (although their influence, obviously, is essential). And the fact is that you can do things with organometallic chemistry that you cannot do using straight-up organic chemistry. Case in point: The venerable Suzuki reaction...unthinkable without palladium! The establishment of the bond between the phenyl rings through a means other than dumb luck seems unthinkable to the organic chemist, but it’s natural for the palladium-equipped metal-organicker. Bromobenzene looks like a potential electrophile at the bromine-bearing carbon, and if you’re familiar with hydroboration you might see phenylboronic acid as a potential nucleophile at the boron-bearing carbon. Catalytic palladium makes it all happen! Organometallic chemistry is full of these mind-bending transformations, and can expand the synthetic toolbox of the organic chemist considerably. To throw another motive into the mix for the non-specialist (or the synthesis-spurning chemist), organometallic chemistry is full of intriguing stories of scientific inquiry and discovery. Exploring how researchers take a new organometallic reaction from “ooh pretty” to strong predictive power is instructive for anyone interested in “how science works,” in a practical sense. We’ll examine a number of classical experiments in organometallic chemistry, both for their value to the field and their contributions to the general nature of scientific inquiry. What kinds of problems should we be able to address as we move forward? Here’s a bulleted list of the most commonly encountered types of problems in an organometallic chemistry course: Describe the structure of an organometallic complex… Predict the product of the given reaction conditions… Draw a reasonable mechanism based on evidence… Devise a synthetic route to synthesize a target organometallic compound… Explain the observation(s)… Predict the results of a series of experiments… The first four are pretty standard organic-esque problems, but it’s the last two, more general classes that really make organometallic chemistry compelling. Just imagine putting yourself in the shoes of the pioneers and making the same predictions they did! There you have it, a short introduction to organometallic chemistry and why it’s worth studying. Of course, we’ll use the remainder of space in the blog to fully describe what organometallic chemistry really is…but it’s helpful to keep these motives in mind as you study. Keep a thirst for predictive power, and it’s hard to go wrong with organometallic chemistry!
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Courses/Bellarmine_University/BU%3A_Chem_103_(Christianson)/Phase_3%3A_Atoms_and_Molecules_-_the_Underlying_Reality/10%3A_Molecular_Structure_and_Geometry/10.3%3A_VSEPR_Geometry
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Skills to Develop To use the VSEPR model to predict molecular geometries. To predict whether a molecule has a dipole moment. The Lewis electron-pair approach can be used to predict the number and types of bonds between the atoms in a substance, and it indicates which atoms have lone pairs of electrons. This approach gives no information about the actual arrangement of atoms in space, however. We continue our discussion of structure and bonding by introducing the valence-shell electron-pair repulsion (VSEPR) model (pronounced “vesper”), which can be used to predict the shapes of many molecules and polyatomic ions. Keep in mind, however, that the VSEPR model, like any model, is a limited representation of reality; the model provides no information about bond lengths or the presence of multiple bonds. Molecular Geometry The specific three dimensional arrangement of atoms in molecules is referred to as molecular geometry . We can describe molecular geometry in terms of the bond distances, angles, and relative arrangements in space (Figure \(\PageIndex{1}\)). A bond angle is the angle between any two bonds that include a common atom, usually measured in degrees. A bond distance (or bond length) is the distance between the nuclei of two bonded atoms along the straight line joining the nuclei. Bond distances are measured in Ångstroms (1 Å = 10 –10 m) or picometers (1 pm = 10 –12 m, 100 pm = 1 Å). Figure \(\PageIndex{1}\): Bond distances (lengths) and angles are shown for the formaldehyde molecule, H 2 CO. There are various instrumental techniques such as X-Ray crystallography and other experimental techniques which can be used to tell us where the atoms are located in a molecule. Using advanced techniques, very complicated structures for proteins, enzymes, DNA, and RNA have been determined. Molecular geometry is critical to the chemistry of vision, smell, taste, drug reactions, and enzyme controlled reactions, to name a few. Example \(\PageIndex{1}\): Carbon Tetrachloride The Lewis structure of carbon tetrachloride provides information about connectivities, provides information about valence orbitals, and provides information about bond character. However, the Lewis structure provides no information about the shape of the molecule, which is defined by t he bond angles and the bond lengths . For carbon tetrachloride, each C-Cl bond length is 1.78Å and each Cl-C-Cl bond angle is 109.5°. Hence, carbon tetrachloride is tetrahedral in structure: The VSEPR Model The VSEPR model can predict the structure of nearly any molecule or polyatomic ion in which the central atom is a nonmetal, as well as the structures of many molecules and polyatomic ions with a central metal atom. The premise of the VSEPR theory is that electron pairs located in bonds and lone pairs repel each other and will therefore adopt the geometry that places electron pairs as far apart from each other as possible. This theory is very simplistic and does not account for the subtleties of orbital interactions that influence molecular shapes; however, the simple VSEPR counting procedure accurately predicts the three-dimensional structures of a large number of compounds, which cannot be predicted using the Lewis electron-pair approach. We can use the VSEPR model to predict the geometry of most polyatomic molecules and ions by focusing only on the number of electron pairs around the central atom , ignoring all other valence electrons present. According to this model, valence electrons in the Lewis structure form groups , which may consist of a single bond, a double bond, a triple bond, a lone pair of electrons, or even a single unpaired electron, which in the VSEPR model is counted as a lone pair. Because electrons repel each other electrostatically, the most stable arrangement of electron groups (i.e., the one with the lowest energy) is the one that minimizes repulsions. Groups are positioned around the central atom in a way that produces the molecular structure with the lowest energy, as illustrated in Figure \(\PageIndex{2}\). Figure \(\PageIndex{2}\): Electron Geometries for Species with Two to Six Electron Groups. Groups are placed around the central atom in a way that produces a molecular structure with the lowest energy, that is, the one that minimizes repulsions. In the VSEPR model, the molecule or polyatomic ion is often given an AX m E n designation, where A is the central atom, X is a bonded atom, E is a nonbonding valence electron group (usually a lone pair of electrons), and m and n are integers. Each group around the central atom is designated as a bonding pair (BP) or lone (nonbonding) pair (LP). From the BP and LP interactions we can predict both the relative positions of the atoms and the bond angles . Using this information, we can describe the molecular geometry , the arrangement of the bonded atoms in a molecule or polyatomic ion. Predicting Electron Pair Geometry and Molecular Structure The following procedure uses VSEPR theory to determine the electron pair geometries and the molecular structures: Draw the Lewis structure of the molecule or polyatomic ion. Count the number of regions of electron density (lone pairs and bonds) around the central atom. A single, double, or triple bond counts as one region of electron density. Identify the electron-pair geometry based on the number of regions of electron density: linear, trigonal planar, tetrahedral, trigonal bipyramidal, or octahedral (Figure \(\PageIndex{2}\)). Use the number of lone pairs to assign an AX m E n designation and determine the molecular geometry. Identify the LP–LP, LP–BP, or BP–BP interactions and predict deviations from ideal bond angles based on the principle that lone pairs occupy more space than multiple bonds, which occupy more space than single bonds. We will illustrate the use of this procedure with several examples, beginning with atoms with two electron groups. In our discussion we will refer to Figure \(\PageIndex{2}\) and Figure \(\PageIndex{3}\), which summarize the common molecular geometries and idealized bond angles of molecules and ions with two to six electron groups. Figure \(\PageIndex{3}\): Common Molecular Geometries for Species with Two to Six Electron Groups. Lone pairs are shown using a dashed line. Two Electron Groups Our first example is a molecule with two bonded atoms and no lone pairs of electrons, \(BeH_2\). AX 2 : BeH 2 1. The central atom, beryllium, contributes two valence electrons, and each hydrogen atom contributes one. The Lewis electron structure is 2. There are two electron groups around the central atom. We see from Figure \(\PageIndex{2}\) that the arrangement that minimizes repulsions places the groups 180° apart. 3. Both groups around the central atom are bonding pairs (BP). Thus BeH 2 is designated as AX 2 . 4. From Figure \(\PageIndex{3}\) we see that with two bonding pairs, the molecular geometry that minimizes repulsions in BeH 2 is linear . AX 2 : CO 2 1. The central atom, carbon, contributes four valence electrons, and each oxygen atom contributes six. The Lewis electron structure is 2. The carbon atom forms two double bonds. Each double bond is a group, so there are two electron groups around the central atom. Like BeH 2 , the arrangement that minimizes repulsions places the groups 180° apart. 3. Once again, both groups around the central atom are bonding pairs (BP), so CO 2 is designated as AX 2 . 4. VSEPR only recognizes groups around the central atom. Thus the lone pairs on the oxygen atoms do not influence the molecular geometry. With two bonding pairs on the central atom and no lone pairs, the molecular geometry of CO 2 is linear (Figure \(\PageIndex{3}\)). The structure of CO 2 is shown in Figure \(\PageIndex{2}\).1. Three Electron Groups AX 3 : BCl 3 1. The central atom, boron, contributes three valence electrons, and each chlorine atom contributes seven valence electrons. The Lewis electron structure is 2. There are three electron groups around the central atom. To minimize repulsions, the groups are placed 120° apart (Figure \(\PageIndex{2}\)). 3. All electron groups are bonding pairs (BP), so the structure is designated as AX 3 . 4. From Figure \(\PageIndex{3}\) we see that with three bonding pairs around the central atom, the molecular geometry of BCl 3 is trigonal planar , as shown in Figure \(\PageIndex{2}\). AX 3 : CO 3 2− 1. The central atom, carbon, has four valence electrons, and each oxygen atom has six valence electrons. As you learned previously, the Lewis electron structure of one of three resonance forms is represented as 2. The structure of CO 3 2− is a resonance hybrid. It has three identical bonds, each with a bond order of \(1 \frac{1}{3}\). We minimize repulsions by placing the three groups 120° apart (Figure \(\PageIndex{2}\)). 3. All electron groups are bonding pairs (BP). With three bonding groups around the central atom, the structure is designated as AX 3 . 4. We see from Figure \(\PageIndex{3}\) that the molecular geometry of CO 3 2− is trigonal planar with bond angles of 120°. In our next example we encounter the effects of lone pairs and multiple bonds on molecular geometry for the first time. AX 2 E: SO 2 1. The central atom, sulfur, has 6 valence electrons, as does each oxygen atom. With 18 valence electrons, the Lewis electron structure is shown below. 2. There are three electron groups around the central atom, two double bonds and one lone pair. We initially place the groups in a trigonal planar arrangement to minimize repulsions (Figure \(\PageIndex{2}\)). 3. There are two bonding pairs and one lone pair, so the structure is designated as AX 2 E. This designation has a total of three electron pairs, two X and one E. Because a lone pair is not shared by two nuclei, it occupies more space near the central atom than a bonding pair (Figure \(\PageIndex{4}\)). Thus bonding pairs and lone pairs repel each other electrostatically in the order BP–BP < LP–BP < LP–LP. In SO 2 , we have one BP–BP interaction and two LP–BP interactions. 4. The molecular geometry is described only by the positions of the nuclei, not by the positions of the lone pairs. Thus with two nuclei and one lone pair the shape is bent , or V shaped , which can be viewed as a trigonal planar arrangement with a missing vertex (Figures 9.2.2.1 and 9.2.3). The O-S-O bond angle is expected to be less than 120° because of the extra space taken up by the lone pair. Figure \(\PageIndex{4}\): The Difference in the Space Occupied by a Lone Pair of Electrons and by a Bonding Pair As with SO 2 , this composite model of electron distribution and negative electrostatic potential in ammonia shows that a lone pair of electrons occupies a larger region of space around the nitrogen atom than does a bonding pair of electrons that is shared with a hydrogen atom. Like lone pairs of electrons, multiple bonds occupy more space around the central atom than a single bond, which can cause other bond angles to be somewhat smaller than expected. This is because a multiple bond has a higher electron density than a single bond, so its electrons occupy more space than those of a single bond. For example, in a molecule such as CH 2 O (AX 3 ), whose structure is shown below, the double bond repels the single bonds more strongly than the single bonds repel each other. This causes a deviation from ideal geometry (an H–C–H bond angle of 116.5° rather than 120°). Four Electron Groups One of the limitations of Lewis structures is that they depict molecules and ions in only two dimensions. With four electron groups, we must learn to show molecules and ions in three dimensions. AX 4 : CH 4 1. The central atom, carbon, contributes four valence electrons, and each hydrogen atom has one valence electron, so the full Lewis electron structure is 2. There are four electron groups around the central atom. As shown in Figure \(\PageIndex{2}\), repulsions are minimized by placing the groups in the corners of a tetrahedron with bond angles of 109.5°. 3. All electron groups are bonding pairs, so the structure is designated as AX 4 . 4. With four bonding pairs, the molecular geometry of methane is tetrahedral (Figure \(\PageIndex{3}\)). AX 3 E: NH 3 1. In ammonia, the central atom, nitrogen, has five valence electrons and each hydrogen donates one valence electron, producing the Lewis electron structure 2. There are four electron groups around nitrogen, three bonding pairs and one lone pair. Repulsions are minimized by directing each hydrogen atom and the lone pair to the corners of a tetrahedron. 3. With three bonding pairs and one lone pair, the structure is designated as AX 3 E. This designation has a total of four electron pairs, three X and one E. We expect the LP–BP interactions to cause the bonding pair angles to deviate significantly from the angles of a perfect tetrahedron. 4. There are three nuclei and one lone pair, so the molecular geometry is trigonal pyramidal . In essence, this is a tetrahedron with a vertex missing (Figure \(\PageIndex{3}\)). However, the H–N–H bond angles are less than the ideal angle of 109.5° because of LP–BP repulsions (Figure \(\PageIndex{3}\) and Figure \(\PageIndex{4}\)). AX 2 E 2 : H 2 O 1. Oxygen has six valence electrons and each hydrogen has one valence electron, producing the Lewis electron structure 2. There are four groups around the central oxygen atom, two bonding pairs and two lone pairs. Repulsions are minimized by directing the bonding pairs and the lone pairs to the corners of a tetrahedron Figure \(\PageIndex{2}\). 3. With two bonding pairs and two lone pairs, the structure is designated as AX 2 E 2 with a total of four electron pairs. Due to LP–LP, LP–BP, and BP–BP interactions, we expect a significant deviation from idealized tetrahedral angles. 4. With two hydrogen atoms and two lone pairs of electrons, the structure has significant lone pair interactions. There are two nuclei about the central atom, so the molecular shape is bent , or V shaped , with an H–O–H angle that is even less than the H–N–H angles in NH 3 , as we would expect because of the presence of two lone pairs of electrons on the central atom rather than one. This molecular shape is essentially a tetrahedron with two missing vertices. Five Electron Groups In previous examples it did not matter where we placed the electron groups because all positions were equivalent. In some cases, however, the positions are not equivalent. We encounter this situation for the first time with five electron groups. AX 5 : PCl 5 1. Phosphorus has five valence electrons and each chlorine has seven valence electrons, so the Lewis electron structure of PCl 5 is 2. There are five bonding groups around phosphorus, the central atom. The structure that minimizes repulsions is a trigonal bipyramid , which consists of two trigonal pyramids that share a base (Figure \(\PageIndex{2}\)): 3. All electron groups are bonding pairs, so the structure is designated as AX 5 . There are no lone pair interactions. 4. The molecular geometry of PCl 5 is trigonal bipyramidal , as shown in Figure \(\PageIndex{3}\). The molecule has three atoms in a plane in equatorial positions and two atoms above and below the plane in axial positions. The three equatorial positions are separated by 120° from one another, and the two axial positions are at 90° to the equatorial plane. The axial and equatorial positions are not chemically equivalent, as we will see in our next example. AX 4 E: SF 4 1. The sulfur atom has six valence electrons and each fluorine has seven valence electrons, so the Lewis electron structure is With an expanded valence, this species is an exception to the octet rule. 2. There are five groups around sulfur, four bonding pairs and one lone pair. With five electron groups, the lowest energy arrangement is a trigonal bipyramid, as shown in Figure \(\PageIndex{2}\). 3. We designate SF 4 as AX 4 E; it has a total of five electron pairs. However, because the axial and equatorial positions are not chemically equivalent, where do we place the lone pair? If we place the lone pair in the equatorial position, we have three LP–BP repulsions at 90°. If we place it in the axial position, we have two 90° LP–BP repulsions at 90°. With fewer 90° LP–BP repulsions, we can predict that the structure with the lone pair of electrons in the equatorial position is more stable than the one with the lone pair in the axial position. We also expect a deviation from ideal geometry because a lone pair of electrons occupies more space than a bonding pair. Figure \(\PageIndex{5}\): Illustration of the Area Shared by Two Electron Pairs versus the Angle between Them At 90°, the two electron pairs share a relatively large region of space, which leads to strong repulsive electron–electron interactions. 4. With four nuclei and one lone pair of electrons, the molecular structure is based on a trigonal bipyramid with a missing equatorial vertex; it is described as a seesaw . The F axial –S–F axial angle is 173° rather than 180° because of the lone pair of electrons in the equatorial plane. AX 3 E 2 : BrF 3 1. The bromine atom has seven valence electrons, and each fluorine has seven valence electrons, so the Lewis electron structure is Once again, we have a compound that is an exception to the octet rule. 2. There are five groups around the central atom, three bonding pairs and two lone pairs. We again direct the groups toward the vertices of a trigonal bipyramid. 3. With three bonding pairs and two lone pairs, the structural designation is AX 3 E 2 with a total of five electron pairs. Because the axial and equatorial positions are not equivalent, we must decide how to arrange the groups to minimize repulsions. If we place both lone pairs in the axial positions, we have six LP–BP repulsions at 90°. If both are in the equatorial positions, we have four LP–BP repulsions at 90°. If one lone pair is axial and the other equatorial, we have one LP–LP repulsion at 90° and three LP–BP repulsions at 90°: Structure (c) can be eliminated because it has a LP–LP interaction at 90°. Structure (b), with fewer LP–BP repulsions at 90° than (a), is lower in energy. However, we predict a deviation in bond angles because of the presence of the two lone pairs of electrons. 4. The three nuclei in BrF 3 determine its molecular structure, which is described as T shaped . This is essentially a trigonal bipyramid that is missing two equatorial vertices. The F axial –Br–F axial angle is 172°, less than 180° because of LP–BP repulsions (Figure \(\PageIndex{2}\).1). Because lone pairs occupy more space around the central atom than bonding pairs, electrostatic repulsions are more important for lone pairs than for bonding pairs. AX 2 E 3 : I 3 − 1. Each iodine atom contributes seven electrons and the negative charge one, so the Lewis electron structure is 2. There are five electron groups about the central atom in I 3 − , two bonding pairs and three lone pairs. To minimize repulsions, the groups are directed to the corners of a trigonal bipyramid. 3. With two bonding pairs and three lone pairs, I 3 − has a total of five electron pairs and is designated as AX 2 E 3 . We must now decide how to arrange the lone pairs of electrons in a trigonal bipyramid in a way that minimizes repulsions. Placing them in the axial positions eliminates 90° LP–LP repulsions and minimizes the number of 90° LP–BP repulsions. The three lone pairs of electrons have equivalent interactions with the three iodine atoms, so we do not expect any deviations in bonding angles. 4. With three nuclei and three lone pairs of electrons, the molecular geometry of I 3 − is linear. This can be described as a trigonal bipyramid with three equatorial vertices missing. The ion has an I–I–I angle of 180°, as expected. Six Electron Groups Six electron groups form an octahedron , a polyhedron made of identical equilateral triangles and six identical vertices (Figure \(\PageIndex{2}\).) AX 6 : SF 6 1. The central atom, sulfur, contributes six valence electrons, and each fluorine atom has seven valence electrons, so the Lewis electron structure is With an expanded valence, this species is an exception to the octet rule. 2. There are six electron groups around the central atom, each a bonding pair. We see from Figure \(\PageIndex{2}\) that the geometry that minimizes repulsions is octahedral . 3. With only bonding pairs, SF 6 is designated as AX 6 . All positions are chemically equivalent, so all electronic interactions are equivalent. 4. There are six nuclei, so the molecular geometry of SF 6 is octahedral. AX 5 E: BrF 5 1. The central atom, bromine, has seven valence electrons, as does each fluorine, so the Lewis electron structure is With its expanded valence, this species is an exception to the octet rule. 2. There are six electron groups around the Br, five bonding pairs and one lone pair. Placing five F atoms around Br while minimizing BP–BP and LP–BP repulsions gives the following structure: 3. With five bonding pairs and one lone pair, BrF 5 is designated as AX 5 E; it has a total of six electron pairs. The BrF 5 structure has four fluorine atoms in a plane in an equatorial position and one fluorine atom and the lone pair of electrons in the axial positions. We expect all F axial –Br–F equatorial angles to be less than 90° because of the lone pair of electrons, which occupies more space than the bonding electron pairs. 4. With five nuclei surrounding the central atom, the molecular structure is based on an octahedron with a vertex missing. This molecular structure is square pyramidal . The F axial –B–F equatorial angles are 85.1°, less than 90° because of LP–BP repulsions. AX 4 E 2 : ICl 4 − 1. The central atom, iodine, contributes seven electrons. Each chlorine contributes seven, and there is a single negative charge. The Lewis electron structure is 2. There are six electron groups around the central atom, four bonding pairs and two lone pairs. The structure that minimizes LP–LP, LP–BP, and BP–BP repulsions is 3. ICl 4 − is designated as AX 4 E 2 and has a total of six electron pairs. Although there are lone pairs of electrons, with four bonding electron pairs in the equatorial plane and the lone pairs of electrons in the axial positions, all LP–BP repulsions are the same. Therefore, we do not expect any deviation in the Cl–I–Cl bond angles. 4. With five nuclei, the ICl4− ion forms a molecular structure that is square planar , an octahedron with two opposite vertices missing. The relationship between the number of electron groups around a central atom, the number of lone pairs of electrons, and the molecular geometry is summarized in Figure \(\PageIndex{6}\). Figure \(\PageIndex{6}\): The molecular structures are identical to the electron-pair geometries when there are no lone pairs present (first column). For a particular number of electron pairs (row), the molecular structures for one or more lone pairs are determined based on modifications of the corresponding electron-pair geometry. Example \(\PageIndex{1}\) Using the VSEPR model, predict the molecular geometry of each molecule or ion. PF 5 (phosphorus pentafluoride, a catalyst used in certain organic reactions) H 3 O + (hydronium ion) Given: two chemical species Asked for: molecular geometry Strategy: Draw the Lewis electron structure of the molecule or polyatomic ion. Determine the electron group arrangement around the central atom that minimizes repulsions. Assign an AX m E n designation; then identify the LP–LP, LP–BP, or BP–BP interactions and predict deviations in bond angles. Describe the molecular geometry. Solution: A The central atom, P, has five valence electrons and each fluorine has seven valence electrons, so the Lewis structure of PF 5 is B There are five bonding groups about phosphorus. The structure that minimizes repulsions is a trigonal bipyramid (Figure 9.2.6). C All electron groups are bonding pairs, so PF 5 is designated as AX 5 . Notice that this gives a total of five electron pairs. With no lone pair repulsions, we do not expect any bond angles to deviate from the ideal. D The PF 5 molecule has five nuclei and no lone pairs of electrons, so its molecular geometry is trigonal bipyramidal. A The central atom, O, has six valence electrons, and each H atom contributes one valence electron. Subtracting one electron for the positive charge gives a total of eight valence electrons, so the Lewis electron structure is B There are four electron groups around oxygen, three bonding pairs and one lone pair. Like NH 3 , repulsions are minimized by directing each hydrogen atom and the lone pair to the corners of a tetrahedron. C With three bonding pairs and one lone pair, the structure is designated as AX 3 E and has a total of four electron pairs (three X and one E). We expect the LP–BP interactions to cause the bonding pair angles to deviate significantly from the angles of a perfect tetrahedron. D There are three nuclei and one lone pair, so the molecular geometry is trigonal pyramidal , in essence a tetrahedron missing a vertex. However, the H–O–H bond angles are less than the ideal angle of 109.5° because of LP–BP repulsions: Exercise \(\PageIndex{1}\) Using the VSEPR model, predict the molecular geometry of each molecule or ion. XeO 3 PF 6 − NO 2 + Answer a trigonal pyramidal Answer b octahedral Answer c linear Example \(\PageIndex{2}\) Predict the molecular geometry of each molecule. XeF 2 SnCl 2 Given: two chemical compounds Asked for: molecular geometry Strategy: Use the strategy given in Example 9.2.1. Solution: A Xenon contributes eight electrons and each fluorine seven valence electrons, so the Lewis electron structure is B There are five electron groups around the central atom, two bonding pairs and three lone pairs. Repulsions are minimized by placing the groups in the corners of a trigonal bipyramid. C From B, XeF 2 is designated as AX 2 E 3 and has a total of five electron pairs (two X and three E). With three lone pairs about the central atom, we can arrange the two F atoms in three possible ways: both F atoms can be axial, one can be axial and one equatorial, or both can be equatorial: The structure with the lowest energy is the one that minimizes LP–LP repulsions. Both (b) and (c) have two 90° LP–LP interactions, whereas structure (a) has none. Thus both F atoms are in the axial positions, like the two iodine atoms around the central iodine in I 3 − . All LP–BP interactions are equivalent, so we do not expect a deviation from an ideal 180° in the F–Xe–F bond angle. D With two nuclei about the central atom, the molecular geometry of XeF 2 is linear. It is a trigonal bipyramid with three missing equatorial vertices. A The tin atom donates 4 valence electrons and each chlorine atom donates 7 valence electrons. With 18 valence electrons, the Lewis electron structure is B There are three electron groups around the central atom, two bonding groups and one lone pair of electrons. To minimize repulsions the three groups are initially placed at 120° angles from each other. C From B we designate SnCl 2 as AX 2 E. It has a total of three electron pairs, two X and one E. Because the lone pair of electrons occupies more space than the bonding pairs, we expect a decrease in the Cl–Sn–Cl bond angle due to increased LP–BP repulsions. D With two nuclei around the central atom and one lone pair of electrons, the molecular geometry of SnCl 2 is bent, like SO 2 , but with a Cl–Sn–Cl bond angle of 95°. The molecular geometry can be described as a trigonal planar arrangement with one vertex missing. Exercise \(\PageIndex{2}\) Predict the molecular geometry of each molecule. SO 3 XeF 4 Answer a trigonal planar Answer b square planar Molecules with No Single Central Atom The VSEPR model can be used to predict the structure of somewhat more complex molecules with no single central atom by treating them as linked AX m E n fragments. We will demonstrate with methyl isocyanate (CH 3 –N=C=O), a volatile and highly toxic molecule that is used to produce the pesticide Sevin. In 1984, large quantities of Sevin were accidentally released in Bhopal, India, when water leaked into storage tanks. The resulting highly exothermic reaction caused a rapid increase in pressure that ruptured the tanks, releasing large amounts of methyl isocyanate that killed approximately 3800 people and wholly or partially disabled about 50,000 others. In addition, there was significant damage to livestock and crops. We can treat methyl isocyanate as linked AX m E n fragments beginning with the carbon atom at the left, which is connected to three H atoms and one N atom by single bonds. The four bonds around carbon mean that it must be surrounded by four bonding electron pairs in a configuration similar to AX 4 . We can therefore predict the CH 3 –N portion of the molecule to be roughly tetrahedral, similar to methane: The nitrogen atom is connected to one carbon by a single bond and to the other carbon by a double bond, producing a total of three bonds, C–N=C. For nitrogen to have an octet of electrons, it must also have a lone pair: Because multiple bonds are not shown in the VSEPR model, the nitrogen is effectively surrounded by three electron pairs. Thus according to the VSEPR model, the C–N=C fragment should be bent with an angle less than 120°. The carbon in the –N=C=O fragment is doubly bonded to both nitrogen and oxygen, which in the VSEPR model gives carbon a total of two electron pairs. The N=C=O angle should therefore be 180°, or linear. The three fragments combine to give the following structure: We predict that all four nonhydrogen atoms lie in a single plane, with a C–N–C angle of approximately 120°. The experimentally determined structure of methyl isocyanate confirms our prediction (Figure \(\PageIndex{8}\)). Figure \(\PageIndex{8}\): The Experimentally Determined Structure of Methyl Isocyanate Certain patterns are seen in the structures of moderately complex molecules. For example, carbon atoms with four bonds (such as the carbon on the left in methyl isocyanate) are generally tetrahedral. Similarly, the carbon atom on the right has two double bonds that are similar to those in CO 2 , so its geometry, like that of CO 2 , is linear. Recognizing similarities to simpler molecules will help you predict the molecular geometries of more complex molecules. Example \(\PageIndex{3}\) Use the VSEPR model to predict the molecular geometry of propyne (H 3 C–C≡CH), a gas with some anesthetic properties. Given: chemical compound Asked for: molecular geometry Strategy: Count the number of electron groups around each carbon, recognizing that in the VSEPR model, a multiple bond counts as a single group. Use Figure \(\PageIndex{3}\) to determine the molecular geometry around each carbon atom and then deduce the structure of the molecule as a whole. Solution: Because the carbon atom on the left is bonded to four other atoms, we know that it is approximately tetrahedral. The next two carbon atoms share a triple bond, and each has an additional single bond. Because a multiple bond is counted as a single bond in the VSEPR model, each carbon atom behaves as if it had two electron groups. This means that both of these carbons are linear, with C–C≡C and C≡C–H angles of 180°. Exercise \(\PageIndex{3}\) Predict the geometry of allene (H 2 C=C=CH 2 ), a compound with narcotic properties that is used to make more complex organic molecules. Answer The terminal carbon atoms are trigonal planar, the central carbon is linear, and the C–C–C angle is 180°. 3D Geometries of \(AB_n\) molecules Understanding molecular geometry in three-dimensional space is an essential skill for chemists because geometry is so critical to molecular properties and function. With the VSEPR process, you should be able to name the correct geometry for a molecule, but you should also be able to visualize what that geometry looks like in real space. Figure \(\PageIndex{9}\) shows some three-dimensional depictions of geometries where a central atom \(A\) is bonded to two or more \(X\) atoms. Try to keep these images in mind when considering the geometries of real molecules! 6 5 4 3 2 AX6 octahedral AX5 trigonal bipyramidal AX4 tetrahedral AX3 trigonal planar AX2 linear 1 lone pair of electrons 1 lone pair of electrons 1 lone pair of electrons 1 lone pair of electrons 1 lone pair of electrons AX5E square pyramidal AX4E distorted tetrahedron AX3E pyramidal AX2E nonlinear AXE linear 2 lone pairs of electrons 2 lone pairs of electrons 2 lone pairs of electrons 2 lone pairs of electrons NaN AX4E2 square planar AX3E2 T-shaped AX2E2 bent AXE2 linear NaN Figure \(\PageIndex{9}\): Three-dimensional animations of common VSEPR molecular geometries Summary Lewis electron structures give no information about molecular geometry , the arrangement of bonded atoms in a molecule or polyatomic ion, which is crucial to understanding the chemistry of a molecule. The valence-shell electron-pair repulsion (VSEPR) model allows us to predict which of the possible structures is actually observed in most cases. It is based on the assumption that pairs of electrons occupy space, and the lowest-energy structure is the one that minimizes electron pair–electron pair repulsions. In the VSEPR model, the molecule or polyatomic ion is given an AX m E n designation, where A is the central atom, X is a bonded atom, E is a nonbonding valence electron group (usually a lone pair of electrons), and m and n are integers. Each group around the central atom is designated as a bonding pair (BP) or lone (nonbonding) pair (LP). From the BP and LP interactions we can predict both the relative positions of the atoms and the angles between the bonds, called the bond angles . From this we can describe the molecular geometry . The VSEPR model can be used to predict the shapes of many molecules and polyatomic ions, but it gives no information about bond lengths and the presence of multiple bonds. A combination of VSEPR and a bonding model, such as Lewis electron structures, is necessary to understand the presence of multiple bonds. Contributors Mike Blaber ( Florida State University ) Robyn Rindge (Class of '98) who now works for PDI Dreamworks (look for his name in the credits of Shrek2.). Robyn drew these rotating molecules using Infini-D (MetaCreations). Paul Groves , chemistry teacher at South Pasadena High School and Chemmy Bear Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected] ).
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Courses/Lansing_Community_College/LCC%3A_Chem_151_-_General_Chemistry_I/Text/05%3A_Gases/5.04%3A_The_Ideal_Gas_Equation
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Learning Objectives Derive the ideal gas law from the constituent gas laws To use the ideal gas law to describe the behavior of a gas. In this module, the relationship between pressure, temperature, volume, and amount of a gas are described and how these relationships can be combined to give a general expression that describes the behavior of a gas. Deriving the Ideal Gas Law Any set of relationships between a single quantity (such as \(V\)) and several other variables (\(P\), \(T\), and \(n\)) can be combined into a single expression that describes all the relationships simultaneously. The three individual expressions were derived previously: Boyle’s law \[V \propto \dfrac{1}{P} \;\; \text{@ constant n and T} \nonumber \] Charles’s law \[V \propto T \;\; \text{@ constant n and P} \nonumber \] Avogadro’s law \[V \propto n \;\; \text{@ constant T and P} \nonumber \] Combining these three expressions gives \[V \propto \dfrac{nT}{P} \label{10.4.1} \] which shows that the volume of a gas is proportional to the number of moles and the temperature and inversely proportional to the pressure. This expression can also be written as \[V= {\rm Cons.} \left( \dfrac{nT}{P} \right) \label{10.4.2} \] By convention, the proportionality constant in Equation \(\ref{10.4.1}\) is called the gas constant, which is represented by the letter \(R\). Inserting R into Equation \(\ref{10.4.2}\) gives \[ V = \dfrac{RnT}{P} = \dfrac{nRT}{P} \label{10.4.3} \] Clearing the fractions by multiplying both sides of Equation \(\ref{10.4.4}\) by \(P\) gives \[PV = nRT \label{10.4.4} \] This equation is known as the ideal gas law . An ideal gas is defined as a hypothetical gaseous substance whose behavior is independent of attractive and repulsive forces and can be completely described by the ideal gas law. In reality, there is no such thing as an ideal gas, but an ideal gas is a useful conceptual model that allows us to understand how gases respond to changing conditions. As we shall see, under many conditions, most real gases exhibit behavior that closely approximates that of an ideal gas. The ideal gas law can therefore be used to predict the behavior of real gases under most conditions. The ideal gas law does not work well at very low temperatures or very high pressures, where deviations from ideal behavior are most commonly observed. Significant deviations from ideal gas behavior commonly occur at low temperatures and very high pressures. Before we can use the ideal gas law, however, we need to know the value of the gas constant R. Its form depends on the units used for the other quantities in the expression. If V is expressed in liters (L), P in atmospheres (atm), T in kelvins (K), and n in moles (mol), then \[R = 0.08206 \dfrac{\rm L\cdot atm}{\rm K\cdot mol} \label{10.4.5} \] Because the product PV has the units of energy, R can also have units of J/(K•mol): \[R = 8.3145 \dfrac{\rm J}{\rm K\cdot mol}\label{10.4.6} \] Standard Conditions of Temperature and Pressure Scientists have chosen a particular set of conditions to use as a reference: 0°C (273.15 K) and \(\rm1\; bar = 100 \;kPa = 10^5\;Pa\) pressure, referred to as standard temperature and pressure ( STP ). \[\text{STP:} \hspace{2cm} T=273.15\;{\rm K}\text{ and }P=\rm 1\;bar=10^5\;Pa \nonumber \] Please note that STP was defined differently in the past. The old definition was based on a standard pressure of 1 atm. We can calculate the volume of 1.000 mol of an ideal gas under standard conditions using the variant of the ideal gas law given in Equation \(\ref{10.4.4}\): \[V=\dfrac{nRT}{P}\label{10.4.7} \] Thus the volume of 1 mol of an ideal gas is 22.71 L at STP and 22.41 L at 0°C and 1 atm , approximately equivalent to the volume of three basketballs. The molar volumes of several real gases at 0°C and 1 atm are given in Table 10.3, which shows that the deviations from ideal gas behavior are quite small. Thus the ideal gas law does a good job of approximating the behavior of real gases at 0°C and 1 atm. The relationships described in Section 10.3 as Boyle’s, Charles’s, and Avogadro’s laws are simply special cases of the ideal gas law in which two of the four parameters (P, V, T, and n) are held fixed. Gas Molar Volume (L) He 22.434 Ar 22.397 H2 22.433 N2 22.402 O2 22.397 CO2 22.260 NH3 22.079 Applying the Ideal Gas Law The ideal gas law allows us to calculate the value of the fourth variable for a gaseous sample if we know the values of any three of the four variables ( P , V , T , and n ). It also allows us to predict the final state of a sample of a gas (i.e., its final temperature, pressure, volume, and amount) following any changes in conditions if the parameters ( P , V , T , and n ) are specified for an initial state. Some applications are illustrated in the following examples. The approach used throughout is always to start with the same equation—the ideal gas law—and then determine which quantities are given and which need to be calculated. Let’s begin with simple cases in which we are given three of the four parameters needed for a complete physical description of a gaseous sample. Example \(\PageIndex{1}\) The balloon that Charles used for his initial flight in 1783 was destroyed, but we can estimate that its volume was 31,150 L (1100 ft 3 ), given the dimensions recorded at the time. If the temperature at ground level was 86°F (30°C) and the atmospheric pressure was 745 mmHg, how many moles of hydrogen gas were needed to fill the balloon? Given: volume, temperature, and pressure Asked for: amount of gas Strategy: Solve the ideal gas law for the unknown quantity, in this case n . Make sure that all quantities are given in units that are compatible with the units of the gas constant. If necessary, convert them to the appropriate units, insert them into the equation you have derived, and then calculate the number of moles of hydrogen gas needed. Solution: A We are given values for P , T , and V and asked to calculate n . If we solve the ideal gas law (Equation \(\ref{10.4.4}\)) for \(n\), we obtain \[\rm745\;mmHg\times\dfrac{1\;atm}{760\;mmHg}=0.980\;atm \nonumber \] B P and T are given in units that are not compatible with the units of the gas constant [ R = 0.08206 (L•atm)/(K•mol)]. We must therefore convert the temperature to kelvins and the pressure to atmospheres: \[T=273+30=303{\rm K}\nonumber \] Substituting these values into the expression we derived for n , we obtain \[\begin{align*} n &=\dfrac{PV}{RT} \\[4pt] &=\rm\dfrac{0.980\;atm\times31150\;L}{0.08206\dfrac{atm\cdot L}{\rm mol\cdot K}\times 303\;K} \\[4pt] &=1.23\times10^3\;mol \end{align*} \nonumber \] Exercise \(\PageIndex{1}\) Suppose that an “empty” aerosol spray-paint can has a volume of 0.406 L and contains 0.025 mol of a propellant gas such as CO 2 . What is the pressure of the gas at 25°C? Answer 1.5 atm In Example \(\PageIndex{1}\), we were given three of the four parameters needed to describe a gas under a particular set of conditions, and we were asked to calculate the fourth. We can also use the ideal gas law to calculate the effect of changes in any of the specified conditions on any of the other parameters, as shown in Example \(\PageIndex{5}\). General Gas Equation When a gas is described under two different conditions, the ideal gas equation must be applied twice - to an initial condition and a final condition. This is: \[\begin{array}{cc}\text{Initial condition }(i) & \text{Final condition} (f) \\ P_iV_i=n_iRT_i & P_fV_f=n_fRT_f\end{array} \nonumber \] Both equations can be rearranged to give: \[R=\dfrac{P_iV_i}{n_iT_i} \hspace{1cm} R=\dfrac{P_fV_f}{n_fT_f} \nonumber \] The two equations are equal to each other since each is equal to the same constant \(R\). Therefore, we have: \[\dfrac{P_iV_i}{n_iT_i}=\dfrac{P_fV_f}{n_fT_f}\label{10.4.8} \] The equation is called the general gas equation . The equation is particularly useful when one or two of the gas properties are held constant between the two conditions. In such cases, the equation can be simplified by eliminating these constant gas properties. Example \(\PageIndex{2}\) Suppose that Charles had changed his plans and carried out his initial flight not in August but on a cold day in January, when the temperature at ground level was −10°C (14°F). How large a balloon would he have needed to contain the same amount of hydrogen gas at the same pressure as in Example \(\PageIndex{1}\)? Given: temperature, pressure, amount, and volume in August; temperature in January Asked for: volume in January Strategy: Use the results from Example \(\PageIndex{1}\) for August as the initial conditions and then calculate the change in volume due to the change in temperature from 30°C to −10°C. Begin by constructing a table showing the initial and final conditions. Simplify the general gas equation by eliminating the quantities that are held constant between the initial and final conditions, in this case \(P\) and \(n\). Solve for the unknown parameter. Solution: A To see exactly which parameters have changed and which are constant, prepare a table of the initial and final conditions: Initial (August) Final (January) \(T_i=30\, °C = 303\, K\) \(T_f=−10\,°C = 263\, K\) \(P_i= 0.980 \, atm\) \(P_f= 0.980\, atm\) \(n_i=1.23 × 10^3\, mol\) \(n_f= 1.23 × 10^3\, mol\) \(V_i=31150\, L\) \(V_f=?\) B Both \(n\) and \(P\) are the same in both cases (\(n_i=n_f,P_i=P_f\)). Therefore, Equation \ref{10.4.8} can be simplified to: \[\dfrac{V_i}{T_i}=\dfrac{V_f}{T_f} \nonumber \] This is the relationship first noted by Charles. C Solving the equation for \(V_f\), we get: \[\begin{align*} V_f &=V_i\times\dfrac{T_f}{T_i} \\[4pt] &=\rm31150\;L\times\dfrac{263\;K}{303\;K} \\[4pt] &=2.70\times10^4\;L \end{align*} \nonumber \] It is important to check your answer to be sure that it makes sense, just in case you have accidentally inverted a quantity or multiplied rather than divided. In this case, the temperature of the gas decreases. Because we know that gas volume decreases with decreasing temperature, the final volume must be less than the initial volume, so the answer makes sense. We could have calculated the new volume by plugging all the given numbers into the ideal gas law, but it is generally much easier and faster to focus on only the quantities that change. Exercise \(\PageIndex{2}\) At a laboratory party, a helium-filled balloon with a volume of 2.00 L at 22°C is dropped into a large container of liquid nitrogen (T = −196°C). What is the final volume of the gas in the balloon? Answer 0.52 L Example \(\PageIndex{1}\) illustrates the relationship originally observed by Charles. We could work through similar examples illustrating the inverse relationship between pressure and volume noted by Boyle ( PV = constant) and the relationship between volume and amount observed by Avogadro ( V / n = constant). We will not do so, however, because it is more important to note that the historically important gas laws are only special cases of the ideal gas law in which two quantities are varied while the other two remain fixed. The method used in Example \(\PageIndex{1}\) can be applied in any such case, as we demonstrate in Example \(\PageIndex{2}\) (which also shows why heating a closed container of a gas, such as a butane lighter cartridge or an aerosol can, may cause an explosion). Example \(\PageIndex{3}\) Aerosol cans are prominently labeled with a warning such as “Do not incinerate this container when empty.” Assume that you did not notice this warning and tossed the “empty” aerosol can in Exercise \(\PageIndex{1}\) (0.025 mol in 0.406 L, initially at 25°C and 1.5 atm internal pressure) into a fire at 750°C. What would be the pressure inside the can (if it did not explode)? Given: initial volume, amount, temperature, and pressure; final temperature Asked for: final pressure Strategy: Follow the strategy outlined in Example \(\PageIndex{2}\). Solution: Prepare a table to determine which parameters change and which are held constant: Initial Final \(V_i=0.406\;\rm L\) \(V_f=0.406\;\rm L\) \(n_i=0.025\;\rm mol\) \(n_f=0.025\;\rm mol\) \(T_i=\rm25\;^\circ C=298\;K\) \(T_i=\rm750\;^\circ C=1023\;K\) \(P_i=1.5\;\rm atm\) \(P_f=?\) Both \(V\) and \(n\) are the same in both cases (\(V_i=V_f,n_i=n_f\)). Therefore, Equation can be simplified to: \[ \dfrac{P_i}{T_i} = \dfrac{P_f}{T_f} \nonumber \] By solving the equation for \(P_f\), we get: \[\begin{align*} P_f &=P_i\times\dfrac{T_f}{T_i} \\[4pt] &=\rm1.5\;atm\times\dfrac{1023\;K}{298\;K} \\[4pt] &=5.1\;atm \end{align*} \nonumber \] This pressure is more than enough to rupture a thin sheet metal container and cause an explosion! Exercise \(\PageIndex{3}\) Suppose that a fire extinguisher, filled with CO 2 to a pressure of 20.0 atm at 21°C at the factory, is accidentally left in the sun in a closed automobile in Tucson, Arizona, in July. The interior temperature of the car rises to 160°F (71.1°C). What is the internal pressure in the fire extinguisher? Answer 23.4 atm In Examples \(\PageIndex{1}\) and \(\PageIndex{2}\) , two of the four parameters ( P , V , T , and n ) were fixed while one was allowed to vary, and we were interested in the effect on the value of the fourth. In fact, we often encounter cases where two of the variables P , V , and T are allowed to vary for a given sample of gas (hence n is constant), and we are interested in the change in the value of the third under the new conditions. Example \(\PageIndex{4}\) We saw in Example \(\PageIndex{1}\) that Charles used a balloon with a volume of 31,150 L for his initial ascent and that the balloon contained 1.23 × 10 3 mol of H 2 gas initially at 30°C and 745 mmHg. Suppose that Gay-Lussac had also used this balloon for his record-breaking ascent to 23,000 ft and that the pressure and temperature at that altitude were 312 mmHg and −30°C, respectively. To what volume would the balloon have had to expand to hold the same amount of hydrogen gas at the higher altitude? Given: initial pressure, temperature, amount, and volume; final pressure and temperature Asked for: final volume Strategy: Follow the strategy outlined in Example \(\PageIndex{3}\). Solution: Begin by setting up a table of the two sets of conditions: Initial Final \(P_i=745\;\rm mmHg=0.980\;atm\) \(P_f=312\;\rm mmHg=0.411\;atm\) \(T_i=\rm30\;^\circ C=303\;K\) \(T_f=\rm750-30\;^\circ C=243\;K\) \(n_i=\rm1.2\times10^3\;mol\) \(n_i=\rm1.2\times10^3\;mol\) \(V_i=\rm31150\;L\) \(V_f=?\) By eliminating the constant property (\(n\)) of the gas, Equation \(\ref{10.4.8}\) is simplified to: \[\dfrac{P_iV_i}{T_i}=\dfrac{P_fV_f}{T_f} \nonumber \] By solving the equation for \(V_f\), we get: \[\begin{align*} V_f &=V_i\times\dfrac{P_i}{P_f}\dfrac{T_f}{T_i} \\[4pt] &=\rm3.115\times10^4\;L\times\dfrac{0.980\;atm}{0.411\;atm}\dfrac{243\;K}{303\;K} \\[4pt] &=5.96\times10^4\;L \end{align*} \nonumber \] Does this answer make sense? Two opposing factors are at work in this problem: decreasing the pressure tends to increase the volume of the gas, while decreasing the temperature tends to decrease the volume of the gas. Which do we expect to predominate? The pressure drops by more than a factor of two, while the absolute temperature drops by only about 20%. Because the volume of a gas sample is directly proportional to both T and 1/ P , the variable that changes the most will have the greatest effect on V . In this case, the effect of decreasing pressure predominates, and we expect the volume of the gas to increase, as we found in our calculation. We could also have solved this problem by solving the ideal gas law for V and then substituting the relevant parameters for an altitude of 23,000 ft: Except for a difference caused by rounding to the last significant figure, this is the same result we obtained previously. There is often more than one “right” way to solve chemical problems. Exercise \(\PageIndex{4}\) A steel cylinder of compressed argon with a volume of 0.400 L was filled to a pressure of 145 atm at 10°C. At 1.00 atm pressure and 25°C, how many 15.0 mL incandescent light bulbs could be filled from this cylinder? (Hint: find the number of moles of argon in each container.) Answer 4.07 × 10 3 Using the Ideal Gas Law to Calculate Gas Densities and Molar Masses The ideal gas law can also be used to calculate molar masses of gases from experimentally measured gas densities. To see how this is possible, we first rearrange the ideal gas law to obtain \[\dfrac{n}{V}=\dfrac{P}{RT}\label{10.4.9} \] The left side has the units of moles per unit volume (mol/L). The number of moles of a substance equals its mass (\(m\), in grams) divided by its molar mass (\(M\), in grams per mole): \[n=\dfrac{m}{M}\label{10.4.10} \] Substituting this expression for \(n\) into Equation \(\ref{10.4.9}\) gives \[\dfrac{m}{MV}=\dfrac{P}{RT}\label{10.4.11} \] Because \(m/V\) is the density \(d\) of a substance, we can replace \(m/V\) by \(d\) and rearrange to give \[\rho=\dfrac{m}{V}=\dfrac{MP}{RT}\label{10.4.12} \] The distance between particles in gases is large compared to the size of the particles, so their densities are much lower than the densities of liquids and solids. Consequently, gas density is usually measured in grams per liter (g/L) rather than grams per milliliter (g/mL). Example \(\PageIndex{5}\) Calculate the density of butane at 25°C and a pressure of 750 mmHg. Given: compound, temperature, and pressure Asked for: density Strategy: Calculate the molar mass of butane and convert all quantities to appropriate units for the value of the gas constant. Substitute these values into Equation \(\ref{10.4.12}\) to obtain the density. Solution: A The molar mass of butane (C 4 H 10 ) is \[M=(4)(12.011) + (10)(1.0079) = 58.123 \rm g/mol \nonumber \] Using 0.08206 (L•atm)/(K•mol) for R means that we need to convert the temperature from degrees Celsius to kelvins ( T = 25 + 273 = 298 K) and the pressure from millimeters of mercury to atmospheres: \[P=\rm750\;mmHg\times\dfrac{1\;atm}{760\;mmHg}=0.987\;atm \nonumber \] B Substituting these values into Equation \(\ref{10.4.12}\) gives \[\rho=\rm\dfrac{58.123\;g/mol\times0.987\;atm}{0.08206\dfrac{L\cdot atm}{K\cdot mol}\times298\;K}=2.35\;g/L \nonumber \] Exercise \(\PageIndex{5}\): Density of Radon Radon (Rn) is a radioactive gas formed by the decay of naturally occurring uranium in rocks such as granite. It tends to collect in the basements of houses and poses a significant health risk if present in indoor air. Many states now require that houses be tested for radon before they are sold. Calculate the density of radon at 1.00 atm pressure and 20°C and compare it with the density of nitrogen gas, which constitutes 80% of the atmosphere, under the same conditions to see why radon is found in basements rather than in attics. Answer radon, 9.23 g/L; N 2 , 1.17 g/L A common use of Equation \(\ref{10.4.12}\) is to determine the molar mass of an unknown gas by measuring its density at a known temperature and pressure. This method is particularly useful in identifying a gas that has been produced in a reaction, and it is not difficult to carry out. A flask or glass bulb of known volume is carefully dried, evacuated, sealed, and weighed empty. It is then filled with a sample of a gas at a known temperature and pressure and reweighed. The difference in mass between the two readings is the mass of the gas. The volume of the flask is usually determined by weighing the flask when empty and when filled with a liquid of known density such as water. The use of density measurements to calculate molar masses is illustrated in Example \(\PageIndex{6}\). Example \(\PageIndex{6}\) The reaction of a copper penny with nitric acid results in the formation of a red-brown gaseous compound containing nitrogen and oxygen. A sample of the gas at a pressure of 727 mmHg and a temperature of 18°C weighs 0.289 g in a flask with a volume of 157.0 mL. Calculate the molar mass of the gas and suggest a reasonable chemical formula for the compound. Given: pressure, temperature, mass, and volume Asked for: molar mass and chemical formula Strategy: Solve Equation \(\ref{10.4.12}\) for the molar mass of the gas and then calculate the density of the gas from the information given. Convert all known quantities to the appropriate units for the gas constant being used. Substitute the known values into your equation and solve for the molar mass. Propose a reasonable empirical formula using the atomic masses of nitrogen and oxygen and the calculated molar mass of the gas. Solution: A Solving Equation \(\ref{10.4.12}\) for the molar mass gives \[M=\dfrac{mRT}{PV}=\dfrac{dRT}{P} \nonumber \] Density is the mass of the gas divided by its volume: \[\rho=\dfrac{m}{V}=\dfrac{0.289\rm g}{0.157\rm L}=1.84 \rm g/L\nonumber \] B We must convert the other quantities to the appropriate units before inserting them into the equation: \[T=18+273=291 K\nonumber \] \[P=727 \, mmHg \times \dfrac{1\rm atm} {760\rm mmHg} =0.957\rm atm \nonumber \] The molar mass of the unknown gas is thus \[M=\rm\dfrac{1.84\;g/L\times0.08206\dfrac{L\cdot atm}{K\cdot mol}\times291\;K}{0.957\;atm}=45.9 g/mol\nonumber \] C The atomic masses of N and O are approximately 14 and 16, respectively, so we can construct a list showing the masses of possible combinations: \[M({\rm NO})=14 + 16=30 \rm\; g/mol\nonumber \] \[M({\rm N_2O})=(2)(14)+16=44 \rm\;g/mol\nonumber \] \[M({\rm NO_2})=14+(2)(16)=46 \rm\;g/mol\nonumber \] The most likely choice is NO 2 which is in agreement with the data. The red-brown color of smog also results from the presence of NO 2 gas. Exercise \(\PageIndex{6}\) You are in charge of interpreting the data from an unmanned space probe that has just landed on Venus and sent back a report on its atmosphere. The data are as follows: pressure, 90 atm; temperature, 557°C; density, 58 g/L. The major constituent of the atmosphere (>95%) is carbon. Calculate the molar mass of the major gas present and identify it. Answer 44 g/mol; \(CO_2\) Summary The ideal gas law is derived from empirical relationships among the pressure, the volume, the temperature, and the number of moles of a gas; it can be used to calculate any of the four properties if the other three are known. Ideal gas equation : \(PV = nRT\), where \(R = 0.08206 \dfrac{\rm L\cdot atm}{\rm K\cdot mol}=8.3145 \dfrac{\rm J}{\rm K\cdot mol}\) General gas equation : \(\dfrac{P_iV_i}{n_iT_i}=\dfrac{P_fV_f}{n_fT_f}\) Density of a gas: \(\rho=\dfrac{MP}{RT}\) The empirical relationships among the volume, the temperature, the pressure, and the amount of a gas can be combined into the ideal gas law , PV = nRT . The proportionality constant, R , is called the gas constant and has the value 0.08206 (L•atm)/(K•mol), 8.3145 J/(K•mol), or 1.9872 cal/(K•mol), depending on the units used. The ideal gas law describes the behavior of an ideal gas , a hypothetical substance whose behavior can be explained quantitatively by the ideal gas law and the kinetic molecular theory of gases. Standard temperature and pressure (STP) is 0°C and 1 atm. The volume of 1 mol of an ideal gas at STP is 22.41 L, the standard molar volume . All of the empirical gas relationships are special cases of the ideal gas law in which two of the four parameters are held constant. The ideal gas law allows us to calculate the value of the fourth quantity ( P , V , T , or n ) needed to describe a gaseous sample when the others are known and also predict the value of these quantities following a change in conditions if the original conditions (values of P , V , T , and n ) are known. The ideal gas law can also be used to calculate the density of a gas if its molar mass is known or, conversely, the molar mass of an unknown gas sample if its density is measured.
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Courses/Modesto_Junior_College/Chemistry_142%3A_Pre-General_Chemistry_(Brzezinski)/CHEM_142%3A_Text_(Brzezinski)/04%3A_Atoms_and_Elements/4.02%3A_The_Nuclear_Atom
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Learning Objectives Explain the observations that led to Thomson's discovery of the electron. Describe Thomson's "plum pudding" model of the atom and the evidence for it. Draw a diagram of Thomson's "plum pudding" model of the atom and explain why it has this name. Describe Rutherford's gold foil experiment and explain how this experiment altered the "plum pudding" model. Draw a diagram of the Rutherford model of the atom and label the nucleus and the electron cloud. Dalton's Atomic Theory held up well to a lot of the different chemical experiments that scientists performed to test it. In fact, for almost 100 years, it seemed as if Dalton's Atomic Theory was the whole truth. However, in 1897, a scientist named J. J. Thomson conducted some research that suggested that Dalton's Atomic Theory was not the entire story. He suggested that the small, negatively charged particles making up the cathode ray were actually pieces of atoms. He called these pieces "corpuscles," although today we know them as electrons . Thanks to his clever experiments and careful reasoning, J. J. Thomson is credited with the discovery of the electron. Electrons and Plums The electron was discovered by J. J. Thomson in 1897. The existence of protons was also known, as was the fact that atoms were neutral in charge. Since the intact atom had no net charge and the electron and proton had opposite charges, the next step after the discovery of subatomic particles was to figure out how these particles were arranged in the atom. This was a difficult task because of the incredibly small size of the atom. Therefore, scientists set out to design a model of what they believed the atom could look like. The goal of each atomic model was to accurately represent all of the experimental evidence about atoms in the simplest way possible. Following the discovery of the electron, J.J. Thomson developed what became known as the " plum pudding " model in 1904. Plum pudding is an English dessert similar to a blueberry muffin. In Thomson's plum pudding model of the atom, the electrons were embedded in a uniform sphere of positive charge like blueberries stuck into a muffin. The positive matter was thought to be jelly-like or similar to a thick soup. The electrons were somewhat mobile. As they got closer to the outer portion of the atom, the positive charge in the region was greater than the neighboring negative charges, and the electron would be pulled back more toward the center region of the atom. However, this model of the atom soon gave way to a new model developed by New Zealander Ernest Rutherford (1871-1937) about five years later. Thomson did still receive many honors during his lifetime, including being awarded the Nobel Prize in Physics in 1906 and a knighthood in 1908. Atoms and Gold In 1911, Rutherford and coworkers Hans Geiger and Ernest Marsden initiated a series of groundbreaking experiments that would completely change the accepted model of the atom. They bombarded very thin sheets of gold foil with fast moving alpha particles . Alpha particles, a type of natural radioactive particle, are positively charged particles with a mass about four times that of a hydrogen atom. According to the accepted atomic model, in which an atom's mass and charge are uniformly distributed throughout the atom, the scientists expected that all of the alpha particles would pass through the gold foil with only a slight deflection or none at all. Surprisingly, while most of the alpha particles were indeed not deflected, a very small percentage (about 1 in 8000 particles) bounced off the gold foil at very large angles. Some were even redirected back toward the source. No prior knowledge had prepared them for this discovery. In a famous quote, Rutherford exclaimed that it was "as if you had fired a 15-inch [artillery] shell at a piece of tissue and it came back and hit you." Rutherford needed to come up with an entirely new model of the atom in order to explain his results. Because the vast majority of the alpha particles had passed through the gold, he reasoned that most of the atom was empty space. In contrast, the particles that were highly deflected must have experienced a tremendously powerful force within the atom. He concluded that all of the positive charge and the majority of the mass of the atom must be concentrated in a very small space in the atom's interior, which he called the nucleus. The nucleus is the tiny, dense, central core of the atom and is composed of protons and neutrons. Rutherford's atomic model became known as the nuclear model . In the nuclear atom, the protons and neutrons, which comprise nearly all of the mass of the atom, are located in the nucleus at the center of the atom. The electrons are distributed around the nucleus and occupy most of the volume of the atom. It is worth emphasizing just how small the nucleus is compared to the rest of the atom. If we could blow up an atom to be the size of a large professional football stadium, the nucleus would be about the size of a marble. Rutherford's model proved to be an important step towards a full understanding of the atom. However, it did not completely address the nature of the electrons and the way in which they occupy the vast space around the nucleus. It was not until some years later that a full understanding of the electron was achieved. This proved to be the key to understanding the chemical properties of elements. Atomic Nucleus The nucleus (plural, nuclei) is a positively charged region at the center of the atom. It consists of two types of subatomic particles packed tightly together. The particles are protons, which have a positive electric charge, and neutrons, which are neutral in electric charge. Outside of the nucleus, an atom is mostly empty space, with orbiting negative particles called electrons whizzing through it. The figure below shows these parts of the atom. The nucleus of the atom is extremely small. Its radius is only about 1/100,000 of the total radius of the atom. Electrons have virtually no mass, but protons and neutrons have a lot of mass for their size. As a result, the nucleus has virtually all the mass of an atom. Given its great mass and tiny size, the nucleus is very dense. If an object the size of a penny had the same density as the nucleus of an atom, its mass would be greater than 30 million tons! Holding it all Together Particles with opposite electric charges attract each other. This explains why negative electrons orbit the positive nucleus. Particles with the same electric charge repel each other. This means that the positive protons in the nucleus push apart from one another. So why doesn't the nucleus fly apart? An even stronger force—called the strong nuclear force —holds protons and neutrons together in the nucleus. Summary Atoms are the ultimate building blocks of all matter. The modern atomic theory establishes the concepts of atoms and how they compose matter. Bombardment of gold foil with alpha particles showed that some particles were deflected. The nuclear model of the atom consists of a small and dense positively charged interior surrounded by a cloud of electrons.
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Courses/Alma_College/Organic_Chemistry_I_(Alma_College)/12%3A_Alkynes_-_An_Introduction_to_Organic_Synthesis/12.07%3A_Reduction_of_Alkynes
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Objectives After completing this section, you should be able to write equations for the catalytic hydrogenation of alkynes to alkanes and cis alkenes. identify the reagent and catalyst required to produce a given alkane or cis alkene from a given alkyne. identify the product formed from the reaction of a given alkyne with hydrogen and a specified catalyst. identify the alkyne that must be used to produce a given alkane or cis alkene by catalytic hydrogenation. write the equation for the reduction of an alkyne with an alkali metal and liquid ammonia. predict the structure of the product formed when a given alkyne is reduced with an alkali metal and liquid ammonia. identify the alkyne that must be used to produce a given alkene by reduction with an alkali metal and ammonia. Key Terms Make certain that you can define, and use in context, the key terms below. anion radical Lindlar catalyst Study Notes The Lindlar catalyst allows a chemist to reduce a triple bond in the presence of a double bond. Thus but Hydrogenation and the Relative Stability of Hydrocarbons Like alkenes, alkynes readily undergo catalytic hydrogenation partially to cis -alkenes or fully to alkanes depending on the reaction employed. The catalytic addition of hydrogen to 2-butyne provides heat of reaction data that reflect the relative thermodynamic stabilities of these hydrocarbons, as shown above. From the heats of hydrogenation, shown in blue in units of kcal/mole, it would appear that alkynes are thermodynamically less stable than alkenes to a greater degree than alkenes are less stable than alkanes. The standard bond energies for carbon-carbon bonds confirm this conclusion. Thus, a double bond is stronger than a single bond, but not twice as strong. The difference (63 kcal/mole) may be regarded as the strength of the π-bond component. Similarly, a triple bond is stronger than a double bond, but not 50% stronger. Here the difference (54 kcal/mole) may be taken as the strength of the second π-bond. The 9 kcal/mole weakening of this second π-bond is reflected in the heat of hydrogenation numbers (36.7 - 28.3 = 8.4). Alkynes can undergo reductive hydrogenation reactions similar to alkenes. Due to the presence of two pi bonds within alkynes, reductive hydrogenation reactions can be partial to form an alkene or complete to form an alkane. Since partial reductive hydrogenation of an alkyne produces an alkene, the stereochemistry provided by the reaction's mechanism determines whether a cis- or trans- alkene is formed. The three most significant alkyne reduction reactions are summarized below: Catalytic Hydrogenation of an Alkyne Much like alkenes, alkynes can be fully hydrogenated into alkanes with the help of a platinum, palladium, or nickel catalyst. Because the reaction is catalyzed on the surface of the metal, it is common for these catalysts to be dispersed on carbon (Pd/C) or finely dispersed, such as Raney-Ni . The presence of two pi bonds in the alkyne causes two equivalents of H 2 to be added during the reaction. An alkene intermediate is formed during the reaction but is not isolated. For catalytic hydrogenation, the Pt, Pd, or Ni catalysts are so effective in promoting the addition of hydrogen to carbon-carbon triple bonds that the alkene intermediate formed cannot be isolated. A less efficient Lindlar's catalyst , permits alkynes to be converted to alkenes without further reduction to an alkane. Lindlar’s Catalyst transforms an alkyne to a cis -alkene because both hydrogen atoms are added to the same side of the alkyne (syn) as shown in the in the previous chapter. Lindlar's catalyst is prepared by deactivating (or poisoning) a conventional palladium catalyst. Lindlar’s catalyst has three components: palladium-calcium carbonate, lead acetate, and quinoline. The quinoline serves to prevent complete hydrogenation of the alkyne to an alkane. Hydrogenation of an Alkyne to a Trans -Alkene The anti-addition of hydrogen to an alkyne pi bond occurs through reaction with sodium or lithium metal dissolved in ammonia. This reaction, also called a dissolving metal reduction , involves a radical mechanism, and produces a trans -alkene product. Mechanism Sodium metal is a powerful reducing agent due to the presence of a 3s 1 electron in its valence shell. Sodium metal easily gives up this electron to become Na + . The mechanism start with a sodium atom donating an electron to the alkyne to create an intermediate with a negative charge and an unpaired electron called a radical anion. Next the amine solvent protonates the anion to create a vinyl radical. A second sodium atom then donates an electron to pair the radical to form a vinyl anion. This vinyl anion intermediate rapid interconverts between cis and trans conformations and determines the stereochemistry of the reaction. The trans -vinyl anion is more stable due to reduced steric crowding and is preferentially formed. Finally, the protonation of the trans -vinyl anion creates the trans -alkene product. Step 1: Electron Donation Step 2: Protonation Step 3: Electron Donation Step 4: Protonation Exercise \(\PageIndex{1}\) Using any alkyne how would you prepare the following compounds: pentane, trans -4-methyl-2-pentene, cis- 4-methyl-2-pentene. Answer
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Courses/University_of_Illinois_Springfield/CHE_124%3A_General_Chemistry_for_the_Health_Professions_(Morsch_and_Andrews)/13%3A_Unsaturated_and_Aromatic_Hydrocarbons/13.4%3A_Chemical_Properties_of_Alkenes
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Skills to Develop To write equations for the addition reactions of alkenes with hydrogen, halogens, and water Alkenes are valued mainly for addition reactions, in which one of the bonds in the double bond is broken. Each of the carbon atoms in the bond can then attach another atom or group while remaining joined to each other by a single bond. Perhaps the simplest addition reaction ishydrogenation—a reaction with hydrogen (H 2 ) in the presence of a catalyst such as nickel (Ni) or platinum (Pt). The product is an alkane having the same carbon skeleton as the alkene. Alkenes also readily undergo halogenation —the addition of halogens. Indeed, the reaction with bromine (Br 2 ) can be used to test for alkenes. Bromine solutions are brownish red. When we add a Br 2 solution to an alkene, the color of the solution disappears because the alkene reacts with the bromine: Another important addition reaction is that between an alkene and water to form an alcohol. This reaction, called hydration , requires a catalyst—usually a strong acid, such as sulfuric acid (H 2 SO 4 ): The hydration reaction is discussed later, where we deal with this reaction in the synthesis of alcohols. Example \(\PageIndex{1}\) Write the equation for the reaction between CH 3 CH=CHCH 3 and each substance. H 2 (Ni catalyst) Br 2 H 2 O (H 2 SO 4 catalyst) SOLUTION In each reaction, the reagent adds across the double bond. Exercise \(\PageIndex{1}\) Write the equation for each reaction. CH 3 CH 2 CH=CH 2 with H 2 (Ni catalyst) CH 3 CH=CH 2 with Cl 2 CH 3 CH 2 CH=CHCH 2 CH 3 with H 2 O (H 2 SO 4 catalyst) Concept Review Exercises What is the principal difference in properties between alkenes and alkanes? How are they alike? If C 12 H 24 reacts with HBr in an addition reaction, what is the molecular formula of the product? Answers Alkenes undergo addition reactions; alkanes do not. Both burn. C 12 H 24 Br 2 Key Takeaway Alkenes undergo addition reactions, adding such substances as hydrogen, bromine, and water across the carbon-to-carbon double bond. Exercises Complete each equation. (CH 3 ) 2 C=CH 2 + Br 2 → \(\mathrm{CH_2\textrm{=C}(CH_3)CH_2CH_3 + H_2 \xrightarrow{Ni}}\) Complete each equation. \(\mathrm{CH_2\textrm{=CHCH=C}H_2 + 2H_2\xrightarrow{Ni}}\) \(\mathrm{(CH_3)_2\textrm{C=C}(CH_3)_2 + H_2O \xrightarrow{H_2SO_4}}\) Answer (CH 3 ) 2 CBrCH 2 Br CH 3 CH(CH 3 )CH 2 CH 3
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Courses/Oregon_Institute_of_Technology/OIT_(Lund)%3A_Organic_Chemistry_with_a_Biological_Emphasis_(Soderberg)/09%3A_Phosphate_Transfer_Reactions/9.05%3A_Phosphorylation_of_Carboxylates
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Thus far we have seen hydroxyl oxygens and phosphate oxygens acting as nucleophilic accepting groups in ATP-dependent phosphate transfer reactions. Carboxylate oxygens can also accept phosphate groups from ATP. This typically happens in two different ways. First, the carboxylate can attack the g-phosphate of ATP to accept phosphate, generates a species known as an 'acyl phosphate'. An example is the first part of the reaction catalyzed by glutamine synthase (EC 6.3.1.2): Alternatively, carboxylate groups are often converted into a species referred to as an 'acyl-AMP' . Here, the carboxylate oxygen attacks the \(\alpha \)-phosphate of ATP leading to release of inorganic pyrophosphate. An example is the first part of the reaction catalyzed by the enzyme asparagine synthetase: (EC 6.3.5.4): Draw a curved-arrow mechanism for the phosphate transfer reaction shown below (EC 2.7.2.3), which is from the glycolysis pathway. Note that ADP is on the reactant side and ATP is a product (the opposite of what we have seen so far). Hint: What functional group is the nucleophile? What functional group is the leaving group?
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Courses/Lumen_Learning/Book%3A_General_College_Chemistry_I_(Lumen)/12%3A_10-_Liquids_and_Solids/12.04%3A_Phase_Transitions
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Learning Objectives By the end of this section, you will be able to: Define phase transitions and phase transition temperatures Explain the relation between phase transition temperatures and intermolecular attractive forces Describe the processes represented by typical heating and cooling curves, and compute heat flows and enthalpy changes accompanying these processes We witness and utilize changes of physical state, or phase transitions, in a great number of ways. As one example of global significance, consider the evaporation, condensation, freezing, and melting of water. These changes of state are essential aspects of our earth’s water cycle as well as many other natural phenomena and technological processes of central importance to our lives. In this module, the essential aspects of phase transitions are explored. Vaporization and Condensation When a liquid vaporizes in a closed container, gas molecules cannot escape. As these gas phase molecules move randomly about, they will occasionally collide with the surface of the condensed phase, and in some cases, these collisions will result in the molecules re-entering the condensed phase. The change from the gas phase to the liquid is called condensation . When the rate of condensation becomes equal to the rate of vaporization , neither the amount of the liquid nor the amount of the vapor in the container changes. The vapor in the container is then said to be in equilibrium with the liquid. Keep in mind that this is not a static situation, as molecules are continually exchanged between the condensed and gaseous phases. Such is an example of a dynamic equilibrium , the status of a system in which reciprocal processes (for example, vaporization and condensation) occur at equal rates. The pressure exerted by the vapor in equilibrium with a liquid in a closed container at a given temperature is called the liquid’s vapor pressure (or equilibrium vapor pressure). The area of the surface of the liquid in contact with a vapor and the size of the vessel have no effect on the vapor pressure, although they do affect the time required for the equilibrium to be reached. We can measure the vapor pressure of a liquid by placing a sample in a closed container, like that illustrated in Figure 1, and using a manometer to measure the increase in pressure that is due to the vapor in equilibrium with the condensed phase. Figure 1. In a closed container, dynamic equilibrium is reached when (a) the rate of molecules escaping from the liquid to become the gas (b) increases and eventually (c) equals the rate of gas molecules entering the liquid. When this equilibrium is reached, the vapor pressure of the gas is constant, although the vaporization and condensation processes continue. The chemical identities of the molecules in a liquid determine the types (and strengths) of intermolecular attractions possible; consequently, different substances will exhibit different equilibrium vapor pressures. Relatively strong intermolecular attractive forces will serve to impede vaporization as well as favoring “recapture” of gas-phase molecules when they collide with the liquid surface, resulting in a relatively low vapor pressure. Weak intermolecular attractions present less of a barrier to vaporization, and a reduced likelihood of gas recapture, yielding relatively high vapor pressures. The following example illustrates this dependence of vapor pressure on intermolecular attractive forces. Example 1: Explaining Vapor Pressure in Terms of IMFs Given the shown structural formulas for these four compounds, explain their relative vapor pressures in terms of types and extents of IMFs: [reveal-answer q=”248315″]Show Answer[/reveal-answer] [hidden-answer a=”248315″] Diethyl ether has a very small dipole and most of its intermolecular attractions are London forces. Although this molecule is the largest of the four under consideration, its IMFs are the weakest and, as a result, its molecules most readily escape from the liquid. It also has the highest vapor pressure. Due to its smaller size, ethanol exhibits weaker dispersion forces than diethyl ether. However, ethanol is capable of hydrogen bonding and, therefore, exhibits stronger overall IMFs, which means that fewer molecules escape from the liquid at any given temperature, and so ethanol has a lower vapor pressure than diethyl ether. Water is much smaller than either of the previous substances and exhibits weaker dispersion forces, but its extensive hydrogen bonding provides stronger intermolecular attractions, fewer molecules escaping the liquid, and a lower vapor pressure than for either diethyl ether or ethanol. Ethylene glycol has two −OH groups, so, like water, it exhibits extensive hydrogen bonding. It is much larger than water and thus experiences larger London forces. Its overall IMFs are the largest of these four substances, which means its vaporization rate will be the slowest and, consequently, its vapor pressure the lowest. [/hidden-answer] Check Your Learning At 20 °C, the vapor pressures of several alcohols are given in this table. Explain these vapor pressures in terms of types and extents of IMFs for these alcohols: 0 1 2 3 4 Compound methanol CH3OH ethanol C2H5OH propanol C3H7OH butanol C4H9OH Vapor Pressure at 25 °C 11.9 kPa 5.95 kPa 2.67 kPa 0.56 kPa [reveal-answer q=”740217″]Show Answer[/reveal-answer] [hidden-answer a=”740217″] All these compounds exhibit hydrogen bonding; these strong IMFs are difficult for the molecules to overcome, so the vapor pressures are relatively low. As the size of molecule increases from methanol to butanol, dispersion forces increase, which means that the vapor pressures decrease as observed: P methanol > P ethanol > P propanol > P butanol . [/hidden-answer] As temperature increases, the vapor pressure of a liquid also increases due to the increased average KE of its molecules. Recall that at any given temperature, the molecules of a substance experience a range of kinetic energies, with a certain fraction of molecules having a sufficient energy to overcome IMF and escape the liquid (vaporize). At a higher temperature, a greater fraction of molecules have enough energy to escape from the liquid, as shown in Figure 2. The escape of more molecules per unit of time and the greater average speed of the molecules that escape both contribute to the higher vapor pressure. Figure 2. Temperature affects the distribution of kinetic energies for the molecules in a liquid. At the higher temperature, more molecules have the necessary kinetic energy, KE, to escape from the liquid into the gas phase. Boiling Points When the vapor pressure increases enough to equal the external atmospheric pressure, the liquid reaches its boiling point. The boiling point of a liquid is the temperature at which its equilibrium vapor pressure is equal to the pressure exerted on the liquid by its gaseous surroundings. For liquids in open containers, this pressure is that due to the earth’s atmosphere. The normal boiling point of a liquid is defined as its boiling point when surrounding pressure is equal to 1 atm (101.3 kPa). Figure 3 shows the variation in vapor pressure with temperature for several different substances. Considering the definition of boiling point, these curves may be seen as depicting the dependence of a liquid’s boiling point on surrounding pressure. Figure 3. The boiling points of liquids are the temperatures at which their equilibrium vapor pressures equal the pressure of the surrounding atmosphere. Normal boiling points are those corresponding to a pressure of 1 atm (101.3 kPa.) Example 2: A Boiling Point at Reduced Pressure A typical atmospheric pressure in Leadville, Colorado (elevation 10,200 feet) is 68 kPa. Use the graph in Figure 3 to determine the boiling point of water at this elevation. [reveal-answer q=”911686″]Show Answer[/reveal-answer] [hidden-answer a=”911686″] The graph of the vapor pressure of water versus temperature in Figure 3 indicates that the vapor pressure of water is 68 kPa at about 90 °C. Thus, at about 90 °C, the vapor pressure of water will equal the atmospheric pressure in Leadville, and water will boil. [/hidden-answer] Check Your Learning The boiling point of ethyl ether was measured to be 10 °C at a base camp on the slopes of Mount Everest. Use Figure 3 to determine the approximate atmospheric pressure at the camp. [reveal-answer q=”36414″]Show Answer[/reveal-answer] [hidden-answer a=”36414″]Approximately 40 kPa (0.4 atm)[/hidden-answer] The quantitative relation between a substance’s vapor pressure and its temperature is described by the Clausius-Clapeyron equation : where Δ H vap is the enthalpy of vaporization for the liquid, R is the gas constant, and ln A is a constant whose value depends on the chemical identity of the substance. This equation is often rearranged into logarithmic form to yield the linear equation: This linear equation may be expressed in a two-point format that is convenient for use in various computations, as demonstrated in the example exercises that follow. If at temperature T 1 , the vapor pressure is P 1 , and at temperature T 2 , the vapor pressure is P 2 , the corresponding linear equations are: Since the constant, ln A , is the same, these two equations may be rearranged to isolate ln A and then set them equal to one another: which can be combined into: Example 3: Estimating Enthalpy of Vaporization Isooctane (2,2,4-trimethylpentane) has an octane rating of 100. It is used as one of the standards for the octane-rating system for gasoline. At 34.0 °C, the vapor pressure of isooctane is 10.0 kPa, and at 98.8 °C, its vapor pressure is 100.0 kPa. Use this information to estimate the enthalpy of vaporization for isooctane. [reveal-answer q=”738838″]Show Answer[/reveal-answer] [hidden-answer a=”738838″] The enthalpy of vaporization, Δ H vap , can be determined by using the Clausius-Clapeyron equation: Since we have two vapor pressure-temperature values (T 1 = 34.0 °C = 307.2 K, P 1 = 10.0 kPa and T 2 = 98.8 °C = 372.0 K, P 2 = 100 kPa), we can substitute them into this equation and solve for Δ H vap . Rearranging the Clausius-Clapeyron equation and solving for Δ H vap yields: Note that the pressure can be in any units, so long as they agree for both P values, but the temperature must be in kelvin for the Clausius-Clapeyron equation to be valid. [/hidden-answer] Check Your Learning At 20.0 °C, the vapor pressure of ethanol is 5.95 kPa, and at 63.5 °C, its vapor pressure is 53.3 kPa. Use this information to estimate the enthalpy of vaporization for ethanol. [reveal-answer q=”869747″]Show Answer[/reveal-answer] [hidden-answer a=”869747″]47,782 J/mol = 47.8 kJ/mol[/hidden-answer] Example 4: Estimating Temperature (or Vapor Pressure) For benzene (C 6 H 6 ), the normal boiling point is 80.1 °C and the enthalpy of vaporization is 30.8 kJ/mol. What is the boiling point of benzene in Denver, where atmospheric pressure = 83.4 kPa? [reveal-answer q=”411401″]Show Answer[/reveal-answer] [hidden-answer a=”411401″] If the temperature and vapor pressure are known at one point, along with the enthalpy of vaporization, Δ H vap, then the temperature that corresponds to a different vapor pressure (or the vapor pressure that corresponds to a different temperature) can be determined by using the Clausius-Clapeyron equation: Since the normal boiling point is the temperature at which the vapor pressure equals atmospheric pressure at sea level, we know one vapor pressure-temperature value (T 1 = 80.1 °C = 353.3 K, P 1 = 101.3 kPa, Δ H vap = 30.8 kJ/mol) and want to find the temperature (T 2 ) that corresponds to vapor pressure P 2 = 83.4 kPa. We can substitute these values into the Clausius-Clapeyron equation and then solve for T 2 . Rearranging the Clausius-Clapeyron equation and solving for T 2 yields: [/hidden-answer] Check Your Learning For acetone (CH 3 ) 2 CO, the normal boiling point is 56.5 °C and the enthalpy of vaporization is 31.3 kJ/mol. What is the vapor pressure of acetone at 25.0 °C? [reveal-answer q=”389951″]Show Answer[/reveal-answer] [hidden-answer a=”389951″]30.1 kPa[/hidden-answer] Enthalpy of Vaporization Vaporization is an endothermic process. The cooling effect can be evident when you leave a swimming pool or a shower. When the water on your skin evaporates, it removes heat from your skin and causes you to feel cold. The energy change associated with the vaporization process is the enthalpy of vaporization, Δ H vap . For example, the vaporization of water at standard temperature is represented by: As described in the chapter on thermochemistry, the reverse of an endothermic process is exothermic. And so, the condensation of a gas releases heat: Example 5: Using Enthalpy of Vaporization Figure 4. Evaporation of sweat helps cool the body. (credit: “Kullez”/Flickr) One way our body is cooled is by evaporation of the water in sweat (Figure 4). In very hot climates, we can lose as much as 1.5 L of sweat per day. Although sweat is not pure water, we can get an approximate value of the amount of heat removed by evaporation by assuming that it is. How much heat is required to evaporate 1.5 L of water (1.5 kg) at T = 37 °C (normal body temperature); Δ H vap = 43.46 kJ/mol at 37 °C. [reveal-answer q=”90685″]Show Answer[/reveal-answer] [hidden-answer a=”90685″] We start with the known volume of sweat (approximated as just water) and use the given information to convert to the amount of heat needed: Thus, 3600 kJ of heat are removed by the evaporation of 1.5 L of water. [/hidden-answer] Check Your Learning How much heat is required to evaporate 100.0 g of liquid ammonia, NH 3 , at its boiling point if its enthalpy of vaporization is 4.8 kJ/mol? [reveal-answer q=”366740″]Show Answer[/reveal-answer] [hidden-answer a=”366740″]28 kJ[/hidden-answer] Melting and Freezing When we heat a crystalline solid, we increase the average energy of its atoms, molecules, or ions and the solid gets hotter. At some point, the added energy becomes large enough to partially overcome the forces holding the molecules or ions of the solid in their fixed positions, and the solid begins the process of transitioning to the liquid state, or melting. At this point, the temperature of the solid stops rising, despite the continual input of heat, and it remains constant until all of the solid is melted. Only after all of the solid has melted will continued heating increase the temperature of the liquid (Figure 5). Figure 5. (a) This beaker of ice has a temperature of −12.0 °C. (b) After 10 minutes the ice has absorbed enough heat from the air to warm to 0 °C. A small amount has melted. (c) Thirty minutes later, the ice has absorbed more heat, but its temperature is still 0 °C. The ice melts without changing its temperature. (d) Only after all the ice has melted does the heat absorbed cause the temperature to increase to 22.2 °C. (credit: modification of work by Mark Ott) If we stop heating during melting and place the mixture of solid and liquid in a perfectly insulated container so no heat can enter or escape, the solid and liquid phases remain in equilibrium. This is almost the situation with a mixture of ice and water in a very good thermos bottle; almost no heat gets in or out, and the mixture of solid ice and liquid water remains for hours. In a mixture of solid and liquid at equilibrium, the reciprocal process of melting and freezing occur at equal rates, and the quantities of solid and liquid therefore remain constant. The temperature at which the solid and liquid phases of a given substance are in equilibrium is called the melting point of the solid or the freezing point of the liquid. Use of one term or the other is normally dictated by the direction of the phase transition being considered, for example, solid to liquid (melting) or liquid to solid (freezing). The enthalpy of fusion and the melting point of a crystalline solid depend on the strength of the attractive forces between the units present in the crystal. Molecules with weak attractive forces form crystals with low melting points. Crystals consisting of particles with stronger attractive forces melt at higher temperatures. The amount of heat required to change one mole of a substance from the solid state to the liquid state is the enthalpy of fusion, ΔH fus of the substance. The enthalpy of fusion of ice is 6.0 kJ/mol at 0 °C. Fusion (melting) is an endothermic process: The reciprocal process, freezing, is an exothermic process whose enthalpy change is −6.0 kJ/mol at 0 °C: Sublimation and Deposition Figure 6. Sublimation of solid iodine in the bottom of the tube produces a purple gas that subsequently deposits as solid iodine on the colder part of the tube above. (credit: modification of work by Mark Ott) Some solids can transition directly into the gaseous state, bypassing the liquid state, via a process known as sublimation. At room temperature and standard pressure, a piece of dry ice (solid CO 2 ) sublimes, appearing to gradually disappear without ever forming any liquid. Snow and ice sublime at temperatures below the melting point of water, a slow process that may be accelerated by winds and the reduced atmospheric pressures at high altitudes. When solid iodine is warmed, the solid sublimes and a vivid purple vapor forms (Figure 6). The reverse of sublimation is called deposition, a process in which gaseous substances condense directly into the solid state, bypassing the liquid state. The formation of frost is an example of deposition. Like vaporization, the process of sublimation requires an input of energy to overcome intermolecular attractions. The enthalpy of sublimation, ΔH sub , is the energy required to convert one mole of a substance from the solid to the gaseous state. For example, the sublimation of carbon dioxide is represented by: Likewise, the enthalpy change for the reverse process of deposition is equal in magnitude but opposite in sign to that for sublimation: Consider the extent to which intermolecular attractions must be overcome to achieve a given phase transition. Converting a solid into a liquid requires that these attractions be only partially overcome; transition to the gaseous state requires that they be completely overcome. As a result, the enthalpy of fusion for a substance is less than its enthalpy of vaporization. This same logic can be used to derive an approximate relation between the enthalpies of all phase changes for a given substance. Though not an entirely accurate description, sublimation may be conveniently modeled as a sequential two-step process of melting followed by vaporization in order to apply Hess’s Law. Viewed in this manner, the enthalpy of sublimation for a substance may be estimated as the sum of its enthalpies of fusion and vaporization, as illustrated in Figure 7. For example: Figure 7. For a given substance, the sum of its enthalpy of fusion and enthalpy of vaporization is approximately equal to its enthalpy of sublimation. Heating and Cooling Curves In the chapter on thermochemistry, the relation between the amount of heat absorbed or related by a substance, q , and its accompanying temperature change, Δ T , was introduced: where m is the mass of the substance and c is its specific heat. The relation applies to matter being heated or cooled, but not undergoing a change in state. When a substance being heated or cooled reaches a temperature corresponding to one of its phase transitions, further gain or loss of heat is a result of diminishing or enhancing intermolecular attractions, instead of increasing or decreasing molecular kinetic energies. While a substance is undergoing a change in state, its temperature remains constant. Figure 8 shows a typical heating curve. Consider the example of heating a pot of water to boiling. A stove burner will supply heat at a roughly constant rate; initially, this heat serves to increase the water’s temperature. When the water reaches its boiling point, the temperature remains constant despite the continued input of heat from the stove burner. This same temperature is maintained by the water as long as it is boiling. If the burner setting is increased to provide heat at a greater rate, the water temperature does not rise, but instead the boiling becomes more vigorous (rapid). This behavior is observed for other phase transitions as well: For example, temperature remains constant while the change of state is in progress. Figure 8. A typical heating curve for a substance depicts changes in temperature that result as the substance absorbs increasing amounts of heat. Plateaus in the curve (regions of constant temperature) are exhibited when the substance undergoes phase transitions. Example 6: Total Heat Needed to Change Temperature and Phase for a Substance How much heat is required to convert 135 g of ice at −15 °C into water vapor at 120 °C? [reveal-answer q=”280448″]Show Answer[/reveal-answer] [hidden-answer a=”280448″] The transition described involves the following steps: Heat ice from −15 °C to 0 °C Melt ice Heat water from 0 °C to 100 °C Boil water Heat steam from 100 °C to 120 °C The heat needed to change the temperature of a given substance (with no change in phase) is: q = m c Δ T (see previous chapter on thermochemistry). The heat needed to induce a given change in phase is given by q = n Δ H . Using these equations with the appropriate values for specific heat of ice, water, and steam, and enthalpies of fusion and vaporization, we have: Converting the quantities in J to kJ permits them to be summed, yielding the total heat required: [/hidden-answer] Check Your Learning What is the total amount of heat released when 94.0 g water at 80.0 °C cools to form ice at −30.0 °C? [reveal-answer q=”29956″]Show Answer[/reveal-answer] [hidden-answer a=”29956″]40.5 kJ[/hidden-answer] Key Concepts and Summary Phase transitions are processes that convert matter from one physical state into another. There are six phase transitions between the three phases of matter. Melting, vaporization, and sublimation are all endothermic processes, requiring an input of heat to overcome intermolecular attractions. The reciprocal transitions of freezing, condensation, and deposition are all exothermic processes, involving heat as intermolecular attractive forces are established or strengthened. The temperatures at which phase transitions occur are determined by the relative strengths of intermolecular attractions and are, therefore, dependent on the chemical identity of the substance. Key Equations Exercises Heat is added to boiling water. Explain why the temperature of the boiling water does not change. What does change? Heat is added to ice at 0 °C. Explain why the temperature of the ice does not change. What does change? What feature characterizes the dynamic equilibrium between a liquid and its vapor in a closed container? Identify two common observations indicating some liquids have sufficient vapor pressures to noticeably evaporate? Identify two common observations indicating some solids, such as dry ice and mothballs, have vapor pressures sufficient to sublime? What is the relationship between the intermolecular forces in a liquid and its vapor pressure? What is the relationship between the intermolecular forces in a solid and its melting temperature? Why does spilled gasoline evaporate more rapidly on a hot day than on a cold day? Carbon tetrachloride, CCl 4 , was once used as a dry cleaning solvent, but is no longer used because it is carcinogenic. At 57.8 °C, the vapor pressure of CCl 4 is 54.0 kPa, and its enthalpy of vaporization is 33.05 kJ/mol. Use this information to estimate the normal boiling point for CCl 4 . When is the boiling point of a liquid equal to its normal boiling point? How does the boiling of a liquid differ from its evaporation? Use the information in Figure 3 to estimate the boiling point of water in Denver when the atmospheric pressure is 83.3 kPa. A syringe at a temperature of 20 °C is filled with liquid ether in such a way that there is no space for any vapor. If the temperature is kept constant and the plunger is withdrawn to create a volume that can be occupied by vapor, what would be the approximate pressure of the vapor produced? Explain the following observations: It takes longer to cook an egg in Ft. Davis, Texas (altitude, 5000 feet above sea level) than it does in Boston (at sea level). Perspiring is a mechanism for cooling the body. The enthalpy of vaporization of water is larger than its enthalpy of fusion. Explain why. Explain why the molar enthalpies of vaporization of the following substances increase in the order CH 4 < C 2 H 6 < C 3 H 8 , even though all three substances experience the same dispersion forces when in the liquid state. Explain why the enthalpies of vaporization of the following substances increase in the order CH 4 < NH 3 < H 2 O, even though all three substances have approximately the same molar mass. The enthalpy of vaporization of CO 2 ( l ) is 9.8 kJ/mol. Would you expect the enthalpy of vaporization of CS 2 ( l ) to be 28 kJ/mol, 9.8 kJ/mol, or −8.4 kJ/mol? Discuss the plausibility of each of these answers. The hydrogen fluoride molecule, HF, is more polar than a water molecule, H 2 O (for example, has a greater dipole moment), yet the molar enthalpy of vaporization for liquid hydrogen fluoride is lesser than that for water. Explain. Ethyl chloride (boiling point, 13 °C) is used as a local anesthetic. When the liquid is sprayed on the skin, it cools the skin enough to freeze and numb it. Explain the cooling effect of liquid ethyl chloride. Which contains the compounds listed correctly in order of increasing boiling points? N 2 < CS 2 < H 2 O < KCl H 2 O < N 2 < CS 2 < KCl N 2 < KCl < CS 2 < H 2 O CS 2 < N 2 < KCl < H 2 O KCl < H 2 O < CS 2 < N 2 How much heat is required to convert 422 g of liquid H 2 O at 23.5 °C into steam at 150 °C? Evaporation of sweat requires energy and thus take excess heat away from the body. Some of the water that you drink may eventually be converted into sweat and evaporate. If you drink a 20-ounce bottle of water that had been in the refrigerator at 3.8 °C, how much heat is needed to convert all of that water into sweat and then to vapor?? (Note: Your body temperature is 36.6 °C. For the purpose of solving this problem, assume that the thermal properties of sweat are the same as for water.) Titanium tetrachloride, TiCl 4 , has a melting point of −23.2 °C and has a Δ H fusion = 9.37 kJ/mol. How much energy is required to melt 263.1 g TiCl 4 ? For TiCl 4 , which will likely have the larger magnitude: Δ H fusion or Δ H vaporization ? Explain your reasoning. [reveal-answer q=”600180″]Show Selected Answers[/reveal-answer] [hidden-answer a=”600180″] 2. The heat is absorbed by the ice, providing the energy required to partially overcome intermolecular attractive forces in the solid and causing a phase transition to liquid water. The solution remains at 0 °C until all the ice is melted. Only the amount of water existing as ice changes until the ice disappears. Then the temperature of the water can rise. 4. We can see the amount of liquid in an open container decrease and we can smell the vapor of some liquids. 6. The vapor pressure of a liquid decreases as the strength of its intermolecular forces increases. 8. As the temperature increases, the average kinetic energy of the molecules of gasoline increases and so a greater fraction of molecules have sufficient energy to escape from the liquid than at lower temperatures. 10. When the pressure of gas above the liquid is exactly 1 atm. 12. Follow an imaginary horizontal line at 83.3 kPa to the curve representing the vapor pressure of water. Then drop a vertical line to the temperature axis. The intersection is at approximately 95 °C. 14. (a) At 5000 feet, the atmospheric pressure is lower than at sea level, and water will therefore boil at a lower temperature. This lower temperature will cause the physical and chemical changes involved in cooking the egg to proceed more slowly, and a longer time is required to fully cook the egg. (b) As long as the air surrounding the body contains less water vapor than the maximum that air can hold at that temperature, perspiration will evaporate, thereby cooling the body by removing the heat of vaporization required to vaporize the water. 16. Dispersion forces increase with molecular mass or size. As the number of atoms composing the molecules in this homologous series increases, so does the extent of intermolecular attraction via dispersion forces and, consequently, the energy required to overcome these forces and vaporize the liquids. 18. The boiling point of CS 2 is higher than that of CO 2 partially because of the higher molecular weight of CS 2 ; consequently, the attractive forces are stronger in CS 2 . It would be expected, therefore, that the heat of vaporization would be greater than that of 9.8 kJ/mol for CO 2 . A value of 28 kJ/mol would seem reasonable. A value of −8.4 kJ/mol would indicate a release of energy upon vaporization, which is clearly implausible. 20. The thermal energy (heat) needed to evaporate the liquid is removed from the skin. 22. Heat needed to bring this amount of water to the normal boiling point: Δ H 1 = mC s ΔT = (422 g)(4.184 J/g °C)(100.0 − 23.5) = 135,000 J Heat needed to vaporize this amount of water: Δ H 2 = n Δ H vap = (23.4 mol)(40,650 J/mol) = 951,000 J Heat to needed to increase the temperature of the steam: Δ H 3 = mC s ΔT = (422 g)(2.09 J/g °C)(150 − 100) = 44,100 J. Adding Δ H 1 , Δ H 2 , and Δ H 3 : 135,000 J + 951,000 J + 44,100 J = 1,130,000 J = 1130 kJ. 24. (a) . Heat required to melt this amount of TiCl 4 is n Δ H fusion = 1.385 mol 9.37 kJ/mol = 13.0kJ. (b) It is likely that the heat of vaporization will have a larger magnitude since in the case of vaporization the intermolecular interactions have to be completely overcome, while melting weakens or destroys only some of them. [/hidden-answer] Glossary boiling point: temperature at which the vapor pressure of a liquid equals the pressure of the gas above it Clausius-Clapeyron equation: mathematical relationship between the temperature, vapor pressure, and enthalpy of vaporization for a substance condensation: change from a gaseous to a liquid state deposition: change from a gaseous state directly to a solid state dynamic equilibrium: state of a system in which reciprocal processes are occurring at equal rates freezing: change from a liquid state to a solid state freezing point: temperature at which the solid and liquid phases of a substance are in equilibrium; see also melting point melting: change from a solid state to a liquid state melting point: temperature at which the solid and liquid phases of a substance are in equilibrium; see also freezing point normal boiling point: temperature at which a liquid’s vapor pressure equals 1 atm (760 torr) sublimation: change from solid state directly to gaseous state vapor pressure: (also, equilibrium vapor pressure) pressure exerted by a vapor in equilibrium with a solid or a liquid at a given temperature vaporization: change from liquid state to gaseous state CC licensed content, Shared previously Chemistry. Provided by : OpenStax College. Located at : http://openstaxcollege.org . License : CC BY: Attribution . License Terms : Download for free at https://openstaxcollege.org/textbooks/chemistry/get
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Courses/Saint_Marys_College_Notre_Dame_IN/CHEM_121L%3A_Principles_of_Chemistry_I_Laboratory/08%3A_Hess's_Law
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Learning Objectives To determine the enthalpy, \(\Delta H\), of several reactions experimentally. To use Hess's Law to determine the enthalpy, \(\Delta H\), for a reaction The study of energy is a vital function of chemistry. Over 90 percent of the energy we produce comes from chemical reactions such as the combustion of fossil fuels. This study of energy and its transformations is known as thermochemistry. As chemical bonds break and form in a chemical reaction, energy in the form of heat is either released or absorbed. Under conditions of constant pressure, the energy change of a reaction is called the heat of reaction or enthalpy change, \(\Delta H\). The amount of heat either absorbed or released during a reaction can be measured if the reaction is carried out in a container that insulates the reaction from its surroundings. By convention if heat is released, the reaction is termed exothermic and \(\Delta H\) is negative. If heat is absorbed, the reaction is termed endothermic and \(\Delta H\) is positive. In this lab, you will determine experimentally the value for \(\Delta H\) for the acid-base neutralization reaction involving hydrochloric acid and sodium hydroxide. Exothermic: \(\Delta H\) < 0 Endothermic: \(\Delta H\) > 0 We have defined the reactants as part of the system. The solvent (water in this case) and the calorimeter itself are the surroundings. By measuring the heat absorbed by the solvent and the calorimeter, we can infer the heat change for the reaction. This assumes, of course, that there is no heat loss. Using an apparatus called a calorimeter, the heat lost or absorbed by a reaction in the calorimeter can be calculated using the following equation: \(\Delta H\) reaction = \(\Delta H\) calorimeter + \(\Delta H\) solution For our purposes, we will assume that the heat gained or lost by the calorimeter itself is negligible. Thus, we will simply use the heat lost or gained by the solution. The enthalpy change, \(\Delta H\), is determined by measuring the temperature change, \(\Delta T\), and the mass of the system. The product of the mass, the specific heat and temperature change of the system will give the enthalpy change for the system. The heat capacity of 1 g of a substance is called its specific heat capacity (4.18 J/g°C for water). \(\Delta H\) = (specific heat of water) x (grams of solution) x (\(\Delta T\)) In this experiment we will measure the heat of neutralization of the following reaction: \[\ce{H^{+} (aq)} + \ce{OH^{-} (aq)} \rightarrow \ce{H_{2}^O} (l) \label{1}\] In addition we will measure the heats of reactions 2 and 3, shown below. Knowing the heat of reaction for 4, we will use Hess’s Law and reaction 2-4 to calculate the heat of reaction for 5. \[\ce{MgO (s)} + 2\space\ce{HCl (aq)} \rightarrow \ce{Mg^{2+} (aq)} + 2\space\ce{Cl^{-} (aq)} + \ce{H_{2}O (l)} \label{2}\] \[\ce{Mg (s)} + 2\space\ce{HCl (aq)} \rightarrow \ce{Mg^{2+} (aq)} + 2\space\ce{Cl^{-} (aq)} + \ce{H_{2} (g)} \label{3}\] \[\ce{H_{2} (g)} + \frac{1}{2}\space \ce{O_{2} (g)} \rightarrow \ce{H_{2}^O (l)} \label{4}\] \[\ce{Mg (s)} + \frac {1}{2}\space\ce{O_{2} (g)} \rightarrow 2\space\ce{MgO (s)} \label{5}\] Pre Lab Video Safety and Waste Disposal All solutions can be disposed of down the drain. Procedure Heat of Neutralization: \(\ce{H^{+} (aq)} + \ce{OH^{-} (aq)} \rightarrow \ce{H_{2}^O} (l) \) Step 1 Obtain a calorimeter consisting of a double Styrofoam cup, a magnetic stirrer and magnet, and the LabQuest with the temperature sensor. Step 2 Pour 25.0 mL of 2M sodium hydroxide (NaOH), measured with your graduated cylinder, into your calorimeter. Step 3 Measure 25.0 mL of 2M hydrochloric acid (HCl) in clean, graduated cylinder. Clean and dry the temperature sensor and support it in the NaOH solution. Step 4 Begin gently stirring the solution with the magnetic stirrer. Step 5 Tap the “Start” button. Allow several data points to be collected for 15 seconds to set the initial temperature. Step 6 With the run ongoing, add the acid (HCl) solution to the base (NaOH), Step 7 When the temperature starts to decline, tap the “stop” button. Click the “analyze” menu and then selecting “Statistics”, “Temperature” to record the maximum temperature reached and the initial temperature. Step 8 Empty your calorimeter (the materials can go down the drain), rinse, and dry all parts completely. Step 9 Repeat the procedure at least once more. If your data looks substantially different between the two runs, repeat a third time. Accurate results will be rewarded. Reaction #2: \(\ce{MgO (s)} + 2\space\ce{HCl (aq)} \rightarrow \ce{Mg^{2+} (aq)} + 2\space\ce{Cl^{-} (aq)} + \ce{H_{2}O (l)}\) Step 1 Measure out 50.0 mL of 1M HCl into the calorimeter. CAUTION: This is a different HCl concentration than the one used in Reaction # 1. Handle the HCl solution with care. It can cause painful burns if it comes in contact with the skin. Step 2 Weigh out about 0.50 g of magnesium oxide, MgO, on a piece of weighing paper. Record the exact mass used in your data table. CAUTION: Avoid inhaling magnesium oxide dust. Step 3 Tap the start button and record the temperature for 15 seconds. Then add the white magnesium oxide powder to the solution. Use the magnetic stirrer to gently to stir the cup contents until a maximum temperature has been reached. Record the minimum and maximum temperatures. The waste can go down the drain. Step 4 Repeat the procedure once more. Step 5 Clean and dry the calorimeter cup, the stir bar, and the thermometer. Reaction #3: \(\ce{Mg (s)} + 2\space\ce{HCl (aq)} \rightarrow \ce{Mg^{2+} (aq)} + 2\space\ce{Cl^{-} (aq)} + \ce{H_{2} (g)}\) Step 1 Repeat the above experiment using ~0.2 g of Mg. ribbon rather than the MgO powder. The magnesium ribbon has been pre-cut to the proper length. Be sure to record the measured mass of the magnesium and the initial and final temperatures. CAUTION: Do not breathe the vapors produced by this reaction! The waste can go down the drain. Processing the Data All calculations should be neatly presented in your laboratory notebook. Step 1 For each reaction the heat gained or lost by the solution is found using: \(q_{soln}\) = \(C_{p}•m•\Delta T\). [Cp = \(4.18\frac {J}{g^{o}C}\) and we assume that the solution has the same density as water,\(1\frac {g}{mL}\)] Step 2 Calculate the heat gained or lost by the reaction, \(q_{rnx}\) = \(-q_{soln}\) Step 3 Calculate \(\Delta H_{rxn}\) for each reaction. \(\Delta H_{rxn}\) = \(\frac {q_{rxn}}{moles}\). Step 4 Using Hess's law and your experimental \(\Delta H_{rxn}\) values for reactions 2 and 3, and the calculated value from the table in the back of the book for reaction 4 (\(\ce{H_{2} (g)} + \frac{1}{2}\space \ce{O_{2} (g)} \rightarrow \ce{H_{2}^O (l)}\)), to calculate the \(\Delta H_{rxn}\) for reaction 5.
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Courses/Lumen_Learning/First_Year_Seminar_(Lumen)/06%3A_Module_4-_Grit_Goal_Setting_and_Motivation/6.07%3A_Learn_What_You_Dont_Want
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Jamie Edwards For a long time, my plan had always been to be a kindergarten teacher. But when I began my undergraduate degree I fell into that ever-growing pool of college students who changed their major three times before graduation. I was swayed by family members, my peers, and the economy, but I eventually realized that I was investing my education in the wrong areas for the wrong reasons. It shouldn’t just be about salaries and job security. I needed to find that personal attachment. At eighteen, it’s hard to see your entire life spread out before you. College may feel like a free-for-all at times, but the reality is that it’s one of the most defining times of our lives. It should never be squandered. I started to imagine my life beyond college—what I found important and the type of lifestyle I wanted in the end. I started thinking about the classes that I was actually interested in—the ones that I looked forward to each week and arrived early to just so I could get a seat up front. A turning point for me was when I took the advice of a campus mentor and enrolled in a career exploration course. I learned more about myself in that class than I had in my entire three years at college prior to taking it. It showed me that my passion was something I had always thought about but never thought about as a career. In high school, I could sit in the Guidance Office for hours on end. I enjoyed listening to others—hearing and helping people work through their struggles. I had seen firsthand how detrimental the absence of career classes can be to someone’s future. Through this realization and my participation in my career exploration class, I saw a viable future in the Higher Education Administration field. As I dove deeper, I was opened to an incredible amount of unique and diverse opportunities to work with students. My main approach was to get a taste anything to do with student services: I shadowed a career counselor in a career services office, attended graduate school fairs and informational sessions, discussed the Higher Education Administration Program with several staff at my college, and most importantly, I talked with my internship coordinator (my mentor). From there, I completed an internship in my prospective field, which gave me a wealth of insight and skills that directly related to my future career goals. From where I sit now—my former personal and professional struggles in tow—I offer up some pieces of advice that were crucial to getting me where I am today. Whether you’re an undecided major who is looking for guidance or a student with a clearly defined career path, I suggest the following: Find a mentor —For me, everything began there. Without my mentor, I wouldn’t have done any of the other items I’m about to suggest. Finding the right mentor is crucial. Look for someone who can complement your personality (typically someone who’s the opposite of you). My advice would be to look beyond your direct supervisor for mentorship. It’s important to create an open forum with your mentor, because there may be a conflict of interest as you discuss work issues and other job opportunities. Potential mentors to consider are an instructor on campus, your academic advisor, a professional currently working in your prospective field, someone you admire in your community, or anyone in your network of friends or family that you feel comfortable discussing your future goals with. Enroll in a Career Exploration/Planning course, or something similar —Even if you do not see the effects of this course immediately (such as dramatically changing your major), you will notice the impact down the road. Making educated career choices and learning job readiness skills will always pay off in the end. Through my career exploration class, I learned how to relate my personality and values to potential career fields. These self-assessments changed my entire thought process, and I see that influence daily. Beyond changing the way you think, the knowledge you gain about effective job search strategies is invaluable. Learning how to write purposeful résumés and cover letters, finding the right approach to the interview process, and recognizing your strengths and weaknesses are just a few of the benefits you can gain from these type of courses. Complete a Job Shadow and/or Informational Interview —No amount of online research is going to give you the same experience as seeing a job at the front line. In a job shadow or an informational interview, you’re able to explore options with no commitment and see how your in-class experience can carry over to a real world setting. Additionally, you’re expanding your professional network by having that personal involvement. You never know how the connections you make might benefit you in the future. My only regret about job shadowing in college is that I didn’t do it sooner. Do an Internship —A main source of frustration for recent grads is the inability to secure an entry-level position without experience. “How do I get a job to gain experience when I can’t get a job without experience?” This is how: do an internship or two! Most colleges even have a course where you can obtain credit for doing it! Not only will you earn credits towards graduation, but you’ll gain the necessary experience to put on your résumé and discuss in future interviews. Having completed four internships throughout my college career, I can’t say they were all great. However, I don’t regret a single one. The first one showed me the type of field I didn’t want to work in. The second confirmed that I was heading in the right direction with my career. My third and fourth internships introduced me to completely different areas of higher education which broadened my knowledge and narrowed my search simultaneously. My takeaway is that sometimes you have to learn what you don’t want in order to find out what you do want. The more informed you are about career options through real life conversations and experiences, the better prepared you will be for your future and the more confident you will be in your career decisions. Always explore your options because even if you learn you hate it, at least you’re one step close to finding what you love. CC licensed content, Shared previously Foundations of Academic Success: Words of Wisdom. Authored by : Thomas C. Priester. Located at : http://textbooks.opensuny.org/foundations-of-academic-success/ . Project : Open SUNY Textbooks. License : CC BY-NC-SA: Attribution-NonCommercial-ShareAlike
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Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/zz%3A_Back_Matter/20%3A_Glossary
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Words (or words that have the same definition) The definition is case sensitive (Optional) Image to display with the definition [Not displayed in Glossary, only in pop-up on pages] (Optional) Caption for Image (Optional) External or Internal Link (Optional) Source for Definition (Eg. "Genetic, Hereditary, DNA ...") (Eg. "Relating to genes or heredity") NaN The infamous double helix https://bio.libretexts.org/ CC-BY-SA; Delmar Larsen Word(s) Definition Image Caption Link Source Sample Word 1 Sample Definition 1 NaN NaN NaN NaN
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Bookshelves/General_Chemistry/General_Chemistry_Supplement_(Eames)/Periodic_Trends
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Periodic trends are specific patterns that are present in the periodic table that illustrate different aspects of a certain element, including its size and its electronic properties. Major periodic trends include: electronegativity , ionization energy , electron affinity , atomic radius , melting point, and metallic character . Periodic trends, arising from the arrangement of the periodic table, provide chemists with an invaluable tool to quickly predict an element's properties. These trends exist because of the similar atomic structure of the elements within their respective group families or periods, and because of the periodic nature of the elements. Characteristics of Metals Characteristics of Nonmetals Effective Nuclear Charge Electron Affinity Ionization Energy Ionization energy is the energy needed to remove an electron from an atom or ion. Unlike atomic radii, we can and do measure ionization energies in the gas phase, when the atom or ion is not interacting with anything else. Sizes of Atoms and Ions Slater's Rules for Effective Nuclear Charge Effective nuclear charge determines the size and energy of orbitals, which determine most properties of atoms. Slater's rules give a simple approximation of effective nuclear charge, which depends on the number of electrons that might get between, so it depends on the electron we are looking at. For any electron, to find the effective nuclear charge it feels, we need to know how many other electrons might get in the way, and how much time it spends near the nucleus.
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Courses/Lumen_Learning/Book%3A_Writing_II_(Lumen)/04%3A_Unit_1%3A_Reading_Strand_A_%E2%80%93_%E2%80%9CPower_of_One%E2%80%9D/04.1%3A_Strand_A_Reading%3A_A_White_Heron_by_Sarah_Orne_Jewett
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by Sarah Orne Jewett About This Reading This story was written in the 1880s, a time when naturalists collected specimens for museums and private collections. As the supplemental reading “ Killing Birds ” states, this remains a current practice for some ornithologists. The woods were already filled with shadows one June evening, just before eight o’clock, though a bright sunset still glimmered faintly among the trunks of the trees. A little girl was driving home her cow, a plodding, dilatory, provoking creature in her behavior, but a valued companion for all that. They were going away from whatever light there was, and striking deep into the woods, but their feet were familiar with the path, and it was no matter whether their eyes could see it or not. There was hardly a night the summer through when the old cow could be found waiting at the pasture bars; on the contrary, it was her greatest pleasure to hide herself away among the huckleberry bushes, and though she wore a loud bell she had made the discovery that if one stood perfectly still it would not ring. So Sylvia had to hunt for her until she found her, and call Co’ ! Co’ ! with never an answering Moo, until her childish patience was quite spent. If the creature had not given good milk and plenty of it, the case would have seemed very different to her owners. Besides, Sylvia had all the time there was, and very little use to make of it. Sometimes in pleasant weather it was a consolation to look upon the cow’s pranks as an intelligent attempt to play hide and seek, and as the child had no playmates she lent herself to this amusement with a good deal of zest. Though this chase had been so long that the wary animal herself had given an unusual signal of her whereabouts, Sylvia had only laughed when she came upon Mistress Moolly at the swamp-side, and urged her affectionately homeward with a twig of birch leaves. The old cow was not inclined to wander farther, she even turned in the right direction for once as they left the pasture, and stepped along the road at a good pace. She was quite ready to be milked now, and seldom stopped to browse. Sylvia wondered what her grandmother would say because they were so late. It was a great while since she had left home at half-past five o’clock, but everybody knew the difficulty of making this errand a short one. Mrs. Tilley had chased the hornéd torment too many summer evenings herself to blame any one else for lingering, and was only thankful as she waited that she had Sylvia, nowadays, to give such valuable assistance. The good woman suspected that Sylvia loitered occasionally on her own account; there never was such a child for straying about out-of-doors since the world was made! Everybody said that it was a good change for a little maid who had tried to grow for eight years in a crowded manufacturing town, but, as for Sylvia herself, it seemed as if she never had been alive at all before she came to live at the farm. She thought often with wistful compassion of a wretched geranium that belonged to a town neighbor. “‘Afraid of folks,'” old Mrs. Tilley said to herself, with a smile, after she had made the unlikely choice of Sylvia from her daughter’s houseful of children, and was returning to the farm. “‘Afraid of folks,’ they said! I guess she won’t be troubled no great with ’em up to the old place!” When they reached the door of the lonely house and stopped to unlock it, and the cat came to purr loudly, and rub against them, a deserted pussy, indeed, but fat with young robins, Sylvia whispered that this was a beautiful place to live in, and she never should wish to go home. “ White Heron ” by birder62. CC-0 . The companions followed the shady wood-road, the cow taking slow steps and the child very fast ones. The cow stopped long at the brook to drink, as if the pasture were not half a swamp, and Sylvia stood still and waited, letting her bare feet cool themselves in the shoal water, while the great twilight moths struck softly against her. She waded on through the brook as the cow moved away, and listened to the thrushes with a heart that beat fast with pleasure. There was a stirring in the great boughs overhead. They were full of little birds and beasts that seemed to be wide awake, and going about their world, or else saying good-night to each other in sleepy twitters. Sylvia herself felt sleepy as she walked along. However, it was not much farther to the house, and the air was soft and sweet. She was not often in the woods so late as this, and it made her feel as if she were a part of the gray shadows and the moving leaves. She was just thinking how long it seemed since she first came to the farm a year ago, and wondering if everything went on in the noisy town just the same as when she was there, the thought of the great red-faced boy who used to chase and frighten her made her hurry along the path to escape from the shadow of the trees. Suddenly this little woods-girl is horror-stricken to hear a clear whistle not very far away. Not a bird’s-whistle, which would have a sort of friendliness, but a boy’s whistle, determined, and somewhat aggressive. Sylvia left the cow to whatever sad fate might await her, and stepped discreetly aside into the bushes, but she was just too late. The enemy had discovered her, and called out in a very cheerful and persuasive tone, “Halloa, little girl, how far is it to the road?” and trembling Sylvia answered almost inaudibly, “A good ways.” She did not dare to look boldly at the tall young man, who carried a gun over his shoulder, but she came out of her bush and again followed the cow, while he walked alongside. “I have been hunting for some birds,” the stranger said kindly, “and I have lost my way, and need a friend very much. Don’t be afraid,” he added gallantly. “Speak up and tell me what your name is, and whether you think I can spend the night at your house, and go out gunning early in the morning.” Sylvia was more alarmed than before. Would not her grandmother consider her much to blame? But who could have foreseen such an accident as this? It did not seem to be her fault, and she hung her head as if the stem of it were broken, but managed to answer “Sylvy,” with much effort when her companion again asked her name. Mrs. Tilley was standing in the doorway when the trio came into view. The cow gave a loud moo by way of explanation. “Yes, you’d better speak up for yourself, you old trial! Where’d she tucked herself away this time, Sylvy?” But Sylvia kept an awed silence; she knew by instinct that her grandmother did not comprehend the gravity of the situation. She must be mistaking the stranger for one of the farmer-lads of the region. The young man stood his gun beside the door, and dropped a lumpy game-bag beside it; then he bade Mrs. Tilley good-evening, and repeated his wayfarer’s story, and asked if he could have a night’s lodging. “Put me anywhere you like,” he said. “I must be off early in the morning, before day; but I am very hungry, indeed. You can give me some milk at any rate, that’s plain.” “Dear sakes, yes,” responded the hostess, whose long slumbering hospitality seemed to be easily awakened. “You might fare better if you went out to the main road a mile or so, but you’re welcome to what we’ve got. I’ll milk right off, and you make yourself at home. You can sleep on husks or feathers,” she proffered graciously. “I raised them all myself. There’s good pasturing for geese just below here towards the ma’sh. Now step round and set a plate for the gentleman, Sylvy!” And Sylvia promptly stepped. She was glad to have something to do, and she was hungry herself. It was a surprise to find so clean and comfortable a little dwelling in this New England wilderness. The young man had known the horrors of its most primitive housekeeping, and the dreary squalor of that level of society which does not rebel at the companionship of hens. This was the best thrift of an old-fashioned farmstead, though on such a small scale that it seemed like a hermitage. He listened eagerly to the old woman’s quaint talk, he watched Sylvia’s pale face and shining gray eyes with ever growing enthusiasm, and insisted that this was the best supper he had eaten for a month, and afterward the new-made friends sat down in the door-way together while the moon came up. Soon it would be berry-time, and Sylvia was a great help at picking. The cow was a good milker, though a plaguy thing to keep track of, the hostess gossiped frankly, adding presently that she had buried four children, so Sylvia’s mother, and a son (who might be dead) in California were all the children she had left. “Dan, my boy, was a great hand to go gunning,” she explained sadly. “I never wanted for pa’tridges or gray squer’ls while he was to home. He’s been a great wand’rer, I expect, and he’s no hand to write letters. There, I don’t blame him, I’d ha’ seen the world myself if it had been so I could. “Sylvy takes after him,” the grandmother continued affectionately, after a minute’s pause. “There ain’t a foot o’ ground she don’t know her way over, and the wild creaturs counts her one o’ themselves. Squer’ls she’ll tame to come an’ feed right out o’ her hands, and all sorts o’ birds. Last winter she got the jay-birds to bangeing here, and I believe she’d ‘a’ scanted herself of her own meals to have plenty to throw out amongst ’em, if I hadn’t kep’ watch. Anything but crows, I tell her, I’m willin’ to help support — though Dan he had a tamed one o’ them that did seem to have reason same as folks. It was round here a good spell after he went away. Dan an’ his father they didn’t hitch, — but he never held up his head ag’in after Dan had dared him an’ gone off.” “ Great White Heron ” provided by the National Gallery of Art. CC-0. The guest did not notice this hint of family sorrows in his eager interest in something else. “So Sylvy knows all about birds, does she?” he exclaimed, as he looked round at the little girl who sat, very demure but increasingly sleepy, in the moonlight. “I am making a collection of birds myself. I have been at it ever since I was a boy.” (Mrs. Tilley smiled.) “There are two or three very rare ones I have been hunting for these five years. I mean to get them on my own ground if they can be found.” “Do you cage ’em up?” asked Mrs. Tilley doubtfully, in response to this enthusiastic announcement. “Oh no, they’re stuffed and preserved, dozens and dozens of them,” said the ornithologist, “and I have shot or snared every one myself. I caught a glimpse of a white heron a few miles from here on Saturday, and I have followed it in this direction. They have never been found in this district at all. The little white heron, it is,” and he turned again to look at Sylvia with the hope of discovering that the rare bird was one of her acquaintances. But Sylvia was watching a hop-toad in the narrow footpath. “You would know the heron if you saw it,” the stranger continued eagerly. “A queer tall white bird with soft feathers and long thin legs. And it would have a nest perhaps in the top of a high tree, made of sticks, something like a hawk’s nest.” Sylvia’s heart gave a wild beat; she knew that strange white bird, and had once stolen softly near where it stood in some bright green swamp grass, away over at the other side of the woods. There was an open place where the sunshine always seemed strangely yellow and hot, where tall, nodding rushes grew, and her grandmother had warned her that she might sink in the soft black mud underneath and never be heard of more. Not far beyond were the salt marshes just this side the sea itself, which Sylvia wondered and dreamed much about, but never had seen, whose great voice could sometimes be heard above the noise of the woods on stormy nights. “I can’t think of anything I should like so much as to find that heron’s nest,” the handsome stranger was saying. “I would give ten dollars to anybody who could show it to me,” he added desperately, “and I mean to spend my whole vacation hunting for it if need be. Perhaps it was only migrating, or had been chased out of its own region by some bird of prey.” Mrs. Tilley gave amazed attention to all this, but Sylvia still watched the toad, not divining, as she might have done at some calmer time, that the creature wished to get to its hole under the door-step, and was much hindered by the unusual spectators at that hour of the evening. No amount of thought, that night, could decide how many wished-for treasures the ten dollars, so lightly spoken of, would buy. The next day the young sportsman hovered about the woods, and Sylvia kept him company, having lost her first fear of the friendly lad, who proved to be most kind and sympathetic. He told her many things about the birds and what they knew and where they lived and what they did with themselves. And he gave her a jack-knife, which she thought as great a treasure as if she were a desert-islander. All day long he did not once make her troubled or afraid except when he brought down some unsuspecting singing creature from its bough. Sylvia would have liked him vastly better without his gun; she could not understand why he killed the very birds he seemed to like so much. But as the day waned, Sylvia still watched the young man with loving admiration. She had never seen anybody so charming and delightful; the woman’s heart, asleep in the child, was vaguely thrilled by a dream of love. Some premonition of that great power stirred and swayed these young creatures who traversed the solemn woodlands with soft-footed silent care. They stopped to listen to a bird’s song; they pressed forward again eagerly, parting the branches — speaking to each other rarely and in whispers; the young man going first and Sylvia following, fascinated, a few steps behind, with her gray eyes dark with excitement. She grieved because the longed-for white heron was elusive, but she did not lead the guest, she only followed, and there was no such thing as speaking first. The sound of her own unquestioned voice would have terrified her — it was hard enough to answer yes or no when there was need of that. At last evening began to fall, and they drove the cow home together, and Sylvia smiled with pleasure when they came to the place where she heard the whistle and was afraid only the night before. II. Half a mile from home, at the farther edge of the woods, where the land was highest, a great pine-tree stood, the last of its generation. Whether it was left for a boundary mark, or for what reason, no one could say; the woodchoppers who had felled its mates were dead and gone long ago, and a whole forest of sturdy trees, pines and oaks and maples, had grown again. But the stately head of this old pine towered above them all and made a landmark for sea and shore miles and miles away. Sylvia knew it well. She had always believed that whoever climbed to the top of it could see the ocean; and the little girl had often laid her hand on the great rough trunk and looked up wistfully at those dark boughs that the wind always stirred, no matter how hot and still the air might be below. Now she thought of the tree with a new excitement, for why, if one climbed it at break of day, could not one see all the world, and easily discover from whence the white heron flew, and mark the place, and find the hidden nest? What a spirit of adventure, what wild ambition! What fancied triumph and delight and glory for the later morning when she could make known the secret! It was almost too real and too great for the childish heart to bear. All night the door of the little house stood open and the whippoorwills came and sang upon the very step. The young sportsman and his old hostess were sound asleep, but Sylvia’s great design kept her broad awake and watching. She forgot to think of sleep. The short summer night seemed as long as the winter darkness, and at last when the whippoorwills ceased, and she was afraid the morning would after all come too soon, she stole out of the house and followed the pasture path through the woods, hastening toward the open ground beyond, listening with a sense of comfort and companionship to the drowsy twitter of a half-awakened bird, whose perch she had jarred in passing. Alas, if the great wave of human interest which flooded for the first time this dull little life should sweep away the satisfactions of an existence heart to heart with nature and the dumb life of the forest! There was the huge tree asleep yet in the paling moonlight, and small and silly Sylvia began with utmost bravery to mount to the top of it, with tingling, eager blood coursing the channels of her whole frame, with her bare feet and fingers, that pinched and held like bird’s claws to the monstrous ladder reaching up, up, almost to the sky itself. First she must mount the white oak tree that grew alongside, where she was almost lost among the dark branches and the green leaves heavy and wet with dew; a bird fluttered off its nest, and a red squirrel ran to and fro and scolded pettishly at the harmless housebreaker. Sylvia felt her way easily. She had often climbed there, and knew that higher still one of the oak’s upper branches chafed against the pine trunk, just where its lower boughs were set close together. There, when she made the dangerous pass from one tree to the other, the great enterprise would really begin. “ Tree Pine Trunk ” by strannik22. CC-0 . She crept out along the swaying oak limb at last, and took the daring step across into the old pine-tree. The way was harder than she thought; she must reach far and hold fast, the sharp dry twigs caught and held her and scratched her like angry talons, the pitch made her thin little fingers clumsy and stiff as she went round and round the tree’s great stem, higher and higher upward. The sparrows and robins in the woods below were beginning to wake and twitter to the dawn, yet it seemed much lighter there aloft in the pine-tree, and the child knew she must hurry if her project were to be of any use. The tree seemed to lengthen itself out as she went up, and to reach farther and farther upward. It was like a great main-mast to the voyaging earth; it must truly have been amazed that morning through all its ponderous frame as it felt this determined spark of human spirit wending its way from higher branch to branch. Who knows how steadily the least twigs held themselves to advantage this light, weak creature on her way! The old pine must have loved his new dependent. More than all the hawks, and bats, and moths, and even the sweet voiced thrushes, was the brave, beating heart of the solitary gray-eyed child. And the tree stood still and frowned away the winds that June morning while the dawn grew bright in the east. Sylvia’s face was like a pale star, if one had seen it from the ground, when the last thorny bough was past, and she stood trembling and tired but wholly triumphant, high in the tree-top. Yes, there was the sea with the dawning sun making a golden dazzle over it, and toward that glorious east flew two hawks with slow-moving pinions. How low they looked in the air from that height when one had only seen them before far up, and dark against the blue sky. Their gray feathers were as soft as moths; they seemed only a little way from the tree, and Sylvia felt as if she too could go flying away among the clouds. Westward, the woodlands and farms reached miles and miles into the distance; here and there were church steeples, and white villages, truly it was a vast and awesome world The birds sang louder and louder. At last the sun came up bewilderingly bright. Sylvia could see the white sails of ships out at sea, and the clouds that were purple and rose-colored and yellow at first began to fade away. Where was the white heron’s nest in the sea of green branches, and was this wonderful sight and pageant of the world the only reward for having climbed to such a giddy height? Now look down again, Sylvia, where the green marsh is set among the shining birches and dark hemlocks; there where you saw the white heron once you will see him again; look, look! a white spot of him like a single floating feather comes up from the dead hemlock and grows larger, and rises, and comes close at last, and goes by the landmark pine with steady sweep of wing and outstretched slender neck and crested head. And wait! wait! do not move a foot or a finger, little girl, do not send an arrow of light and consciousness from your two eager eyes, for the heron has perched on a pine bough not far beyond yours, and cries back to his mate on the nest and plumes his feathers for the new day! The child gives a long sigh a minute later when a company of shouting cat-birds comes also to the tree, and vexed by their fluttering and lawlessness the solemn heron goes away. She knows his secret now, the wild, light, slender bird that floats and wavers, and goes back like an arrow presently to his home in the green world beneath. Then Sylvia, well satisfied, makes her perilous way down again, not daring to look far below the branch she stands on, ready to cry sometimes because her fingers ache and her lamed feet slip. Wondering over and over again what the stranger would say to her, and what he would think when she told him how to find his way straight to the heron’s nest. “Sylvy, Sylvy!” called the busy old grandmother again and again, but nobody answered, and the small husk bed was empty and Sylvia had disappeared. The guest waked from a dream, and remembering his day’s pleasure hurried to dress himself that it might sooner begin. He was sure from the way the shy little girl looked once or twice yesterday that she had at least seen the white heron, and now she must really be made to tell. Here she comes now, paler than ever, and her worn old frock is torn and tattered, and smeared with pine pitch. The grandmother and the sportsman stand in the door together and question her, and the splendid moment has come to speak of the dead hemlock-tree by the green marsh. But Sylvia does not speak after all, though the old grandmother fretfully rebukes her, and the young man’s kind, appealing eyes are looking straight in her own. He can make them rich with money; he has promised it, and they are poor now. He is so well worth making happy, and he waits to hear the story she can tell. No, she must keep silence! What is it that suddenly forbids her and makes her dumb? Has she been nine years growing and now, when the great world for the first time puts out a hand to her, must she thrust it aside for a bird’s sake? The murmur of the pine’s green branches is in her ears, she remembers how the white heron came flying through the golden air and how they watched the sea and the morning together, and Sylvia cannot speak; she cannot tell the heron’s secret and give its life away. Dear loyalty, that suffered a sharp pang as the guest went away disappointed later in the day, that could have served and followed him and loved him as a dog loves! Many a night Sylvia heard the echo of his whistle haunting the pasture path as she came home with the loitering cow. She forgot even her sorrow at the sharp report of his gun and the sight of thrushes and sparrows dropping silent to the ground, their songs hushed and their pretty feathers stained and wet with blood. Were the birds better friends than their hunter might have been, — who can tell? Whatever treasures were lost to her, woodlands and summer-time, remember! Bring your gifts and graces and tell your secrets to this lonely country child! CC licensed content, Shared previously White Heron. Provided by : Pixabay. Located at : https://pixabay.com/en/egret-great-egret-florida-birds-1040448/ . License : CC0: No Rights Reserved . License Terms : Pixabay License Great White Heron. Authored by : Robert Havell after John James Audubon. Provided by : National Gallery of Art. Located at : https://images.nga.gov/?service=asset&action=show_zoom_window_popup&language=en&asset=34813&location=grid&asset_list=60603,34813,111442,32862&basket_item_id=undefined . License : CC0: No Rights Reserved Tree Pine Trunk. Authored by : strannik22. Provided by : Pixabay . Located at : https://pixabay.com/en/tree-pine-trunk-491507/ . License : CC0: No Rights Reserved . License Terms : Pixabay License Public domain content A White Heron. Authored by : Sarah Orne Jewett. License : Public Domain: No Known Copyright
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Courses/Modesto_Junior_College/Chemistry_142%3A_Pre-General_Chemistry_(Brzezinski)/CHEM_142%3A_Text_(Brzezinski)/06%3A_Chemical_Compounds/6.06%3A_Lewis_Structures_of_Ionic_Compounds-_Electrons_Transferred/6.6.02%3A_Writing_Lewis_Structures_for_Covalent_Compounds
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↵ Learning Objectives Draw Lewis structures for covalent compounds. The following procedure can be used to construct Lewis electron structures for more complex molecules and ions. How-to: Constructing Lewis electron structures 1. Determine the total number of valence electrons in the molecule or ion. Add together the valence electrons from each atom. (Recall that the number of valence electrons is indicated by the position of the element in the periodic table.) If the species is a polyatomic ion, remember to add or subtract the number of electrons necessary to give the total charge on the ion. For CO 3 2 − , for example, we add two electrons to the total because of the −2 charge. 2. Arrange the atoms to show specific connections. When there is a central atom, it is usually the least electronegative element in the compound. Chemists usually list this central atom first in the chemical formula (as in CCl 4 and CO 3 2 − , which both have C as the central atom), which is another clue to the compound’s structure. Hydrogen and the halogens are almost always connected to only one other atom, so they are usually terminal rather than central. 3. Place a bonding pair of electrons between each pair of adjacent atoms to give a single bond. In H 2 O, for example, there is a bonding pair of electrons between oxygen and each hydrogen. 4. Beginning with the terminal atoms, add enough electrons to each atom to give each atom an octet (two for hydrogen). These electrons will usually be lone pairs. 5. If any electrons are left over, place them on the central atom. We will explain later that some atoms are able to accommodate more than eight electrons. 6. If the central atom has fewer electrons than an octet, use lone pairs from terminal atoms to form multiple (double or triple) bonds to the central atom to achieve an octet. This will not change the number of electrons on the terminal atoms. 7. Final check Always make sure all valence electrons are accounted for and that each atom has an octet of electrons, except for hydrogen (with two electrons). The central atom is usually the least electronegative element in the molecule or ion; hydrogen and the halogens are usually terminal. Now let’s apply this procedure to some particular compounds, beginning with one we have already discussed. Example \(\PageIndex{1}\): Water Write the Lewis Structure for H 2 O. Solution Steps for Writing Lewis Structures Example \(\PageIndex{1}\) 1. Determine the total number of valence electrons in the molecule or ion. Each H atom (group 1) has 1 valence electron, and the O atom (group 16) has 6 valence electrons, for a total of 8 valence electrons. 2. Arrange the atoms to show specific connections. H O H Because H atoms are almost always terminal, the arrangement within the molecule must be HOH. 3. Place a bonding pair of electrons between each pair of adjacent atoms to give a single bond. 4. Beginning with the terminal atoms, add enough electrons to each atom to give each atom an octet (two for hydrogen). Placing one bonding pair of electrons between the O atom and each H atom gives H -O- H with 4 electrons left over. Each H atom has a full valence shell of 2 electrons. 5. If any electrons are left over, place them on the central atom. Adding the remaining 4 electrons to the oxygen (as two lone pairs) gives the following structure: 6. If the central atom has fewer electrons than an octet, use lone pairs from terminal atoms to form multiple (double or triple) bonds to the central atom to achieve an octet. Not necessary. 7. Final check. The Lewis structure gives oxygen an octet and each hydrogen 2 electrons. Example \(\PageIndex{2}\) Write the Lewis structure for the \(CH_2O\) molecule Solution Steps for Writing Lewis Structures Example \(\PageIndex{2}\) 1. Determine the total number of valence electrons in the molecule or ion. Each hydrogen atom (group 1) has 1 valence electron, carbon (group 14) has 4 valence electrons, and oxygen (group 16) has 6 valence electrons, for a total of [(2)(1) + 4 + 6] = 12 valence electrons. 2. Arrange the atoms to show specific connections. Because carbon is less electronegative than oxygen and hydrogen is normally terminal, C must be the central atom. 3. Place a bonding pair of electrons between each pair of adjacent atoms to give a single bond. Placing a bonding pair of electrons between each pair of bonded atoms gives the following: 6 electrons are used, and 6 are left over. 4. Beginning with the terminal atoms, add enough electrons to each atom to give each atom an octet (two for hydrogen). Adding all 6 remaining electrons to oxygen (as three lone pairs) gives the following: Although oxygen now has an octet and each hydrogen has 2 electrons, carbon has only 6 electrons. 5. If any electrons are left over, place them on the central atom. Not necessary. There are no electrons left to place on the central atom. 6. If the central atom has fewer electrons than an octet, use lone pairs from terminal atoms to form multiple (double or triple) bonds to the central atom to achieve an octet. To give carbon an octet of electrons, we use one of the lone pairs of electrons on oxygen to form a carbon–oxygen double bond: 7. Final check Both the oxygen and the carbon now have an octet of electrons, so this is an acceptable Lewis electron structure. The O has two bonding pairs and two lone pairs, and C has four bonding pairs. This is the structure of formaldehyde, which is used in embalming fluid. Exercise \(\PageIndex{1}\) Write Lewis electron structures for CO 2 and SCl 2 , a vile-smelling, unstable red liquid that is used in the manufacture of rubber. Answer CO 2 . Answer SCl 2 . The United States Supreme Court has the unenviable task of deciding what the law is. This responsibility can be a major challenge when there is no clear principle involved or where there is a new situation not encountered before. Chemistry faces the same challenge in extending basic concepts to fit a new situation. Drawing of Lewis structures for polyatomic ions uses the same approach, but tweaks the process a little to fit a somewhat different set of circumstances. Writing Lewis Structures for Polyatomic Ions (CK-12) Recall that a polyatomic ion is a group of atoms that are covalently bonded together and which carry an overall electrical charge. The ammonium ion, \(\ce{NH_4^+}\), is formed when a hydrogen ion \(\left( \ce{H^+} \right)\) attaches to the lone pair of an ammonia \(\left( \ce{NH_3} \right)\) molecule in a coordinate covalent bond. When drawing the Lewis structure of a polyatomic ion, the charge of the ion is reflected in the number of total valence electrons in the structure. In the case of the ammonium ion: \(1 \: \ce{N}\) atom \(= 5\) valence electrons \(4 \: \ce{H}\) atoms \(= 4 \times 1 = 4\) valence electrons subtract 1 electron for the \(1+\) charge of the ion total of 8 valence electrons in the ion It is customary to put the Lewis structure of a polyatomic ion into a large set of brackets, with the charge of the ion as a superscript outside of the brackets. Exercise \(\PageIndex{2}\) Draw the Lewis electron dot structure for the sulfate ion. Answer ( CK12 License) Exceptions to the Octet Rule (BC Campus) As important and useful as the octet rule is in chemical bonding, there are some well-known violations. This does not mean that the octet rule is useless—quite the contrary. As with many rules, there are exceptions, or violations. There are three violations to the octet rule. Odd-electron molecules represent the first violation to the octet rule. Although they are few, some stable compounds have an odd number of electrons in their valence shells. With an odd number of electrons, at least one atom in the molecule will have to violate the octet rule. Examples of stable odd-electron molecules are NO , NO 2 , and ClO 2 . The Lewis electron dot diagram for NO is as follows: Although the O atom has an octet of electrons, the N atom has only seven electrons in its valence shell. Although NO is a stable compound, it is very chemically reactive, as are most other odd-electron compounds. Electron-deficient molecules represent the second violation to the octet rule. These stable compounds have less than eight electrons around an atom in the molecule. The most common examples are the covalent compounds of beryllium and boron. For example, beryllium can form two covalent bonds, resulting in only four electrons in its valence shell: Boron commonly makes only three covalent bonds, resulting in only six valence electrons around the B atom. A well-known example is BF 3 : The third violation to the octet rule is found in those compounds with more than eight electrons assigned to their valence shell. These are called expanded valence shell molecules. Such compounds are formed only by central atoms in the third row of the periodic table or beyond that have empty d orbitals in their valence shells that can participate in covalent bonding. One such compound is PF 5 . The only reasonable Lewis electron dot diagram for this compound has the P atom making five covalent bonds: Formally, the P atom has 10 electrons in its valence shell. Example \(\PageIndex{3}\): Octet Violations Identify each violation to the octet rule by drawing a Lewis electron dot diagram. ClO SF 6 Solution a. With one Cl atom and one O atom, this molecule has 6 + 7 = 13 valence electrons, so it is an odd-electron molecule. A Lewis electron dot diagram for this molecule is as follows: b. In SF 6 , the central S atom makes six covalent bonds to the six surrounding F atoms, so it is an expanded valence shell molecule. Its Lewis electron dot diagram is as follows: Exercise \(\PageIndex{3}\): Xenon Difluoride Identify the violation to the octet rule in XeF 2 by drawing a Lewis electron dot diagram. Answer The Xe atom has an expanded valence shell with more than eight electrons around it. Summary Lewis dot symbols provide a simple rationalization of why elements form compounds with the observed stoichiometries. A plot of the overall energy of a covalent bond as a function of internuclear distance is identical to a plot of an ionic pair because both result from attractive and repulsive forces between charged entities. In Lewis electron structures, we encounter bonding pairs , which are shared by two atoms, and lone pairs , which are not shared between atoms. Lewis structures for polyatomic ions follow the same rules as those for other covalent compounds. There are three violations to the octet rule: odd-electron molecules, electron-deficient molecules, and expanded valence shell molecules.
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Courses/Brevard_College/CHE_103_Principles_of_Chemistry_I/08%3A_Chemical_Reactions/8.04%3A_Classifying_Chemical_Reactions
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The reactions we have examined in the previous sections can be classified into a few simple types. Organizing reactions in this way is useful because it will assist us in predicting the products of unknown reactions. There are many different classifications of chemical reactions, but here we will focus on the following types: synthesis , decomposition , single replacement and double replacement . In addition, we will see that some of these reactions involve changes in the oxidation numbers of the reactants and products; these will be referred to as oxidation-reduction , or “ redox ” reactions (see figure below) Figure 8.1. Classification of Chemical reactions The first type of reaction we will consider is a synthesis reaction (also called a combination reaction ). In a synthesis reaction, elements or compounds undergo reaction and combine to form a single new substance. The reaction of sodium metal with chlorine gas to give sodium chloride is an example of a synthesis reaction where both reactants are elements. 2 Na (s) + Cl 2 (g) → 2 NaCl (s) In this reaction, sodium metal and chlorine gas have combined to yield (synthesize) the more complex molecule, sodium chloride. In decomposition reactions, a single compound will break down to form two or more new substances. The substances formed can be elements, compounds, or a mixture of both elements and compounds. Two simple examples of decomposition reactions are shown below. Cu 2 S (s) → 2 Cu (s) + S (s) CaCO 3 (s) → CaO (s) + CO 2 (g) In a single-replacement reaction (also called a single-displacement reaction) an element and a compound will react so that their elements are switched. In other words, an element will typically displace another element from within a compound. As a general rule, metals will replace metals in compounds and non-metals will typically replace non-metals . An example of a single replacement reaction is shown below. Zn (s) + CuCl 2 (s) → ZnCl 2 (s) + Cu (s) In this example, elemental zinc has displaced the metal, copper, from copper(II) chloride to form zinc chloride and elemental copper. In the reactants, zinc was elemental and in the products, it is present within the compound, zinc chloride. Likewise, copper was present in a compound in the reactants and is elemental in the products. In another example, iron metal will react with an aqueous solution of copper sulfate to give copper metal and iron(II) sulfate. Fe (s) + CuSO 4 (aq) → Cu (s) + FeSO 4 (aq) In this reaction, elemental iron replaces copper in a compound with sulfate anion and elemental copper metal is formed; metal replaces metal. The tendency of metals to replace other metals in single-replacement reactions is often referred to as an activity series, as we will learn next semester. A double-replacement reaction (or double-displacement) two ionic compounds in aqueous solution switch anions and form two new compounds. In order for a chemical reaction to occur in a double-replacement reaction, one of the new compounds that is formed must be insoluble in water, forming a solid precipitate . If both of the new compounds which are formed are water-soluble, then no reaction has occurred (as we will learn later). An example of a double-replacement reaction is shown below. BaCl 2 (aq) + Na 2 SO 4 (aq) → BaSO 4 (s) + 2 NaCl (aq) In this reaction, solid barium sulfate is formed as a precipitate. This is a chemical change and this is a valid chemical reaction. Paul R. Young , Professor of Chemistry, University of Illinois at Chicago, Wiki: AskTheNerd; PRY askthenerd.com - pyoung uic.edu; ChemistryOnline.com
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Courses/Widener_University/CHEM_145%3A_FA22_Van_Bramer/08%3A_Chemical_Bonding_and_Molecular_Geometry/8.06%3A_Strengths_of_Ionic_and_Covalent_Bonds
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Learning Objectives Describe the energetics of covalent and ionic bond formation and breakage Use the Born-Haber cycle to compute lattice energies for ionic compounds Use average covalent bond energies to estimate enthalpies of reaction A bond’s strength describes how strongly each atom is joined to another atom, and therefore how much energy is required to break the bond between the two atoms. In this section, you will learn about the bond strength of covalent bonds, and then compare that to the strength of ionic bonds, which is related to the lattice energy of a compound. Bond Strength: Covalent Bonds Stable molecules exist because covalent bonds hold the atoms together. We measure the strength of a covalent bond by the energy required to break it, that is, the energy necessary to separate the bonded atoms. Separating any pair of bonded atoms requires energy; the stronger a bond, the greater the energy required to break it. The energy required to break a specific covalent bond in one mole of gaseous molecules is called the bond energy or the bond dissociation energy. The bond energy for a diatomic molecule, \(D_{X–Y}\), is defined as the standard enthalpy change for the endothermic reaction: \[XY_{(g)}⟶X_{(g)}+Y_{(g)}\;\;\; D_{X−Y}=ΔH° \label{7.6.1}\] For example, the bond energy of the pure covalent H–H bond, \(\Delta_{H–H}\), is 436 kJ per mole of H–H bonds broken: \[H_{2(g)}⟶2H_{(g)} \;\;\; D_{H−H}=ΔH°=436kJ \label{EQ2}\] Breaking a bond always require energy to be added to the molecule. Correspondingly, making a bond always releases energy. Molecules with three or more atoms have two or more bonds. The sum of all bond energies in such a molecule is equal to the standard enthalpy change for the endothermic reaction that breaks all the bonds in the molecule. For example, the sum of the four C–H bond energies in CH 4 , 1660 kJ, is equal to the standard enthalpy change of the reaction: The average C–H bond energy, \(D_{C–H}\), is 1660/4 = 415 kJ/mol because there are four moles of C–H bonds broken per mole of the reaction. Although the four C–H bonds are equivalent in the original molecule, they do not each require the same energy to break; once the first bond is broken (which requires 439 kJ/mol), the remaining bonds are easier to break. The 415 kJ/mol value is the average, not the exact value required to break any one bond. The strength of a bond between two atoms increases as the number of electron pairs in the bond increases. Generally, as the bond strength increases, the bond length decreases. Thus, we find that triple bonds are stronger and shorter than double bonds between the same two atoms; likewise, double bonds are stronger and shorter than single bonds between the same two atoms. Average bond energies for some common bonds appear in Table \(\PageIndex{2}\), and a comparison of bond lengths and bond strengths for some common bonds appears in Table \(\PageIndex{2}\). When one atom bonds to various atoms in a group, the bond strength typically decreases as we move down the group. For example, C–F is 439 kJ/mol, C–Cl is 330 kJ/mol, and C–Br is 275 kJ/mol. Bond Bond Energy Unnamed: 2 Bond.1 Bond Energy.1 Unnamed: 5 Bond.2 Bond Energy.2 H–H 436 NaN C–S 260 NaN F–Cl 255.0 H–C 415 NaN C–Cl 330 NaN F–Br 235.0 H–N 390 NaN C–Br 275 NaN Si–Si 230.0 H–O 464 NaN C–I 240 NaN Si–P 215.0 H–F 569 NaN N–N 160 NaN Si–S 225.0 H–Si 395 NaN \(\mathrm{N=N}\) 418 NaN Si–Cl 359.0 H–P 320 NaN \(\mathrm{N≡N}\) 946 NaN Si–Br 290.0 H–S 340 NaN N–O 200 NaN Si–I 215.0 H–Cl 432 NaN N–F 270 NaN P–P 215.0 H–Br 370 NaN N–P 210 NaN P–S 230.0 H–I 295 NaN N–Cl 200 NaN P–Cl 330.0 C–C 345 NaN N–Br 245 NaN P–Br 270.0 \(\mathrm{C=C}\) 611 NaN O–O 140 NaN P–I 215.0 \(\mathrm{C≡C}\) 837 NaN \(\mathrm{O=O}\) 498 NaN S–S 215.0 C–N 290 NaN O–F 160 NaN S–Cl 250.0 \(\mathrm{C=N}\) 615 NaN O–Si 370 NaN S–Br 215.0 \(\mathrm{C≡N}\) 891 NaN O–P 350 NaN Cl–Cl 243.0 C–O 350 NaN O–Cl 205 NaN Cl–Br 220.0 \(\mathrm{C=O}\) 741 NaN O–I 200 NaN Cl–I 210.0 \(\mathrm{C≡O}\) 1080 NaN F–F 160 NaN Br–Br 190.0 C–F 439 NaN F–Si 540 NaN Br–I 180.0 C–Si 360 NaN F–P 489 NaN I–I 150.0 C–P 265 NaN F–S 285 NaN NaN NaN Bond Bond Length (Å) Bond Energy (kJ/mol) C–C 1.54 345 \(\mathrm{C=C}\) 1.34 611 \(\mathrm{C≡C}\) 1.20 837 C–N 1.43 290 \(\mathrm{C=N}\) 1.38 615 \(\mathrm{C≡N}\) 1.16 891 C–O 1.43 350 \(\mathrm{C=O}\) 1.23 741 \(\mathrm{C≡O}\) 1.13 1080 We can use bond energies to calculate approximate enthalpy changes for reactions where enthalpies of formation are not available. Calculations of this type will also tell us whether a reaction is exothermic or endothermic. An exothermic reaction (Δ H negative, heat produced) results when the bonds in the products are stronger than the bonds in the reactants. An endothermic reaction (Δ H positive, heat absorbed) results when the bonds in the products are weaker than those in the reactants. The enthalpy change, Δ H , for a chemical reaction is approximately equal to the sum of the energy required to break all bonds in the reactants (energy “in”, positive sign) plus the energy released when all bonds are formed in the products (energy “out,” negative sign). This can be expressed mathematically in the following way: \[\Delta H=\sum D_{\text{bonds broken}}− \sum D_{\text{bonds formed}} \label{EQ3}\] In this expression, the symbol \(\Sigma\) means “the sum of” and D represents the bond energy in kilojoules per mole, which is always a positive number. The bond energy is obtained from a table and will depend on whether the particular bond is a single, double, or triple bond. Thus, in calculating enthalpies in this manner, it is important that we consider the bonding in all reactants and products. Because D values are typically averages for one type of bond in many different molecules, this calculation provides a rough estimate, not an exact value, for the enthalpy of reaction. Consider the following reaction: \[\ce{H_{2(g)} + Cl_{2(g)}⟶2HCl_{(g)}} \label{EQ4}\] or \[\ce{H–H_{(g)} + Cl–Cl_{(g)}⟶2H–Cl_{(g)}} \label{\EQ5}\] To form two moles of HCl, one mole of H–H bonds and one mole of Cl–Cl bonds must be broken. The energy required to break these bonds is the sum of the bond energy of the H–H bond (436 kJ/mol) and the Cl–Cl bond (243 kJ/mol). During the reaction, two moles of H–Cl bonds are formed (bond energy = 432 kJ/mol), releasing 2 × 432 kJ; or 864 kJ. Because the bonds in the products are stronger than those in the reactants, the reaction releases more energy than it consumes: \[\begin {align*} ΔH&= \sum \mathrm{D_{bonds\: broken}}− \sum \mathrm{D_{bonds\: formed}}\\[4pt] &=\mathrm{[D_{H−H}+D_{Cl−Cl}]−2D_{H−Cl}}\\[4pt] &=\mathrm{[436+243]−2(432)=−185\:kJ} \end {align*}\] This excess energy is released as heat, so the reaction is exothermic. Table T2 gives a value for the standard molar enthalpy of formation of HCl(g), \(ΔH^\circ_\ce f\), of –92.307 kJ/mol. Twice that value is –184.6 kJ, which agrees well with the answer obtained earlier for the formation of two moles of HCl. Example \(\PageIndex{1}\): Using Bond Energies to Approximate Enthalpy Changes Methanol, CH 3 OH, may be an excellent alternative fuel. The high-temperature reaction of steam and carbon produces a mixture of the gases carbon monoxide, CO, and hydrogen, H 2 , from which methanol can be produced. Using the bond energies in Table \(\PageIndex{2}\), calculate the approximate enthalpy change, Δ H , for the reaction here: \[CO_{(g)}+2H2_{(g)}⟶CH_3OH_{(g)}\] Solution First, we need to write the Lewis structures of the reactants and the products: From this, we see that Δ H for this reaction involves the energy required to break a C–O triple bond and two H–H single bonds, as well as the energy produced by the formation of three C–H single bonds, a C–O single bond, and an O–H single bond. We can express this as follows (via Equation \ref{EQ3}): \[\begin {align*} ΔH&= \sum D_{bonds\: broken}− \sum D_{bonds\: formed}\\ ΔH&=\mathrm{[D_{C≡O}+2(D_{H−H})]−[3(D_{C−H})+D_{C−O}+D_{O−H}]} \end {align*}\] Using the bond energy values in Table \(\PageIndex{2}\), we obtain: \[\begin {align*} ΔH&=[1080+2(436)]−[3(415)+350+464]\\ &=\ce{−107\:kJ} \end {align*}\] We can compare this value to the value calculated based on \(ΔH^\circ_\ce f\) data from Appendix G: \[\begin {align*} ΔH&=[ΔH^\circ_{\ce f}\ce{CH3OH}(g)]−[ΔH^\circ_{\ce f}\ce{CO}(g)+2×ΔH^\circ_{\ce f}\ce{H2}]\\ &=[−201.0]−[−110.52+2×0]\\ &=\mathrm{−90.5\:kJ} \end {align*}\] Note that there is a fairly significant gap between the values calculated using the two different methods. This occurs because D values are the average of different bond strengths; therefore, they often give only rough agreement with other data. Exercise \(\PageIndex{1}\) Ethyl alcohol, CH 3 CH 2 OH, was one of the first organic chemicals deliberately synthesized by humans. It has many uses in industry, and it is the alcohol contained in alcoholic beverages. It can be obtained by the fermentation of sugar or synthesized by the hydration of ethylene in the following reaction: Using the bond energies in Table \(\PageIndex{2}\), calculate an approximate enthalpy change, Δ H , for this reaction. Answer –35 kJ Summary The strength of a covalent bond is measured by its bond dissociation energy, that is, the amount of energy required to break that particular bond in a mole of molecules. Multiple bonds are stronger than single bonds between the same atoms. The enthalpy of a reaction can be estimated based on the energy input required to break bonds and the energy released when new bonds are formed. For ionic bonds, the lattice energy is the energy required to separate one mole of a compound into its gas phase ions. Lattice energy increases for ions with higher charges and shorter distances between ions. Lattice energies are often calculated using the Born-Haber cycle, a thermochemical cycle including all of the energetic steps involved in converting elements into an ionic compound. Key Equations Bond energy for a diatomic molecule: \(\ce{XY}(g)⟶\ce{X}(g)+\ce{Y}(g)\hspace{20px}\ce{D_{X–Y}}=ΔH°\) Enthalpy change: Δ H = ƩD bonds broken – ƩD bonds formed Lattice energy for a solid MX: \(\ce{MX}(s)⟶\ce M^{n+}(g)+\ce X^{n−}(g)\hspace{20px}ΔH_\ce{lattice}\) Lattice energy for an ionic crystal: \(ΔH_\ce{lattice}=\mathrm{\dfrac{C(Z^+)(Z^-)}{R_o}}\) Footnotes This question is taken from the Chemistry Advanced Placement Examination and is used with the permission of the Educational Testing Service. Glossary bond energy (also, bond dissociation energy) energy required to break a covalent bond in a gaseous substance Born-Haber cycle thermochemical cycle relating the various energetic steps involved in the formation of an ionic solid from the relevant elements lattice energy (Δ H lattice ) energy required to separate one mole of an ionic solid into its component gaseous ions
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Courses/Harper_College/CHM_110%3A_Fundamentals_of_Chemistry/02%3A_Radiation-_Pros_and_Cons/2.09%3A_The_Bohr_Model_-_Atoms_with_Orbits
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Learning Objectives Define an energy level in terms of the Bohr model. Discuss how the Bohr model can be used to explain atomic spectra. Electric light bulbs contain a very thin wire in them that emits light when heated. The wire is called a filament. The particular wire used in light bulbs is made of tungsten. A wire made of any metal would emit light under these circumstances, but tungsten was chosen because the light it emits contains virtually every frequency and therefore, the light emitted by tungsten appears white. A wire made of some other element would emit light of some color that was not convenient for our uses. Every element emits light when energized by heating or passing electric current through it. Elements in solid form begin to glow when they are heated sufficiently, and elements in gaseous form emit light when electricity passes through them. This is the source of light emitted by neon signs and is also the source of light in a fire. Each Element Has a Unique Spectrum The light frequencies emitted by atoms are mixed together by our eyes so that we see a blended color. Several physicists, including Angstrom in 1868 and Balmer in 1875, passed the light from energized atoms through glass prisms in such a way that the light was spread out so they could see the individual frequencies that made up the light. The emission spectrum (or atomic spectrum ) of a chemical element is the unique pattern of light obtained when the element is subjected to heat or electricity. When hydrogen gas is placed into a tube and electric current passed through it, the color of emitted light is pink. But when the color is spread out, we see that the hydrogen spectrum is composed of four individual frequencies. The pink color of the tube is the result of our eyes blending the four colors. Every atom has its own characteristic spectrum; no two atomic spectra are alike. The image below shows the emission spectrum of iron. Because each element has a unique emission spectrum, elements can be defined using them. You may have heard or read about scientists discussing what elements are present in the sun or some more distant star, and after hearing that, wondered how scientists could know what elements were present in a place no one has ever been. Scientists determine what elements are present in distant stars by analyzing the light that comes from stars and finding the atomic spectrum of elements in that light. If the exact four lines that compose hydrogen's atomic spectrum are present in the light emitted from the star, that element contains hydrogen. Bohr's Model of the Atom By 1913, the concept of the atom had evolved from Dalton's indivisible spheres idea, to J. J. Thomson's plum pudding model, and then to Rutherford's nuclear atom theory. Rutherford, in addition to carrying out the brilliant experiment that demonstrated the presence of the atomic nucleus, also proposed that the electrons circled the nucleus in a planetary type motion. The solar system or planetary model of the atom was attractive to scientists because it was similar to something with which they were already familiar, namely the solar system. Unfortunately, there was a serious flaw in the planetary model. It was already known that when a charged particle (such as an electron) moves in a curved path, it gives off some form of light and loses energy in doing so. This is, after all, how we produce TV signals. If the electron circling the nucleus in an atom loses energy, it would necessarily have to move closer to the nucleus as it loses energy, and would eventually crash into the nucleus. Furthermore, Rutherford's model was unable to describe how electrons give off light forming each element's unique atomic spectrum. These difficulties cast a shadow on the planetary model and indicated that, eventually, it would have to be replaced. In 1913, the Danish physicist Niels Bohr proposed a model of the electron cloud of an atom in which electrons orbit the nucleus and were able to produce atomic spectra. Understanding Bohr's model requires some knowledge of electromagnetic radiation (or light). Energy Levels Bohr's key idea in his model of the atom is that electrons occupy definite orbitals that require the electron to have a specific amount of energy. In order for an electron to be in the electron cloud of an atom, it must be in one of the allowable orbitals and it must have the precise energy required for that orbit. Orbits closer to the nucleus would require smaller amounts of energy for an electron, and orbits farther from the nucleus would require the electron to have a greater amount of energy. The possible orbits are known as energy levels . One of the weaknesses of Bohr's model was that he could not offer a reason why only certain energy levels or orbits were allowed. Bohr hypothesized that the only way electrons could gain or lose energy would be to move from one energy level to another, thus gaining or losing precise amounts of energy. The energy levels are quantized , meaning that only specific amounts are possible. It would be like a ladder that had rungs only at certain heights. The only way you can be on that ladder is to be on one of the rungs, and the only way you could move up or down would be to move to one of the other rungs. Suppose we had such a ladder with 10 rungs. Other rules for the ladder are that only one person can be on a rung in the normal state, and the ladder occupants must be on the lowest rung available. If the ladder had five people on it, they would be on the lowest five rungs. In this situation, no person could move down because all of the lower rungs are full. Bohr worked out rules for the maximum number of electrons that could be in each energy level in his model, and required that an atom in its normal state (ground state) had all electrons in the lowest energy levels available. Under these circumstances, no electron could lose energy because no electron could move down to a lower energy level. In this way, Bohr's model explained why electrons circling the nucleus did not emit energy and spiral into the nucleus. Bohr's Model and Atomic Spectra The evidence used to support Bohr's model came from the atomic spectra. He suggested that an atomic spectrum is made by the electrons in an atom moving energy levels. The electrons typically have the lowest energy possible, called the ground state . If the electrons are given energy (through heat, electricity, light, etc.) the electrons in an atom could absorb energy by jumping to a higher energy level, or excited state . The electrons then give off the energy in the form of a piece of light—called a photon —that they had absorbed, to fall back to a lower energy level. The energy emitted by electrons dropping back to lower energy levels will always be precise amounts of energy, because the differences in energy levels are precise. This explains why you see specific lines of light when looking at an atomic spectrum—each line of light matches a specific "step down" that an electron can take in that atom. This also explains why each element produces a different atomic spectrum. Because each element has different acceptable energy levels for its electrons, the possible steps each element's electrons can take differ from all other elements. Summary Bohr's model suggests each atom has a set of unchangeable energy levels, and electrons in the electron cloud of that atom must be in one of those energy levels. Bohr's model suggests that the atomic spectra of atoms is produced by electrons gaining energy from some source, jumping up to a higher energy level, then immediately dropping back to a lower energy level and emitting the energy difference between the two energy levels. The existence of the atomic spectra is support for Bohr's model of the atom. Bohr's model was only successful in calculating energy levels for the hydrogen atom. Vocabulary Emission spectrum (or atomic spectrum ) - The unique pattern of light given off by an element when it is given energy. Energy levels - Possible orbits that an electron can have in the electron cloud of an atom. Ground state - To be in the lowest energy level possible. Excited state - To be in a higher energy level. Photon - A piece of electromagnetic radiation, or light, with a specific amount of energy. Quantized - To have a specific amount.
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Bookshelves/Analytical_Chemistry/Molecular_and_Atomic_Spectroscopy_(Wenzel)/2%3A_Ultraviolet_Visible_Absorption_Spectroscopy/2.3%3A_Effect_of_Non-bonding_Electrons
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Compare the UV absorption spectrum of benzene and pyridine. Benzene has a set of conjugated \(\pi\) - bonds and the lowest energy transition would be a \(\pi\) - \(\pi\) * transition as shown in Figure \(\PageIndex{17}\). The UV/VIS absorption spectrum for benzene is shown in Figure \(\PageIndex{18}\). Benzene absorbs radiation in the vacuum ultraviolet over the range from 160-208 nm with a \(\lambda\) max value of about 178 nm. Pyridine has a similar conjugation of double bonds comparable to what occurs in benzene. For pyridine, the lowest energy transition involves the n - \(\pi\) * orbitals and this will be much lower in energy than the \(\pi\) - \(\pi\) * transition in pyridine or benzene. The UV/VIS absorption spectrum of pyridine is shown in Figure \(\PageIndex{20}\). The shift toward higher wavelengths when compared to benzene is quite noticeable in the spectrum of pyridine, where the peaks from 320-380 nm represent the n - \(\pi\) * transition and the peak at about 240 nm is a \(\pi\) - \(\pi\) * transition. Note that intensity and therefore the molar absorptivity of the n - \(\pi\) * transition is lower than that of the \(\pi\) - \(\pi\) * transition. This is usually the case with organic compounds. Dye molecules absorb in the visible portion of the spectrum. They absorb wavelengths complementary to the color of the dye. Most \(\pi\) - \(\pi\) * transitions in organic molecules are in the ultraviolet portion of the spectrum unless the system is highly conjugated. Visible absorption is achieved in dye molecules by having a combination of conjugation and non-bonding electrons. Azo dyes with the N=N group are quite common, one example of which is shown in Figure \(\PageIndex{21}\).
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Bookshelves/Organic_Chemistry/Organic_Chemistry_I_(Liu)/10%3A_Alkenes_and_Alkynes/10.05%3A_Reaction_of_Alkenes-_Hydrogenation
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When alkenes react with hydrogen gas in the presence of a variety of metal catalysts, a hydrogen molecule will be added to the double bond in the way that each carbon atom bonded with one hydrogen atom, such addition reaction is called hydrogenation . Catalysts are must-have for hydrogenation, so the reaction can also be called catalytic hydrogenation . The commonly applied metal catalysts involve palladium and platinum. Palladium, which is used as a powder absorbed on charcoal to maximize the surface area, is the most common catalyst that is referred to as palladium on charcoal (Pd/carbon). Platinum, which is used usually as oxide PtO2, is also employed frequently and referred to as Adams catalyst. These metal catalysts are not soluble in the reaction mixture and therefore are described as heterogeneous catalysts. The heterogeneous catalyst can be easily filtrated out of the reaction mixture after reaction, and then be recycled and reused. The hydrogenation reaction does not take place without catalyst because of the enormous activation energy. The catalysts lower down the activation energy by weakening the H-H bond, and make the reaction feasible at room temperature. The details of the mechanism of catalytical hydrogenation are not completely clear. What was understood was that hydrogen gas is adsorbed on the surface of the metal, and the alkene also complexes with the metal by overlapping its π orbitals with vacant orbitals of the metal. The reaction occur on the surface of the metal catalyst, with both hydrogen atoms added from the same side of the alkene, to give alkane as the product that diffuses away from the metal surface. This mode of addition that the atoms added from the same side of the alkene is called the syn addition . Example:
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Courses/University_of_California_Davis/Chem_4B%3A_General_Chemistry_for_Majors_II_(Larsen)/Chem_4B_Textbook/Unit_I%3A_Chemical_Thermodynamics/II%3A_Principles_of_Thermodynamics_(Entropy_and_Gibbs_Energy)/2.03%3A_Entropy_and_Heat%3A_Experimental_Basis_of_the_Second_Law_of_Thermodynamics
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Learning Objectives The Learning Objective of this Module is to understand the relationship between internal energy and entropy. Thermodynamic Definition of Entropy Experiments show that the magnitude of ΔS vap is 80–90 J/(mol•K) for a wide variety of liquids with different boiling points. However, liquids that have highly ordered structures due to hydrogen bonding or other intermolecular interactions tend to have significantly higher values of \(ΔS_{vap}\). For instance, ΔS vap for water is 102 J/(mol•K). Another process that is accompanied by entropy changes is the formation of a solution. As illustrated in Figure \(\PageIndex{1}\), the formation of a liquid solution from a crystalline solid (the solute) and a liquid solvent is expected to result in an increase in the number of available microstates of the system and hence its entropy. Indeed, dissolving a substance such as NaCl in water disrupts both the ordered crystal lattice of \(\ce{NaCl}\) and the ordered hydrogen-bonded structure of water, leading to an increase in the entropy of the system. At the same time, however, each dissolved \(\ce{Na^{+}}\) ion becomes hydrated by an ordered arrangement of at least six water molecules, and the Cl − ions also cause the water to adopt a particular local structure. Both of these effects increase the order of the system, leading to a decrease in entropy. The overall entropy change for the formation of a solution therefore depends on the relative magnitudes of these opposing factors. In the case of an \(\ce{NaCl}\) solution, disruption of the crystalline \(\ce{NaCl}\) structure and the hydrogen-bonded interactions in water is quantitatively more important, so \(ΔS_{soln} > 0\). Dissolving \(\ce{NaCl}\) in water results in an increase in the entropy of the system. Each hydrated ion, however, forms an ordered arrangement with water molecules, which decreases the entropy of the system. The magnitude of the increase is greater than the magnitude of the decrease, so the overall entropy change for the formation of an NaCl solution is positive. Example \(\PageIndex{1}\) Predict which substance in each pair has the higher entropy and justify your answer. 1 mol of NH 3 (g) or 1 mol of He(g), both at 25°C 1 mol of Pb(s) at 25°C or 1 mol of Pb(l) at 800°C Given : amounts of substances and temperature Asked for : higher entropy Strategy : From the number of atoms present and the phase of each substance, predict which has the greater number of available microstates and hence the higher entropy. Solution : Both substances are gases at 25°C, but one consists of He atoms and the other consists of NH 3 molecules. With four atoms instead of one, the NH 3 molecules have more motions available, leading to a greater number of microstates. Hence we predict that the NH 3 sample will have the higher entropy. The nature of the atomic species is the same in both cases, but the phase is different: one sample is a solid, and one is a liquid. Based on the greater freedom of motion available to atoms in a liquid, we predict that the liquid sample will have the higher entropy. Exercise \(\PageIndex{1}\) Predict which substance in each pair has the higher entropy and justify your answer. 1 mol of He(g) at 10 K and 1 atm pressure or 1 mol of He(g) at 250°C and 0.2 atm a mixture of 3 mol of H 2 (g) and 1 mol of N 2 (g) at 25°C and 1 atm or a sample of 2 mol of NH 3 (g) at 25°C and 1 atm Answer a 1 mol of He(g) at 250°C and 0.2 atm (higher temperature and lower pressure indicate greater volume and more microstates) Answer b a mixture of 3 mol of H 2 (g) and 1 mol of N 2 (g) at 25°C and 1 atm (more molecules of gas are present)
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Courses/College_of_the_Canyons/Chem_201%3A_General_Chemistry_I_OER/13%3A_Solution/14.02%3A_Electrolytes
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Learning Objectives Define and give examples of electrolytes Distinguish between the physical and chemical changes that accompany dissolution of ionic and covalent electrolytes Relate electrolyte strength to solute-solvent attractive forces When some substances are dissolved in water, they undergo either a physical or a chemical change that yields ions in solution. These substances constitute an important class of compounds called electrolytes . Substances that do not yield ions when dissolved are called nonelectrolytes . If the physical or chemical process that generates the ions is essentially 100% efficient (all of the dissolved compound yields ions), then the substance is known as a strong electrolyte . If only a relatively small fraction of the dissolved substance undergoes the ion-producing process, it is called a weak electrolyte . Substances may be identified as strong, weak, or nonelectrolytes by measuring the electrical conductance of an aqueous solution containing the substance. To conduct electricity, a substance must contain freely mobile, charged species. Most familiar is the conduction of electricity through metallic wires, in which case the mobile, charged entities are electrons. Solutions may also conduct electricity if they contain dissolved ions, with conductivity increasing as ion concentration increases. Applying a voltage to electrodes immersed in a solution permits assessment of the relative concentration of dissolved ions, either quantitatively, by measuring the electrical current flow, or qualitatively, by observing the brightness of a light bulb included in the circuit (Figure \(\PageIndex{1}\)). Ionic Electrolytes Water and other polar molecules are attracted to ions, as shown in Figure \(\PageIndex{2}\). The electrostatic attraction between an ion and a molecule with a dipole is called an ion-dipole attraction . These attractions play an important role in the dissolution of ionic compounds in water. When ionic compounds dissolve in water, the ions in the solid separate and disperse uniformly throughout the solution because water molecules surround and solvate the ions, reducing the strong electrostatic forces between them. This process represents a physical change known as dissociation . Under most conditions, ionic compounds will dissociate nearly completely when dissolved, and so they are classified as strong electrolytes. Let us consider what happens at the microscopic level when we add solid KCl to water. Ion-dipole forces attract the positive (hydrogen) end of the polar water molecules to the negative chloride ions at the surface of the solid, and they attract the negative (oxygen) ends to the positive potassium ions. The water molecules penetrate between individual K + and Cl − ions and surround them, reducing the strong interionic forces that bind the ions together and letting them move off into solution as solvated ions, as Figure shows. The reduction of the electrostatic attraction permits the independent motion of each hydrated ion in a dilute solution, resulting in an increase in the disorder of the system as the ions change from their fixed and ordered positions in the crystal to mobile and much more disordered states in solution. This increased disorder is responsible for the dissolution of many ionic compounds, including KCl, which dissolve with absorption of heat. In other cases, the electrostatic attractions between the ions in a crystal are so large, or the ion-dipole attractive forces between the ions and water molecules are so weak, that the increase in disorder cannot compensate for the energy required to separate the ions, and the crystal is insoluble. Such is the case for compounds such as calcium carbonate (limestone), calcium phosphate (the inorganic component of bone), and iron oxide (rust). Why do Ionic Solids Dissolve in Water (Ion-Dipole IMF)?: https://youtu.be/yz1Ml0Q8b_I Covalent Electrolytes Pure water is an extremely poor conductor of electricity because it is only very slightly ionized—only about two out of every 1 billion molecules ionize at 25 °C. Water ionizes when one molecule of water gives up a proton to another molecule of water, yielding hydronium and hydroxide ions. \[\ce{H_2O (l)+ H_2O (l) \rightleftharpoons H_3O^{+} (aq) + OH^{−} (aq)} \label{11.3.2}\] In some cases, we find that solutions prepared from covalent compounds conduct electricity because the solute molecules react chemically with the solvent to produce ions. For example, pure hydrogen chloride is a gas consisting of covalent HCl molecules. This gas contains no ions. However, when we dissolve hydrogen chloride in water, we find that the solution is a very good conductor. The water molecules play an essential part in forming ions: Solutions of hydrogen chloride in many other solvents, such as benzene, do not conduct electricity and do not contain ions. Hydrogen chloride is an acid , and so its molecules react with water, transferring H + ions to form hydronium ions (\(H_3O^+\)) and chloride ions (Cl − ): This reaction is essentially 100% complete for HCl (i.e., it is a strong acid and, consequently, a strong electrolyte). Likewise, weak acids and bases that only react partially generate relatively low concentrations of ions when dissolved in water and are classified as weak electrolytes. The reader may wish to review the discussion of strong and weak acids provided in the earlier chapter of this text on reaction classes and stoichiometry. Summary Substances that dissolve in water to yield ions are called electrolytes. Electrolytes may be covalent compounds that chemically react with water to produce ions (for example, acids and bases), or they may be ionic compounds that dissociate to yield their constituent cations and anions, when dissolved. Dissolution of an ionic compound is facilitated by ion-dipole attractions between the ions of the compound and the polar water molecules. Soluble ionic substances and strong acids ionize completely and are strong electrolytes, while weak acids and bases ionize to only a small extent and are weak electrolytes. Nonelectrolytes are substances that do not produce ions when dissolved in water. Glossary dissociation physical process accompanying the dissolution of an ionic compound in which the compound’s constituent ions are solvated and dispersed throughout the solution electrolyte substance that produces ions when dissolved in water ion-dipole attraction electrostatic attraction between an ion and a polar molecule nonelectrolyte substance that does not produce ions when dissolved in water strong electrolyte substance that dissociates or ionizes completely when dissolved in water weak electrolyte substance that ionizes only partially when dissolved in water
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Bookshelves/Organic_Chemistry/Understanding_Organic_Chemistry_Through_Computation_(Boaz_and_Pearce)/02%3A_Bond_Lengths_and_Resonance/2.03%3A_Computational_Instructions
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Start by creating a folder on your desktop and naming it Resonance. Next, you should download the supplementary files associated with this exercise and move them to a folder that you just created. Contained within the supplementary files you will find a subfolder labeled DMA, which stands for dimethyl acetamide. Open DMA_coord.xyz in Avogadro to examine the structure of dimethyl acetamide. The starting structure of DMA should look like Figure 4 and show a functional group known as an amide, where a nitrogen is adjacent to a carbon-oxygen double bond. This file contains the input structure that we are going to tell our computational chemistry software package to optimize at the DFT level of theory. To optimize the structure of DMA we need to have an input file that tells Orca, the computational chemistry program that we are using, what we want to calculate and on what structure Orca should run the calculation. You can view this input file by clicking on DMA.inp within the DMA subfolder that you downloaded as part of the supplementary files. As shown in Figure 5, the input file is a brief text file with instructions for your computer. The first line of the input file, which starts with a number sign (#), is a comment indicating what the script is trying to accomplish. In this case, we are optimizing the geometry of DMA. The second line which begins with an exclamation point (!) tells Orca the functional of the calculation (B3LYP), the basis set (DEF2-SVP), and what we are asking the program to calculate (OPT-geometry optimization). Line 3, which beings with an asterisk (*), tells Orca what file contains the coordinate files for our molecule. The phrase xyzfile tells the computer that the file you are using has the molecule described using XYZ cartesian coordinates. There are then a series of two numbers which provide information about charge and spin of the chemical structure. The first number is the net charge of molecule(s) in the XYZ file. In this case it is zero because DMA is a neutral molecule. The second number is the net spin multiplicity (S) of the molecule that you are calculating, which in this case is 1 (singlet). We will talk more about the spin multiplicity of molecules in a later exercise, but for now you can think about this meaning that all electrons are paired so that they have a partner of the opposite spin. Finally, this line has the name of the coordinates file for DMA that we examined earlier. Because Orca does not have a graphical user interface (GUI), we will need to tell the computer to run the calculation using the command prompt of the computer. To do this, right click the start button on your PC and search for the command prompt. First, we need to tell the computer to look on the C drive and we do this by typing C: and hitting enter in the command prompt. Next, we need to tell the computer where the input script and the coordinates file are to run the calculation. We do this by typing cd (space) and pasting the file path. When you hit enter, the computer will paste a new line indicating that the current directory has changed, as shown in Figure 6A. To find the file path of your input script, right click on the input script within the DMA subfolder and select properties. The file path will appear under location, and you can highlight and copy this file path as shown in Figure 6B. Next, we will run the calculation by typing orca DMA.inp > DMA.out and pressing enter in the command prompt. At first, it may not appear like anything is happening, but the folder on your desktop labeled DMA will quickly become populated with the output of your calculation. Depending upon the speed of your computer, the calculation will take from 1-5 minutes, and upon completion the command prompt will print another line indicating that it is ready for the next command (Figure 7) You can examine your results by opening the final geometry file DMA.xyz in Avogadro. To measure the bond lengths and angles, click on the measure button (looks like a ruler) and click on the atoms whose bond lengths and angles you want to measure (Figure 8). References Hanwell, M. D.; Curtis, D. E.; Lonie, D. C.; Vandermeersch, T.; Zurek, E.; Hutchison, G. R. Avogadro: An Advanced Semantic Chemical Editor, Visualization, and Analysis Platform. J Cheminform 2012 , 4 (1), 17. https://doi.org/10.1186/1758-2946-4-17 . Avogadro: An Open-Source Molecular Builder and Visualization Tool. http://avogadro.cc/ . Neese, F. The ORCA Program System. WIREs Computational Molecular Science 2012 , 2 (1), 73–78. https://doi.org/10.1002/wcms.81 . Neese, F. Software Update: The ORCA Program System, Version 4.0. WIREs Computational Molecular Science 2018 , 8 (1), e1327. https://doi.org/10.1002/wcms.1327 . Neese, F.; Wennmohs, F.; Becker, U.; Riplinger, C. The ORCA Quantum Chemistry Program Package. J. Chem. Phys. 2020 , 152 (22), 224108. https://doi.org/10.1063/5.0004608 .
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Courses/BridgeValley_Community_and_Technical_College/Fundamentals_of_Chemistry/06%3A_Molecules_and_Compounds/6.12%3A_Molecular_Shapes
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Learning Objective Determine the shape of simple molecules. Molecules have shapes. There is an abundance of experimental evidence to that effect, from their physical properties to their chemical reactivity. Small molecules—molecules with a single central atom—have shapes that can be easily predicted. The basis of molecular shapes is called valence shell electron pair repulsion (VSEPR). VSEPR says that electron pairs, being composed of negatively charged particles, repel each other to get as far away from each other as possible. VSEPR makes a distinction between electron group geometry , which expresses how electron groups (bonds and nonbonding electron pairs) are arranged, and molecular geometry , which expresses how the atoms in a molecule are arranged. However, the two geometries are related. There are two types of electron groups : any type of bond—single, double, or triple—and lone electron pairs. When applying VSEPR to simple molecules, the first thing to do is to count the number of electron groups around the central atom. Remember that a multiple bond counts as only one electron group. Any molecule with only two atoms is linear. A molecule whose central atom contains only two electron groups orients those two groups as far apart from each other as possible—180° apart. When the two electron groups are 180° apart, the atoms attached to those electron groups are also 180° apart, so the overall molecular shape is linear. Examples include BeH 2 and CO 2 : A molecule with three electron groups orients the three groups as far apart as possible. They adopt the positions of an equilateral triangle—120° apart and in a plane. The shape of such molecules is trigonal planar . An example is BF 3 : Some substances have a trigonal planar electron group distribution, but have atoms bonded to only two of the three electron groups. An example is GeF 2 : From an electron group geometry perspective, GeF 2 has a trigonal planar shape, but its real shape is dictated by the positions of the atoms. This shape is called bent or angular . A molecule with four electron groups about the central atom orients the four groups in the direction of a tetrahedron, as shown in Figure \(\PageIndex{1}\): Tetrahedral Geometry. If there are four atoms attached to these electron groups, then the molecular shape is also tetrahedral . Methane (CH 4 ) is an example. This diagram of CH 4 illustrates the standard convention of displaying a three-dimensional molecule on a two-dimensional surface. The straight lines are in the plane of the page, the solid wedged line is coming out of the plane toward the reader, and the dashed wedged line is going out of the plane away from the reader. NH 3 is an example of a molecule whose central atom has four electron groups, but only three of them are bonded to surrounding atoms. Although the electron groups are oriented in the shape of a tetrahedron, from a molecular geometry perspective, the shape of NH 3 is trigonal pyramidal . H 2 O is an example of a molecule whose central atom has four electron groups, but only two of them are bonded to surrounding atoms. Although the electron groups are oriented in the shape of a tetrahedron, the shape of the molecule is bent or angular . A molecule with four electron groups about the central atom, but only one electron group bonded to another atom is linear because there are only two atoms in the molecule. Double or triple bonds count as a single electron group. CH 2 O has the following Lewis electron dot diagram. The central C atom has three electron groups around it because the double bond counts as one electron group. The three electron groups repel each other to adopt a trigonal planar shape: (The lone electron pairs on the O atom are omitted for clarity.) The molecule will not be a perfect equilateral triangle because the C–O double bond is different from the two C–H bonds, but both planar and triangular describe the appropriate approximate shape of this molecule. Example \(\PageIndex{1}\) What is the approximate shape of each molecule? \(\ce{PCl3}\) \(\ce{NOF}\) Solution The first step is to draw the Lewis electron dot diagram of the molecule. a: For PCl 3 , the electron dot diagram is as follows: The lone electron pairs on the Cl atoms are omitted for clarity. The P atom has four electron groups with three of them bonded to surrounding atoms, so the molecular shape is trigonal pyramidal. b: The electron dot diagram for NOF is as follows: The N atom has three electron groups on it, two of which are bonded to other atoms. The molecular shape is bent. Exercise \(\PageIndex{1}\) What is the approximate molecular shape of \(\ce{CH2Cl2}\)? Answer Tetrahedral Table \(\PageIndex{1}\) summarizes the shapes of molecules based on their number of electron groups and surrounding atoms. Number of Electron Groups on Central Atom Number of Surrounding Atoms Molecular Shape any 1 linear 2 2 linear 3 3 trigonal planar 3 2 bent 4 4 tetrahedral 4 3 trigonal pyramidal 4 2 bent Summary The approximate shape of a molecule can be predicted from the number of electron groups and the number of surrounding atoms.
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Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/02%3A_Atoms_and_The_Atomic_Theory/2.3%3A_The_Nuclear_Atom
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Learning Objectives To become familiar with the components and structure of the atom. Dissecting the Atom Once scientists concluded that all matter contains negatively charged electrons, it became clear that atoms, which are electrically neutral, must also contain positive charges to balance the negative ones. Thomson proposed that the electrons were embedded in a uniform sphere that contained both the positive charge and most of the mass of the atom, much like raisins in plum pudding or chocolate chips in a cookie (Figure \(\PageIndex{1}\)). In a single famous experiment, however, Rutherford showed unambiguously that Thomson’s model of the atom was incorrect. Rutherford aimed a stream of α particles at a very thin gold foil target (part (a) in Figure \(\PageIndex{2}\)) and examined how the α particles were scattered by the foil. Gold was chosen because it could be easily hammered into extremely thin sheets, minimizing the number of atoms in the target. If Thomson’s model of the atom were correct, the positively-charged α particles should crash through the uniformly distributed mass of the gold target like cannonballs through the side of a wooden house. They might be moving a little slower when they emerged, but they should pass essentially straight through the target (part (b) in Figure \(\PageIndex{2}\)). To Rutherford’s amazement, a small fraction of the α particles were deflected at large angles, and some were reflected directly back at the source (part (c) in Figure \(\PageIndex{2}\)). According to Rutherford, “It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you.” Rutherford’s results were not consistent with a model in which the mass and positive charge are distributed uniformly throughout the volume of an atom. Instead, they strongly suggested that both the mass and positive charge are concentrated in a tiny fraction of the volume of an atom, which Rutherford called the nucleus. It made sense that a small fraction of the α particles collided with the dense, positively charged nuclei in either a glancing fashion, resulting in large deflections, or almost head-on, causing them to be reflected straight back at the source. Although Rutherford could not explain why repulsions between the positive charges in nuclei that contained more than one positive charge did not cause the nucleus to disintegrate, he reasoned that repulsions between negatively charged electrons would cause the electrons to be uniformly distributed throughout the atom’s volume.Today it is known that strong nuclear forces, which are much stronger than electrostatic interactions, hold the protons and the neutrons together in the nucleus. For this and other insights, Rutherford was awarded the Nobel Prize in Chemistry in 1908. Unfortunately, Rutherford would have preferred to receive the Nobel Prize in Physics because he considered physics superior to chemistry. In his opinion, “All science is either physics or stamp collecting.” The historical development of the different models of the atom’s structure is summarized in Figure \(\PageIndex{3}\). Rutherford established that the nucleus of the hydrogen atom was a positively charged particle, for which he coined the name proton in 1920. He also suggested that the nuclei of elements other than hydrogen must contain electrically neutral particles with approximately the same mass as the proton. The neutron, however, was not discovered until 1932, when James Chadwick (1891–1974, a student of Rutherford; Nobel Prize in Physics, 1935) discovered it. As a result of Rutherford’s work, it became clear that an α particle contains two protons and neutrons, and is therefore the nucleus of a helium atom. Rutherford’s model of the atom is essentially the same as the modern model, except that it is now known that electrons are not uniformly distributed throughout an atom’s volume. Instead, they are distributed according to a set of principles described by Quantum Mechanics. Figure \(\PageIndex{4}\) shows how the model of the atom has evolved over time from the indivisible unit of Dalton to the modern view taught today. The Nuclear Atom The precise physical nature of atoms finally emerged from a series of elegant experiments carried out between 1895 and 1915. The most notable of these achievements was Ernest Rutherford's famous 1911 alpha-ray scattering experiment, which established that Almost all of the mass of an atom is contained within a tiny (and therefore extremely dense) nucleus which carries a positive electric charge whose value identifies each element and is known as the atomic number of the element. Almost all of the volume of an atom consists of empty space in which electrons, the fundamental carriers of negative electric charge, reside. The extremely small mass of the electron (1/1840 the mass of the hydrogen nucleus) causes it to behave as a quantum particle, which means that its location at any moment cannot be specified; the best we can do is describe its behavior in terms of the probability of its manifesting itself at any point in space. It is common (but somewhat misleading) to describe the volume of space in which the electrons of an atom have a significant probability of being found as the electron cloud . The latter has no definite outer boundary, so neither does the atom. The radius of an atom must be defined arbitrarily, such as the boundary in which the electron can be found with 95% probability. Atomic radii are typically 30-300 pm. The nucleus is itself composed of two kinds of particles. Protons are the carriers of positive electric charge in the nucleus; the proton charge is exactly the same as the electron charge, but of opposite sign. This means that in any [electrically neutral] atom, the number of protons in the nucleus (often referred to as the nuclear charge ) is balanced by the same number of electrons outside the nucleus. The other nuclear particle is the neutron . As its name implies, this particle carries no electrical charge. Its mass is almost the same as that of the proton. Most nuclei contain roughly equal numbers of neutrons and protons, so we can say that these two particles together account for almost all the mass of the atom. Note Because the electrons of an atom are in contact with the outside world, it is possible for one or more electrons to be lost, or some new ones to be added. The resulting electrically-charged atom is called an ion . Atoms consist of electrons, protons, and neutrons. This is an oversimplification that ignores the other subatomic particles that have been discovered, but it is sufficient for discussion of chemical principles. Some properties of these subatomic particles are summarized in Table \(\PageIndex{1}\) which illustrates three important points: Electrons and protons have electrical charges that are identical in magnitude but opposite in sign. Relative charges of −1 and +1 are assigned to the electron and proton, respectively. Neutrons have approximately the same mass as protons but no charge. They are electrically neutral. The mass of a proton or a neutron is about 1836 times greater than the mass of an electron. Protons and neutrons constitute the bulk of the mass of atoms. The discovery of the electron and the proton was crucial to the development of the modern model of the atom and provides an excellent case study in the application of the scientific method. In fact, the elucidation of the atom’s structure is one of the greatest detective stories in the history of science. Particle Mass (g) Atomic Mass (amu) Electrical Charge (coulombs) Relative Charge electron \(9.109 \times 10^{-28}\) 0.0005486 −1.602 × 10−19 −1 neutron \(1.675 \times 10^{-24}\) 1.008665 0 0 proton \(1.673 \times 10^{-24}\) 1.007276 +1.602 × 10−19 +1 * For a review of using scientific notation and units of measurement, see Essential Skills 1 (Section 1.9). * For a review of using scientific notation and units of measurement, see Essential Skills 1 (Section 1.9). * For a review of using scientific notation and units of measurement, see Essential Skills 1 (Section 1.9). * For a review of using scientific notation and units of measurement, see Essential Skills 1 (Section 1.9). * For a review of using scientific notation and units of measurement, see Essential Skills 1 (Section 1.9). The Nuclear Atom: https://youtu.be/eqoyZuv1tWA Summary The atom consists of discrete particles that govern its chemical and physical behavior. Atoms, the smallest particles of an element that exhibit the properties of that element, consist of negatively charged electrons around a central nucleus composed of more massive positively charged protons and electrically neutral neutrons. Radioactivity is the emission of energetic particles and rays (radiation) by some substances. Three important kinds of radiation are α particles (helium nuclei), β particles (electrons traveling at high speed), and γ rays (similar to x-rays but higher in energy). Conceptual Problems Describe the experiment that provided evidence that the proton is positively charged. What observation led Rutherford to propose the existence of the neutron? What is the difference between Rutherford’s model of the atom and the model chemists use today? If cathode rays are not deflected when they pass through a region of space, what does this imply about the presence or absence of a magnetic field perpendicular to the path of the rays in that region? Describe the outcome that would be expected from Rutherford’s experiment if the charge on α particles had remained the same but the nucleus were negatively charged. If the nucleus were neutral, what would have been the outcome? Describe the differences between an α particle, a β particle, and a γ ray. Which has the greatest ability to penetrate matter? Problems Please be sure you are familiar with the topics discussed in Essential Skills 1 ( Section 1.9 ) before proceeding to the Numerical Problems. Using the data in Table 1.3 and the periodic table (see Chapter 32 ), calculate the percentage of the mass of a silicon atom that is due to a. electrons. b. protons. Using the data in Table 1.3 and the periodic table (see Chapter 32 ), calculate the percentage of the mass of a helium atom that is due to a. electrons. b. protons. The radius of an atom is approximately 10 4 times larger than the radius of its nucleus. If the radius of the nucleus were 1.0 cm, what would be the radius of the atom in centimeters? in miles? The total charge on an oil drop was found to be 3.84 × 10 −18 coulombs. What is the total number of electrons contained in the drop?
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Courses/Centre_College/CHE_332%3A_Inorganic_Chemistry/08%3A_Coordination_Chemistry_-_Electronic_Spectra/8.03%3A_Electronic_Spectra_of_Coordination_Compounds/8.3.02%3A_Symmetry_labels_for_split_terms
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While the free ion terms symbols are derived using Russell-Saunders terms, we only briefly discussed how the term symbols may be found from a character table upon ligand-field splitting. The symmetry labels (called Mulliken Labels) that are used in character tables to describe the symmetry of wavefunctions were described in a previous section ( Section 4.3.3 ). Just as orbitals are wavefunctions that are described by the irreducible representations that are part of a character table, so are the terms. When we are discussing the terms of an octahedral complex, we can apply what we already know about octahedral symmetry and symmetry labels to understand how the terms labels are assigned. Upon close inspection of any of the Tanabe-Sugano diagrams shown previously, or in the Resources Section, you may notice that each free ion term (listed at the right of the diagram) splits into a term or set of terms that retain the same multiplicity (the left superscript). Under an octahedral field, all terms have symbols \(A, B, E\) or \(T\) because they are singly, doubly, or triply degenerate. While some of these terms have subscripts "1" or "2", all the terms also have "\(g\)" subscripts due to the symmetry of the \(d\) orbitals. Some of the symbols might also look familiar, as they are similar to the labels you know for splitting of \(d\) orbitals, like \(E_g\) and \(T_{2g}\). It is useful to know that under an octahedral field, the \(D, F, G, H, I\) terms split, but the terms \(S, P\) do not split. From inspection of the \(O_h\) character table, we find that free ion terms will split as follows: \[\begin{array}{|l|l|} \hline \textbf{Free Ion Term} & \textbf{Terms under } O_h\\ \hline \text{S} & A_{1g} \\ P & T_{1g} \\ D & T_{2g}+E_g \\ F & A_{2g}+T_{2g}+T_{1g} \\ G & A_{1g}+E_g+T_{2g}+T_{1g} \\ H & E_g+T_{1g}+T_{1g}+T_{2g} \\ I & A_{1g}+A_{2g}+E_g+T_{1g}+T_{2g}+T_{2g} \\ \hline \end{array} \nonumber \] The term symbols under an octahedral field have symmetry that matches symmetry of other elements within the molecule, including the \(d\)-orbitals. A reminder of their symmetry meanings is below (see also Section 4.3.3 ). Review of Mulliken Labels Each wavefunction (including terms) can be described by a Mulliken label (symmetry label). The labels used for terms in an octahedral ligand field are listed in the table below. \(\begin{array}{l|l} \hline \textbf{Mulliken Labels} & \textbf{meaning}\\ \hline \text{A or B} & \text{singly degenerate} \\ E & \text{doubly degenerate} \\ T & \text{triply degenerate} \\ \hline \textbf{Subscripts} & \textbf{meaning} \\ \hline 1 & \text{symmetric to } \sigma_v \text{ or perpendicular } C_2 \\ 2 & \text{anti-symmetric to } \sigma_v \text{ or perpendicular } C_2 \\ g & \text{symmetric to inversion center} \\ u & \text{anti-symmetric to inversion center} \\ \hline \end{array} \) We can use our knowledge of symmetry to derive the term labels for a given electron configuration under an octahedral field. Let's focus only on the capital letter that is assigned as the term symbol. Assignment of the octahedral term subscripts will not be described here, but are similar to assignment of these same subscripts to orbitals with matching symmetry. For example, the \(d^1\) electron configuration has a free ion ground state term of \(^2D\). According to the table shown above, these terms will split into \(E_g\) and \(T_{2g}\) terms. Which would be its ground state term under an octahedral field? And which is the excited state term? To answer these questions we need to look at the electron configurations of the ground and excited states in an octahedral field. It is convenient to know that the term symbol under an octahedral field indicates the degeneracies of the related electron configurations . Let's take a look at the electron configurations of the split \(d\)-orbitals to determine which term is the ground state. In the ground state of a \(d^1\) under an octahedral ligand field, there would be one electron in the \(t_{2g}\) orbitals, thus the electron configuration is \(t_{2g}^1e_g^0\). For simplicity we can consider only the cases where \(m_s\) values are \(+\frac{1}{2}\). There are three ways to put the electron in \(t_{2g}\) therefore, the ground state is triply degenerate, with a symbol "\(T\)". This is illustrated in the left panel of Figure \(\PageIndex{1}\). In the excited state of a \(d^1\) under an octahedral ligand field, there would be one electron in the \(e_{g}\) orbitals, or a configuration of \(t_2g^0e_g^1\). There are two ways to arrange the electron with \(m_s=+\frac{1}{2}\) in the \(e_{g}\) orbitals (Figure \(\PageIndex{1}\), left). Therefore, the excited state is doubly degenerate, with a symbol "\(E\)". In summary, the \(^2D\) free ion term for \(d^1\) splits into a lower-energy \(^2T_{2g}\) and a higher-energy \(^2E_g\) term under an octahedral field (Figure \(\PageIndex{1}\), left). We could use a similar treatment to derive the relative energies and terms of a \(d^9\) ion in an octahedral field. The ground state term of \(d^9\) is the same as for \(d^1\): it is \(^2D\). If we look at the ground state and excited state electron configurations of a \(d^9\) ion, we find that the ground state configuration is \(t_{2g}^6 e_g^3\), which is doubly degenerate. On the other hand, the excited state has an electron configuration of \(t_{2g}^5 e_g^4\), and has triple degeneracy. In summary, in the case of \(d^9\), the \(^2D\) free ion term splits into a lower energy \(^2E_g\) term and a higher-energy \(^2T_{2g}\) term. This is exactly the opposite of \(d^1\)! The \(d^1\) and \(d^9\) cases are opposite because they are related by what is said to be a "positive hole" concept. The \(d^9\) configuration could be derived from a \(d^{10}\) configuration with one positive "hole" (an electron removed). The positive hole is similar to the lone electron in \(d^1\) except that in the ground state, the hole is in \(e_g\), while the electron in \(d^1\) is in \(t_{2g}\); thus the two situations are exact opposites. Just as \(d^1\) and \(d^9\) are related by the hole concept, so are \(d^4\) and \(d^6\). If you consider that \(d^1\) is the addition of one electron to the totally symmetric \(d^0\) case, then perhaps you can see how \(d^6\) is similar in that it is the addition of one electron to the totally symmetric \(d^5\) case. Thus, \(d^1\) and \(d^6\) have similar splittings under the octahedral field. Similarly, \(d^9\) is created by removing one electron from the totally symmetric case of \(d^{10}\) and \(d^4\) is created by removing one electron from \(d^5\). Thus \(d^4\) and \(d^9\) have identical splittings for their ground state free ion terms, which in these cases are a \(D\) term. Due to the relationships between \(d^1, d^4, d^6,\) and \(d^9\), the same Orgel Diagram is used to show term splitting for these cases (Figure \(\PageIndex{2}\)). Orgel diagrams are yet another form of correlation diagrams. Just as there are relationships between \(d^1,d^4,d^6,\) and \(d^9\), there are similar relationships between \(d^2,d^3,d^7,\) and \(d^8\), and there is an Orgel diagram that can represent the splitting of their relevant free-ion terms. Tetrahedral splitting Because tetrahedral field splitting has the opposite pattern to octahedral field splitting of the \(d\)-orbitals, there are relationships between tetrahedral and octahedral terms. In general, for a tetrahedral field with \(d^n\) electrons, the Tanabe-Sugano diagram of an octahedral diagram for \(d^{10-n}\) can be used to interpret the tetrahedral case. The Orgel diagrams are labeled in general terms for this reason. They can be applied to both tetrahedral and octahedral cases. Jahn-Teller distortions We can also see what would happen to the terms in the case of Jahn-Teller distortion. In the case of a Jahn-Teller distortion (most common for \(d^9\) and high-spin \(d^4\) where the \(e_g\) orbitals are asymmetrically occupied), we can use the degeneracy of orbital electron configurations to identify term labels. Let's walk through an example using the most simple case: a \(d^1\) metal ion (Figure \(\PageIndex{3}\)). Jahn-teller usually causes a tetragonal distortion of an octahedron. This would be a change from \(O_h\) to \(D_{4h}\) symmetry. The \(t_{2g}\) orbitals (the set \(d_{xy}, d_{xz}, d_{yz}\) are split into \(e_g\) (\(d_{xz},d_{yz}\)) and \(b_{2g}\) (\(d_{xy}\)) by this symmetry change. Thus, we can expect any \(T_{2g}\) terms to be split similarly into \(E_g\) and \(B_{2g}\) terms by Jahn-Teller distortion. The \(e_g\) orbitals under \(O_h\) are split into \(a_1g\) (\(d^{z^2}\)) and \(b_1g\) (\(d_{x^2-y^2}\)) under \(D_{4h}\). Thus, we can expect \(E_g\) terms to be split into \(B_{1g}\) and \(A_{1g}\) terms by Jahn-Teller distortion (Figure \(\PageIndex{4}\)). Just as \(d^1\) and \(d^9\) gave opposite order of their terms under an octahedral field, they also give opposite order of terms in the Jahn-Teller distortion.
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Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Carboxylic_Acids/Properties_of_Carboxylic_Acids/Carboxyl_Derivatives/Carboxylic_Derivatives_-_Physical_Properties
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The important classes of organic compounds known as alcohols, phenols, ethers, amines and halides consist of alkyl and/or aryl groups bonded to hydroxyl, alkoxyl, amino and halo substituents respectively. If these same functional groups are attached to an acyl group (RCO–) their properties are substantially changed, and they are designated as carboxylic acid derivatives . Carboxylic acids have a hydroxyl group bonded to an acyl group, and their functional derivatives are prepared by replacement of the hydroxyl group with substituents, such as halo, alkoxyl, amino and acyloxy. Some examples of these functional derivatives were displayed earlier . The following table lists some representative derivatives and their boiling points. An aldehyde and ketone of equivalent molecular weight are also listed for comparison. Boiling points are given for 760 torr (atmospheric pressure), and those listed as a range are estimated from values obtained at lower pressures. The relatively high boiling point of carboxylic acids is due to extensive hydrogen bonded dimerization. Similar hydrogen bonding occurs between molecules of 1º and 2º-amides (amides having at least one N–H bond), and the first three compounds in the table serve as hydrogen bonding examples. Formula IUPAC Name Molecular Weight Boiling Point Water Solubility CH3(CH2)2CO2H butanoic acid 88.0 164 ºC very soluble CH3(CH2)2CONH2 butanamide 87.0 216-220 ºC soluble CH3CH2CONHCH3 N-methylpropanamide 87.0 205 -210 ºC soluble CH3CON(CH3)2 N,N-dimethylethanamide 87.0 166 ºC very soluble HCON(CH3)CH2CH3 N-ethyl, N-methylmethanamide 87.0 170-180 ºC very soluble CH3(CH2)3CN pentanenitrile 83.0 141 ºC slightly soluble CH3CO2CHO ethanoic methanoic anhydride 88.0 105-112 ºC reacts with water CH3CH2CO2CH3 methyl propanoate 88.0 80 ºC slightly soluble CH3CO2C2H5 ethyl ethanoate 88.0 77 ºC moderately soluble CH3CH2COCl propanoyl chloride 92.5 80 ºC reacts with water CH3(CH2)3CHO pentanal 86.0 103 ºC slightly soluble CH3(CH2)2COCH3 2-pentanone 86.0 102 ºC slightly soluble The last nine entries in the above table cannot function as hydrogen bond donors, so hydrogen bonded dimers and aggregates are not possible. The relatively high boiling points of equivalent 3º-amides and nitriles are probably due to the high polarity of these functions. Indeed, if hydrogen bonding is not present, the boiling points of comparable sized compounds correlate reasonably well with their dipole moments. Contributors William Reusch, Professor Emeritus ( Michigan State U. ), Virtual Textbook of Organic Chemistry
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Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/The_Live_Textbook_of_Physical_Chemistry_(Peverati)/12%3A_Phase_Equilibrium/12.01%3A_Phase_Stability
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We have already encountered the gas, liquid, and solid phases and already discussed some of their properties. These terms are intuitive since these are the three most common states of matter.\(^1\) For this reason, we have previously used the terms without the necessity of formally defining their meaning. However, a formal definition of “phase” is necessary to discuss several concepts in this chapter and the following ones: Definition: Phase A region of the system with homogeneous chemical composition and physical state. Let’s now use the total differential of the chemical potential and the definition of molar Gibbs free energy for one component: \[\begin{equation} \begin{aligned} d\mu &= \left( \dfrac{\partial \mu}{\partial T} \right)_P dT + \left( \dfrac{\partial \mu}{\partial P} \right)_T dP \\ d\mu &= -SdT+\overline{V}dP, \end{aligned} \end{equation} \label{12.1.1} \] to write: \[ \left( \dfrac{\partial \mu}{\partial T} \right)_P=-S \qquad \left( \dfrac{\partial \mu}{\partial P} \right)_T =\overline{V}. \label{12.1.2} \] We can use these definitions to study the dependence of the chemical potential with respect to changes in pressure and temperature. If we plot \(\mu\) as a function of \(T\) using the first coefficient in Equation \ref{12.1.2}, we obtain the diagram in Figure \(\PageIndex{1}\). The diagram presents three curves, each corresponding to one of the three most common states of matter – solid, liquid, and gas. As we saw in several previous chapters, the entropy of a phase is almost constant with respect to temperature,\(^2\) and therefore the three curves are essentially straight, with negative angular coefficients \(-S\). This also explains why the solid phase has a basically flat line since, according to the third law, the entropy of a perfect solid is zero and close to zero if the solid is not perfect. The difference between the three lines’ angular coefficients is explained by the fact that each of these states has a different value of entropy: \[ \left( \dfrac{\partial \mu_{\text{solid}}}{\partial T} \right)_P =-S_{\text{s}} \qquad \left( \dfrac{\partial \mu_{\text{liquid}}}{\partial T} \right)_P =-S_{\text{l}} \qquad \left( \dfrac{\partial \mu_{\text{gas}}}{\partial T} \right)_P =-S_{\text{g}}, \label{12.1.3} \] and since the entropy of a gas is always bigger than the entropy of a liquid, which in turn, is yet bigger than the entropy of a solid (\(S_{\text{g}} \gg S_{\text{l}}>S_{\text{s}}\)), we obtain three lines with different angular coefficients that intersect each other. At each temperature, the phase with the lowest chemical potential will be the most stable (see red segments in Figure \(\PageIndex{1}\)). At each intersection between two lines, the two phases have the same chemical potential, representing the temperature at which they coexist. This temperature is the temperature at which the phase change happens. Recalling from general chemistry, at the junction between the solid and the liquid lines, the fusion (fus) process occurs, and the corresponding temperature is called the melting point \(T_{\text{m}}\). At the junction between the liquid and the gas lines, the vaporization (vap) process happens, and the corresponding temperature is called the boiling point \(T_{\text{b}}\). Depending on the substance and the pressure at which the process happens, the solid line might intersect the gas line before the liquid line. When that occurs, the liquid phase is never observed, and only the sublimation (subl) process happens at the sublimation point \(T_{\text{subl}}\). The effects of pressure on this diagram can be studied using the second coefficient in Equation \ref{12.1.2}. For the majority of substances, \(\overline{V}_{\text{g}} \gg \overline{V}_{\text{l}} > \overline{V}_{\text{s}}\), hence the curves will shift to lower values when the pressure is reduced, as in Figure \(\PageIndex{2}\). Notice also that since \(\overline{V}_{\text{l}} \cong \overline{V}_{\text{s}}\), the shifts for both the solid and liquid lines is much smaller than the shift for the gas line. These shifts also translate to different values of the junctions, which means the phase changes will occur at different temperatures. Therefore both the melting point and the boiling point in general increase when pressure is increased (and vice versa). Notice how the change for the melting point is always much smaller than the change for the boiling point. Water is a noticeable exception to these trend because \(\overline{V}_{\mathrm{H}_2\mathrm{O,l}} < \overline{V}_{\text{ice}}\). This explains the experimental observation that increasing the pressure on ice causes the ice to melt\(^3\) Considering the intersections between two lines, two phases are in equilibrium with each other at each of these points. Therefore their chemical potentials must be equal: For two or more phases to be in equilibrium, their chemical potential must be equal: \[ \mu_{\alpha} = \mu_{\beta}. \label{12.1.4} \] If we now change either the temperature or the pressure, the location of the intersection will be shifted (see again Figure \(\PageIndex{2}\) and the discussion above). For infinitesimal changes in variables, the new location will be: \[ \mu_{\alpha} + d\mu_{\alpha}= \mu_{\beta}+d\mu_{\beta}, \label{12.1.5} \] which using Equation \ref{12.1.4}, simply becomes: \[ d\mu_{\alpha}= d\mu_{\beta}. \label{12.1.6} \] Replacing the differential with the definition of chemical potential in Equation \ref{12.1.1}, we obtain: \[\begin{equation} \begin{aligned} -S_{\alpha}dT+\overline{V}_{\alpha} &= -S_{\beta}dT+\overline{V}_{\beta} \\ \underbrace{\left(S_{\beta}-S_{\alpha}\right)}_{\Delta S} dT &= \underbrace{\left( \overline{V}_{\beta}-\overline{V}_{\alpha}\right)}_{\Delta \overline{V}}, \end{aligned} \end{equation} \label{12.1.7} \] which can be rearranged into: \[ \dfrac{dP}{dT}=\dfrac{\Delta S}{\Delta \overline{V}}. \label{12.1.8} \] This equation is known as the Clapeyron equation , and it is the mathematical relation at the basis of the pressure-temperature phase diagrams. Plotting the results of Equation \ref{12.1.8} on a \(PT\) phase diagram for common substances results in three lines representing the equilibrium between two different phases. These diagrams are useful to study the relationship between the phases of a substance. Other states of matter—such as plasma—are possible, but they are not usually observed at the values of temperature and pressure that classical thermodynamics is usually applied to. Discussion of these extreme cases is beyond the scope of this textbook.︎ Think, for example, at the integral \(\int SdT\), for which we can assume \(S\) independent of temperature to obtain \(S\Delta T\). In practice, the entropy increases slightly with the temperature. Therefore the curves in Figure \(\PageIndex{1}\) are slightly concave downwards (remember that they are obtained from values of \(-S\), so if \(S\) increase with \(T\), the curves bend downwards.)︎ Despite the effect being minimal, it is one of the contributing causes to the fact that we can skate on ice, but we can’t on stone. If we increase our pressure on ice by reducing our footprints’ surface area using thin skates, ice will slightly melt under our own weight, creating a thin liquid film on which we can skate because of the reduced friction.
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Bookshelves/General_Chemistry/Chemistry_2e_(OpenStax)/14%3A_Acid-Base_Equilibria/14.04%3A_Hydrolysis_of_Salt_Solutions
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Learning Objectives By the end of this section, you will be able to: Predict whether a salt solution will be acidic, basic, or neutral Calculate the concentrations of the various species in a salt solution Describe the acid ionization of hydrated metal ions Salts with Acidic Ions Salts are ionic compounds composed of cations and anions, either of which may be capable of undergoing an acid or base ionization reaction with water. Aqueous salt solutions, therefore, may be acidic, basic, or neutral, depending on the relative acid-base strengths of the salt's constituent ions. For example, dissolving ammonium chloride in water results in its dissociation, as described by the equation \[\ce{NH4Cl(s) \rightleftharpoons NH4^{+}(aq) + Cl^{-}(aq)} \nonumber \] The ammonium ion is the conjugate acid of the base ammonia, NH 3 ; its acid ionization (or acid hydrolysis) reaction is represented by \[\ce{NH4^{+}(aq) + H2O(l) \rightleftharpoons H3O^{+}(aq) + NH3(aq)} \quad K_{ a }=K_{ w } / K_{ b } \nonumber \] Since ammonia is a weak base, K b is measurable and K a > 0 (ammonium ion is a weak acid). The chloride ion is the conjugate base of hydrochloric acid, and so its base ionization (or base hydrolysis ) reaction is represented by \[\ce{Cl^{-}(aq) + H2O(l) \rightleftharpoons HCl(aq) + OH^{-}(aq)} \quad K_{ b }=K_{ w } / K_{ a } \nonumber \] Since \(\ce{HCl}\) is a strong acid, K a is immeasurably large and K b ≈ 0 (chloride ions don’t undergo appreciable hydrolysis). Thus, dissolving ammonium chloride in water yields a solution of weak acid cations (\(NH_4^{+}\)) and inert anions (\(Cl^{-})\), resulting in an acidic solution. Example \(\PageIndex{1}\): Calculating the pH of an Acidic Salt Solution Aniline is an amine that is used to manufacture dyes. It is isolated as anilinium chloride, a salt prepared by the reaction of the weak base aniline and hydrochloric acid. What is the pH of a 0.233 M solution of anilinium chloride \[\ce{C6H5NH3^{+}(aq) + H2O(l) \rightleftharpoons H3O^{+}(aq) + C6H5NH2(aq)} \nonumber \] Solution The K a for anilinium ion is derived from the K b for its conjugate base, aniline (see Appendix H): \[K_{ a }=\frac{K_{ w }}{K_{ b }}=\frac{1.0 \times 10^{-14}}{4.3 \times 10^{-10}}=2.3 \times 10^{-5} \nonumber \] Using the provided information, an ICE table for this system is prepared: Substituting these equilibrium concentration terms into the K a expression gives \[\begin{align*} K_a & =\dfrac{\left[ \ce{C6H5NH2} \right]\left[ \ce{H3O^{+}} \right] }{\left[ \ce{C6H5NH3^{+}}\right]} \\[4pt] 2.3 \times 10^{-5} & = \dfrac{(x)(x)}{0.233-x} \end{align*} \nonumber \] Assuming \(x \ll 0.233\), the equation is simplified and solved for \(x\): \[\begin{align*} & 2.3 \times 10^{-5}= \dfrac{x^2}{0.233} \\[4pt] & x=0.0023 ~\text{M} \end{align*} \nonumber \] The ICE table defines \(x\) as the hydronium ion molarity, and so the pH is computed as \[\text{pH} =-\log \left[ \ce{H3O^{+}} \right]=-\log (0.0023)=2.64 \nonumber \] Exercise \(\PageIndex{1}\) What is the hydronium ion concentration in a 0.100- M solution of ammonium nitrate, \(\ce{NH4NO3}\), a salt composed of the ions \(NH_4^{+}\) and \(NO_3^{-}\) and. Which is the stronger acid \(\ce{C6H5NH3+}\) or \(\ce{NH4+}\)? Answer \([\ce{H3O^{+}}] = 7.5 \times 10^{−6} ~\text{M}\); is the stronger acid. Salts with Basic Ions As another example, consider dissolving sodium acetate in water: \[\ce{NaCH_3CO_2(s) \rightleftharpoons Na^{+}(aq) + CH_3CO_2^{-}(aq)} \nonumber \] The sodium ion does not undergo appreciable acid or base ionization and has no effect on the solution pH. This may seem obvious from the ion's formula, which indicates no hydrogen or oxygen atoms, but some dissolved metal ions function as weak acids, as addressed later in this section. The acetate ion, \(\ce{CH3CO2^{-}(aq)}\), is the conjugate base of acetic acid, \(\ce{CH3CO2H}\), and so its base ionization (or base hydrolysis ) reaction is represented by \[\ce{CH3CO2^{-}(aq) + H2O(l) \rightleftharpoons CH_3 CO_2 H (aq) + OH -(aq)} \quad K_{ b }=K_{ w } / K_{ a } \nonumber \] Because acetic acid is a weak acid, its K a is measurable and K b > 0 (acetate ion is a weak base). Dissolving sodium acetate in water yields a solution of inert cations (Na + ) and weak base anions \(( \ce{CH3CO2^{-}})\), resulting in a basic solution. Example \(\PageIndex{2}\): Equilibrium in a Solution of a Salt of a Weak Acid and a Strong Base Determine the acetic acid concentration in a solution with and \([\ce{OH^{-}}] = 0.050~\text{M}\) and \([\ce{OH^{-}}] = 2.5 \times 10^{−6} M\) at equilibrium. The reaction is: \[\ce{CH_3CO_2^{-}(aq) + H2O(l) \rightleftharpoons CH_3CO_2H(aq) + OH^{-}(aq)} \nonumber \] Solution The provided equilibrium concentrations and a value for the equilibrium constant will permit calculation of the missing equilibrium concentration. The process in question is the base ionization of acetate ion, for which \[\ce{CH_3CO_2^{-}(aq) + H2O(l) \rightleftharpoons CH_3CO_2H(aq) + OH^{-}(aq)} \nonumber \] Substituting the available values into the K b expression gives \[\begin{align*} K_{ b } &=\frac{\left[ CH_3 CO_2 H \right]\left[ \ce{OH^{-}} \right]}{\left[ \ce{CH3CO2^{-}} \right]}=5.6 \times 10^{-10} \\[4pt] &=\frac{\left[ \ce{CH3CO2H} \right]\left(2.5 \times 10^{-6}\right)}{(0.050)}=5.6 \times 10^{-10} \end{align*} \nonumber \] Solving the above equation for the acetic acid molarity yields \([\ce{CH3CO2H}] = 1.1 \times 10^{−5}~ \text{M}\). Exercise \(\PageIndex{2}\) What is the pH of a 0.083- M solution of \(\ce{NaCN}\)? Answer 11.11 Salts with Acidic and Basic Ions Some salts are composed of both acidic and basic ions, and so the pH of their solutions will depend on the relative strengths of these two species. Likewise, some salts contain a single ion that is amphiprotic, and so the relative strengths of this ion’s acid and base character will determine its effect on solution pH. For both types of salts, a comparison of the K a and K b values allows prediction of the solution’s acid-base status, as illustrated in the following example exercise. Example \(\PageIndex{3}\): Determining the Acidic or Basic Nature of Salts Determine whether aqueous solutions of the following salts are acidic, basic, or neutral: KBr NaHCO 3 Na 2 HPO 4 NH 4 F Solution Consider each of the ions separately in terms of its effect on the pH of the solution, as shown here: (a) The \(\ce{K^{+}}\) cation is inert and will not affect pH. The bromide ion is the conjugate base of a strong acid, and so it is of negligible base strength (no appreciable base ionization). The solution is neutral. (b) The \(\ce{Na^{+}}\) cation is inert and will not affect the pH of the solution; while the \(\ce{HCO3^{−}}\) anion is amphiprotic. The \(K_a\) of \(\ce{HCO3^{-}}\) is \(4.7 \times 10^{−11}\), and its \(K_b\) is: \[\frac{1.0 \times 10^{-14}}{4.3 \times 10^{-7}}=2.3 \times 10^{-8} \nonumber \] Since \(K_b \gg K_a\), the solution is basic. (c) The \(\ce{Na^{+}}\) cation is inert and will not affect the pH of the solution, while the \(\ce{HPO4^{2−}}\) anion is amphiprotic. The \(K_a\) of \(\ce{HPO4^{2−}}\) is \(4.2 \times 10^{−13}\), and its \(K_b\) is: \[\frac{1.0 \times 10^{-14}}{6.2 \times 10^{-8}}=1.6 \times 10^{-7} \nonumber \] Because \(K_b \gg K_a\), the solution is basic. (d) The \(\ce{NH4^{+}}\) ion is acidic (see above discussion) and the \(\ce{F^{−}}\) ion is basic (conjugate base of the weak acid \(\ce{HF}\)). Comparing the two ionization constants: \(K_a\) of \(\ce{NH4^{+}}\) is 5.6 × 10−10 and the \(K_b\) of \(\ce{F^{−}}\) is \(1.6 \times 10^{−11}\), so the solution is acidic, since \(K_a > K_b\). Exercise \(\PageIndex{3}\) Determine whether aqueous solutions of the following salts are acidic, basic, or neutral: K 2 CO 3 CaCl 2 KH 2 PO 4 (NH 4 ) 2 CO 3 Answer (a) basic; (b) neutral; (c) acidic; (d) basic The Ionization of Hydrated Metal Ions Unlike the group 1 and 2 metal ions of the preceding examples (Na + , Ca 2 + , etc.), some metal ions function as acids in aqueous solutions. These ions are not just loosely solvated by water molecules when dissolved, instead they are covalently bonded to a fixed number of water molecules to yield a complex ion (see chapter on coordination chemistry). As an example, the dissolution of aluminum nitrate in water is typically represented as \[\ce{Al(NO3)3(s) \rightleftharpoons Al^{3+}(aq) +3 NO3^{-}(aq)} \nonumber \] However, the aluminum(III) ion actually reacts with six water molecules to form a stable complex ion, and so the more explicit representation of the dissolution process is \[\ce{Al(NO3)3(s) +6 H2O(l) \rightleftharpoons Al(H2O)6^{3+}(aq) +3 NO_3^{-}(aq)} \nonumber \] As shown in Figure \(\PageIndex{1}\), the ions involve bonds between a central Al atom and the O atoms of the six water molecules. Consequently, the bonded water molecules' O–H bonds are more polar than in nonbonded water molecules, making the bonded molecules more prone to donation of a hydrogen ion: \[\ce{Al(H2O)6^{3+}(aq) + H2O(l) \rightleftharpoons H3O^{+}(aq) + Al(H2O)5(OH)^{2+}(aq)} \quad \quad K_{ a }=1.4 \times 10^{-5} \nonumber \] The conjugate base produced by this process contains five other bonded water molecules capable of acting as acids, and so the sequential or step-wise transfer of protons is possible as depicted in few equations below: \[\begin{align*} \ce{Al(H2O)_6^{3+}(aq) + H2O(l) &\rightleftharpoons H3O^{+}(aq) + Al(H2O)_5(OH)^{2+}(aq)} \\[4pt] \ce{Al(H2O)_5(OH)^{2+}(aq) + H2O(l) &\rightleftharpoons H3O^{+}(aq) + Al(H2O)_4(OH)_2^{+}(aq)} \\[4pt] \ce{Al(H2O)_4(OH)_2^{+}(aq) + H2O(l) &\rightleftharpoons H3O^{+}(aq) + Al(H2O)_3(OH)_3(aq)} \end{align*} \nonumber \] This is an example of a polyprotic acid, the topic of discussion in a later section of this chapter. Aside from the alkali metals (group 1) and some alkaline earth metals (group 2), most other metal ions will undergo acid ionization to some extent when dissolved in water. The acid strength of these complex ions typically increases with increasing charge and decreasing size of the metal ions. The first-step acid ionization equations for a few other acidic metal ions are shown below: \[\begin{align*} \ce{Fe(H2O)_6^{3+}(aq) + H2O(l) &\rightleftharpoons H3O^{+}(aq) + Fe(H2O)_5(OH)^{2+}(aq)} && p K_{ a }=2.74 \\[4pt] \ce{Cu(H2O)_6^{2+}(aq) + H2O(l) &\rightleftharpoons H3O^{+}(aq) + Cu(H2O)_5(OH)^{+}(aq)} && p K_{ a }= \sim 6.3 \\[4pt] \ce{Zn(H2O)_4^{2+}(aq) + H2O(l) &\rightleftharpoons H3O^{+}(aq) + Zn(H2O)_3(OH)^{+}(aq)} && p K_{ a }=9.6 \end{align*} \nonumber \] Example \(\PageIndex{4}\): Hydrolysis of [Al(H 2 O) 6 ] 3+ Calculate the pH of a 0.10- M solution of aluminum chloride, which dissolves completely to give the hydrated aluminum ion (\(\ce{[Al(H2O)6]^{3+}}\)) in solution. Solution The equation for the reaction and K a are: \[\ce{Al(H2O)6^{3+}(aq) + H2O(l) \rightleftharpoons H3O^{+}(aq) + Al(H2O)5(OH)^{2+}(aq)} \quad K_{ a }=1.4 \times 10^{-5} \nonumber \] An ICE table with the provided information is Substituting the expressions for the equilibrium concentrations into the equation for the ionization constant yields: \[\begin{align*} K_{ a } &=\frac{\left[ \ce{H3O^{+}} \right]\left[ \ce{Al(H2O)5(OH)^{2+}} \right]}{\left[ \ce{Al(H2O)6^{3+}} \right]}\\[4pt] &=\frac{(x)(x)}{0.10-x}=1.4 \times 10^{-5} \end{align*} \nonumber \] Assuming \(x \ll 0.10\) and solving the simplified equation gives: \[x=1.2 \times 10^{-3} ~\text{M} \nonumber \] The ICE table defined \(x\) as equal to the hydronium ion concentration, and so the pH is calculated to be \[[\ce{H3O^{+}}]=0+x=1.2 \times 10^{-3}~\text{M} \nonumber \] \[\text{pH} =-\log \left[ \ce{H3O^{+}} \right]=2.92 \nonumber \] Therefore this is an acidic solution. Exercise \(\PageIndex{4}\) What is \([\ce{Al(H2O)5(OH)^{2+}}]\) in a 0.15-M solution of \(\ce{Al(NO3)3}\) that contains enough of the strong acid \(\ce{HNO3}\) to bring [\(\ce{H3O^{+}}\)] to 0.10 M ? Answer \(2.1 \times 10^{−5} M\)
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Courses/Mount_Aloysius_College/CHEM_100%3A_General_Chemistry_(O'Connor)/18%3A_Amino_Acids_Proteins_and_Enzymes/18.08%3A_Enzyme_Inhibition
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↵ Learning Objectives Explain what an enzyme inhibitor is. Distinguish between reversible and irreversible inhibitors. Distinguish between competitive and noncompetitive inhibitors. Previously, we noted that enzymes are inactivated at high temperatures and by changes in pH. These are nonspecific factors that would inactivate any enzyme. The activity of enzymes can also be regulated by more specific inhibitors. Many compounds are poisons because they bind covalently to particular enzymes or kinds of enzymes and inactivate them (Table \(\PageIndex{1}\)). Poison Formula Example of Enzyme Inhibited Action arsenate \(\ce{AsO4^{3−}}\) glyceraldehyde 3-phosphate dehydrogenase substitutes for phosphate iodoacetate \(\ce{ICH2COO^{−}}\) triose phosphate dehydrogenase binds to cysteine \(\ce{SH}\) group diisopropylfluoro-phosphate (DIFP; a nerve poison) NaN acetylcholinesterase binds to serine \(\ce{OH}\) group Irreversible Inhibition: Poisons An irreversible inhibitor inactivates an enzyme by bonding covalently to a particular group at the active site. The inhibitor-enzyme bond is so strong that the inhibition cannot be reversed by the addition of excess substrate. The nerve gases, especially Diisopropyl fluorophosphate (DIFP), irreversibly inhibit biological systems by forming an enzyme-inhibitor complex with a specific OH group of serine situated at the active sites of certain enzymes. The peptidases trypsin and chymotrypsin contain serine groups at the active site and are inhibited by DIFP . Reversible Inhibition A reversible inhibitor inactivates an enzyme through noncovalent, more easily reversed, interactions. Unlike an irreversible inhibitor, a reversible inhibitor can dissociate from the enzyme. Reversible inhibitors include competitive inhibitors and noncompetitive inhibitors. (There are additional types of reversible inhibitors.) A competitive inhibitor is any compound that bears a structural resemblance to a particular substrate and thus competes with that substrate for binding at the active site of an enzyme. The inhibitor is not acted on by the enzyme but does prevent the substrate from approaching the active site. The degree to which a competitive inhibitor interferes with an enzyme’s activity depends on the relative concentrations of the substrate and the inhibitor. If the inhibitor is present in relatively large quantities, it will initially block most of the active sites. But because the binding is reversible, some substrate molecules will eventually bind to the active site and be converted to product. Increasing the substrate concentration promotes displacement of the inhibitor from the active site. Competitive inhibition can be completely reversed by adding substrate so that it reaches a much higher concentration than that of the inhibitor. Studies of competitive inhibition have provided helpful information about certain enzyme-substrate complexes and the interactions of specific groups at the active sites. As a result, pharmaceutical companies have synthesized drugs that competitively inhibit metabolic processes in bacteria and certain cancer cells. Many drugs are competitive inhibitors of specific enzymes. A classic example of competitive inhibition is the effect of malonate on the enzyme activity of succinate dehydrogenase (Figure \(\PageIndex{1}\)). Malonate and succinate are the anions of dicarboxylic acids and contain three and four carbon atoms, respectively. The malonate molecule binds to the active site because the spacing of its carboxyl groups is not greatly different from that of succinate. However, no catalytic reaction occurs because malonate does not have a CH 2 CH 2 group to convert to CH=CH. This reaction will also be discussed in connection with the Krebs cycle and energy production. To Your Health: Penicillin Chemotherapy is the strategic use of chemicals (that is, drugs) to destroy infectious microorganisms or cancer cells without causing excessive damage to the other, healthy cells of the host. From bacteria to humans, the metabolic pathways of all living organisms are quite similar, so the search for safe and effective chemotherapeutic agents is a formidable task. Many well-established chemotherapeutic drugs function by inhibiting a critical enzyme in the cells of the invading organism. An antibiotic is a compound that kills bacteria; it may come from a natural source such as molds or be synthesized with a structure analogous to a naturally occurring antibacterial compound. Antibiotics constitute no well-defined class of chemically related substances, but many of them work by effectively inhibiting a variety of enzymes essential to bacterial growth. Penicillin, one of the most widely used antibiotics in the world, was fortuitously discovered by Alexander Fleming in 1928, when he noticed antibacterial properties in a mold growing on a bacterial culture plate. In 1938, Ernst Chain and Howard Florey began an intensive effort to isolate penicillin from the mold and study its properties. The large quantities of penicillin needed for this research became available through development of a corn-based nutrient medium that the mold loved and through the discovery of a higher-yielding strain of mold at a United States Department of Agriculture research center near Peoria, Illinois. Even so, it was not until 1944 that large quantities of penicillin were being produced and made available for the treatment of bacterial infections. Penicillin functions by interfering with the synthesis of cell walls of reproducing bacteria. It does so by inhibiting an enzyme—transpeptidase—that catalyzes the last step in bacterial cell-wall biosynthesis. The defective walls cause bacterial cells to burst. Human cells are not affected because they have cell membranes, not cell walls. Several naturally occurring penicillins have been isolated. They are distinguished by different R groups connected to a common structure: a four-member cyclic amide (called a lactam ring) fused to a five-member ring. The addition of appropriate organic compounds to the culture medium leads to the production of the different kinds of penicillin. The penicillins are effective against gram-positive bacteria (bacteria capable of being stained by Gram’s stain) and a few gram-negative bacteria (including the intestinal bacterium Escherichia coli ). They are effective in the treatment of diphtheria, gonorrhea, pneumonia, syphilis, many pus infections, and certain types of boils. Penicillin G was the earliest penicillin to be used on a wide scale. However, it cannot be administered orally because it is quite unstable; the acidic pH of the stomach converts it to an inactive derivative. The major oral penicillins—penicillin V, ampicillin, and amoxicillin—on the other hand, are acid stable. Some strains of bacteria become resistant to penicillin through a mutation that allows them to synthesize an enzyme—penicillinase—that breaks the antibiotic down (by cleavage of the amide linkage in the lactam ring). To combat these strains, scientists have synthesized penicillin analogs (such as methicillin) that are not inactivated by penicillinase. Some people (perhaps 5% of the population) are allergic to penicillin and therefore must be treated with other antibiotics. Their allergic reaction can be so severe that a fatal coma may occur if penicillin is inadvertently administered to them. Fortunately, several other antibiotics have been discovered. Most, including aureomycin and streptomycin, are the products of microbial synthesis. Others, such as the semisynthetic penicillins and tetracyclines, are made by chemical modifications of antibiotics; and some, like chloramphenicol, are manufactured entirely by chemical synthesis. They are as effective as penicillin in destroying infectious microorganisms. Many of these antibiotics exert their effects by blocking protein synthesis in microorganisms. Initially, antibiotics were considered miracle drugs, substantially reducing the number of deaths from blood poisoning, pneumonia, and other infectious diseases. Some seven decades ago, a person with a major infection almost always died. Today, such deaths are rare. Seven decades ago, pneumonia was a dreaded killer of people of all ages. Today, it kills only the very old or those ill from other causes. Antibiotics have indeed worked miracles in our time, but even miracle drugs have limitations. Not long after the drugs were first used, disease organisms began to develop strains resistant to them. In a race to stay ahead of resistant bacterial strains, scientists continue to seek new antibiotics. The penicillins have now been partially displaced by related compounds, such as the cephalosporins and vancomycin. Unfortunately, some strains of bacteria have already shown resistance to these antibiotics. Some reversible inhibitors are noncompetitive. A noncompetitive inhibitor can combine with either the free enzyme or the enzyme-substrate complex because its binding site on the enzyme is distinct from the active site. Binding of this kind of inhibitor alters the three-dimensional conformation of the enzyme, changing the configuration of the active site with one of two results. Either the enzyme-substrate complex does not form at its normal rate, or, once formed, it does not yield products at the normal rate. Because the inhibitor does not structurally resemble the substrate, the addition of excess substrate does not reverse the inhibitory effect. Feedback inhibition is a normal biochemical process that makes use of noncompetitive inhibitors to control some enzymatic activity. In this process, the final product inhibits the enzyme that catalyzes the first step in a series of reactions. Feedback inhibition is used to regulate the synthesis of many amino acids. For example, bacteria synthesize isoleucine from threonine in a series of five enzyme-catalyzed steps. As the concentration of isoleucine increases, some of it binds as a noncompetitive inhibitor to the first enzyme of the series (threonine deaminase), thus bringing about a decrease in the amount of isoleucine being formed (Figure \(\PageIndex{2}\)). Summary An irreversible inhibitor inactivates an enzyme by bonding covalently to a particular group at the active site. A reversible inhibitor inactivates an enzyme through noncovalent, reversible interactions. A competitive inhibitor competes with the substrate for binding at the active site of the enzyme. A noncompetitive inhibitor binds at a site distinct from the active site.
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Courses/University_of_North_Carolina_Charlotte/CHEM_2141%3A__Survey_of_Physical_Chemistry/05%3A_Helmholtz_and_Gibbs_Energies/5.06%3A_Thermodynamic_Functions_have_Natural_Variables
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The fundamental thermodynamic equations follow from five primary thermodynamic definitions and describe internal energy, enthalpy, Helmholtz energy, and Gibbs energy in terms of their natural variables. Here they will be presented in their differential forms. Introduction The fundamental thermodynamic equations describe the thermodynamic quantities U, H, G, and A in terms of their natural variables. The term "natural variable" simply denotes a variable that is one of the convenient variables to describe U, H, G, or A. When considered as a whole, the four fundamental equations demonstrate how four important thermodynamic quantities depend on variables that can be controlled and measured experimentally. Thus, they are essentially equations of state, and using the fundamental equations, experimental data can be used to determine sought-after quantities like \(G\) or \(H\). First Law of Thermodynamics The first law of thermodynamics is represented below in its differential form \[ dU = đq+đw \nonumber \] where \(U\) is the internal energy of the system, \(q\) is heat flow of the system, and \(w\) is the work of the system. The "đ" symbol represent inexact differentials and indicates that both \(q\) and \(w\) are path functions. Recall that \(U\) is a state function. The first law states that internal energy changes occur only as a result of heat flow and work done. It is assumed that w refers only to PV work, where \[ w = -\int{pdV} \nonumber \] The fundamental thermodynamic equation for internal energy follows directly from the first law and the principle of Clausius: \[ dU = đq + đw \nonumber \] \[ dS = \dfrac{\delta q_{rev}}{T} \nonumber \] we have \[ dU = TdS + \delta w \nonumber \] Since only \(PV\) work is performed, \[ dU = TdS - pdV \label{DefU} \] The above equation is the fundamental equation for \(U\) with natural variables of entropy \(S\) and volume\(V\). Principle of Clausius The Principle of Clausius states that the entropy change of a system is equal to the ratio of heat flow in a reversible process to the temperature at which the process occurs. Mathematically this is written as \[ dS = \dfrac{\delta q_{rev}}{T} \nonumber \] where \(S\) is the entropy of the system, \(q_{rev}\) is the heat flow of a reversible process, and \(T\) is the temperature in Kelvin. Enthalpy Mathematically, enthalpy is defined as \[ H = U + pV \label{DefEnth} \] where \(H\) is enthalpy of the system, p is pressure, and V is volume. The fundamental thermodynamic equation for enthalpy follows directly from it deffinition (Equation \(\ref{DefEnth}\)) and the fundamental equation for internal energy (Equation \(\ref{DefU}\)) : \[ dH = dU + d(pV) \nonumber \] \[ = dU + pdV + VdP \nonumber \] \[ dU = TdS - pdV \nonumber \] \[ dH = TdS - pdV + pdV + Vdp \nonumber \] \[ dH = TdS + Vdp \nonumber \] The above equation is the fundamental equation for H. The natural variables of enthalpy are S and p, entropy and pressure. Gibbs Energy The mathematical description of Gibbs energy is as follows \[ G = U + pV - TS = H - TS \label{Defgibbs} \] where \(G\) is the Gibbs energy of the system. The fundamental thermodynamic equation for Gibbs Energy follows directly from its definition \(\ref{Defgibbs}\) and the fundamental equation for enthalpy \(\ref{DefEnth}\): \[ dG = dH - d(TS) \nonumber \] \[ = dH - TdS - SdT \nonumber \] Since \[ dH = TdS + VdP \nonumber \] \[ dG = TdS + VdP - TdS - SdT \nonumber \] \[ dG = VdP - SdT \label{EqGibbs1} \] The above equation is the fundamental equation for G. The natural variables of Gibbs energy are P and T. Helmholtz Energy Mathematically, Helmholtz energy is defined as \[ A = U - TS \label{DefHelm} \] where \(A\) is the Helmholtz energy of the system, which is often written as the symbol \(F\). The fundamental thermodynamic equation for Helmholtz energy follows directly from its definition (Equation \(\ref{DefHelm}\)) and the fundamental equation for internal energy (Equation \(\ref{DefU}\)): \[ dA = dU - d(TS) \nonumber \] \[ = dU - TdS - SdT \nonumber \] Since \[ dU = TdS - pdV \nonumber \] \[ dA = TdS - pdV -TdS - SdT \nonumber \] \[ dA = -pdV - SdT \label{EqHelm1} \] Equation \(\ref{EqHelm1}\) is the fundamental equation for A with natural variables of \(V\) and \(T\). For the definitions to hold, it is assumed that only PV work is done and that only reversible processes are used. These assumptions are required for the first law and the principle of Clausius to remain valid. Also, these equations do not account include n, the number of moles, as a variable. When \(n\) is included, the equations appear different, but the essence of their meaning is captured without including the n-dependence. Chemical Potential The fundamental equations derived above were not dependent on changes in the amounts of species in the system. Below the n-dependent forms are presented 1 ,4 . \[ dU = TdS - PdV + \sum_{i=1}^{N}\mu_idn_i \nonumber \] \[ dH = TdS + VdP + \sum_{i=1}^{N}\mu_idn_i \nonumber \] \[ dG = -SdT + Vdp + \sum_{i=1}^{N}\mu_idn_i \nonumber \] \[ dA = -SdT - PdV + \sum_{i=1}^{N}\mu_idn_i \nonumber \] where μ i is the chemical potential of species i and dn i is the change in number of moles of substance i. Importance/Relevance of Fundamental Equations The differential fundamental equations describe U, H, G, and A in terms of their natural variables. The natural variables become useful in understanding not only how thermodynamic quantities are related to each other, but also in analyzing relationships between measurable quantities (i.e. P, V, T) in order to learn about the thermodynamics of a system. Below is a table summarizing the natural variables for U, H, G, and A: Thermodynamic Quantity Natural Variables U (internal energy) S, V H (enthalpy) S, P G (Gibbs energy) T, P A (Helmholtz energy) T, V References DOI: 10.1063/1.1749582 DOI: 10.1063/1.1749549 DOI:10.1103/PhysRev.3.273 A Treatise on Physical Chemistry, 3rd ed.; Taylor, H. S. and Glasstone, S., Eds.; D. Van Nostrand Company: New York, 1942; Vol. 1; p 454-485. Problems If the assumptions made in the derivations above were not made, what would effect would that have? Try to think of examples were these assumptions would be violated. Could the definitions, principles, and laws used to derive the fundamental equations still be used? Why or why not? For what kind of system does the number of moles not change? This said, do the fundamental equations without n-dependence apply to a wide range of processes and systems? Answers If it was not assumed that PV-work was the only work done, then the work term in the second law of thermodynamics equation would include other terms (e.g. for electrical work, mechanical work). If reversible processes were not assumed, the Principle of Clausius could not be used. One example of such situations could the movement of charged particles towards a region of like charge (electrical work) or an irreversible process like combustion of hydrocarbons or friction. In general, a closed system of non-reacting components would fit this description. For example, the number of moles would not change for a closed system in which a gas is sealed (to prevent leaks) in a container and allowed to expand/is contracted.
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Bookshelves/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)
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This book aims to teach undergraduates introductory concepts of organic and biochemistry in one semester, assuming introductory general chemistry knowledge. It is intended for health sciences, like pre-nursing, psychology, kinesiology, and non-science majors. This book will hopefully provide enough knowledge to prepare the students to understand basic organic and biochemistry used in their professional subjects, take additional chemistry courses if needed, and appreciate the impact of chemistry in everyday life. HOMEWORK is associated, and solutions are accessible through instructors. Front Matter 1: Bonding in organic compounds 2: Nomenclature and physical properties of organic compounds 3: Stereochemistry 4: Organic reactions 5: Carbohydrates 6: Lipids 7: Proteins 8: Nucleic acids 9: Food to energy metabolic pathways Back Matter
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Courses/De_Anza_College/CHEM_10%3A_Introduction_to_Chemistry_(Parajon_Puenzo)/05%3A_Chemical_Bonds/5.02%3A_Lewis_(Electron-Dot)_Symbols
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Learning Objective Draw a Lewis electron dot diagram for an atom. Know the importance of Lewis dot in bonding. At the beginning of the 20 th century, an American physical chemist G. N. Lewis (1875–1946) devised a system of symbols—now called Lewis electron dot symbols (often shortened to Lewis dot symbols ) that can be used for predicting the number of bonds formed by most elements in their compounds. Lewis Dot symbols convenient representation of valence electrons allows scientists to keep track of valence electrons during bond formation consists of the atomic symbol for the element plus a dot for each valence electron To write an element’s Lewis dot symbol, we place dots representing its valence electrons, one at a time, around the element’s chemical symbol. Up to four dots are placed above, below, to the left, and to the right of the symbol (in any order, as long as elements with four or fewer valence electrons have no more than one dot in each position). The next dots, for elements with more than four valence electrons, are again distributed one at a time, each paired with one of the first four. For example, the element sulfur has six valence electrons (note roman numeral above group on the periodic table) and its Lewis symbol would be: Fluorine, for example, has seven valence electrons, so its Lewis dot symbol is constructed as follows: Lewis eventually published his theory of chemical bonding in 1916. Formation of chemical bonds to complete the requirement of eight electrons for the atom becomes a natural tendency. Lewis dot symbols of the first two periods are given here to illustrate this point. In fact, the entire group (column) of elements have the same Lewis dot symbols, because they have the same number of valence electrons. By going through the periodic table, we see that the Lewis electron dot symbols of atoms will never have more than eight dots around the atomic symbo see Table \(\PageIndex{1}\). 0 1 2 3 4 5 6 7 \(\ce{H\cdot}\) NaN NaN NaN NaN NaN NaN \(\textrm{He:}\) \(\underset{\:}{\ce{Li\cdot}}\) \(\underset{\:}{\ce{\cdot Be \cdot}}\) \(\ce{ \cdot \underset{\:}{\overset{\Large{\cdot}}{B}} \cdot}\) \(\ce{ \cdot \underset{\Large{\cdot}}{\overset{\Large{\cdot}}{C}} \cdot}\) \( \underset{\Large{\cdot\,}} {\overset{\Large{\cdot}} {\textrm{:N}\cdot}} \) \( \underset{\Large{\cdot\cdot\,}} {\overset{\Large{\cdot}} {\textrm{:O}\cdot}} \) \( \underset{\Large{\cdot\cdot}} {\overset{\Large{\cdot\cdot}} {\textrm{:F}\cdot}} \) \( \underset{\Large{\cdot\cdot}} {\overset{\Large{\cdot\cdot}}{\textrm{:Ne:}}} \) \(\ce{Na}\) \(\ce{K}\) \(\ce{Rb}\) \(\ce{Cs}\) \(\ce{Mg}\) \(\ce{Ca}\) \(\ce{Sr}\) \(\ce{Ba}\) \(\ce{Al}\) \(\ce{Ga}\) \(\ce{In}\) \(\ce{Tl}\) \(\ce{Si}\) \(\ce{Ge}\) \(\ce{Sn}\) \(\ce{Pb}\) \(\ce{P}\) \(\ce{As}\) \(\ce{Sb}\) \(\ce{Bi}\) \(\ce{S}\) \(\ce{Se}\) \(\ce{Te}\) \(\ce{Po}\) \(\ce{Cl}\) \(\ce{Br}\) \(\ce{I}\) \(\ce{At}\) \(\ce{Ar}\) \(\ce{Kr}\) \(\ce{At}\) \(\ce{Rn}\) Example \(\PageIndex{1}\): Lewis Dot Diagrams What is the Lewis electron dot diagram for each element? aluminum selenium Solution The valence electron configuration for aluminum is 3 s 2 3 p 1 . So it would have three dots around the symbol for aluminum, two of them paired to represent the 3 s electrons: \[\dot{Al:} \nonumber \] The valence electron configuration for selenium is 4 s 2 4 p 4 . In the highest-numbered shell, the n = 4 shell, there are six electrons. Its electron dot diagram is as follows: \[\mathbf{\cdot }\mathbf{\dot{\underset{.\: .}Se}}\mathbf{:} \nonumber \] Exercise \(\PageIndex{1}\) What is the Lewis electron dot diagram for each element? phosphorus argon Answer a \[\mathbf{\cdot }\mathbf{\dot{\underset{.}P}}\mathbf{:} \nonumber \] Answer b \[\mathbf{:}\mathbf{\ddot{\underset{.\, .}Ar}}\mathbf{:} \nonumber \] Summary Lewis electron dot diagrams use dots to represent valence electrons around an atomic symbol. Lewis dot symbols can be used to predict the number of bonds formed by elements in a compound.
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Courses/Willamette_University/WU%3A_Chem_199_-_Better_Living_Through_Chemistry/02%3A_Chemicals_in_Water/2.02%3A_How_Molecules_Dissolve/2.2.02%3A_Molarity
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Learning Objectives Describe the fundamental properties of solutions Calculate solution concentrations using molarity Perform dilution calculations using the dilution equation In preceding sections, we focused on the composition of substances: samples of matter that contain only one type of element or compound. However, mixtures—samples of matter containing two or more substances physically combined—are more commonly encountered in nature than are pure substances. Similar to a pure substance, the relative composition of a mixture plays an important role in determining its properties. The relative amount of oxygen in a planet’s atmosphere determines its ability to sustain aerobic life. The relative amounts of iron, carbon, nickel, and other elements in steel (a mixture known as an “alloy”) determine its physical strength and resistance to corrosion. The relative amount of the active ingredient in a medicine determines its effectiveness in achieving the desired pharmacological effect. The relative amount of sugar in a beverage determines its sweetness (Figure \(\PageIndex{1}\)). In this section, we will describe one of the most common ways in which the relative compositions of mixtures may be quantified. Solutions We have previously defined solutions as homogeneous mixtures, meaning that the composition of the mixture (and therefore its properties) is uniform throughout its entire volume. Solutions occur frequently in nature and have also been implemented in many forms of manmade technology. We will explore a more thorough treatment of solution properties in the chapter on solutions and colloids, but here we will introduce some of the basic properties of solutions. The relative amount of a given solution component is known as its concentration . Often, though not always, a solution contains one component with a concentration that is significantly greater than that of all other components. This component is called the solvent and may be viewed as the medium in which the other components are dispersed, or dissolved . Solutions in which water is the solvent are, of course, very common on our planet. A solution in which water is the solvent is called an aqueous solution . A solute is a component of a solution that is typically present at a much lower concentration than the solvent. Solute concentrations are often described with qualitative terms such as dilute (of relatively low concentration) and concentrated (of relatively high concentration). Concentrations may be quantitatively assessed using a wide variety of measurement units, each convenient for particular applications. Molarity ( M ) is a useful concentration unit for many applications in chemistry. Molarity is defined as the number of moles of solute in exactly 1 liter (1 L) of the solution: \[M=\mathrm{\dfrac{mol\: solute}{L\: solution}} \label{3.4.2} \] Example \(\PageIndex{1}\): Calculating Molar Concentrations A 355-mL soft drink sample contains 0.133 mol of sucrose (table sugar). What is the molar concentration of sucrose in the beverage? Solution Since the molar amount of solute and the volume of solution are both given, the molarity can be calculated using the definition of molarity. Per this definition, the solution volume must be converted from mL to L: \[\begin{align*} M &=\dfrac{mol\: solute}{L\: solution} \\[4pt] &=\dfrac{0.133\:mol}{355\:mL\times \dfrac{1\:L}{1000\:mL}} \\[4pt] &= 0.375\:M \label{3.4.1} \end{align*} \] Exercise \(\PageIndex{1}\) A teaspoon of table sugar contains about 0.01 mol sucrose. What is the molarity of sucrose if a teaspoon of sugar has been dissolved in a cup of tea with a volume of 200 mL? Answer 0.05 M Example \(\PageIndex{2}\): Deriving Moles and Volumes from Molar Concentrations How much sugar (mol) is contained in a modest sip (~10 mL) of the soft drink from Example \(\PageIndex{1}\)? Solution In this case, we can rearrange the definition of molarity to isolate the quantity sought, moles of sugar. We then substitute the value for molarity that we derived in Example 3.4.2, 0.375 M : \[M=\mathrm{\dfrac{mol\: solute}{L\: solution}} \label{3.4.3} \] \[ \begin{align*} \mathrm{mol\: solute} &= \mathrm{ M\times L\: solution} \label{3.4.4} \\[4pt] \mathrm{mol\: solute} &= \mathrm{0.375\:\dfrac{mol\: sugar}{L}\times \left(10\:mL\times \dfrac{1\:L}{1000\:mL}\right)} &= \mathrm{0.004\:mol\: sugar} \label{3.4.5} \end{align*} \] Exercise \(\PageIndex{2}\) What volume (mL) of the sweetened tea described in Example \(\PageIndex{1}\) contains the same amount of sugar (mol) as 10 mL of the soft drink in this example? Answer 80 mL Example \(\PageIndex{3}\): Calculating Molar Concentrations from the Mass of Solute Distilled white vinegar (Figure \(\PageIndex{2}\)) is a solution of acetic acid, \(CH_3CO_2H\), in water. A 0.500-L vinegar solution contains 25.2 g of acetic acid. What is the concentration of the acetic acid solution in units of molarity? Solution As in previous examples, the definition of molarity is the primary equation used to calculate the quantity sought. In this case, the mass of solute is provided instead of its molar amount, so we must use the solute’s molar mass to obtain the amount of solute in moles: \[\mathrm{\mathit M=\dfrac{mol\: solute}{L\: solution}=\dfrac{25.2\: g\: \ce{CH3CO2H}\times \dfrac{1\:mol\: \ce{CH3CO2H}}{60.052\: g\: \ce{CH3CO2H}}}{0.500\: L\: solution}=0.839\: \mathit M} \label{3.4.6} \] \[M=\mathrm{\dfrac{0.839\:mol\: solute}{1.00\:L\: solution}} \nonumber \] Nov 29, 2019, 5:24 PM \[\mathrm{\mathit M=\dfrac{mol\: solute}{L\: solution}=0.839\:\mathit M} \label{3.4.7} \] Exercise \(\PageIndex{3}\) Calculate the molarity of 6.52 g of \(CoCl_2\) (128.9 g/mol) dissolved in an aqueous solution with a total volume of 75.0 mL. Answer 0.674 M Example \(\PageIndex{4}\): Determining the Mass of Solute in a Given Volume of Solution How many grams of NaCl are contained in 0.250 L of a 5.30- M solution? Solution The volume and molarity of the solution are specified, so the amount (mol) of solute is easily computed as demonstrated in Example \(\PageIndex{3}\): \[M=\mathrm{\dfrac{mol\: solute}{L\: solution}} \label{3.4.9} \] \[\mathrm{mol\: solute= \mathit M\times L\: solution} \label{3.4.10} \] \[\mathrm{mol\: solute=5.30\:\dfrac{mol\: NaCl}{L}\times 0.250\:L=1.325\:mol\: NaCl} \label{3.4.11} \] Finally, this molar amount is used to derive the mass of NaCl: \[\mathrm{1.325\: mol\: NaCl\times\dfrac{58.44\:g\: NaCl}{mol\: NaCl}=77.4\:g\: NaCl} \label{3.4.12} \] Exercise \(\PageIndex{4}\) How many grams of \(CaCl_2\) (110.98 g/mol) are contained in 250.0 mL of a 0.200-M solution of calcium chloride? Answer 5.55 g \(CaCl_2\) When performing calculations stepwise, as in Example \(\PageIndex{3}\), it is important to refrain from rounding any intermediate calculation results, which can lead to rounding errors in the final result. In Example \(\PageIndex{4}\), the molar amount of NaCl computed in the first step, 1.325 mol, would be properly rounded to 1.32 mol if it were to be reported; however, although the last digit (5) is not significant, it must be retained as a guard digit in the intermediate calculation. If we had not retained this guard digit, the final calculation for the mass of NaCl would have been 77.1 g, a difference of 0.3 g. In addition to retaining a guard digit for intermediate calculations, we can also avoid rounding errors by performing computations in a single step (Example \(\PageIndex{5}\)). This eliminates intermediate steps so that only the final result is rounded. Example \(\PageIndex{5}\): Determining the Volume of Solution In Example \(\PageIndex{3}\), we found the typical concentration of vinegar to be 0.839 M . What volume of vinegar contains 75.6 g of acetic acid? Solution First, use the molar mass to calculate moles of acetic acid from the given mass: \[\mathrm{g\: solute\times\dfrac{mol\: solute}{g\: solute}=mol\: solute} \label{3.4.13} \] Then, use the molarity of the solution to calculate the volume of solution containing this molar amount of solute: \[\mathrm{mol\: solute\times \dfrac{L\: solution}{mol\: solute}=L\: solution} \label{3.4.14} \] Combining these two steps into one yields: \[\mathrm{g\: solute\times \dfrac{mol\: solute}{g\: solute}\times \dfrac{L\: solution}{mol\: solute}=L\: solution} \label{3.4.15} \] \[\mathrm{75.6\:g\:\ce{CH3CO2H}\left(\dfrac{mol\:\ce{CH3CO2H}}{60.05\:g}\right)\left(\dfrac{L\: solution}{0.839\:mol\:\ce{CH3CO2H}}\right)=1.50\:L\: solution} \label{3.4.16} \] Exercise \(\PageIndex{5}\): What volume of a 1.50-M KBr solution contains 66.0 g KBr? Answer 0.370 L Dilution of Solutions Dilution is the process whereby the concentration of a solution is lessened by the addition of solvent. For example, we might say that a glass of iced tea becomes increasingly diluted as the ice melts. The water from the melting ice increases the volume of the solvent (water) and the overall volume of the solution (iced tea), thereby reducing the relative concentrations of the solutes that give the beverage its taste (Figure \(\PageIndex{2}\)). Dilution is also a common means of preparing solutions of a desired concentration. By adding solvent to a measured portion of a more concentrated stock solution , we can achieve a particular concentration. For example, commercial pesticides are typically sold as solutions in which the active ingredients are far more concentrated than is appropriate for their application. Before they can be used on crops, the pesticides must be diluted. This is also a very common practice for the preparation of a number of common laboratory reagents (Figure \(\PageIndex{3}\)). A simple mathematical relationship can be used to relate the volumes and concentrations of a solution before and after the dilution process. According to the definition of molarity, the molar amount of solute in a solution is equal to the product of the solution’s molarity and its volume in liters: \[n=ML \nonumber \] Expressions like these may be written for a solution before and after it is diluted: \[n_1=M_1L_1 \nonumber \] \[n_2=M_2L_2 \nonumber \] where the subscripts “1” and “2” refer to the solution before and after the dilution, respectively. Since the dilution process does not change the amount of solute in the solution, n 1 = n 2 . Thus, these two equations may be set equal to one another: \[M_1L_1=M_2L_2 \nonumber \] This relation is commonly referred to as the dilution equation. Although we derived this equation using molarity as the unit of concentration and liters as the unit of volume, other units of concentration and volume may be used, so long as the units properly cancel per the factor-label method. Reflecting this versatility, the dilution equation is often written in the more general form: \[C_1V_1=C_2V_2 \nonumber \] where \(C\) and \(V\) are concentration and volume, respectively. Example \(\PageIndex{6}\): Determining the Concentration of a Diluted Solution If 0.850 L of a 5.00- M solution of copper nitrate, Cu(NO 3 ) 2 , is diluted to a volume of 1.80 L by the addition of water, what is the molarity of the diluted solution? Solution We are given the volume and concentration of a stock solution, V 1 and C 1 , and the volume of the resultant diluted solution, V 2 . We need to find the concentration of the diluted solution, C 2 . We thus rearrange the dilution equation in order to isolate C 2 : \[C_1V_1=C_2V_2 \nonumber \] \[C_2=\dfrac{C_1V_1}{V_2} \nonumber \] Since the stock solution is being diluted by more than two-fold (volume is increased from 0.85 L to 1.80 L), we would expect the diluted solution’s concentration to be less than one-half 5 M . We will compare this ballpark estimate to the calculated result to check for any gross errors in computation (for example, such as an improper substitution of the given quantities). Substituting the given values for the terms on the right side of this equation yields: \[C_2=\mathrm{\dfrac{0.850\:L\times 5.00\:\dfrac{mol}{L}}{1.80\: L}}=2.36\:M \nonumber \] This result compares well to our ballpark estimate (it’s a bit less than one-half the stock concentration, 5 M ). Exercise \(\PageIndex{6}\) What is the concentration of the solution that results from diluting 25.0 mL of a 2.04-M solution of CH3OH to 500.0 mL? Answer 0.102 M \(CH_3OH\) Example \(\PageIndex{7}\): Volume of a Diluted Solution What volume of 0.12 M HBr can be prepared from 11 mL (0.011 L) of 0.45 M HBr? Solution We are given the volume and concentration of a stock solution, V 1 and C 1 , and the concentration of the resultant diluted solution, C 2 . We need to find the volume of the diluted solution, V 2 . We thus rearrange the dilution equation in order to isolate V 2 : \[C_1V_1=C_2V_2 \nonumber \] \[V_2=\dfrac{C_1V_1}{C_2} \nonumber \] Since the diluted concentration (0.12 M ) is slightly more than one-fourth the original concentration (0.45 M ), we would expect the volume of the diluted solution to be roughly four times the original volume, or around 44 mL. Substituting the given values and solving for the unknown volume yields: \[V_2=\dfrac{(0.45\:M)(0.011\: \ce L)}{(0.12\:M)} \nonumber \] \[V_2=\mathrm{0.041\:L} \nonumber \] The volume of the 0.12- M solution is 0.041 L (41 mL). The result is reasonable and compares well with our rough estimate. Exercise \(\PageIndex{7}\) A laboratory experiment calls for 0.125 M \(HNO_3\). What volume of 0.125 M \(HNO_3\) can be prepared from 0.250 L of 1.88 M \(HNO_3\)? Answer 3.76 L Example \(\PageIndex{8}\): Volume of a Concentrated Solution Needed for Dilution What volume of 1.59 M KOH is required to prepare 5.00 L of 0.100 M KOH? Solution We are given the concentration of a stock solution, C 1 , and the volume and concentration of the resultant diluted solution, V 2 and C 2 . We need to find the volume of the stock solution, V 1 . We thus rearrange the dilution equation in order to isolate V 1 : \[C_1V_1=C_2V_2 \nonumber \] \[V_1=\dfrac{C_2V_2}{C_1} \nonumber \] Since the concentration of the diluted solution 0.100 M is roughly one-sixteenth that of the stock solution (1.59 M ), we would expect the volume of the stock solution to be about one-sixteenth that of the diluted solution, or around 0.3 liters. Substituting the given values and solving for the unknown volume yields: \[V_1=\dfrac{(0.100\:M)(5.00\:\ce L)}{1.59\:M} \nonumber \] \[V_1=0.314\:\ce L \nonumber \] Thus, we would need 0.314 L of the 1.59- M solution to prepare the desired solution. This result is consistent with our rough estimate. Exercise \(\PageIndex{8}\) What volume of a 0.575-M solution of glucose, C 6 H 12 O 6 , can be prepared from 50.00 mL of a 3.00-M glucose solution? Answer 0.261 Summary Solutions are homogeneous mixtures. Many solutions contain one component, called the solvent, in which other components, called solutes, are dissolved. An aqueous solution is one for which the solvent is water. The concentration of a solution is a measure of the relative amount of solute in a given amount of solution. Concentrations may be measured using various units, with one very useful unit being molarity, defined as the number of moles of solute per liter of solution. The solute concentration of a solution may be decreased by adding solvent, a process referred to as dilution. The dilution equation is a simple relation between concentrations and volumes of a solution before and after dilution. Key Equations \(M=\mathrm{\dfrac{mol\: solute}{L\: solution}}\) C 1 V 1 = C 2 V 2 Glossary aqueous solution solution for which water is the solvent concentrated qualitative term for a solution containing solute at a relatively high concentration concentration quantitative measure of the relative amounts of solute and solvent present in a solution dilute qualitative term for a solution containing solute at a relatively low concentration dilution process of adding solvent to a solution in order to lower the concentration of solutes dissolved describes the process by which solute components are dispersed in a solvent molarity ( M ) unit of concentration, defined as the number of moles of solute dissolved in 1 liter of solution solute solution component present in a concentration less than that of the solvent solvent solution component present in a concentration that is higher relative to other components
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Courses/University_of_Toronto/UTSC%3A_First-Year_Chemistry_Textbook_(Winter_2025)/03%3A_Electronic_Structure_and_Periodic_Properties/3.04%3A_Development_of_Quantum_Theory
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Learning Objectives By the end of this section, you will be able to: Extend the concept of wave–particle duality that was observed in electromagnetic radiation to matter as well Understand the general idea of the quantum mechanical description of electrons in an atom, and that it uses the notion of three-dimensional wave functions, or orbitals, that define the distribution of probability to find an electron in a particular part of space List and describe traits of the four quantum numbers that form the basis for completely specifying the state of an electron in an atom Bohr’s model explained the experimental data for the hydrogen atom and was widely accepted, but it also raised many questions. Why did electrons orbit at only fixed distances defined by a single quantum number \(n\) = 1, 2, 3, and so on, but never in between? Why did the model work so well describing hydrogen and one-electron ions, but could not correctly predict the emission spectrum for helium or any larger atoms? To answer these questions, scientists needed to completely revise the way they thought about matter. Behavior in the Microscopic World We know how matter behaves in the macroscopic world—objects that are large enough to be seen by the naked eye follow the rules of classical physics. A billiard ball moving on a table will behave like a particle: It will continue in a straight line unless it collides with another ball or the table cushion, or is acted on by some other force (such as friction). The ball has a well-defined position and velocity (or a well-defined momentum, \(p = mv\), defined by mass \(m\) and velocity \(v\)) at any given moment. In other words, the ball is moving in a classical trajectory. This is the typical behavior of a classical object. When waves interact with each other, they show interference patterns that are not displayed by macroscopic particles such as the billiard ball. For example, interacting waves on the surface of water can produce interference patterns similar to those shown on Figure \(\PageIndex{1}\). This is a case of wave behavior on the macroscopic scale, and it is clear that particles and waves are very different phenomena in the macroscopic realm. As technological improvements allowed scientists to probe the microscopic world in greater detail, it became increasingly clear by the 1920s that very small pieces of matter follow a different set of rules from those we observe for large objects. The unquestionable separation of waves and particles was no longer the case for the microscopic world. One of the first people to pay attention to the special behavior of the microscopic world was Louis de Broglie . He asked the question: If electromagnetic radiation can have particle-like character, can electrons and other submicroscopic particles exhibit wavelike character? In his 1925 doctoral dissertation, de Broglie extended the wave–particle duality of light that Einstein used to resolve the photoelectric-effect paradox to material particles. He predicted that a particle with mass \(m\) and velocity \(v\) (that is, with linear momentum \(p\)) should also exhibit the behavior of a wave with a wavelength value \(λ\), given by this expression in which \(h\) is the familiar Planck’s constant: \[\lambda=\frac{h}{m v}=\frac{h}{p} \nonumber \] This is called the de Broglie wavelength . Unlike the other values of \(λ\) discussed in this chapter, the de Broglie wavelength is a characteristic of particles and other bodies, not electromagnetic radiation (note that this equation involves velocity [\(v\), m/s], not frequency [\(\nu\), Hz]. Although these two symbols appear nearly identical, they mean very different things). Where Bohr had postulated the electron as being a particle orbiting the nucleus in quantized orbits, de Broglie argued that Bohr’s assumption of quantization can be explained if the electron is considered not as a particle, but rather as a circular standing wave such that only an integer number of wavelengths could fit exactly within the orbit (Figure \(\PageIndex{2}\)). For a circular orbit of radius \(r\), the circumference is \(2πr\), and so de Broglie’s condition is: \[2 \pi r=n \lambda, \quad \quad n=1,2,3, \ldots \nonumber \] Shortly after de Broglie proposed the wave nature of matter, two scientists at Bell Laboratories, C. J. Davisson and L. H. Germer , demonstrated experimentally that electrons can exhibit wavelike behavior by showing an interference pattern for electrons travelling through a regular atomic pattern in a crystal. The regularly spaced atomic layers served as slits, as used in other interference experiments. Since the spacing between the layers serving as slits needs to be similar in size to the wavelength of the tested wave for an interference pattern to form, Davisson and Germer used a crystalline nickel target for their “slits,” since the spacing of the atoms within the lattice was approximately the same as the de Broglie wavelengths of the electrons that they used. Figure \(\PageIndex{3}\) shows an interference pattern. It is strikingly similar to the interference patterns for light shown in Electromagnetic Energy for light passing through two closely spaced, narrow slits. The wave–particle duality of matter can be seen in Figure \(\PageIndex{3}\) by observing what happens if electron collisions are recorded over a long period of time. Initially, when only a few electrons have been recorded, they show clear particle-like behavior, having arrived in small localized packets that appear to be random. As more and more electrons arrived and were recorded, a clear interference pattern that is the hallmark of wavelike behavior emerged. Thus, it appears that while electrons are small localized particles, their motion does not follow the equations of motion implied by classical mechanics, but instead it is governed by some type of a wave equation. Thus the wave–particle duality first observed with photons is actually a fundamental behavior intrinsic to all quantum particles. Link to Learning View the Dr. Quantum – Double Slit Experiment cartoon for an easy-to-understand description of wave–particle duality and the associated experiments. Example \(\PageIndex{1}\): Calculating the Wavelength of a Particle If an electron travels at a velocity of \(1.000 \times 10^7\, \text{m s}^{–1}\) and has a mass of \(9.109 \times 10^{–28}\,\text{g}\), what is its wavelength? Solution We can use de Broglie’s equation to solve this problem, but we first must do a unit conversion of Planck’s constant. You learned earlier that \(1\, \text{J} = 1 \text{kg m}^2/\text{s}^2\). Thus, we can write \(h = 6.626 \times 10^{–34}\,\text{J} \cdot \text{s}\) as \(6.626 \times 10^{–34}\,\text{kg m}^2/\text{s}\). \[\begin{align*} \lambda &=\frac{h}{m v}\\[4pt] &=\frac{6.626 \times 10^{-34} \,\text{kg m}^2 / \text{s}}{\left(9.109 \times 10^{-31} \,\text{kg} \right)\left(1.000 \times 10^7\, \text{m} / \text{s} \right)} \\[4pt] &=7.274 \times 10^{-11} m \end{align*} \] This is a small value, but it is significantly larger than the size of an electron in the classical (particle) view. This size is the same order of magnitude as the size of an atom. This means that electron wavelike behavior is going to be noticeable in an atom. Exercise \(\PageIndex{1}\) Calculate the wavelength of a softball with a mass of 100 g traveling at a velocity of 35 m s –1 , assuming that it can be modeled as a single particle. Answer 1.9 10 –34 m. We never think of a thrown softball having a wavelength, since this wavelength is so small it is impossible for our senses or any known instrument to detect (strictly speaking, the wavelength of a real baseball would correspond to the wavelengths of its constituent atoms and molecules, which, while much larger than this value, would still be microscopically tiny). The de Broglie wavelength is only appreciable for matter that has a very small mass and/or a very high velocity. Werner Heisenberg considered the limits of how accurately we can measure properties of an electron or other microscopic particles. He determined that there is a fundamental limit to how accurately one can measure both a particle’s position and its momentum simultaneously. The more accurately we measure the momentum of a particle, the less accurately we can determine its position at that time, and vice versa. This is summed up in what we now call the Heisenberg uncertainty principle : It is fundamentally impossible to determine simultaneously and exactly both the momentum and the position of a particle . For a particle of mass \( m\) moving with velocity \(v_x\) in the \(x\) direction (or equivalently with momentum \(p_x\)), the product of the uncertainty in the position, \(Δx\), and the uncertainty in the momentum, \(Δp_x\), must be greater than or equal to \(ℏ/2\) (where \(\hbar=ℎ/2\), the value of Planck’s constant divided by \(2π\)). \[\Delta x \times \Delta p_x=(\Delta x)(m \Delta v) \geq \frac{\hbar}{2} \nonumber \] This equation allows us to calculate the limit to how precisely we can know both the simultaneous position of an object and its momentum. For example, if we improve our measurement of an electron’s position so that the uncertainty in the position (\(Δx\)) has a value of, say, 1 pm (10 –12 m, about 1% of the diameter of a hydrogen atom), then our determination of its momentum must have an uncertainty with a value of at least \[\left[\Delta p=m \Delta v=\frac{\hbar}{(2 \Delta x)}\right]=\frac{\left(1.055 \times 10^{-34}\, \text{kg}\, \text{m}^2 / \text{s} \right)}{\left(2 \times 1 \times 10^{-12}\, \text{m} \right)}=5 \times 10^{-23}\, \text{kg}\, \text{m} / \text{s} \nonumber \] The value of \(\hbar\) is not large, so the uncertainty in the position or momentum of a macroscopic object like a baseball is too insignificant to observe. However, the mass of a microscopic object such as an electron is small enough that the uncertainty can be large and significant. It should be noted that Heisenberg’s uncertainty principle is not just limited to uncertainties in position and momentum, but it also links other dynamical variables. For example, when an atom absorbs a photon and makes a transition from one energy state to another, the uncertainty in the energy and the uncertainty in the time required for the transition are similarly related, as: \[\Delta E \Delta t \geq \frac{\hbar}{2} \nonumber \] Heisenberg’s principle imposes ultimate limits on what is knowable in science. The uncertainty principle can be shown to be a consequence of wave–particle duality, which lies at the heart of what distinguishes modern quantum theory from classical mechanics. Link to Learning Read this article that describes a recent macroscopic demonstration of the uncertainty principle applied to microscopic objects. The Quantum–Mechanical Model of an Atom Shortly after de Broglie published his ideas that the electron in a hydrogen atom could be better thought of as being a circular standing wave instead of a particle moving in quantized circular orbits, Erwin Schrödinger extended de Broglie’s work by deriving what is today known as the Schrödinger equation. When Schrödinger applied his equation to hydrogen-like atoms, he was able to reproduce Bohr’s expression for the energy and, thus, the Rydberg formula governing hydrogen spectra. Schrödinger described electrons as three-dimensional stationary waves, or wavefunctions , represented by the Greek letter psi, \(ψ\). A few years later, Max Born proposed an interpretation of the wavefunction \(ψ\) that is still accepted today: Electrons are still particles, and so the waves represented by \(ψ\) are not physical waves but, instead, are complex probability amplitudes. The square of the magnitude of a wavefunction \(|\psi|^2\) describes the probability of the quantum particle being present near a certain location in space. This means that wavefunctions can be used to determine the distribution of the electron’s density with respect to the nucleus in an atom. In the most general form, the Schrödinger equation can be written as: \[\hat{H} \psi=E \psi \nonumber \] \(\hat{H}\) is the Hamiltonian operator, a set of mathematical operations representing the total energy of the quantum particle (such as an electron in an atom), ψ is the wavefunction of this particle that can be used to find the special distribution of the probability of finding the particle, and \(E\) is the actual value of the total energy of the particle. Schrödinger’s work, as well as that of Heisenberg and many other scientists following in their footsteps, is generally referred to as quantum mechanics . Link to Learning You may also have heard of Schrödinger because of his famous thought experiment. This story explains the concepts of superposition and entanglement as related to a cat in a box with poison. Understanding Quantum Theory of Electrons in Atoms The goal of this section is to understand the electron orbitals (location of electrons in atoms), their different energies, and other properties. The use of quantum theory provides the best understanding to these topics. This knowledge is a precursor to chemical bonding. As was described previously, electrons in atoms can exist only on discrete energy levels but not between them. It is said that the energy of an electron in an atom is quantized, that is, it can be equal only to certain specific values and can jump from one energy level to another but not transition smoothly or stay between these levels. The energy levels are labeled with an \(n\) value, where \(n = 1, 2, 3, ….\) Generally speaking, the energy of an electron in an atom is greater for greater values of \(n\). This number, n , is referred to as the principal quantum number. The principal quantum number defines the location of the energy level. It is essentially the same concept as the n in the Bohr atom description. Another name for the principal quantum number is the shell number. The shells of an atom can be thought of concentric circles radiating out from the nucleus. The electrons that belong to a specific shell are most likely to be found within the corresponding circular area. The further we proceed from the nucleus, the higher the shell number, and so the higher the energy level (Figure \(\PageIndex{4}\)). The positively charged protons in the nucleus stabilize the electronic orbitals by electrostatic attraction between the positive charges of the protons and the negative charges of the electrons. So the further away the electron is from the nucleus, the greater the energy it has. This quantum mechanical model for where electrons reside in an atom can be used to look at electronic transitions, the events when an electron moves from one energy level to another. If the transition is to a higher energy level, energy is absorbed, and the energy change has a positive value. To obtain the amount of energy necessary for the transition to a higher energy level, a photon is absorbed by the atom. A transition to a lower energy level involves a release of energy, and the energy change is negative. This process is accompanied by emission of a photon by the atom. The following equation summarizes these relationships and is based on the hydrogen atom: \[\begin{align*} \Delta E &=E_{\text {final }}-E_{\text {initial }}\\[4pt] &=-2.18 \times 10^{-18}\left(\frac{1}{n_{ f }^2}-\frac{1}{n_{ i }^2}\right) J \end{align*} \] The values \(n_f\) and \(n_i\) are the final and initial energy states of the electron. The principal quantum number is one of three quantum numbers used to characterize an orbital. An atomic orbital is a general region in an atom within which an electron is most probable to reside. The quantum mechanical model specifies the probability of finding an electron in the three-dimensional space around the nucleus and is based on solutions of the Schrödinger equation. In addition, the principal quantum number defines the energy of an electron in a hydrogen or hydrogen-like atom or an ion (an atom or an ion with only one electron) and the general region in which discrete energy levels of electrons in a multi-electron atoms and ions are located. Another quantum number is \(l\), the secondary (angular momentum) quantum number . It is an integer that may take the values, \(l = 0, 1, 2, …, n – 1\). This means that an orbital with \(n = 1\) can have only one value of \(l\), \(l = 0\), whereas \(n = 2\) permits \(l = 0\) and \(l = 1\), and so on. Whereas the principal quantum number, \(n\), defines the general size and energy of the orbital, the secondary quantum number \(l\) specifies the shape of the orbital. Orbitals with the same value of \(l\) define a subshell . Orbitals with \(l = 0\) are called s orbitals and they make up the s subshells. The value \(l = 1\) corresponds to the p orbitals. For a given \(n\), \(p\) orbitals constitute a \(p\) subshell (e.g., 3p if n = 3). The orbitals with \(l = 2\) are called the d orbital s , followed by the f-, g-, and h- orbitals for \(l\) = 3, 4, and 5. There are certain distances from the nucleus at which the probability density of finding an electron located at a particular orbital is zero. In other words, the value of the wavefunction \(ψ\) is zero at this distance for this orbital. Such a value of radius \(r\) is called a radial node . The number of radial nodes in an orbital is \(n – l – 1\). Consider the examples in Figure \(\PageIndex{5}\). The orbitals depicted are of the s type, thus \(l = 0\) for all of them. It can be seen from the graphs of the probability densities that there are 1 – 0 – 1 = 0 places where the density is zero (nodes) for 1 s ( n = 1), 2 – 0 – 1 = 1 node for 2 s , and 3 – 0 – 1 = 2 nodes for the 3 s orbitals. The s subshell electron density distribution is spherical and the p subshell has a dumbbell shape. The d and f orbitals are more complex. These shapes represent the three-dimensional regions within which the electron is likely to be found. The magnetic quantum number , \(m_l\), specifies the relative spatial orientation of a particular orbital. Generally speaking, \(m_l\) are rangeed from \(-l\) to \(+l\), i.e. \[–l, –(l – 1), …, 0, …, (l – 1), l\). \nonumber \] The total number of possible orbitals with the same value of \(l\) (that is, in the same subshell) is \(2l + 1\). Thus, there is one s-orbital in an s subshell (\(l = 0\)), there are three p-orbitals in a p subshell (\(l = 1\)), five d-orbitals in a d subshell (l = 2), seven f-orbitals in an f subshell (\(l = 3\)), and so forth. The principal quantum number defines the general value of the electronic energy. The angular momentum quantum number determines the shape of the orbital. And the magnetic quantum number specifies orientation of the orbital in space, as can be seen in Figure \(\PageIndex{6}\). Figure \(\PageIndex{7}\) illustrates the energy levels for various orbitals. The number before the orbital name (such as 2 s , 3 p , and so forth) stands for the principal quantum number, \(n\). The letter in the orbital name defines the subshell with a specific angular momentum quantum number l = 0 for s orbitals, 1 for p orbitals, 2 for d orbitals. Finally, there are more than one possible orbitals for \(l ≥ 1\), each corresponding to a specific value of m l . In the case of a hydrogen atom or a one-electron ion (such as He + , Li 2 + , and so on), energies of all the orbitals with the same \(n\) are the same. This is called a degeneracy, and the energy levels for the same principal quantum number, \(n\), are called degenerate orbitals . However, in atoms with more than one electron, this degeneracy is eliminated by the electron–electron interactions, and orbitals that belong to different subshells have different energies, as shown on Figure \(\PageIndex{7}\). Orbitals within the same subshell are still degenerate and have the same energy. While the three quantum numbers discussed in the previous paragraphs work well for describing electron orbitals, some experiments showed that they were not sufficient to explain all observed results. It was demonstrated in the 1920s that when hydrogen-line spectra are examined at extremely high resolution, some lines are actually not single peaks but, rather, pairs of closely spaced lines. This is the so-called fine structure of the spectrum, and it implies that there are additional small differences in energies of electrons even when they are located in the same orbital. These observations led Samuel Goudsmit and George Uhlenbeck to propose that electrons have a fourth quantum number. They called this the spin quantum number , or \(m_s\). The other three quantum numbers, \(n\), \(l\), and \(m_l\), are properties of specific atomic orbitals that also define in what part of the space an electron is most likely to be located. Orbitals are a result of solving the Schrödinger equation for electrons in atoms. The electron spin is a different kind of property. It is a completely quantum phenomenon with no analogues in the classical realm. In addition, it cannot be derived from solving the Schrödinger equation and is not related to the normal spatial coordinates (such as the Cartesian x , y , and z ). Electron spin describes an intrinsic electron "rotation" or "spinning." Each electron acts as a tiny magnet or a tiny rotating object with an angular momentum, or as a loop with an electric current, even though this rotation or current cannot be observed in terms of spatial coordinates. The magnitude of the overall electron spin can only have one value, and an electron can only “spin” in one of two quantized states. One is termed the \(α\) state, with the z component of the spin being in the positive direction of the z axis. This corresponds to the spin quantum number \(m_s=+1/2\). The other is called the \(β state\), with the z component of the spin being negative and \(m_s=-1/2\). Any electron, regardless of the atomic orbital it is located in, can only have one of those two values of the spin quantum number. The energies of electrons having \(m_s=+1/2\) and \(m_s=-1/2\) are different if an external magnetic field is applied. Figure \(\PageIndex{8}\) illustrates this phenomenon. An electron acts like a tiny magnet. Its moment is directed up (in the positive direction of the z axis) for the \(m_s=+1/2\) spin quantum number and down (in the negative z direction) for the spin quantum number of \(m_s=-1/2\). A magnet has a lower energy if its magnetic moment is aligned with the external magnetic field (the left electron on Figure \(\PageIndex{8}\)) and a higher energy for the magnetic moment being opposite to the applied field. This is why an electron with \(m_s=+1/2\) has a slightly lower energy in an external field in the positive z direction, and an electron with \(m_s=-1/2\) has a slightly higher energy in the same field. This is true even for an electron occupying the same orbital in an atom. A spectral line corresponding to a transition for electrons from the same orbital but with different spin quantum numbers has two possible values of energy; thus, the line in the spectrum will show a fine structure splitting. The Pauli Exclusion Principle An electron in an atom is completely described by four quantum numbers: \(n\), \(l\), \(m_l\), and \(m_s\). The first three quantum numbers define the orbital and the fourth quantum number describes the intrinsic electron property called spin. An Austrian physicist Wolfgang Pauli formulated a general principle that gives the last piece of information that we need to understand the general behavior of electrons in atoms. The Pauli exclusion principle can be formulated as follows: No two electrons in the same atom can have exactly the same set of all the four quantum numbers. What this means is that two electrons can share the same orbital (the same set of the quantum numbers n , l , and m l ) only if their spin quantum numbers m s have different values. Since the spin quantum number can only have two values no more than two electrons can occupy the same orbital (and if two electrons are located in the same orbital, they must have opposite spins). Therefore, any atomic orbital can be populated by only zero, one, or two electrons. The properties and meaning of the quantum numbers of electrons in atoms are briefly summarized in Table \(\PageIndex{1}\). Name Symbol Allowed values Physical meaning principal quantum number \(n\) 1, 2, 3, 4, …. shell, the general region for the value of energy for an electron on the orbital angular momentum or azimuthal quantum number \(l\) \(0 ≤ l ≤ n – 1\) subshell, the shape of the orbital magnetic quantum number \(m_l\) \(– l ≤ m_l ≤ l\) orientation of the orbital spin quantum number \(m_s\) \(+\frac{1}{2}\) and \(-\frac{1}{2}\) direction of the intrinsic quantum “spinning” of the electron Example \(\PageIndex{2}\): Working with Shells and Subshells Indicate the number of subshells, the number of orbitals in each subshell, and the values of \(l\) and \(m_l\) for the orbitals in the \(n = 4\) shell of an atom. Solution For \(n = 4\), \(l\) can have values of 0, 1, 2, and 3. Thus, s, p, d, and f subshells are found in the n = 4 shell of an atom. For \(l = 0\) (the s subshell), \(m_l\) can only be 0. Thus, there is only one 4s orbital. For \(l = 1\) (p-type orbitals), m can have values of –1, 0, +1, so we find three 4p orbitals. For l = 2 (d-type orbitals), \(m_l\) can have values of –2, –1, 0, +1, +2, so we have five 4d orbitals. When \(l = 3\) (f-type orbitals), \(m_l\) can have values of –3, –2, –1, 0, +1, +2, +3, and we can have seven 4f orbitals. Thus, we find a total of 16 orbitals in the \(n = 4\) shell of an atom. Exercise \(\PageIndex{2}\) Identify the subshell in which electrons with the following quantum numbers are found: n = 3, l = 1 n = 5, l = 3 n = 2, l = 0 Answer 3 p 5 f 2s Example \(\PageIndex{3}\): Maximum Number of Electrons Calculate the maximum number of electrons that can occupy a shell with (a) n = 2, (b) n = 5, and (c) n as a variable. Note you are only looking at the orbitals with the specified n value, not those at lower energies. Solution When \(n = 2\), there are four orbitals (a single 2s orbital, and three orbitals labeled 2p). These four orbitals can contain eight electrons. When \(n = 2\), there are four orbitals (a single 2s orbital, and three orbitals labeled 2p). These four orbitals can contain eight electrons. When \(n = 5\), there are five subshells of orbitals that we need to sum: \[\begin{align*} \text{1 orbital labeled} & \text{ 5s} \\[4pt] \text{3 orbitals labeled} &\text{ 5p} \\[4pt] \text{5 orbitals labeled} & \text{ 5d} \\[4pt] \text{7 orbitals labeled} & \text{ 5f}\\[4pt] + \quad \text{9 orbitals labeled} &\text{ 5g} \\[4pt] \hline \text{25 orbitals total} \end{align*} \nonumber \] Again, each orbital holds two electrons, so 50 electrons can fit in this shell. The number of orbitals in any shell \(n\) will equal \(n^2\). There can be up to two electrons in each orbital, so the maximum number of electrons will be \(2 × n^2\). Exercise \(\PageIndex{3}\) If a shell contains a maximum of 32 electrons, what is the principal quantum number, \(n\)? Answer \(n = 4\) Example \(\PageIndex{4}\): Working with Quantum Numbers Complete the following table for atomic orbitals: Orbital \(n\) \(l\) \(m_l\) degeneracy Radial nodes (#) 4f NaN NaN NaN NaN NaN 4.0 1.0 NaN NaN NaN 7.0 NaN 7.0 3.0 5d NaN NaN NaN NaN Solution The table can be completed using the following rules: The orbital designation is \(nl\), where \(l = 0, 1, 2, 3, 4, 5, …\) is mapped to the letter sequence s, p, d, f, g, h, …, The \(m_l\) degeneracy is the number of orbitals within an l subshell, and so is \(2l + 1\) (there is one s orbital, three p orbitals, five d orbitals, seven f orbitals, and so forth). The number of radial nodes is equal to \(n – l – 1\). Orbital \(n\) \(l\) \(m_l\) degeneracy Radial nodes (#) 4f 4 3 7 0 4p 4 1 3 2 7f 7 3 7 3 5d 5 2 5 2 Exercise \(\PageIndex{1}\) How many orbitals have \(l = 2\) and \(n = 3\)? Answer The five degenerate 3 d orbitals
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PROBLEM \(\PageIndex{1}\) In the following drawing, the green spheres represent atoms of a certain element. The purple spheres represent atoms of another element. If the spheres of different elements touch, they are part of a single unit of a compound. The following chemical change represented by these spheres may violate one of the ideas of Dalton's atomic theory. Which one? Answer The starting materials consist of one green sphere and two purple spheres. The products consist of two green spheres and two purple spheres. This violates Dalton’s postulate that that atoms are not created during a chemical change, but are merely redistributed. PROBLEM \(\PageIndex{2}\) Which postulate of Dalton's theory is consistent with the following observation concerning the weights of reactants and products? When 100 grams of solid calcium carbonate is heated, 44 g of CO 2 and 56 g of CaO are produced. Answer Atoms are neither created nor destroyed during a chemical change, but are instead rearranged to yield substances that are different from those present before the change (Based on the Law of Conservation of Mass). PROBLEM \(\PageIndex{3}\) Samples of compound X, Y, and Z are analyzed, with results shown here. Do these data provide example(s) of the law of definite proportions, the law of multiple proportions, neither, or both? What do these data tell you about compounds X, Y, and Z? Compound Description Mass of Carbon Mass of Hydrogen X clear, colorless, liquid with strong odor 1.776 g 0.148 g Y clear, colorless, liquid with strong odor 1.974 g 0.329 g Z clear, colorless, liquid with strong odor 7.812 g 0.651 g Answer X+Z are similar compounds (same ratios of C and H), aligning with the Law of Definite Proportions X+Y and Y+Z are different compounds (differing ratios of C and H), aligning with the Law of Multiple Proportions Click here to see a video of the solution. PROBLEM \(\PageIndex{4}\) How are electrons and protons similar? How are they different? Answer Electrons and protons are both charged subatomic particles. Protons are much larger than electrons (contributing more mass to the overall atom). Changing the number of protons changes the identity of the atom, which changing the number of electrons changes the charge . PROBLEM \(\PageIndex{5}\) How are protons and neutrons similar? How are they different? Answer Protons and neutrons are both located in the nucleus of the atom. Protons and neutrons both contribute to the overall mass of the atom. Protons carry a charge while neutrons are neutral. PROBLEM \(\PageIndex{6}\) Predict and test the behavior of α particles fired at a “plum pudding” model atom. (a) Predict the paths taken by α particles that are fired at atoms with a Thomson’s plum pudding model structure. Explain why you expect the α particles to take these paths. (b) If α particles of higher energy than those in (a) are fired at plum pudding atoms, predict how their paths will differ from the lower-energy α particle paths. Explain your reasoning. (c) Now test your predictions from (a) and (b). Select the “Plum Pudding Atom” tab above. Set “Alpha Particles Energy” to “min,” and select “show traces.” Click on the gun to start firing α particles. Does this match your prediction from (a)? If not, explain why the actual path would be that shown in the simulation. Hit the pause button, or “Reset All.” Set “Alpha Particles Energy” to “max,” and start firing α particles. Does this match your prediction from (b)? If not, explain the effect of increased energy on the actual paths as shown in the simulation. Answer a The plum pudding model indicates that the positive charge is spread uniformly throughout the atom, so we expect the α particles to (perhaps) be slowed somewhat by the positive-positive repulsion, but to follow straight-line paths (i.e., not to be deflected) as they pass through the atoms. Answer b Higher-energy α particles will be traveling faster (and perhaps slowed less) and will also follow straight-line paths through the atoms. Answer c The α particles followed straight-line paths through the plum pudding atom. There was no apparent slowing of the α particles as they passed through the atoms. PROBLEM \(\PageIndex{7}\) Predict and test the behavior of α particles fired at a Rutherford atom model. (a) Predict the paths taken by α particles that are fired at atoms with a Rutherford atom model structure. Explain why you expect the α particles to take these paths. (b) If α particles of higher energy than those in (a) are fired at Rutherford atoms, predict how their paths will differ from the lower-energy α particle paths. Explain your reasoning. (c) Predict how the paths taken by the α particles will differ if they are fired at Rutherford atoms of elements other than gold. What factor do you expect to cause this difference in paths, and why? (d) Now test your predictions from (a), (b), and (c). Select the “Rutherford Atom” tab above. Due to the scale of the simulation, it is best to start with a small nucleus, so select “20” for both protons and neutrons, “min” for energy, show traces, and then start firing α particles. Does this match your prediction from (a)? If not, explain why the actual path would be that shown in the simulation. Pause or reset, set energy to “max,” and start firing α particles. Does this match your prediction from (b)? If not, explain the effect of increased energy on the actual path as shown in the simulation. Pause or reset, select “40” for both protons and neutrons, “min” for energy, show traces, and fire away. Does this match your prediction from (c)? If not, explain why the actual path would be that shown in the simulation. Repeat this with larger numbers of protons and neutrons. What generalization can you make regarding the type of atom and effect on the path of α particles? Be clear and specific. Answer a The Rutherford atom has a small, positively charged nucleus, so most α particles will pass through empty space far from the nucleus and be undeflected. Those α particles that pass near the nucleus will be deflected from their paths due to positive-positive repulsion. The more directly toward the nucleus the α particles are headed, the larger the deflection angle will be. Answer b Higher-energy α particles that pass near the nucleus will still undergo deflection, but the faster they travel, the less the expected angle of deflection Answer c If the nucleus is smaller, the positive charge is smaller and the expected deflections are smaller—both in terms of how closely the α particles pass by the nucleus undeflected and the angle of deflection. If the nucleus is larger, the positive charge is larger and the expected deflections are larger—more α particles will be deflected, and the deflection angles will be larger. Answer d The paths followed by the α particles match the predictions from (a), (b), and (c) Contributors Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected] ). Adelaide Clark, Oregon Institute of Technology
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Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Structure_and_Nomenclature_of_Coordination_Compounds/Isomers/Optical_Isomers_in_Inorganic_Complexes/Enantiomers_in_Octahedral_Complexes_with_Bidentate_Ligands
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Enantiomers are another kind of isomer that occur in octahedral metal complexes. Like the square planar platinum complexes seen before, these compounds consist of metal ions with other atoms or groups bound to them. More information about the binding of ligands to metals is found in the section on Lewis acids and bases . These groups that bind to metal ions are called ligands. One example of a type of octahedral compounds that can form enantiomers is bidentate complexes. In bidentate complexes (from the Greek two teeth) or chelating complexes (from the Greek crab), a ligand binds very tightly to the metal because it holds onto the metal via more than one atom. Ethylenediamine is one example of a bidentate ligand. Bidentate ligands bind very tightly to a metal because they form two bonds with it, rather than just one. In an octahedral complex, the two donor atoms in a bidentate ligand bind cis to each other. They cannot reach all the way around the molecule to bind trans to each other. The spatial relationship between the metal and the two atoms connected to it from the same ligand forms a plane. If more than one bidentate ligand is connected to the metal, the relative orientation of one plane to another creates the possibility of mirror images. A complex containing three bidentate ligands can take on the shape of a left-handed propeller or a right-handed propeller. These shapes are alternatively described as a left-handed screw and a right handed screw. If you can picture turning the shape so that it screws into the page behind it, which direction would you turn the screwdriver? If you would twist the screwdriver clockwise, then you have a right handed screw. If you would twist the screwdriver counter-clockwise, then you have a left handed screw. What do we know about enantiomers? enantiomers have identical physical properties, except... enantiomers have opposite optical rotation. These kinds of complexes were historically important in demonstrating how small molecules and ions bound to metal cations. By showing that some metal complexes were chiral and displayed optical activity, early 20th century workers such as Alfred Werner were able to rule out some competing ideas about the structures of metal compounds. Today, we know that metal complexes play important roles in enzymes in biology, and Werner's work on metal complexes laid the groundwork for how we think about these complexes. In addition, stereochemistry in metal complexes became very important in the late 20th century, especially as pharmaceutical companies looked for catalysts that could aid in the production of one enantiomer of a drug, and not the other, in order to maximize pharmaceutical effectiveness and minimize side effects. Problem SC18.1. Identify the relationships between the following pairs of iron(III) oxalate ions. Problem SC18.2. Indicate whether each of the following compounds is the Δ or Λ isomer and draw its enantiomer. Exercise: How many stereoisomers does an octahedral complex have? (contribution from B.J. Johnson)
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Courses/Indiana_Tech/EWC%3A_CHEM_1020_-_General_Chemistry_I_(Budhi)/05%3A_Gases/5.04%3A_The_Ideal_Gas_Law
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Learning Objectives Derive the ideal gas law from the constituent gas laws To use the ideal gas law to describe the behavior of a gas. In this module, the relationship between pressure, temperature, volume, and amount of a gas are described and how these relationships can be combined to give a general expression that describes the behavior of a gas. Deriving the Ideal Gas Law Any set of relationships between a single quantity (such as \(V\)) and several other variables (\(P\), \(T\), and \(n\)) can be combined into a single expression that describes all the relationships simultaneously. The three individual expressions were derived previously: Boyle’s law \[V \propto \dfrac{1}{P} \;\; \text{@ constant n and T} \nonumber \] Charles’s law \[V \propto T \;\; \text{@ constant n and P} \nonumber \] Avogadro’s law \[V \propto n \;\; \text{@ constant T and P} \nonumber \] Combining these three expressions gives \[V \propto \dfrac{nT}{P} \label{10.4.1} \] which shows that the volume of a gas is proportional to the number of moles and the temperature and inversely proportional to the pressure. This expression can also be written as \[V= {\rm Cons.} \left( \dfrac{nT}{P} \right) \label{10.4.2} \] By convention, the proportionality constant in Equation \(\ref{10.4.1}\) is called the gas constant, which is represented by the letter \(R\). Inserting R into Equation \(\ref{10.4.2}\) gives \[ V = \dfrac{RnT}{P} = \dfrac{nRT}{P} \label{10.4.3} \] Clearing the fractions by multiplying both sides of Equation \(\ref{10.4.4}\) by \(P\) gives \[PV = nRT \label{10.4.4} \] This equation is known as the ideal gas law . An ideal gas is defined as a hypothetical gaseous substance whose behavior is independent of attractive and repulsive forces and can be completely described by the ideal gas law. In reality, there is no such thing as an ideal gas, but an ideal gas is a useful conceptual model that allows us to understand how gases respond to changing conditions. As we shall see, under many conditions, most real gases exhibit behavior that closely approximates that of an ideal gas. The ideal gas law can therefore be used to predict the behavior of real gases under most conditions. The ideal gas law does not work well at very low temperatures or very high pressures, where deviations from ideal behavior are most commonly observed. Significant deviations from ideal gas behavior commonly occur at low temperatures and very high pressures. Before we can use the ideal gas law, however, we need to know the value of the gas constant R. Its form depends on the units used for the other quantities in the expression. If V is expressed in liters (L), P in atmospheres (atm), T in kelvins (K), and n in moles (mol), then \[R = 0.08206 \dfrac{\rm L\cdot atm}{\rm K\cdot mol} \label{10.4.5} \] Because the product PV has the units of energy, R can also have units of J/(K•mol): \[R = 8.3145 \dfrac{\rm J}{\rm K\cdot mol}\label{10.4.6} \] Standard Conditions of Temperature and Pressure Scientists have chosen a particular set of conditions to use as a reference: 0°C (273.15 K) and \(\rm1\; bar = 100 \;kPa = 10^5\;Pa\) pressure, referred to as standard temperature and pressure ( STP ). \[\text{STP:} \hspace{2cm} T=273.15\;{\rm K}\text{ and }P=\rm 1\;bar=10^5\;Pa \nonumber \] Please note that STP was defined differently in the past. The old definition was based on a standard pressure of 1 atm. We can calculate the volume of 1.000 mol of an ideal gas under standard conditions using the variant of the ideal gas law given in Equation \(\ref{10.4.4}\): \[V=\dfrac{nRT}{P}\label{10.4.7} \] Thus the volume of 1 mol of an ideal gas is 22.71 L at STP and 22.41 L at 0°C and 1 atm , approximately equivalent to the volume of three basketballs. The molar volumes of several real gases at 0°C and 1 atm are given in Table 10.3, which shows that the deviations from ideal gas behavior are quite small. Thus the ideal gas law does a good job of approximating the behavior of real gases at 0°C and 1 atm. The relationships described in Section 10.3 as Boyle’s, Charles’s, and Avogadro’s laws are simply special cases of the ideal gas law in which two of the four parameters (P, V, T, and n) are held fixed. Gas Molar Volume (L) He 22.434 Ar 22.397 H2 22.433 N2 22.402 O2 22.397 CO2 22.260 NH3 22.079 Applying the Ideal Gas Law The ideal gas law allows us to calculate the value of the fourth variable for a gaseous sample if we know the values of any three of the four variables ( P , V , T , and n ). It also allows us to predict the final state of a sample of a gas (i.e., its final temperature, pressure, volume, and amount) following any changes in conditions if the parameters ( P , V , T , and n ) are specified for an initial state. Some applications are illustrated in the following examples. The approach used throughout is always to start with the same equation—the ideal gas law—and then determine which quantities are given and which need to be calculated. Let’s begin with simple cases in which we are given three of the four parameters needed for a complete physical description of a gaseous sample. Example \(\PageIndex{1}\) The balloon that Charles used for his initial flight in 1783 was destroyed, but we can estimate that its volume was 31,150 L (1100 ft 3 ), given the dimensions recorded at the time. If the temperature at ground level was 86°F (30°C) and the atmospheric pressure was 745 mmHg, how many moles of hydrogen gas were needed to fill the balloon? Given: volume, temperature, and pressure Asked for: amount of gas Strategy: Solve the ideal gas law for the unknown quantity, in this case n . Make sure that all quantities are given in units that are compatible with the units of the gas constant. If necessary, convert them to the appropriate units, insert them into the equation you have derived, and then calculate the number of moles of hydrogen gas needed. Solution: A We are given values for P , T , and V and asked to calculate n . If we solve the ideal gas law (Equation \(\ref{10.4.4}\)) for \(n\), we obtain \[\rm745\;mmHg\times\dfrac{1\;atm}{760\;mmHg}=0.980\;atm \nonumber \] B P and T are given in units that are not compatible with the units of the gas constant [ R = 0.08206 (L•atm)/(K•mol)]. We must therefore convert the temperature to kelvins and the pressure to atmospheres: \[T=273+30=303{\rm K}\nonumber \] Substituting these values into the expression we derived for n , we obtain \[\begin{align*} n &=\dfrac{PV}{RT} \\[4pt] &=\rm\dfrac{0.980\;atm\times31150\;L}{0.08206\dfrac{atm\cdot L}{\rm mol\cdot K}\times 303\;K} \\[4pt] &=1.23\times10^3\;mol \end{align*} \nonumber \] Exercise \(\PageIndex{1}\) Suppose that an “empty” aerosol spray-paint can has a volume of 0.406 L and contains 0.025 mol of a propellant gas such as CO 2 . What is the pressure of the gas at 25°C? Answer 1.5 atm In Example \(\PageIndex{1}\), we were given three of the four parameters needed to describe a gas under a particular set of conditions, and we were asked to calculate the fourth. We can also use the ideal gas law to calculate the effect of changes in any of the specified conditions on any of the other parameters, as shown in Example \(\PageIndex{5}\). General Gas Equation When a gas is described under two different conditions, the ideal gas equation must be applied twice - to an initial condition and a final condition. This is: \[\begin{array}{cc}\text{Initial condition }(i) & \text{Final condition} (f) \\ P_iV_i=n_iRT_i & P_fV_f=n_fRT_f\end{array} \nonumber \] Both equations can be rearranged to give: \[R=\dfrac{P_iV_i}{n_iT_i} \hspace{1cm} R=\dfrac{P_fV_f}{n_fT_f} \nonumber \] The two equations are equal to each other since each is equal to the same constant \(R\). Therefore, we have: \[\dfrac{P_iV_i}{n_iT_i}=\dfrac{P_fV_f}{n_fT_f}\label{10.4.8} \] The equation is called the general gas equation . The equation is particularly useful when one or two of the gas properties are held constant between the two conditions. In such cases, the equation can be simplified by eliminating these constant gas properties. Example \(\PageIndex{2}\) Suppose that Charles had changed his plans and carried out his initial flight not in August but on a cold day in January, when the temperature at ground level was −10°C (14°F). How large a balloon would he have needed to contain the same amount of hydrogen gas at the same pressure as in Example \(\PageIndex{1}\)? Given: temperature, pressure, amount, and volume in August; temperature in January Asked for: volume in January Strategy: Use the results from Example \(\PageIndex{1}\) for August as the initial conditions and then calculate the change in volume due to the change in temperature from 30°C to −10°C. Begin by constructing a table showing the initial and final conditions. Simplify the general gas equation by eliminating the quantities that are held constant between the initial and final conditions, in this case \(P\) and \(n\). Solve for the unknown parameter. Solution: A To see exactly which parameters have changed and which are constant, prepare a table of the initial and final conditions: Initial (August) Final (January) \(T_i=30\, °C = 303\, K\) \(T_f=−10\,°C = 263\, K\) \(P_i= 0.980 \, atm\) \(P_f= 0.980\, atm\) \(n_i=1.23 × 10^3\, mol\) \(n_f= 1.23 × 10^3\, mol\) \(V_i=31150\, L\) \(V_f=?\) B Both \(n\) and \(P\) are the same in both cases (\(n_i=n_f,P_i=P_f\)). Therefore, Equation \ref{10.4.8} can be simplified to: \[\dfrac{V_i}{T_i}=\dfrac{V_f}{T_f} \nonumber \] This is the relationship first noted by Charles. C Solving the equation for \(V_f\), we get: \[\begin{align*} V_f &=V_i\times\dfrac{T_f}{T_i} \\[4pt] &=\rm31150\;L\times\dfrac{263\;K}{303\;K} \\[4pt] &=2.70\times10^4\;L \end{align*} \nonumber \] It is important to check your answer to be sure that it makes sense, just in case you have accidentally inverted a quantity or multiplied rather than divided. In this case, the temperature of the gas decreases. Because we know that gas volume decreases with decreasing temperature, the final volume must be less than the initial volume, so the answer makes sense. We could have calculated the new volume by plugging all the given numbers into the ideal gas law, but it is generally much easier and faster to focus on only the quantities that change. Exercise \(\PageIndex{2}\) At a laboratory party, a helium-filled balloon with a volume of 2.00 L at 22°C is dropped into a large container of liquid nitrogen (T = −196°C). What is the final volume of the gas in the balloon? Answer 0.52 L Example \(\PageIndex{1}\) illustrates the relationship originally observed by Charles. We could work through similar examples illustrating the inverse relationship between pressure and volume noted by Boyle ( PV = constant) and the relationship between volume and amount observed by Avogadro ( V / n = constant). We will not do so, however, because it is more important to note that the historically important gas laws are only special cases of the ideal gas law in which two quantities are varied while the other two remain fixed. The method used in Example \(\PageIndex{1}\) can be applied in any such case, as we demonstrate in Example \(\PageIndex{2}\) (which also shows why heating a closed container of a gas, such as a butane lighter cartridge or an aerosol can, may cause an explosion). Example \(\PageIndex{3}\) Aerosol cans are prominently labeled with a warning such as “Do not incinerate this container when empty.” Assume that you did not notice this warning and tossed the “empty” aerosol can in Exercise \(\PageIndex{1}\) (0.025 mol in 0.406 L, initially at 25°C and 1.5 atm internal pressure) into a fire at 750°C. What would be the pressure inside the can (if it did not explode)? Given: initial volume, amount, temperature, and pressure; final temperature Asked for: final pressure Strategy: Follow the strategy outlined in Example \(\PageIndex{2}\). Solution: Prepare a table to determine which parameters change and which are held constant: Initial Final \(V_i=0.406\;\rm L\) \(V_f=0.406\;\rm L\) \(n_i=0.025\;\rm mol\) \(n_f=0.025\;\rm mol\) \(T_i=\rm25\;^\circ C=298\;K\) \(T_i=\rm750\;^\circ C=1023\;K\) \(P_i=1.5\;\rm atm\) \(P_f=?\) Both \(V\) and \(n\) are the same in both cases (\(V_i=V_f,n_i=n_f\)). Therefore, Equation can be simplified to: \[ \dfrac{P_i}{T_i} = \dfrac{P_f}{T_f} \nonumber \] By solving the equation for \(P_f\), we get: \[\begin{align*} P_f &=P_i\times\dfrac{T_f}{T_i} \\[4pt] &=\rm1.5\;atm\times\dfrac{1023\;K}{298\;K} \\[4pt] &=5.1\;atm \end{align*} \nonumber \] This pressure is more than enough to rupture a thin sheet metal container and cause an explosion! Exercise \(\PageIndex{3}\) Suppose that a fire extinguisher, filled with CO 2 to a pressure of 20.0 atm at 21°C at the factory, is accidentally left in the sun in a closed automobile in Tucson, Arizona, in July. The interior temperature of the car rises to 160°F (71.1°C). What is the internal pressure in the fire extinguisher? Answer 23.4 atm In Examples \(\PageIndex{1}\) and \(\PageIndex{2}\) , two of the four parameters ( P , V , T , and n ) were fixed while one was allowed to vary, and we were interested in the effect on the value of the fourth. In fact, we often encounter cases where two of the variables P , V , and T are allowed to vary for a given sample of gas (hence n is constant), and we are interested in the change in the value of the third under the new conditions. Example \(\PageIndex{4}\) We saw in Example \(\PageIndex{1}\) that Charles used a balloon with a volume of 31,150 L for his initial ascent and that the balloon contained 1.23 × 10 3 mol of H 2 gas initially at 30°C and 745 mmHg. Suppose that Gay-Lussac had also used this balloon for his record-breaking ascent to 23,000 ft and that the pressure and temperature at that altitude were 312 mmHg and −30°C, respectively. To what volume would the balloon have had to expand to hold the same amount of hydrogen gas at the higher altitude? Given: initial pressure, temperature, amount, and volume; final pressure and temperature Asked for: final volume Strategy: Follow the strategy outlined in Example \(\PageIndex{3}\). Solution: Begin by setting up a table of the two sets of conditions: Initial Final \(P_i=745\;\rm mmHg=0.980\;atm\) \(P_f=312\;\rm mmHg=0.411\;atm\) \(T_i=\rm30\;^\circ C=303\;K\) \(T_f=\rm750-30\;^\circ C=243\;K\) \(n_i=\rm1.2\times10^3\;mol\) \(n_i=\rm1.2\times10^3\;mol\) \(V_i=\rm31150\;L\) \(V_f=?\) By eliminating the constant property (\(n\)) of the gas, Equation \(\ref{10.4.8}\) is simplified to: \[\dfrac{P_iV_i}{T_i}=\dfrac{P_fV_f}{T_f} \nonumber \] By solving the equation for \(V_f\), we get: \[\begin{align*} V_f &=V_i\times\dfrac{P_i}{P_f}\dfrac{T_f}{T_i} \\[4pt] &=\rm3.115\times10^4\;L\times\dfrac{0.980\;atm}{0.411\;atm}\dfrac{243\;K}{303\;K} \\[4pt] &=5.96\times10^4\;L \end{align*} \nonumber \] Does this answer make sense? Two opposing factors are at work in this problem: decreasing the pressure tends to increase the volume of the gas, while decreasing the temperature tends to decrease the volume of the gas. Which do we expect to predominate? The pressure drops by more than a factor of two, while the absolute temperature drops by only about 20%. Because the volume of a gas sample is directly proportional to both T and 1/ P , the variable that changes the most will have the greatest effect on V . In this case, the effect of decreasing pressure predominates, and we expect the volume of the gas to increase, as we found in our calculation. We could also have solved this problem by solving the ideal gas law for V and then substituting the relevant parameters for an altitude of 23,000 ft: Except for a difference caused by rounding to the last significant figure, this is the same result we obtained previously. There is often more than one “right” way to solve chemical problems. Exercise \(\PageIndex{4}\) A steel cylinder of compressed argon with a volume of 0.400 L was filled to a pressure of 145 atm at 10°C. At 1.00 atm pressure and 25°C, how many 15.0 mL incandescent light bulbs could be filled from this cylinder? (Hint: find the number of moles of argon in each container.) Answer 4.07 × 10 3 Using the Ideal Gas Law to Calculate Gas Densities and Molar Masses The ideal gas law can also be used to calculate molar masses of gases from experimentally measured gas densities. To see how this is possible, we first rearrange the ideal gas law to obtain \[\dfrac{n}{V}=\dfrac{P}{RT}\label{10.4.9} \] The left side has the units of moles per unit volume (mol/L). The number of moles of a substance equals its mass (\(m\), in grams) divided by its molar mass (\(M\), in grams per mole): \[n=\dfrac{m}{M}\label{10.4.10} \] Substituting this expression for \(n\) into Equation \(\ref{10.4.9}\) gives \[\dfrac{m}{MV}=\dfrac{P}{RT}\label{10.4.11} \] Because \(m/V\) is the density \(d\) of a substance, we can replace \(m/V\) by \(d\) and rearrange to give \[\rho=\dfrac{m}{V}=\dfrac{MP}{RT}\label{10.4.12} \] The distance between particles in gases is large compared to the size of the particles, so their densities are much lower than the densities of liquids and solids. Consequently, gas density is usually measured in grams per liter (g/L) rather than grams per milliliter (g/mL). Example \(\PageIndex{5}\) Calculate the density of butane at 25°C and a pressure of 750 mmHg. Given: compound, temperature, and pressure Asked for: density Strategy: Calculate the molar mass of butane and convert all quantities to appropriate units for the value of the gas constant. Substitute these values into Equation \(\ref{10.4.12}\) to obtain the density. Solution: A The molar mass of butane (C 4 H 10 ) is \[M=(4)(12.011) + (10)(1.0079) = 58.123 \rm g/mol \nonumber \] Using 0.08206 (L•atm)/(K•mol) for R means that we need to convert the temperature from degrees Celsius to kelvins ( T = 25 + 273 = 298 K) and the pressure from millimeters of mercury to atmospheres: \[P=\rm750\;mmHg\times\dfrac{1\;atm}{760\;mmHg}=0.987\;atm \nonumber \] B Substituting these values into Equation \(\ref{10.4.12}\) gives \[\rho=\rm\dfrac{58.123\;g/mol\times0.987\;atm}{0.08206\dfrac{L\cdot atm}{K\cdot mol}\times298\;K}=2.35\;g/L \nonumber \] Exercise \(\PageIndex{5}\): Density of Radon Radon (Rn) is a radioactive gas formed by the decay of naturally occurring uranium in rocks such as granite. It tends to collect in the basements of houses and poses a significant health risk if present in indoor air. Many states now require that houses be tested for radon before they are sold. Calculate the density of radon at 1.00 atm pressure and 20°C and compare it with the density of nitrogen gas, which constitutes 80% of the atmosphere, under the same conditions to see why radon is found in basements rather than in attics. Answer radon, 9.23 g/L; N 2 , 1.17 g/L A common use of Equation \(\ref{10.4.12}\) is to determine the molar mass of an unknown gas by measuring its density at a known temperature and pressure. This method is particularly useful in identifying a gas that has been produced in a reaction, and it is not difficult to carry out. A flask or glass bulb of known volume is carefully dried, evacuated, sealed, and weighed empty. It is then filled with a sample of a gas at a known temperature and pressure and reweighed. The difference in mass between the two readings is the mass of the gas. The volume of the flask is usually determined by weighing the flask when empty and when filled with a liquid of known density such as water. The use of density measurements to calculate molar masses is illustrated in Example \(\PageIndex{6}\). Example \(\PageIndex{6}\) The reaction of a copper penny with nitric acid results in the formation of a red-brown gaseous compound containing nitrogen and oxygen. A sample of the gas at a pressure of 727 mmHg and a temperature of 18°C weighs 0.289 g in a flask with a volume of 157.0 mL. Calculate the molar mass of the gas and suggest a reasonable chemical formula for the compound. Given: pressure, temperature, mass, and volume Asked for: molar mass and chemical formula Strategy: Solve Equation \(\ref{10.4.12}\) for the molar mass of the gas and then calculate the density of the gas from the information given. Convert all known quantities to the appropriate units for the gas constant being used. Substitute the known values into your equation and solve for the molar mass. Propose a reasonable empirical formula using the atomic masses of nitrogen and oxygen and the calculated molar mass of the gas. Solution: A Solving Equation \(\ref{10.4.12}\) for the molar mass gives \[M=\dfrac{mRT}{PV}=\dfrac{dRT}{P} \nonumber \] Density is the mass of the gas divided by its volume: \[\rho=\dfrac{m}{V}=\dfrac{0.289\rm g}{0.157\rm L}=1.84 \rm g/L\nonumber \] B We must convert the other quantities to the appropriate units before inserting them into the equation: \[T=18+273=291 K\nonumber \] \[P=727 \, mmHg \times \dfrac{1\rm atm} {760\rm mmHg} =0.957\rm atm \nonumber \] The molar mass of the unknown gas is thus \[M=\rm\dfrac{1.84\;g/L\times0.08206\dfrac{L\cdot atm}{K\cdot mol}\times291\;K}{0.957\;atm}=45.9 g/mol\nonumber \] C The atomic masses of N and O are approximately 14 and 16, respectively, so we can construct a list showing the masses of possible combinations: \[M({\rm NO})=14 + 16=30 \rm\; g/mol\nonumber \] \[M({\rm N_2O})=(2)(14)+16=44 \rm\;g/mol\nonumber \] \[M({\rm NO_2})=14+(2)(16)=46 \rm\;g/mol\nonumber \] The most likely choice is NO 2 which is in agreement with the data. The red-brown color of smog also results from the presence of NO 2 gas. Exercise \(\PageIndex{6}\) You are in charge of interpreting the data from an unmanned space probe that has just landed on Venus and sent back a report on its atmosphere. The data are as follows: pressure, 90 atm; temperature, 557°C; density, 58 g/L. The major constituent of the atmosphere (>95%) is carbon. Calculate the molar mass of the major gas present and identify it. Answer 44 g/mol; \(CO_2\) Summary The ideal gas law is derived from empirical relationships among the pressure, the volume, the temperature, and the number of moles of a gas; it can be used to calculate any of the four properties if the other three are known. Ideal gas equation : \(PV = nRT\), where \(R = 0.08206 \dfrac{\rm L\cdot atm}{\rm K\cdot mol}=8.3145 \dfrac{\rm J}{\rm K\cdot mol}\) General gas equation : \(\dfrac{P_iV_i}{n_iT_i}=\dfrac{P_fV_f}{n_fT_f}\) Density of a gas: \(\rho=\dfrac{MP}{RT}\) The empirical relationships among the volume, the temperature, the pressure, and the amount of a gas can be combined into the ideal gas law , PV = nRT . The proportionality constant, R , is called the gas constant and has the value 0.08206 (L•atm)/(K•mol), 8.3145 J/(K•mol), or 1.9872 cal/(K•mol), depending on the units used. The ideal gas law describes the behavior of an ideal gas , a hypothetical substance whose behavior can be explained quantitatively by the ideal gas law and the kinetic molecular theory of gases. Standard temperature and pressure (STP) is 0°C and 1 atm. The volume of 1 mol of an ideal gas at STP is 22.41 L, the standard molar volume . All of the empirical gas relationships are special cases of the ideal gas law in which two of the four parameters are held constant. The ideal gas law allows us to calculate the value of the fourth quantity ( P , V , T , or n ) needed to describe a gaseous sample when the others are known and also predict the value of these quantities following a change in conditions if the original conditions (values of P , V , T , and n ) are known. The ideal gas law can also be used to calculate the density of a gas if its molar mass is known or, conversely, the molar mass of an unknown gas sample if its density is measured.
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Courses/Nassau_Community_College/Organic_Chemistry_I_and_II/05%3A_An_Introduction_to_Organic_Reactions_using_Free_Radical_Halogenation_of_Alkanes/5.02%3A_5.2_Reaction_Mechanism_Notation_and_Symbols
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Learning Objective accurately and precisely use reaction mechanism notation and symbols including curved arrows to show the flow of electrons The Arrow Notation in Mechanisms Since chemical reactions involve the breaking and making of bonds, a consideration of the movement of bonding (and non-bonding) valence shell electrons is essential to this understanding. It is now common practice to show the movement of electrons with curved arrows, and a sequence of equations depicting the consequences of such electron shifts is termed a mechanism . In general, two kinds of curved arrows are used in drawing mechanisms: 0 1 2 A full head on the arrow indicates the movement or shift of an electron pair: NaN NaN A partial head (fishhook) on the arrow indicates the shift of a single electron: NaN NaN The use of these symbols in bond-breaking and bond-making reactions is illustrated below. If a covalent single bond is broken so that one electron of the shared pair remains with each fragment, as in the first example, this bond-breaking is called homolysis . If the bond breaks with both electrons of the shared pair remaining with one fragment, as in the second and third examples, this is called heterolysis . 0 1 2 Bond-Breaking NaN Bond-Making NaN NaN NaN Other Arrow Symbols Chemists also use arrow symbols for other purposes, and it is essential to use them correctly. 0 1 2 The Reaction Arrow The Equilibrium Arrow The Resonance Arrow NaN NaN NaN The following equations illustrate the proper use of these symbols: Reactive Intermediates The products of bond breaking, shown above, are not stable in the usual sense, and cannot be isolated for prolonged study. Such species are referred to as reactive intermediates , and are believed to be transient intermediates in many reactions. The general structures and names of four such intermediates are given below. A pair of widely used terms, related to the Lewis acid-base notation, should also be introduced here. Electrophile: An electron deficient atom, ion or molecule that has an affinity for an electron pair, and will bond to a base or nucleophile. Nucleophile: An atom, ion or molecule that has an electron pair that may be donated in bonding to an electrophile (or Lewis acid). Using these definitions, it is clear that carbocations ( called carbonium ions in the older literature ) are electrophiles and carbanions are nucleophiles. Carbenes have only a valence shell sextet of electrons and are therefore electron deficient. In this sense they are electrophiles, but the non-bonding electron pair also gives carbenes nucleophilic character. As a rule, the electrophilic character dominates carbene reactivity. Carbon radicals have only seven valence electrons, and may be considered electron deficient; however, they do not in general bond to nucleophilic electron pairs, so their chemistry exhibits unique differences from that of conventional electrophiles. Radical intermediates are often called free radicals . The importance of electrophile / nucleophile terminology comes from the fact that many organic reactions involve at some stage the bonding of a nucleophile to an electrophile, a process that generally leads to a stable intermediate or product. Reactions of this kind are sometimes called ionic reactions , since ionic reactants or products are often involved. Some common examples of ionic reactions and their mechanisms may be examined below. The shapes ideally assumed by these intermediates becomes important when considering the stereochemistry of reactions in which they play a role. A simple tetravalent compound like methane, CH 4 , has a tetrahedral configuration. Carbocations have only three bonds to the charge bearing carbon, so it adopts a planar trigonal configuration. Carbanions are pyramidal in shape ( tetrahedral if the electron pair is viewed as a substituent), but these species invert rapidly at room temperature, passing through a higher energy planar form in which the electron pair occupies a p-orbital. Radicals are intermediate in configuration, the energy difference between pyramidal and planar forms being very small. Since three points determine a plane, the shape of carbenes must be planar; however, the valence electron distribution varies. Ionic Reactions The principles and terms introduced in the previous sections can now be summarized and illustrated by the following three examples. Reactions such as these are called ionic or polar reactions, because they often involve charged species and the bonding together of electrophiles and nucleophiles . Ionic reactions normally take place in liquid solutions, where solvent molecules assist the formation of charged intermediates. 0 1 NaN The substitution reaction shown on the left can be viewed as taking place in three steps. The first is an acid-base equilibrium, in which HCl protonates the oxygen atom of the alcohol. The resulting conjugate acid then loses water in a second step to give a carbocation intermediate. Finally, this electrophile combines with the chloride anion nucleophile to give the final product. NaN The addition reaction shown on the left can be viewed as taking place in two steps. The first step can again be considered an acid-base equilibrium, with the pi-electrons of the carbon-carbon double bond functioning as a base. The resulting conjugate acid is a carbocation, and this electrophile combines with the nucleophilic bromide anion. NaN The elimination reaction shown on the left takes place in one step. The bond breaking and making operations that take place in this step are described by the curved arrows. The initial stage may also be viewed as an acid-base interaction, with hydroxide ion serving as the base and a hydrogen atom component of the alkyl chloride as an acid. NaN There are many kinds of molecular rearrangements called isomerizations. The examples shown on the left are from an important class called tautomerization or, more specifically, keto-enol tautomerization. Tautomers are rapidly interconverted constitutional isomers, usually distinguished by a different bonding location for a labile hydrogen atom (colored red here) and a differently located double bond. The equilibrium between tautomers is not only rapid under normal conditions, but it often strongly favors one of the isomers (acetone, for example, is 99.999% keto tautomer). Even in such one-sided equilibria, evidence for the presence of the minor tautomer comes from the chemical behavior of the compound. Tautomeric equilibria are catalyzed by traces of acids or bases that are generally present in most chemical samples. Exercise 1. Add curved arrows to explain the indicated reactivity and classify the reaction as "homolytic cleavage" or "heterolytic cleavage". 2. Add the correct arrow to each expression below using your knowledge of chemistry. 3. Classify the following reactions as substituion, addition, elimination, or tautomerization (an example of isomerization). Answer 1. 2. 3.
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Courses/Lumen_Learning/Book%3A_Western_Civilization_(Lumen)/Ch._08_The_Middle_Ages_in_Europe/09.31%3A_Intellectual_Life
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Learning Objective Describe intellectual life in the Middle Ages Key Points Increased contact with Byzantium and with the Islamic world in Muslim-dominated Spain and Sicily, the Crusades, and the Reconquista allowed Europeans to seek and translate the works of Hellenic and Islamic philosophers and scientists, especially Aristotle. The groundwork for the rebirth of learning was also laid by the process of political consolidation and centralization of the monarchies of Europe, especially of Charlemagne and Otto I. Cathedral schools and universities started to develop, with young men proceeding to university to study the trivium and quadrivium . Scholasticism was a fusing of philosophy and theology by 12th- and 13th-century scholars that tried to employ a systematic approach to truth and reason. Royal and noble courts saw the development of poems and songs spread by traveling minstrels. Legal studies advanced in Western Europe. Algebra was invented, allowing more developed mathematics, and astronomy and medicine advanced. Terms Thomas Aquinas Italian Dominican friar and priest (c. 1225 CE–1274 CE) and an immensely influential philosopher and theologian in the tradition of scholasticism. trivium In medieval universities, the trivium comprised the three subjects that were taught first: grammar, logic, and rhetoric. Corpus Juris Civilis The modern name for a collection of fundamental works in jurisprudence, issued from 529–534 CE by order of Eastern Roman Emperor Justinian I. quadrivium The four subjects, or arts, taught after the trivium. It consisted of arithmetic, geometry, music, and astronomy and was considered preparatory work for the serious study of philosophy and theology. Ptolemy Greco-Roman writer of Alexandria (c. CE 90–c. 168 CE) known as a mathematician, astronomer, geographer, astrologer, and poet. Ptolemy was the author of several scientific treatises, three of which were of continuing importance to later Islamic and European science. Aristotle Greek philosopher and scientist born in Stagirus, northern Greece, in 384 BCE. His writings covered many subjects and constitute the first comprehensive system of Western philosophy. scholasticism Method of critical thought that dominated teaching by the academics (scholastics, or schoolmen) of medieval universities in Europe from about 1100–1700 CE. During the 11th century, developments in philosophy and theology led to increased intellectual activity, sometimes called the renaissance of 12th century. The intellectual problems discussed throughout this period were the relation of faith to reason, the existence and simplicity of God, the purpose of theology and metaphysics, and the issues of knowledge, of universals, and of individuation. Philosophical discourse was stimulated by the rediscovery of Aristotle—more than 3,000 pages of his works would eventually be translated—and his emphasis on empiricism and rationalism. Scholars such as Peter Abelard (d. 1142) and Peter Lombard (d. 1164) introduced Aristotelian logic into theology. Historical Conditions The groundwork for the rebirth of learning was also laid by the process of political consolidation and centralization of the monarchies of Europe. This process of centralization began with Charlemagne, King of the Franks (768–814) and later Holy Roman Emperor (800–814). Charlemagne’s inclination towards education, which led to the creation of many new churches and schools where students were required to learn Latin and Greek, has been called the “Carolingian Renaissance.” A second “renaissance” occurred during the reign of Otto I, King of the Saxons from 936–973 and Holy Roman Emperor from 952. Otto was successful in unifying his kingdom and asserting his right to appoint bishops and archbishops throughout the kingdom. Otto’s assumption of this ecclesiastical power brought him into close contact with the best-educated and ablest class of men in his kingdom. From this close contact, many new reforms were introduced in the Saxon kingdom and in the Holy Roman Empire. Thus, Otto’s reign has also been called a “renaissance.” The renaissance of the twelfth century has been identified as the third and final of the medieval renaissances. Yet the renaissance of the 12th century was far more thoroughgoing than those renaissances that preceded in the Carolingian and Ottonian periods. Conquest of and contact with the Muslim world through the Crusades and the reconquest of Spain also yielded new texts and knowledge. Most notably, contact with Muslims led to the the European rediscovery and translation of Aristotle, whose wide-ranging works influenced medieval philosophy, theology, science, and medicine. Schools and Universities The late-11th and early-12th centuries also saw the rise of cathedral schools throughout Western Europe, signaling the shift of learning from monasteries to cathedrals and towns. Cathedral schools were in turn replaced by the universities established in major European cities. The first universities in Europe included the University of Bologna (1088), the University of Paris (c. 1150, later associated with the Sorbonne), and the University of Oxford (1167). In Europe, young men proceeded to university when they had completed their study of the trivium— the preparatory arts of grammar, rhetoric, and dialectic or logic—and the quadrivium— arithmetic, geometry, music, and astronomy. Mob Quad at Merton College, University of Oxford. Aerial view of Merton College’s Mob Quad, the oldest quadrangle of the university, constructed from 1288-1378. Philosophy and theology fused in scholasticism, an attempt by 12th- and 13th-century scholars to reconcile authoritative texts, most notably Aristotle and the Bible. This movement tried to employ a systemic approach to truth and reason and culminated in the thought of Thomas Aquinas (d. 1274), who wrote the Summa Theologica , or Summary of Theology. The development of medieval universities allowed them to aid materially in the translation and propagation of these texts and started a new infrastructure, which was needed for scientific communities. In fact, the European university put many of these texts at the center of its curriculum, with the result that the “medieval university laid far greater emphasis on science than does its modern counterpart and descendent.” Poems and Stories Royal and noble courts saw the development of chivalry and the ethos of courtly love. This culture was expressed in the vernacular languages rather than Latin, and comprised poems, stories, legends, and popular songs spread by troubadours, or wandering minstrels. Often the stories were written down in the chansons de geste, or “songs of great deeds,” such as “The Song of Roland” or “The Song of Hildebrand.” Secular and religious histories were also produced. Geoffrey of Monmouth (d. c. 1155) composed his Historia Regum Britanniae , a collection of stories and legends about Arthur. Other works were more clearly pure history, such as Otto von Freising’s (d. 1158) Gesta Friderici Imperatoris, detailing the deeds of Emperor Frederick Barbarossa, or William of Malmesbury’s (d. c. 1143) Gesta Regum, on the kings of England. Legal Studies Legal studies advanced during the 12th century. Both secular law and canon law, or ecclesiastical law, were studied in the High Middle Ages. Secular law, or Roman law, was advanced greatly by the discovery of the Corpus Juris Civilis in the 11th century, and by 1100 Roman law was being taught at Bologna. This led to the recording and standardization of legal codes throughout Western Europe. Canon law was also studied, and around 1140 a monk named Gratian, a teacher at Bologna, wrote what became the standard text of canon law—the Decretum . Algebra and Astronomy Among the results of the Greek and Islamic influence on this period in European history were the replacement of Roman numerals with the decimal positional number system and the invention of algebra, which allowed more advanced mathematics. Astronomy advanced following the translation of Ptolemy’s Almagest from Greek into Latin in the late 12th century. Medicine was also studied, especially in southern Italy, where Islamic medicine influenced the school at Salerno. A YouTube element has been excluded from this version of the text. You can view it online here: http://pb.libretexts.org/wcw/?p=298 The Weird Truth about Arabic Numerals. How the world came to use so-called Arabic numerals—from the scholarship of ancient Hindu mathematicians, to Muslim scientist Al-Khwarizmi, to the merchants of medieval Italy. Sources CC licensed content, Shared previously Boundless World History. Authored by : Boundless. Located at : https://www.boundless.com/world-history/textbooks/boundless-world-history-textbook/ . License : CC BY-SA: Attribution-ShareAlike
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Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Nitriles/Reactivity_of_Nitriles/Conversion_of_nitriles_to_1_amines_using_LiAlH4
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Nitriles can be converted to 1° amines by reaction with LiAlH 4 . During this reaction the hydride nucleophile attacks the electrophilic carbon in the nitrile to form an imine anion. Once stabilized by a Lewis acid-base complexation the imine salt can accept a second hydride to form a dianion. The dianion can then be converted to an amine by addition of water. General Reaction Going from reactants to products simplified Example Mechanism 1) Nucleophilic Attack by the Hydride 2) Second nucleophilic attack by the hydride. 3) Protonation by addition of water to give an amine Contributors Prof. Steven Farmer ( Sonoma State University )
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Courses/can/CHEM_210%3A_General_Chemistry_I_(An_Atoms_Up_Approach)/15%3A_Gases_and_Gas_Laws/15.04%3A_Ideal_Gases_and_The_Ideal_Gas_Law
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Learning Objectives Explain all the quantities involved in the ideal gas law. Evaluate the gas constant \(R\) from experimental results. Calculate \(T\), \(V\), \(P\), or \(n\) of the ideal gas law: \(P V = n R T\). Describe the ideal gas law using graphics. The Ideal Gas Law The volume ( V ) occupied by n moles of any gas has a pressure ( P ) at temperature ( T ) in Kelvin. The relationship for these variables, \[P V = n R T\] where R is known as the gas constant, is called the ideal gas law or equation of state . Properties of the gaseous state predicted by the ideal gas law are within 5% for gases under ordinary conditions. In other words, given a set of conditions, we can predict or calculate the properties of a gas to be within 5% by applying the ideal gas law. How to apply such a law for a given set of conditions is the focus of general chemistry. At a temperature much higher than the critical temperature and at low pressures, however, the ideal gas law is a very good model for gas behavior. When dealing with gases at low temperature and at high pressure, correction has to be made in order to calculate the properties of a gas in industrial and technological applications. One of the common corrections made to the ideal gas law is the van der Waal's equation, but there are also other methods dealing with the deviation of gas from ideality. The Gas Constant R Repeated experiments show that at standard temperature (273 K) and pressure (1 atm or 101325 N/m 2 ), one mole ( n = 1) of gas occupies 22.4 L volume. Using this experimental value, you can evaluate the gas constant R , \(\begin{align} R &= \dfrac{P V}{n T} = \mathrm{\dfrac{1\: atm\:\: 22.4\: L}{1\: mol\:\: 273\: K}}\\ \\ &= \mathrm{0.08205\: \dfrac{L\: atm}{mol\cdot K}} \end{align}\) When SI units are desirable, P = 101325 N/m 2 (Pa for pascal) instead of 1 atm. The volume is 0.0224 m 3 . The numerical value and units for R are \(\begin{align} R &= \mathrm{\dfrac{101325\: \dfrac{N}{m^2}\: 0.0224\: m^3}{1\: mol\: 273\: K}}\\ \\ &= \mathrm{8.314\: \dfrac{J}{mol\cdot K}} \end{align}\) Note that \(\mathrm{1\: L\: atm = 0.001\: m^3 \times 101325\: \dfrac{N}{m^2} = 101.325\: J\: (or\: N\: m)}\). Since energy can be expressed in many units, other numerical values and units for R are frequently in use. The Gas Constant For your information, the gas constant can be expressed in the following values and units. \(\begin{align} R &= \mathrm{0.08205\: \dfrac{L\: atm}{mol \cdot K}} &&\textrm{Notes:} \\ &= \mathrm{8.3145\: \dfrac{L\: kPa}{mol\cdot K}} &&\mathrm{1\: atm = 101.32\: kPa} \\ &= \mathrm{8.3145\: \dfrac{J}{mol\cdot K}} &&\mathrm{1\: J = 1\: L\: kPa} \\ &= \mathrm{1.987\: \dfrac{cal}{mol\cdot K}} &&\mathrm{1\: cal = 4.182\: J} \\ &= \mathrm{62.364\: \dfrac{L\: torr}{mol\cdot K}} &&\mathrm{1\: atm = 760\: torr}\\ \end{align}\) The gas constant R is such a universal constant for all gases that its values are usually listed in the "Physical Constants" of textbooks and handbooks. It is also listed in Constants of our HandbookMenu at the left bottom. Although we try to use SI units all the time, the use of atm for pressure is still common. Thus, we often use R = 8.314 J / (mol·K) or 8.3145 J / mol·K. The volume occupied by one mole, n = 1, of substance is called the molar volume , \(V_{\textrm{molar}} = \dfrac{V}{n}\). Using the molar volume notation, the ideal gas law is: \(P V_{\textrm{molar}} = R T\) Applications of the Ideal Gas Law The ideal gas law has four parameters and a constant, R , \(P V = n R T\), and it can be rearranged to give an expression for each of P, V, n or T . For example, \(P = \dfrac{n R T}{V}\) (Boyle's law) \(P = \left(\dfrac{n R}{V}\right) T\) (Charles's law) These equations are Boyle's law and Charles's law respectively. Similar expressions can be derived for V, n and T in terms of other variables. Thus, there are many applications. However, you must make sure that you use the proper numerical value for the gas constant R according to the units you have for the parameters. Furthermore, \(\dfrac{n}{V}\) is number of moles per unit volume, and this quantity has the same units as the concentration ( C ). Thus, the concentration is a function of pressure and temperature, \(C = \dfrac{P}{R T}\) At 1.0 atm pressure and room temperature of 298 K, the concentration of an ideal gas is 0.041 mol/L. Avogadro's law can be further applied to correlate gas density \(\rho\) (weight per unit volume or n M / V ) and molecular mass M of a gas. The following equation is easily derived from the ideal gas law: \[P M =\dfrac{n M}{V}R T\] Thus, we have \(\begin{align} P M &= \dfrac{d R T}{M}\\ \rho &= \dfrac{n M}{V} \leftarrow \textrm{definition, and}\\ \rho &= \dfrac{P M}{R T}\\ M &= \dfrac{d R T}{P} \end{align}\) Example 1 An air sample containing only nitrogen and oxygen gases has a density of 1.3393 g / L at STP. Find the weight and mole percentages of nitrogen and oxygen in the sample. Solution From the density \(\rho\), we can evaluate an average molecular weight (also called molar mass). \(\begin{align} P M &= d R T\\ M &= 22.4 \times d\\ &= \mathrm{22.4\: L/mol \times 1.3393\: g/L}\\ &= \mathrm{30.0\: g / mol} \end{align}\) Assume that we have 1.0 mol of gas, and x mol of which is nitrogen, then (1 - x ) is the amount of oxygen. The average molar mass is the mole weighted average, and thus, \(\mathrm{28.0\, x + 32.0 (1 - x) = 30.0}\) \(\mathrm{- 4\, x = - 2}\) \(\mathrm{x = 0.50\: mol\: of\: N_2,\: and\: 1.0 - 0.50 = 0.50\: mol\: O_2}\) Now, to find the weight percentage, find the amounts of nitrogen and oxygen in 1.0 mol (30.0 g) of the mixture. \(\mathrm{Mass\: of\: 0.5\: mol\: nitrogen = 0.5 \times 28.0 = 14.0\: g}\) \(\mathrm{Mass\: of\: 0.5\: mol\: oxygen = 0.5 \times 32.0 = 16.0\: g}\) \(\mathrm{Percentage\: of\: nitrogen = 100 \times \dfrac{14.0}{30.0} = 46.7 \% }\) \(\mathrm{Percentage\: of\: oxygen = 100 \times \dfrac{16.0}{30.0} = 100 - 46.7 = 53.3 \%}\) DISCUSSION We can find the density of pure nitrogen and oxygen first and evaluate the fraction from the density. \(\rho \mathrm{\: of\: N_2 = \dfrac{28.0}{22.4} = 1.2500\: g/L}\) \(\rho \mathrm{\: of\: O_2 = \dfrac{32.0}{22.4} = 1.4286\: g/L}\) \(\mathrm{1.2500\, x + 1.4286 (1 - x) = 1.3393}\) Solving for x gives \(\mathrm{x = 0.50}\) (same result as above) Exercise 1 Now, repeat the calculations for a mixture whose density is 1.400 g/L. Example 2 What is the density of acetone (\(\ce{C3H6O}\)) vapor at 1.0 atm and 400 K? Solution The molar mass of acetone = 3*12.0 + 6*1.0 + 16.0 = 58.0. Thus, \(\begin{align} \rho &= \dfrac{P M}{R T}\\ \\ &= \mathrm{\dfrac{1.0 \times 58.0\: atm\: \dfrac{g}{mol}} {0.08205\: \dfrac{L\: atm}{mol\: K} \times 400\: K}}\\ \\ &= \mathrm{1.767\: g / L} \end{align}\) Exercise 2 The density of acetone is 1.767 g/L; calculate its molar mass. Confidence Building Questions What does the variable n stand for in the ideal gas law, \(P V = n R T\)? Hint: number of moles of gas in a closed system. Skill: Describe the ideal gas law. A closed system means no energy or mass flow into or out of a system. In a closed system, how many independent variables are there among n, T, V and P for a gas? Note: an independent variable can be of any arbitrary values. Hint: one Skill: The ideal gas equation shows the interdependence of the variables. Only one of them can be varied independently. What is the molar volume of an ideal gas at 2 atm and 1000 K? Hint: 41.0 Skill: Evaluate molar volume at any condition. A certain amount of a gas is enclosed in a container of fixed volume. If you let heat (energy) flow into it, what will increase? (In a multiple choice, you may have volume, pressure, temperature, and any combination of these to choose from.) Hint: Both pressure and temperature will increase. Skill: Explain a closed system and apply ideal gas law. For a certain amount ( n = constant) of gas in a closed system, how does volume V vary with the temperature? In the following, k is a constant depending on n and P . \(V = k T\) \(V = \dfrac{k}{T}\) \(T V = k\) \(V = k T^2\) \(V = k\) Hint: a Skill: Explain Charles's law. Boyle's law is P V = constant. A sketch of P vs V on graph paper is similar to a sketch of the equation x y = 5. What curve(s) does this equation represent? a parabola an ellipse a hyperbola a pair of hyperbolas a straight line a surface Hint: d Skill: Apply the skills acquired in math courses to chemical problem solving. For a certain amount of gas in a closed system, which one of the following equations is valid? Subscripts 1 and 2 refer to specific conditions 1 and 2 respectively. \(P_1 V_1 T_1 = P_2 V_2 T_2\) \(P_1 V_1 T_2 = P_2 V_2 T_1\) \(P_1 V_2 T_1 = P_2 V_1 T_2\) \(P_2 V_1 T_1 = P_1 V_2 T_2\) \(\dfrac{P_1 V_2}{T_1} = \dfrac{P_2 V_1}{T_2}\) Hint: b Skill: Rearrange a mathematical equation. The gas constant R is 8.314 J / mol·K. Convert the numerical value of R so that its units are cal / (mol·K). A unit conversion table will tell you that 1 cal = 4.184 J. Make sure you know where to find it. During the exam, the conversion factor is given, but you should know how to use it. Hint: 1.987 cal/(mol K). Skill: Use conversion factors, for example: \(\mathrm{8.314\: J\times\dfrac{1\: cal}{4.184\: J}=\: ?\: cal}\) At standard temperature and pressure, how many moles of \(\ce{H2}\) are contained in a 1.0 L container? Hint: 0.045 mol/L Discussion: There are many methods for calculating this value. At standard temperature and pressure, how many grams of \(\ce{CO2}\) are contained in a 3.0 L container? Molar mass of \(\mathrm{CO_2 = 44}\) . Hint: 5.89 g in 3 L One method: It contains \(n \mathrm{= \dfrac{1\: atm \times 3\: L}{0.08205\: \dfrac{L\: atm}{mol\cdot K}\times 273\: K}}\) What is the pressure if 1 mole of \(\ce{N2}\) occupies 1 L of volume at 1000 K? Hint: 82.1 atm Discussion: Depending on the numerical value and units of R you use, you will get the pressure in various units. At 1000 K, some of the \(\ce{N2}\) molecules may dissociate. If that is true, the pressure will be higher! What is the temperature if 1 mole of \(\ce{N2}\) occupying 100 L of volume has a pressure of 20 Pa (1 Pa = 1 Nm -2 )? Hint: 240 K Discussion: At T = 240 K, ideal gas law may not apply to \(\ce{CO2}\), because this gas liquifies at a rather high temperature. The ideal gas law is still good for \(\ce{N2}\), \(\ce{H2}\), \(\ce{O2}\) etc, because these gases liquify at much lower temperatures.
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Courses/University_of_California_Davis/UCD_Chem_002A/UCD_Chem_2A/Homework/Exercise_Solutions/Exercise_03
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0 1 2 NaN UC Davis CHE 2A: General Chemistry Instructor: Hayashi Chem 2A Chem 2B Chem 2C Reading Assignments Exercises Worksheets Homework Unit I: Atomic Theory Unit II: Chemical Reactions Unit III: Gases Unit IV: Electronic Structure and Bonding Unit I: Atomic Theory Unit II: Chemical Reactions Unit III: Gases Unit IV: Electronic Structure and Bonding Unit I: Atomic Theory Unit II: Chemical Reactions Unit III: Gases Unit IV: Electronic Structure and Bonding Template:HideTOC 3.1 Write a balanced molecular equation for the dissociation of \(NaCl\) in water. Based on the equation in part a, is \(NaCl\) a strong electrolyte or a weak electrolyte? Hydrofluoric acid is a weak acid, and does not completely dissociate in water. Is it a strong electrolyte or a weak electrolyte? A4.15S \(NaCl \rightarrow Na^+ + Cl^-\) \(NaCl\) is a strong electrolyte Hydrofluoric acid is a weak electrolyte. Hint: How do strong electrolytes dissociate in water vs. weak electrolytes. http://chemwiki.ucdavis.edu/Under_Construction/chem1/Solution_Chemistry/Ions_and_Electrolytes Solution \(NaCl\) dissociates completely in water, therefore NaCl breaks up into one positively charged sodium ion (\(Na^+\)) and one negatively charged chlorine ion (\(Cl_-\)). When a compound in water dissociates completely into ions it is a strong electrolyte. This is because the resulting solution is able to more easily conduct electricity. Hydrofluoric acid does not dissociate completely (weak acid), therefore a solution of \(HF\) and \(H_2O\) would not conduct electricity easily. Therefore it is a weak electrolyte. 3.2 Write the balanced, complete ionic, and net ionic equations for the following reactions : \(CoCl_2 + NaCl \rightarrow\) \(KSO_4 + Ba(NO_3)_2 \rightarrow\) \(NaOH + CaCl_2 \rightarrow\) Answer Balanced: \(CoCl_{2\;(aq)} + NaCl_{(aq)} \rightarrow CoCl_{2\;(aq)} + NaCl_{(aq)}\) Complete ionic: \(Co^{2+}_{(aq)} + 2Cl^-_{(aq)} + Na^+_{(aq)} + Cl^-_{(aq)} \rightarrow Co^{2+}_{(aq)} + 2Cl^-_{(aq)} + Na^+_{(aq)} + Cl^-_{(aq)}\) Net ionic: No reaction Balanced: \(K_2SO_{4\;(aq)} + Ba(NO_3)_{2\;(aq)} \rightarrow BaSO_{4\;(s)} + 2KNO_{3\;(aq)}\) Complete ionic equation: \(2K^+_{(aq)} + SO^{2-}_{4\;(aq)} + Ba^{2+}_{(aq)} +2NO_{3\;(aq)}+ \rightarrow BaSO_{4\;(s)} + 2K^+_{(aq)} +2NO^+_{3\;(aq)}\) Net ionic equation: \(SO^{2-}_{4\;(aq)} + Ba^{2+}_{(aq)} \rightarrow BaSO_{4\;(s)}\) Balanced: \(2NaOH_{(aq)} + CaCl_{2\;(aq)} \rightarrow 2NaCl_{(aq)} + Ca(OH)_{2\;(s)}\) Complete ionic equation: \(2Na^+_{(aq)} + 2OH^-_{(aq)} + Ca^{2+}_{(aq)} + 2Cl^-_{(aq)} \rightarrow 2Na^+_{(aq)} + 2Cl^-_{(aq)} + Ca(OH)_{2\;(s)}\) Net ionic equation: \(Ca^{2+}_{(aq)} + 2OH^-_{(aq)} \rightarrow Ca(OH)_{2\;(s)}\) Hint: Ions not used in the creation of a solid are spectator ions http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Reactions_in_Aqueous_Solutions/Precipitation_Reactions http://chemwiki.ucdavis.edu/Physical...lubility_Rules Solution The equation reads the same forwards as backwards, and all compounds in the balanced equation are soluble. Because the products and reactants are the same, the Net result is that no reaction occurs. All reactants and products are soluble except for BaSO 4(s) , because although most sulfates are soluble, Barium is an exception to this rule. Therefore all other ions are spectator ions except for the ions involved in the synthesis of BaSO 4 (s) The product NaCl is a soluble salt whereas Ca(OH) 2(s) is an insoluble hydroxide. Ca(OH) 2(s) is the only precipitate that forms during this reaction therefore all ions are spectator ions except for thos involved in the synthesis of Ca(OH) 2(s) . 3.3 What amount of 0.100 M \(KOH\) is needed to neutralize 10.00 mL of 0.210 M \(HF\)? Answer 21.0 mL Hint: Set up the balanced equation for this reaction. Solution This reaction is one to one: \[HF + KOH \rightarrow H_2O + KF \] Take the given information and let the equation guide the conversion: \[10.00 \; mL \; HF \times \dfrac{0.210\; mM \;HF}{1 \;mL \;HF} \times \dfrac{1\; mM\; KOH}{1\;mM \;HF} \times \dfrac{1\; mL\; KOH}{ 0.100\; mM\; KOH} = 21.0\; mL\] By the way: Like other sources of fluoride ions, \(KF\) is poisonous, although lethal doses approach gram levels for humans. It is harmful by inhalation and ingestion. It is highly corrosive, and skin contact may cause severe burns. 3.4 A students has 97 mL of 0.80 M HCl which they would like to neutralize with 0.18M NaOH. What volume of NaOH should be used? Answer 0.43 L Hint: Find the amount of moles of HCl present to find the amount of moles of NaOH are needed to neutralize the HCl. http://chemwiki.ucdavis.edu/Wikitexts/ ChemTutor/Solutions Solution First find how many moles of HCl are present in the initial solution, do this through the equation \((0.80\; mol/L) \times (0.097\;L)\) to find that 0.0776 moles of HCl are present. This indicates how many moles of NaOH are needed to neutralize this solution. To find the volume of NaOH needed to neutralize the moles of HCl use the equation (0.0776 moles of HCl) = (0.18 moles of NaOH/ L) * (X amount L) to find that the amount of Liters of NaOH needed is 0.43 L. 3.5 Below are a list of reactions. State whether each is an oxidation-reduction reaction and if it is an oxidation-reduction reaction, identify the reducing and oxidizing agents. \(CO_2(g) + H_2(g) \rightarrow CO_{(g)} + H_2O_{(g)}\) \(K_2SO_{4\, (aq)} + BaI2(aq) \rightarrow BaSO_{4\, (s)} + 2KI_{(aq)}\) \(H_3PO_{4\,(aq)} + Se^{2-}_{(aq)} \rightarrow HPO_4^{2-}(aq) + H_2Se_{(aq)}\) \(Zn (s) + CuSO_4 \rightarrow ZnSO_4(aq) + Cu(s)\) Answer redox; the reducing agent being oxidized is hydrogen and the oxidizing agent being reduced is carbon. not redox not redox redox; zinc is the reducing agent that is oxidized and copper is the oxidizing agent that is reduced. Hint: Take note of the oxidation numbers of each element. Solution Carbon has an oxidation number of 4+ on the left side of the equation and 2+ on the right side. Oxygen’s oxidation number is 2- on both sides of the equation and hydrogen has an oxidation number of 0 on the left and 1+ on the right. This means that carbon is the oxidizing agent that steals electrons from hydrogen and hydrogen is the reducing agent that gives electrons to carbon. The oxidation numbers of each element remain the same on either side of the equation therefore this is not a redox equation. Same as the solution to part b. The oxidation number of zinc is 0 on the left and 2+ on the right, the opposite is true of copper. This means that zinc is the reducing agent and copper is the oxidizing agent. 3.6 Indicate the oxidation state of the element assigned in the compounds below: \(C\) in \(CO_2\) \(S\) in \(CaSO_4\) \(Sn\) in \(SnO_2\) \(Cd\) in \(Cd\) \(Co\) in \(CoCl_2\) Answer Oxidation state of C: +4 Oxidation state of S: +6 Oxidation state of Sn: +4 Oxidation state of Cd: 0 Oxidation state of Co: +2 Hint: http://chemwiki.ucdavis.edu/Analytical_Chemistry/Electrochemistry/Redox_Chemistry/Oxidation_State http://chemwiki.ucdavis.edu/Analytical_Chemistry/Electrochemistry/Oxidation_States_%28Oxidation_Numbers%29 Solution (Oxidation state of C) + (2* Oxidation state of O) = 0 (Oxidation state of C) + (2*-2) = 0 (Oxidation state of C) = +4 (Oxidation state of Ca)+(Oxidation state of S)+(4*Oxidation state of O) = 0 (+2) + (Oxidation state of S) + (4*-2) = 0 (Oxidation state of S) = +6 (Oxidation state of Sn) + (2* Oxidation state of O) = 0 (Oxidation state of Sn) + (2* -2) = 0 (Oxidation state of Sn) = +4 (Oxidation State of Cd) =0 (Oxidation state of Co) + (2* Oxidation state of Cl) = 0 (Oxidation state of Co) + (2*-1) = 0 (Oxidation state of Co) = +2 3.7 When titrating a 4.00 mL volume of the diprotic acid, \(H_2SO_4\) of 0.645 M concentration, what volume of 0.493 M \(NaOH\) is necessary? Answer 10.5 mL Hint: How do a strong base and weak polyprotic acid react together? http://chemwiki.ucdavis.edu/Analytical_Chemistry/Quantitative_Analysis/Titration Solution Write the equation out the neutralization reaction: \[H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O\] So then, \[4.00\;mL\; H_2SO_4 \times \dfrac{0.645\; mmol\; H_2SO_4}{1.00\; mL\; H_2SO_4}\] \[= 2.58\; mmol\; H_2SO_4 \times \dfrac{2\; mmol\; NaOH}{1\; mmol\; H_2SO_4} \times {1\; mL\; NaOH}{0.493\; mmol\; NaOH} = 10.5\; mL\] 3.8 Write down a balanced formula equation for an acid-base reaction that forms potassium nitrate. Answer \[KOH_{(aq)} + HNO_{3\, (aq)} \rightarrow 2KNO_{3\, (aq)} + H_2O_{(l)}\] Hint: \ (H^+\) from the acid and \(OH^-\) from the bas e produces water \ (K^+\) and \(NO_3^-\) comes from eit her the acid or the base. Solution In an acid- base reaction, the products are always water and a salt (which may or may not be soluble). In this case, the salt is \(KNO_3\). Water is produced by the \(H^+\) and \(OH^-\) from the acid and base. The acid and the base in the reaction must also contains \(K^+\) and \(NO_3^-\) to form \(KNO_3\). So the acid is \(HNO_3\) and the base is \(KOH\). 3.9 Balance the following equations in both acidic and basic environments: \(H_2(g) + O_2(g) \rightarrow H_2O(l)\) \(Cr_2O_7^{2-}(aq) + C_2H_5OH (l) \rightarrow Cr^{3+}(aq) + CO_2(g)\) \(Fe^{2+}(aq) + MnO_4^-(aq) \rightarrow Fe^{3+}(aq) + Mn^{2+}(aq)\) \(Zn (s) + NO_3^-(aq) \rightarrow Zn^{2+}(aq) + NO(g)\) \(Al (s) + H_2O (l) + O_2 (g) \rightarrow [Al(OH)_4]^-(aq)\) Balance in only acidic Solutions \( Mn^{2+} + BiO_3^- \rightarrow MnO4^- + Bi^{3+} \) \( ClO_3^- + Cl - \rightarrow Cl_2 + ClO_2 \) Balance in only Basic Solution \(Zn + NO_3^ - \rightarrow Zn(OH)_4^{2-} + NH_3 \) \( Al + OH^- \rightarrow AlO_2^- + H_2 \) Hint : Balancing Redox Reactions Answers a) \(H_2(g) + O_2(g) \rightarrow H_2O(l)\) Acidic Answer: \(2H_{2\;(g)} + O_{2\;(g)} \rightarrow 2H_2O_{(l)}\) Basic Answer: \(2H_{2\;(g)} + O_{2\;(g)} \rightarrow 2H_2O_{(l)}\) b) \(Cr_2O_7^{2-}(aq) + C_2H_5OH (l) \rightarrow Cr^{3+}(aq) + CO_2(g)\) Acidic Answer: \(2Cr_2O_7^{2-}(aq) + C_2H_5OH_{(l)} + 16H^+_{(aq)} \rightarrow 4Cr^{3+}_{(aq)} + 2CO_{2\;(g)} + 11H_2O_{(l)}\) Basic Answer: \(2Cr_2O_7^{2-}(aq) + C_2H_5OH(l) + 5 H_2O_{(l)} \rightarrow 4Cr^{3+}_{(aq)} + 2CO_{2\;(g)} + 16OH^-(aq)\) c) \(Fe^{2+}(aq) + MnO_4^-(aq) \rightarrow Fe^{3+}(aq) + Mn^{2+}(aq)\) Acidic Answer: \(MnO_4^-(aq) + 5Fe^{2+}_{(aq)} + 8H^+_{(aq)} \rightarrow Mn^{2+}_{(aq)} + 5Fe^{3+}_{(aq)} + 4H_2O_{(l)}\) Basic Answer: \(MnO_4^-(aq) + 5Fe^{2+}_{(aq)} + 4H_2O_{(l)} \rightarrow Mn^{2+}_{(aq)} + 5Fe^{3+}_{(aq)} + 8OH^-_{(aq)}\) d) \(Zn (s) + NO_3^-(aq) \rightarrow Zn^{2+}(aq) + NO(g)\) Acidic Answer: \(3Zn(s) + 2NO_3^-(aq) + 8H^+_{(aq)} \rightarrow 3Zn^{2+}_{(aq)} + 2NO_{(g)} + 4H_2O_{(l)}\) Basic Answer: \(3Zn(s) + 2NO_3^-(aq) + 4H_2O_{(l)} \rightarrow 3Zn^{2+}_{(aq)} + 2NO_{(g)} + 8OH^-_{(aq)}\) e) \(Al (s) + H_2O (l) + O_2 (g) \rightarrow [Al(OH)_4]^-(aq)\) Acidic Answer: \(4Al(s) + 3O_{2\;(g)} + 10 H_2O_{(l)} \rightarrow 4[Al(OH)_4]^-(aq) + 4H^+_{(aq)}\) Basic Answer: \(4Al(s) + 3O_{2\;(g)} + 6H_2O_{(l)} + 4OH^-_{(aq)} \rightarrow 4[Al(OH)_4]^-_{(aq)}\) f) Balance in only acidic Solutions \( 14 H^+ + 2 Mn^{2+} + 5 BiO_3^- \rightarrow 2 MnO_4 - + 5 Bi ^{3+} + 7 H_2O \) \( 4 H^+ + 2 ClO_3^- + 2 Cl^- \rightarrow 2 ClO_2 + 2 H_2O + Cl_2 \) g) Balance in in only Basic Solution \( 6 H_2O + NO_3^- + 7 OH^- + 4 Zn \rightarrow NH_3 + 4 Zn(OH)_4^{2-} \) \( 2 Al + 2 H_2O + 2 OH^- \rightarrow 2 AlO_2^- + 3 H_2 \) 3.10 Write the products of these acid-base reactions. Balance the complete reaction. \(HClO_{4\; (aq)} + Mg(OH)_{2\; (aq)} \rightarrow\) \(H_2S_{(aq)} + NaOH_{(aq)} \rightarrow\) Answer \(2HClO_4 (aq) + Mg(OH)_2 (aq) \rightarrow Mg(ClO_4)_2 (aq) + 2H_2O (l)\) \(H_2S(aq) + 2NaOH (aq) \rightarrow Na_2S (aq) + 2H_2O (l)\) Hint See ‘Neutralization’ reaction & solubility rules. http://chemwiki.ucdavis.edu/Physical...and_Bases/Acid Solution Acid + Base -> Salt + Water \(Mg^{2+}\) is soluble in \(ClO_4^-\). Balance the equation. \(Na^+\) is soluble in \(S^{2-}\). Balance the equation. 3.11 Write chemical reactions for the dissociation of the following compounds in water and determine if they are strong electrolytes or weak electrolytes: \(HCl\), \(NaOH\), \(MgCl_2\), \(Al_2S_3\) A.4.25S - strong electrolyte: \(HCl \rightarrow H^+ + Cl^-\) strong electrolyte: \(NaOH \rightarrow Na^+ + OH^-\) strong electrolyte: \(MgCl_2 \rightarrow Mg^{2+} + 2Cl^-\) weak electrolyte: \(Al_2S_3 \rightarrow Al_2S_3\) Hint Strong electrolytes generally dissociate in water more easily than weak electrolytes. http://chemwiki.ucdavis.edu/Wikitext...d_Electrolytes Solution \(HCl\) is a strong acid and dissociates in water completely. It dissociates into the ions \(H^+\) and \(Cl^-\) therefore creating an electrolytic solution and being a strong electrolyte. \(NaOH\) is a strong electrolyte and dissociates in water completely. It dissociates into the ions \(Na^+\) and \(OH^-\), therefore creating an electrolytic solution and being a strong electrolyte. \(MgCl_2\) is a salt that dissociates in water completely. It dissociates into \(Mg^{2+}\) ions and \(Cl^-\) ions, therefore creating an electrolytic solution and being a strong electrolyte. \(Al_2S_3\) is a sulfide compound, Solubility rules indicate that sulfide compounds are only soluble in water when paired with an element from either group 1 or group 2. Aluminum is a group 3 element, therefore \(Al_2S_3\) is insoluble and a weak electrolyte. 3.12 Dichloromethane (\(CH_2Cl_2\)) is a useful solvent for many chemical processes in the organic chemistry lab. In the food industry, dichloromethane can be used to decaffeinated coffee and tea as well as to prepare extracts of hops and other flavorings. Figure: Sample of dichloromethane. from Wikipedia. Dichloromethane's vapor pressure at 25 °C is 57.3 kPa. Convert this pressure to the following pressure units. atm torr psi mm H bar Answer 0.566 atm 430 torr 8.32 psi 430 mm Hg 0.573 bar Hint Know the conversion between two different pressure units such as 1 atm = 14.7psi, 1 atm=760 torr, 1 atm=1.01325 bar, 1atm=101.325k Pa, and 1 torr=1mm Hg. Then do the dimensional analysis. http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Phases_of_Matter/Gases/The_Ideal_Gas_Law Solution 1 atm =101.325 kPa, so (1 atm/101.325 kPa)(57.3 kPa)=0.566atm 1 atm=760 torr, so (760 torr/1 atm)(0.56 6atm)= 430 torr 1 atm=14.7 psi, so (14.7 psi/1 atm)(0.566 atm)= 8.32 psi 1 torr=1mm Hg, so (1mm Hg/1 torr)(430 torr)= 430 mm Hg 1 atm=1.01325 bar, so (1.01325 bar/1atm)(0.566 atm)= 0.573 bar 3.13 An ideal gas is contained in a flask with a volume of 10 mL. If the volume of the container is increased to 50 mL, what is the final pressure, \(P_f\) in the flask? Assume the original pressure is \(P_i\) and the temperature is maintained constant. Answer \(P_f=\dfrac{P_i}{5}\) Hint Consider Boyle’s law and relationship between volume and pressure. http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Phases_of_Matter/Gases/Gas_Laws Solution For a given mass, at constant temperature, the pressure times the volume is a constant. If the same substance under two different sets of condition the law can be expressed as follow. \[P_iV_i=P_fV_f=C \] So \[ P_i \times 10\; mL = P_f \times 50\; mL\] \[P_f=\dfrac{P_i}{5}\] 3.14 A truck tire will blow out if its internal pressure increases to 2 atm. In a morning, a driver filled the tire will air to 1.50 atm at \(18 ^oC\). Driving the truck at noon increases the temperature to \(32 ^oC\). Assuming the volume of the tire is constant, what’s the final condition of the tire? Figure : Overpressurizing a tire can result in catastrophic failure (i.e., "a blown tire"). Answer Since \(P_f= 1.57\; atm < 2\; atm\), the tire will not blow out. Hint Consider the ideal gas law and calculate final internal pressure. http://chemwiki.ucdavis.edu/Physical..._Ideal_Gas_Law Solution Step 1: Write down your given information: \(P_i=1.5\;atm\) \(P_f =?\) \(V_f=V_i\) \(n_f=n_i\) \(R = 0.0820574 \; L*atm*mol^{-1}K^{-1}\) \(T_i= 18^oC\) \(T_f=32 ^oC\) Step 2: Convert as necessary: Temperature into Kelvin: \[T_i= 18^oC + 273\;K = 291 K\] \[T_f=32 ^oC+273\; K =305\; K\] Step 3: Plug in the variables into the ideal gas law and solve for volume: \[V=\dfrac{nRT}{P}\] \[P=\left(\dfrac{nR}{V}\right) T\] \[\dfrac{P}{T}=\dfrac{nR}{V}\] \[P_f=\dfrac{n_fRT_f}{V}\] \[P_f=\dfrac{nR}{V} \times T_f\] \[P_f=\dfrac{P_i}{T_i} \times T_f\] \[P_f = \dfrac{1.50\;atm}{291\;K} \times 305\;K\] Since \(P_f= 1.57\; atm < 2\; atm\), the tire will not blow out. 3.15 Please calculate the density of nitrogen gas at STP. Answer \(\rho =1.251g/L\) Hint Consider the ideal gas law and definition of standard temperature and pressure. http://chemwiki.ucdavis.edu/Physical..._Ideal_Gas_Law Solution STP corresponds to 273 K (0° Celsius) and 1 atm Step 1:Write down your given information: \(P=1\;atm\) \(V=1\;L\) \(n=?\) \(R=0.0820574 \;L*atm*mol^{-1}K^{-1}\) \(T=273\;K\) Step 2: Convert as necessary: Temperature is already given in Kelvin. The pressure is given in units to match those of \(R\) Step 3: Plug in the variables into the appropriate equation: \[V=\dfrac{nRT}{P}\] \[n=\dfrac{VP}{RT}\] \[n=\dfrac{1\; L \times 1\; atm}{0.0820574 L*atm*mol^{-1}K^{-1} \times 273\; K}\] therefore, there are \(n=0.0446\; mol\) of \(N_2\), but we need mass, not moles. \[m=MM \times n\] \[m=28.01\; g \times 0.0446\; mol\] \[m=1.25\; g\] \[\rho=\dfrac{m}{V}\] \[\rho=\dfrac{1.251\; g}{1\; L}\] \[\rho=1.251\; g/L\] 3.16 A chemist notices that one of many organic solvents is leaking from the hood. The rate of effusion of this organic gas is 2.866 mL/sec while the effusion of oxygen gas is 3.640 mL/sec. What is the molar mass of this organic solvent? Answer \(MM= 58.1\; g/mol\) Hint Consider Graham’s Law. http://chemwiki.ucdavis.edu/Textbook_Maps/Map%3A_Zumdahl_9e/05%3A_Gases/5.07_Effusion_and_Diffusion Solution Graham's Law states that the rate of effusion of two different gases at the same conditions are inversely proportional to the square roots of their molar masses as given by the following equation: \[ \dfrac{\text{Rate of effusion of A}}{\text{Rate of effusion of B}} =\sqrt{\dfrac{MM_B}{MM_A}}\] Assuming A is oxygen gas and B is the organic solvent \[\dfrac{3.640\; mL/sec}{2.866\; mL/sec}=\sqrt{\dfrac{MM_B}{36.0\;g/mol}}\] \[MM_B= 58.1\; g/mol\]
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Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Electronic_Spectroscopy/Spin-orbit_Coupling
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Spin-orbit coupling refers to the interaction of a particle's "spin" motion with its "orbital" motion. The Spin-orbit coupling Hamiltonian The magnitude of spin-orbit coupling splitting is measured spectroscopically as \[\begin{align*} H_{so} &=\dfrac{1}{2} hcA \left( (l+s)(l+s+1)-l(l+1)-s(s+1)\right) \\[4pt] &= \dfrac{1}{2} hcA \left(l^2 +s^2 + ls + sl + l +s -l^2 -l -s^2 -s) \right) \\[4pt] &= hcA \textbf{l} \cdot \textbf{s} \end{align*}\] The expression can be modified by realizing that \(j = l + s\). \[H_{so}=\dfrac{1}{2} hcA \biggr(j(j+1)-l(l+1)-s(s+1)\biggr)\] where \(A\) is the magnitude of the spin-orbit coupling in wave numbers. The magnitude of the spin orbit coupling can be calculated in terms of molecule parameters by the substitution \[hcA\,\widehat{L}\cdot\widehat{S}=\dfrac{Z\alpha^2}{2}\dfrac{1}{r^3}\widehat{L}\cdot\widehat{S}\] where \(a\) is the fine structure constant (\(a = 1/137.037\)) and the carrots indicate that \(L\) and \(S\) are operators. The fine structure constant is a dimensionless constant, \(a = \dfrac{e^2}{ác}\). \(Z\) is an effective atomic number. The spin orbit coupling splitting can be calculated from \[E_{so}=\int \Psi^*H_{SO}\Psi\,d\tau=\dfrac{Z}{2(137)^2}\int\Psi^*\dfrac{\widehat{L}\cdot\widehat{S}}{r^3}\Psi\,d\tau\] This expression can be recast to give an spin-orbit coupling energy in terms of molecular parameters \[E_{so}=\dfrac{1}{2} \biggr(j(j+1)-l(l+1)-s(s+1)\biggr)=\dfrac{Z}{2(137)^2}\biggr\langle\dfrac{1}{r^3}\biggr\rangle\] where \[\biggr\langle\dfrac{1}{r^3}\biggr\rangle=\int\Psi^*\biggr(\dfrac{1}{r^3}\biggr)\Psi\,d\tau\] We can evaluate this integral explicitly for a given atomic orbital. For example for Y210 we have \[\Psi_{210}=\dfrac{1}{4\sqrt{2\pi}}\biggr(\dfrac{Z}{a_0}\biggr)^\dfrac{3}{2}\dfrac{Zr}{a_0}e^{-Zr/2a_0}\cos\theta\] so that the integral is \[\biggr\langle\dfrac{1}{r^3}\biggr\rangle=\dfrac{1}{32\pi}\biggr (\dfrac{Z}{a_0}\biggr)^5\int_{0}^{2z}d\phi\int_{0}^{z}\cos^2\theta\sin\theta\,d\theta cos\theta\int_{0}^{\infty}r^2e^{Zr/a_0}\biggr(\dfrac{1}{r^3}\biggr)r^2\,dr\] which integrates to \[\biggr\langle\dfrac{1}{r^3}\biggr\rangle=\dfrac{1}{32\pi}\biggr(\dfrac{Z}{a_0}\biggr)^52\pi\biggr(\dfrac{2}{3}\biggr)\biggr(\dfrac{a_0\,^2}{Z^2}\biggr)=\dfrac{1}{24}\biggr(\dfrac{Z}{a_0}\biggr)^3\] Or \(Z^3/24\) in atomic units. Therefore in atomic units we have \[\biggr\langle\dfrac{1}{r^3}\biggr\rangle=\dfrac{Z^3}{n^3l(l+1/2)(l+1)}\] Therefore, in general the spin-orbit splitting is given by \[E_{so}=\dfrac{Z^4}{2(137)^2n^3}\Biggr(\dfrac{j(j+1)-l(l+1)-s(s+1)}{2l(l+1/2)(l+1)}\Biggr)\] Note that the spin-orbit coupling increases as the fourth power of the effective nuclear charge Z, but only as the third power of the principal quantum number n. This indicates that spin orbit-coupling interactions are significantly larger for atoms that are further down a particular column of the periodic table.
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Courses/Duke_University/CHEM_310L%3A_Physical_Chemistry_I_Laboratory/CHEM310L_-_Physical_Chemistry_I_Lab_Manual/01%3A_Orientation_to_this_course/1.01%3A_Pre-lab_orientation_assignment
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Please complete the following. Plan at least three or four hours in the week prior to classes to complete this assignment. Log into Duke's Canvas platform and familiarize yourself with this course's Canvas site. Read the entire orientation module in this manual. Read the Duke Chemistry Safety Manual . Find the syllabus, download it, and read the entire document carefully. Complete the "Orientation Assignment" that is found at the end of the syllabus. Complete the assignment and submit it on Gradescope prior to the first course meeting. Complete the Matlab Prelab assignment in the next module. It is summarized below: Download and install MatLab on your personal laptop computer. Duke students can get MatLab for free from Duke OIT . Complete the MatLab Basic Tutorials to become familiar with basic functions of the MatLab program. Things to bring to the first lab meeting: Please bring your working computer with MatLab installed with you to our first meeting.
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Bookshelves/Organic_Chemistry/Organic_Chemistry_(OpenStax)/06%3A_An_Overview_of_Organic_Reactions/6.11%3A_A_Comparison_between_Biological_Reactions_and_Laboratory_Reactions
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Beginning in the next chapter, we’ll be seeing a lot of reactions, some that are important in laboratory chemistry yet don’t occur in nature and others that have counterparts in biological pathways. In comparing laboratory reactions with biological reactions, several differences are apparent. For one, laboratory reactions are usually carried out in an organic solvent such as diethyl ether or dichloromethane to dissolve the reactants and bring them into contact, whereas biological reactions occur in the aqueous medium within cells. For another, laboratory reactions often take place over a wide range of temperatures without catalysts, while biological reactions take place at the temperature of the organism and are catalyzed by enzymes. We’ll look at enzymes in more detail in Section 26.11 , but you may already be aware that an enzyme is a large, globular, protein molecule that contains in its structure a protected pocket called its active site . The active site is lined by acidic or basic groups as needed for catalysis and has precisely the right shape to bind and hold a substrate molecule in the orientation necessary for reaction. Figure \(\PageIndex{1}\) shows a molecular model of hexokinase, along with an X-ray crystal structure of the glucose substrate and adenosine diphosphate (ADP) bound in the active site. Hexokinase is an enzyme that catalyzes the initial step of glucose metabolism—the transfer of a phosphate group from ATP to glucose, giving glucose 6-phosphate and ADP. The structures of ATP and ADP were shown at the end of Section 6.8. Note how the hexokinase-catalyzed phosphorylation reaction of glucose is written. It’s common when writing biological equations to show only the structures of the primary reactant and product, while abbreviating the structures of various biological “reagents” and by-products such as ATP and ADP. A curved arrow intersecting the straight reaction arrow indicates that ATP is also a reactant and ADP also a product. Yet a third difference between laboratory and biological reactions is that laboratory reactions are often done using relatively small, simple reagents such as Br 2 , HCl, NaBH 4 , CrO 3 , and so forth, while biological reactions usually involve relatively complex “reagents” called coenzymes . In the hexokinase-catalyzed phosphorylation of glucose just shown, ATP is the coenzyme. As another example, compare the H 2 molecule, a laboratory reagent that adds to a carbon–carbon double bond to yield an alkane, with the reduced nicotinamide adenine dinucleotide (NADH) molecule, a coenzyme that effects an analogous addition of hydrogen to a double bond in many biological pathways. Of all the atoms in the coenzyme, only the one hydrogen atom shown in red is transferred to the double-bond substrate. Don’t be intimidated by the size of the ATP or NADH molecule; most of the structure is there to provide an overall shape for binding to the enzyme and to provide appropriate solubility behavior. When looking at biological molecules, focus on the small part of the molecule where the chemical change takes place. One final difference between laboratory and biological reactions is in their specificity. A catalyst might be used in the laboratory to catalyze the reaction of thousands of different substances, but an enzyme, because it can only bind a specific substrate molecule having a specific shape, will usually catalyze only a specific reaction. It’s this exquisite specificity that makes biological chemistry so remarkable, and that makes life possible. Table \(\PageIndex{1}\) summarizes some of the differences between laboratory and biological reactions. Unnamed: 0 Laboratory reaction Biological reaction Solvent Organic liquid, such as ether Aqueous environment in cells Temperature Wide range; −80 to 150 °C Temperature of organism Catalyst Either none, or very simple Large, complex enzymes needed Reagent size Usually small and simple Relatively complex coenzymes Specificity Little specificity for substrate Very high specificity for substrate
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Ancillary_Materials/Worksheets/Worksheets%3A_Inorganic_Chemistry/Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry/13%3A_Concepts_of_Acidity/13.10%3A_Proton_Donation_and_Structure
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Structure plays a key role in determining how easily a compound can provide protons. For example, note that a number of common acids contain OH groups, such as sulfuric acid and acetic acid. What is it about this group that makes these compounds acidic? Figure AB10.1. Acetic and sulfuric acid. Probably it has something to do with the O-H bond being polar. However, water has an O-H bond, but it is not terribly acidic. A polar O-H bond may be one factor contributing to Bronsted acidity (the formation of H + ). However, it cannot be the only factor. In contrast, some compounds, such as NaOH or KOH, are not very acidic. In fact, they are termed Arhennius bases, which means they are ionic compounds that dissociate in water to give hydroxide ions (HO - ). Remember, HO - is a good Lewis base, because it has lone pairs to donate, and it also has a negative charge that makes it especially nucleophilic. Why do these compounds ionize to form HO - , whereas other compounds containing the OH group ionize to form H + ? The difference has to do with which bond to the electronegative element, oxygen, is most polar. The most polar bond is the one most likely to ionize. In alkali metal hydroxides, such as NaOH, the Na-O bond is most polar. The electronegativity difference between sodium and oxygen is larger than that between oxygen and hydrogen. In fact, sodium is on the very left hand side of the periodic table, whereas oxygen is in the upper right hand corner. A combination like that results in an ionic bond, not a covalent one, so sodium hydroxide should be thought of as Na + and HO - . Figure AB10.2. Bond polarity in sulfuric acid. In main group hydroxyls such as nitric acid or sulfuric acid, the most polar bond is the O-H bond. N-O bonds or S-O bonds are not nearly as polar as O-H bonds. The O-H bond ionizes more easily. However, relative bond polarity is not the only factor here. Acidity is strongly influenced by the structure of the ions produced in each case. More stable ions are produced more easily. If HO - dissociates to form a proton, a O 2- or oxide anion will result. That oxygen has a high nuclear charge and a high electron affinity, but a 2- charge is a lot of negative charge on one small atom. This ion will not be very stable. On the other hand, nitric acid dissociates to form a proton and nitrate ion, NO 3- . This ion has a single negative charge on an element that has a high electron affinity, oxygen. In addition, there is an electronegative atom, nitrogen, attached to that oxygen, and another electronegative element, oxygen, attached to that one. Electronegativity is the ability to draw electronic charge through bonds. That means these other electronegative atoms will draw some negative charge away from the anionic atom. In this way, the negative charge is dispersed and stabilized. Dispersing or spreading charge out helps to stabilize charge. Concentrating more charge in one location is destabilizing. Nearby electronegative atoms can help stabilize negative charge. There is another, related factor that stabilizes the nitrate anion by spreading out the negative charge. Notice that the negative charge could be drawn on two of the three oxygen atoms at once. The negative charge does not have to be on any particular oxygen. It can really be shared by all three. In Lewis terms, we show that fact by drawing all three resonance structures for the nitrate ion. Figure AB10.3. Charge stability in nitrate anion. Resonance or delocalization spreads out charge, stabilizing it. In the subsequent sections, we can look at these and other factors in more detail. Problem AB10.1. Which of the following acids do you think has the lowest pKa? a) HClO 2 or HCl0 4 b) H 3 PO 3 or H 3 PO 4
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Courses/Intercollegiate_Courses/Cheminformatics/02%3A_Representing_Small_Molecules_on_Computers/2.04%3A_Line_Notation
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Learning Objectives: Explain what SMILES, SMARTS and SMIRKS are. Explain what InChI and InChIKey are. Review SMILES specification rules. Compare and contrast SMILES and InChI. Demonstrate how to interpret SMILES, SMARTS, InChI strings into their corresponding chemical structures. Introduction Line notations represent structures as a linear string of characters. They are widely used in Cheminformatics because computers can easily process linear strings of data. Examples of line notations include the Wiswesser Line-Formula Notation (WLN) 1 , Sybyl Line Notation (SLN) 2 , 3 and Representation of structure diagram arranged linearly (ROSDAL) 4 , 5 . Currently, the most widely used linear notations are the Simplified Molecular-Input Line-Entry System (SMILES) 6-9 and the IUPAC Chemical Identifier (InChI) 10-13 , which are described below. In this class we will focus on SMILES and InChI line notation. SMILES The Simplified Molecular-Input Line-Entry System (SMILES) 6-9 is a line notation for describing chemical structures using short ASCII strings. SMILES is like a connection table in that it identifies the nodes and edges of a molecular graph. SMILES was developed in the late 1980s and implemented by Daylight Chemical Information Systems (Santa Fe, NM), but it is still widely used today. A detailed information on SMILES can be found in Chapter 3 14 of the Daylight Theory Manual as well as the SMILES tutorial 15 . SMILES Specification Rules In SMILES, hydrogen are typically implicitly implied and atoms are represented by their atomic symbol enclosed in brackets unless they are elements of the “organic subset” (B, C, N, O, P, S, F, Cl, Br, and I), which do not require brackets unless they are charged. So gold would be [Au] but chlorine would be Cl. If hydrogens are explicitly implied brackets are used. A formal charge is represented by one of the symbols + or -. Single, double, triple, and aromatic bonds are represented by the symbols, -, =, #, and :, respectively. Single and aromatic bonds may be, and usually are, omitted. Here are some examples of SMILES strings. 0 1 2 3 4 Function Symbol NaN Function Symbol single bond* - NaN Positive charge [C+] double bond = NaN Negative charge [C-] triple bond # NaN aromatic carbon c (lower case c) aromatic bond* : NaN NaN NaN Table 2.4.2 shows some common SMILES strings. Note the following conventions Branches are specified by enclosures in parentheses and can be nested or stacked, as shown in these examples. Rings are represented by breaking one single or aromatic bond in each ring, and designating this ring-closure point with a digit immediately following the atoms connected through the broken bond. Atoms in aromatic rings are specified by lower cases letters. Aromatic Rings use lower case c Although the carbon-carbon bonds in these two SMILES are omitted, it is possible to deduce that the omitted bonds are single bonds (for cyclohexane) and aromatic bonds (for benzene). One can also represent an aromatic compound as a non-aromatic, KeKulé structure. For example, the following is a valid SMILES string for benzene. C1=CC=CC=C1 Benzene (C6H6) 0 1 2 3 4 5 6 7 SMILES Name (formula) NaN SMILES Name (formula) NaN SMILES Name(formula) C Methane (CH4) NaN COC Dimethyl ether (CH3OCH3) NaN CC(C)CO Isobutyl alcohol (CH3-CH(CH3)-CH2-OH) CC Ethane (CH3CH3) NaN CCO Ethanol (CH3CH2OH) NaN CC(CCC(=O)N)CN 5-amino-4-methylpentanamide C=C Ethene (CH2CH2) NaN CC=O Acetaldehyde (CH3-CH=O) NaN C1CCCCC1 Cyclohexane (C6H12) C#C Ethyne (CHCH) NaN CC(=O)[O-] Acetate NaN c1ccccc1 Benzene (C6H6) (aromatic representation) C#N Hydrogen Cyanide (HCN) NaN [C-]#N Cyanide anion NaN C1=CC=CC=C1 Benzene (C6H6) (KeKulé representation) Note that aromaticity is not a measurable physical quantit y, but a concept without a unanimous mathematical definition. As a result, different aromaticity detection algorithms often disagree with each other on whether a given molecule is aromatic or not, making it difficult to interchange information between databases that use different aromaticity detection algorithms for SMILES generation. Also note that a ring structure can have multiple potential ring-closure points. For example, a six-membered ring has six bonds, each of which can be a ring-closure point. As a result, a ring compound may be represented by many different but equally valid SMILES strings. Actually, it is very common that there are a lot of SMILES strings that represent the same structure, whether it has a ring or not, because one can start with any atom in a molecule to derive a SMILES string. Therefore, it is necessary to select a “unique SMILES” for a molecule among many possibilities. Because this is done through a process called “canonicalization”, this unique SMILES string is also called the “canonical SMILES”. Isomeric SMILES Isomeric SMILES allow for the specification of the isotopism and stereochemistry of a molecule. Information on isotopism is indicated by the integral atomic mass preceding the atomic symbol. The atomic mass must be specified inside square brackets. For example, C-13 methane can be represented by “[13CH4]”. Configuration around double bonds is specified by “directional bonds” (characters / and \). For example, E- and Z-1,2-difluoroethene can be represented by the following isomeric SMILES: F/C=C/F or F\C=C\F (E)-1,2-difluoroethene (trans isomer) F/C=C\F or F\C=C/F (Z)-1,2-difluoroethene (cis isomer) Configuration around tetrahedral centers are indicated by the symbols “@” or “@@” C[C@@H](C(=O)O)N L-Alanine C[C@H](C(=O)O)N D-Alanine More detailed information on chirality specification can be found in Chapter 3 14 of the Daylight Theory Manual. Limitations of SMILES SMILES is proprietary and it is not an open project. This has led different chemical software developers to use different SMILES generation algorithms, resulting in different SMILES versions for the same compound. Therefore, SMILES strings obtained from different databases or research groups are not interchangeable unless they used the same software to generate the SMILES strings. With an aim to address this interchangeability issue of SMILES, an open-source project has launched to develop an open, standard version of the SMILES language called OpenSMILES . However, the most noticeable community effort in this area is development of InChI, which is described in next section. SMARTS SM iles AR britrary T arget S pecification (SMARTS) notation allows one to search in certain databases (like PubChem) for generic structures. It is a language used for describing molecular patterns. SMARTS is useful for substructure searching, which finds a particular pattern (subgraph) in a molecule. SMARTS are straightforward extensions of SMILES. All SMILES symbols and properties are legal in SMARTS. SMARTS includes logical operators and additional molecular descriptors. Detailed information on SMARTS is given in the SMARTS specification document in the Daylight theory manual and SMARTS tutorial . SMIRKS Another extension of SMILES is SMIRKS, which is a line notation for generic reactions. A generic reaction represents a group of reactions that undergo the same set of atom and bond changes. Note that SMILES and SMARTS can be used to represent reactions, using the “>” symbol between the reactants, products, and agents, as described in the SMILES and SMARTS specification documents. (Therefore, these SMILES and SMARTS that describe reactions are often called reaction SMILES and reaction SMARTS, respectively.) On the other hand, SMIRKS is used to represent types of reactions (e.g., S N 2 reaction). More detailed information on SMIRKS is given in the SMIRKS specification document and SMIRKS tutorial . InChI Since 1919 the International Union of Pure and Applied Chemistry ( IUPAC ) has been the international authority on chemical nomenclature and terminology. IUPAC currently consists of members from 57 national adhering organizations ( NAOs ) whose recommendations are made public through the IUPAC journal Pure and Applied Chemistry and the IUPAC Color books . As we entered the new millennial the leadership of IUPAC recognized the need to extend chemical nomenclature into the digital realm of computer databases and software agents, and in March of 2000 during a meeting at the U.S. Naval Academy started a project with the U.S. National Institute of Standards and Technology ( NIST ), to build a machine readable nomenclature standard, the InChI. It was originally called INChI for IUPAC/NIST Chemical Identifier, but was changed to InChI (International Chemical Identifier) as although it was built with efforts from NIST, it was not appropriate for NIST as a government agency to place its name as a recommendation for the identifier. In 2010 the InChI Trust was formed and development of the standard is continuing the purview of the InChI Trust and the IUPAC InChI subcommittee . InChI is an open, freely available non-proprietary computer generated chemical identifier that is based on a hierarchical layered line notation (see below). The first three layers essentially deal with the information within the simplified connection table, and the additional layers are added as needed, and deal with complexities like isomers, isotopic distributions and the other types of issues brought up in section 2.3 of this chapter, and these layers are extensible. A standard InChI has a predefined number of layers, and these can be extended to non-standard InChI's that can have new layers relating to define additional information, that is what is meant by extensible layes. Unlike SMILES, InChI is a canonical line notation and so is a unique identifier that is built upon a set of nomenclature rules. That is, although there are canonical SMILES built through a canonicalization algorithm, there can be more than one canonicalization algorithm for SMILES, and so you can have more than one SMILES string for the same structure. Students may be familiar with the American Chemical Society's Chemical Abstract Service (CAS) registry number, which is supposed to be a unique identifier based on the registry system, but issues can arise (see below, other identifiers, and problem). Also, a CAS registry number is associated with a compound that has been published in the primary literature or patents, and the CAS system bases its identifiers on the registry system, not the structure of the molecule. That is, InChI is not a registry system, it is a type of nomenclature that describes the structure of a molecule, and you can make an InChI for a molecule that does not exist, as long as you specify its structure. The most recent version of InChI (and its documentation) can be obtained at the InChI Trust Download site . InChI: A Layered Notation The power of a layered notation is that it gets to the essence of what is a molecule? For example, we think water is H 2 O, but if you look at a real sample of water you will note that some of the hydrogens are protium (one proton) and others are deuterium ( a proton and a neutron), and in fact the ratio of deuterium to protium in ground water samples can vary from one region of the US to another, and thus the molar mass of samples of water can vary . In fact IUPAC has now adopted an " interval atomic weight notation system " for some elements whose atomic mass varies across samples, and this can affect physical properties of a sample, like the vapor pressure of water. So water is water, but not all water is the same, and the question becomes, do you care? If you are uploading data to a database and know the isotopic distribution you care, but if you do not know it, you do not care, but in both cases, your data deals with water. Through a layered notation system you can have an isotopic layer to describe your water if you care, but you don't need it if you don't care. This leads to one of the issues that comes up with the layered notation, in that you can have different InChI's for a compound, depending on the kind of information you want in the name, that is how many and what kind of layers you use. This leads to the standard InChI, which is an InChI that has defined layers and is thus canonical, and starts with a number followed by the letter "s" to indicate the version of the standard. Standard InChI Standard InChI version 1.05 was released in January 2017 and has 6 core layers (and several sublayers within the core layers) and starts with InChI=1S/... . Each layer in the InChI string is separated by a "/" and the "main layer" is essentially the connection table. The InChI software generates both standard and nonstandard InChI, with the standard InChI having "fixed options" that ensures interoperability between databases and software agents. The standard InChI (version 1.05) has the following layers: Main Layer Chemical Formula Layer (based on Hill Notation ) Connections- bonds between atoms and may have sublayers, with the last one dealing with mobile hydrogens. Charge Layer Component Charge Protons Stereochemical Layer Double Bond sp 2 (Z/E) Sterochemistry Tetrahedral Sterochemistry Isotopic Layer Fixed Hydrogen Layer (binds mobile hydrogens) Polymer Layer (actually a new experimental layer) and does not affect the content of the earlier layers. Figure \(\PageIndex{1}\): The main layers for a standard InChI of [(R)-carboxy(chloro)methyl]azanium, the protonated form of 2-( 35 Cl)chloro-R-glycine . Note each layer or sublayer is separated by a forward slash [/]. NonStandard InChI Note a nonstandard InChI does not start with InChI=1S/... but with a InChI=1/.... and has additional layers that approach different facets of a molecule's structure or features. In fact a company could create their own lawyer for a nonstandard information and encode into it proprietary information that they wished to keep private. The nonstandard InChI may not be canonical, but can handle facets of information information that a standard can not, in fact for a standard to be canonical different tautomers must have the same InChI, or you have two InChIs for the same molecule So defining specific tautomers is one use of a nonstandard InChI as can be seen in the case of 4,5-Dihydro-1,3-Oxazol-3-lum . Figure 2 shows the two tautomeric forms of this molecule which must have the same standard InChI or it would not be canonical (you would have two InChIs for the same molecules). If you want to define just one of the tautomers, you need to use a nonstandard InChI and add a fixed hydrogen layer (in red). Although these are two ways of drawing the same molecule, one form may be favored over the other in certain environments and so there may be data indicative of the behavior of one of these form and not the other, and thus there may be a need to distinguish between the tautomers. 0 1 InChI=1S/C3H5NO/c1-2-5-3-4-1/h3H,1-2H2/p+1 InChI=1S/C3H5NO/c1-2-5-3-4-1/h3H,1-2H2/p+1 NaN NaN InChI=1/C3H5NO/c1-2-5-3-4-1/h3H,1-2H2/p+1/fC3H6NO/h4H/q+1 InChI=1/C3H5NO/c1-2-5-3-4-1/h3H,1-2H2/p+1/fC3H6NO/h5H/q+1 Figure \(\PageIndex{2}\): The above two structures are tautomeric drawings of the same molecule and thus have the same standard InChI. If you were interested in just one of the structures you could use a nonstandard InChI with a fixed hydrogen layer (in red). Borrowed from section 6.2 of InChI Trust FAQ . Drawbacks of InChI InChIs are not meant to be human readable but to contain molecular information that computers can read within the layers, so unlike SMILES you can't really read even a simple InChI (see Figure 3), never mind a complex one (figure 4). SMILES InChI CC(=O)C InChI=1S/C3H6O/c1-3(2)4/h1-2H3 Acetone Acetone Figure \(\PageIndex{3}\): Canonical SMILES and InChI for Acetone (source: PubChem ) Another drawback of InChI is just like an IUPAC systematic name, they are of variable length and become real long (figure 4). The problem with the variable length is it makes InChI impractical as a database registry number, and the length is often too long for internet search engines to handle. IUPAC Systematic Name InChI magnesium;methyl (3R,21S,22S)-16-ethenyl-11-ethyl-12,17,21,26-tetramethyl-4-oxo-22-[3-oxo-3-[(E,7R,11R)-3,7,11,15-tetramethylhexadec-2-enoxy]propyl]-23,25-diaza-7,24-diazanidahexacyclo[18.2.1.15,8.110,13.115,18.02,6]hexacosa-1,5,8(26),9,11,13(25),14,16,18,20(23)-decaene-3-carboxylate InChI=1S/C55H73N4O5.Mg/c1-13-39-35(8)42-28-44-37(10)41(24-25-48(60)64-27-26-34(7)23-17-22-33(6)21-16-20-32(5)19-15-18-31(3)4)52(58-44)50-51(55(62)63-12)54(61)49-38(11)45(59-53(49)50)30-47-40(14-2)36(9)43(57-47)29-46(39)56-42;/h13,26,28-33,37,41,51H,1,14-25,27H2,2-12H3,(H-,56,57,58,59,61);/q-1;+2/p-1/b34-26+;/t32-,33-,37+,41+,51-;/m1./s1 Chlorophyll A Chlorophyll A Figure \(\PageIndex{4}\): On the top left is the IUPAC systematic name for chlorophyll A and on the right is it's InChI (source: PubChem ). There is an additional issue with the InChI in that some of the characters interfere with web search queries and thus the InChI itself is not appropriate for web searches. To solve these problems a hashed InChI Key has been developed which is of constant length and enables web searches. The hashed key is also of constant length, making it better suited for databases. InChI Keys The InChI suite will generate a hashed version of the InChI, the InChI Key. The hash function generates a standard key of 27 characters that stores information in four parts (see figure 5). The InChIKey may be a standard or nonstandard key as indicated by the version, but all keys are of the same length and format. Figure \(\PageIndex{5}\): InChI key for 2-( 35 Cl)chloro-R-glycine (molecule in figure 1). The hash function is a one-way conversion (figure 6), that is, if you have an InChI you can generate the key, but if you have the key you can not generate the InChI. The key can function as an identifier is you made it registry number where you would need a look up table to know the molecule it is associated with. Figure \(\PageIndex{6}\): The one way InChIKey generation function. If two different chemical compound databases have the same chemical (InChI)they will generate the same standard InChIKeys and thus it is customary for databases and other information sources like Wikipedia chemboxes to generate standard InChIKeys, and they effectively function as a standard "registry number", that is, if two chemicals in different databases have the same standard InChIKey, they are the same chemical. On the other hand if you had a non-standard InChI the non-standard layers would induce variability of the key and so you could not compare across databases. InChIKeys and Web Searching The molecule (R)-2-( 35 Cl)chloroglycine probably does not exist and was created to demonstrate the layers of an InChI and the correlating key. If you do a web search of the entire key (UWPWWENWLZPQGU-WRFRXMDISA-0) you do not get any hits, but if you search just the main layer you get several hits. What you are doing is essentially looking for any molecule with the same simplified connection table, that is, all stereoisomers, or isotopic labels. One of the hits is for [ ( S )-carboxy(chloro)methyl]azanium which is the other isomer. If you go to properties they are all computed and none were deposited to PubChem by vendors or contributors, and so this molecule has probably never been synthesized. Under 5.2 Related Compounds/Exact Same Parent you also get the non-protonated form ( 2-chloro-L-glycine ) , of which there is published information. It is also of interest that the search of first part of the InChIKey also turned up an article in Russian on the L isomer, (you may need to download the pdf to see the actual bonds). InChI OER The InChI Trust runs an Open Education Resource (OER) where you can find material on InChI https://www.inchi-trust.org/oer/ . The InChI OER is a repository where anyone can upload and tag material on InChI, or link to and tag existing material on the use of InChI. Once material is posted within the OER it can be searched through a filter system. Figure \(\PageIndex{7}\): InChI OER tag filter and associated content. The default setting is to show all OER site material, clicking non-OER will extend the filter to include off site material like publications which have records that have been submitted to the OER. InChI Layers Explorer In this activity we will use the InChI OER to obtain an Excel spreadsheet that breaks an InChI into layers, and start to analyze how cheminformatics functionality can be integrated into common tools like spreadsheets. Go to the InChI OER and in the filter click "Spreadsheet" (middle of figure \(\PageIndex{8}\)). This filters the content to items that are tagged "spreadsheet" and also removes any tag that is not associated with one of those content items. Now move down to tag category "File Type" and while holding the <ctrl> key, click Excel (right figure \(\PageIndex{8}\)). You now get a list of excel spreadsheets (figure \(\PageIndex{9}\)). Figure \(\PageIndex{8}\): InChI OER Tag Filter. On the left is the default setting and all content loaded to the site is displayed in the window (right side of figure \(\PageIndex{7}\). In the middle the filter for spreadsheets is activated, and you can see there are two types that have been uploaded, Google Sheets and Excel sheets. On the right both Spreadsheet and Excel have been activated, and so only spreadsheets in Excel are displayed and the content view is reduced to those items that are tagged both "Spreadsheet" and "Excel" (Figure \(\PageIndex{9}\)) Figure \(\PageIndex{9}\): At the time this page was created there were three items uploaded to the InChI OER that were tagged as Excel Spreadsheets. Click on the InChILayersExplorer and you go to it's content page. This page will have a description of the content and a green information box (Figure \(\PageIndex{10}\)), and in the information box is a "Download Publication Files", that allows you to obtain the spreadsheet. Figure \(\PageIndex{10}\): Green Information box for the InChILayersExplorer Now click on the link in the "Download Publications File" field and you will have a copy of the InChI Layers Explorer, which you should open and enable editing. Activity \(\PageIndex{1}\) Using the InChILayersExplorer show the difference between the InChI for (R)-thalidomide and (S)-thalidomide. Note, the goal of this activity is not to answer the question, but to gain an understanding on how the InChILayersExplorer works, which is in effect a "smart spreadsheet" that communicates with database web APIs via webservices functions. One of the skills we hope you can gain from this class is enough familiarity with how code works so if you see new code, you can hack in and figure how it works. Be sure to enable the spreadsheet after you download it. This spreadsheet communicates with the NCI Chemical Resolver (section 2.7. ) 1. Type (R)-thalidomide in the yellow region (type over CoA), OK, it fails, now try the (S) isomer, and it still fails, so now try thalidomide without specifying an isomer. OK, so you have the InChI for thalidomide, but there is nothing in the stereochemical layer, as you have not specified the stereochemistry. These spreadsheet uses the Chemical Identifier Resolver of the NIH which will be covered in section 2.6.2.1.1 ), which can be accessed directly at https://cactus.nci.nih.gov/chemical/structure and is shwon in figure \(\PageIndex{11}\). Now lets start by searching for (R)-thalidomide directly in the resolver (figure \(\PageIndex{11}\)). Figure \(\PageIndex{11}\): NCI/CADD Chemical Resolver set up to find standard InChI for (R)-thalidomid As you may have guessed, neither (R) or (S) works, but "thalidomide" does (incidentally, you have to hit submit, not Structure), and so this resolver will not provide information on the isomers of thalidomide. So now do a web search of (R)-thalidomide, and paste in its key (UEJJHQNACJXSKW-SECBINFHSA-N), and note the stereochemical layer [/t9-/m1/s1] is the only part that is different. Now repeating for (S)-thalidomide. You should get the following results Table \(\PageIndex{3}\) 0 1 2 Compound InChI Key Stereochemical layer thalidomide UEJJHQNACJXSKW-UHFFFAOYSA-N none (R)-thalidomide UEJJHQNACJXSKW-SECBINFHSA-N /t9-/m1/s1 (S)-thalidomide UEJJHQNACJXSKW-VIFPVBQESA-N /t9-/m0/s1 Note, if you click on the merged cells that generates the InChI (Rows 7-8) you see the following code. Figure \(\PageIndex{11}\): Code in spreadsheet that uses WEBSERVICE function to get InChI from NCI/CADD chemical resolver Now open up a browser tab and paste in the following URL: https://cactus.nci.nih.gov/chemical/structure/thalidomide/stdinchi Now go back to the NCI Chemical Resolver and click the dropdown box of the "convert to" field (figure\(\PageIndex{12}\) and try another option, say "TwirlyMol(3D). \(\PageIndex{12}\): Dropdown menu of NCI Chemical resolver showing some of the options. Can you figure out the URL that uses the NCI Chemical Resolver to give the 3D molecule in a webpage? Once you have done this, can you identify a problem that has resulted from these molecular representations. Hint, think of adding two more columns to table \(\PageIndex{3}\), one for 2D and one for 3D images. What is the issue when you draw the 3D image that does not arise when you draw the 2D? References and Further Reading (1) Wiswesser, W. J. J. Chem. Inf. Comput. Sci. 1982 , 22 , 88. (2) Ash, S.; Cline, M. A.; Homer, R. W.; Hurst, T.; Smith, G. B. J. Chem. Inf. Comput. Sci. 1997 , 37 , 71. (3) Homer, R. W.; Swanson, J.; Jilek, R. J.; Hurst, T.; Clark, R. D. J. Chem Inf. Model. 2008 , 48 , 2294. (4) Barnard, J. M.; Jochum, C. J.; Welford, S. M. Acs Symposium Series 1989 , 400 , 76. (5) Rohbeck, H. G. In Software Development in Chemistry 5 ; Gmehling, J., Ed.; Springer Berlin Heidelberg: 1991, p 49. (6) Weininger, D. J. Chem. Inf. Comput. Sci. 1988 , 28 , 31. (7) Weininger, D.; Weininger, A.; Weininger, J. L. J. Chem. Inf. Comput. Sci. 1989 , 29 , 97. (8) Weininger, D. J. Chem. Inf. Comput. Sci. 1990 , 30 , 237. (9) SMILES: Simplified Molecular Input Line Entry System ( http://www.daylight.com/smiles/ ) (Accessed on 6/30/2015). (10) Heller, S.; McNaught, A.; Stein, S.; Tchekhovskoi, D.; Pletnev, I. J. Cheminform. 2013 , 5 , 7. (11) Heller, S.; McNaught, A.; Pletnev, I.; Stein, S.; Tchekhovskoi, D. J. Cheminform. 2015 , 7 , 23. (12) The IUPAC International Chemical Identifier (InChI) ( http://www.iupac.org/home/publications/e-resources/inchi.html ) (Accessed on 6/29/2015). (13) InChI Trust ( http://www.inchi-trust.org/ ) (Accessed on 6/29/2015). (14) Daylight Theory Manual, Chapter 3: SMILES - A Simplified Chemical Language ( http://www.daylight.com/dayhtml/doc/theory/theory.smiles.html ) (Accessed on 6/23/2015). (15) Daylight SMILES Tutorial ( http://www.daylight.com/dayhtml_tutorials/languages/smiles/index.html ) (Accessed on 6/23/2015). Contributors Robert E. Belford (University of Arkansas Little Rock; Department of Chemistry). The breadth, depth and veracity of this work is the responsibility of Robert E. Belford, [email protected] . You should contact him if you have any concerns. This material has both original contributions, and content built upon prior contributions of the LibreTexts Community and other resources, including but not limited to: Sunghwan Kim Material Adapted from 2017 Cheminformatics OLCC
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Courses/Los_Angeles_Trade_Technical_College/Chem_51/19%3A_Nuclear_Chemistry/19.10%3A_Biological_Effects_of_Radiation
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Skills to Develop Describe the biological impact of ionizing radiation Define units for measuring radiation exposure Explain the operation of common tools for detecting radioactivity List common sources of radiation exposure in the US The increased use of radioisotopes has led to increased concerns over the effects of these materials on biological systems (such as humans). All radioactive nuclides emit high-energy particles or electromagnetic waves. When this radiation encounters living cells, it can cause heating, break chemical bonds, or ionize molecules. The most serious biological damage results when these radioactive emissions fragment or ionize molecules. For example, alpha and beta particles emitted from nuclear decay reactions possess much higher energies than ordinary chemical bond energies. When these particles strike and penetrate matter, they produce ions and molecular fragments that are extremely reactive. The damage this does to biomolecules in living organisms can cause serious malfunctions in normal cell processes, taxing the organism’s repair mechanisms and possibly causing illness or even death (Figure \(\PageIndex{1}\)). Figure \(\PageIndex{1}\) : Radiation can harm biological systems by damaging the DNA of cells. If this damage is not properly repaired, the cells may divide in an uncontrolled manner and cause cancer. Ionizing vs. Nonionizing Radiation There is a large difference in the magnitude of the biological effects of nonionizing radiation (for example, light and microwaves) and ionizing radiation , emissions energetic enough to knock electrons out of molecules (for example, α and β particles, γ rays, X-rays, and high-energy ultraviolet radiation) (Figure \(\PageIndex{2}\)). Figure \(\PageIndex{2}\) : Lower frequency, lower-energy electromagnetic radiation is nonionizing, and higher frequency, higher-energy electromagnetic radiation is ionizing. Energy absorbed from nonionizing radiation speeds up the movement of atoms and molecules, which is equivalent to heating the sample. Although biological systems are sensitive to heat (as we might know from touching a hot stove or spending a day at the beach in the sun), a large amount of nonionizing radiation is necessary before dangerous levels are reached. Ionizing radiation, however, may cause much more severe damage by breaking bonds or removing electrons in biological molecules, disrupting their structure and function. The damage can also be done indirectly, by first ionizing H 2 O (the most abundant molecule in living organisms), which forms a H 2 O + ion that reacts with water, forming a hydronium ion and a hydroxyl radical: Because the hydroxyl radical has an unpaired electron, it is highly reactive. (This is true of any substance with unpaired electrons, known as a free radical.) This hydroxyl radical can react with all kinds of biological molecules (DNA, proteins, enzymes, and so on), causing damage to the molecules and disrupting physiological processes. Examples of direct and indirect damage are shown in Figure \(\PageIndex{3}\). Figure \(\PageIndex{3}\) : Ionizing radiation can (a) directly damage a biomolecule by ionizing it or breaking its bonds, or (b) create an H 2 O + ion, which reacts with H 2 O to form a hydroxyl radical, which in turn reacts with the biomolecule, causing damage indirectly. Biological Effects of Exposure to Radiation Radiation can harm either the whole body (somatic damage) or eggs and sperm (genetic damage). Its effects are more pronounced in cells that reproduce rapidly, such as the stomach lining, hair follicles, bone marrow, and embryos. This is why patients undergoing radiation therapy often feel nauseous or sick to their stomach, lose hair, have bone aches, and so on, and why particular care must be taken when undergoing radiation therapy during pregnancy. Different types of radiation have differing abilities to pass through material (Figure \(\PageIndex{4}\)). A very thin barrier, such as a sheet or two of paper, or the top layer of skin cells, usually stops alpha particles. Because of this, alpha particle sources are usually not dangerous if outside the body, but are quite hazardous if ingested or inhaled (see the Chemistry in Everyday Life feature on Radon Exposure). Beta particles will pass through a hand, or a thin layer of material like paper or wood, but are stopped by a thin layer of metal. Gamma radiation is very penetrating and can pass through a thick layer of most materials. Some high-energy gamma radiation is able to pass through a few feet of concrete. Certain dense, high atomic number elements (such as lead) can effectively attenuate gamma radiation with thinner material and are used for shielding. The ability of various kinds of emissions to cause ionization varies greatly, and some particles have almost no tendency to produce ionization. Alpha particles have about twice the ionizing power of fast-moving neutrons, about 10 times that of β particles, and about 20 times that of γ rays and X-rays. Figure \(\PageIndex{4}\): The ability of different types of radiation to pass through material is shown. From least to most penetrating, they are alpha < beta < neutron < gamma. For many people, one of the largest sources of exposure to radiation is from radon gas (Rn-222). Radon-222 is an α emitter with a half–life of 3.82 days. It is one of the products of the radioactive decay series of U-238, which is found in trace amounts in soil and rocks. The radon gas that is produced slowly escapes from the ground and gradually seeps into homes and other structures above. Since it is about eight times more dense than air, radon gas accumulates in basements and lower floors, and slowly diffuses throughout buildings (Figure \(\PageIndex{5}\)). < Figure \(\PageIndex{5}\): Radon-222 seeps into houses and other buildings from rocks that contain uranium-238, a radon emitter. The radon enters through cracks in concrete foundations and basement floors, stone or porous cinderblock foundations, and openings for water and gas pipes. Radon is found in buildings across the country, with amounts depending on where you live. The average concentration of radon inside houses in the US (1.25 pCi/L) is about three times the levels found in outside air, and about one in six houses have radon levels high enough that remediation efforts to reduce the radon concentration are recommended. Exposure to radon increases one’s risk of getting cancer (especially lung cancer), and high radon levels can be as bad for health as smoking a carton of cigarettes a day. Radon is the number one cause of lung cancer in nonsmokers and the second leading cause of lung cancer overall. Radon exposure is believed to cause over 20,000 deaths in the US per year. Effects of Long-term Radiation Exposure on the Human Body The effects of radiation depend on the type, energy, and location of the radiation source, and the length of exposure. As shown in Figure \(\PageIndex{8}\), the average person is exposed to background radiation, including cosmic rays from the sun and radon from uranium in the ground (see the Chemistry in Everyday Life feature on Radon Exposure); radiation from medical exposure, including CAT scans, radioisotope tests, X-rays, and so on; and small amounts of radiation from other human activities, such as airplane flights (which are bombarded by increased numbers of cosmic rays in the upper atmosphere), radioactivity from consumer products, and a variety of radionuclides that enter our bodies when we breathe (for example, carbon-14) or through the food chain (for example, potassium-40, strontium-90, and iodine-131). Figure \(\PageIndex{8}\) : The total annual radiation exposure for a person in the US is about 620 mrem. The various sources and their relative amounts are shown in this bar graph. (source: U.S. Nuclear Regulatory Commission). A short-term, sudden dose of a large amount of radiation can cause a wide range of health effects, from changes in blood chemistry to death. Short-term exposure to tens of rems of radiation will likely cause very noticeable symptoms or illness; a dose of about 500 rems is estimated to have a 50% probability of causing the death of the victim within 30 days of exposure. Exposure to radioactive emissions has a cumulative effect on the body during a person’s lifetime, which is another reason why it is important to avoid any unnecessary exposure to radiation. Health effects of short-term exposure to radiation are shown in Table \(\PageIndex{2}\). Exposure (rem) Health Effect Time to Onset (without treatment) 5–10 changes in blood chemistry — 50 nausea hours 55 fatigue — 70 vomiting — 75 hair loss 2–3 weeks 90 diarrhea — 100 hemorrhage — 400 possible death within 2 months 1000 destruction of intestinal lining — NaN internal bleeding — NaN death 1–2 weeks 2000 damage to central nervous system — NaN loss of consciousness; minutes NaN death hours to days It is impossible to avoid some exposure to ionizing radiation. We are constantly exposed to background radiation from a variety of natural sources, including cosmic radiation, rocks, medical procedures, consumer products, and even our own atoms. We can minimize our exposure by blocking or shielding the radiation, moving farther from the source, and limiting the time of exposure. Summary We are constantly exposed to radiation from a variety of naturally occurring and human-produced sources. This radiation can affect living organisms. Ionizing radiation is the most harmful because it can ionize molecules or break chemical bonds, which damages the molecule and causes malfunctions in cell processes. It can also create reactive hydroxyl radicals that damage biological molecules and disrupt physiological processes. Radiation can cause somatic or genetic damage, and is most harmful to rapidly reproducing cells. Types of radiation differ in their ability to penetrate material and damage tissue, with alpha particles the least penetrating but potentially most damaging and gamma rays the most penetrating. Various devices, including Geiger counters, scintillators, and dosimeters, are used to detect and measure radiation, and monitor radiation exposure. We use several units to measure radiation: becquerels or curies for rates of radioactive decay; gray or rads for energy absorbed; and rems or sieverts for biological effects of radiation. Exposure to radiation can cause a wide range of health effects, from minor to severe, and including death. We can minimize the effects of radiation by shielding with dense materials such as lead, moving away from the source, and limiting time of exposure. Footnotes 1 Source: US Environmental Protection Agency Glossary becquerel (Bq) SI unit for rate of radioactive decay; 1 Bq = 1 disintegration/s curie (Ci) larger unit for rate of radioactive decay frequently used in medicine; 1 Ci = 3.7 × 10 10 disintegrations/s Geiger counter instrument that detects and measures radiation via the ionization produced in a Geiger-Müller tube gray (Gy) SI unit for measuring radiation dose; 1 Gy = 1 J absorbed/kg tissue ionizing radiation radiation that can cause a molecule to lose an electron and form an ion millicurie (mCi) larger unit for rate of radioactive decay frequently used in medicine; 1 Ci = 3.7 × 10 10 disintegrations/s nonionizing radiation radiation that speeds up the movement of atoms and molecules; it is equivalent to heating a sample, but is not energetic enough to cause the ionization of molecules radiation absorbed dose (rad) SI unit for measuring radiation dose, frequently used in medical applications; 1 rad = 0.01 Gy radiation dosimeter device that measures ionizing radiation and is used to determine personal radiation exposure relative biological effectiveness (RBE) measure of the relative damage done by radiation roentgen equivalent man (rem) unit for radiation damage, frequently used in medicine; 1 rem = 1 Sv scintillation counter instrument that uses a scintillator—a material that emits light when excited by ionizing radiation—to detect and measure radiation sievert (Sv) SI unit measuring tissue damage caused by radiation; takes into account energy and biological effects of radiation Contributors Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected] ).
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Courses/Erie_Community_College/ECC%3A_Introduction_to_General_Organic_and_Biochemistry_(Sorrentino)/Text/12%3A_Organic_Chemistry_-_Alkanes/12.12%3A_Organic_Chemistry%3A_Alkanes
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To ensure that you understand the material in this chapter, you should review the meanings of the following bold terms in the summary and ask yourself how they relate to the topics in the chapter. Organic chemistry is the chemistry of carbon compounds, and inorganic chemistry is the chemistry of all the other elements. Carbon atoms can form stable covalent bonds with other carbon atoms and with atoms of other elements, and this property allows the formation the tens of millions of organic compounds. Hydrocarbons contain only hydrogen and carbon atoms. Hydrocarbons in which each carbon atom is bonded to four other atoms are called alkanes or saturated hydrocarbons . They have the general formula C n H 2 n + 2 . Any given alkane differs from the next one in a series by a CH 2 unit. Any family of compounds in which adjacent members differ from each other by a definite factor is called a homologous series . Carbon atoms in alkanes can form straight chains or branched chains. Two or more compounds having the same molecular formula but different structural formulas are isomers of each other. There are no isomeric forms for the three smallest alkanes; beginning with C 4 H 10 , all other alkanes have isomeric forms. A structural formula shows all the carbon and hydrogen atoms and how they are attached to one another. A condensed structural formula shows the hydrogen atoms right next to the carbon atoms to which they are attached. A line-angle formula is a formula in which carbon atoms are implied at the corners and ends of lines. Each carbon atom is understood to be attached to enough hydrogen atoms to give each carbon atom four bonds. The IUPAC System of Nomenclature provides rules for naming organic compounds. An alkyl group is a unit formed by removing one hydrogen atom from an alkane. The physical properties of alkanes reflect the fact that alkane molecules are nonpolar. Alkanes are insoluble in water and less dense than water. Alkanes are generally unreactive toward laboratory acids, bases, oxidizing agents, and reducing agents. They do burn (undergo combustion reactions ). Alkanes react with halogens by substituting one or more halogen atoms for hydrogen atoms to form halogenated hydrocarbons . An alkyl halide (haloalkane) is a compound resulting from the replacement of a hydrogen atom of an alkane with a halogen atom. Cycloalkanes are hydrocarbons whose molecules are closed rings rather than straight or branched chains. A cyclic hydrocarbon is a hydrocarbon with a ring of carbon atoms
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Courses/Riverland_Community_College/CHEM_1000_-_Introduction_to_Chemistry_(Riverland)/04%3A_Atoms_Elements_and_the_Periodic_Table/4.03%3A_The_Nuclear_Atom
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Learning Objectives Explain the observations that led to Thomson's discovery of the electron. Describe Thomson's "plum pudding" model of the atom and the evidence for it. Draw a diagram of Thomson's "plum pudding" model of the atom and explain why it has this name. Describe Rutherford's gold foil experiment and explain how this experiment altered the "plum pudding" model. Draw a diagram of the Rutherford model of the atom and label the nucleus and the electron cloud. Dalton's Atomic Theory held up well to a lot of the different chemical experiments that scientists performed to test it. In fact, for almost 100 years, it seemed as if Dalton's Atomic Theory was the whole truth. However, in 1897, a scientist named J. J. Thomson conducted some research that suggested that Dalton's Atomic Theory was not the entire story. He suggested that the small, negatively charged particles making up the cathode ray were actually pieces of atoms. He called these pieces "corpuscles," although today we know them as electrons . Thanks to his clever experiments and careful reasoning, J. J. Thomson is credited with the discovery of the electron. Electrons and Plums The electron was discovered by J. J. Thomson in 1897. The existence of protons was also known, as was the fact that atoms were neutral in charge. Since the intact atom had no net charge and the electron and proton had opposite charges, the next step after the discovery of subatomic particles was to figure out how these particles were arranged in the atom. This was a difficult task because of the incredibly small size of the atom. Therefore, scientists set out to design a model of what they believed the atom could look like. The goal of each atomic model was to accurately represent all of the experimental evidence about atoms in the simplest way possible. Following the discovery of the electron, J.J. Thomson developed what became known as the " plum pudding " model in 1904. Plum pudding is an English dessert similar to a blueberry muffin. In Thomson's plum pudding model of the atom, the electrons were embedded in a uniform sphere of positive charge like blueberries stuck into a muffin. The positive matter was thought to be jelly-like or similar to a thick soup. The electrons were somewhat mobile. As they got closer to the outer portion of the atom, the positive charge in the region was greater than the neighboring negative charges, and the electron would be pulled back more toward the center region of the atom. However, this model of the atom soon gave way to a new model developed by New Zealander Ernest Rutherford (1871-1937) about five years later. Thomson did still receive many honors during his lifetime, including being awarded the Nobel Prize in Physics in 1906 and a knighthood in 1908. Atoms and Gold In 1911, Rutherford and coworkers Hans Geiger and Ernest Marsden initiated a series of groundbreaking experiments that would completely change the accepted model of the atom. They bombarded very thin sheets of gold foil with fast moving alpha particles . Alpha particles, a type of natural radioactive particle, are positively charged particles with a mass about four times that of a hydrogen atom. According to the accepted atomic model, in which an atom's mass and charge are uniformly distributed throughout the atom, the scientists expected that all of the alpha particles would pass through the gold foil with only a slight deflection or none at all. Surprisingly, while most of the alpha particles were indeed not deflected, a very small percentage (about 1 in 8000 particles) bounced off the gold foil at very large angles. Some were even redirected back toward the source. No prior knowledge had prepared them for this discovery. In a famous quote, Rutherford exclaimed that it was "as if you had fired a 15-inch [artillery] shell at a piece of tissue and it came back and hit you." Rutherford needed to come up with an entirely new model of the atom in order to explain his results. Because the vast majority of the alpha particles had passed through the gold, he reasoned that most of the atom was empty space. In contrast, the particles that were highly deflected must have experienced a tremendously powerful force within the atom. He concluded that all of the positive charge and the majority of the mass of the atom must be concentrated in a very small space in the atom's interior, which he called the nucleus. The nucleus is the tiny, dense, central core of the atom and is composed of protons and neutrons. Rutherford's atomic model became known as the nuclear model . In the nuclear atom, the protons and neutrons, which comprise nearly all of the mass of the atom, are located in the nucleus at the center of the atom. The electrons are distributed around the nucleus and occupy most of the volume of the atom. It is worth emphasizing just how small the nucleus is compared to the rest of the atom. If we could blow up an atom to be the size of a large professional football stadium, the nucleus would be about the size of a marble. Rutherford's model proved to be an important step towards a full understanding of the atom. However, it did not completely address the nature of the electrons and the way in which they occupy the vast space around the nucleus. It was not until some years later that a full understanding of the electron was achieved. This proved to be the key to understanding the chemical properties of elements. Atomic Nucleus The nucleus (plural, nuclei) is a positively charged region at the center of the atom. It consists of two types of subatomic particles packed tightly together. The particles are protons, which have a positive electric charge, and neutrons, which are neutral in electric charge. Outside of the nucleus, an atom is mostly empty space, with orbiting negative particles called electrons whizzing through it. The figure below shows these parts of the atom. The nucleus of the atom is extremely small. Its radius is only about 1/100,000 of the total radius of the atom. Electrons have virtually no mass, but protons and neutrons have a lot of mass for their size. As a result, the nucleus has virtually all the mass of an atom. Given its great mass and tiny size, the nucleus is very dense. If an object the size of a penny had the same density as the nucleus of an atom, its mass would be greater than 30 million tons! Holding it all Together Particles with opposite electric charges attract each other. This explains why negative electrons orbit the positive nucleus. Particles with the same electric charge repel each other. This means that the positive protons in the nucleus push apart from one another. So why doesn't the nucleus fly apart? An even stronger force—called the strong nuclear force —holds protons and neutrons together in the nucleus. Summary Atoms are the ultimate building blocks of all matter. The modern atomic theory establishes the concepts of atoms and how they compose matter. Bombardment of gold foil with alpha particles showed that some particles were deflected. The nuclear model of the atom consists of a small and dense positively charged interior surrounded by a cloud of electrons.
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Courses/Smith_College/CHM_118%3A_Advanced_General_Chemistry_(Strom)/08%3A_Kinetics/8.06%3A_Collision_Theory
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Learning Objectives By the end of this section, you will be able to: Use the postulates of collision theory to explain the effects of physical state, temperature, and concentration on reaction rates Define the concepts of activation energy and transition state Use the Arrhenius equation in calculations relating rate constants to temperature We should not be surprised that atoms, molecules, or ions must collide before they can react with each other. Atoms must be close together to form chemical bonds. This simple premise is the basis for a very powerful theory that explains many observations regarding chemical kinetics, including factors affecting reaction rates. Collision theory is based on the following postulates: The rate of a reaction is proportional to the rate of reactant collisions: \[\text { reaction rate } \propto \frac{\# \text { collisions }}{\text { time }} \nonumber \] The reacting species must collide in an orientation that allows contact between the atoms that will become bonded together in the product. The collision must occur with adequate energy to permit mutual penetration of the reacting species’ valence shells so that the electrons can rearrange and form new bonds (and new chemical species). We can see the importance of the two physical factors noted in postulates 2 and 3, the orientation and energy of collisions, when we consider the reaction of carbon monoxide with oxygen: \[\ce{2 CO (g) + O_2(g) \longrightarrow 2 CO_2(g)} \nonumber \] Carbon monoxide is a pollutant produced by the combustion of hydrocarbon fuels. To reduce this pollutant, automobiles have catalytic converters that use a catalyst to carry out this reaction. It is also a side reaction of the combustion of gunpowder that results in muzzle flash for many firearms. If carbon monoxide and oxygen are present in sufficient amounts, the reaction will occur at high temperature and pressure. The first step in the gas-phase reaction between carbon monoxide and oxygen is a collision between the two molecules: \[\ce{CO (g) + O_2(g) \longrightarrow CO_2(g) + O (g)} \nonumber \] Although there are many different possible orientations the two molecules can have relative to each other, consider the two presented in Figure \(\PageIndex{1}\). In the first case, the oxygen side of the carbon monoxide molecule collides with the oxygen molecule. In the second case, the carbon side of the carbon monoxide molecule collides with the oxygen molecule. The second case is clearly more likely to result in the formation of carbon dioxide, which has a central carbon atom bonded to two oxygen atoms (\ce{O=C=O}\). This is a rather simple example of how important the orientation of the collision is in terms of creating the desired product of the reaction. If the collision does take place with the correct orientation, there is still no guarantee that the reaction will proceed to form carbon dioxide. In addition to a proper orientation, the collision must also occur with sufficient energy to result in product formation. When reactant species collide with both proper orientation and adequate energy, they combine to form an unstable species called an activated complex or a transition state . These species are very short lived and usually undetectable by most analytical instruments. In some cases, sophisticated spectral measurements have been used to observe transition states. Collision theory explains why most reaction rates increase as concentrations increase. With an increase in the concentration of any reacting substance, the chances for collisions between molecules are increased because there are more molecules per unit of volume. More collisions mean a faster reaction rate, assuming the energy of the collisions is adequate. Activation Energy and the Arrhenius Equation The minimum energy necessary to form a product during a collision between reactants is called the activation energy ( E a ) . How this energy compares to the kinetic energy provided by colliding reactant molecules is a primary factor affecting the rate of a chemical reaction. If the activation energy is much larger than the average kinetic energy of the molecules, the reaction will occur slowly since only a few fast-moving molecules will have enough energy to react. If the activation energy is much smaller than the average kinetic energy of the molecules, a large fraction of molecules will be adequately energetic and the reaction will proceed rapidly. Figure \(\PageIndex{2}\) shows how the energy of a chemical system changes as it undergoes a reaction converting reactants to products according to the equation \[\ce{A+B \longrightarrow C+D} \nonumber \] These reaction diagrams are widely used in chemical kinetics to illustrate various properties of the reaction of interest. Viewing the diagram from left to right, the system initially comprises reactants only, \(\ce{A + B}\). Reactant molecules with sufficient energy can collide to form a high-energy activated complex or transition state. The unstable transition state can then subsequently decay to yield stable products, C + D . The diagram depicts the reaction's activation energy, E a , as the energy difference between the reactants and the transition state. Using a specific energy, the enthalpy (see chapter on thermochemistry), the enthalpy change of the reaction, Δ H , is estimated as the energy difference between the reactants and products. In this case, the reaction is exothermic (Δ H < 0) since it yields a decrease in system enthalpy. The Arrhenius equation relates the activation energy and the rate constant, k , for many chemical reactions: \[k=A e^{-E_{ a } / R T} \nonumber \] In this equation, \(R\) is the ideal gas constant, which has a value 8.314 J/mol/K, \(T\) is temperature on the Kelvin scale, \(E_a\) is the activation energy in joules per mole, e is the constant 2.7183, and A is a constant called the frequency factor , which is related to the frequency of collisions and the orientation of the reacting molecules. Postulates of collision theory are nicely accommodated by the Arrhenius equation. The frequency factor, A , reflects how well the reaction conditions favor properly oriented collisions between reactant molecules. An increased probability of effectively oriented collisions results in larger values for A and faster reaction rates. The exponential term, e −Ea/RT , describes the effect of activation energy on reaction rate. According to kinetic molecular theory (see chapter on gases), the temperature of matter is a measure of the average kinetic energy of its constituent atoms or molecules. The distribution of energies among the molecules composing a sample of matter at any given temperature is described by the plot shown in Figure \(\PageIndex{3a}\). Two shaded areas under the curve represent the numbers of molecules possessing adequate energy ( RT ) to overcome the activation barriers ( E a ). A lower activation energy results in a greater fraction of adequately energized molecules and a faster reaction. The exponential term also describes the effect of temperature on reaction rate. A higher temperature represents a correspondingly greater fraction of molecules possessing sufficient energy ( RT ) to overcome the activation barrier ( E a ), as shown in Figure \(\PageIndex{3b}\). This yields a greater value for the rate constant and a correspondingly faster reaction rate. A convenient approach for determining \(E_a\) for a reaction involves the measurement of k at two or more different temperatures and using an alternate version of the Arrhenius equation that takes the form of a linear equation \[\begin{align*} \ln k & =\left(\frac{-E_{ a }}{R}\right)\left(\frac{1}{T}\right) + \ln A \\[4pt] y & =m x+b \end{align*} \] A plot of \(\ln k\) versus \(\frac{1}{T}\) is linear with a slope equal to \(\frac{-E_a}{R}\) and a y-intercept equal to \(\ln A\). Example \(\PageIndex{1}\): Determination of E a The variation of the rate constant with temperature for the decomposition of \(\ce{HI(g)}\) to \(\ce{H2(g)}\) and \(\ce{I2(g)}\) is given here. What is the activation energy for the reaction? \[\ce{2 HI (g) \longrightarrow H_2(g) + I_2(g)} \nonumber \] T (K) k (L/mol/s) 555 3.52 × 10−7 575 1.22 × 10−6 645 8.59 × 10−5 700 1.16 × 10−3 781 3.95 × 10−2 Solution Use the provided data to derive values of \(\frac{1}{T}\) and \(\ln k\): \(\frac{1}{T}\) \(\ln k\) 1.80 × 10−3 −14.860 1.74 × 10−3 −13.617 1.55 × 10−3 −9.362 1.43 × 10−3 −6.759 1.28 × 10−3 −3.231 Figure \(\PageIndex{4}\) is a graph of \(\ln k\) versus \(\frac{1}{T}\). In practice, the equation of the line (slope and y -intercept) that best fits these plotted data points would be derived using a statistical process called regression. This is helpful for most experimental data because a perfect fit of each data point with the line is rarely encountered. For the data here, the fit is nearly perfect and the slope may be estimated using any two of the provided data pairs. Using the first and last data points permits estimation of the slope. \[\begin{align*} \text { Slope }= & \frac{\Delta(\ln k)}{\Delta\left(\frac{1}{T}\right)} \\[4pt] = & \frac{(-14.860)-(-3.231)}{\left(1.80 \times 10^{-3} K^{-1}\right)-\left(1.28 \times 10^{-3} K^{-1}\right)} \\[4pt] = & \frac{-11.629}{0.52 \times 10^{-3} K^{-1}}=-2.2 \times 10^4 K \\[4pt] = & -\frac{E_{ a }}{R} \\[4pt] E_{ a }= & - \text { slope } \times R=-\left(-2.2 \times 10^4 K \times 8.314 J mol^{-1} K^{-1}\right) \\[4pt] & 1.8 \times 10^5 J mol^{-1} \text { or } 180 kJ mol^{-1} \end{align*} \] Alternative approach: A more expedient approach involves deriving activation energy from measurements of the rate constant at just two temperatures. In this approach, the Arrhenius equation is rearranged to a convenient two-point form: \[\ln \frac{k_1}{k_2}=\frac{E_{ a }}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right) \nonumber \] Rearranging this equation to isolate activation energy yields: \[E_{ a }=-R\left(\frac{\ln k_2-\ln k_1}{\left(\frac{1}{T_2}\right)-\left(\frac{1}{T_1}\right)}\right) \nonumber \] Any two data pairs may be substituted into this equation—for example, the first and last entries from the above data table: \[E_{ a }=-8.314 J mol^{-1} K^{-1}\left(\frac{-3.231-(-14.860)}{1.28 \times 10^{-3} K^{-1}-1.80 \times 10^{-3} K^{-1}}\right) \nonumber \] and the result is E a = 1.8 10 5 J mol −1 or 180 kJ mol −1 This approach yields the same result as the more rigorous graphical approach used above, as expected. In practice, the graphical approach typically provides more reliable results when working with actual experimental data. Exercise \(\PageIndex{1}\) The rate constant for the rate of decomposition of \(\ce{N2O5}\) to \(\ce{NO}\) and \(\ce{O2}\) in the gas phase is 1.66 L/mol/s at 650 K and 7.39 L/mol/s at 700 K: \[\ce{2 N2O5(g) -> 4 NO(g) + 3 O2(g)} \nonumber \] Assuming the kinetics of this reaction are consistent with the Arrhenius equation, calculate the activation energy for this decomposition. Answer 1.1 10 5 J mol −1 or 110 kJ mol −1
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Courses/Monterey_Peninsula_College/CHEM_30A%3A_Introduction_to_Chemistry_for_Health_Sciences/08%3A_Moles_and_Stoichiometry/8.01%3A_The_Mole
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Learning Objectives To define the mole unit. Figure \(\PageIndex{1}\) shows that we need 2 hydrogen atoms and 1 oxygen atom to make 1 water molecule. If we want to make 2 water molecules, we will need 4 hydrogen atoms and 2 oxygen atoms. If we want to make 5 molecules of water, we need 10 hydrogen atoms and 5 oxygen atoms. The ratio of atoms we will need to make any number of water molecules is the same: 2 hydrogen atoms to 1 oxygen atom. One problem we have, however, is that it is extremely difficult, if not impossible, to organize atoms one at a time. As stated in the introduction, we deal with billions of atoms at a time. How can we keep track of so many atoms (and molecules) at a time? We do it by using mass rather than by counting individual atoms. A hydrogen atom has a mass of approximately 1 u. An oxygen atom has a mass of approximately 16 u. The ratio of the mass of an oxygen atom to the mass of a hydrogen atom is therefore approximately 16:1. If we have 2 atoms of each element, the ratio of their masses is approximately 32:2, which reduces to 16:1—the same ratio. If we have 12 atoms of each element, the ratio of their total masses is approximately (12 × 16):(12 × 1), or 192:12, which also reduces to 16:1. If we have 100 atoms of each element, the ratio of the masses is approximately 1,600:100, which again reduces to 16:1. As long as we have equal numbers of hydrogen and oxygen atoms, the ratio of the masses will always be 16:1. The same consistency is seen when ratios of the masses of other elements are compared. For example, the ratio of the masses of silicon atoms to equal numbers of hydrogen atoms is always approximately 28:1, while the ratio of the masses of calcium atoms to equal numbers of lithium atoms is approximately 40:7. So we have established that the masses of atoms are constant with respect to each other, as long as we have the same number of each type of atom. Consider a more macroscopic example. If a sample contains 40 g of Ca, this sample has the same number of atoms as there are in a sample of 7 g of Li. What we need, then, is a number that represents a convenient quantity of atoms so we can relate macroscopic quantities of substances. Clearly even 12 atoms are too few because atoms themselves are so small. We need a number that represents billions and billions of atoms. Chemists use the term mole to represent a large number of atoms or molecules. Just as a dozen implies 12 things, a mole (abbreviated as mol ) represents 6.022 × 10 23 things . The number 6.022 × 10 23 , called Avogadro’s number after the 19th-century chemist Amedeo Avogadro, is the number we use in chemistry to represent macroscopic amounts of atoms and molecules. Thus, if we have 6.022 × 10 23 Na atoms, we say we have 1 mol of Na atoms. If we have 2 mol of Na atoms, we have 2 × (6.022 × 10 23 ) Na atoms, or 1.2044 × 10 24 Na atoms. Similarly, if we have 0.5 mol of benzene (C 6 H 6 ) molecules, we have 0.5 × (6.022 × 10 23 ) C 6 H 6 molecules, or 3.011 × 10 23 C 6 H 6 molecules. A mole represents a very large number! If 1 mol of quarters were stacked in a column, it could stretch back and forth between Earth and the sun 6.8 billion times. Notice that we are applying the mole unit to different types of chemical entities . The word mole represents a number of things—6.022 × 10 23 of them—but does not by itself specify what “they” are. The chemical entities can be atoms , molecules , formula units and ions . This specific information needs to be specified accurately. Most students find this confusing hence, we need to review the composition of elements, covalent and ionic compounds. Most elements are made up of individual atoms , such as helium. However, some elements consist of molecules , such as the diatomic elements, nitrogen, hydrogen, oxygen, etc. One mole of He consists of 6.022 × 10 23 He atoms but one mole of nitrogen contains 6.022 × 10 23 N 2 molecules . The basic units of covalent (molecular) compounds are molecules as well. The molecules of "compounds" consist of different kinds of atoms while the molecules of "elements" consist of only one type of atom. For example, the molecules of ammonia (NH 3 ) consist of nitrogen and hydrogen atoms while N 2 molecules have N atoms only. Compounds that are ionic, like NaCl, are represented by ionic formulas. One mole of NaCl, for example, refers to 6.022 × 10 23 formula units of NaCl. And, one formula unit of NaCl consists of one sodium ion and one chloride ion . Figure 6.1.2 summarizes the basic units of elements, covalent and ionic compounds Conversion Between Moles and Atoms, Molecules and Ions Using our unit conversion techniques learned in Chapter 1, we can use the mole relationship and the chemical formula to convert back and forth between the moles and the number of chemical entities (atoms, molecules or ions). Because 1 N 2 molecule contains 2 N atoms, 1 mol of N 2 molecules (6.022 × 10 23 molecules) has 2 mol of N atoms. Using formulas to indicate how many atoms of each element we have in a substance, we can relate the number of moles of molecules to the number of moles of atoms. For example, in 1 mol of ethanol (C 2 H 6 O), we can construct the following relationships (Table \(\PageIndex{1}\)): 1 Molecule of \(C_2H_6O\) Has 1 Mol of \(C_2H_6O\) Has Molecular Relationships 2 C atoms 2 mol of C atoms \(\mathrm{\dfrac{2\: mol\: C\: atoms}{1\: mol\: C_2H_6O\: molecules}}\) or \(\mathrm{\dfrac{1\: mol\: C_2H_6O\: molecules}{2\: mol\: C\: atoms}}\) 6 H atoms 6 mol of H atoms \(\mathrm{\dfrac{6\: mol\: H\: atoms}{1\: mol\: C_2H_6O\: molecules}}\) or \(\mathrm{\dfrac{1\: mol\: C_2H_6O\: molecules}{6\: mol\: H\: atoms}}\) 1 O atom 1 mol of O atoms \(\mathrm{\dfrac{1\: mol\: O\: atoms}{1\: mol\: C_2H_6O\: molecules}}\) or \(\mathrm{\dfrac{1\: mol\: C_2H_6O\: molecules}{1\: mol\: O\: atoms}}\) The following example illustrates how we can use these relationships as conversion factors. Example \(\PageIndex{1}\) If a sample consists of 2.5 mol of ethanol (C 2 H 6 O), how many moles of carbon atoms, hydrogen atoms, and oxygen atoms does it have? Solution Using the relationships in Table \(\PageIndex{1}\), we apply the appropriate conversion factor for each element: Note how the unit mol C 2 H 6 O molecules cancels algebraically. Similar equations can be constructed for determining the number of H and O atoms: \(\mathrm{2.5\: mol\: C_2H_6O\: molecules\times\dfrac{6\: mol\: H\: atoms}{1\: mol\: C_2H_6O\: molecules}=15\: mol\: H\: atoms}\) \(\mathrm{2.5\: mol\: C_2H_6O\: molecules\times\dfrac{1\: mol\: O\: atoms}{1\: mol\: C_2H_6O\: molecules}=2.5\: mol\: O\: atoms}\) Exercise \(\PageIndex{1}\) If a sample contains 6.75 mol of Na 2 SO 4 , how many moles of sodium atoms, sulfur atoms, and oxygen atoms does it have? Answer 13.5 mol Na, 6.75 mol S and 27 mol O. We can use Avogadro's number as a conversion factor, or ratio, in dimensional analysis problems. For example, if we are dealing with element X, the mole relationship is expressed as follows: \[\text{1 mol X} = 6.022 \times 10^{23} \text{ X atoms} \nonumber \] We can convert this relationship into two possible conversion factors shown below: \(\mathrm{\dfrac{1\: mol\: X\: }{6.022\times 10^{23}\: X\: atoms}}\) or \(\mathrm{\dfrac{6.022\times 10^{23}\: X\: atoms}{1\: mol\: X\: }}\) If the number of "atoms of element X" is given, we can convert it into "moles of X" by multiplying the given value with the conversion factor at the left. However, if the number of "mol of X" is given, the appropriate conversion factor to use is the one at the right. If we are dealing with a molecular compound (such as C 4 H 10 ), the mole relationship is expressed as follows: \[\text{1 mol C4H10} = 6.022 \times 10^{23} \text{ C4H10 molecules} \nonumber \] If working with ionic compounds (such as NaCl), the mole relationship is expressed as follows: \[\text{1 mol NaCl} = 6.022 \times 10^{23} \text{ NaCl formula units} \nonumber \] Example \(\PageIndex{2}\) How many formula units are present in 2.34 mol of NaCl? How many ions are in 2.34 mol? Solution Typically in a problem like this, we start with what we are given and apply the appropriate conversion factor. Here, we are given a quantity of 2.34 mol of NaCl, to which we can apply the definition of a mole as a conversion factor : \(\mathrm{2.34\: mol\: NaCl\times\dfrac{6.022\times10^{23}\: NaCl\: units}{1\: mol\: NaCl}=1.41\times10^{24}\: NaCl\: units}\) Because there are two ions per formula unit, there are \(\mathrm{1.41\times10^{24}\: NaCl\: units\times\dfrac{2\: ions}{NaCl\: units}=2.82\times10^{24}\: ions}\) in the sample. Exercise \(\PageIndex{2}\) How many molecules are present in 16.02 mol of C 4 H 10 ? How many atoms are in 16.02 mol? Answer 9.647 x 10 24 molecules, 1.351 x 10 26 atoms. Key Takeaway A mole is 6.022 × 10 23 things.
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Courses/Chippewa_Valley_Technical_College/CVTC_Basic_Chemistry/01%3A_Matter_Atoms_and_Elements/1.13%3A_Chapter_1_Exercises/1.13.02%3A_Elements_Atoms_and_the_Periodic_Table_(Exercises)
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These are homework exercises to accompany Chapter 2 of the Ball et al. " The Basics of GOB Chemistry " Textmap. The Elements Exercises Which of the following substances are elements? sodium milk gold water air liquefied nitrogen Which of the following substances are elements? paper electricity neon carbon wood concrete Write the chemical symbol for each element. silver sulfur nitrogen neon Write the chemical symbol for each element. bromine oxygen lithium boron Explain why it is improper to write CO as the chemical symbol for cobalt. Explain why it is improper to write NO as the chemical symbol for nobelium. Complete the following table. Element Symbol Element Name F NaN Fe NaN I NaN Cr NaN C NaN P NaN Complete the following table. Element Symbol Element Name Mg NaN Mn NaN Ca NaN Cl NaN K NaN Pt NaN Answers element not an element element not an element not an element element 2. not an element not an element element element not an element not an element Ag S N Ne 4. Br O Li B By convention, the second letter in an element’s symbol is always lowercase. 6. By convention, the second letter in an element’s symbol is always lowercase. Additionally, NO represents a compound. 7. Element Symbol Element Name F fluorine Fe iron I iodine Cr chromium C carbon P phosphorus 8. Empty DataFrame Columns: [(Element Symbol, Mg, Mn, Ca, Cl, K, Pt), (Element Name, magnesium, manganese, calcium, chlorine, potassium, platinum)] Index: [] Atomic Theory Exercises Which of the following elements exist as diatomic molecules? helium hydrogen iodine gold Which of the following elements exist as diatomic molecules? chlorine potassium silver oxygen Why is it proper to represent the elemental form of helium as He but improper to represent the elemental form of hydrogen as H? Why is it proper to represent the elemental form of chlorine as Cl 2 but improper to represent the elemental form of calcium as Ca 2 ? Answers no yes yes no 2. yes no no yes 3. Hydrogen exists as a diatomic molecule in its elemental form; helium does not exist as a diatomic molecule. 4. Chlorine exists as a diatomic molecule in its elemental form; calcium does not exist as a diatomic molecule. The Structure of Atoms Exercises Which is smaller—an electron or a helium atom? Which is larger—a proton or an atom of lead? Which subatomic particle has a positive charge? Which subatomic particle has a negative charge? Which subatomic particle is electrically neutral? Does it exist inside or outside the nucleus? Protons are among the (most, least) massive subatomic particles, and they are found (inside, outside) the nucleus. Electrons are among the (most, least) massive subatomic particles, and they are found (inside, outside) the nucleus. Describe why Rutherford used the term planetary model to describe his model of atomic structure. Why is the planetary model not an appropriate way to describe the structure of an atom? What happened to most of the alpha particles in Rutherford’s experiment? Explain why that happened. Electrons account for the (majority, minority) of the (mass, volume) of an atom. Answers An electron is smaller. 2. An atom of lead is larger. proton; electron 4. neutron; inside the nucleus most; inside 6. least; outside Electrons are in orbit about the nucleus. 8. Electrons do not have specific circular orbits about the nucleus. 9. Most of the alpha particles went through the metal sheet because atoms are mostly empty space. 10. minority; mass Nuclei of Atoms Exercises How many protons are in the nucleus of each element? radon tungsten chromium beryllium How many protons are in the nucleus of each element? sulfur uranium calcium lithium What are the atomic numbers of the elements in Exercise 1? What are the atomic numbers of the elements in Exercise 2? How many electrons are in neutral atoms of the elements in Exercise 1? How many electrons are in neutral atoms of the elements in Exercise 2? Complete the following table. Number of Protons Number of Neutrons Element Name Isotope Symbol 80.0 120.0 NaN NaN NaN NaN NaN \(\mathrm{^{55}_{26}Fe}\) NaN 2.0 hydrogen NaN Complete the following table. Number of Protons Number of Neutrons Element Name Isotope Symbol NaN NaN NaN \(\mathrm{^{3}_{2}He}\) 95.0 153.0 NaN NaN NaN 21.0 potassium NaN State the number of protons, neutrons, and electrons in neutral atoms of each isotope. 131 I 40 K 201 Hg 19 F State the number of protons, neutrons, and electrons in neutral atoms of each isotope. 3 H 133 Cs 56 Fe 207 Pb What is the mass number of a gallium atom that has 38 neutrons in it? What is the mass number of a uranium atom that has 143 neutrons in it? Complete each sentence. 48 Ti has _____ neutrons. 40 Ar has _____ neutrons. 3 H has _____ neutrons. Complete each sentence. 18 O has _____ neutrons. 60 Ni has _____ neutrons. 127 I has _____ neutrons. Answers 86 74 24 4 2. 16 92 20 3 86, 74, 24, and 4 4. 16, 92, 20, 3 86, 74, 24, and 4 6. 16, 92, 20, 3 Number of Protons Number of Neutrons Element Name Isotope Symbol 80 120 mercury \(\mathrm{^{200}_{80}Hg}\) 26 29 iron \(\mathrm{^{55}_{26}Fe}\) 1 2 hydrogen \(\mathrm{^{3}_{1}H}\) 8. Empty DataFrame Columns: [(Number of Protons, 2, 95, 19), (Number of Neutrons, 1, 153, 21), (Element Name, helium, americium, potassium), (Isotope Symbol, \(\mathrm{^{3}_{2}He}\), \(\mathrm{^{248}_{95}Am}\), \(\mathrm{^{40}_{19}K}\))] Index: [] protons: 53; neutrons: 78; electrons: 53 protons: 19; neutrons: 21; electrons: 19 protons: 80; neutrons: 121; electrons: 80 protons: 9; neutrons: 10; electrons: 9 10. protons: 1; neutrons: 2; electrons: 1 protons: 55; neutrons: 78; electrons: 55 protons: 26; neutrons: 30; electrons: 26 protons: 82; neutrons: 125; electrons: 82 11. 69 12. 235 13. 26 22 2 14. 10 32 74 Atomic Masses Exercises What is the atomic mass of zinc in atomic mass units? What is the atomic mass of barium in atomic mass units? What is the average mass of a single magnesium atom in grams? What is the average mass of a single calcium atom in grams? What is the mass of 1.000 × 10 24 aluminum atoms in grams? What is the mass of 5.000 × 10 23 carbon atoms in grams? 7. Which has more mass—1 tungsten atom or 11 oxygen atoms? 8. Which has more mass—1 magnesium atom or 6 helium atoms? 9. Determine the atomic mass of lithium, given the isotopic composition: 92.4% lithium-7 (mass 7.016 u) and 7.60% lithium-6 (mass 6.015 u). 10. Determine the atomic mass of neon, given the isotopic composition: 90.48% neon-20 (mass 19.992 u), 0.27% neon-21 (mass 20.994 u), and 9.25% neon-22 (mass 21.991 u). 11. Determine the atomic mass of magnesium, given the isotopic composition: 78.70% magnesium-24 (mass 23.98 u), 10.13% magnesium-25 (mass 24.99 u) and 11.17% magnesium-26 (mass 25.98). 12. Determine the atomic mass of bromine, given the isotopic composition: 50.69% bromine-79 (mass 78.9183 u), 49.31% and bromine-81 (mass 80.9163 u). Answers 65.4 u 2. 137.33 u 4.038 × 10 −23 g 4. 6.657 × 10 −23 g 44.81 g 6. 9.974 g 7. 1 tungsten atom 8. 1 magnesium atom 9. 6.94 u 10. 20.18 u 11. 24.30 u 12. 79.90 u The Periodic Table Exercises Which elements have chemical properties similar to those of magnesium? sodium fluorine calcium barium selenium Which elements have chemical properties similar to those of lithium? sodium calcium beryllium barium potassium Which elements have chemical properties similar to those of chlorine? sodium fluorine calcium iodine sulfur Which elements have chemical properties similar to those of carbon? silicon oxygen germanium barium argon Which elements are alkali metals? sodium magnesium aluminum potassium calcium Which elements are alkaline earth metals? sodium magnesium aluminum potassium calcium Which elements are halogens? oxygen fluorine chlorine sulfur carbon Which elements are noble gases? helium hydrogen oxygen neon chlorine Which pairs of elements are located in the same period? H and Li H and He Na and S Na and Rb Which pairs of elements are located in the same period? V and Nb K and Br Na and P Li and Mg In each pair of atoms, which atom has the greater atomic radius? H and Li N and P Cl and Ar Al and Cl In each pair of atoms, which atom has the greater atomic radius? H and He N and F Cl and Br Al and B Scandium is a (metal, nonmetal, semimetal) and is a member of the (main group elements, transition metals). Silicon is a (metal, nonmetal, semimetal) and is a member of the (main group elements, transition metals). Answers no no yes yes no 2. yes no no no yes no yes no yes no 4. yes no yes no no yes no no yes no 6. no yes no no yes no yes yes no no 8. yes no no yes no no yes yes no 10. no yes yes no 11. Li P Cl Al 12. H N Br Al 13. metal; transition metals 14. semimetal; main group elements " Elements, Atoms, and the Periodic Table" by LibreTexts is licensed under CC BY-NC-SA .
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Courses/Louisville_Collegiate_School/General_Chemistry/LibreTexts_Louisville_Collegiate_School_Chapters_05%3A_Thermochemistry/LibreTexts%2F%2FLouisville_Collegiate_School%2F%2FChapters%2F%2F05%3A_Thermochemistry%2F%2F5.1%3A_Energy_Basics
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Learning Objectives Define energy, distinguish types of energy, and describe the nature of energy changes that accompany chemical and physical changes Distinguish the related properties of heat, thermal energy, and temperature Define and distinguish specific heat and heat capacity, and describe the physical implications of both Perform calculations involving heat, specific heat, and temperature change Chemical changes and their accompanying changes in energy are important parts of our everyday world (Figure \(\PageIndex{1}\)). The macronutrients in food (proteins, fats, and carbohydrates) undergo metabolic reactions that provide the energy to keep our bodies functioning. We burn a variety of fuels (gasoline, natural gas, coal) to produce energy for transportation, heating, and the generation of electricity. Industrial chemical reactions use enormous amounts of energy to produce raw materials (such as iron and aluminum). Energy is then used to manufacture those raw materials into useful products, such as cars, skyscrapers, and bridges. Over 90% of the energy we use comes originally from the sun. Every day, the sun provides the earth with almost 10,000 times the amount of energy necessary to meet all of the world’s energy needs for that day. Our challenge is to find ways to convert and store incoming solar energy so that it can be used in reactions or chemical processes that are both convenient and nonpolluting. Plants and many bacteria capture solar energy through photosynthesis. We release the energy stored in plants when we burn wood or plant products such as ethanol. We also use this energy to fuel our bodies by eating food that comes directly from plants or from animals that got their energy by eating plants. Burning coal and petroleum also releases stored solar energy: These fuels are fossilized plant and animal matter. This chapter will introduce the basic ideas of an important area of science concerned with the amount of heat absorbed or released during chemical and physical changes—an area called thermochemistry . The concepts introduced in this chapter are widely used in almost all scientific and technical fields. Food scientists use them to determine the energy content of foods. Biologists study the energetics of living organisms, such as the metabolic combustion of sugar into carbon dioxide and water. The oil, gas, and transportation industries, renewable energy providers, and many others endeavor to find better methods to produce energy for our commercial and personal needs. Engineers strive to improve energy efficiency, find better ways to heat and cool our homes, refrigerate our food and drinks, and meet the energy and cooling needs of computers and electronics, among other applications. Understanding thermochemical principles is essential for chemists, physicists, biologists, geologists, every type of engineer, and just about anyone who studies or does any kind of science. Energy Energy can be defined as the capacity to supply heat or do work. One type of work ( w ) is the process of causing matter to move against an opposing force. For example, we do work when we inflate a bicycle tire—we move matter (the air in the pump) against the opposing force of the air surrounding the tire. Like matter, energy comes in different types. One scheme classifies energy into two types: potential energy , the energy an object has because of its relative position, composition, or condition, and kinetic energy , the energy that an object possesses because of its motion. Water at the top of a waterfall or dam has potential energy because of its position; when it flows downward through generators, it has kinetic energy that can be used to do work and produce electricity in a hydroelectric plant (Figure \(\PageIndex{2}\)). A battery has potential energy because the chemicals within it can produce electricity that can do work. Energy can be converted from one form into another, but all of the energy present before a change occurs always exists in some form after the change is completed. This observation is expressed in the law of conservation of energy: during a chemical or physical change, energy can be neither created nor destroyed, although it can be changed in form. (This is also one version of the first law of thermodynamics, as you will learn later.) When one substance is converted into another, there is always an associated conversion of one form of energy into another. Heat is usually released or absorbed, but sometimes the conversion involves light, electrical energy, or some other form of energy. For example, chemical energy (a type of potential energy) is stored in the molecules that compose gasoline. When gasoline is combusted within the cylinders of a car’s engine, the rapidly expanding gaseous products of this chemical reaction generate mechanical energy (a type of kinetic energy) when they move the cylinders’ pistons. According to the law of conservation of matter (seen in an earlier chapter), there is no detectable change in the total amount of matter during a chemical change. When chemical reactions occur, the energy changes are relatively modest and the mass changes are too small to measure, so the laws of conservation of matter and energy hold well. However, in nuclear reactions, the energy changes are much larger (by factors of a million or so), the mass changes are measurable, and matter-energy conversions are significant. This will be examined in more detail in a later chapter on nuclear chemistry. To encompass both chemical and nuclear changes, we combine these laws into one statement: The total quantity of matter and energy in the universe is fixed. Thermal Energy, Temperature, and Heat Thermal energy is kinetic energy associated with the random motion of atoms and molecules. Temperature is a quantitative measure of “hot” or “cold.” When the atoms and molecules in an object are moving or vibrating quickly, they have a higher average kinetic energy (KE), and we say that the object is “hot.” When the atoms and molecules are moving slowly, they have lower KE , and we say that the object is “cold” (Figure \(\PageIndex{3}\)). Assuming that no chemical reaction or phase change (such as melting or vaporizing) occurs, increasing the amount of thermal energy in a sample of matter will cause its temperature to increase. And, assuming that no chemical reaction or phase change (such as condensation or freezing) occurs, decreasing the amount of thermal energy in a sample of matter will cause its temperature to decrease. Measuring Energy and Heat Capacity Historically, energy was measured in units of calories (cal) . A calorie is the amount of energy required to raise one gram of water by 1 degree C (1 kelvin). However, this quantity depends on the atmospheric pressure and the starting temperature of the water. The ease of measurement of energy changes in calories has meant that the calorie is still frequently used. The Calorie (with a capital C), or large calorie, commonly used in quantifying food energy content, is a kilocalorie. The SI unit of heat, work, and energy is the joule. A joule (J) is defined as the amount of energy used when a force of 1 newton moves an object 1 meter. It is named in honor of the English physicist James Prescott Joule. One joule is equivalent to 1 kg m 2 /s 2 , which is also called 1 newton–meter. A kilojoule (kJ) is 1000 joules. To standardize its definition, 1 calorie has been set to equal 4.184 joules. We now introduce two concepts useful in describing heat flow and temperature change. The heat capacity ( C ) of a body of matter is the quantity of heat ( q ) it absorbs or releases when it experiences a temperature change (Δ T ) of 1 degree Celsius (or equivalently, 1 kelvin) \[C=\dfrac{q}{ΔT} \label{5.2.1} \] Heat capacity is determined by both the type and amount of substance that absorbs or releases heat. It is therefore an extensive property—its value is proportional to the amount of the substance. For example, consider the heat capacities of two cast iron frying pans. The heat capacity of the large pan is five times greater than that of the small pan because, although both are made of the same material, the mass of the large pan is five times greater than the mass of the small pan. More mass means more atoms are present in the larger pan, so it takes more energy to make all of those atoms vibrate faster. The heat capacity of the small cast iron frying pan is found by observing that it takes 18,150 J of energy to raise the temperature of the pan by 50.0 °C \[C_{\text{small pan}}=\mathrm{\dfrac{18,140\; J}{50.0\; °C} =363\; J/°C} \label{5.2.2} \] The larger cast iron frying pan, while made of the same substance, requires 90,700 J of energy to raise its temperature by 50.0 °C. The larger pan has a (proportionally) larger heat capacity because the larger amount of material requires a (proportionally) larger amount of energy to yield the same temperature change: \[C_{\text{large pan}}=\mathrm{\dfrac{90,700\; J}{50.0\;°C}=1814\; J/°C} \label{5.2.3} \] The specific heat capacity ( c ) of a substance, commonly called its “specific heat,” is the quantity of heat required to raise the temperature of 1 gram of a substance by 1 degree Celsius (or 1 kelvin): \[c = \dfrac{q}{\mathrm{m\Delta T}} \label{5.2.4} \] Specific heat capacity depends only on the kind of substance absorbing or releasing heat. It is an intensive property—the type, but not the amount, of the substance is all that matters. For example, the small cast iron frying pan has a mass of 808 g. The specific heat of iron (the material used to make the pan) is therefore: \[c_\ce{iron}=\mathrm{\dfrac{18,140\; J}{(808\; g)(50.0\;°C)} = 0.449\; J/g\; °C} \label{5.2.5} \] The large frying pan has a mass of 4040 g. Using the data for this pan, we can also calculate the specific heat of iron: \[c_\ce{iron}=\mathrm{\dfrac{90,700\; J}{(4,040\; g)(50.0\;°C)}=0.449\; J/g\; °C} \label{5.2.6} \] Although the large pan is more massive than the small pan, since both are made of the same material, they both yield the same value for specific heat (for the material of construction, iron). Note that specific heat is measured in units of energy per temperature per mass and is an intensive property, being derived from a ratio of two extensive properties (heat and mass). The molar heat capacity, also an intensive property, is the heat capacity per mole of a particular substance and has units of J/mol °C (Figure \(\PageIndex{7}\)). Liquid water has a relatively high specific heat (about 4.2 J/g °C); most metals have much lower specific heats (usually less than 1 J/g °C). The specific heat of a substance varies somewhat with temperature. However, this variation is usually small enough that we will treat specific heat as constant over the range of temperatures that will be considered in this chapter. Specific heats of some common substances are listed in Table \(\PageIndex{1}\). Substance Symbol (state) Specific Heat (J/g °C) helium He(g) 5.193 water H2O(l) 4.184 ethanol C2H6O(l) 2.376 ice H2O(s) 2.093 (at −10 °C) water vapor H2O(g) 1.864 nitrogen N2(g) 1.040 air NaN 1.007 oxygen O2(g) 0.918 aluminum Al(s) 0.897 carbon dioxide CO2(g) 0.853 argon Ar(g) 0.522 iron Fe(s) 0.449 copper Cu(s) 0.385 lead Pb(s) 0.130 gold Au(s) 0.129 silicon Si(s) 0.712 If we know the mass of a substance and its specific heat, we can determine the amount of heat, q , entering or leaving the substance by measuring the temperature change before and after the heat is gained or lost: \[\begin{align*} q &= \ce{(specific\: heat)×(mass\: of\: substance)×(temperature\: change)}\label{5.2.7}\\q&=c×m×ΔT \\[4pt] &=c×m×(T_\ce{final}−T_\ce{initial})\end{align*} \] In this equation, \(c\) is the specific heat of the substance, m is its mass, and Δ T (which is read “delta T”) is the temperature change, T final − T initial . If a substance gains thermal energy, its temperature increases, its final temperature is higher than its initial temperature, T final − T initial has a positive value, and the value of q is positive. If a substance loses thermal energy, its temperature decreases, the final temperature is lower than the initial temperature, T final − T initial has a negative value, and the value of q is negative. Example \(\PageIndex{1}\): Measuring Heat A flask containing \(\mathrm{8.0 \times 10^2\; g}\) of water is heated, and the temperature of the water increases from 21 °C to 85 °C. How much heat did the water absorb? Solution To answer this question, consider these factors: the specific heat of the substance being heated (in this case, water) the amount of substance being heated (in this case, 800 g) the magnitude of the temperature change (in this case, from 21 °C to 85 °C). The specific heat of water is 4.184 J/g °C, so to heat 1 g of water by 1 °C requires 4.184 J. We note that since 4.184 J is required to heat 1 g of water by 1 °C, we will need 800 times as much to heat 800 g of water by 1 °C. Finally, we observe that since 4.184 J are required to heat 1 g of water by 1 °C, we will need 64 times as much to heat it by 64 °C (that is, from 21 °C to 85 °C). This can be summarized using the equation: \[\begin{align*} q&=c×m×ΔT \\[4pt] &=c×m×(T_\ce{final}−T_\ce{initial}) \\[4pt] &=\mathrm{(4.184\:J/\cancel{g}°C)×(800\:\cancel{g})×(85−21)°C}\\[4pt] &=\mathrm{(4.184\:J/\cancel{g}°\cancel{C})×(800\:\cancel{g})×(64)°\cancel{C}}\\[4pt] &=\mathrm{210,000\: J(=210\: kJ)} \end{align*} \nonumber \] Because the temperature increased, the water absorbed heat and \(q\) is positive. Exercise \(\PageIndex{1}\) How much heat, in joules, must be added to a \(\mathrm{5.00 \times 10^2 \;g}\) iron skillet to increase its temperature from 25 °C to 250 °C? The specific heat of iron is 0.451 J/g °C. Answer \(\mathrm{5.05 \times 10^4\; J}\) Note that the relationship between heat, specific heat, mass, and temperature change can be used to determine any of these quantities (not just heat) if the other three are known or can be deduced. Example \(\PageIndex{2}\): Determining Other Quantities A piece of unknown metal weighs 348 g. When the metal piece absorbs 6.64 kJ of heat, its temperature increases from 22.4 °C to 43.6 °C. Determine the specific heat of this metal (which might provide a clue to its identity). Solution Since mass, heat, and temperature change are known for this metal, we can determine its specific heat using the relationship: \[\begin{align*} q&=c \times m \times \Delta T \\[4pt] &=c \times m \times (T_\ce{final}−T_\ce{initial}) \end{align*} \nonumber \] Substituting the known values: \[6,640\; \ce J=c \times \mathrm{(348\; g) \times (43.6 − 22.4)\; °C} \nonumber \] Solving: \[c=\mathrm{\dfrac{6,640\; J}{(348\; g) \times (21.2°C)} =0.900\; J/g\; °C} \nonumber \] Comparing this value with the values in Table \(\PageIndex{1}\), this value matches the specific heat of aluminum, which suggests that the unknown metal may be aluminum. Exercise \(\PageIndex{2}\) A piece of unknown metal weighs 217 g. When the metal piece absorbs 1.43 kJ of heat, its temperature increases from 24.5 °C to 39.1 °C. Determine the specific heat of this metal, and predict its identity. Answer \(c = \mathrm{0.45 \;J/g \;°C}\); the metal is likely to be iron from checking Table \(\PageIndex{1}\). Solar Thermal Energy Power Plants The sunlight that reaches the earth contains thousands of times more energy than we presently capture. Solar thermal systems provide one possible solution to the problem of converting energy from the sun into energy we can use. Large-scale solar thermal plants have different design specifics, but all concentrate sunlight to heat some substance; the heat “stored” in that substance is then converted into electricity. The Solana Generating Station in Arizona’s Sonora Desert produces 280 megawatts of electrical power. It uses parabolic mirrors that focus sunlight on pipes filled with a heat transfer fluid (HTF) (Figure \(\PageIndex{8}\)). The HTF then does two things: It turns water into steam, which spins turbines, which in turn produces electricity, and it melts and heats a mixture of salts, which functions as a thermal energy storage system. After the sun goes down, the molten salt mixture can then release enough of its stored heat to produce steam to run the turbines for 6 hours. Molten salts are used because they possess a number of beneficial properties, including high heat capacities and thermal conductivities. The 377-megawatt Ivanpah Solar Generating System, located in the Mojave Desert in California, is the largest solar thermal power plant in the world (Figure \(\PageIndex{9}\)). Its 170,000 mirrors focus huge amounts of sunlight on three water-filled towers, producing steam at over 538 °C that drives electricity-producing turbines. It produces enough energy to power 140,000 homes. Water is used as the working fluid because of its large heat capacity and heat of vaporization. Summary Energy is the capacity to do work (applying a force to move matter). Kinetic energy (KE) is the energy of motion; potential energy is energy due to relative position, composition, or condition. When energy is converted from one form into another, energy is neither created nor destroyed (law of conservation of energy or first law of thermodynamics). Matter has thermal energy due to the KE of its molecules and temperature that corresponds to the average KE of its molecules. Heat is energy that is transferred between objects at different temperatures; it flows from a high to a low temperature. Chemical and physical processes can absorb heat (endothermic) or release heat (exothermic). The SI unit of energy, heat, and work is the joule (J). Specific heat and heat capacity are measures of the energy needed to change the temperature of a substance or object. The amount of heat absorbed or released by a substance depends directly on the type of substance, its mass, and the temperature change it undergoes. Key Equations \(q=c×m×ΔT=c×m×(T_\ce{final}−T_\ce{initial})\) Glossary calorie (cal) unit of heat or other energy; the amount of energy required to raise 1 gram of water by 1 degree Celsius; 1 cal is defined as 4.184 J endothermic process chemical reaction or physical change that absorbs heat energy capacity to supply heat or do work exothermic process chemical reaction or physical change that releases heat heat ( q ) transfer of thermal energy between two bodies heat capacity ( C ) extensive property of a body of matter that represents the quantity of heat required to increase its temperature by 1 degree Celsius (or 1 kelvin) joule (J) SI unit of energy; 1 joule is the kinetic energy of an object with a mass of 2 kilograms moving with a velocity of 1 meter per second, 1 J = 1 kg m 2 /s and 4.184 J = 1 cal kinetic energy energy of a moving body, in joules, equal to \(\dfrac{1}{2}mv^2\) (where m = mass and v = velocity) potential energy energy of a particle or system of particles derived from relative position, composition, or condition specific heat capacity ( c ) intensive property of a substance that represents the quantity of heat required to raise the temperature of 1 gram of the substance by 1 degree Celsius (or 1 kelvin) temperature intensive property of matter that is a quantitative measure of “hotness” and “coldness” thermal energy kinetic energy associated with the random motion of atoms and molecules thermochemistry study of measuring the amount of heat absorbed or released during a chemical reaction or a physical change work ( w ) energy transfer due to changes in external, macroscopic variables such as pressure and volume; or causing matter to move against an opposing force
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Courses/Oregon_Institute_of_Technology/OIT_(Lund)%3A_Organic_Chemistry_with_a_Biological_Emphasis_(Soderberg)/08%3A_Nucleophilic_Substitution_Reactions/8.E%3A_Nucleophilic_Substitution_Reactions_(Exercises)
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P8.1: Rank the following molecules in order of how fast they would be expected to react with \(CH_3SNa\) in acetone. (\(CH_3SNa\) is simply the sodium salt of \(CH_3S^-\). \(Na^+\) is a spectator ion.) P8.2: Draw line structures representing the most stable cation with the given molecular formula: \(C_3H_7^+\) \(C_4H_9^+\) \(C_3H_8N^+\) \(C_4H_7^+\) P8.3: For each pair of carbocations below, choose the one that is more stable, and explain your reasoning. P8.4: Arrange the following species in order of increasing nucleophilicity in protic solvent: P8.5: Predict the organic products of the following nucleophilic substitution reactions, all of which are carried out in polar aprotic solvent. Show stereochemistry at chiral carbons. Hints: \(Na_2CO_3\), sodium carbonate, is a weak base. For part (f): What is the conjugate acid of \(NH_2^-\)? What is the \(pK_a\) of this conjugate acid, and what is the \(pK_a\) of a terminal alkyne? a. b. c. d. e. f. P8.6: Which of the reactions in the previous problem has a unimolecular rate determining step? Explain. P8.7: From the following pairs, select the compound that would react more rapidly with bromomethane in acetone solvent. water or hydroxide ion \(CH_3S^-\) or \(CH_3OH\) \(CH_2S^-\) or \(CH_3SH\) acetate ion or hydroxide ion diethyl sulfide or diethyl ether dimethylamine or diethylether trimethylamine or 2,2-dimethylpropane P8.8: Methyl iodide (0.10 mole) is added to a solution that contains 0.10 mole \(NaOCH_3\) and 0.10 mole \(NaSCH_3\). Predict the most abundant neutral organic product that would form, and explain your reasoning. Assume that you isolate a mixture the major product (which you predicted in part) along with a smaller amount of a different nucleophilic substitution product. Explain briefly but specifically how you could use \(^1H-NMR\) to determine the ratio of the two products in the mixture. P8.9: For each pair of compounds, predict which will more rapidly undergo solvolysis in methanol solution. P8.10: Predict the solvolysis product(s) of each of the reactions below. Consider both regiochemistry and stereochemistry. a. b. c. d. Draw a complete curved-arrow mechanism for the formation of the secondary allylilc alcohol product in part (a). P8.11: Show starting compounds that would lead to the following products through nucleophilic substitution reactions. a. b. c. d. P8.12: The fused ring compound shown below is very unreactive to nucleophilic substitution, even with a powerful nucleophile.. Explain. (Hint – consider bond geometry - a model will be very helpful!) P8.13: Laboratory synthesis of isopentenyl diphosphate - the 'building block' molecule used by nature for the construction of isoprenoid molecules (section 1.3A) - was accomplished by first converting isopentenyl alcohol into an alkyl tosylate then displacing the tosylate group with an inorganic pyrophosphate nucleophile. Based on this verbal description, draw a mechanism for the second (nucleophilic substitution) step, showing starting and ending compounds for the step and curved arrows for electron movement. P8.14: Choline, an important neutotransmitter in the nervous system, is formed from 2-(N,N-dimethylamino)ethanol: Besides the enzyme and the starting compound, what other important biomolecule do you expect plays a part in the reaction? Draw a mechanism for the reaction. Briefly explain how \(^1H-NMR\) could be used to distinguish between the substrate and the product of this reaction. P8.15: The following is a reaction in the biosynthesis of morphine in opium poppies. (Science 1967, 155, 170; J. Biol. Chem 1995, 270, 31091). Draw a complete mechanism, assuming an \(S_N1\) pathway. What would you expect to be the most noticeable difference between the IR spectrum of the product and that of the substrate? This reaction is an example of the regiospecificity of enzymatic nucleophilic substitution reactions noted earlier in the chapter. Draw two alternate nucleophilic, ring-closing steps for this reaction (leading to different products from what is shown above), and explain why these alternate pathways are both less favorable than the actual reaction catalyzed by the enzyme. P8.16: The enzymatic reaction below, which is part of the metabolism of nucleic acids, proceeds by an \(S_N1\) mechanism. The new bond formed in the substitution is indicated. Predict the structures of the two substrates A and B. Draw a complete mechanism, and use resonance drawings to illustrate how both the carbocation intermediate and the leaving group are stabilized. P8.17: Below is the first step of the reaction catalyzed by anthranilate synthase, an enzyme involved in biosynthesis of the amino acid tryptophan. This reaction is somewhat unusual in that the leaving group is a hydroxide anion, which is of course is normally thought to be a very poor leaving group. However, studies show that an \(Mg^{+2}\) ion is bound in the active site close to the hydroxide. Explain how the presence of the magnesium ion contributes to the viability of hydroxide as a leaving group. Draw a complete mechanism for the reaction, assuming an \(S_N1\) pathway. P8.18: The reaction below is part of the biosynthesis of peptidoglycan, a major component of bacterial cell walls. Is it likely to proceed by a nucleophilic substitution mechanism? Explain. P8.19: Compare the reaction below, catalyzed by the enzyme AMP-DMAPP transferase, to the protein prenyltransferase reaction we learned about in section 8.8, the mechanism of which, as we discussed, is thought to be mostly \(S_N2\)-like with some \(S_N1\)-like character. Is the AMP-DMAPP transferase reaction below likely to have more or less \(S_N1\)-like character compared to the protein prenyltransferase reaction? Explain. Given your answer to part (a), which reaction is likely to be more dramatically slowed down when a fluorinated isoprenoid substrate analog is substituted for the natural substrate? Explain. P8.20: In a classic experiment in physical organic chemistry, (\(R\))-2-iodooctane was allowed to react (non-enzymatically) with a radioactive isotope of iodide ion, and the researchers monitored how fast the radioactive iodide was incorporated into the alkane (the rate constant of incorporation, \(k_i\)) and also how fast optical activity was lost (the rate constant of racemization, \(k_r\)). They found that the rate of racemization was, within experimental error, equal to twice the rate of incorporation. Discuss the significance of this result - what does it say about the actual mechanism of the reaction?
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Bookshelves/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/2%3A_Atoms_Isotopes_and_Mols
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2.10 Conversions Between Grams, Mol, and Atoms (Video) 2.11 Periodic Law in the Periodic Table (Video) 2.12 Different Groups in the Periodic Table 2.13 Predicting the Charge of Atoms Using the Periodic Table (Video) 2.1 Fundamental Experiments in Chemistry (Video) 2.2 The Nuclear Atom (Video) 2.3 How Elements Are Represented on the Periodic Table (Video) 2.4 Ions of Atoms (Video) 2.5 Isotopes of Elements (Video) 2.6 Determining the Number of Electrons, Protons, and Neutrons in an Atom (Video) 2.7 Mass Numbers and Atomic Mass of Elements (Video) 2.8 Finding the Averaged Atomic Weight of an Element (Video) 2.9 Determining the Molar Mass of a Molecule (Video)
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Courses/Louisville_Collegiate_School/General_Chemistry/LibreTexts_Louisville_Collegiate_School_Chapters_18%3A_Representative_Metals_Metalloids_and_Nonmetals/LibreTexts%2F%2FLouisville_Collegiate_School%2F%2FChapters%2F%2F18%3A_Representative_Metals%2C_Metalloids%2C_and_Nonmetals%2F%2F18.06%3A_Occurrence%2C_Preparation%2C_and_Properties_of_Carbonates
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Learning Objectives Describe the preparation, properties, and uses of some representative metal carbonates The chemistry of carbon is extensive; however, most of this chemistry is not relevant to this chapter. The other aspects of the chemistry of carbon will appear in the chapter covering organic chemistry. In this chapter, we will focus on the carbonate ion and related substances. The metals of groups 1 and 2, as well as zinc, cadmium, mercury, and lead(II), form ionic carbonates —compounds that contain the carbonate anions, \(\ce{CO3^2-}\). The metals of group 1, magnesium, calcium, strontium, and barium also form hydrogen carbonates —compounds that contain the hydrogen carbonate anion, \(\ce{HCO3-}\), also known as the bicarbonate anion . With the exception of magnesium carbonate, it is possible to prepare carbonates of the metals of groups 1 and 2 by the reaction of carbon dioxide with the respective oxide or hydroxide. Examples of such reactions include: \[ \begin{align} \ce{Na2O}(s)+\ce{CO2}(g) &⟶ \ce{Na2CO3}(s)\\[4pt] \ce{Ca(OH)2}(s)+\ce{CO2}(g) &⟶\ce{CaCO3}(s)+\ce{H2O}(l) \end{align} \nonumber \] The carbonates of the alkaline earth metals of group 12 and lead(II) are not soluble. These carbonates precipitate upon mixing a solution of soluble alkali metal carbonate with a solution of soluble salts of these metals. Examples of net ionic equations for the reactions are: \[ \begin{align} \ce{Ca^2+}(aq)+\ce{CO3^2-}(aq) &⟶\ce{CaCO3}(s) \\[4pt] \ce{Pb^2+}(aq)+\ce{CO3^2-}(aq) &⟶\ce{PbCO3}(s) \end{align} \nonumber \] Pearls and the shells of most mollusks are calcium carbonate. Tin(II) or one of the trivalent or tetravalent ions such as Al 3 + or Sn 4 + behave differently in this reaction as carbon dioxide and the corresponding oxide form instead of the carbonate. Alkali metal hydrogen carbonates such as NaHCO 3 and CsHCO 3 form by saturating a solution of the hydroxides with carbon dioxide. The net ionic reaction involves hydroxide ion and carbon dioxide: \[\ce{OH-}(aq)+\ce{CO2}(aq)⟶\ce{HCO3-}(aq) \nonumber \] It is possible to isolate the solids by evaporation of the water from the solution. Although they are insoluble in pure water, alkaline earth carbonates dissolve readily in water containing carbon dioxide because hydrogen carbonate salts form. For example, caves and sinkholes form in limestone when CaCO 3 dissolves in water containing dissolved carbon dioxide: \[\ce{CaCO3}(s)+\ce{CO2}(aq)+\ce{H2O}(l)⟶\ce{Ca^2+}(aq)+\ce{2HCO3-}(aq) \nonumber \] Hydrogen carbonates of the alkaline earth metals remain stable only in solution; evaporation of the solution produces the carbonate. Stalactites and stalagmites, like those shown in Figure \(\PageIndex{1}\), form in caves when drops of water containing dissolved calcium hydrogen carbonate evaporate to leave a deposit of calcium carbonate. The two carbonates used commercially in the largest quantities are sodium carbonate and calcium carbonate. In the United States, sodium carbonate is extracted from the mineral trona, Na 3 (CO 3 )(HCO 3 )(H 2 O) 2 . Following recrystallization to remove clay and other impurities, heating the recrystallized trona produces Na 2 CO 3 : \[\ce{2Na3(CO3)(HCO3)(H2O)2}(s)⟶\ce{3Na2CO3}(s)+\ce{5H2O}(l)+\ce{CO2}(g) \nonumber \] Carbonates are moderately strong bases. Aqueous solutions are basic because the carbonate ion accepts hydrogen ion from water in this reversible reaction: \[\ce{CO3^2-}(aq)+\ce{H2O}(l)⇌\ce{HCO3-}(aq)+\ce{OH-}(aq) \nonumber \] Carbonates react with acids to form salts of the metal, gaseous carbon dioxide, and water. The reaction of calcium carbonate, the active ingredient of the antacid Tums, with hydrochloric acid (stomach acid), as shown in Figure \(\PageIndex{2}\), illustrates the reaction: \[\ce{CaCO3}(s)+\ce{2HCl}(aq)⟶\ce{CaCl2}(aq)+\ce{CO2}(g)+\ce{H2O}(l) \nonumber \] Other applications of carbonates include glass making—where carbonate ions serve as a source of oxide ions—and synthesis of oxides. Hydrogen carbonates are amphoteric because they act as both weak acids and weak bases. Hydrogen carbonate ions act as acids and react with solutions of soluble hydroxides to form a carbonate and water: \[\ce{KHCO3}(aq)+\ce{KOH}(aq)⟶\ce{K2CO3}(aq)+\ce{H2O}(l) \nonumber \] With acids, hydrogen carbonates form a salt, carbon dioxide, and water. Baking soda (bicarbonate of soda or sodium bicarbonate) is sodium hydrogen carbonate. Baking powder contains baking soda and a solid acid such as potassium hydrogen tartrate (cream of tartar), KHC 4 H 4 O 6 . As long as the powder is dry, no reaction occurs; immediately after the addition of water, the acid reacts with the hydrogen carbonate ions to form carbon dioxide: \[\ce{HC4H4O6-}(aq)+\ce{HCO3-}(aq)⟶\ce{C4H4O6^2-}(aq)+\ce{CO2}(g)+\ce{H2O}(l) \nonumber \] Dough will trap the carbon dioxide, causing it to expand during baking, producing the characteristic texture of baked goods. Summary The usual method for the preparation of the carbonates of the alkali and alkaline earth metals is by reaction of an oxide or hydroxide with carbon dioxide. Other carbonates form by precipitation. Metal carbonates or hydrogen carbonates such as limestone (CaCO 3 ), the antacid Tums (CaCO 3 ), and baking soda (NaHCO 3 ) are common examples. Carbonates and hydrogen carbonates decompose in the presence of acids and most decompose on heating. Glossary bicarbonate anion salt of the hydrogen carbonate ion, \(\ce{HCO3-}\) carbonate salt of the anion \(\ce{CO3^2-}\); often formed by the reaction of carbon dioxide with bases hydrogen carbonate salt of carbonic acid, H 2 CO 3 (containing the anion \(\ce{HCO3-}\)) in which one hydrogen atom has been replaced; an acid carbonate; also known as bicarbonate ion
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Courses/Athabasca_University/Chemistry_360%3A_Organic_Chemistry_II/Chapter_21%3A_Carboxylic_Acid_Derivatives%3A_Nucleophilic_Acyl_Substitution_Reactions/21.04_Chemistry_of_Acid_Halides
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Objectives After completing this section, you should be able to identify the reagent normally used to convert a carboxylic acid to an acid bromide. write equations to show how an acid halide may be converted into each of the following: a carboxylic acid, an ester, an amide. write a detailed mechanisms for the reaction of an acid halide with each of the following: water, an alcohol, ammonia, a primary or secondary amine. identify the product formed when a given acid halide reacts with any of the following reagents: water, an alcohol, a primary or secondary amine. identify the acid halide, the reagents, or both, needed to prepare a given carboxylic acid, ester or amide. identify the product formed when a given acid halide reacts with water, a given alcohol, ammonia, or a given primary or secondary amine. identify lithium aluminum hydride as a reagent for reducing acid halides to primary alcohols, and explain the limited practical value of this reaction. identify the partial reduction of an acid halide using lithium tri‑ tert ‑butoxyaluminum to form an aldehyde. write an equation to describe the formation of a tertiary alcohol by the reaction of an acid halide with a Grignard reagent. write a detailed mechanism for the reaction of an acid halide with a Grignard reagent. identify the product formed from the reaction of a given acid halide with a given Grignard reagent. identify the acid halide, the Grignard reagent, or both, needed to prepare a given tertiary alcohol. write an equation to illustrate the reaction of an acid halide with a lithium diorganocopper reagent. identify the product formed from the reaction of a given acid halide with a given lithium diorganocopper reagent. identify the acid halide, the lithium diorganocopper reagent, or both, that must be used to prepare a given ketone. Study Notes This figure provides a convenient general summary of a few of the reactions described in Section 21.4. Note that LiAlH 4 is a common reagent for hydride [H − ] reduction of and acid chloride to an alcohol. Formation of Acid Halides Carboxylic acids react with thionyl chloride (\(SOCl_2\)) to form acid chlorides. During the reaction the hydroxyl group of the carboxylic acid is converted to a chlorosulfite intermediate making it a better leaving group. The chloride anion produced during the reaction acts a nucleophile. Earlier (Section 10.5) we saw that primary and secondary alcohols react with PBr 3 to afford the corresponding alkyl bromide. In a similar fashion acid bromides can be formed from the carboxylic acid. Nucleophilic Acyl Substitution Mechanism If you understand the mechanism of a typical nucleophilic acyl substitution, the reaction of an acyl halide with water, an alcohol or ammonia should not present you with any difficulty. X = Cl, Br :Nu = H 2 O, ROH, NH 2 R, NHR 2 etc. Acid chlorides react with water to form carboxylic acids. General reaction Example 21.4.1 Mechanism 1) Nucleophilic attack by water 2) Leaving group is removed 3) Deprotonation Acid chlorides react with carboxylic acids to form anhydrides Acid chlorides react with alcohols to form esters Example 21.4.2 Acid chlorides react with ammonia, 1 o amines and 2 o amines to form amides. Example 21.4.3 Acid Chlorides can be Reduced to form 1 o Alcohols Grignard reagents convert acid chloride to 3 o alcohols Organocuprate reagents convert acid chlorides to ketones. Example 21.4.4 Exercises Exercise 21.4.1 Draw the mechanism for the following reaction. Answer Exercise 21.4.2 Propose a synthesis of the following molecules from an acid chloride and an amide. Answer a) Acetyl chloride and dimethylamine b) Benzoyl chloride and ethylamine c) Acetyl chloride and ammonia
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Courses/Madera_Community_College/MacArthur_Chemistry_3A_v_1.2/10%3A_Aqueous_Solutions/10.02%3A_Measures_of_Concentration/10.2.01%3A_Percent_Solutions
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There are human cultures that do not recognize numbers above three. Anything greater than that is simply referred to as "much" or "many". Although this form of calculation may seem very limited, American culture does the same thing to a certain degree. For example, there are several ways to express the amount of solute in a solution in a quantitative manner. The concentration of a solution is a measure of the amount of solute that has been dissolved in a given amount of solvent or solution. A concentrated solution is one that has a relatively large amount of dissolved solute. A dilute solution is one that has a relatively small amount of dissolved solute. However, those terms are vague, and it is often necessary to express concentration with numbers. Percent Solutions One way to describe the concentration of a solution is by the percent of a solute in the solvent. The percent can further be determined in one of two ways: (1) the ratio of the mass of the solute divided by the mass of the solution or (2) the ratio of the volume of the solute divided by the volume of the solution. Mass Percent When the solute in a solution is a solid, a convenient way to express the concentration is by mass percent \(\left( \frac{\text{mass}}{\text{mass}} \right)\), which is the grams of solute per \(100 \: \text{g}\) of solution. \[\text{Percent by mass} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100\%\nonumber \] Suppose that a solution was prepared by dissolving \(25.0 \: \text{g}\) of sugar into \(100 \: \text{g}\) of water. The percent by mass would be calculated by: \[\text{Percent by mass} = \frac{25 \: \text{g sugar}}{125 \: \text{g solution}} \times 100\% = 20\% \: \text{sugar}\nonumber \] Sometimes you may want to make up a particular mass of solution of a given percent by mass, and need to calculate what mass of the solvent to use. For example, you need to make \(3000 \: \text{g}\) of a \(5\%\) solution of sodium chloride. You can rearrange and solve for the mass of solute: \[\text{mass of solute} = \frac{\text{percent by mass}}{100\%} \times \text{mass of solution} = \frac{5\%}{100\%} \times 3000 \: \text{g} = 150 \: \text{g} \: \ce{NaCl}\nonumber \] You would need to weigh out \(150 \: \text{g}\) of \(\ce{NaCl}\) and add it to \(2850 \: \text{g}\) of water. Notice that it was necessary to subtract the mass of the \(\ce{NaCl}\) \(\left( 150 \: \text{g} \right)\) from the mass of solution \(\left( 3000 \: \text{g} \right)\) to calculate the mass of the water that would need to be added. Volume Percent The percentage of solute in a solution can more easily be determined by volume when the solute and solvent are both liquids. The volume of the solute divided by the volume of the solution, expressed as a percent, yields the percent by volume \(\left( \frac{\text{volume}}{\text{volume}} \right)\) of the solution. If a solution is made by adding \(40 \: \text{mL}\) of ethanol to \(200 \: \text{mL}\) of water, the percent by volume is: \[\begin{align*} \text{Percent by volume} &= \frac{\text{volume of solute}}{\text{volume of solution}} \times 100\% \\[8px] &= \frac{40 \: \text{mL ethanol}}{240 \: \text{mL solution}} \times 100\% \\[8px] &= 16.7\% \: \text{ethanol} \end{align*}\nonumber \] Frequently, ingredient labels on food products and medicines have amounts listed as percentages (see figure below). Summary The concentration of a solution is a measure of the amount of solute that has been dissolved in a given amount of solvent or solution. A concentrated solution has a relatively large amount of dissolved solute. A dilute solution has a relatively small amount of dissolved solute. Techniques for calculation of percent mass and percent volume solution concentrations are described.
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Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/15%3A_Principles_of_Chemical_Equilibrium/15.2%3A_The_Equilibrium_Constant_Expression
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Learning Objectives To know the relationship between the equilibrium constant and the rate constants for the forward and reverse reactions. To write an equilibrium constant expression for any reaction. Because an equilibrium state is achieved when the forward reaction rate equals the reverse reaction rate, under a given set of conditions there must be a relationship between the composition of the system at equilibrium and the kinetics of a reaction (represented by rate constants). We can show this relationship using the system described in Equation \ref{15.2.1}, the decomposition of \(N_2O_4\) to \(NO_2\). Both the forward and reverse reactions for this system consist of a single elementary reaction, so the reaction rates are as follows: \[\text{forward rate} = k_f[N_2O_4] \label{15.2.1}\] and \[\text{reverse rate} = k_r[NO_2]^2 \label{15.2.2}\] At equilibrium, the forward rate equals the reverse rate (definition of equilibrium): \[ k_f[N_2O_4] = k_r[NO_2]^2 \label{15.2.3}\] so \[\dfrac{k_f}{k_r}=\dfrac{[NO_2]^2}{[N_2O_4]} \label{15.2.4}\] The ratio of the rate constants gives us a new constant, the equilibrium constant (\(K\)), which is defined as follows: \[K=\dfrac{k_f}{k_r} \label{15.2.5}\] Hence there is a fundamental relationship between chemical kinetics and chemical equilibrium: under a given set of conditions, the composition of the equilibrium mixture is determined by the magnitudes of the rate constants for the forward and the reverse reactions. The equilibrium constant is equal to the rate constant for the forward reaction divided by the rate constant for the reverse reaction. Table \(\PageIndex{1}\) lists the initial and equilibrium concentrations from five different experiments using the reaction system described by Equation \ref{15.1}. At equilibrium the magnitude of the quantity \([NO_2]^2/[N_2O_4]\) is essentially the same for all five experiments. In fact, no matter what the initial concentrations of \(NO_2\) and \(N_2O_4\) are, at equilibrium the quantity \([NO_2]^2/[N_2O_4]\) will always be \(6.53 \pm 0.03 \times 10^{−3}\) at 25°C, which corresponds to the ratio of the rate constants for the forward and reverse reactions. That is, at a given temperature, the equilibrium constant for a reaction always has the same value, even though the specific concentrations of the reactants and products vary depending on their initial concentrations. Unnamed: 0_level_0 Initial Concentrations Concentrations at Equilibrium Unnamed: 5_level_0 Experiment [\(N_2O_4\)] (M) [\(NO_2\)] (M) [\(N_2O_4\)] (M) [\(NO_2\)] (M) \(K = [NO_2]^2/[N_2O_4]\) 1 0.050 0.000 0.0417 0.0165 \(6.54 \times 10^{−3}\) 2 0.000 0.100 0.0417 0.0165 \(6.54 \times 10^{−3}\) 3 0.075 0.000 0.0647 0.0206 \(6.56 \times 10^{−3}\) 4 0.000 0.075 0.0304 0.0141 \(6.54 \times 10^{−3}\) 5 0.025 0.075 0.0532 0.0186 \(6.50 \times 10^{−3}\) Introduction to Dynamic Equilibrium: https://youtu.be/4AJbFuzW2cs Developing an Equilibrium Constant Expression In 1864, the Norwegian chemists Cato Guldberg (1836–1902) and Peter Waage (1833–1900) carefully measured the compositions of many reaction systems at equilibrium. They discovered that for any reversible reaction of the general form \[aA+bB \rightleftharpoons cC+dD \label{15.2.6}\] where A and B are reactants, C and D are products, and a, b, c, and d are the stoichiometric coefficients in the balanced chemical equation for the reaction, the ratio of the product of the equilibrium concentrations of the products (raised to their coefficients in the balanced chemical equation) to the product of the equilibrium concentrations of the reactants (raised to their coefficients in the balanced chemical equation) is always a constant under a given set of conditions. This relationship is known as the law of mass action and can be stated as follows: \[K=\dfrac{[C]^c[D]^d}{[A]^a[B]^b} \label{15.2.7}\] where \(K\) is the equilibrium constant for the reaction. Equation \ref{15.2.6} is called the equilibrium equation, and the right side of Equation \ref{15.2.7} is called the equilibrium constant expression. The relationship shown in Equation \ref{15.2.7} is true for any pair of opposing reactions regardless of the mechanism of the reaction or the number of steps in the mechanism. The equilibrium constant can vary over a wide range of values. The values of \(K\) shown in Table \(\PageIndex{2}\), for example, vary by 60 orders of magnitude. Because products are in the numerator of the equilibrium constant expression and reactants are in the denominator, values of K greater than \(10^3\) indicate a strong tendency for reactants to form products. In this case, chemists say that equilibrium lies to the right as written, favoring the formation of products. An example is the reaction between \(H_2\) and \(Cl_2\) to produce \(HCl\), which has an equilibrium constant of \(1.6 \times 10^{33}\) at 300 K. Because \(H_2\) is a good reductant and \(Cl_2\) is a good oxidant, the reaction proceeds essentially to completion. In contrast, values of \(K\) less than \(10^{-3}\) indicate that the ratio of products to reactants at equilibrium is very small. That is, reactants do not tend to form products readily, and the equilibrium lies to the left as written, favoring the formation of reactants. Reaction Temperature (K) Equilibrium Constant (K) \(\ce{S(s) + O2(g) \rightleftharpoons SO2(g)}\) 300 \(4.4 \times 10^{53}\) \(\ce{2H2(g) + O2(g) \rightleftharpoons 2H2O (g)}\) 500 \(2.4 \times 10^{47}\) \(\ce{H2(g) + Cl2(g) \rightleftharpoons 2 HCl(g)}\) 300 \(1.6 \times 10^{33}\) \(\ce{H2(g) + Br2(g) \rightleftharpoons 2HBr(g)}\) 300 \(4.1 \times 10^{18}\) \(\ce{2NO(g) + O2(g) \rightleftharpoons 2NO2(g)}\) 300 \(4.2 \times 10^{13}\) \(\ce{3H2(g) + N2(g) \rightleftharpoons 2NH3(g)}\) 300 \(2.7 \times 10^{8}\) \(\ce{H2(g) + D2(g) \rightleftharpoons 2HD(g)}\) 100 \(1.92\) \(\ce{H2(g) + I2(g) \rightleftharpoons 2HI(g)}\) 300 \(2.9 \times 10^{−1}\) \(\ce{I2(g) \rightleftharpoons 2I(g)}\) 800 \(4.6 \times^{ 10−7}\) \(\ce{Br2(g) \rightleftharpoons 2Br(g)}\) 1000 \(4.0 \times 10^{−7}\) \(\ce{Cl2(g) \rightleftharpoons 2Cl (g)}\) 1000 \(1.8 \times 10^{−9}\) \(\ce{F2(g) \rightleftharpoons 2F(g)}\) 500 \(7.4 \times 10^{−13}\) *Equilibrium constants vary with temperature. The K values shown are for systems at the indicated temperatures. *Equilibrium constants vary with temperature. The K values shown are for systems at the indicated temperatures. *Equilibrium constants vary with temperature. The K values shown are for systems at the indicated temperatures. You will also notice in Table \(\PageIndex{2}\) that equilibrium constants have no units, even though Equation \ref{15.2.7} suggests that the units of concentration might not always cancel because the exponents may vary. In fact, equilibrium constants are calculated using “effective concentrations,” or activities, of reactants and products, which are the ratios of the measured concentrations to a standard state of 1 M. As shown in Equation \ref{15.2.8}, the units of concentration cancel, which makes \(K\) unitless as well: \[ \dfrac{[A]_{measured}}{[A]_{standard\; state}}=\dfrac{\cancel{M}}{\cancel{M}} = \dfrac{\cancel{\frac{mol}{L}}}{\cancel{\frac{mol}{L}}} \label{15.2.8}\] Many reactions have equilibrium constants between 1000 and 0.001 (\(10^3 \ge K \ge 10^{−3}\)), neither very large nor very small. At equilibrium, these systems tend to contain significant amounts of both products and reactants, indicating that there is not a strong tendency to form either products from reactants or reactants from products. An example of this type of system is the reaction of gaseous hydrogen and deuterium, a component of high-stability fiber-optic light sources used in ocean studies, to form HD: \[H_{2(g)}+D_{2(g)} \rightleftharpoons 2HD_{(g)} \label{15.2.9}\] The equilibrium constant expression for this reaction is \[K= \dfrac{[HD]^2}{[H_2][D_2]}\] with \(K\) varying between 1.9 and 4 over a wide temperature range (100–1000 K). Thus an equilibrium mixture of \(H_2\), \(D_2\), and \(HD\) contains significant concentrations of both product and reactants. Figure \(\PageIndex{3}\) summarizes the relationship between the magnitude of K and the relative concentrations of reactants and products at equilibrium for a general reaction, written as \[\text{reactants} \rightleftharpoons \text{products}.\] Because there is a direct relationship between the kinetics of a reaction and the equilibrium concentrations of products and reactants (Equations \ref{15.2.8} and \ref{15.2.7}), when \(k_f \gg k_r\), \(K\) is a large number, and the concentration of products at equilibrium predominate. This corresponds to an essentially irreversible reaction. Conversely, when \(k_f \ll k_r\), \(K\) is a very small number, and the reaction produces almost no products as written. Systems for which \(k_f ≈ k_r\) have significant concentrations of both reactants and products at equilibrium. A large value of the equilibrium constant \(K\) means that products predominate at equilibrium; a small value means that reactants predominate at equilibrium. Example \(\PageIndex{1}\) Write the equilibrium constant expression for each reaction. \(N_{2(g)}+3H_{2(g)} \rightleftharpoons 2NH_{3(g)}\) \(CO_{(g)}+\frac{1}{2}O_{2(g)} \rightleftharpoons CO_{2(g)}\) \(2CO_{2(g)} \rightleftharpoons 2CO_{(g)}+O_{2(g)}\) Given : balanced chemical equations Asked for : equilibrium constant expressions Strategy : Refer to Equation \ref{15.2.7}. Place the arithmetic product of the concentrations of the products (raised to their stoichiometric coefficients) in the numerator and the product of the concentrations of the reactants (raised to their stoichiometric coefficients) in the denominator. Solution : The only product is ammonia, which has a coefficient of 2. For the reactants, \(N_2\) has a coefficient of 1 and H2 has a coefficient of 3. The equilibrium constant expression is as follows: \[\dfrac{[NH_3]^2}{[N_2][H_2]^3}\] The only product is carbon dioxide, which has a coefficient of 1. The reactants are \(CO\), with a coefficient of 1, and \(O_2\), with a coefficient of \(\frac{1}{2}\). Thus the equilibrium constant expression is as follows: \[\dfrac{[CO_2]}{[CO][O_2]^{1/2}}\] This reaction is the reverse of the reaction in part b, with all coefficients multiplied by 2 to remove the fractional coefficient for \(O_2\). The equilibrium constant expression is therefore the inverse of the expression in part b, with all exponents multiplied by 2: \[\dfrac{[CO]^2[O_2]}{[CO_2]^2}\] Exercise \(\PageIndex{1}\) Write the equilibrium constant expression for each reaction. \(N_2O_{(g)} \rightleftharpoons N_{2(g)}+\frac{1}{2}O_{2(g)}\) \(2C_8H_{18(g)}+25O_{2(g)} \rightleftharpoons 16CO_{2(g)}+18H_2O_{(g)}\) \(H_{2(g)}+I_{2(g)} \rightleftharpoons 2HI_{(g)}\) Answer: \(K=\dfrac{[N_2][O_2]^{1/2}}{[N_2O]}\) \(K=\dfrac{[CO_2]^{16}[H_2O]^{18}}{[C_8H_{18}]^2[O_2]^{25}}\) \(K=\dfrac{[HI]^2}{[H_2][I_2]}\) Example \(\PageIndex{2}\) Predict which systems at equilibrium will (a) contain essentially only products, (b) contain essentially only reactants, and (c) contain appreciable amounts of both products and reactants. \(H_{2(g)}+I_{2(g)} \rightleftharpoons 2HI_{(g)}\;\;\; K_{(700K)}=54\) \(2CO_{2(g)} \rightleftharpoons 2CO_{(g)}+O_{2(g)}\;\;\; K_{(1200K)}=3.1 \times 10^{−18}\) \(PCl_{5(g)} \rightleftharpoons PCl_{3(g)}+Cl_{2(g)}\;\;\; K_{(613K)}=97\) \(2O_{3(g)} \rightleftharpoons 3O_{2(g)} \;\;\; K_{(298 K)}=5.9 \times 10^{55}\) Given : systems and values of \(K\) Asked for : composition of systems at equilibrium Strategy : Use the value of the equilibrium constant to determine whether the equilibrium mixture will contain essentially only products, essentially only reactants, or significant amounts of both. Solution : Only system 4 has \(K \gg 10^3\), so at equilibrium it will consist of essentially only products. System 2 has \(K \ll 10^{−3}\), so the reactants have little tendency to form products under the conditions specified; thus, at equilibrium the system will contain essentially only reactants. Both systems 1 and 3 have equilibrium constants in the range \(10^3 \ge K \ge 10^{−3}\), indicating that the equilibrium mixtures will contain appreciable amounts of both products and reactants. Exercise \(\PageIndex{2}\) Hydrogen and nitrogen react to form ammonia according to the following balanced chemical equation: \[3H_{2(g)}+N_{2(g)} \rightleftharpoons 2NH_{3(g)}\] Values of the equilibrium constant at various temperatures were reported as \(K_{25°C} = 3.3 \times 10^8\), \(K_{177°C} = 2.6 \times 10^3\), and \(K_{327°C} = 4.1\). At which temperature would you expect to find the highest proportion of \(H_2\) and \(N_2\) in the equilibrium mixture? Assuming that the reaction rates are fast enough so that equilibrium is reached quickly, at what temperature would you design a commercial reactor to operate to maximize the yield of ammonia? Answer : 327°C, where \(K\) is smallest 25°C Variations in the Form of the Equilibrium Constant Expression Because equilibrium can be approached from either direction in a chemical reaction, the equilibrium constant expression and thus the magnitude of the equilibrium constant depend on the form in which the chemical reaction is written. For example, if we write the reaction described in Equation 15.2.6 in reverse, we obtain the following: \[cC+dD \rightleftharpoons aA+bB \label{15.2.10}\] The corresponding equilibrium constant \(K′\) is as follows: \[K'=\dfrac{[A]^a[B]^b}{[C]^c[D]^d} \label{15.2.11}\] This expression is the inverse of the expression for the original equilibrium constant, so \(K′ = 1/K\). That is, when we write a reaction in the reverse direction, the equilibrium constant expression is inverted. For instance, the equilibrium constant for the reaction \(N_2O_4\) \rightleftharpoons 2NO_2\) is as follows: \[K=\dfrac{[NO_2]^2}{[N_2O_4]} \label{15.2.12}\] but for the opposite reaction, \(2 NO_2 \rightleftharpoons N_2O_4\), the equilibrium constant K′ is given by the inverse expression: \[K'=\dfrac{[N_2O_4]}{[NO_2]^2} \label{15.2.13}\] Consider another example, the formation of water: \(2H_{2(g)}+O_{2(g)} \rightleftharpoons 2H_2O_{(g)}\). Because \(H_2\) is a good reductant and \(O_2\) is a good oxidant, this reaction has a very large equilibrium constant (\(K = 2.4 \times 10^{47}\) at 500 K). Consequently, the equilibrium constant for the reverse reaction, the decomposition of water to form \(O_2\) and \(H_2\), is very small: \(K′ = 1/K = 1/(2.4 \times 10^{47}) = 4.2 \times 10^{−48}\). As suggested by the very small equilibrium constant, and fortunately for life as we know it, a substantial amount of energy is indeed needed to dissociate water into \(H_2\) and \(O_2\). The equilibrium constant for a reaction written in reverse is the inverse of the equilibrium constant for the reaction as written originally. Writing an equation in different but chemically equivalent forms also causes both the equilibrium constant expression and the magnitude of the equilibrium constant to be different. For example, we could write the equation for the reaction \[2NO_2 \rightleftharpoons N_2O_4\] as \[NO_2 \rightleftharpoons \frac{1}{2}N_2O_4\] with the equilibrium constant K″ is as follows: \[ K′′=\dfrac{[N_2O_4]^{1/2}}{[NO_2]} \label{15.2.14}\] The values for K′ (Equation \ref{15.2.13}) and K″ are related as follows: \[ K′′=(K')^{1/2}=\sqrt{K'} \label{15.2.15}\] In general, if all the coefficients in a balanced chemical equation were subsequently multiplied by \(n\), then the new equilibrium constant is the original equilibrium constant raised to the \(n^{th}\) power. Example \(\PageIndex{3}\): The Haber Process At 745 K, K is 0.118 for the following reaction: \[\ce{N2(g) + 3H2(g) <=> 2NH3(g)} \nonumber\] What is the equilibrium constant for each related reaction at 745 K? \(2NH_{3(g)} \rightleftharpoons N2(g)+3H_{2(g)}\) \(\frac{1}{2}N_{2(g)}+\frac{3}{2}H_{2(g)} \rightleftharpoons NH_{3(g)}\) Given : balanced equilibrium equation, K at a given temperature, and equations of related reactions Asked for : values of \(K\) for related reactions Strategy : Write the equilibrium constant expression for the given reaction and for each related reaction. From these expressions, calculate \(K\) for each reaction. Solution : The equilibrium constant expression for the given reaction of \(N_{2(g)}\) with \(H_{2(g)}\) to produce \(NH_{3(g)}\) at 745 K is as follows: \[K=\dfrac{[NH_3]^2}{[N_2][H_2]^3}=0.118\] This reaction is the reverse of the one given, so its equilibrium constant expression is as follows: \[K'=\dfrac{1}{K}=\dfrac{[N_2][H_2]^3}{[NH_3]^2}=\dfrac{1}{0.118}=8.47\] In this reaction, the stoichiometric coefficients of the given reaction are divided by 2, so the equilibrium constant is calculated as follows: \[K′′=\dfrac{[NH_3]}{[N_2]^{1/2}[H_2]^{3/2}}=K^{1/2}=\sqrt{K}=\sqrt{0.118} = 0.344\] Exercise \(\PageIndex{3}\) At 527°C, the equilibrium constant for the reaction \[2SO_{2(g)}+O_{2(g)} \rightleftharpoons 2SO_{3(g)}\] is \(7.9 \times 10^4\). Calculate the equilibrium constant for the following reaction at the same temperature: \[SO_{3(g)} \rightleftharpoons SO_{2(g)}+\frac{1}{2}O_{2(g)}\] Answer : \(3.6 \times 10^{−3}\) Determining the Equilibrium Expression: https://youtu.be/ZK9cMIWFerY Summary The law of mass action describes a system at equilibrium in terms of the concentrations of the products and the reactants. For a system involving one or more gases, either the molar concentrations of the gases or their partial pressures can be used. Definition of equilibrium constant in terms of forward and reverse rate constants: \[K=\dfrac{k_f}{k_r} \nonumber\] Equilibrium constant expression (law of mass action): \[K=\dfrac{[C]^c[D]^d}{[A]^a[B]^b} \nonumber\] Equilibrium constant expression for reactions involving gases using partial pressures: \[K_p=\dfrac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b} \nonumber\] Relationship between \(K_p\) and \(K\): \[K_p = K(RT)^{Δn} \nonumber \] The ratio of the rate constants for the forward and reverse reactions at equilibrium is the equilibrium constant (K), a unitless quantity. The composition of the equilibrium mixture is therefore determined by the magnitudes of the forward and reverse rate constants at equilibrium. Under a given set of conditions, a reaction will always have the same \(K\). For a system at equilibrium, the law of mass action relates \(K\) to the ratio of the equilibrium concentrations of the products to the concentrations of the reactants raised to their respective powers to match the coefficients in the equilibrium equation. The ratio is called the equilibrium constant expression. When a reaction is written in the reverse direction, \(K\) and the equilibrium constant expression are inverted. For gases, the equilibrium constant expression can be written as the ratio of the partial pressures of the products to the partial pressures of the reactants, each raised to a power matching its coefficient in the chemical equation. An equilibrium constant calculated from partial pressures (\(K_p\)) is related to \(K\) by the ideal gas constant (\(R\)), the temperature (\(T\)), and the change in the number of moles of gas during the reaction. An equilibrium system that contains products and reactants in a single phase is a homogeneous equilibrium; a system whose reactants, products, or both are in more than one phase is a heterogeneous equilibrium. When a reaction can be expressed as the sum of two or more reactions, its equilibrium constant is equal to the product of the equilibrium constants for the individual reactions.
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Courses/Sacramento_City_College/SCC%3A_Chem_420_-_Organic_Chemistry_I/09%3A_Reactions_of_Alkenes
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Learning Objectives After reading this chapter and completing ALL the exercises, a student can b e able to draw the general Electrophilic Addition Reaction (EAR) mechanism for an alkene - refer to section 9.1 predict the products/specify the reagents for EAR of hydrohalic acids (HX) with symmetrical alkenes - refer to section 9.2 predict the products/specify the reagents for EAR of hydrohalic acids (HX) with asymmetrical alkenes using Markovnikov's Rule for Regioselectivity - refer to section 9.3 apply the principles of regioselectivity and stereoselectivity to the addition reactions of alkenes - refer to sections 9.3 - 9.14 predict the products, specify the reagents, and discern most efficient reaction for hydration of alkenes (acid catalyzed hydration; or oxymercuration/demercuration; or hydroboration/oxidation) - refer to sections 9.4, 9.5, and 9.6 respectively discern the stereochemical differences between the EAR of chiral and achiral alkenes - refer to sections 9.7 and 9.8 predict the products/specify the reagents for halogenation and hydrohalogenation of alkenes - refer to sections 9.9 and 9.10 respectively recognize organic oxidation and reduction reactions - refer to sections 9.11 and 9.12 predict the products/specify the reagents for hydrogenation (reduction) of alkenes - refer to section 9.11 predict the products/specify the reagents for epoxidation of alkenes - refer to section 9.12 predict the products/specify the reagents for dihydroxylation of alkenes - refer to sections 9.13 and 9.14 predict the products/specify the reagents for oxidative cleavage of alkenes - refer to section 9.15 predict the products of carbene additions to alkenes - refer to section 9.16 predict the polymer/specify the monomer for radical, chain -growth polymers of alkenes - refer to section 9.17 discuss an example biological addition reactions - refer to section 9.18 9.1: Electrophilic Addition Reactions (EARs) Electrophilic addition reactions can occur in compounds containing pi bonds like the alkenes. Depending on the structure of the alkene and the specific reagents, the reactions can be regioselective and/or stereoselective. 9.2: Addition of Hydrogen Halides to Symmetrical Alkenes The regioselective reaction of the carbon-carbon double bond in alkenes with hydrohalogens (HX) is a controlled by carbocation stability. Consequently, the symmetry of the alkene must be considered for this mechanistic pathway. 9.3: Alkene Asymmetry and Markovnikov's Rule The regioselectivity of electrophilic addition reactions is determined by carbocation stability and is summarized by Markovnikov's Rule. 9.4: Hydration- Acid Catalyzed Addition of Water Electrophilic hydration is the act of adding electrophilic hydrogen from a non-nucleophilic strong acid (a reusable catalyst, examples of which include sulfuric and phosphoric acid) and applying appropriate temperatures to break the alkene's double bond. After a carbocation is formed, water bonds with the carbocation to form a 1º, 2º, or 3º alcohol on the alkane. 9.5: Hydration- Oxymercuration-Demercuration Oxymercuration is a stereospecific, regioselective electrophilic addition raction because there are no carbocation rearrangements due to stabilization of the reactive intermediate. The Markovnikov products are reliably synthesized by this pathway. 9.6: Hydration- Hydroboration-Oxidation Hydroboration-oxidation of alkenes has been a very valuable laboratory method for the stereoselectivity and regioselectivity of alkenes that are the non-Markovnikov products for alkene hydration. 9.7: Stereochemistry of Reactions - Hydration of Achiral Alkenes For achiral alkenes, the symmetrical trigonal planar geometry of the carbocation leads to equivalent synthesis of both R and S products giving a racemic (50/50). 9.8: Stereochemistry of Reactions - Hydration of Chiral Alkenes Chiral alkenes form electrophilic addition products in non 50:50 ratios due to the differences in steric effects between the enantiomeric starting materials. 9.9: Addition of Halogens Halogens can act as electrophiles due to polarizability of their covalent bond and react with the pi bond of alkenes. This electrophilic addition mechanism is stereospecific. The orientation of the electrophile during a stereospecific electrophilic addition reaction will determine the stereochemistry of the product(s). 9.10: Formation of Halohydrins When the halogenation reaction of alkenes is performed in a nucleophilic solvent like water or alcohol, then the solvent becomes the nucleophile to give halohydrin or haloalkoxy products. 9.11: Reduction of Alkenes - Catalytic Hydrogenation Catalytic hydrogenation of a carbon-carbon double bond is a reduction reaction. The alkene orientation required for interaction with the surface of the catalyst means that t his reaction is stereospecific. 9.12: Oxidation of Alkenes - Epoxidation Oxidation of alkenes is introduced and the epoxidation of alkenes is discussed. 9.13: Dihydroxylation of Alkenes Alkenes can react to produce glycols (two adjacent hydroxyl groups) through either an anti- or syn- addition mechanism that is stereospecific. 9.14: Opening of Epoxides - Acidic versus Basic Conditions Ring-opening reactions can proceed by either SN2 or SN1 mechanisms, depending on the nature of the epoxide and on the reaction conditions. 9.15: Oxidative Cleavage of Alkenes Oxidative cleavage of alkenes can occur by several different pathways. The most common reagents and pathways are discussed in this section. 9.16: Addition of Carbenes to Alkenes - Cyclopropane Synthesis A carbene such as methlyene will react with an alkene breaking the pi bond to form a cyclopropane. Since methylene is highly reactive, it is prepared in situ immediately preceding the addition of the alkene. 9.17: Radical Chain-Growth Polymerization All the monomers from which addition polymers are made are alkenes or functionally substituted alkenes. The free radical mechanism for chain growth polymers is explained. 9.18: 9.18 Biological Additions of Radicals to Alkenes Some electrophilic addition reactions that take place in nature and the role of enzymes in such processes are introduced. 9.19: Additional Exercises This section has additional exercises for the key learning objectives of this chapter. 9.20: Solutions to Additional Exercises This section has the solutions to the additional exercises from the previous section.
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Courses/Taft_College/CHEM_1510%3A_Introductory_College_Chemistry/15%3A_Solutions/15.02%3A_Solutions_-_Homogeneous_Mixtures
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Learning Objectives Learn terminology involving solutions. Explain the significance of the statement "like dissolves like." Explain why certain substances dissolve in other substances. The major component of a solution is called the solvent. The minor component of a solution is called the solute . By major and minor we mean whichever component has the greater or lesser presence by mass or by moles. Sometimes this becomes confusing, especially with substances with very different molar masses. However, here we will confine the discussion to solutions for which the major component and the minor component are obvious. Solutions exist for every possible phase of the solute and the solvent. Salt water, for example, is a solution of solid \(\ce{NaCl}\) in liquid water, while air is a solution of a gaseous solute (\(\ce{O2}\)) in a gaseous solvent (\(\ce{N2}\)). In all cases, however, the overall phase of the solution is the same phase as the solvent. Table \(\PageIndex{1}\) lists some common types of solutions, with examples of each. Solvent Phase Solute Phase Example gas gas air liquid gas carbonated beverages liquid liquid ethanol (C2H5OH) in H2O (alcoholic beverages) liquid solid salt water solid gas H2 gas absorbed by Pd metal solid liquid Hg(ℓ) in dental fillings solid solid steel alloys Example \(\PageIndex{1}\): Sugar and Water A solution is made by dissolving 1.00 g of sucrose (\(\ce{C12H22O11}\)) in 100.0 g of liquid water. Identify the solvent and solute in the resulting solution. Solution Either by mass or by moles, the obvious minor component is sucrose , so it is the solute . Water —the majority component—is the solvent . The fact that the resulting solution is the same phase as water also suggests that water is the solvent. Exercise \(\PageIndex{1}\) A solution is made by dissolving 3.33 g of \(\ce{HCl(g)}\) in 40.0 g of liquid methyl alcohol (\(\ce{CH3OH}\)). Identify the solvent and solute in the resulting solution. Answer solute: \(\ce{HCl(g)}\) solvent: \(\ce{CH3OH}\) Like Dissolves Like A simple way to predict which compounds will dissolve in other compounds is the phrase "like dissolves like". What this means is that polar compounds dissolve polar compounds, nonpolar compounds dissolve nonpolar compounds, but polar and nonpolar do not dissolve in each other. Even some nonpolar substances dissolve in water but only to a limited degree. Have you ever wondered why fish are able to breathe? Oxygen gas, a nonpolar molecule, does dissolve in water—it is this oxygen that the fish take in through their gills. The reason we can enjoy carbonated sodas is also due to a nonpolar compound that dissolves in water. Pepsi-cola and all the other sodas have carbon dioxide gas, \(\ce{CO_2}\), a nonpolar compound, dissolved in a sugar-water solution. In this case, to keep as much gas in solution as possible, the sodas are kept under pressure. This general trend of "like dissolves like" is summarized in the following table: Solute (Polarity of Compound) Solvent (Polarity of Compound) Dominant Intermolecular Force Is Solution Formed? Polar Polar Dipole-Dipole Force and/or Hydrogen Bond yes Non-polar Non-polar Dispersion Force yes Polar Non-polar NaN no Non-polar Polar NaN no Ionic Polar Ion-Dipole yes Ionic Non-polar NaN no Note that every time charged particles (ionic compounds or polar substances) are mixed, a solution is formed. When particles with no charges (nonpolar compounds) are mixed, they will form a solution. However, if substances with charges are mixed with other substances without charges, a solution does not form. When an ionic compound is considered "insoluble", it doesn't necessarily mean the compound is completely untouched by water. All ionic compounds dissolve to some extent. An insoluble compound just doesn't dissolve in any noticeable or appreciable amount. What is it that makes a solute soluble in some solvents but not others? The answer is intermolecular interactions. The intermolecular interactions include London dispersion forces, dipole-dipole interactions, and hydrogen bonding (as described in Chapter 10 ). From experimental studies, it has been determined that if molecules of a solute experience the same intermolecular forces that the solvent does, the solute will likely dissolve in that solvent. So, NaCl—a very polar substance because it is composed of ions—dissolves in water, which is very polar, but not in oil, which is generally nonpolar. Nonpolar wax dissolves in nonpolar hexane, but not in polar water. Example \(\PageIndex{2}\): Polar and Nonpolar Solvents Would \(\ce{I2}\) be more soluble in \(\ce{CCl4}\) or \(\ce{H2O}\)? Explain your answer. Solution \(\ce{I2}\) is nonpolar. Of the two solvents, \(\ce{CCl4}\) is nonpolar and \(\ce{H2O}\) is polar, so \(\ce{I2}\) would be expected to be more soluble in \(\ce{CCl4}\). Exercise \(\PageIndex{2}\) Would \(\ce{C3H7OH}\) be more soluble in \(\ce{CCl4}\) or \(\ce{H2O}\)? Explain your answer. Answer \(\ce{H2O}\), because both experience hydrogen bonding. Example \(\PageIndex{3}\) Water is considered a polar solvent. Which substances should dissolve in water? methanol (\(\ce{CH3OH}\)) sodium sulfate (\ce{Na2SO4}\)) octane (\(\ce{C8H18}\)) Solution Because water is polar, substances that are polar or ionic will dissolve in it. Because of the OH group in methanol, we expect its molecules to be polar. Thus, we expect it to be soluble in water. As both water and methanol are liquids, the word miscible can be used in place of soluble . Sodium sulfate is an ionic compound, so we expect it to be soluble in water. Like other hydrocarbons, octane is nonpolar, so we expect that it would not be soluble in water. Exercise \(\PageIndex{3}\): Toluene Toluene (\(\ce{C6H5CH3}\)) is widely used in industry as a nonpolar solvent. Which substances should dissolve in toluene? water (H 2 O) sodium sulfate (Na 2 SO 4 ) octane (C 8 H 18 ) Answer Octane (\(\ce{C8H18}\)) will dissolve. It is also non-polar. Summary Solutions are composed of a solvent (major component) and a solute (minor component). “Like dissolves like” is a useful rule for deciding if a solute will be soluble in a solvent.
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Ancillary_Materials/Worksheets/Worksheets%3A_Inorganic_Chemistry/Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry/13%3A_Concepts_of_Acidity/13.14%3A_Direction_of_Proton_Transfer
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Bronsted-Lowry acidity is a special case of Lewis acidity. In Lewis acidity, an electron donor shares electrons with an electron acceptor, forming a bond. In some cases, the electron acceptor is a proton. If the proton accepts electrons from the donor, but does not relinquish its bond to its previous partner, the interaction is called a hydrogen bond. If the proton exchanges a new bond with the donor for an old bond with its previous partner -- if it releases a pair of electrons to its partned as it accepts a pair of electrons from the new donor -- the event is called proton transfer. The compound that provided the proton is called a Bronsted-Lowry acid. The compound that donated the new bond to the proton is called the Bronsted-Lowry base. Because the proton allowed its former partner to take the pair of electrons from their former bond, that partner becomes a Lewis base. In a proton transfer, the proton moves from one Lewis base to another. The proton could conceivably move back to its original partner, however. The original partner could simply donate its pair of electrons to the proton again. It would displace the new partner and win the proton back. This situation is called reversibility. Reactions can often move back and forth. In order to convey this idea, when illustrating or writing about these reactions, a pair of opposing arrows are used to show that the reaction can go from left to right as written, as well as from right to left. In a reversible reaction, the change that occurs in the reaction can be undone. The reaction can move forwards and backwards. In most cases, the reaction settles out on one side or the other. Either the reaction goes mostly forward or it goes mostly backward. The point where the reaction settles is termed the equilibrium. At equilibrium, the reactants that come together may mostly be converted into new products. Looking at an equation or diagram of the reaction, , the equilibrium is said to "lie to the right", because the products of the reaction are usually written on the right hand side of the reaction arrow or equilibrium arrow. Some reactions "lie to the left", meaning very little of the original reactants are ultimately converted to the products show. The equilibrium is the balance established between the products formed and the original reactants in a reversible reaction. In a proton transfer, the equilibrium is determined by how tightly the proton is held by each Bronsted acid. The proton will simply remain bonded to whichever compound binds it more tightly. If the difference in binding is great, the equilibrium will lie far to the left or far to the right. If the difference in proton binding is small, there will be a mixture, in which the proton could be in either position. Remember, most reactions involve zillions of molecules. There is plenty of room for mixtures. We can predict where the proton will end up by looking at pKas. A higher pKa means the proton is more tightly held. By comparing the pKa's of the Bronsted acids on both sides of the equation, we can determine which compound will retain the proton. The equilibrium will lie towards the compound with the higher pKa. This idea is illustrated in the equilibrium between hydronium ion and ammonium ion. Figure AB14.1. Direction of proton transfer between hydronium and ammonium ion. As another example, if hydrogen chloride is dissolved in water, the HCl may give up its proton to the water. Water has a lone pair and can act as a base. However, in doing so, the water will form hydronium ion, H 3 O + . Hydronium ion is Bronsted acidic and can provide a proton to something else that has a lone pair, such as a chloride ion. This reaction could go back and forth. Where will it settle out? HCl has a pKa of -8. Hydronium ion has a pKa of 0. The equilibrium in the reaction described above lies to the right, towards the hydronium ion produced when the hydrogen chloride dissociates. The proton will remain on the oxygen. Problem AB14.1. Write an equation for the proton transfer reactions that could happen in each of the following mixtures. Use structures in your equations. Predict the position of the equilibrium in each case. a) HF plus water b) CH 3 CO 2 H plus ammonia c) phenol (C 6 H 5 OH) plus sodium carbonate d) HCN plus acetonitrile Problem AB14.2. Use curved arrows to show the proton transfer reaction between the following compounds. Predict the products for these proton transfer reactions. Use pKa to determine whether each reaction is reactant favored OR mixture OR product favored.
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Bookshelves/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/15%3A_Metabolic_Cycles/15.06%3A_Homeostasis
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Learning Outcomes Describe how homeostasis and equilibrium are different. Remove one stone and the whole arch collapses. The same is true for the human body. All the systems work together to maintain stability or homeostasis. Disrupt one system, and the whole body may be affected. Homeostasis All of the organs and organ systems of the human body work together like a well-oiled machine. This is because they are closely regulated by the nervous and endocrine systems. The nervous system controls virtually all body activities, and the endocrine system secretes hormones that regulate these activities. Functioning together, the organ systems supply body cells with all the substances they need and eliminate their wastes. They also keep temperature, pH, and other conditions at just the right levels to support life processes. Maintaining Homeostasis The process in which organ systems work to maintain a stable internal environment is called homeostasis . Keeping a stable internal environment requires constant adjustments. Here are just three of the many ways that human organ systems help the body maintain homeostasis: Respiratory system: A high concentration of carbon dioxide in the blood triggers faster breathing. The lungs exhale more frequently, which removes carbon dioxide from the body more quickly. Excretory system: A low level of water in the blood triggers retention of water by the kidneys. The kidneys produce more concentration urine, so less water is lost from the body. Endocrine system: A high concentration of sugar in the blood triggers secretion of insulin by an endocrine gland called the pancreas. Insulin is a hormone that helps cells absorb sugar from the blood. So how does your body maintain homeostasis? The regulation of your internal environment is done primarily through negative feedback. Negative feedback is a response to a stimulus that keeps a variable close to a set value (see figure below). Essentially, it "shuts off" or "turns off" a system when it varies from a set value. For example, your body has an internal thermostat. During a winter day, in your house a thermostat senses the temperature in a room and responds by turning on or off the heater. Your body acts in much the same way. When body temperature rises, receptors in the skin and the brain sense the temperature change. The temperature change triggers a command from the brain. This command can cause several responses. If you are too hot, the skin makes sweat and blood vessels near the skin surface dilate. This response helps decrease body temperature. Another example of negative feedback has to do with blood glucose levels. When glucose (sugar) levels in the blood are too high, the pancreas secretes insulin to stimulate the absorption of glucose and the conversion of glucose into glycogen, which is stored in the liver. As blood glucose levels decrease, less insulin is produced. When glucose levels are too low, another hormone called glucagon is produced, which causes the liver to convert glycogen back to glucose. Positive Feedback Some processes in the body are regulated by positive feedback. Positive feedback is when a response to an event increases the likelihood of the event to continue. An example of positive feedback is milk production in nursing mothers. As the baby drinks her mother's milk, the hormone prolactin, a chemical signal, is released. The more the baby suckles, the more prolactin is released, which causes more milk to be produced. Other examples of positive feedback include contractions during childbirth. When constrictions in the uterus push a baby into the birth canal, additional contractions occur. Failure of Homeostasis Many homeostatic mechanisms such as these work continuously to maintain stable conditions in the human body. Sometimes, however, the mechanisms fail. When they do, cells may not get everything they need, or toxic wastes may accumulate in the body. If homeostasis is not restored, the imbalance may lead to disease or even death. Homeostasis vs. Equilibrium Homeostasis requires an input of energy to maintain a specific condition necessary for life. Disturbances to homeostasis must be responded to in order to avoid death or disease. For example, a body needs to maintain a certain internal temperature. Go outside in cold weather - body shivers to maintain its body temperature. Dynamic equilibrium is maintaining a specific condition that minimizes the system's energy, depending on the circumstances. A disturbance to an equilibrium is responded to in order to shift the process to reestablish an equilibrium. For example, if a warm object (say a metal bowl) is placed outside in cold weather - the transfer of heat occurs and the temperature of the bowl equilibrates to the outside temperature. If this happened to a person, it would not be good.
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Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/05%3A_Chemical_Kinetics_Reaction_Mechanisms_and_Chemical_Equilibrium/5.06%3A_Mechanisms_and_Elementary_Processes
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To see what we mean by an elementary process, let us consider some possible mechanisms for the base hydrolysis of methyl iodide: \[\ce{CH_3I + OH^{-} \to CH_3OH + I^{-}}. \nonumber \] In this reaction, a carbon–iodide bond is broken and a carbon–oxygen bond is formed. While any number of reaction sequences sum to this overall equation, we can write down three that are reasonably simple and plausible. The \(\ce{C \bond{-} I}\) could be broken first and the \(\ce{C \bond{-} OH}\) bond formed thereafter. Alternatively, the \(\ce{C\bond{-}OH}\) bond could be formed first and the \(\ce{C \bond{-} I}\) bond broken thereafter. In the first case, we have an intermediate species, \(\ce{CH^{+}_3}\), of reduced coordination number, and in the second we have an intermediate, \(\ce{HO \bond{-} CH3 \bond{-} I^{-}}\), of increased coordination number . Finally, we can suppose that the bond-forming and bond-breaking steps occur simultaneously, so that no intermediate species is formed at all. Heterolytic bond-breaking precedes bond-making \[ \begin{align*} \ce{CH_3I} &\to \ce{CH^+_3 + I^{-}} \\[4pt] \ce{CH^{+}_3 + OH^{-}} &\to \ce{CH_3OH} \end{align*} \tag{mechanism a} \] Bond-making precedes bond-breaking \[\ce{CH_3I + OH^{-} \to HO\bond{...}CH_3\bond{...} I^{-}} \\[4pt] \ce{HO \bond{...} CH_3 \bond{...} I^{-} -> CH_3OH + I^{-}} \tag{mechanism b} \] Bond-breaking and bond-making are simultaneous \[\ce{CH_3I + OH^{-} -> [ HO \bond{...} CH_3 \bond{...} I^{-} ]^{\ddagger} -> CH3OH + I^{-}} \tag{mechanism c} \] The distinction between mechanism (b) and mechanism (c) is that an intermediate is formed in the former but not in the latter. Nevertheless, mechanism (c) clearly involves an intermediate structure in which both the incoming and the leaving group are bonded to the central carbon atom. The distinction between mechanisms (b) and (c) depends on the nature of the intermediate structure. In mechanism (b), we suppose that the intermediate is a bona fide chemical entity; once a molecule of it is formed, that molecule has a finite lifetime. In (c), we suppose that the intermediate structure is transitory; it does not correspond to a molecule with an independent existence. For this distinction to be meaningful, we must have a criterion that establishes the shortest lifetime we are willing to associate with “real molecules.” It might seem that any minimum lifetime we pick must be wholly arbitrary. Fortunately this is not the case; there is a natural definition for a minimum molecular lifetime. The definition arises from the fact that molecules undergo vibrational motions. If a collection of atoms retains a particular relative orientation for such a short time that it never undergoes a motion that we would recognize as a vibration, it lacks an essential characteristic of a normal molecule. This means that the period of a high-frequency molecular vibration (roughly \({10}^{-14}\) s) is the shortest time that a collection of atoms can remain together and still have all of the characteristics of a molecule. If a structure persists for more than a few vibrations, it is reasonable to call it a molecule, albeit a possibly very unstable one. In mechanism (c) the structure designated \(\ce{[HO \bond{...}CH_3 \bond{...}I^{-}]^{\ddagger}}\) depicts a transitory arrangement of the constituent atoms. The atomic arrangement does not persist long enough for the \(\ce{HO\bond{-}CH_3}\) bond or the \(\ce{CH_3\bond{-}I}\) bond to undergo vibrational motion. A structure with these characteristics is called an activated complex or a transition state for the reaction, and a superscript double dagger, \(\mathrm{\ddagger}\), is conventionally used to signal that a structure has this character. The distinction between a bona fide intermediate and a transition state is clear enough in principle, but it can be very difficult to establish experimentally. These considerations justify our earlier definition: An elementary reaction is one in which there are no intermediates. Any atomic arrangement that occurs during an elementary reaction does not persist long enough to vibrate before the arrangement goes on to become products or reverts to reactants. An elementary reaction is one in which there are no intermediates. We can distinguish a small number of possible kinds of elementary reaction: termolecular elementary reactions, bimolecular elementary reactions, and unimolecular reactions. A single molecule can spontaneously rearrange to a new structure or break into smaller pieces. Two molecules can react to form one or more products. Three molecules can react to produce products. Or we can imagine that some larger number of molecules reacts. We refer to these possibilities as unimolecular , bimolecular , termolecular , and higher-molecularity processes. The stoichiometry of many reactions is so complicated as to preclude the possibility that they could occur as a single elementary process. For example, the reaction \[\ce{3Fe^{2+} + HCrO_4^{-} + 7H^{+} -> 3Fe^{3+} + Cr^{3+} + 4H2O} \nonumber \] can not plausibly occur in a single collision of three ferrous ions, one chromate ion, and seven hydronium ions. It is just too unlikely that all of these species could find themselves in the same place, at the same time, in the proper orientation, and with sufficient energy to react. In such cases, the stoichiometric mechanism must be a series of elementary steps. For this reaction, a skeletal representation of one plausible series is \[\begin{align*} \ce{Fe(II) + Cr(VI)} &\to \ce{Fe(III) + Cr(V)} \\[4pt] \ce{Fe(II) + Cr(V)} &\to \ce{Fe(III) + Cr(IV)} \\[4pt] \ce{Fe(II) + Cr(IV)} &\to \ce{Fe(III) + Cr(III)} \end{align*} \nonumber \]
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Courses/Lumen_Learning/Book%3A_Microeconomics-2_(Lumen)/05%3A_Module-_Supply_and_Demand/5.21%3A_Reading-_Economic_Systems
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Legoland, Billund, Denmark: Picture of a Planned Economy? Types of Economies In the modern world today, there is a range of economic systems, from market economies to planned (or command) economies. Market Economies A market is any situation that brings together buyers and sellers of goods or services. Buyers and sellers can be either individuals or businesses. In a market economy , economic decision-making happens through markets. Market economies are based on private enterprise: the means of production (resources and businesses) are owned and operated by private individuals or groups of private individuals. Businesses supply goods and services based on demand. Which goods and services are supplied depends on what is demanded by consumers or other businesses. A person’s income is based on his or her ability to convert resources (especially labor) into something that society values. The more society values the person’s output, the higher the income they will earn (think Lady Gaga or LeBron James). Examples of free-market economies include Hong Kong, Singapore, Australia, and the United States. Free Markets In a market economy , decisions about what products are available and at what prices are determined through the interaction of supply and demand. A competitive market is one in which there is a large number of buyers and sellers, so that no one can control the market price. A free market is one in which the government does not intervene in any way. A free and competitive market economy is the ideal type of market economy, because what is supplied is exactly what consumers demand. Price controls are an example of a market that is not free. When government intervenes, the market outcomes will be different from those that would occur in a free and competitive market model. When markets are less than perfectly competitive (e.g., monopolistic), the market outcomes will also differ. Planned (or Command) Economies Command economies operate very differently. In a command economy , economic effort is devoted to goals passed down from a ruler or ruling class. Ancient Egypt was a good example: A large part of economic life was devoted to building pyramids (like the one at the left), for the pharaohs. Medieval manor life is another example: The lord provided the land for growing crops and protection in the event of war. In return, vassals provided labor and soldiers to do the lord’s bidding. In the last century, communism emphasized command economies. In a command economy, resources and businesses are owned by the government. The government decides what goods and services will be produced and what prices will be charged for them. The government decides what methods of production will be used and how much workers will be paid. Some necessities like health care and education are provided for free, as long as the state determines that you need them. Currently, North Korea and Cuba have command economies. The primary distinction between a free and command economy is the degree to which the government determines what can be produced and what prices will be charged. In a free market, these determinations are made by the collective decisions of the market itself (which is comprised of producers and consumers). Producers and consumers make rational decisions about what will satisfy their self-interest and maximize profits, and the market responds accordingly. In a planned economy, the government makes most decisions about what will be produced and what the prices will be, and the market must follow that plan. Most economies in the real world are mixed; they combine elements of command and market systems. The U.S. economy is positioned toward the market-oriented end of the spectrum. Many countries in Europe and Latin America, while primarily market-oriented, have a greater degree of government involvement in economic decisions than in the U.S. economy. China and Russia, while they are closer now to having a market-oriented system than several decades ago, remain closer to the command-economy end of the spectrum. The following Crash Course video provides additional information about the broad economic choices that countries make when they decide between planned and market economies. The narrators talk fast, so you’ll need to listen closely and possibly watch the video a second time! A YouTube element has been excluded from this version of the text. You can view it online here: http://pb.libretexts.org/micro2/?p=114 Economic systems determine the following: What to produce? Who to produce it? Who gets it? In a planned economy, government controls the factors of production: In a true communist economy, there is no private property—everyone owns the factors of production. This type of planned economy is called a command economy In a socialist economy, there is some private property and some private control of industry. In a free-market (capitalist) economy, individuals own the factors of production: Businesses produce products. Consumers choose the products they prefer causing the companies that product them to make more profit. Even in free markets, governments will Maintain the rule of law Create public goods and services such as roads and education Step in when the market gets things wrong (e.g., setting minimum wage, establishing environmental standards) In reality, economies are neither completely free-market nor completely planned. Neither exists in “pure” form, since all societies and governments regulate their economies to varying degrees. Throughout this course we will consider a number of ways in which the U.S. government influences and controls the economy. Self Check: Economic Systems Answer the question(s) below to see how well you understand the topics covered in the previous section. This short quiz does not count toward your grade in the class, and you can retake it an unlimited number of times. You’ll have more success on the Self Check if you’ve completed the Reading in this section. Use this quiz to check your understanding and decide whether to (1) study the previous section further or (2) move on to the next section. https://assessments.lumenlearning.co...sessments/1558 CC licensed content, Original Reading: Economic Systems. Authored by : Steven Greenlaw and Lumen Learning. License : CC BY: Attribution CC licensed content, Shared previously Principles of Microeconomics 1.4. Authored by : OpenStax College. Provided by : Rice University. Located at : http://cnx.org/contents/[email protected]:n_POCARx@12/How-Economies-Can-Be-Organized . License : CC BY: Attribution . License Terms : Download for free at http://cnx.org/content/col11627/latest LEGO World - Fan Zone. Authored by : Brickset. Located at : https://www.flickr.com/photos/brickset/8491998046/ . License : CC BY: Attribution Sphinx. Authored by : David Stanley. Located at : https://www.flickr.com/photos/davidstanleytravel/8590203473/ . License : CC BY: Attribution All rights reserved content Economic Systems and Macroeconomics: Crash Course Economics #3. Provided by : CrashCourse. Located at : https://youtu.be/B43YEW2FvDs . License : All Rights Reserved . License Terms : Standard YouTube license
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Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/04%3A_The_Distribution_of_Gas_Velocities/4.09%3A_Pressure_Variations_for_Macroscopic_Samples
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At \(300\) K, the standard deviation of \(N_2\) speeds is about 40% of the average speed. Clearly the relative variation among molecular speeds in a sample of ordinary gas is very large. Why do we not observe macroscopic effects from this variation? In particular, if we measure the pressure at a small area of the container wall, why do we not observe pressure variations that reflect the wide variety of speeds with which molecules strike the wall? Qualitatively, the answer is obvious. A single molecule whose scalar velocity is \(v\) contributes \(P_1\left(v\right)={mv^2}/{3V}\) to the pressure on the walls of its container. (See problem 20.) When we measure pressure, we measure an average squared velocity. Even if we measure the pressure over a very small area and a very short time, the number of molecules striking the wall during the time of the measurement is very large. Consequently, the average speed of the molecules hitting the wall during any one such measurement is very close to the average speed in any other such measurement. We are now able to treat this question quantitatively. For \(N_2\) gas at \(300\) K and \(1\) bar, roughly \(\mathrm{3\times }{\mathrm{10}}^{\mathrm{15}}\) molecules collide with a square millimeter of wall every microsecond. (See problem 12.) The standard deviation of the velocity of an \(N_2\) molecule is \(201\ \mathrm{m\ }{\mathrm{s}}^{\mathrm{-1}}\). Using the central limit theorem, the standard deviation of the average of \(3\times {10}^{15}\) molecular speeds is \[\frac{201\, m\,s^{-1}}{\sqrt{3 \times 10^{15}}} \approx 4 \times 10^{-6} \mathrm{ms}^{-1} \nonumber \] The distribution of the average of \(3\times {10}^{15}\) molecular speeds is very narrow indeed. Similarly, when molecular velocities follow the Maxwell-Boltzmann distribution function, we can show that the expected value of the pressure for a single-molecule collision is \(\left\langle P_1\left(v\right)\right\rangle =kT/V\). (See problem 21.) The variance of the distribution of these individual pressure measurements is \(\sigma^2_{P_1\left(v\right)}={2k^2T^2}/{3V^2}\), so that the magnitude of the standard deviation is comparable to that of the average: \[\sigma_{P_{1}(v)} /\left\langle P_{1}(v)\right\rangle=\sqrt{2 / 3} \nonumber \] For the distribution of averages of \(3\times {10}^{15}\) pressure contributions, we find \[\begin{aligned} P_{a v g} &=\left\langle P_{1}(v)\right\rangle \\ &=\sqrt{3 / 2} \sigma_{P_{1}(v)} \end{aligned} \nonumber \] \[\sigma_{avg}=\dfrac{\sigma_{P_1\left(v\right)}}{\sqrt{3\times 10^{15}}} \nonumber \] and \[\frac{\sigma_{avg}}{P_{avg}}\approx 1.5\times {10}^{-8} \nonumber \]
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Courses/Eastern_Mennonite_University/EMU%3A_Chemistry_for_the_Life_Sciences_(Cessna)/09%3A_Solutions
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Solutions are all around us. Air, for example, is a solution. If you live near a lake, a river, or an ocean, that body of water is not pure H 2 O but most probably a solution. Much of what we drink—for example, soda, coffee, tea, and milk—is at least in part a solution. Solutions are a large part of everyday life. A lot of the chemistry occurring around us happens in solution. In fact, much of the chemistry that occurs in our own bodies takes place in solution, and many solutions—such as the Ringer’s lactate IV solution—are important for our health. In our understanding of chemistry, we need to understand a little bit about solutions. In this chapter, you will learn about the special characteristics of solutions, how solutions are characterized, and some of their properties.
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Bookshelves/Organic_Chemistry/Organic_Chemistry_II_(Morsch_et_al.)/22%3A_Carbonyl_Alpha-Substitution_Reactions/22.03%3A_Alpha_Halogenation_of_Aldehydes_and_Ketones
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Objectives After completing this section, you should be able to write an equation to illustrate the alpha halogenation of aldehydes and ketones. identify the product formed from the alpha halogenation of a given aldehyde or ketone. identify the carbonyl compound, the reagents, or both, needed to prepare a given α ‑halogenated aldehyde or ketone. illustrate the importance of the alpha halogenation of carbonyl compounds as an intermediate step in the synthesis of α , β ‑unsaturated aldehydes and ketones. write a detailed mechanism for the acid‑catalyzed halogenation of a ketone. describe the evidence provided by kinetic experiments supporting the suggestion that the acid‑catalyzed, alpha halogenation of ketones proceeds via the rate‑determining formation of an enol. Study Notes Note: α ‑bromo ketones are a good starting material to generate α , β ‑unsaturated ketones by dehydrobromination. Aldehydes and ketones can substitute an α-hydrogen for a halogen atom in the presence of an acid. This reaction takes place using acid catalyzed tautomerization to form a nucleophilic enol, which then reacts with an electrophilic halogen (Cl 2 , Br 2 or I 2 ). Because an enol intermediate is used, a racemic mixture of products can be produced. A particularly useful variation of this reaction uses bromine in an acetic acid solvent. The α -bromo substituted product can then be readily transformed into an α , β ‑unsaturated carbonyl through reaction with pyridine and heat. General Reaction ( α substitution) Example 22.3.1 Acid Catalyzed Mechanism The mechanism begins with protonation of the carbonyl oxygen followed by removal of an α -hydrogen to form the enol. Lone pair electrons from the enol oxygen move to form a carbonyl while the pi electrons from the double bond attack the halogen forming an oxonium ion intermediate with a C-X sigma bond in the α -position. Deprotonation of the oxonium ion intermediate provides the α- halogen substituted product and regenerates the acid catalyst. 1) Protonation by the acid catalyst 2) Removal of an α -hydrogen to form the enol. This step is slow and represent the rate determine step. 3) Nucleophilic attack on the halogen 4) Deprotonation Experimental Evidence of the Enol Intermediate This reaction was the focus of one of the first mechanistic investigations in organic chemistry. In the early 1900's chemist Arthur Lapworth showed that the rates of chlorination, bromination, and iodination of acetone were all the same. Also, it was shown that the rates for all three halogenation reactions were first-order with respect to acetone and the acid catalyst but independent of the halogen concentration (overall second-order for the mechanism). The rate law expression for the α -halogenation of a ketone can be given by: rate = k [ketone] [H + ] This implies that the halogen participates in the mechanism through a fast step which occurs after the rate-determining step. These observations led Lapworth to theorize that the rate-determining step of the mechanism involves converting acetone to a more reactive form. The fact that the substitution occurs on the α -carbon led Lapworth to propose that the more reactive form was an enol tautomer of acetone. Synthetic Uses for α -Halogenated Carbonyls The product of an α- bromination can be converted to an α , β ‑unsaturated carbonyl by reaction with pyridine and heat which causes the elimination of H and Br. This reaction takes place by an E2 elimination mechanism and creates a C=C double bond which is conjugated with the carbonyl. In order to promote an E2 reaction, a sterically hindered base, pyridine, is often used. An example of this reaction involves the α- bromination of 2-methylcyclopentanone to form 2-bromo-2-methylcyclopentanone. Because enol tautomers prefer to form on the more substituted α -carbon, α -bromination also occurs on the more substituted α -carbon. Although the enol intermediate causes a racemic mixture of the α -brominated compound to form, it is irrelevant because the chiral carbon is subsequently converted to an achiral alkene. Subsequent reaction with pyridine and heat forms the α , β ‑unsaturated ketone, 2-methyl-2-cyclopentenone. Deuterium Exchange More evidence for the formation of an enol intermediate was provided using a reaction called deuterium exchange. Deuterium is an isotope of hydrogen which contains one proton and one neutron. Due to the acidic nature of α-hydrogens, they can be exchanged with deuterium by reaction with the isotopic form of water, D 2 O (deuterium oxide-heavy water). The process is accelerated by addition of the deuterium equivalent of a strong acid, such as deuterium chloride (DCl), which quickly reacts with D 2 O to form D 3 O + , the deuterium equivalent of hydronium (H 3 O + ). If an excess of D 2 O is used, the exchange process continues to the end result of all α-hydrogens present in a given compound being replaced with deuterium. Deuterium exchange is an effective method for introducing an isotopic label into a molecule. Also, deuterium does not appear in 1 H NMR, so deuterium exchange can help determine peak assignments. General Reaction (Deuterium exchange) Example 22.3.2 Mechanism in Acidic Conditions The mechanism for deuterium exchange is virtually the same as keto-enol tautomerism under acidic conditions, as shown in Section 22.2 . The only difference is that when the keto tautomer is reformed a deuterium is placed in the α -position. 1) Formation of an enol 2) α -Deuteration 3) De-deuteration to form the keto tautomer Example \(\PageIndex{1}\) A simple method for determining the number of α -hydrogen in a compound is through reaction of D 3 O + . The reaction product is then isolated and its molecular weight is determined by mass spectrometry. For example, if cyclopentanone is reacted with D 3 O + , the isolated product has a molecular weight of 88 g/mol. Please explain how this method works and how many α -hydrogens cyclopentanone is predicted to have. Solution During acid catalyzed deuterium exchange each α -hydrogen in the compound is replaced with a deuterium. For each proton (AW = 1) replaced with a deuterium (AW = 2) the molecular weight of the compound is increased by one. Since cyclopentanone has a molecular weight of 84 g/mol and the isolated product has a molecular weight of 88 g/mol it can be predicted that cyclopentanone has four α -hydrogens. Kinetic investigations into the mechanism of this reaction provided further evidence for the formation of a reactive enol intermediate. It was shown that the rate of deuterium exchange was the same as the rate of halogenation for ketones. This implies that both reactions have a common intermediate involved in the rate determining step of their mechanism, an enol. Exercises 1) Please draw the products of the following reactions 2) Draw out the mechanism for the following reaction. 3) How might you form 2-hepten-4-one from 4-heptanone? 4) Show the products of the following reactions: 5) The following compound was reacted with D 3 O + . The only signals that could be found in the 1H NMR spectrum of the product were at 3.9 ppm (3H) and 6.6-6.9 ppm (4H). Please explain the results of the NMR. Solutions 1) 2) 3) [1) Br 2 , H 3 O + ; 2) Pyridine, Heat] 4) 5) A deuterium exchange reaction occurred. All of the alpha-hydrogens in the molecule have been exchanged with deuterium. Because detueriums do not appear in a typical 1 H NMR, only the remaining hydrogens appear.
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Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Time_Dependent_Quantum_Mechanics_and_Spectroscopy_(Tokmakoff)/04%3A_Irreversible_Relaxation/4.01%3A_Introduction_to_Dissipative_Dynamics
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How does irreversible behavior, a hallmark of chemical systems, arise from the deterministic Time Dependent Schrödinger Equation? We will answer this question specifically in the context of quantum transitions from a given energy state of the system to energy states its surroundings. Qualitatively, such behavior can be expected to arise from destructive interference between oscillatory solutions of the system and the set of closely packed manifold of energy states of the bath. To illustrate this point, consider the following calculation for the probability amplitude for an initial state of the system coupled to a finite but growing number of randomly chosen states belonging to the bath. Here, even with only 100 or 1000 states, recurrences in the initial state amplitude are suppressed by destructive interference between paths. Clearly in the limit that the accepting state are truly continuous, the initial amplitude prepared in \(\ell\) will be spread through an infinite number of continuum states. We will look at this more closely by describing the relaxation of an initially prepared state as a result of coupling to a continuum of states of the surroundings. This is common to all dissipative processes in which the surroundings to the system of interest form a continuous band of states. To begin, let us define a continuum. We are familiar with eigenfunctions being characterized by quantized energy levels, where only discrete values of the energy are allowed. However, this is not a general requirement. Discrete levels are characteristic of particles in bound potentials, but free particles can take on a continuous range of energies given by their momentum, \[ E = \dfrac{\langle p^2 \rangle}{2m}.\] The same applies to dissociative potential energy surfaces, and bound potentials in which the energy exceeds the binding energy. For instance, photoionization or photodissociation of a molecule involves a light field coupling a bound state into a continuum. Other examples are common in condensed matter. The intermolecular motions of a liquid, the lattice vibrations of a crystal, or the allowed energies within the band structure of a metal or semiconductor are all examples of a continuum. For a discrete state imbedded in such a continuum, the Golden Rule gives the probability of transition from the system state \(| \ell \rangle\) to a continuum state \(| \ell \rangle\) as: \[\overline {w} _ {k \ell} = \frac {\partial \overline {P} _ {k \ell}} {\partial t} = \frac {2 \pi} {\hbar} \left| V _ {k \ell} \right|^{2} \rho \left( E _ {k} = E _ {\ell} \right)\] The transition rate \(\overline {\mathcal {W}} _ {k \ell}\) is constant in time, when \(\left| V _ {k \ell} \right|^{2}\) is constant in time, which will be true for short time intervals. Under these conditions integrating the rate equation on the left gives \[\begin{align} \overline {P} _ {k \ell} &= \overline {w} _ {k \ell} \left( t - t _ {0} \right) \\[4pt] \overline {P} _ {\ell \ell} &= 1 - \overline {P} _ {k \ell}. \end{align}\] The probability of transition to the continuum of bath states varies linearly in time. As we noted, this will clearly only work for times such that \[P _ {k} (t) - P _ {k} ( 0 ) \gg1.\] What long time behavior do we expect? A time independent rate with population governed by \[\overline {w} _ {k \ell} = \partial \overline {P} _ {k \ell} / \partial t\] is a hallmark of first order kinetics and exponential relaxation. In fact, for exponential relaxation out of a state \(\ell\), the short time behavior looks just like the first order result: \[\begin{align} \overline {P} _ {\ell \ell} (t) &= \exp \left( - \overline {w} _ {k \ell} t \right) \\ &= 1 - \overline {w} _ {k \ell} t + \ldots\label{3.4} \end{align} \] So we might believe that \(\overline {\mathcal {W}} _ {k \ell}\) represents the tangent to the relaxation behavior at \(t - 0\). The problem we had previously was we did not account for depletion of initial state. In fact, we will see that when we look a touch more carefully, that the long time relaxation behavior of state \(\ell\) is exponential and governed by the golden rule rate. The decay of the initial state is irreversible because there is feedback with a distribution of destructively interfering phases. Let’s formulate this problem a bit more carefully. We will look at transitions to a continuum of states \(\{k \}\) from an initial state \(\ell\) under a constant perturbation. These together form a complete set; so for \[H (t) = H _ {0} + V (t)\] with \(H _ {0} | n \rangle = E _ {n} | n \rangle\). \[1 = \sum _ {n} | n \rangle \langle n | = | \ell \rangle \left\langle \ell \left| + \sum _ {k} \right| k \right\rangle \langle k | \label{3.5}\] As we go on, you will see that we can identify \(\ell\) with the “system” and \(\{k \}\) with the “bath” when we partition \[H _ {0} = H _ {S} + H _ {B}.\] Now let’s make some simplifying assumptions. For transitions into the continuum, we will assume that transitions only occur between \(\ell\) and states of the continuum, but that there are no interactions between states of the continuum: \(\left\langle k | V | k^{\prime} \right\rangle = 0\). This can be rationalized by thinking of this problem as a discrete set of states interacting with a continuum of normal modes. Moreover, we will assume that the coupling of the initial to continuum states is a constant for all states \(k\): \(\langle \ell | V | k \rangle = \left\langle \ell | V | k^{\prime} \right\rangle = \cdots\). For reasons that we will see later, we will also keep the diagonal matrix element \(\langle \ell | V | \ell \rangle = 0\). With these assumptions, we can summarize the Hamiltonian for our problem as \begin{aligned} H(t) &=H_{0}+V(t) \\ H_{0} &=|\ell\rangle E_{\ell}\left\langle\ell\left|+\sum_{k}\right| k\right\rangle E_{k}\langle k| \\ V(t) &=\sum_{k}\left[|k\rangle V_{k \ell}\langle\ell|+| \ell\rangle V_{2 k}\langle k|\right]+|\ell\rangle V_{\ell \ell}\langle\ell| \label{3.6}\end{aligned} We are seeking a more accurate description of the occupation of the initial and continuum states, for which we will use the interaction picture expansion coefficients \[b _ {k} (t) = \left\langle k \left| U _ {I} \left( t , t _ {0} \right) \right| \ell \right\rangle \label{3.7}\] Earlier, we saw that the exact solution to \(U_I\) was: \[U _ {I} \left( t , t _ {0} \right) = 1 - \frac {i} {\hbar} \int _ {t _ {0}}^{t} d \tau V _ {I} ( \tau ) U _ {I} \left( \tau , t _ {0} \right) \label{3.8}\] This form was not very practical, since \(U_I\) is a function of itself. For first-order perturbation theory, we set the final term in this equation \(U_I\), \(U _ {I} \left( \tau , t _ {0} \right) \rightarrow 1\). Here, in order to keep the feedback between \( |\ell \rangle \) and the continuum states, we keep it as is. \[b _ {k} (t) = \langle k | \ell \rangle - \frac {i} {\hbar} \int _ {t _ {0}}^{t} d \tau \left\langle k \left| V _ {I} ( \tau ) U _ {I} \left( \tau , t _ {0} \right) \right| \ell \right\rangle \label{3.9}\] Inserting Equation \ref{3.7}, and recognizing \(k \neq l\), \[b _ {k} (t) = - \frac {i} {\hbar} \sum _ {n} \int _ {t _ {0}}^{t} d \tau e^{i \omega _ {h n} \tau} V _ {k n} b _ {n} ( \tau ) \label{3.10}\] Note, \(V_{kn}\) is not a function of time. Equation \ref{3.10} expresses the occupation of state \(k\) in terms of the full history of the system from \(t _ {0} \rightarrow t\) with amplitude flowing back and forth between the states n. Equation \ref{3.10} is just the integral form of the coupled differential equations that we used before: \[i \hbar \frac {\partial b _ {k}} {\partial t} = \sum _ {n} e^{i \omega _ {b n} t} V _ {k n} b _ {n} (t) \label{3.11}\] These exact forms allow for feedback between all the states, in which the amplitudes \(b_k\) depend on all other states. Since you only feed from \(\ell\) into \(k\), we can remove the summation in Equation \ref{3.10} and express the complex amplitude of a state within the continuum as \[b _ {k} = - \frac {i} {\hbar} V _ {k \ell} \int _ {t _ {0}}^{t} d \tau e^{i \omega _ {k} \tau} b _ {\ell} ( \tau ) \label{3.12}\] We want to calculate the rate of leaving \(| \ell \rangle\), including feeding from continuum back into initial state. From Equation \ref{3.11} we can separate terms involving the continuum and the initial state: \[i \hbar \frac {\partial} {\partial t} b _ {\ell} = \sum _ {k \neq \ell} e^{i \omega _ {\mu} t} V _ {\ell k} b _ {k} + V _ {\ell \ell} b _ {\ell} \label{3.13}\] Now substituting Equation \ref{3.12} into Equation \ref{3.13}, and setting \(t_0 =0\): \[\frac {\partial b _ {\ell}} {\partial t} = - \frac {1} {\hbar^{2}} \sum _ {k \neq \ell} \left| V _ {k \ell} \right|^{2} \int _ {0}^{t} b _ {\ell} ( \tau ) e^{i \omega _ {k} ( \tau - t )} d \tau - \frac {i} {\hbar} V _ {\ell \ell} b _ {\ell} (t) \label{3.14}\] This is an integro-differential equation that describes how the time-development of \(b_ℓ\) depends on the entire history of the system. Note we have two time variables for the two propagation routes: \[\left. \begin{array} {l} {\tau : | \ell \rangle \rightarrow | k \rangle} \\ {t : | k \rangle \rightarrow | \ell \rangle} \end{array} \right. \label{3.15}\] The next assumption is that \(b_ℓ\) varies slowly relative to \(\omega_{kℓ}\), so we can remove it from integral. This is effectively a weak coupling statement: \(\hbar \omega _ {k \ell} \gg V _ {k \ell}\). \(b\) is a function of time, but since it is in the interaction picture it evolves slowly compared to the \(\omega_{kℓ}\) oscillations in the integral. \[\frac {\partial b _ {\ell}} {\partial t} = b _ {\ell} \left[ - \frac {1} {\hbar^{2}} \sum _ {k \neq \ell} \left| V _ {k \ell} \right|^{2} \int _ {0}^{t} e^{i \omega _ {k} ( \tau - t )} d \tau - \frac {i} {\hbar} V _ {\ell \ell} \right] \label{3.16}\] Now, we want the long time evolution of \(b\), for times \(\omega _ {k \ell} t > > 1\), we will investigate the integration limit \(t \rightarrow \infty\). Note Complex integration of Equation \ref{3.16}: Defining \(t^{\prime} = \tau - t\) \[\int _ {0}^{t} e^{i \omega _ {k l} ( \tau - t )} d \tau = - \int _ {0}^{t} e^{i \omega _ {k l} t^{\prime}} d t^{\prime} \label{3.17}\] The integral \(\lim _ {T \rightarrow \infty} \int _ {0}^{T} e^{i \omega t^{\prime}} d t^{\prime}\) is purely oscillatory and not well behaved. The strategy to solve this is to integrate: \[\begin{align} \lim _ {\varepsilon \rightarrow 0^{+}} \int _ {0}^{\infty} e^{( i \omega + \varepsilon ) t^{\prime}} d t^{\prime} & = \lim _ {\varepsilon \rightarrow 0^{+}} \frac {1} {i \omega + \varepsilon} \\ & = \lim _ {\varepsilon \rightarrow 0^{+}} \left( \frac {\varepsilon} {\omega^{2} + \varepsilon^{2}} + i \frac {\omega} {\omega^{2} + \varepsilon^{2}} \right) \\ & = \pi \delta ( \omega ) - i \mathbb {P} \frac {1} {\omega} \label{3.19} \end{align}\] (This expression is valid when used under an integral) In the final term we have written in terms of the Cauchy Principle Part: \[\mathbb {P} \left( \frac {1} {x} \right) = \left\{\begin{array} {l l} {\frac {1} {x}} & {x \neq 0} \\ {0} & {x = 0} \end{array} \right. \label{3.20}\] Using Equation \ref{3.19}, Equation \ref{3.16} becomes \ref{3.21} \[\frac {\partial b _ {\ell}} {\partial t} = b _ {\ell} [ - \underbrace {\frac {\pi} {\hbar^{2}} \sum _ {k \neq \ell} \left| V _ {k \ell} \right|^{2} \delta \left( \omega _ {k \ell} \right)} _ {\text {term} 1} - \frac {i} {\hbar} \left( \underbrace {V _ {\ell \ell} + \mathbb {P} \sum _ {k \neq \ell} \frac {\left| V _ {k \ell} \right|^{2}} {E _ {k} - E _ {\ell}} )} _ {\text {term} 2} \right] \label{3.21}\] Note that Term 1 is just the Golden Rule rate, written explicitly as a sum over continuum states instead of an integral \[\sum _ {k \neq \ell} \delta \left( \omega _ {k \ell} \right) \Rightarrow \hbar \rho \left( E _ {k} = E _ {\ell} \right) \label{3.22}\] \[\overline {w} _ {k \ell} = \int d E _ {k} \rho \left( E _ {k} \right) \left[ \frac {2 \pi} {\hbar} \left| V _ {k l} \right|^{2} \delta \left( E _ {k} - E _ {\ell} \right) \right] \label{3.23}\] Term 2 is just the correction of the energy of \(E_ℓ\) from second-order time-independent perturbation theory, \[\Delta E _ {\ell} = \langle \ell | V | \ell \rangle + \sum _ {k \neq l} \frac {\langle k | V | \left. \ell \right|^{2}} {E _ {k} - E _ {\ell}} \label{3.25} \] So, the time evolution of \(b _ {\ell}\) is governed by a simple first-order differential equation \[\frac{\partial b_{\ell}}{\partial t}=b_{\ell}\left(-\frac{\bar{w}_{k \ell}}{2}-\frac{i}{\hbar} \Delta E_{\ell}\right)\] Which can be solved with \(b _ {\ell} ( 0 ) = 1\) to give \[b _ {\ell} (t) = \exp \left( - \frac {\overline {w} _ {k l} t} {2} - \frac {i} {\hbar} \Delta E _ {\ell} t \right) \label{3.26}\] We see that one has exponential decay of amplitude of \(b _ {\ell}\)! This is a manner of irreversible relaxation from coupling to the continuum. Now, since there may be additional interferences between paths, we switch from the interaction picture back to Schrödinger Picture, \[c _ {\ell} (t) = \exp \left[ - \left( \frac {\overline {w} _ {k \ell}} {2} + i \frac {E _ {\ell}^{\prime}} {\hbar} \right) t \right] \label{3.27}\] with the corrected energy \[E _ {\ell}^{\prime} \equiv E _ {\ell} + \Delta E \label{3.28}\] and \[P _ {\ell} = \left| c _ {\ell} \right|^{2} = \exp \left[ - \overline {w} _ {k \ell} t \right] \label{3.29}\] The solutions to the Time Dependent Schrödinger Equation are expected to be complex and oscillatory. What we see here is a real dissipative component and an imaginary dispersive component. The probability decays exponentially from initial state. Fermi’s Golden Rule rate tells you about long times! Now, what is the probability of appearing in any of the states \(|k \rangle\)? Using Equation \ref{3.12}: \[b _ {k} (t) = - \frac {i} {\hbar} \int _ {0}^{t} V _ {k \ell} e^{i \omega _ {k l} \tau} b _ {\ell} ( \tau ) d \tau\] \[= V _ {k \ell} \frac {1 - \exp \left( - \frac {\overline {w} _ {k \ell}} {2} t - \frac {i} {h} \left( E _ {\ell}^{\prime} - E _ {k} \right) t \right)} {E _ {k} - E _ {\ell}^{\prime} + i \hbar \overline {w} _ {k \ell} / 2}\] \[= V _ {k \ell} \frac {1 - c _ {\ell} (t)} {E _ {k} - E _ {\ell}^{\prime} + i \hbar \overline {w} _ {k \ell} / 2}\] (3.30) If we investigate the long time limit (\(t \rightarrow \infty\)), noting that \(P _ {k \ell} = \left| b _ {k} \right|^{2}\), we find \[P _ {k l} = \frac {\left| V _ {k l} \right|^{2}} {\left( E _ {k} - E _ {i}^{\prime} \right)^{2} + \Gamma^{2} / 4} \label{3.31}\] with \[\Gamma \equiv \overline {w} _ {k \ell} \cdot \hbar \label{3.32}\] The probability distribution for occupying states within the continuum is described by a Lorentzian distribution with maximum probability centered at the corrected energy of the initial state \(E _ {\ell}^{\prime}\). The width of the distribution is given by the relaxation rate, which is proxy for \(\left| V _ {k \ell} \right|^{2} \rho \left( E _ {\ell} \right)\), the coupling to the continuum and density of states. Readings Cohen-Tannoudji, C.; Diu, B.; Lalöe, F., Quantum Mechanics. Wiley-Interscience: Paris, 1977; p. 1344. Merzbacher, E., Quantum Mechanics. 3rd ed.; Wiley: New York, 1998; p. 510. Nitzan, A., Chemical Dynamics in Condensed Phases. Oxford University Press: New York, 2006; p. 305. 4. Schatz, G. C.; Ratner, M. A., Quantum Mechanics in Chemistry. Dover Publications: Mineola, NY, 2002; Ch. 9.
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Bookshelves/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/15%3A_Organic_Acids_and_Bases_and_Some_of_Their_Derivatives/15.12%3A_Amines_as_Bases
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Learning Objectives Name the typical reactions that take place with amines. Describe heterocyclic amines. Recall that ammonia (NH 3 ) acts as a base because the nitrogen atom has a lone pair of electrons that can accept a proton. Amines also have a lone electron pair on their nitrogen atoms and can accept a proton from water to form substituted ammonium (NH 4 + ) ions and hydroxide (OH − ) ions: As a specific example, methylamine reacts with water to form the methylammonium ion and the OH − ion. Nearly all amines, including those that are not very soluble in water, will react with strong acids to form salts soluble in water. Amine salts are named like other salts: the name of the cation is followed by the name of the anion. Example \(\PageIndex{1}\) What are the formulas of the acid and base that react to form [CH 3 NH 2 CH 2 CH 3 ] + CH 3 COO − ? Solution The cation has two groups—methyl and ethyl—attached to the nitrogen atom. It comes from ethylmethylamine (CH 3 NHCH 2 CH 3 ). The anion is the acetate ion. It comes from acetic acid (CH 3 COOH). Exercise \(\PageIndex{1}\) What are the formulas of the acid and base that react to form \(\ce{(CH3CH2CH2)3NH^{+}I^{−}}\)? To Your Health: Amine Salts as Drugs Salts of aniline are properly named as anilinium compounds, but an older system, still in use for naming drugs, identifies the salt of aniline and hydrochloric acid as “aniline hydrochloride.” These compounds are ionic—they are salts—and the properties of the compounds (solubility, for example) are those characteristic of salts. Many drugs that are amines are converted to hydrochloride salts to increase their solubility in aqueous solution. Heterocyclic Amines Looking back at the various cyclic hydrocarbons discussed previously, we see that all the atoms in the rings of these compounds are carbon atoms. In other cyclic compounds, called heterocyclic compounds (Greek heteros , meaning “other”), nitrogen, oxygen, sulfur, or some other atom is incorporated in the ring. Many heterocyclic compounds are important in medicine and biochemistry. Some compose part of the structure of the nucleic acids, which in turn compose the genetic material of cells and direct protein synthesis. Many heterocyclic amines occur naturally in plants. Like other amines, these compounds are basic. Such a compound is an alkaloid , a name that means “like alkalis.” Many alkaloids are physiologically active, including the familiar drugs caffeine, nicotine, and cocaine. To Your Health: Three Well-Known Alkaloids Caffeine is a stimulant found in coffee, tea, and some soft drinks. Its mechanism of action is not well understood, but it is thought to block the activity of adenosine, a heterocyclic base that acts as a neurotransmitter, a substance that carries messages across a tiny gap (synapse) from one nerve cell (neuron) to another cell. The effective dose of caffeine is about 200 mg, corresponding to about two cups of strong coffee or tea. Nicotine acts as a stimulant by a different mechanism; it probably mimics the action of the neurotransmitter acetylcholine. People ingest this drug by smoking or chewing tobacco. Its stimulant effect seems transient, as this initial response is followed by depression. Nicotine is highly toxic to animals. It is especially deadly when injected; the lethal dose for a human is estimated to be about 50 mg. Nicotine has also been used in agriculture as a contact insecticide. Cocaine acts as a stimulant by preventing nerve cells from taking up dopamine, another neurotransmitter, from the synapse. High levels of dopamine are therefore available to stimulate the pleasure centers of the brain. The enhancement of dopamine action is thought to be responsible for cocaine’s “high” and its addictive properties. After the binge, dopamine is depleted in less than an hour. This leaves the user in a pleasureless state and (often) craving more cocaine. Cocaine is used as the salt cocaine hydrochloride and in the form of broken lumps of the free (unneutralized) base, which is called crack cocaine . \[\underbrace{\ce{C17H21O4N}}_{\text{cocaine (freebase)}} + \ce{HCl ->} \underbrace{\ce{C17H21O4NH^{+}Cl^{-}}}_{\text{cocaine hydrochloride}} \nonumber \] Because it is soluble in water, cocaine hydrochloride is readily absorbed through the watery mucous membranes of the nose when it is snorted. Crack cocaine is more volatile than cocaine hydrochloride. It vaporizes at the temperature of a burning cigarette. When smoked, cocaine reaches the brain in 15 s. Summary Amines are bases; they react with acids to form salts. Salts of aniline are properly named as anilinium compounds, but an older system is used to name drugs: the salts of amine drugs and hydrochloric acid are called “hydrochlorides.” Heterocyclic amines are cyclic compounds with one or more nitrogen atoms in the ring.
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Courses/Clackamas_Community_College/CH_112%3A_Chemistry_for_Health_Sciences/10%3A_Acids_and_Bases/10.05%3A_The_Strengths_of_Acids_and_Bases
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Learning Objectives Describe the difference between strong and weak acids and bases. Describe how a chemical reaction reaches chemical equilibrium. Define the pH scale and use it to describe acids and bases. Acids and bases do not all demonstrate the same degree of chemical activity in solution. Different acids and bases have different strengths. Strong and Weak Acids Let us consider the strengths of acids first. A small number of acids ionize completely in aqueous solution. For example, when HCl dissolves in water, every molecule of HCl separates into a hydronium ion and a chloride ion: \[\ce{HCl(g) + H2O(l) ->[\sim 100\%] H_3O^{+}(aq) + Cl^{-} (aq)} \label{Eq1} \] HCl(aq) is one example of a strong acid , which is a compound that is essentially 100% ionized in aqueous solution. There are very few strong acids. The important ones are listed in Table \(\PageIndex{1}\). Acids Bases HCl LiOH HBr NaOH HI KOH HNO3 Mg(OH)2 H2SO4 Ca(OH)2 HClO4 NaN By analogy, a strong base is a compound that is essentially 100% ionized in aqueous solution. As with acids, there are only a few strong bases, which are also listed in Table \(\PageIndex{1}\). If an acid is not listed in Table \(\PageIndex{1}\), it is likely a weak acid , which is a compound that is not 100% ionized in aqueous solution. Similarly, a weak base is a compound that is not 100% ionized in aqueous solution. For example, acetic acid (\(\ce{HC2H3O2}\)) is a weak acid. The ionization reaction for acetic acid is as follows: \[\ce{HC2H3O2(aq) + H2O(ℓ) \rightarrow H3O^{+}(aq) + C2H3O^{−}2(aq)} \label{Eq2} \] Depending on the concentration of HC 2 H 3 O 2 , the ionization reaction may occur only for 1%–5% of the acetic acid molecules. Looking Closer: Household Acids and Bases Many household products are acids or bases. For example, the owner of a swimming pool may use muriatic acid to clean the pool. Muriatic acid is another name for hydrochloric acid [\(\ce{HCl(aq)}\)]. Vinegar has already been mentioned as a dilute solution of acetic acid [\(\ce{HC2H3O2(aq)}\)]. In a medicine chest, one may find a bottle of vitamin C tablets; the chemical name of vitamin C is ascorbic acid (\(\ce{HC6H7O6}\)). One of the more familiar household bases is ammonia (NH 3 ), which is found in numerous cleaning products. As we mentioned previously, ammonia is a base because it increases the hydroxide ion concentration by reacting with water: \[\ce{NH3(aq) + H2O(ℓ) \rightarrow NH^{+}4(aq) + OH^{−}(aq)} \label{Eq3} \] Many soaps are also slightly basic because they contain compounds that act as Brønsted-Lowry bases, accepting protons from water and forming excess hydroxide ions. This is one reason that soap solutions are slippery. Perhaps the most dangerous household chemical is the lye-based drain cleaner. Lye is a common name for sodium hydroxide, although it is also used as a synonym for potassium hydroxide. Lye is an extremely caustic chemical that can react with grease, hair, food particles, and other substances that may build up and form a clog in a pipe. Unfortunately, lye can also attack tissues and other substances in our bodies. Thus, when we use lye-based drain cleaners, we must be very careful not to touch any of the solid drain cleaner or spill the water it was poured into. Safer, nonlye drain cleaners use peroxide compounds to react on the materials in the clog and clear the drain. Chemical Equilibrium in Weak Acids and Bases The behavior of weak acids and bases illustrates a key concept in chemistry. Does the chemical reaction describing the ionization of a weak acid or base just stop when the acid or base is done ionizing? Actually, no. Rather, the reverse process—the reformation of the molecular form of the acid or base—occurs, ultimately at the same rate as the ionization process. For example, the ionization of the weak acid \(\ce{HC2H3O2 (aq)}\) is as follows: \[\ce{HC2H3O2(aq) + H2O(ℓ) \rightarrow H3O^{+}(aq) + C2H3O^{−}2(aq)} \label{Eq4} \] The reverse process also begins to occur: \[\ce{H3O^{+}(aq) + C2H3O^{−}2(aq) \rightarrow HC2H3O2(aq) + H2O(ℓ)} \label{Eq5} \] Eventually, there is a balance between the two opposing processes, and no additional change occurs. The chemical reaction is better represented at this point with a double arrow: \[\ce{HC2H3O2(aq) + H2O(ℓ) <=> H3O^{+}(aq) + C2H3O^{-}2(aq)} \label{Eq6} \] The \(\rightleftharpoons\) implies that both the forward and reverse reactions are occurring, and their effects cancel each other out. A process at this point is considered to be at chemical equilibrium (or equilibrium). It is important to note that the processes do not stop. They balance out each other so that there is no further net change; that is, chemical equilibrium is a dynamic equilibrium . Example \(\PageIndex{1}\): Partial Ionization Write the equilibrium chemical equation for the partial ionization of each weak acid or base. HNO 2 (aq) C 5 H 5 N(aq) Solution HNO 2 (aq) + H 2 O(ℓ) ⇆ NO 2 − (aq) + H 3 O + (aq) C 5 H 5 N(aq) + H 2 O(ℓ) ⇆ C 5 H 5 NH + (aq) + OH − (aq) Exercise \(\PageIndex{1}\) Write the equilibrium chemical equation for the partial ionization of each weak acid or base. \(HF_{(aq)}\) \(AgOH_{(aq)}\) CH 3 NH 2 (aq) Answer a. HF(aq) + H 2 O(ℓ) ⇆ F − (aq) + H 3 O + (aq) b. AgOH(aq) ⇆ Ag + (aq) + OH − (aq) c. CH 3 NH 2 (aq) + H 2 O(ℓ) ⇆ CH 3 NH 3 + (aq) + OH − (aq) Acid Ionization Constant, \(K_\text{a}\) The ionization for a general weak acid, \(\ce{HA}\), can be written as follows: \[\ce{HA} \left( aq \right) \rightleftharpoons \ce{H^+} \left( aq \right) + \ce{A^-} \left( aq \right) \nonumber \] Because the acid is weak, an equilibrium expression can be written. An acid ionization constant \(\left( K_\text{a} \right)\) is the equilibrium constant for the ionization of an acid. \[K_\text{a} = \frac{\left[ \ce{H^+} \right] \left[ \ce{A^-} \right]}{\left[ \ce{HA} \right]} \nonumber \] The acid ionization represents the fraction of the original acid that has been ionized in solution. Therefore, the numerical value of \(K_\text{a}\) is a reflection of the strength of the acid. Weak acids with relatively higher \(K_\text{a}\) values are stronger than acids with relatively lower \(K_\text{a}\) values. Because strong acids are essentially \(100\%\) ionized, the concentration of the acid in the denominator is nearly zero and the \(K_\text{a}\) value approaches infinity. For this reason, \(K_\text{a}\) values are generally reported for weak acids only. The table below is a listing of acid ionization constants for several acids. Note that polyprotic acids have a distinct ionization constant for each ionization step, with each successive ionization constant being smaller than the previous one. Name of Acid Ionization Equation \(K_\text{a}\) Sulfuric acid \(\ce{H_2SO_4} \rightleftharpoons \ce{H^+} + \ce{HSO_4^-}\) \(\ce{HSO_4} \rightleftharpoons \ce{H^+} + \ce{SO_4^{2-}}\) very large \(1.3 \times 10^{-2}\) Hydrofluoric acid \(\ce{HF} \rightleftharpoons \ce{H^+} + \ce{F^-}\) \(7.1 \times 10^{-4}\) Nitrous acid \(\ce{HNO_2} \rightleftharpoons \ce{H^+} + \ce{NO_2^-}\) \(4.5 \times 10^{-4}\) Benzoic acid \(\ce{C_6H_5COOH} \rightleftharpoons \ce{H^+} + \ce{C_6H_5COO^-}\) \(6.5 \times 10^{-5}\) Acetic acid \(\ce{CH_3COOH} \rightleftharpoons \ce{H^+} + \ce{CH_3COO^-}\) \(1.8 \times 10^{-5}\) Carbonic acid \(\ce{H_2CO_3} \rightleftharpoons \ce{H^+} + \ce{HCO_3^-}\) \(\ce{HCO_3^-} \rightleftharpoons \ce{H^+} + \ce{CO_3^{2-}}\) \(4.2 \times 10^{-7}\) \(4.8 \times 10^{-11}\) Hydrofluoric acid \(HF_{(aq)}\) reacts directly with glass (very few chemicals react with glass). Hydrofluoric acid is used in glass etching. Strong and Weak Bases and Base Ionization Constant, K b As with acids, bases can either be strong or weak, depending on their extent of ionization. A strong base is a base, which ionizes completely in an aqueous solution. The most common strong bases are soluble metal hydroxide compounds such as potassium hydroxide. Some metal hydroxides are not as strong simply because they are not as soluble. Calcium hydroxide is only slightly soluble in water, but the portion that does dissolve also dissociates into ions. A weak base is a base that ionizes only slightly in an aqueous solution. Recall that a base can be defined as a substance, which accepts a hydrogen ion from another substance. When a weak base such as ammonia is dissolved in water, it accepts an \(\ce{H^+}\) ion from water, forming the hydroxide ion and the conjugate acid of the base, the ammonium ion. \[\ce{NH_3} \left( aq \right) + \ce{H_2O} \left( l \right) \rightleftharpoons \ce{NH_4^+} \left( aq \right) + \ce{OH^-} \left( aq \right) \nonumber \] The equilibrium greatly favors the reactants and the extent of ionization of the ammonia molecule is very small. An equilibrium expression can be written for the reactions of weak bases with water. Because the concentration of water is extremely large and virtually constant, the water is not included in the expression. A base ionization constant \(\left( K_\text{b} \right)\) is the equilibrium constant for the ionization of a base. For ammonia the expression is: \[K_\text{b} = \frac{\left[ \ce{NH_4^+} \right] \left[ \ce{OH^-} \right]}{\left[ \ce{NH_3} \right]} \nonumber \] The numerical value of \(K_\text{b}\) is a reflection of the strength of the base. Weak bases with relatively higher \(K_\text{b}\) values are stronger than bases with relatively lower \(K_\text{b}\) values. Table \(\PageIndex{3}\) is a listing of base ionization constants for several weak bases. Name of Base Ionization Equation \(K_\text{b}\) Methylamine \(\ce{CH_3NH_2} + \ce{H_2O} \rightleftharpoons \ce{CH_3NH_3^+} + \ce{OH^-}\) \(5.6 \times 10^{-4}\) Ammonia \(\ce{NH_3} + \ce{H_2O} \rightleftharpoons \ce{NH_4^+} + \ce{OH^-}\) \(1.8 \times 10^{-5}\) Pyridine \(\ce{C_5H_5N} + \ce{H_2O} \rightleftharpoons \ce{C_5H_5NH^+} + \ce{OH^-}\) \(1.7 \times 10^{-9}\) Acetate ion \(\ce{CH_3COO^-} + \ce{H_2O} \rightleftharpoons \ce{CH_3COOH} + \ce{OH^-}\) \(5.6 \times 10^{-10}\) Fluoride ion \(\ce{F^-} + \ce{H_2O} \rightleftharpoons \ce{HF} + \ce{OH^-}\) \(1.4 \times 10^{-11}\) Urea \(\ce{H_2NCONH_2} + \ce{H_2O} \rightleftharpoons \ce{H_2NCONH_3^+} + \ce{OH^-}\) \(1.5 \times 10^{-14}\) The Ion-Product of Water As we have already seen, H 2 O can act as an acid or a base. Within any given sample of water, some \(\ce{H2O}\) molecules are acting as acids, and other \(\ce{H2O}\) molecules are acting as bases. The chemical equation is as follows: \[\color{red}{\underbrace{\ce{H2O}}_{\text{acid}}} + \color{blue}{\underbrace{\ce{H2O}}_{\text{base}}} \color{black} \ce{<=> H3O^{+} + OH^{−}} \label{Auto} \] Similar to a weak acid, the autoionization of water is an equilibrium process, and is more properly written as follows: \[\ce{H2O(ℓ) + H2O(ℓ) <=> H3O^{+}(aq) + OH^{-}(aq)} \label{Eq7} \] We often use the simplified form of the reaction: \[\ce{H2O(l) <=> H+(aq) + OH−(aq)} \nonumber \] The equilibrium constant for the autoionization of water is referred to as the ion-product for water and is given the symbol Kw. \[K_w = [\ce{H^{+}}][\ce{OH^{-}}] \nonumber \] The ion-product of water ( K w) is the mathematical product of the concentration of hydrogen ions and hydroxide ions. Note that H 2 O is not included in the ion-product expression because it is a pure liquid. The value of Kw is very small, in accordance with a reaction that favors the reactants. At 25 o C, the experimentally determined value of \(K_w\) in pure water is 1.0×10 −14 . \[K_w = [\ce{H^{+}}][\ce{OH^{−}}] = 1.0 \times 10^{−14} \nonumber \] In a sample of pure water, the concentrations of hydrogen and hydroxide ions are equal to one another. Pure water or any other aqueous solution in which this ratio holds is said to be neutral. To find the molarity of each ion, the square root of Kw is taken. [H + ] = [OH − ] = 1.0×10 −7 The product of these two concentrations is 1.0×10 −14 \[\color{red}{\ce{[H^{+}]}} \color{black}{\times} \color{blue}{\ce{[OH^{-}]}} \color{black} = (1.0 \times 10^{-7})( 1.0 \times 10^{-7}) = 1.0 \times 10^{-14} \nonumber \] For acids, the concentration of H + or [H + ]) is greater than 1.0×10 −7 M For bases, the concentration of OH − or [OH − ] is greater than 1.0×10 −7 M. Aqueous HCl is an example of acidic solution. Hydrogen chloride (HCl) ionizes to produce H + and Cl − ions upon dissolving in water. This increases the concentration of H + ions in the solution. According to Le Chatelier's principle , the equilibrium represented by \[\ce{H2O(l) <=> H^{+}(aq) + OH^{−}(aq)} \nonumber \] \[\ce{HCl(g) -> H^{+}(aq) + Cl^{−}(aq)} \nonumber \] is forced to the left , towards the reactant. As a result, the concentration of the hydroxide ion decreases . Now, consider KOH (aq), a basic solution . Solid potassium hydroxide (KOH) dissociates in water to yield potassium ions and hydroxide ions. KOH(s) → K + (aq) + OH − (aq) The increase in concentration of the OH − ions will cause a decrease in the concentration of the H + ions. No matter whether the aqueous solution is an acid, a base, or neutral:and the ion-product of [H + ][OH − ] remains constant. For acidic solutions, [H + ]) is greater than [OH − ]. For basic solutions, [OH−] is greater than [H + ]. For neutral solutions, [H 3 O + ] = [OH−] = 1.0×10 −7 M This means that if you know \(\ce{[H^{+}]}\) for a solution, you can calculate what \(\ce{[OH^{−}]}\)) has to be for the product to equal \(1.0 \times 10^{−14}\), or if you know \(\ce{[OH^{−}]}\)), you can calculate \(\ce{[H^{+}]}\). This also implies that as one concentration goes up, the other must go down to compensate so that their product always equals the value of \(K_w\). \[K_w = \color{red}{\ce{[H_3O^+]}} \color{blue}{\ce{[OH^{-}]}} \color{black} = 1.0 \times 10^{-14} \label{eq10} \] Example \(\PageIndex{2}\) Hydrochloric acid (HCl) is a strong acid, meaning it is 100% ionized in solution. What is the [H + ] and the [OH − ] in a solution of 2.0×10 −3 M HCl? Solution Step 1: List the known values and plan the problem. Known [HCl] = 2.0×10 −3 M Kw = 1.0×10 −14 Unknown [H + ]=?M [OH − ]=?M Because HCl is 100% ionized, the concentration of H+ ions in solution will be equal to the original concentration of HCl. Each HCl molecule that was originally present ionizes into one H+ ion and one Cl− ion. The concentration of OH− can then be determined from the [H+] and Kw. Step 2: Solve. [H + ]=2.0×10 −3 M Kw = [H + ][OH − ] = 1.0×10 −14 [OH − ] = Kw/[H + ] = 1.0×10 −14 /2.0×10 −3 = 5.0×10 −12 M Step 3: Think about your result. The [H + ] is much higher than the [OH − ] because the solution is acidic. As with other equilibrium constants, the unit for Kw is customarily omitted. Sodium hydroxide (NaOH) is a strong base. What is the [H + ] and the [OH − ] in a 0.001 M NaOH solution at 25 °C? Answer [OH − ] = 0.001M or 1 x 10 -3 M; [H + ]=1×10 − 11 M. The pH Scale One qualitative measure of the strength of an acid or a base solution is the pH scale, which is based on the concentration of the hydronium (or hydrogen) ion in aqueous solution. \[pH = -\log[H^+] \nonumber \] or \[pH = -\log[H_3O^+] \nonumber \] Figure \(\PageIndex{3}\) illus trates this relationship, along with some examples of various solutions. Because hydrogen ion concentrations are generally less than one (for example \(1.3 \times 10^{-3}\,M\)), th e log of the number will be a negative number. To make pH even easier to work with, pH is defined as the negative log of \([H_3O^+]\) , which will give a positive value for pH. A neutral (neither acidic nor basic) solution has a pH of 7. A pH below 7 means that a solution is acidic, with lower values of pH corresponding to increasingly acidic solutions. A pH greater than 7 indicates a basic solution, with higher values of pH corresponding to increasingly basic solutions. Thus, given the pH of several solutions, you can state which ones are acidic, which ones are basic, and which are more acidic or basic than others. These are summarized in Table \(\PageIndex{4}. Table \(\PageIndex{4}\): Acidic, Basic and Neutral pH Values Empty DataFrame Columns: [(Classification, acidic, neutral, basic), (Relative Ion Concentrations, [H+] > [OH−], [H+] = [OH−], [H+] < [OH−]), (pH at 25 °C, pH < 7, pH = 7, pH > 7)] Index: [] Example \(\PageIndex{3}\) Find the pH, given the \([H^+]\) of the following: 1 ×10 -3 M 2.5 ×10 -11 M 4.7 ×10 -9 M Solution pH = - log [H 3 O + ] Substitute the known quantity into the equation and solve. Use a scientific calculator for b and c. pH = - log [ 1 × 10 −3 ] = 3. 0 (1 decimal place since 1 has 1 significant figure) pH = - log [ 2.5 ×10 -11 ] = 10. 60 (2 decimal places since 2.5 has 2 significant figures) pH = - log [ 4.7 ×10 -9 ] = 8. 33 (2 decimal places since 4.7 has 2 significant figures) Note on significant figures: Because the number(s) before the decimal point in the pH value relate to the power on 10, the number of digits after the decimal point (underlined) is what determines the number of significant figures in the final answer. Find the pH, given [H + ] of the following: 5.8 ×10 -4 M 1.0×10 -7 M Answer a. 3.24 b. 7.00 Table \(\PageIndex{5}\) lists the pH of several common solutions. The most acidic among the listed solutions is battery acid with the lowest pH value (0.3). The most basic is 1M NaOH solution with the highest pH value of 14.0. Notice that some biological fluids (stomach acid and urine) are nowhere near neutral. You may also notice that many food products are slightly acidic. They are acidic because they contain solutions of weak acids. If the acid components of these foods were strong acids, the food would likely be inedible. Solution pH battery acid 0.3 stomach acid 1–2 lemon or lime juice 2.1 vinegar 2.8–3.0 Coca-Cola 3 wine 2.8–3.8 beer 4–5 coffee 5 milk 6 urine 6 pure H2O 7 (human) blood 7.3–7.5 sea water 8 antacid (milk of magnesia) 10.5 NH3 (1 M) 11.6 bleach 12.6 NaOH (1 M) 14.0 Example \(\PageIndex{4}\) Label each solution as acidic, basic, or neutral based only on the stated \(pH\). milk of magnesia, pH = 10.5 pure water, pH = 7 wine, pH = 3.0 Solution With a pH greater than 7, milk of magnesia is basic. (Milk of magnesia is largely Mg(OH) 2 .) Pure water, with a pH of 7, is neutral. With a pH of less than 7, wine is acidic. Identify each substance as acidic, basic, or neutral based only on the stated \(pH\). human blood with \(pH\) = 7.4 household ammonia with \(pH\) = 11.0 cherries with \(pH\) = 3.6 Answer a. slightly basic b. basic c. acidic Acid Rain Normal rainwater has a pH between 5 and 6 due to the presence of dissolved CO 2 which forms carbonic acid: \[\ce{H2O (l) + CO2(g) ⟶ H2CO3(aq)} \label{14} \] \[\ce{H2CO3(aq) \rightleftharpoons H^+(aq) + HCO3^- (aq)} \label{15} \] Acid rain is rainwater that has a pH of less than 5, due to a variety of nonmetal oxides, including CO 2 , SO 2 , SO 3 , NO , and NO 2 being dissolved in the water and reacting with it to form not only carbonic acid, but sulfuric acid and nitric acid. The formation and subsequent ionization of sulfuric acid are shown here: \[\ce{H2O (l) + SO3(g) ⟶ H2SO4(aq)} \label{16} \] \[\ce{H2SO4(aq) ⟶ H^+(aq) + HSO4^- (aq)} \label{17} \] Carbon dioxide is naturally present in the atmosphere because we and most other organisms produce it as a waste product of metabolism. Carbon dioxide is also formed when fires release carbon stored in vegetation or when we burn wood or fossil fuels. Sulfur trioxide in the atmosphere is naturally produced by volcanic activity, but it also stems from burning fossil fuels, which have traces of sulfur, and from the process of “roasting” ores of metal sulfides in metal-refining processes. Oxides of nitrogen are formed in internal combustion engines where the high temperatures make it possible for the nitrogen and oxygen in air to chemically combine. Acid rain is a particular problem in industrial areas where the products of combustion and smelting are released into the air without being stripped of sulfur and nitrogen oxides. In North America and Europe until the 1980s, it was responsible for the destruction of forests and freshwater lakes, when the acidity of the rain actually killed trees, damaged soil, and made lakes uninhabitable for all but the most acid-tolerant species. Acid rain also corrodes statuary and building facades that are made of marble and limestone (Figure \(\PageIndex{4}\)). Regulations limiting the amount of sulfur and nitrogen oxides that can be released into the atmosphere by industry and automobiles have reduced the severity of acid damage to both natural and manmade environments in North America and Europe. It is now a growing problem in industrial areas of China and India. Key Takeaways Acids and bases can be strong or weak depending on the extent of ionization in solution. Most chemical reactions reach equilibrium at which point there is no net change. The ion-product of [H + ][OH − ] in an aqueous solution remains constant. A p H value is simply the negative of the logarithm of the H + ion concentration (-log[H + ]). The pH scale is used to succinctly communicate the acidity or basicity of a solution. A solution is acidic if p H < 7. A solution is basic if p H > 7. A solution is neutral if p H = 7.
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Courses/Nassau_Community_College/CHE200_-_Introduction_to_Organic_Chemistry_(Resch)/03%3A_Conformations_and_Stereochemistry/3.04%3A_Chirality_and_stereoisomers
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We turn now to concept of chirality that formed the basis of the story about Louis Pasteur in the beginning of this chapter. Recall that the term chiral , from the Greek work for 'hand', refers to anything which cannot be superimposed on its own mirror image. Your hands, of course, are chiral - you cannot superimpose your left hand on your right, and you cannot fit your left hand into a right-handed glove (which is also a chiral object). Another way of saying this is that your hands do not have a mirror plane of symmetry : you cannot find any plane which bisects your hand in such a way that one side of the plane is a mirror image of the other side. Chiral objects do not have a plane of symmetry . Your face, on the other hand is achiral - lacking chirality - because, some small deviations notwithstanding, you could superimpose your face onto its mirror image. If someone were to show you a mirror image photograph of your face, you could line the image up, point-for-point, with your actual face. Your face has a plane of symmetry, because the left side is the mirror image of the right side. What Pasteur, Biot, and their contemporaries did not yet fully understand when Pasteur made his discovery of molecular chirality was the source of chirality at the molecular level. It stood to reason that a chiral molecule is one that does not contain a plane of symmetry, and thus cannot be superimposed on its mirror image. We now know that chiral molecules contain one or more chiral centers , which are almost always tetrahedral ( sp 3 -hybridized) carbons with four different substituents. Consider the cartoon molecule A below: a tetrahedral carbon, with four different substituents denoted by balls of four different colors (for the time being, don't worry about exactly what these substituents could be - we will see real examples very soon). Another image The mirror image of A, which we will call B, is drawn on the right side of the figure, and an imaginary mirror is in the middle. Notice that every point on A lines up through the mirror with the same point on B: in other words, if A looked in the mirror, it would see B looking back. Now, if we flip compound A over and try to superimpose it point for point on compound B, we find that we cannot do it: if we superimpose any two colored balls, then the other two are misaligned. A is not superimposable on its mirror image (B), thus by definition A is a chiral molecule. It follows that B also is not superimposable on its mirror image (A), and thus it is also a chiral molecule. Also notice in the figure below (and convince yourself with models) that neither A nor B has an internal plane of symmetry. A and B are stereoisomers : molecules with the same molecular formula and the same bonding arrangement, but a different arrangement of atoms in space . There are two types of stereoisomers: enantiomers and diastereomers. Enantiomers are pairs of stereoisomers which are mirror images of each other: thus, A and B are enantiomers. It should be self-evident that a chiral molecule will always have one (and only one) enantiomer: enantiomers come in pairs. Enantiomers have identical physical properties (melting point, boiling point, density, and so on). However, enantiomers do differ in how they interact with polarized light (we will learn more about this soon) and they may also interact in very different ways with other chiral molecules - proteins, for example. We will begin to explore this last idea in later in this chapter, and see many examples throughout the remainder of our study of biological organic chemistry. Diastereomers are stereoisomers which are not mirror images of each other. For now, we will concentrate on understanding enantiomers, and come back to diastereomers later. We defined a chiral center as a tetrahedral carbon with four different substituents. If, instead, a tetrahedral carbon has two identical substituents (two black atoms in the cartoon figure below), then of course it still has a mirror image ( everything has a mirror image, unless we are talking about a vampire!) However, it is superimposable on its mirror image, and has a plane of symmetry. This molecule is achiral (lacking chirality). Using the same reasoning, we can see that a trigonal planar ( sp 2 -hybridized) carbon is also not a chiral center. Notice that structure E can be superimposed on F, its mirror image - all you have to do is pick E up, flip it over, and it is the same as F. This molecule has a plane of symmetry, and is achiral. Let's apply our general discussion to real molecules. For now, we will limit our discussion to molecules with a single chiral center. It turns out that tartaric acid, the subject of our chapter introduction, has two chiral centers, so we will come back to it later. Consider 2-butanol, drawn in two dimensions below. Carbon #2 is a chiral center: it is sp 3 -hybridized and tetrahedral (even though it is not drawn that way above), and the four things attached to is are different: a hydrogen, a methyl (-CH 3 ) group, an ethyl (-CH 2 CH 3 ) group, and a hydroxyl (OH) group. Let's draw the bonding at C 2 in three dimensions, and call this structure A. We will also draw the mirror image of A, and call this structure B. When we try to superimpose A onto B, we find that we cannot do it. A and B are both chiral molecules, and they are enantiomers of each other. 2-propanol, unlike 2-butanol, is not a chiral molecule. Carbon #2 is bonded to two identical substituents (methyl groups), and so it is not a chiral center. Notice that 2-propanol is superimposable on its own mirror image. When we look at very simple molecules like 2-butanol, it is not difficult to draw out the mirror image and recognize that it is not superimposable. However, with larger, more complex molecules, this can be a daunting challenge in terms of drawing and three-dimensional visualization. The easy way to determine if a molecule is chiral is simply to look for the presence of one or more chiral centers: molecules with chiral centers will (almost always) be chiral. We insert the 'almost always' caveat here because it is possible to come up with the exception to this rule - we will have more to say on this later, but don't worry about it for now. Here's another trick to make your stereochemical life easier: if you want to draw the enantiomer of a chiral molecule, it is not necessary to go to the trouble of drawing the point-for-point mirror image, as we have done up to now for purposes of illustration. Instead, keep the carbon skeleton the same, and simply reverse the solid and dashed wedge bonds on the chiral carbon: that accomplishes the same thing. You should use models to convince yourself that this is true, and also to convince yourself that swapping any two substituents about the chiral carbon will result in the formation of the enantiomer. Here are four more examples of chiral biomolecules, each one shown as a pair of enantiomers, with chiral centers marked by red dots. Here are some examples of achiral biomolecules - convince yourself that none of them contain a chiral center: When looking for chiral centers, it is important to recognize that the question of whether or not the dashed/solid wedge drawing convention is used is irrelevant. Chiral molecules are sometimes drawn without using wedges (although obviously this means that stereochemical information is being omitted). Conversely, wedges may be used on carbons that are not chiral centers – look, for example, at the drawings of glycine and citrate in the figure above. Can a chiral center be something other than a tetrahedral carbon with four different substituents? The answer to this question is 'yes' - however, these alternative chiral centers are very rare in the context of biological organic chemistry, and outside the scope of our discussion here. You may also have wondered about amines: shouldn't we consider a secondary or tertiary amine to be a chiral center, as they are tetrahedral and attached to four different substituents, if the lone-pair electrons are counted as a 'substituent'? Put another way, isn't an amine non-superimposable on its mirror image? The answer: yes it is, in the static picture, but in reality, the nitrogen of an amine is rapidly and reversibly inverting, or turning inside out, at room temperature. If you have trouble picturing this, take an old tennis ball and cut it in half. Then, take one of the concave halves and flip it inside out, then back again: this is what the amine is doing. The end result is that the two 'enantiomers' if the amine are actually two rapidly interconverting forms of the same molecule, and thus the amine itself is not a chiral center. This inversion process does not take place on a tetrahedral carbon, which of course has no lone-pair electrons. Exercise 3.8 Locate all of the chiral centers (there may be more than one in a molecule). Remember, hydrogen atoms bonded to carbon usually are not drawn in the line structure convention - but they are still there! Exercise 3.9 a) Draw two enantiomers of i) mevalonate and ii) serine. b) Are the two 2-butanol structures below enantiomers? Exercise 3.10 Label the molecules below as chiral or achiral, and locate all chiral centers. Solutions to exercises Khan Academy video tutorials Chirality Enantiomers
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Courses/University_of_Arkansas_Little_Rock/00001%3AF23_Gen_Chem_2_Lab/Ancillary_Material/Drafts
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Graphing Tutorials and Videos The following tutorials and videos guide you through the use of Google Sheets. Organizing Google Drive Sheet Skills Shortcut to move file in drive Instructions Tabs Sheet Skills Video Create, Duplicate, Rename, and Organize Tabs Instructions Formulas Basics Sheet Skills Formulas Basics: Use formulas, reference cells Instructions Simple Mathematical Functions Sheet Skills Video Simple Mathematical Functions: Addition, Subtraction, Multiplication, Division, Exponents, Parentheses Instructions 9 Add texts here. Do not delete this text first. Logarithms Sheet Skills Video Logarithms: Log and LN Instructions Charts Charts Basics Sheet Skills Video Making and editing a chart, titles Instructions Trendline and Equations Sheet Skills Making a line of best fit and types of equations Answer Multiple Series Sheet Skills Video Creating a chart to compare data Answer
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Courses/University_of_Arkansas_Little_Rock/Chem_1403%3A_General_Chemistry_2/Homework/17%3A_Aqueous_Equilibria
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Common Ion Effect Exercise \(\PageIndex{1.a}\) Consider a solution of 0.20 M HF, K a = 6.8x10 -4 . \(HF\rightleftharpoons H^+ + F^-\) What is the % Ionization of the pure HF solution? Would you expect that adding NaF would increase or decrease the % ionization What is the % Ionization if the solution was also 0.20 M NaF? Answer a. For pure 0.20 M HF, [HA] i <100K a so: \[ [H^+]=[F^-]=\sqrt{K_a[HA]_i} =\sqrt{(6.8x10^{-4})(0.20M)}= 0.012M \\ so \\ \% I=\frac{[A^-]}{[HA]_i}(100)=\frac{0.012}{.2}(100)=5.8 \text{% ionized} \nonumber\] Answer b. Adding fluoride will push the reaction to the left and suppress ionization Answer c. 0 1 2 3 4 R HF(aq) ⇌ H+(aq) + F-(aq) I 0.20 M NaN 0 0.20 M C -x NaN +x +x E 0.20-x NaN x 0.20+x \[K_{a}=\frac{x\left ( 0.20+x \right )}{0.20-x}=\approxeq \frac{x(0.2)}{0.2}=6.8*10^{-4}\nonumber\] \[x=\left [ H^{+} \right ]=6.8*10^{-4}\nonumber\] \[\text{% ionization}=\frac{6.8x10^{-4}}{.2}=0.34 \text{% ionized} \nonumber \] So the common ion reduced the % ionized from 5.8% to 0.34% Buffer Solutions Exercise \(\PageIndex{2.a}\) The Henderson-Hasselbalch equation is \(\left [ H^{+} \right ]=K_{a}+\frac{\left [salt \right ]}{\left [ acid \right ]}\) \(pH=pK_{a}-log\frac{\left [salt \right ]}{\left [ acid \right ]}\) \(pH=pK_{a}+log\frac{\left [ salt \right ]}{\left [ acid \right ]}\) \(pH=pK_{a}+log\frac{\left [ acid \right ]}{\left [ salt \right ]}\) \(pH=log\frac{\left [ acid \right ]}{\left [ salt\right ]}\) Answer c. \(pH=pK_{a}+log\frac{\left [ salt \right ]}{\left [ acid \right ]}\) Exercise \(\PageIndex{2.b}\) Consider a solution of 0.20 M HF and 0.20 M NaF, K a = 6.8x10 -4 in a 1 liter container. \(HF\rightleftharpoons H^+ + F^-\) What is the pH of the solution? What is the change in pH if you add 0.02mol NaOH to one liter of water neutral? What is the change in pH if you add 0.02mol NaOH to one liter of the above solution that is 0.20M HF and 0.20M NaF? Answer a. 0 1 2 3 4 R HF(aq) ⇌ H+(aq) + F-(aq) I 0.20 M NaN 0 0.20 M C -x NaN +x +x E 0.20-x NaN x 0.20+x \[K_{a}=\frac{x\left ( 0.20+x \right )}{0.20-x}=\approxeq \frac{x(0.2)}{0.2}=6.8*10^{-4}\nonumber\] \[x=\left [ H^{+} \right ]=6.8*10^{-4}\nonumber\] \[pH=-log6.8x10^{-4}=3.17 \nonumber \] Answer b. pOH=-log0.02=1.70, so pH = 12.3. so the change in pH is 12.3=7=5.3 Answer c. The 0.02mol NaOH converts 0.02mol of HF to A - , so after reaction [HF]= 0.20-0.02 = 0.18M and [A - ]= 0.20+0.02M=2.02M \[ph=pK_a+log{F^-}{HA}=-log(6.8x10^{-4})+log\frac{2.02}{0.18} = 3.26 \nonumber \] so the change in pH is 3.26-3.17 = 0.09 Exercise \(\PageIndex{2.c}\) What is the ratio of HCO 3 - to H 2 CO 3 in our blood that has a pH=7.4? Ka 1 =4.3x10 -7 Answer \[pH=PK_a+ log \frac{[A^-]}{[HA]} \\ \; \\ pH-pK_a=log \frac{[A^-]}{[HA]} \;\;\therefore \;\;\frac{[A^-]}{[HA]}=10^{pH-pK_a} \nonumber\] \[pK_a=-log4.3x10^{-7}=6.37 \nonumber\] \[\frac{\left [ HCO_{3}^{-} \right ]}{\left [ H_{2}CO_{3} \right ]}=10^{7.4-6.367} = 10.8 \nonumber\] Exercise \(\PageIndex{2.d}\) How many grams of NH 4 Cl must be added to 1.00L of 0.10M NH 3 to form a buffer that has a pH=9.0? Kb=1.8x10 -5 (Regardless the volume change.) Answer \[pOH=pKb+log\frac{salt}{base} \;\; \therefore \;\; \frac{salt}{base}=10^{pOH-pK_b} \\ \; \\ \; \; and \;[salt]=[base]10^{pOH-pK_b} \\ now \; pOH=14-pH=14-9=5 \;\; and \;\;pK_b=-log(1.8x10^{-5})= 4.74 \\ \; \\ \text{ammonium chloride is the salt and so }[NH_4Cl]= [0.1M]10^{5-4.74}=0.18M\] so 1 liter needs 0.18mol, \[0.18mol*53.45gNH_4Cl/mol=9.62g \nonumber \] Exercise \(\PageIndex{2.e}\) Which one of the following pairs cannot be mixed to form a buffer solution? H 3 PO 4 , KH 2 PO 4 KOH, HF NaC 2 H 3 O 2 , HCl NH 3 , NH 4 Cl RbOH, HBr Answer e. RbOH, HBr Exercise \(\PageIndex{2.f}\) Consider a solution containing 0.100M fluoride ion and 0.126M hydrogen fluoride. The concentration of hydrogen fluoride after addition of 5.00mL if 0.0100M HCl to 25.0mL of this solution is ______. Answer First calculate the moles of each species \[mol_{F^{-}}= 0.100M*0.0250L=2.50*10^{-3}mol \nonumber \] \[mol_{HF}=0.126M*0.0250L=3.15*10^{-3}mol \nonumber \] \[mol_{HCl}=mol_{H^{+}}=0.0100M*0.00500L=5.00*10^{-5}mol \nonumber \] Now calculate how much of each species is left over, noting all the H + from the HCl reacted with the fluoride \[x=mol_{H^{+}}=5.00*10^{-5}mol \nonumber \] 0 1 2 3 4 R F-(aq) + H+(aq) ⇌ HF(aq) I 2.50*10-3 mol 5.00*10-5 mol NaN 3.15*10-3 mol C -x -x NaN +x E 2.50*10-3 - x 0 NaN 3.15*10-3 + x \[F^{-}=\left ( 2.50*10^{-3}mol \right )-\left ( 5.00*10^{-5}mol \right )=2.45*10^{-3}mol \nonumber \] \[HF=\left ( 3.15*10^{-3}mol \right )+\left ( 5.00*10^{-5}mol \right )=3.20*10^{-3}mol \nonumber \] To get the HF concentration you need to divide the moles by the total volume, and take into account the dilution factor. \[\left [ HF \right ]=\frac{3.20*10^{-3}mol}{0.0250L+0.00500L}=0.107M \nonumber \] Exercise \(\PageIndex{2.g}\) Of the following, which solution has the greatest buffering capacity? 0.821 M HF and 0.217 M NaF 0.821 M HF and 0.909 M NaF 0.100 M HF and 0.217 M NaF 0.121 M HF and 0.667 M NaF They are all buffer solutions and would all have the same capacity. Answer b. 0.821 M HF and 0.909 M NaF Exercise \(\PageIndex{2.h}\) The K a of acetic acid is 1.7*10 -5 . What is the pH of a buffer prepared by combining 50.0mL of 1.00M potassium acetate and 50.0mL of 1.00M acetic acid? Answer First note you by adding equal volumes that you diluted each solution in half \[\left [K CH_{3}COO \right ]=\frac{1.00M*50.0mL}{50.0mL+50.0mL}=0.500M \nonumber \] \[\left [ CH_{3}COOH \right ]=\frac{1.00M*50.0mL}{50.0mL+50.0mL}=0.500M \nonumber \] Second, note how by adding equal volumes of equal concentrations of the acid and salt, that pH=pK a . \[pH=pK_{a}+log\frac{\left [ CH_{3}COOK \right ]}{\left [ CH_{3}COOH \right ]}=-log\left ( 1.7*10^{-5} \right )+log\frac{\cancel{0.500M}}{\cancel{0.500M}}=4.77 \nonumber \] Exercise \(\PageIndex{2.i}\) The K b of ammonia is 1.8*10 -5 . What is the pH of a buffer prepared by combining 50.0mL of 1.00M ammonia and 50.0mL of 1.00M ammonium nitrate? Answer Note, as you have equal mole of the base and its salt, and so pOH=pK_b=-log(1.8x10^{-5}=4.74, so pH=14-4.74=9.26 Taking into account dilution: \[\left [ NH_{4}NO_{3} \right ]=\frac{1.00M*50mL}{50.0mL+50.0mL}=0.500M \nonumber \] \[\left [ NH_{3} \right ]=\frac{1.00M*50mL}{50.0mL+50.0mL}=0.500M \nonumber \] \[pOH=pK_{b}+log\frac{\left [ NH_jj{4}NO_{3} \right ]}{\left [ NH_{3} \right ]}=-log\left ( 1.8*10^{-5} \right )+log\frac{\cancel{0.500M}}{\cancel{0.500M}}=4.77 \nonumber \] f Exercise \(\PageIndex{2.j}\) Calculate the pH of a solution prepared by dissolving 0.25 mol of benzoic acid (C 7 H 5 O 2 H) and 0.15 mol of sodium benzoate (NaC 7 H 5 O 2 ) in 1.00L of solution. K a = 6.5*10 -5 for benzoic acid. Answer \[pH=pK_{a}+log\frac{salt}{acid}=-log(6.5*10^{-5})+log\frac{0.150}{0.250}=4.19+(-0.22)=3.97 \nonumber \] Exercise \(\PageIndex{2.k}\) Determine the pH of a solution prepared by adding 0.45 mol of solid KOAc to 1.00L of 2.00M HOAc. K a = 1.8*10 -5 of HOAc. Answer \[pH=pK_{a}+log\frac{\left [ CH_{3}COO^{-} \right ]}{\left [ CH_{3}COOH \right ]}=-log\left ( 1.8*10^{-5} \right )+log\frac{0.45mol}{2.00mol}=4.10 \nonumber \] Exercise \(\PageIndex{2.l}\) The pH of a solution is prepared by dissolving 0.35 mol of solid CH 3 NH 3 Cl (methylamine hydrochloride) in 1.00L of 1.10 M CH 3 NH 2 (methylamine) is _____. The K b for methylamine is 4.4*10 -4 . Answer \[pOH=pK_{b}+log\frac{\left [ CH_{3}NH_{3}^{+} \right ]}{\left [ CH_{3}NH_{2} \right ]}=-log\left ( 4.4*10^{-4} \right )+log\frac{0.35mol}{1.10mol}=2.86 \nonumber \] \[pH=14-pOH=14-2.86=11.14 \nonumber \] Exercise \(\PageIndex{2.m}\) Which of the following substances, when added to a solution of hydrofluoric acid, could be used to prepare a buffer solution? HCl NaNO 3 NaF NaCl NaBr Answer c. NaF Exercise \(\PageIndex{2.n}\) Which of the following could not be added to a solution of sodium acetate to prepare a buffer? sodium acetate acetic acid hydrochloric acid nitric acid more than one of these answers is correct Answer a. sodium acetate Exercise \(\PageIndex{2.o}\) Calculate the fluoride ion concentration (in M) in a 1.0L aqueous solution containing 0.40 mol of HF and 0.10 mol of HCl. (Ka for HF = 6.8*10 -4 ). Answer 0 1 2 3 4 R HF(aq) ⇌ H+(aq) + F-(aq) I 0.40 mol NaN 0.10 mol 0 C -x NaN +x +x E 0.40-x NaN 0.10+x x \[K_{a}=\frac{\left [ H^{+} \right ]\left [ F^{-} \right ]}{\left [ HF \right ]}=\frac{\left [ 0.10+x \right ]\left [ x \right ]}{\left [ 0.40-x \right ]}=6.8*10^{-4}\nonumber\] \[\frac{0.10(x)}{0.40}=6.8*10^{-4}\nonumber\] \[x=\frac{0.40}{0.10}*6.8*10^{-4}=2.7*10^{-3}M\nonumber\] Exercise \(\PageIndex{2.p}\) Calculate the pH of 1.0L aqueous solution containing 0.30mol of HF and 0.10mol of HCl. (K a for HF = 6.8*10 -4 ) Answer 0 1 2 3 4 R HF(aq) ⇌ H+(aq) + F-(aq) I 0.30 mol NaN 0.10 mol 0 C -x NaN +x +x E 0.30-x NaN 0.10+x x \[K_{a}=\frac{\left [ H^{+} \right ]\left [ F^{-} \right ]}{\left [ HF \right ]}=\frac{\left [ 0.10+x \right ]\left [ x \right ]}{\left [ 0.30-x \right ]}=6.8*10^{-4}\nonumber\] \[\frac{0.10(x)}{0.43}=6.8*10^{-4}\nonumber\] \[x=\frac{0.40}{0.10}*6.8*10^{-4}=2.7*10^{-3}M\nonumber\] \[pH=-log\left [ H^{+} \right ]=-log\left [ 0.10+2.7*10^{-3}M \right ]=0.99 \nonumber \] Acid-Base Titrations Textbook: Section 17.3 Exercise \(\PageIndex{3.a}\) How many milliliters of 0.0750M NaOH are required to titrate 50.0ml of 0.025M HCl? Answer \[OH^{-}(aq)+H^{+}\rightleftharpoons H_{2}O(l) \nonumber \] \[\frac{50.0mL*0.025M}{0.075M}=16.7mL \nonumber \] Exercise \(\PageIndex{3.b}\) How many milliliters of the same NaOH are needed to titrate 25.0ml of HCl solution that has 1.85g of HCl per liter? Answer \[OH^{-}(aq)+H^{+}\rightleftharpoons H_{2}O(l) \nonumber \] \[\frac{\frac{1.85g/L}{36.45g/mol}*25.0mL}{0.075M}=16.9mL \nonumber \] Exercise \(\PageIndex{3.c}\) How many milliliters of the same NaOH are needed to titrate 20.0ml of 0.050M HBr? Answer \[OH^{-}(aq)+HBr\rightleftharpoons H_{2}O(l)+Br^{-}(aq) \nonumber \] \[\frac{0.050M*20.0mL}{0.075M}=13.3mL \nonumber \] Exercise \(\PageIndex{3.d}\) Calculate the pH of the solution formed when 20.0ml of 0.100M NaOH is added to 40.0ml of 0.100M of HC 2 H 3 O 2 . Ka=1.8x10 -5 Answer You are neutralizing a weak acid (acetic acid) with a strong base (NaOH). You first need to find the equivalence point and determine if they are in stoichiometric proportions, or if one is in excess. This in effect tells you which region of the titration curve you are in. The vol. base required to neutralize all the acid is \[M_aV_a=M_bV_b \; \therefore \; V_b=V_a(\frac{M_a}{M_b})= 40.0mL(\frac{1.0M}{1.0M})= 40.0mL\] This makes total sense as you have equal concentrations, so it will take equal volumes to neutralize each other. You have added half the amount of base required to neutralize the weak acid, and so you are at a half titer, and the Henderson Hasselbach equation becomes pH = pK a , as half of the weak acid has been neutralized, so \[pH=-log\left ( 1.8*10^{-5}M \right )=4.74 \nonumber \] Of course you could have used eq. 17.3.8 , noting that the ration of the salt to the weak acid is the ratio of the volume of strong base added to the volume still needed to get to the equivalence point: \[ pH = pK_a + log{V_b}{V_{eq}-V_b}= pK_a+ log\frac{20}{40-20}= pK_a+ log\frac{20}{20}=pK_a=-log\left ( 1.8*10^{-5}M \right )=4.74 \nonumber \] or if you want to spend a real lot of time: \[ [HA] = \frac{n_{HA_{initial}}-n_{HA_{consumed}}}{V_{total}} =\frac{M_aV_a(initial)-M_bV_b}{V_a+V_b} \; \; and [A^-]=\frac{M_bV_b}{V_a+V_b} \nonumber \] Then the Henderson Hasselbach eq becomes \[ pH = pK_a + log\frac{\frac{M_bV_b}{V_a+V_b}}{\frac{M_aV_a(initial)-M_bV_b}{V_a+V_b} } = -log\left ( 1.8*10^{-5}M \right )+ log\frac {\frac{0.1M*0.400l-0.1M*0.20l}{0.40l+0.20l}}{\frac{0.1M*0.20l}{0.40l+0.20l}} \\ \; \\ -log\left ( 1.8*10^{-5}M \right )+ log\frac {\frac{0.04molel-0.02mol}{0.60l}}{\frac{0.02mol}{0.60l}} \\ \; \\ = -log\left ( 1.8*10^{-5}M \right )+ log\frac {\frac{0.02mol}{0.60l}}{\frac{0.02mol}{0.60l}}= -log\left ( 1.8*10^{-5}M \right ) =4.74\] Exercise \(\PageIndex{3.e}\) Calculate the pH at the equivalence point in the titration of 50.0ml of 0.20M HC 2 H 3 O 2 with 0.05M NaOH. Ka=1.8x10 -5 Answer All of the weak acid has been neutralized to its salt, and so you need to calculate it's concentration after dilution. \[\frac{50.0mL*0.20M}{0.05}M=200.0mL \nonumber \] \[\left [ C_{2}H_{3}O_{2}^{-} \right ]=\frac{0.20M*0.050L}{\left ( 0.050+0.200 \right )L}=0.040M \nonumber \] \[C_{2}H_{3}O_{2}^{-}(aq)+H_{2}O(l)\rightleftharpoons HC_{2}H_{3}O_{2}(aq)+OH^{-}(aq) \nonumber \] Now calculate K b ' (of the conjugate base, the salt) \[K_{b}=\frac{K_{w}}{K_{a}}=\frac{10^{-14}}{1.8*10^{-5}}=5.56x10^{-10} \] Since \([C_{2}H_{3}O_{2}^{-}] > 100K_b, \; \; pOH=-log(\sqrt{K_b'[A^-]})\) \[pOH=-log(\sqrt{5.56x10^{-10}(0.040M)}= -log(4.716x10^{-6})=5.33 \nonumber \] \[pH=14-5.33=8.67\] To solve in two steps: \[pOH = -log\sqrt{K_b'[A^-]}=-log\sqrt{\frac{K_w}{K_a}(\frac{M_aV_a}{V_a+V_b})} \\ \; \\ pOH=-log\sqrt{(\frac{1x10^{-14}}{1.8x10^{-5}})( \frac{(0.20M)0.050l}{0.050l+0.200l})}=5.33 \\ \; \\ pH=14-pOH=14-5.33 = 8.67\] Exercise \(\PageIndex{3.f}\) Determine the pH when 50.0ml of 0.010M NaOH is added to 40.0ml of 0.010M HC 2 H 3 O 2 . Answer First, calculate the volume of base required to neutralize the acid, the equivalence point. \[M_bV_b=M_aV_a \therefore V_B=V_a(\frac{M_a}{M_b}) = 40mL(\frac{0.10M}{0.10M}) = 40 mL\] Since the volume added is greater than the volume needed to neutralize the acid we are in the excess base region. \[ [OH^-]_{excess} = \frac {M_b(V_b(total)-V_b(equiv))}{V_b+V_a}=\frac{0.010M(50.0ml-40.0ml)}{50.0+40.0}=0.0011M\] pOH=-log0.0011=2.95 pH=14-pOH=14-2.95-11.05 Exercise \(\PageIndex{3.g}\) Consider the titration of 25.0mL of 0.723M HClO 4 with 0.273M KOH. What is the H 3 O + concentration before any KOH is added? What is the H 3 O + concentration after addition of 10.0mL of KOH? What is the H 3 O + concentration after addition of 66.2mL of KOH? What is the H 3 O + concentration after addition of 80.0mL of KOH? Answer a. \[\left [ H_{3}O^{+} \right ]=\left [ HClO_{4} \right ]=0.723mol \nonumber \] Answer b. \[moles\,HClO_{4}=0.723M*0.0250L=0.018075mol \nonumber \] \[moles\,KOH=0.273M*0.0100L=0.00273mol \nonumber \] \[moles\,H_{3}O^{+}=moles\,HClO_{4}-moles\,KOH=0.018075mol-0.00273mol=0.015345mol \nonumber \] \[\left [ H_{3}O^{+} \right ]=\frac{0.015345mol}{0.0250L+0.0100L}=0.438M \nonumber \] Answer c. \[moles\,HClO_{4}=0.723M*0.0250L=0.018075mol \nonumber \] \[moles\,KOH=0.273M*0.0662L=0.0180725mol \nonumber \] The solution is neutral because the acid and base have the same amount of moles. So the concentration of H 3 O + is 1.00*10 -7 . Answer d. \[moles\,HClO_{4}=0.723M*0.0250L=0.018075mol \nonumber \] \[moles\,KOH=0.273M*0.0800L=0.02184mol \nonumber \] \[excess\,moles=moles\,KOH-moles\,HClO_{4}=0.02184mol-0.018075mol=0.003765mol \nonumber \] \[\left [ OH^{-} \right ]=\frac{0.003765mol}{0.0800L+0.0250L}=0.0359M \nonumber \] \[\left [ H_{3}O^{+} \right ]=\frac{10^{-14}}{\left [ OH^{-} \right ]}=\frac{10^{-14}}{0.0359M}=2.79*10^{-13}M \nonumber \] Exercise \(\PageIndex{3.h}\) A _____ yields a titration curve with an initial pH of 1.00, an equivalence point at pH 7.00, and a relatively long, nearly vertical middle section. strong acid titrated by a strong base strong base titrated by a strong acid weak acid titrated by a strong base weak base titrated by a strong acid weak base titrated by a weak acid Answer a. strong acid titrated by a strong base Exercise \(\PageIndex{3.i}\) An initial pH of 13.00, an equivalence point at pH 7.00 and a relatively long, nearly vertical middle section corresponds to a titration curve for _____. strong acid titrated by a strong base strong base titrated by strong acid weak acid titrated by strong acid weak base titrated by strong acid weak base titrated by weak acid Answer b. strong base titrated by strong acid Exercise \(\PageIndex{3.j}\) The pH of a solution prepared by mixing 45mL of 0.183M KOH with a 65mL of 0 .145M HCl is_____? Answer These are both strong \[moles\,KOH=0.183*0.045L=0.008235mol \nonumber \] \[moles\,HCl=0.145*0.065L=0.009425mol \nonumber \] There is more acid than base, so pH is determined by the excess acid \[moles\,H^{+}_{excess}=moles\,HCl-moles\,KOH=0.009425mol-0.008235mol=0.00119mol \nonumber \] \[\left [ H^{+} \right ]=\frac{0.00119mol}{0.045+0.065}=0.01082M \nonumber \] \[pH=-log\left [ H^{+} \right ]=-log(0.01082M)=1.97 \nonumber \] Titration of Weak Acid by Strong Base Exercise \(\PageIndex{3.1.a}\) 25mL of 0.15 M CH 3 COOH K a = 1.8*10 -5 titrated with 0.15M NaOH What is the pH at V B (volume of base) = 0? Answer \[pH=-log\left [ H^{+} \right ]=-log\sqrt{K_{a}\left [ HA \right ]_{i}}\nonumber \] \[pH=-log\sqrt{(1.8*10^{-5})*0.15M}=-log(0.0016431677)=2.78\nonumber \] Exercise \(\PageIndex{3.1.b}\) What is the pH if 25mL of 0.15 M CH 3 COOH (K a = 1.8*10 -5 ) is titrated with 10.0mL of 0.15M NaOH Answer First calculate the volume to reach the equivlalence point and then determine which region of the titration curve you are in \[M_aV_a=M_bV_b \therefore V_b(equiv)= V_a\frac{M_a}{M_b}= 25mL\frac{0.15M}{0.15M}=25mL\] We are in the buffer region, and have neutralized 10/25ths, and have 15/25ths to go.so \[pH=pK_{a}+log\frac{\left [ A^{-} \right ]}{\left [ HA \right ]}= -log1.8*10^{-5} + log{10}{15}=4.47 \nonumber \] Exercise \(\PageIndex{3.1.c}\) 25mL of 0.15 M CH 3 COOH K a = 1.8*10 -5 titrated with 0.15M NaOH What is the pH at V B at half equivalence (12.5mL)? Answer \[pH=pKa\nonumber \] \[pH=-log\left ( 1.8*10^{-5} \right )=4.74\nonumber \] Exercise \(\PageIndex{3.1.d}\) What is the pH if 25mL of 0.15 M CH 3 COOH (K a = 1.8*10 -5 ) titrated with 15 mL of 0.15M NaOH Answer First calculate the volume to reach the equivalence point and then determine which region of the titration curve you are in \[M_aV_a=M_bV_b \therefore V_b(equiv)= V_a\frac{M_a}{M_b}= 25mL\frac{0.15M}{0.15M}=25mL\] We are in the buffer region, and have neutralized 15/25ths, and have 10/25ths to go.so \[pH=pK_{a}+log\frac{\left [ A^{-} \right ]}{\left [ HA \right ]}= -log1.8*10^{-5} + log{15}{10}=4.92 \nonumber \] Exercise \(\PageIndex{3.1.e}\) What is the pH at the equivalence point if 25mL of 0.15 M CH 3 COOH (K a = 1.8*10 -5 ) is titrated with 0.15M NaOH Answer First calculate the volume to reach the equivalence point and then determine which region of the titration curve you are in \[M_aV_a=M_bV_b \therefore V_b(equiv)= V_a\frac{M_a}{M_b}= 25mL\frac{0.15M}{0.15M}=25mL\] \[\left [ OH^{-} \right ]=\sqrt{K_{b}'\left [ A^{-} \right ]} = \sqrt{\frac{K_w}{K_a}[A^-]} \\ \; \\ [OH^-]=\sqrt{\frac{K_w}{K_a}[\frac{M_bV_b}{V_a+V_b}]} =\sqrt{\frac{10^{-14}}{1.8x10^{-5}}[\frac{(0.15M)(25ml}{25ml+25ml}]} =6.45x10^{-6} \nonumber \] \[pOH=-log[OH^-]=-log6.45x10^{-6} =5.19 \\ \; \\ pH=14-pOH=14-5.19=8.81 \] Exercise \(\PageIndex{3.1.f}\) What is the pH when 25mL of 0.15 M CH 3 COOH (K a = 1.8*10 -5 ) is titrated with 30.0 mL of0.15M NaOH Answer First calculate the volume to reach the equivalence point and then determine which region of the titration curve you are in \[M_aV_a=M_bV_b \therefore V_b(equiv)= V_a\frac{M_a}{M_b}= 25mL\frac{0.15M}{0.15M}=25mL\] So there are 5 mL of excess base \[[OH^-]=\frac{(0.15M)(5mL)}{25 mL + 30 mL} = 0.0136\] \[pH=14-pOH=14-(-log(0.0136))=12.13 Exercise \(\PageIndex{3.1.g}\) What is the molarity of an HOAc solution if 25.5mL of this solution required 37.5mL of 0.175M NaOH to reach the equivalence point? Answer \[M_{a}V_{a}=M_{b}V_{b}\nonumber \] \[M_{a}=\frac{M_{b}V_{b}}{V_{a}}=\frac{0.175M*0.0375L}{0.0255L}=0.257M\nonumber \] Exercise \(\PageIndex{3.1.h}\) Consider an experiment where 35.0mL of 0.175M HOAc is titrated when 0.25M in NaOH. What is the pH at equivalence point for this titration? The Ka for HOAc is 1.8 * 10 -5 . Answer First calculate the volume to reach the equivalence point and then determine which region of the titration curve you are in \[M_aV_a=M_bV_b \therefore V_b(equiv)= V_a\frac{M_a}{M_b}= 35mL\frac{0.175M}{0.25M}=24.5mL\] \[\left [ OH^{-} \right ]=\sqrt{K_{b}'\left [ A^{-} \right ]} = \sqrt{\frac{K_w}{K_a}[A^-]} \\ \; \\ [OH^-]=\sqrt{\frac{K_w}{K_a}[\frac{M_bV_b}{V_a+V_b}]} =\sqrt{\frac{10^{-14}}{1.8x10^{-5}}[\frac{(0.175M)(35ml)}{24.5ml+35ml}]} =7.56x10^{-6} \nonumber \] \[pOH=-log[OH^-]=-log7.56x10^{-6} =5.12 \\ \; \\ pH=14-pOH=14-5.12=8.88 Exercise \(\PageIndex{3.1.i}\) 50.50mL of 0.116 M HF is titrated with 0.1200 M NaOH. (Ka for HF is 6.8 * 10 -4 ) How many mL of the base are required to reach the equivalence point? What is the pH after 50.50mL of base has been added? Answer a. \[M_{1}V_{1}=M_{2}V_{2}\nonumber \] \[V_{2}=\frac{M_{1}V_{1}}{M_{2}}=\frac{0.116M*0.0505L}{0.1200M}=48.82mL\nonumber \] Answer b. The amount of excess base is 50.50mL-48.82mL=1.68mL \[ [OH^-] = \frac{M_bV_b(excess)}{V_T}= \frac{0.1200M(1.68mL)}{50.50mL+50.50mL}=0.0019M \nonumber\] \[pOH=-log0.0019=2.70 \;\; pH=14-pOH=14-2.7=11.300\] Exercise \(\PageIndex{3.1.j}\) 65.50mL of 0.1610 M HF is titrated where 0.1200 M NaOH. What is the pH at the equivalence point? (Ka for HF is 6.8 * 10 -4 ) Answer First calculate the volume to reach the equivalence point. \[M_aV_a=M_bV_b \therefore V_b(equiv)= V_a\frac{M_a}{M_b}= 65.50mL\frac{0.1610M}{0.1200M}=87.88mL\] \[\left [ OH^{-} \right ]=\sqrt{K_{b}'\left [ A^{-} \right ]} = \sqrt{\frac{K_w}{K_a}[A^-]} \\ \; \\ [OH^-]=\sqrt{\frac{K_w}{K_a}[\frac{M_bV_b}{V_a+V_b}]} =\sqrt{\frac{10^{-14}}{6.8x10^{-4}}[\frac{(0.1200M)(87.88ml)}{65.50ml+87.88ml}]} =1.0x10^{-6} \nonumber \] \[pOH=-log[OH^-]=-log1.00x10^{-6} =6.00 \\ \; \\ pH=14-pOH=14-6=8.00 Titration of Weak Base with Strong Acid Exercise \(\PageIndex{3.2.a}\) What is the pH if 25mL of 0.15M NH 3 (K b = 1.8*10 -5 ) is titrated with 0.00 mL of 0.15M HCl What is the pH at V A (volume of acid) = 0? Answer First calculate the volume to reach the equivalence point and then determine which region of the titration curve you are in \[M_aV_a=M_bV_b \therefore V_b(equiv)= V_a\frac{M_a}{M_b}= 65.50mL\frac{0.1610M}{0.1200M}=87.88mL\] \[\left [ OH^{-} \right ]=\sqrt{K_{b}*\left [ NH_{3} \right ]}=\sqrt{1.8*10^{-5}*\left [ 0.15 \right ]}\nonumber\] \[pOH=-log\left ( \sqrt{1.8*10^{-5}*\left [ 0.15 \right ]} \right )=2.78\nonumber\] \[pH=14-pOH=14-2.78=11.22\nonumber\] Exercise \(\PageIndex{3.s.b}\) What is the pH if 25mL of 0.15M NH 3 (K b = 1.8*10 -5 ) is titrated with 10 mL of 0.15M HCl? Answer First calculate the volume to reach the equivalence point and then determine which region of the titration curve you are in \[M_aV_a=M_bV_b \therefore V_b(equiv)= V_a\frac{M_a}{M_b}= 250mL\frac{0.15M}{0.15M}=25mL\] You are in the buffer region, so use the Henderson Hasselbach eq. of a base titrated with an acid. You have titrated 10/25ths of the base and have 15/25ths to go, os the ratio of salt to base is 10 to 15 or 2 to 3 \[pOH=pK_{b}+log\frac{base}{salt}\nonumber\] \[pOH=-log(1.8*10^{-5})+log\frac{2}{3} = 4.74-0.176=4.57\nonumber\] \[pH=14-pOH=14-4.57=9.43\nonumber\] Exercise \(\PageIndex{3.2.c}\) What is the pH when 25mL of 0.15M NH 3 (K b = 1.8*10 -5 ) titrated with 12.5 mL of 0.15M HCl Answer First calculate the volume to reach the equivalence point and then determine which region of the titration curve you are in \[M_aV_a=M_bV_b \therefore V_b(equiv)= V_a\frac{M_a}{M_b}= 250mL\frac{0.15M}{0.15M}=25mL\] You are at half equivalence, so [B]=[BH + ] (you have turned have the base to its conjugate acid) and pOH=pK b . \[pOH=pK_{b}\nonumber\] \[pOH=-log(1.8*10^{-5})=4.74\nonumber\] \[pH=14-pOH=14-4.74=9.26\nonumber\] Exercise \(\PageIndex{3.2.d}\) What is the pH if 25mL of 0.15M NH 3 (K b = 1.8*10 -5 ) is titrated with 15 mL of 0.15M HCl Answer First calculate the volume to reach the equivalence point and then determine which region of the titration curve you are in \[M_aV_a=M_bV_b \therefore V_b(equiv)= V_a\frac{M_a}{M_b}= 250mL\frac{0.15M}{0.15M}=25mL\] You are in the buffer region, so use the Henderson Hasselbach eq. of a base titrated with an acid. You have titrated 15/25ths of the base and have 10/25ths to go, os the ratio of salt to base is 15 to 10 or 3 to 2 \[pOH=pK_{b}+log\frac{base}{salt}\nonumber\] \[pOH=-log(1.8*10^{-5})+log\frac{3}{2} = 4.74+0.176=4.92\nonumber\] \[pH=14-pOH=14-4.92=9.08\nonumber\] Exercise \(\PageIndex{3.2.e}\) What is the pH when 25mL of 0.15M NH 3 (K b = 1.8*10 -5 ) titrated with 25 mL of 0.15M HCl? Answer You have neutralized all the base and converted it to its salt \[ [H^+]=\sqrt{K_{a}'\left [ B \right ]} = \sqrt{\frac{K_w}{K_b}[B]} \\ \; \\ [H^+]=\sqrt{\frac{K_w}{K_b}[\frac{M_aV_a}{V_a+V_b}]} =\sqrt{\frac{10^{-14}}{1.8x10^{-5}}[\frac{(0.15M)(25ml)}{25ml+25ml}]} =6.45x10^{-6} \nonumber \] pH=-log[H + ]=-log(6.45x10 -6 )=5.19 Exercise \(\PageIndex{3.2.f}\) What is the pH when 25m.0L of 0.15M NH 3 (K b = 1.8*10 - 5 ) is titrated with 30.0 mL of0.15M HCl? Answer First calculate the volume to reach the equivalence point and then determine which region of the titration curve you are in \[M_aV_a=M_bV_b \therefore V_b(equiv)= V_a\frac{M_a}{M_b}= 250mL\frac{0.15M}{0.15M}=25mL\] You are in the excess acid range and have added 5 mL more than is needed to neutralize the base \[ [H^+]=\frac{\text{moles acid excess}}{V_T} \nonumber \] \[pH=-log\frac{0.0050(0.15)}{0.025+0.03}=1.87\nonumber\] Solubility of Salts Textbook: Section 17.4 Exercise \(\PageIndex{4.a}\) In which of the following aqueous solutions would you expect AgCl to have the lowest solubility? pure water 0.020 M BaCl 2 0.015 M NaCl 0.020 M AgNO 3 0.020 M KCl Answer b. 0.020 M BaCl 2 , note the chloride salts have a common ion with the silver chloride equilibrium (AgCl(s) <-> Ag^+(aq) + Cl^-(aq)), and so push it to the left (towards the precipitate). This is the common ion effect and it is inhibiting the ionization. Since BaCl 2 provides the most chloride, it pushes it to the left the most Exercise \(\PageIndex{4.b}\) Given the following table K sp values, determine which compound listed has the greatest solubility. 0 1 Compound Ksp CdCO3 5.2*10-12 Cd(OH)2 2.5*10-14 AgI 8.3*10-17 Fe(OH)3 4.0*10-38 ZnCO3 1.4*10-11 CdCO 3 Cd(OH) 2 AgI Fe(OH) 3 ZnCO 3 Answer b. Cd(OH) 2 note: options (a), (c) and (e) are the same formula (x 2 =K sp ) and (e) would be the most soluble because it has the largest K, so you only need to check (b), (d) and (e) e) \[x^2=K_{sp} \therefore x=\sqrt{K_{sp}}=\sqrt{1.4x10^{-11}}=3.74x10^{-6} \nonumber\] b) \[x(2x)^2=4X^3=K_{sp} \therefore x= \left ( \frac{K_{sp}}{4} \right )^\frac{1}{3} = \left ( \frac{2.5x10^{-14}}{4} \right )^\frac{1}{3} = 1.84x10^{-5} \nonumber\] d) \[x(3x)^3=27x^4=K_{sp} \therefore x= \left ( \frac{K_{sp}}{27} \right )^\frac{1}{4} = \left ( \frac{4.0x10^{-38}}{27} \right )^\frac{1}{4} =1.96x10^{-10}\] Exercise \(\PageIndex{4.c}\) The solubility of which one of the following will not be affected by the pH of the solution? Na 3 PO 4 NaF KNO 3 AlCl 3 MnS Answer c. KNO 3 , it is the salt of a strong acid (HNO 3 ) and a strong base (KOH). Phosphate, fluoride and hydrogen sulfide are all basic salts (look at their K a 's), Al is a Lewis acid. Exercise \(\PageIndex{4.d}\) Write the expression relating solubility (x) to K sp for silver sulfide. Answer \[Ag_{2}S(s)\rightleftharpoons 2Ag^{+}(aq)+S^{2-}(aq)\nonumber\] \[K_{sp}=\left [ Ag^{+} \right ]^{2}\left [ S^{2-} \right ]=(2x)^{2}(x)=4x^{3}\nonumber\] \[x=(\frac{K_{sp}}{4})^{\frac{1}{3}}\nonumber\] Exercise \(\PageIndex{4.e}\) In which aqueous system is PbI 2 least soluble? H 2 O 0.5 M HI 0.2 M HI 1.0 M HNO 3 0.8 M KI Answer e. 0.8 M KI, this is the common ion effect, you are looking at the equilibrium of PbI 2 and adding a soluble iodide salt or acid (anything that adds iodide) will inhibit ionization. The more iodide you add the greater the effect, and (e) adds the most Exercise \(\PageIndex{4.f}\) Of the substances below, _____ will decrease the solubility of Pb(OH) 2 in a saturated solution. NaNO 3 H 2 O 2 HNO 3 NaOH NaCl Answer d. NaOH, the common hydroxide inhibits the ionization and causes the lead(II)hydroxide to precipitate out Exercise \(\PageIndex{4.g}\) What is the Pb +2 concentration for a saturated solution of PbSO 4 ? K sp =6.3x10 -7 Answer \[K_{sp}=\left [ Pb^{2+} \right ]\left [ SO_{4}^{2-} \right ]=\left [ x \right ]\left [ x \right ]=x^{2}\nonumber\] \[x=\sqrt{K_{sp}}=\sqrt{6.3*10^{-7}}=7.94*10^{-4}M\nonumber\] Exercise \(\PageIndex{4.h}\) What is the Pb +2 concentration if 0.10mol of Na 2 SO 4 is added to 1L of a saturated solution of PbSO 4 ? Answer \[K_{sp}=\left [ Pb^{2+} \right ]\left [ SO_{4}^{2-} \right ]=\left [ x \right ]\left [ 0.10M \right ]\nonumber\] \[x=\frac{K_{sp}}{\left [ 0.10M \right ]}=\frac{6.3*10^{-7}}{\left [ 0.10M \right ]}=6.3*10^{-6}M\nonumber\] Exercise \(\PageIndex{4.i}\) What is the solubility of Ca 3 (PO 4 ) 2 ? K sp =2.0x10 -29 Answer \[Ca_{3}\left ( PO_{4}\right )_{2}(aq)\rightarrow 3Ca^{2+}(aq)+2PO_{4}^{3-}(aq)\nonumber\] \[K_{sp}=\left [ Ca^{2+} \right ]^{3}\left [ PO_{4}^{3-} \right ]^{2}=\left [ 3x \right ]^{3}\left [ 2x \right ]^{2}=x^{5}(27)(4)\nonumber\] \[x^{5}=\frac{K_{sp}}{108} x=\left ( \frac{K_{sp}}{108} \right )^{\frac{1}{5}}\nonumber\] \[x=\left ( \frac{2.0*10^{-29}}{108} \right )^{\frac{1}{5}}=7.1*10^{-7}M\nonumber\] Exercise \(\PageIndex{4.j}\) What is the Ca 2 + concentration for a saturated solution of Ca 3 (PO 4 ) 2 ? Answer \[Ca_{3}\left ( PO_{4}\right )_{2}(aq)\rightarrow 3Ca^{2+}(aq)+2PO_{4}^{3-}(aq)\nonumber\] \[K_{sp}=\left [ Ca^{2+} \right ]^{3}\left [ PO_{4}^{3-} \right ]^{2}=\left [ 3x \right ]^{3}\left [ 2x \right ]^{2}=x^{5}(27)(4)\nonumber\] \[x^{5}=\frac{K_{sp}}{108} x=\left ( \frac{K_{sp}}{108} \right )^{\frac{1}{5}}\nonumber\] \[x=\left ( \frac{2.0*10^{-29}}{108} \right )^{\frac{1}{5}}=7.1*10^{-7}M\nonumber\] \[\left [ Ca^{2+} \right ]=3*(7.1*10^{-7}M)=2.1*10^{-6}M\nonumber\] Exercise \(\PageIndex{4.k}\) What is the PO 4 3 - concentration for a saturated solution of Ca 3 (PO 4 ) 2 ? Answer \[Ca_{3}\left ( PO_{4}\right )_{2}(aq)\rightarrow 3Ca^{2+}(aq)+2PO_{4}^{3-}(aq)\nonumber\] \[K_{sp}=\left [ Ca^{2+} \right ]^{3}\left [ PO_{4}^{3-} \right ]^{2}=\left [ 3x \right ]^{3}\left [ 2x \right ]^{2}=x^{5}(27)(4)\nonumber\] \[x^{5}=\frac{K_{sp}}{108} x=\left ( \frac{K_{sp}}{108} \right )^{\frac{1}{5}}\nonumber\] \[x=\left ( \frac{2.0*10^{-29}}{108} \right )^{\frac{1}{5}}=7.1*10^{-7}M\nonumber\] \[\left [ PO_{4}^{3-} \right ]=2*(7.1*10^{-7}M)=1.4*10^{-6}M\nonumber\] Exercise \(\PageIndex{4.l}\) What is the K sp for MgF 2 if F - =0.00236M? Answer \[MgF_{2}(aq)\rightarrow Mg^{+2}(aq)+2F^{-}(aq)\nonumber\] \[K_{sp}=\left [ Mg^{+2} \right ]\left [ F^{-} \right ]^{2}=\left [ x \right ]\left [ 2x \right ]^{2}\nonumber\] \[[F^-]=2x=0.00236M\nonumber\] \[[Mg+2]=x=\frac{0.00236M}{2}=0.00118M\nonumber\] \[K_{sp}=\left [ 0.00118M \right ]\left [ 0.00236 \right ]^{2}=6.6*10^{-9}\nonumber\] Exercise \(\PageIndex{4.m}\) If the molar solubility of CaF 2 at 25°C is 1.25x10 - 3 mol/L, what is the K sp at this temperature? Answer \[CaF_{2}(s)\rightleftharpoons Ca^{2+}(aq)+2F^{-}(aq)\nonumber\] \[K_{sp}=\left [ Ca^{2+} \right ]\left [ F^{-} \right ]^{2}=\left [ X \right ]\left [ 2X \right ]^{2}\nonumber\] \[K_{sp}=\left [ X \right ]4\left [ X \right ]^{2}=4\left [ X \right ]^{3}=4\left [ 1.25*10^{-5} \right ]^{3}=7.81*10^{-9}\nonumber\] Exercise \(\PageIndex{4.n}\) A saturated solution of NaF contains 4.0 g of salt in 100.0ml of water at 15°C, what is the solubility product for NaF? Answer \[\frac{\frac{4.0g}{42g/mol}}{0.10L}=0.952M\nonumber\] \[NaF(s)\rightleftharpoons Na^{+}(aq)+F^{-}(aq)\nonumber\] \[K_{sp}=\left [ Na^{+} \right ]\left [ F^{-} \right ]=0.952^{2}=0.91\nonumber\] Note: sodium fluoride is a soluble salt and we usually do not use solubilty constants for soluble salts. Exercise \(\PageIndex{4.o}\) The K sp for AgBr is 5.0x10 -13 at 25°C, what is the molar solubility of AgBr? Answer \[AgBr(s)\rightleftharpoons Ag^{+}(aq)+Br^{-}(aq)\nonumber\] \[K_{sp}=\left [ Ag^{+} \right ]\left [ Br^{-} \right ]=5.0*10^{-13}\nonumber\] \[x^{2}=5.0*10^{-13}\nonumber\] \[x=7.1*10^{-7}M\nonumber\] Exercise \(\PageIndex{4.p}\) Calculate the molar solubility of AgBr in 0.10M NaBr solution? Answer 0 1 2 3 4 R AgBr(s) ⇌ Ag+(aq) + Br-(aq) I --- NaN 0 0.10M C --- NaN x x E --- NaN x .10+x \[K_{sp}=\left [ Ag^{+} \right ]\left [ Br^{-} \right ]=5.0*10^{-13}=\left(0.10+x\right)x \approxeq (0.10)x\nonumber\] \[x=5.0*10^{-12}\nonumber\] Exercise \(\PageIndex{4.q}\) Calculate the solubility of Mn(OH) 2 in grams per liter when buffered at pH = 9.50? K sp = 5.0x10 -13 Answer \[\left [ OH^{-} \right ]=10^{-\left ( 14-9.5 \right )}=3.16*10^{-5}\nonumber\] \[Mn(OH)_{2}(s)\rightleftharpoons Mn^{2+}(aq)+2OH^{-}(aq)\nonumber = x(2x)^2\] \[Mn^{+2}=x \; , \, OH^-=2x\] \[K_{sp}=x[OH^-]^2 \; \therefore \; x=\frac{K_{sp}}{[OH^-]^2}=\frac{5.0x10^{-13}}{[3.16x10^{-5}]^2}=5.07x10^{-4}M \nonumber \] \[ S=5.07x10^{-4}\frac{mol}{l} \left( \frac{88.95g}{mol}\right) = 0.045g/l \nonumber \] Exercise \(\PageIndex{4.r}\) What is the molar solubility of MgC 2 O 4 ? (K sp for MgC 2 O 4 is 8.6*10 -5 ) Answer \[MgC_{2}O_{4}\rightleftharpoons Mg^{2+}+C_{2}O_{4}^{2-}\nonumber\] \[K_{sp}=\left [ Mg^{2+} \right ]\left [ C_{2}O_{4}^{2-} \right ]\nonumber\] \[K_{sp}=(s)*(s)=s^2\nonumber\] \[s^{2}=K_{sp}\nonumber\] \[s=\sqrt{K_{sp}}=\sqrt{8.6*10^{-5}}=9.27*10^{-3}mol/L\nonumber\] Exercise \(\PageIndex{4.s}\) Calculate the concentration (in M) of iodine ions in a saturated solution of lead(II) iodine (K sp = 1.39*10 -8 ) Answer \[PbI_{2}(s)\rightleftharpoons Pb^{2+}(aq)+2I^{-}(aq)\nonumber\] \[K_{sp}=\left [ Pb^{2+} \right ]\left [ I^{-} \right ]^{2}\nonumber\] \[1.39*10^{-8}=x*(2x)^{2}\nonumber\] \[1.39*10^{-8}=4x^{3}\nonumber\] \[x^{3}=3.475*10^{-9}\nonumber\] \[x=(3.475*10^{-9})^{1/3}=1.515*10^{-3}\nonumber\] \[\left [ I^{-} \right ]=2x=2*1.515*10^{-3}=3.03*10^{-3}\nonumber\] Exercise \(\PageIndex{4.t}\) Calculate the molar solubility of silver carbonate (K sp = 6.15*10 -12 ) Answer \[Ag_{2}CO_{3}(s)\rightleftharpoons 2Ag^{+}(aq)+CO_{3}^{2-}(aq)\nonumber\] \[K_{sp}=\left [ Ag^{+} \right ]^{2}\left [ CO_{3}^{2-} \right ]=(2x)^{2}*x\nonumber\] \[K_{sp}=(2x)^{2}*x\nonumber\] \[K_{sp}=4x^{3}\nonumber\] \[x^3=\frac{K_{sp}}{4} \nonumber \] \[x=\left( \frac{K_{sp}}{4} \right)^{1/3} =\left( \frac{6.15*10^{-12}}{4} \right)^{1/3} = 1.15*10^{-4}\nonumber \] Exercise \(\PageIndex{4.u}\) What is the bromide concentration (in M) in a saturated solutions of mercury(II) bromide. K sp = 8.0*10 -20 . Answer \[HgBr_{2}(s)\rightleftharpoons Hg^{2+}(aq)+2Br^{-}(aq)\nonumber\] \[K_{sp}=\left [ Hg^{2+} \right ]\left [ Br^{-} \right ]^{2}\nonumber\] \[ [Hc^{+2}]=x, \; \; [Br^-]=2x \nonumber \] \[K_{sp}=x*(2x)^{2}\nonumber\] \[K_{sp}=4x^{3}\nonumber\] \[x^3=\frac{K_{sp}}{4} \nonumber \] \[x=\left( \frac{K_{sp}}{4} \right)^{1/3} =\left( \frac{8.0*10^{-20}}{4} \right)^{1/3} = 2.71*10^{-7}\nonumber \] \[\left [ Br^{-} \right ]=2s=2*(2.71*10^{-7})=5.4*10^{-7}M\nonumber\] Exercise \(\PageIndex{4.v}\) The solubility of AuCl 3 (as Au 3 + and Cl - ) in water at 298K is 3.3*10 -7 M. K sp for AuCl 3 is _____. Answer \[AuCl_{3}(s)\rightleftharpoons Au^{3+}(aq)+3Cl^{-}(aq)\nonumber\] \[K_{sp}=\left [ Au^{3+} \right ]\left [ Cl^{-} \right ]^{3}\nonumber\] \[K_{sp}=x(3x)^3=27x^4, \;\;\; and \;\;\; x=3.37x10^{-7} \nonumber \] \[K_{sp}=27\left( 3.37x10^{-7} \right)^4=3.2*10^{-25} \nonumber \] Exercise \(\PageIndex{4.w}\) What is the molar solubility of PbS? (K sp (PbS) = 8.0*10 -28 ) Answer \[PbS(s)\rightleftharpoons Pb^{+}(aq)+S^{-}(aq) \nonumber\] \[K_{sp}=\left [ Pb^{+} \right ]\left [ S^{-} \right ] \nonumber\] \[K_{sp}=x^2 \nonumber\] \[x=\sqrt{K_{sp}}=\sqrt{8x10^{-28}}= 2.8*10^{-14}\nonumber\] Exercise \(\PageIndex{4.x}\) Calculate the maximum concentration (in M) of lead ions in a solution containing 0.181M of sulfide. The K sp for lead sulfide is 3.4*10 -28 . Answer \[PbS(s)\rightleftharpoons Pb^{+2}(aq)+S^{-2}(aq) \nonumber\] \[K_{sp}=\left [ Pb^{+2} \right ]\left [ S^{-2} \right ] =x^2\nonumber\] \[ Pb^{+2}]=[S^{-2}]=x \nonumber \] \[x=\frac{K_{sp}}{[S^{-2}]}=\frac{3.4*10^{-28}}{0.181}=1.9*10^{-27}M\nonumber\] Exercise \(\PageIndex{4.y}\) The last two problems dealt with the concentration of lead in a saturated solution of lead sulfide, and in the presence of 0.181M of sulfide. (K sp for lead sulfide is 3.4*10 -28 ). Explain how the common ion effect affects the solubility of lead in water, and how this could be used to remove lead, and why it may not be a good idea. Answer lead(II)sulfide is an extremely insoluble salt, having a solubility of 2.8 x 10 - 14 M or 2.8 fM, while in the presence of 0.181M sulfide, it drops to 1.9 rontoM (1.9x10 -27 , which is virtually nonexistent. The issue is when you use a counter ion to suppress the ionization of a toxic compound like lead, you need to be sure the counter ion is safe. Sulfate is commonly used, as are phosphates. Precipitation Reactions Complex Ion Equilibria 17.6 Exercise \(\PageIndex{4.r}\) Calculate the concentration of Cu 2 + at equilibrium when NH 3 is added to a 0.010M CuCl 2 solution to produce an equilibrium concentration of [NH 3 ] = 0.02M. Neglect the volume change when the ammonia is added. K f =5x10 12 Answer K f =5x10 12 , which is very very large, so assume almost all the copper is converted to the complex, also note, \[K_{f}=\frac{\left [ Cu(NH_{3})_{4}^{2+} \right ]}{\left [ Cu^{2+} \right ]\left [ NH_{3} \right ]^{4}}\nonumber\] \[\left [ Cu^{2+} \right ]=\frac{\left [ Cu(NH_{3})_{4}^{2+} \right ]}{K_{f}\left [ NH_{3} \right ]^{4}}\nonumber\] \[\left [ Cu^{2+} \right ]=\frac{\left [ 0.01 \right ]}{5*10^{12}\left [ .02 \right ]^{4}}=1.2*10^{-8}\nonumber\] 17.6 Solubility and Complex Ions Robert E. Belford (University of Arkansas Little Rock; Department of Chemistry). The breadth, depth and veracity of this work is the responsibility of Robert E. Belford, [email protected] . You should contact him if you have any concerns. This material has both original contributions, and content built upon prior contributions of the LibreTexts Community and other resources, including but not limited to: Liliane Poirot
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Courses/Pacific_Union_College/Quantum_Chemistry/06%3A_The_Hydrogen_Atom
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The solution of the Schrödinger equation (wave equation) for the hydrogen atom uses the fact that the Coulomb potential produced by the nucleus is isotropic (it is radially symmetric in space and only depends on the distance to the nucleus). Although the resulting energy eigenfunctions (the orbitals) are not necessarily isotropic themselves, their dependence on the angular coordinates follows completely generally from this isotropy of the underlying potential: the eigenstates of the Hamiltonian (that is, the energy eigenstates) can be chosen as simultaneous eigenstates of the angular momentum operator. This corresponds to the fact that angular momentum is conserved in the orbital motion of the electron around the nucleus. Therefore, the energy eigenstates may be classified by two angular momentum quantum numbers, ℓ and m (both are integers). The angular momentum quantum number ℓ = 0, 1, 2, ... determines the magnitude of the angular momentum. The magnetic quantum number m = −ℓ, ..., +ℓ determines the projection of the angular momentum on the (arbitrarily chosen) z-axis. 6.1: The Schrodinger Equation for the Hydrogen Atom Can Be Solved Exactly The solution of the Schrödinger equation (wave equation) for the hydrogen atom uses the fact that the Coulomb potential produced by the nucleus is isotropic (it is radially symmetric in space and only depends on the distance to the nucleus). Although the resulting energy eigenfunctions (the orbitals) are not necessarily isotropic themselves, their dependence on the angular coordinates follows completely generally from this isotropy. 6.2: The Wavefunctions of a Rigid Rotator are Called Spherical Harmonics The solutions to the hydrogen atom Schrödinger equation are functions that are products of a spherical harmonic functions and a radial function. 6.3: The Three Components of Angular Momentum Cannot be Measured Simultaneously with Arbitrary Precision The angular momentum operator is one of several related operators analogous to classical angular momentum. The angular momentum operator plays a central role in the theory of atomic physics and other quantum problems involving rotational symmetry. Two orthogonal components of angular momentum (e.g., \(L_x\) and \(L_y\)) are complementary and cannot be simultaneously known or measured. It is, however, possible to simultaneously measure or specify \(L^2\) and any one component of \(L\). 6.4: Hydrogen Atomic Orbitals Depend upon Three Quantum Numbers In solving the Schrödinger equation of the hydrogen atom, we have encountered three quantum numbers. The quantum numbers are not independent; the choice of nn limits the choice of ll, which in turn limits the choice of mm. A fourth quantum number, ss, does not follow directly from solving the Schrödinger equation but is to do with spin (discussed later). 6.5: s-orbitals are Spherically Symmetric The hydrogen atom wavefunctions are called atomic orbitals. An atomic orbital is a function that describes one electron in an atom. The radial probability distribution is introduced in this section. 6.6: Orbital Angular Momentum and the p-Orbitals The physical quantity known as angular momentum plays a dominant role in the understanding of the electronic structure of atoms. 6.7: The Helium Atom Cannot Be Solved Exactly The second element in the periodic table provides our first example of a quantum-mechanical problem which cannot be solved exactly. Nevertheless, as we will show, approximation methods applied to helium can give accurate solutions in perfect agreement with experimental results. In this sense, it can be concluded that quantum mechanics is correct for atoms more complicated than hydrogen. By contrast, the Bohr theory failed miserably in attempts to apply it beyond the hydrogen atom. 6.E: The Hydrogen Atom (Exercises) These are homework exercises to accompany Chapter 6 of McQuarrie and Simon's "Physical Chemistry: A Molecular Approach" Textmap.
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Courses/Arkansas_Northeastern_College/CH14133%3A_Chemistry_for_General_Education/13%3A_Solutions/13.09%3A_Freezing_Point_Depression_and_Boiling_Point_Elevation
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Learning Objectives Explain what the term "colligative" means, and list the colligative properties. Indicate what happens to the boiling point and the freezing point of a solvent when a solute is added to it. Calculate boiling point elevations and freezing point depressions for a solution. People who live in colder climates have seen trucks put salt on the roads when snow or ice is forecast. Why is this done? As a result of the information you explore in this section, you will understand why these events occur. You will also learn to calculate exactly how much of an effect a specific solute can have on the boiling point or freezing point of a solution. The example given in the introduction is an example of a colligative property. Colligative properties are properties that differ based on the concentration of solute in a solvent, but not on the type of solute. What this means for the example above is that people in colder climates do not necessarily need salt to get the same effect on the roads—any solute will work. However, the higher the concentration of solute, the more these properties will change. Boiling Point Elevation Water boils at \(100^\text{o} \text{C}\) at \(1 \: \text{atm}\) of pressure, but a solution of saltwater does not . When table salt is added to water, the resulting solution has a higher boiling point than the water did by itself. The ions form an attraction with the solvent particles that prevents the water molecules from going into the gas phase. Therefore, the saltwater solution will not boil at \(100^\text{o} \text{C}\). In order for the saltwater solution to boil, the temperature must be raised about \(100^\text{o} \text{C}\). This is true for any solute added to a solvent; the boiling point will be higher than the boiling point of the pure solvent (without the solute). In other words, when anything is dissolved in water, the solution will boil at a higher temperature than pure water would. The boiling point elevation due to the presence of a solute is also a colligative property. That is, the amount of change in the boiling point is related to the number of particles of solute in a solution and is not related to the chemical composition of the solute. A \(0.20 \: \text{m}\) solution of table salt and a \(0.20 \: \text{m}\) solution of hydrochloric acid would have the same effect on the boiling point. Freezing Point Depression The effect of adding a solute to a solvent has the opposite effect on the freezing point of a solution as it does on the boiling point. A solution will have a lower freezing point than a pure solvent. The freezing point is the temperature at which the liquid changes to a solid. At a given temperature, if a substance is added to a solvent (such as water), the solute-solvent interactions prevent the solvent from going into the solid phase. The solute-solvent interactions require the temperature to decrease further in order to solidify the solution. A common example is found when salt is used on icy roadways. Salt is put on roads so that the water on the roads will not freeze at the normal \(0^\text{o} \text{C}\) but at a lower temperature, as low as \(-9^\text{o} \text{C}\). The de-icing of planes is another common example of freezing point depression in action. A number of solutions are used, but commonly a solution such as ethylene glycol, or a less toxic monopropylene glycol, is used to de-ice an aircraft. The aircrafts are sprayed with the solution when the temperature is predicted to drop below the freezing point. The freezing point depression is the difference in the freezing points of the solution from the pure solvent. This is true for any solute added to a solvent; the freezing point of the solution will be lower than the freezing point of the pure solvent (without the solute). Thus, when anything is dissolved in water, the solution will freeze at a lower temperature than pure water would. The freezing point depression due to the presence of a solute is also a colligative property. That is, the amount of change in the freezing point is related to the number of particles of solute in a solution and is not related to the chemical composition of the solute. A \(0.20 \: \text{m}\) solution of table salt and a \(0.20 \: \text{m}\) solution of hydrochloric acid would have the same effect on the freezing point. Comparing the Freezing and Boiling Point of Solutions Recall that covalent and ionic compounds do not dissolve in the same way. Ionic compounds break up into cations and anions when they dissolve. Covalent compounds typically do not break up. For example a sugar/water solution stays as sugar and water, with the sugar molecules staying as molecules. Remember that colligative properties are due to the number of solute particles in the solution. Adding 10 molecules of sugar to a solvent will produce 10 solute particles in the solution. When the solute is ionic, such as \(\ce{NaCl}\) however, adding 10 formulas of solute to the solution will produce 20 ions (solute particles) in the solution. Therefore, adding enough \(\ce{NaCl}\) solute to a solvent to produce a \(0.20 \: \text{m}\) solution will have twice the effect of adding enough sugar to a solvent to produce a \(0.20 \: \text{m}\) solution. Colligative properties depend on the number of solute particles in the solution. "\(i\)" is the number of particles that the solute will dissociate into upon mixing with the solvent. For example, sodium chloride, \(\ce{NaCl}\), will dissociate into two ions so for \(\ce{NaCl}\), \(i = 2\); for lithium nitrate, \(\ce{LiNO_3}\), \(i = 2\); and for calcium chloride, \(\ce{CaCl_2}\), \(i = 3\). For covalent compounds, \(i\) is always equal to 1. By knowing the molality of a solution and the number of particles a compound will dissolve to form, it is possible to predict which solution in a group will have the lowest freezing point. To compare the boiling or freezing points of solutions, follow these general steps: Label each solute as ionic or covalent. If the solute is ionic, determine the number of ions in the formula. Be careful to look for polyatomic ions. Multiply the original molality (\(\text{m}\)) of the solution by the number of particles formed when the solution dissolves. This will give you the total concentration of particles dissolved. Compare these values. The higher total concentration will result in a higher boiling point and a lower freezing point. Example \(\PageIndex{1}\) Rank the following solutions in water in order of increasing (lowest to highest) freezing point: \(0.1 \: \text{m} \: \ce{NaCl}\) \(0.1 \: \text{m} \: \ce{C_6H_{12}O_6}\) \(0.1 \: \text{m} \: \ce{CaI_2}\) Solution To compare freezing points, we need to know the total concentration of all particles when the solute has been dissolved. \(0.1 \: \text{m} \: \ce{NaCl}\): This compound is ionic (metal with nonmetal), and will dissolve into 2 parts. The total final concentration is: \(\left( 0.1 \: \text{m} \right) \left( 2 \right) = 0.2 \: \text{m}\) \(0.1 \: \text{m} \: \ce{C_6H_{12}O_6}\): This compound is covalent (nonmetal with nonmetal), and will stay as 1 part. The total final concentration is: \(\left( 0.1 \: \text{m} \right) \left(1 \right) = 0.1 \: \text{m}\) \(0.1 \: \text{m} \: \ce{CaI_2}\): This compound is ionic (metal with nonmetal), and will dissolve into 3 parts. The total final concentration is: \(\left( 0.1 \: \text{m} \right) \left( 3 \right) = 0.3 \: \text{m}\) Remember, the greater the concentration of particles, the lower the freezing point will be. \(0.1 \: \text{m} \: \ce{CaI_2}\) will have the lowest freezing point, followed by \(0.1 \: \text{m} \: \ce{NaCl}\), and the highest of the three solutions will be \(0.1 \: \text{m} \: \ce{C_6H_{12}O_6}\), but all three of them will have a lower freezing point than pure water. The boiling point of a solution is higher than the boiling point of a pure solvent, and the freezing point of a solution is lower than the freezing point of a pure solvent. However, the amount to which the boiling point increases or the freezing point decreases depends on the amount of solute that is added to the solvent. A mathematical equation is used to calculate the boiling point elevation or the freezing point depression. The boiling point elevation is the amount that the boiling point temperature increases compared to the original solvent. For example, the boiling point of pure water at \(1.0 \: \text{atm}\) is \(100^\text{o} \text{C}\) while the boiling point of a \(2\%\) saltwater solution is about \(102^\text{o} \text{C}\). Therefore, the boiling point elevation would be \(2^\text{o} \text{C}\). The freezing point depression is the amount that the freezing temperature decreases . Both the boiling point elevation and the freezing point depression are related to the molality of the solution. Looking at the formula for the boiling point elevation and freezing point depression, we see similarities between the two. The equation used to calculate the increase in the boiling point is: \[\Delta T_b = k_b \cdot \text{m} \cdot i \label{BP} \] Where: \(\Delta T_b =\) the amount the boiling point increases. \(k_b =\) the boiling point elevation constant which depends on the solvent (for water, this number is \(0.515^\text{o} \text{C/m}\)). \(\text{m} =\) the molality of the solution. \(i =\) the number of particles formed when that compound dissolves (for covalent compounds, this number is always 1). The following equation is used to calculate the decrease in the freezing point: \[\Delta T_f = k_f \cdot \text{m} \cdot i \label{FP} \] Where: \(\Delta T_f =\) the amount the freezing temperature decreases. \(k_f =\) the freezing point depression constant which depends on the solvent (for water, this number is \(1.86^\text{o} \text{C/m}\)). \(\text{m} =\) the molality of the solution. \(i =\) the number of particles formed when that compound dissolves (for covalent compounds, this number is always 1). Example \(\PageIndex{2}\): Adding Antifreeze to Protein Engines Antifreeze is used in automobile radiators to keep the coolant from freezing. In geographical areas where winter temperatures go below the freezing point of water, using pure water as the coolant could allow the water to freeze. Since water expands when it freezes, freezing coolant could crack engine blocks, radiators, and coolant lines. The main component in antifreeze is ethylene glycol, \(\ce{C_2H_4(OH)_2}\). What is the concentration of ethylene glycol in a solution of water, in molality, if the freezing point dropped by \(2.64^\text{o} \text{C}\)? The freezing point constant, \(k_f\), for water is \(1.86^\text{o} \text{C/m}\). Solution Use the equation for freezing point depression of solution (Equation \(\ref{FP}\)): \[\Delta T_f = k_f \cdot \text{m} \cdot i \nonumber \] Substituting in the appropriate values we get: \[2.64^\text{o} \text{C} = \left( 1.86^\text{o} \text{C/m} \right) \left( \text{m} \right) \left( 1 \right) \nonumber \] Solve for \(\text{m}\) by dividing both sides by \(1.86^\text{o} \text{C/m}\). \[\text{m} = 1.42 \nonumber \] Example \(\PageIndex{3}\): Adding Salt to Elevate Boiling Temperature A solution of \(10.0 \: \text{g}\) of sodium chloride is added to \(100.0 \: \text{g}\) of water in an attempt to elevate the boiling point. What is the boiling point of the solution? \(k_b\) for water is \(0.52^\text{o} \text{C/m}\). Solution Use the equation for boiling point elevation of solution (Equation \(\ref{BP}\)): \[\Delta T_b = k_b \cdot \text{m} \cdot i \nonumber \] We need to be able to substitute each variable into this equation. \(k_b = 0.52^\text{o} \text{C/m}\) \(\text{m}\): We must solve for this using stoichiometry. Given: \(10.0 \: \text{g} \: \ce{NaCl}\) and \(100.0 \: \text{g} \: \ce{H_2O}\) Find: \(\text{mol} \: \ce{NaCl}/\text{kg} \: \ce{H_2O}\). Ratios: molar mass of \(\ce{NaCl}\), \(1000 \: \text{g} = 1 \: \text{kg}\) \[\dfrac{10.0 \: \cancel{\text{g} \: \ce{NaCl}}}{100.0 \: \cancel{\text{g} \: \ce{H_2O}}} \cdot \dfrac{1 \: \text{mol} \: \ce{NaCl}}{58.45 \: \cancel{\text{g} \: \ce{NaCl}}} \cdot \dfrac{1000 \: \cancel{\text{g} \: \ce{H_2O}}}{1 \: \text{kg} \: \ce{H_2O}} = 1.71 \: \text{m} \nonumber \] For \(\ce{NaCl}\), \(i = 2\) Substitute these values into the equation \(\Delta T_b = k_b \cdot \text{m} \cdot i\). We get: \[\Delta T_b = \left( 0.52 \dfrac{^\text{o} \text{C}}{\cancel{\text{m}}} \right) \left( 1.71 \: \cancel{\text{m}} \right) \left( 2 \right) = 1.78^\text{o} \text{C} \nonumber \] Water normally boils at \(100^\text{o} \text{C}\), but our calculation shows that the boiling point increased by \(1.78^\text{o} \text{C}\). Our new boiling point is \(101.78^\text{o} \text{C}\). Note: Since sea water contains roughly 28.0 g of NaCl per liter, this saltwater solution is approximately four times more concentrated than sea water (all for a 2° C rise of boiling temperature). Summary Colligative properties are properties that are due only to the number of particles in solution, and are not related to the chemical properties of the solute. Boiling points of solutions are higher than the boiling points of the pure solvents. Freezing points of solutions are lower than the freezing points of the pure solvents. Ionic compounds split into ions when they dissolve, forming more particles. Covalent compounds stay as complete molecules when they dissolve. Vocabulary Colligative property - A property that is due only to the number of particles in solution, and not the type of the solute. Boiling point elevation - The amount that the boiling point of a solution increases from the boiling point of the pure solvent. Freezing point depression - The amount that the freezing point of a solution decreases from the freezing point of the pure solvent.
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Courses/Grand_Rapids_Community_College/CHM_120%3A_Survey_of_General_Chemistry_(Crandell)/04%3A_Lewis_Structures_Geometry_and_Polarity_of_Molecules/4.02%3A_Molecular_Structure_and_Polarity
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Learning Objectives By the end of this section, you will be able to: Predict the structures of small molecules using valence shell electron pair repulsion (VSEPR) theory Explain the concepts of polar covalent bonds and molecular polarity Assess the polarity of a molecule based on its bonding and structure Thus far, we have used two-dimensional Lewis structures to represent molecules. However, molecular structure is actually three-dimensional, and it is important to be able to describe molecular bonds in terms of their distances, angles, and relative arrangements in space (Figure \(\PageIndex{1}\)). A bond angle is the angle between any two bonds that include a common atom, usually measured in degrees. A bond distance (or bond length) is the distance between the nuclei of two bonded atoms along the straight line joining the nuclei. Bond distances are measured in Ångstroms (1 Å = 10 –10 m) or picometers (1 pm = 10 –12 m, 100 pm = 1 Å). VSEPR Theory Valence shell electron-pair repulsion theory (VSEPR theory) enables us to predict the molecular structure, including approximate bond angles around a central atom, of a molecule from an examination of the number of bonds and lone electron pairs in its Lewis structure. The VSEPR model assumes that electron pairs in the valence shell of a central atom will adopt an arrangement that minimizes repulsions between these electron pairs by maximizing the distance between them. The electrons in the valence shell of a central atom form either bonding pairs of electrons, located primarily between bonded atoms, or lone pairs. The electrostatic repulsion of these electrons is reduced when the various regions of high electron density assume positions as far from each other as possible. VSEPR theory predicts the arrangement of electron pairs around each central atom and, usually, the correct arrangement of atoms in a molecule. We should understand, however, that the theory only considers electron-pair repulsions. Other interactions, such as nuclear-nuclear repulsions and nuclear-electron attractions, are also involved in the final arrangement that atoms adopt in a particular molecular structure. As a simple example of VSEPR theory, let us predict the structure of a gaseous BeF 2 molecule. The Lewis structure of BeF 2 (Figure \(\PageIndex{2}\)) shows only two electron pairs around the central beryllium atom. With two bonds and no lone pairs of electrons on the central atom, the bonds are as far apart as possible, and the electrostatic repulsion between these regions of high electron density is reduced to a minimum when they are on opposite sides of the central atom. The bond angle is 180° (Figure \(\PageIndex{2}\)). Figure \(\PageIndex{3}\) illustrates this and other electron-pair geometries that minimize the repulsions among regions of high electron density (bonds and/or lone pairs). Two regions of electron density around a central atom in a molecule form a linear geometry; three regions form a trigonal planar geometry; four regions form a tetrahedral geometry; five regions form a trigonal bipyramidal geometry; and six regions form an octahedral geometry. Electron-pair Geometry versus Molecular Structure It is important to note that electron-pair geometry around a central atom is not the same thing as its molecular structure. The electron-pair geometries shown in Figure \(\PageIndex{3}\) describe all regions where electrons are located, bonds as well as lone pairs. Molecular structure describes the location of the atoms , not the electrons. We differentiate between these two situations by naming the geometry that includes all electron pairs the electron-pair geometry . The structure that includes only the placement of the atoms in the molecule is called the molecular structure . The electron-pair geometries will be the same as the molecular structures when there are no lone electron pairs around the central atom, but they will be different when there are lone pairs present on the central atom. For example, the methane molecule, CH 4 , which is the major component of natural gas, has four bonding pairs of electrons around the central carbon atom; the electron-pair geometry is tetrahedral, as is the molecular structure (Figure \(\PageIndex{4}\)). On the other hand, the ammonia molecule, NH 3 , also has four electron pairs associated with the nitrogen atom, and thus has a tetrahedral electron-pair geometry. One of these regions, however, is a lone pair, which is not included in the molecular structure, and this lone pair influences the shape of the molecule (Figure \(\PageIndex{5}\)). As seen in Figure \(\PageIndex{5}\), small distortions from the ideal angles in Figure \(\PageIndex{3}\) can result from differences in repulsion between various regions of electron density. VSEPR theory predicts these distortions by establishing an order of repulsions and an order of the amount of space occupied by different kinds of electron pairs. The order of electron-pair repulsions from greatest to least repulsion is: \[\text { lone pair-lone pair > lone pair-bonding pair > bonding pair-bonding pair } \nonumber \] This order of repulsions determines the amount of space occupied by different regions of electrons. A lone pair of electrons occupies a larger region of space than the electrons in a triple bond; in turn, electrons in a triple bond occupy more space than those in a double bond, and so on. The order of sizes from largest to smallest is: \[\text { lone pair }>\text { triple bond }>\text { double bond }>\text { single bond } \nonumber \] Consider formaldehyde, H 2 CO, which is used as a preservative for biological and anatomical specimens (Figure \(\PageIndex{1}\)). This molecule has regions of high electron density that consist of two single bonds and one double bond. The basic geometry is trigonal planar with 120° bond angles, but we see that the double bond causes slightly larger angles (121°), and the angle between the single bonds is slightly smaller (118°). In the ammonia molecule, the three hydrogen atoms attached to the central nitrogen are not arranged in a flat, trigonal planar molecular structure, but rather in a three-dimensional trigonal pyramid (Figure \(\PageIndex{5}\)) with the nitrogen atom at the apex and the three hydrogen atoms forming the base. The ideal bond angles in a trigonal pyramid are based on the tetrahedral electron pair geometry. Again, there are slight deviations from the ideal because lone pairs occupy larger regions of space than do bonding electrons. The H–N–H bond angles in NH 3 are slightly smaller than the 109.5° angle in a regular tetrahedron (Figure \(\PageIndex{3}\)) because the lone pair-bonding pair repulsion is greater than the bonding pair-bonding pair repulsion (Figure \(\PageIndex{5}\)). Figure \(\PageIndex{6}\) illustrates the ideal molecular structures, which are predicted based on the electron-pair geometries for various combinations of lone pairs and bonding pairs. According to VSEPR theory, the terminal atom locations (Xs in Figure \(\PageIndex{6}\)) are equivalent within the linear, trigonal planar, and tetrahedral electron-pair geometries (the first three rows of the table). It does not matter which X is replaced with a lone pair because the molecules can be rotated to convert positions. For trigonal bipyramidal electron-pair geometries, however, there are two distinct X positions, as shown in Figure \(\PageIndex{7}\) an axial position (if we hold a model of a trigonal bipyramid by the two axial positions, we have an axis around which we can rotate the model) and an equatorial position (three positions form an equator around the middle of the molecule). As shown in Figure \(\PageIndex{6}\), the axial position is surrounded by bond angles of 90°, whereas the equatorial position has more space available because of the 120° bond angles. In a trigonal bipyramidal electron-pair geometry, lone pairs always occupy equatorial positions because these more spacious positions can more easily accommodate the larger lone pairs. Theoretically, we can come up with three possible arrangements for the three bonds and two lone pairs for the ClF 3 molecule (Figure \(\PageIndex{7}\)). The stable structure is the one that puts the lone pairs in equatorial locations, giving a T-shaped molecular structure. When a central atom has two lone electron pairs and four bonding regions, we have an octahedral electron-pair geometry. The two lone pairs are on opposite sides of the octahedron (180° apart), giving a square planar molecular structure that minimizes lone pair-lone pair repulsions (Figure \(\PageIndex{6}\)). Predicting Electron Pair Geometry and Molecular Structure The following procedure uses VSEPR theory to determine the electron pair geometries and the molecular structures: Write the Lewis structure of the molecule or polyatomic ion. Count the number of regions of electron density (lone pairs and bonds) around the central atom. A single, double, or triple bond counts as one region of electron density. Identify the electron-pair geometry based on the number of regions of electron density: linear, trigonal planar, tetrahedral, trigonal bipyramidal, or octahedral (Figure \(\PageIndex{6}\), first column). Use the number of lone pairs to determine the molecular structure (Figure \(\PageIndex{6}\)). If more than one arrangement of lone pairs and chemical bonds is possible, choose the one that will minimize repulsions, remembering that lone pairs occupy more space than multiple bonds, which occupy more space than single bonds. In trigonal bipyramidal arrangements, repulsion is minimized when every lone pair is in an equatorial position. In an octahedral arrangement with two lone pairs, repulsion is minimized when the lone pairs are on opposite sides of the central atom. The following examples illustrate the use of VSEPR theory to predict the molecular structure of molecules or ions that have no lone pairs of electrons. In this case, the molecular structure is identical to the electron pair geometry. Example \(\PageIndex{1}\): Predicting Electron-pair Geometry and Molecular Structure: CO 2 and BCl 3 Predict the electron-pair geometry and molecular structure for each of the following: carbon dioxide, CO 2 , a molecule produced by the combustion of fossil fuels boron trichloride, BCl 3 , an important industrial chemical Solution (a) We write the Lewis structure of CO 2 as: This shows us two regions of high electron density around the carbon atom—each double bond counts as one region, and there are no lone pairs on the carbon atom. Using VSEPR theory, we predict that the two regions of electron density arrange themselves on opposite sides of the central atom with a bond angle of 180°. The electron-pair geometry and molecular structure are identical, and CO 2 molecules are linear. (b) We write the Lewis structure of BCl 3 as: Thus we see that BCl 3 contains three bonds, and there are no lone pairs of electrons on boron. The arrangement of three regions of high electron density gives a trigonal planar electron-pair geometry. The B–Cl bonds lie in a plane with 120° angles between them. BCl 3 also has a trigonal planar molecular structure (Figure \(\PageIndex{8}\)). The electron-pair geometry and molecular structure of BCl 3 are both trigonal planar. Note that the VSEPR geometry indicates the correct bond angles (120°), unlike the Lewis structure shown above. Exercise \(\PageIndex{1}\) Carbonate, is a common polyatomic ion found in various materials from eggshells to antacids. What are the electron-pair geometry and molecular structure of this polyatomic ion? Answer The electron-pair geometry is trigonal planar and the molecular structure is trigonal planar. Due to resonance, all three C–O bonds are identical. Whether they are single, double, or an average of the two, each bond counts as one region of electron density. Example \(\PageIndex{2}\): Predicting Electron-pair Geometry and Molecular Structure: Ammonium Two of the top 50 chemicals produced in the United States, ammonium nitrate and ammonium sulfate, both used as fertilizers, contain the ammonium ion. Predict the electron-pair geometry and molecular structure of the cation. Solution We write the Lewis structure of as: We can see that \(\ce{NH4^{+}}\) contains four bonds from the nitrogen atom to hydrogen atoms and no lone pairs. We expect the four regions of high electron density to arrange themselves so that they point to the corners of a tetrahedron with the central nitrogen atom in the middle (Figure \(\PageIndex{6}\)). Therefore, the electron pair geometry of Figure \(\PageIndex{9}\). Exercise \(\PageIndex{2}\) Identify a molecule with trigonal bipyramidal molecular structure. Answer Any molecule with five electron pairs around the central atoms including no lone pairs will be trigonal bipyramidal. PF 5 is a common example. The next several examples illustrate the effect of lone pairs of electrons on molecular structure. Example \(\PageIndex{3}\): Predicting Electron-pair Geometry and Molecular Structure: Lone Pairs on the Central Atom Predict the electron-pair geometry and molecular structure of a water molecule. Solution The Lewis structure of H 2 O indicates that there are four regions of high electron density around the oxygen atom: two lone pairs and two chemical bonds: We predict that these four regions are arranged in a tetrahedral fashion (Figure \(\PageIndex{10}\)), as indicated in Figure \(\PageIndex{6}\). Thus, the electron-pair geometry is tetrahedral and the molecular structure is bent with an angle slightly less than 109.5°. In fact, the bond angle is 104.5°. Exercise \(\PageIndex{3}\) The hydronium ion, H 3 O + , forms when acids are dissolved in water. Predict the electron-pair geometry and molecular structure of this cation. Answer electron pair geometry: tetrahedral; molecular structure: trigonal pyramidal Example \(\PageIndex{4}\): Predicting Electron-pair Geometry and Molecular Structure: SF 4 Sulfur tetrafluoride, SF 4 , is extremely valuable for the preparation of fluorine-containing compounds used as herbicides (i.e., SF 4 is used as a fluorinating agent). Predict the electron-pair geometry and molecular structure of a SF 4 molecule. Solution The Lewis structure of SF 4 indicates five regions of electron density around the sulfur atom: one lone pair and four bonding pairs: We expect these five regions to adopt a trigonal bipyramidal electron-pair geometry. To minimize lone pair repulsions, the lone pair occupies one of the equatorial positions. The molecular structure (Figure \(\PageIndex{11}\)) is that of a seesaw (Figure \(\PageIndex{6}\)). Exercise \(\PageIndex{4}\) Predict the electron pair geometry and molecular structure for molecules of XeF 2 . Answer The electron-pair geometry is trigonal bipyramidal. The molecular structure is linear. Example \(\PageIndex{5}\): Predicting Electron-pair Geometry and Molecular Structure: XeF 4 Of all the noble gases, xenon is the most reactive, frequently reacting with elements such as oxygen and fluorine. Predict the electron-pair geometry and molecular structure of the XeF 4 molecule. Solution The Lewis structure of XeF 4 indicates six regions of high electron density around the xenon atom: two lone pairs and four bonds: These six regions adopt an octahedral arrangement (Figure \(\PageIndex{6}\)), which is the electron-pair geometry. To minimize repulsions, the lone pairs should be on opposite sides of the central atom (Figure \(\PageIndex{12}\)). The five atoms are all in the same plane and have a square planar molecular structure. Exercise \(\PageIndex{5}\) In a certain molecule, the central atom has three lone pairs and two bonds. What will the electron pair geometry and molecular structure be? Answer electron pair geometry: trigonal bipyramidal; molecular structure: linear Molecular Structure for Multicenter Molecules When a molecule or polyatomic ion has only one central atom, the molecular structure completely describes the shape of the molecule. Larger molecules do not have a single central atom, but are connected by a chain of interior atoms that each possess a “local” geometry. The way these local structures are oriented with respect to each other also influences the molecular shape, but such considerations are largely beyond the scope of this introductory discussion. For our purposes, we will only focus on determining the local structures. Example \(\PageIndex{6}\): Predicting Structure in Multicenter Molecules The Lewis structure for the simplest amino acid, glycine, H 2 NCH 2 CO 2 H, is shown here. Predict the local geometry for the nitrogen atom, the two carbon atoms, and the oxygen atom with a hydrogen atom attached: Solution Consider each central atom independently. The electron-pair geometries: nitrogen––four regions of electron density; tetrahedral carbon ( C H 2 )––four regions of electron density; tetrahedral carbon ( C O 2 )—three regions of electron density; trigonal planar oxygen ( O H)—four regions of electron density; tetrahedral The local structures: nitrogen––three bonds, one lone pair; trigonal pyramidal carbon ( C H 2 )—four bonds, no lone pairs; tetrahedral carbon ( C O 2 )—three bonds (double bond counts as one bond), no lone pairs; trigonal planar oxygen ( O H)—two bonds, two lone pairs; bent (109°) Exercise \(\PageIndex{6}\) Another amino acid is alanine, which has the Lewis structure shown here. Predict the electron-pair geometry and local structure of the nitrogen atom, the three carbon atoms, and the oxygen atom with hydrogen attached: Answer electron-pair geometries: nitrogen––tetrahedral; carbon ( C H)—tetrahedral; carbon ( C H 3 )—tetrahedral; carbon ( C O 2 )—trigonal planar; oxygen ( O H)—tetrahedral; local structures: nitrogen—trigonal pyramidal; carbon ( C H)—tetrahedral; carbon ( C H 3 )—tetrahedral; carbon ( C O 2 )—trigonal planar; oxygen ( O H)—bent (109°) Link to Learning The molecular shape simulator lets you build various molecules and practice naming their electron-pair geometries and molecular structures. Example \(\PageIndex{7}\): Molecular Simulation Using molecular shape simulator allows us to control whether bond angles and/or lone pairs are displayed by checking or unchecking the boxes under “Options” on the right. We can also use the “Name” checkboxes at bottom-left to display or hide the electron pair geometry (called “electron geometry” in the simulator) and/or molecular structure (called “molecular shape” in the simulator). Build the molecule HCN in the simulator based on the following Lewis structure: \[\ce{H-C#N} \nonumber \] Click on each bond type or lone pair at right to add that group to the central atom. Once you have the complete molecule, rotate it to examine the predicted molecular structure. What molecular structure is this? Solution The molecular structure is linear. Exercise \(\PageIndex{7}\) Build a more complex molecule in the simulator. Identify the electron-group geometry, molecular structure, and bond angles. Then try to find a chemical formula that would match the structure you have drawn. Answer Answers will vary. For example, an atom with four single bonds, a double bond, and a lone pair has an octahedral electron-group geometry and a square pyramidal molecular structure. XeOF 4 is a molecule that adopts this structure. Molecular Polarity and Dipole Moment As discussed previously, polar covalent bonds connect two atoms with differing electronegativities, leaving one atom with a partial positive charge (δ+) and the other atom with a partial negative charge (δ–), as the electrons are pulled toward the more electronegative atom. This separation of charge gives rise to a bond dipole moment . The magnitude of a bond dipole moment is represented by the Greek letter mu ( µ ) and is given by the formula shown here, where Q is the magnitude of the partial charges (determined by the electronegativity difference) and r is the distance between the charges: \[\mu= Qr \nonumber \] This bond moment can be represented as a vector, a quantity having both direction and magnitude (Figure \(\PageIndex{13}\)). Dipole vectors are shown as arrows pointing along the bond from the less electronegative atom toward the more electronegative atom. A small plus sign is drawn on the less electronegative end to indicate the partially positive end of the bond. The length of the arrow is proportional to the magnitude of the electronegativity difference between the two atoms. Figure \(\PageIndex{13}\): (a) There is a small difference in electronegativity between C and H, represented as a short vector. (b) The electronegativity difference between B and F is much larger, so the vector representing the bond moment is much longer. A whole molecule may also have a separation of charge, depending on its molecular structure and the polarity of each of its bonds. If such a charge separation exists, the molecule is said to be a polar molecule (or dipole); otherwise the molecule is said to be nonpolar. The dipole moment measures the extent of net charge separation in the molecule as a whole. We determine the dipole moment by adding the bond moments in three-dimensional space, taking into account the molecular structure. For diatomic molecules, there is only one bond, so its bond dipole moment determines the molecular polarity. Homonuclear diatomic molecules such as Br 2 and N 2 have no difference in electronegativity, so their dipole moment is zero. For heteronuclear molecules such as CO, there is a small dipole moment. For HF, there is a larger dipole moment because there is a larger difference in electronegativity. When a molecule contains more than one bond, the geometry must be taken into account. If the bonds in a molecule are arranged such that their bond moments cancel (vector sum equals zero), then the molecule is nonpolar. This is the situation in CO 2 (Figure \(\PageIndex{14}\)). Each of the bonds is polar, but the molecule as a whole is nonpolar. From the Lewis structure, and using VSEPR theory, we determine that the CO 2 molecule is linear with polar C=O bonds on opposite sides of the carbon atom. The bond moments cancel because they are pointed in opposite directions. In the case of the water molecule (Figure \(\PageIndex{14}\)), the Lewis structure again shows that there are two bonds to a central atom, and the electronegativity difference again shows that each of these bonds has a nonzero bond moment. In this case, however, the molecular structure is bent because of the lone pairs on O, and the two bond moments do not cancel. Therefore, water does have a net dipole moment and is a polar molecule (dipole). The OCS molecule has a structure similar to CO 2 , but a sulfur atom has replaced one of the oxygen atoms. To determine if this molecule is polar, we draw the molecular structure. VSEPR theory predicts a linear molecule: The C-O bond is considerably polar. Although C and S have very similar electronegativity values, S is slightly more electronegative than C, and so the C-S bond is just slightly polar. Because oxygen is more electronegative than sulfur, the oxygen end of the molecule is the negative end. Chloromethane, CH 3 Cl, is a tetrahedral molecule with three slightly polar C-H bonds and a more polar C-Cl bond. The relative electronegativities of the bonded atoms is H < C < Cl, and so the bond moments all point toward the Cl end of the molecule and sum to yield a considerable dipole moment (the molecules are relatively polar). For molecules of high symmetry such as BF 3 (trigonal planar), CH 4 (tetrahedral), PF 5 (trigonal bipyramidal), and SF 6 (octahedral), all the bonds are of identical polarity (same bond moment) and they are oriented in geometries that yield nonpolar molecules (dipole moment is zero). Molecules of less geometric symmetry, however, may be polar even when all bond moments are identical. For these molecules, the directions of the equal bond moments are such that they sum to give a nonzero dipole moment and a polar molecule. Examples of such molecules include hydrogen sulfide, H 2 S (nonlinear), and ammonia, NH 3 (trigonal pyramidal). To summarize, to be polar, a molecule must: Contain at least one polar covalent bond. Have a molecular structure such that the sum of the vectors of each bond dipole moment does not cancel. Properties of Polar Molecules Polar molecules tend to align when placed in an electric field with the positive end of the molecule oriented toward the negative plate and the negative end toward the positive plate (Figure \(\PageIndex{15}\)). We can use an electrically charged object to attract polar molecules, but nonpolar molecules are not attracted. Also, polar solvents are better at dissolving polar substances, and nonpolar solvents are better at dissolving nonpolar substances. Link to Learning The molecule polarity simulation provides many ways to explore dipole moments of bonds and molecules. Example \(\PageIndex{8}\): Polarity Simulations Open the molecule polarity simulation and select the “Three Atoms” tab at the top. This should display a molecule ABC with three electronegativity adjustors. You can display or hide the bond moments, molecular dipoles, and partial charges at the right. Turning on the Electric Field will show whether the molecule moves when exposed to a field, similar to Figure \(\PageIndex{15}\). Use the electronegativity controls to determine how the molecular dipole will look for the starting bent molecule if: A and C are very electronegative and B is in the middle of the range. A is very electronegative, and B and C are not. Solution Molecular dipole moment points immediately between A and C. Molecular dipole moment points along the A–B bond, toward A. Exercise \(\PageIndex{8}\) Determine the partial charges that will give the largest possible bond dipoles. Answer The largest bond moments will occur with the largest partial charges. The two solutions above represent how unevenly the electrons are shared in the bond. The bond moments will be maximized when the electronegativity difference is greatest. The controls for A and C should be set to one extreme, and B should be set to the opposite extreme. Although the magnitude of the bond moment will not change based on whether B is the most electronegative or the least, the direction of the bond moment will.
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Bookshelves/Inorganic_Chemistry/Inorganic_Chemistry_(Saito)/05%3A_Chemistry_of_Main-Group_Metals/5.01%3A_Group_1_Metals
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Group 1 metals are called alkali metals . Alkali metals are abundant in minerals and sea water. Especially the content of sodium, Na, in the Earth's crust is fourth after Al, Fe, and Ca. Although the existence of sodium or potassium ions was recognized for many years, a number of attempts to isolate the metals from aqueous solutions of their salts failed because of their high reactivity with water. Potassium (1807) and subsequently sodium were isolated by the electrolysis of molten salt of KOH or NaOH by H. Davy in the 19th century. Lithium Li was discovered as a new element (1817), and Davy soon isolated it by molten salt electrolysis of Li 2 O. Rubidium, Rb and Cesium, Cs, were discovered as new elements by spectroscopy in 1861. Francium, Fr, was discovered using a radiochemical technique in 1939. Its natural abundance is very low. Unnamed: 0 mp (°C) bp (°C) d(20 °C) (g cm-3) E0 (V) M+ + e- I (kJ mol-1) Li 181 1342 0.534 -3.04 520 Na 98 883 0.968 -2.71 496 K 63 759 0.856 -2.93 419 Rb 39 688 1.532 -2.98 403 Cs 28 671 1.900 -3.03 376 As shown in Table \(\PageIndex{1}\), melting-points, boiling points, and densities of alkali metals are low, and they are soft metals. Since the outer shell contains only one s-electron, the ionization energy is very low, and mono cations of alkali metals form easily. Qualitative analysis of alkali metals is possible by means of flame reactions using characteristic luminescence lines. Especially the orange D-line of sodium is used in the sodium lamp. Alkali metals are oxidized by water evolving hydrogen gas due to their low reduction potentials. Except lithium, the heavier alkali metals react violently with water, and sufficient caution should be exercised in their handling. Exercise \(\PageIndex{1}\) Describe the reactivity of alkali metals in water. Answer The reactivity of lithium is the lowest, sodium reacts violently, and potassium, rubidium, and cesium react explosively. Alkali metals are also highly reactive to oxygen or halogens. As alkali metals are very reducing, they are used widely as reducing agents. Because of the high affinity of alkali metals to halogens, they are important in organic and inorganic syntheses which produce alkali metal halides as the result of condensation and metathesis reactions. Although it is generally difficult to dissolve metals in solvents to make atomic dipersions, alkali metals can be dispersed in liquid ammonia solutions, amalgams, and as cryptand (Figure \(\PageIndex{1}\)), naphthalene, or benzophenone (C 6 H 5 ) 2 CO complexes. Ammonia boils at -33.35 °C but liquid ammonia can be easily handled. Alkali metals dissolve readily in liquid ammonia and dilute solutions are blue but concentrated ones show a bronze color. The metal is recovered when ammonia is evaporated from metal solutions. Alkali metal solutions show the same color irrespective of the kind of alkali metals as the color is due to the solvated electrons. Namely, the dissolution is accompanied by the separation of the alkali metal atoms into metal cations and electrons solvated by ammonia, according to the following equation. \[M + n\; NH_{3} \rightarrow M^{+} [e^{-} (NH_{3})] \nonumber \] The liquid ammonia solution of an alkali metal is conductive and paramagnetic. The highly reducing solution is used for special reduction reactions or syntheses of alkali metal complexes and polyhalides.
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Bookshelves/General_Chemistry/Chemistry_2e_(OpenStax)
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This text is designed for the two-semester general chemistry course. For many students, this course provides the foundation to a career in chemistry, while for others, this may be their only college-level science course. As such, this textbook provides an important opportunity for students to learn the core concepts of chemistry and understand how those concepts apply to their lives and the world around them. The text has been developed to meet the scope and sequence of most general chemistry courses. Front Matter 1: Essential Ideas 2: Atoms, Molecules, and Ions 3: Composition of Substances and Solutions 4: Stoichiometry of Chemical Reactions 5: Thermochemistry 6: Electronic Structure and Periodic Properties 7: Chemical Bonding and Molecular Geometry 8: Advanced Theories of Covalent Bonding 9: Gases 10: Liquids and Solids 11: Solutions and Colloids 12: Kinetics 13: Fundamental Equilibrium Concepts 14: Acid-Base Equilibria 15: Equilibria of Other Reaction Classes 16: Thermodynamics 17: Electrochemistry 18: Representative Metals, Metalloids, and Nonmetals 19: Transition Metals and Coordination Chemistry 20: Organic Chemistry 21: Nuclear Chemistry 22: Appendices Back Matter Thumbnail: Lead Iodide (CC BY 2.0 Generic; Paige Powers via Flickr )
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Courses/Fullerton_College/Beginning_Chemistry_(Chan)/07%3A_Stoichiometry_and_the_Mole
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We have already established that quantities are important in science, especially in chemistry. It is important to make accurate measurements of a variety of quantities when performing experiments. However, it is also important to be able to relate one measured quantity to another, unmeasured quantity. In this chapter, we will consider how we manipulate quantities to relate them to each other. 7.1: Introduction 7.2: Stoichiometry Quantities of substances can be related to each other using balanced chemical equations. 7.3: The Mole The mole is a key unit in chemistry. The molar mass of a substance, in grams, is numerically equal to one atom's or molecule's mass in atomic mass units. 7.4: The Mole in Chemical Reactions Balanced chemical reactions are balanced in terms of moles. A balanced chemical reaction gives equivalences in moles that allow stoichiometry calculations to be performed. 7.5: Mole-Mass and Mass-Mass Calculations Mole quantities of one substance can be related to mass quantities using a balanced chemical equation. Mass quantities of one substance can be related to mass quantities using a balanced chemical equation. In all cases, quantities of a substance must be converted to moles before the balanced chemical equation can be used to convert to moles of another substance. 7.6: Yields Theoretical yield is the calculated yield using the balanced chemical reaction. Actual yield is what is actually obtained in a chemical reaction. Percent yield is a comparison of the actual yield with the theoretical yield. 7.7: Limiting Reagents The limiting reagent is the reactant that produces the least amount of product. Mass-mass calculations can determine how much product is produced, and how much of the other reactants remain. 7.E: Stoichiometry and the Mole (Exercises) These are exercises and select solutions to accompany Chapter 5 of the "Beginning Chemistry" Textmap formulated around the Ball et al. textbook.
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Bookshelves/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/IV%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_2/06%3A_Electrophilic_Addition_to_Alkenes/6.13%3A_Ziegler-Natta_Polymerization
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Polyethylene and polypropylene are two enormously important materials on the market. The fact that their use continues to persist despite legitimate environmental concerns is a testament to how useful these materials have become over the years. Polyethylene and polypropylene can be thought of as polymers of the very simple alkenes, ethene and propene. In fact, that's exactly where these materials come from. Exercise \(\PageIndex{1}\) Draw the first intermediate formed after the initiation step in the cationic polymerization of: ethene propene Unlike other polymers of alkenes that we have looked at, polyethylene and polypropylene are not polymerised via cationic methods. Instead, these monomers are enchained through a process called " Ziegler-Natta polymerisation ." This process is named after a German and an Italian chemist who are independently credited with its development in the 1950's. In Ziegler-Natta polymerisation, monomers are treated with a catalyst, such as a mixture of titanium chloride (or related compounds, like oxovanadium chloride) with triethylaluminum (or trimethylaluminum). Other components are often added, such as magnesium chloride, to modify the catalyst and improve performance. The mixture described here produces a heterogeneous catalyst; it is an insoluble solid. Now, the catalyst isn't really titanium chloride, because all of these components react together to make something new. Exactly what they make may be hard to determine. It's a complicated gmish. Despite the complicated catalyst mixture, we do know a little bit about the mechanism of reaction. Partly this information comes from studies of model compounds. Model compounds are simpler than the industrial catalysts, but they still have some structural features in common with their hard-working cousins. They have enough in common to be able to carry out polymerization catalysis, although maybe not as well as the industrial heavyweights. So, what do we think happens? It seems pretty clear that one of the things that the trialkylaluminum does is provide an alkyl group to titanium. That shouldn't be too surprising. The triethylaluminum, like ethyllithium or ethylmagnesium bromide, ought to be a source of nucleophilic ethyl groups. The titanium tetrachloride ought to be a pretty good electrophile, complete with halide leaving groups. We can imagine at least one of those chlorides getting replaced by an ethyl ligand. Exercise \(\PageIndex{2}\) Provide a mechanism for the ethylation of titanium tetrachloride with triethylaluminum. The next step, presumably, could be the binding of an alkene ligand to the transition metal. Now we are looking at an organo transition metal compound. We should be thinking about organometallic reaction mechanisms. For example, a 1,2-insertion of the alkene into the metal-carbon bond would provide a new metal alkyl. From there, and using these same elementary steps in succession, it is easy to imagine how polymerization of propene might occur. Bnding of an additional alkene, followed by the 1,2-insertion of the propene into the metal-carbon bond, results in formation of the propene dimer (with a methyl end group). The mechanism of Ziegler-Natta polymerization involves alkene binding and insertion into metal-carbon bonds Let's pause for a moment and look a little more closely at the role of the aluminum compound. This topic is peripheral to electrophilic addition although important to the subject of catalysis. In catalysis, "promoters" and "supports" are sometimes added to improve catalyst function. They might do so in a number of ways. They may take part directly in the reaction, providing additional Lewis acidic or Lewis basic sites, in much the same way that amino acid residues surrounding the active site of an enzyme may help catalyse a reaction. They may play a more subtle role, affecting physical properties of the catalyst (such as its solubility) or even tuning up the chemical properties of the catalyst. For example, maybe the promoter adds a little more electron density to the catalyst, making it a little less electrophilic. That might make the catalyst more stable; maybe it becomes more selective, reacting more carefully instead of with wild abandon. Maybe it makes the catalyst last longer. Although classic Ziegler-Natta polymerization involves heterogeneous catalysis, lots of variations have been developed, including model systems to study the basics of the reaction as well as other, working homogeneous catalysts. One very successful variation, developed by Walter Kaminsky at University of Hamburg, uses Cp 2 ZrCl 2 as catalyst and methylalumoxane (MAO) as a promoter. This zirconium species, in which the zirconium atom is wedged between two cyclopentadienly ligands, is commonly called a "zirconocene". MAO is another poorly-defined species. It is obtained by treating trimethylaluminum with a trace of moisture. If you remember anything about Grignard reagents or alkyllithiums, you might think that isn't such a good idea. In truth, it is an even worse idea with trimethylaluminum with either of those other two metal alkyls. The trimethylaluminum is quickly decomposed into something else, a poorly-defined species called "methylalumoxane". What we know about the structure of MAO may be a little bit fuzzy. Once again, some light can be shed on the subject via model studies. In some beautiful work done in Andrew Barron 's lab at Rice University, alkyl aluminum oxide clusters were obtained via the careful treatment of tri( t butyl)aluminum with water. Aluminum oxide clusters resulted, containing two, three, four or six aluminum atoms, depending on the reaction conditions. A drawing of one example, a hexamer, is shown below. Some of the alkyl groups have been replaced by oxide ligands. We can imagine something similar would happen with trimethylaluminum. The aluminum forms bonds with oxygen, which bridges between different aluminum atoms. The structure is probably oligomeric itself, forming large clusters of aluminum oxide, although it must retain some methyl groups as well. Exercise \(\PageIndex{3}\) Provide a mechanism for the formation of an oxy-bridged aluminum dimer via treatement of trimethylaluminum with water. Exercise \(\PageIndex{4}\) Why does tri( t butyl)aluminum with water produce well-defined structures, whereas trimethylaluminum with water leads to a mess? Just as in the original Ziegler-Natta catalyst, the "extra stuff" plays an important role. The MAO may tune up the qualities of the zirconium catalyst, in addition to providing an alkyl group. Kaminsky's "zirconocene" catalysts are used commercially to produce polypropylene. They have been particularly important in developing ways to control the stereochemistry of the reaction. Consider a polypropylene chain. Each methyl group that hangs from the zig-zagging backbone of the polymer could have two possible orientations. It could be coming forward, shown with a wedge. It could be going backward, shown with a dash. The stereochemical relationship between those wedges and dashes is called "tacticity". Tacticity basically comes in three flavours: random, alternating and same. A random stereochemical arrangment is described as an "atactic" polymer. If all the methyl groups are on the same side in a regular zig-zag projection of the backbone (either all wedges or all dashes) the arrangement is described as "isotactic". If instead the methyl groups alternate (wedge-dash-wedge-dash), the arranement is called "syndiotactic". In a collaboration with Hans-Herbert Brintzinger at Konstanz University, Kaminsky developed zirconocene catalysts that could control the tacticity of the polymer chain. Use of the original zirconocene, Cp 2 ZrCl 2 , resulted in an atactic polymer. However, modified zirconocene catalysts selectively make either the isotactic or the syndiotactic polypropylene. We will look at two examples. The first one is sometimes described as a C 2 catalyst and leads to formation of isotactic polypropylene. The second is sometimes described as a C s catalyst and leads to formation of syndiotactic polypropylene. (The labels, C s and C 2 , are symmetry point groups that describe the shape of the catalyst, but we won't go into that idea any further.) Exercise \(\PageIndex{5}\) One of these catalysts (C 2 vs. C s ) is chiral; the other one isn't. Which is which? Let's look at the C 2 catalyst. We'll strip it down to just the zirconocene part, leaving off the chlorides. The chlorides are likely replaced by methyl groups or else lost via dissociation (presumably leaving Cp' 2 ZrCl + ion). We will rotate the catalyst fragment to look at it from the "front": the more open part of the zirconocene "wedge", which is the direction from which a newly coordinating propene would approach. Here is the view from in front of the wedge. This is the surface that the propene will interact with as it approaches. In our discussion, we will use "quadrant analysis", a standard tool for trying to analyse stereocontrol in transition metal catalysis. In quadrant analysis, we try to imagine differences in steric barriers in each of four quadrants around the metal centre. How will the arrangement of bulky groups influence the approach of a substrate? In the C 2 catalyst we are using, it looks like there will be more room in the upper right and lower left quadrants. The upper left and lower right are blocked by those rings. When the propene is approaching, the polymer chain will present the largest obstacle, because it is extending a significant distance away from the zirconium atom. To minimize steric interactions,the chain may extend into the relatively empty upper right quadrant. Alternatively, the polymer change could extend into the relatively open lower left quadrant, but that would really give us the same drawing, just rotated by 180 degrees. Exercise \(\PageIndex{6}\) Make a drawing of the complex with the polymer in the lower left quadrant. As we think about bringing the propene into the empty coordination site next to the polymer chain, there are two questions we need to consider about orientation. The first is about which end of the alkene to bring into the wedge. The propene has two different ends: one end sports two hydrogen atoms, whereas the other end has a hydrogen and a methyl. It seems likely that the propene will fit best if the narrow end, the one with the two hydrogen atoms, extends into that narrow wedge. Now we need to think about which of the propene's two faces will coordinate to the zirconium. To think about the faces of propene, hold your hand out flat, with the thumb forming a right angle with the rest of your hand. The back of your hand represents one face of the propene; the palm of your hand represents the other. The propene will enter in such a way as to minimise steric interactions. It looks like the easiest way is as shown below. The methyl group is placed in the lower left quadrant. You can think of it as keeping the methyl group away from the upper ring or keeping it away from the polymer chain, which is also in an upper quadrant. If your left hand is propene, we have coordinated the back of your hand to the zirconium, with the thumb pointing down. If we had coordinated your palm, the thumb would be pointed up. Once again, if we started with the polymer in the lower left quadrant, the entire drawing just rotates 180 degrees. It is still the same face that coordinates (the back of your left hand). Exercise \(\PageIndex{7}\) Make a drawing of the complex with the polymer in the lower left quadrant and the propene coordinating to the right. The next event is 1,2-insertion of the alkene into the metal-alkyl bond. That leaves us with the following structure. Notice that we have formed a new stereocentre. Because the methyl group of coordinated propene was pointing down, and the hydrogen adjacent to it was therefore up, then in the new stereocenter the methyl is still in the lower of two possible positions and the hydrogen is in the upper of two possible positions. Furthermore, the methyl is pushed back and to the left because the alkyl came from the right. In order to help keep track of that insertion step, here is a drawing with colour labels. The narrow end of the propene is now attached to the zirconium via a sigma bond. The wider end of the proene has formed the new chiral center. The carbon that used to form a sigma bond to zirconium is now just another carbon along the growing polymer chain. The youngest part of the polymer is found at the growing end. When another propene approaches to occupy the empty position, it will coordinate using the same face as the previous propene. The 1,2-insertion produces a new chiral centre. The C 2 catalyst is producing an isotactic polymer chain. The is an example of "site control" of polymerization. The chiral C 2 catalyst has influenced the stereochemistry of the growing chain. Exercise \(\PageIndex{8}\) Assign configuration ( R or S ) to each of the chiral centres along the polymer chain in the above drawing to confirm isotacticity. Keep in mind that "polymer" stands for a long chain of carbons. Now let's look at the C s catalyst. We'll strip it down like we did before. Where will a growing polymer chain go? Obviously it should go into one of the upper quadrants. It isn't obvious whether it should be upper right or upper left. This time, it makes a difference, because the picture would not be the same. Exercise \(\PageIndex{9}\) Draw the catalyst site with the polymer chain in the upper left quadrant. However, once the propene approaches, a clear preference occurs. The chiral centre on the polymer chain is shown with the hydrogen up towards the Cp ring, since it is small and won't cause too much steric interaction. The polymer chain is extending forward, into the big, open space of the wedge. The leaves the methyl pointing to one side of the wedge. Which way will the propene approach? Will it come in on the same side as the methyl, or the opposite side? Probably the opposite side, as shown below. Researchers suspect the methyl group points "down" in the above drawing, rather than "up", because the polymer chain is a bigger steric obstacle than the lower aromatic ring, and the polymer chain is in an upper quadrant. Once again, the approach of the propene is sterochemically controlled, although this time the stereochemistry of an existing chiral centre in the growing polymer chain influenced how things proceeded. Once again, upon 1,2-insertion, a new chiral centre is formed. The C s catalyst is forming a syndiotactic polymer chain. Because of the influence of an existing choral centre on the stereochemical outcome of the reaction, this catalyst is considered to work through "chain-end control". The catalyst site simply amplifies the influence of that chiral centre upon the chiral centre that forms next. It does so by bringing the reactants together into a small space where the steric differences of two subtly different pathways become more important. Exercise \(\PageIndex{10}\) Label the configurations (R or S) of the chiral centres in the above drawing to confirm syndiotacticity.
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Courses/American_River_College/Chemistry_305_(S21_Zarzana)/Map%3A_Introductory_Chemistry_(Tro)/15%3A_Chemical_Equilibrium/15.06%3A_Calculating_and_Using_Equilibrium_Constants
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Learning Objectives To understand how different phases affect equilibria. When the products and reactants of an equilibrium reaction form a single phase, whether gas or liquid, the system is a homogeneous equilibrium. In such situations, the concentrations of the reactants and products can vary over a wide range. In contrast, a system whose reactants, products, or both are in more than one phase is a heterogeneous equilibrium, such as the reaction of a gas with a solid or liquid. As noted in the previous section, the equilibrium constant expression is actually a ratio of activities. To simplify the calculations in general chemistry courses, the activity of each substance in the reaction is often approximated using a ratio of the molarity of a substance compared to the standard state of that substance. For substances that are liquids or solids, the standard state is just the concentration of the substance within the liquid or solid. Because the molar concentrations of pure liquids and solids normally do not vary greatly with temperature, the ratio of the molarity to the standard state for substances that are liquids or solids always has a value of 1. For example, for a compound such as CaF 2 (s), the term going into the equilibrium expression is [CaF 2 ]/[CaF 2 ] which cancels to unity. Thus, when the activities of the solids and liquids (including solvents) are incorporated into the equilibrium expression, they do not change the value. Consider the following reaction, which is used in the final firing of some types of pottery to produce brilliant metallic glazes: \[\ce{CO2(g) + C(s) \rightleftharpoons 2CO(g)} \label{Eq14.4.1} \] The glaze is created when metal oxides are reduced to metals by the product, carbon monoxide. The equilibrium constant expression for this reaction is as follows: \[K=\dfrac{a_{\ce{CO}}^2}{a_{\ce{CO2}}a_{C}}=\dfrac{[\ce{CO}]^2}{[\ce{CO2}][1]}=\dfrac{[\ce{CO}]^2}{[\ce{CO_2}]}\label{Eq14.4.2} \] The equilibrium constant for this reaction can also be written in terms of the partial pressures of the gases: \[K_p=\dfrac{(P_{CO})^2}{P_{CO_2}} \label{Eq14.4.3} \] Incorporating all the constant values into \(K′\) or \(K_p\) allows us to focus on the substances whose concentrations change during the reaction. Although the activities of pure liquids or solids are not written explicitly in the equilibrium constant expression, these substances must be present in the reaction mixture for chemical equilibrium to occur. Whatever the concentrations of \(\ce{CO}\) and \(\ce{CO_2}\), the system described in Equation \(\ref{Eq14.4.1}\) will reach chemical equilibrium only if a stoichiometric amount of solid carbon or excess solid carbon has been added so that some is still present once the system has reached equilibrium. As shown in Figure \(\PageIndex{1}\), it does not matter whether 1 g or 100 g of solid carbon is present; in either case, the composition of the gaseous components of the system will be the same at equilibrium. Example \(\PageIndex{1}\) Write each expression for \(K\), incorporating all constants, and \(K_p\) for the following equilibrium reactions. \(\ce{PCl3(l) + Cl2(g) <=> PCl5(s)}\) \(\ce{Fe3O4(s) + 4H2(g) <=> 3Fe(s) + 4H2O(g)}\) Given : balanced equilibrium equations. Asked for : expressions for \(K\) and \(K_p\). Strategy : Find \(K\) by writing each equilibrium constant expression as the ratio of the concentrations of the products and reactants, each raised to its coefficient in the chemical equation. Then express \(K_p\) as the ratio of the partial pressures of the products and reactants, each also raised to its coefficient in the chemical equation. Solution This reaction contains a pure solid (\(PCl_5\)) and a pure liquid (\(PCl_3\)). Their activities are equal to 1, so when incorporated into the equilibrium constant expression, they do not change the value. So \[K=\dfrac{1}{(1)[Cl_2]} \nonumber \] and \[K_p=\dfrac{1}{(1)P_{Cl_2}} \nonumber \] This reaction contains two pure solids (\(Fe_3O_4\) and \(Fe\)), which are each assigned a value of 1 in the equilibrium constant expressions: \[K=\dfrac{(1)[H_2O]^4}{(1)[H_2]^4} \nonumber \] and \[K_p=\dfrac{(1)(P_{H_2O})^4}{(1)(P_{H_2})^4} \nonumber \] Exercise \(\PageIndex{1}\) Write the expressions for \(K\) and \(K_p\) for the following reactions. \(\ce{CaCO3(s) <=> CaO(s) + CO2(g)}\) \( \underset{glucose}{\ce{C6H12O6(s)}} + \ce{6O2(g) <=> 6CO2(g) + 6H2O(g)}\) Answer a \(K = [\ce{CO_2}]\) and \(K_p = P_{\ce{CO_2}}\) Answer b \(K=\dfrac{[CO_2]^6[H_2O]^6}{[O_2]^6}\) and \(K_p=\dfrac{(P_{CO_2})^6(P_{H_2O})^6}{(P_{O_2})^6}\) For reactions carried out in solution, the solvent is assumed to be pure, and therefore is assigned an activity equal to 1 in the equilibrium constant expression. The activities of the solutes are approximated by their molarities. The result is that the equilibrium constant expressions appear to only depend upon the concentrations of the solutes. The activities of pure solids, pure liquids, and solvents are defined as having a value of '1'. Often, it is said that these activities are "left out" of equilibrium constant expressions. This is an unfortunate use of words. The activities are not "left out" of equilibrium constant expressions. Rather, because they have a value of '1', they do not change the value of the equilibrium constant when they are multiplied together with the other terms. The activities of the solutes are approximated by their molarities. Summary An equilibrated system that contains products and reactants in a single phase is a homogeneous equilibrium; a system whose reactants, products, or both are in more than one phase is a heterogeneous equilibrium.
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Courses/City_College_of_San_Francisco/Chemistry_101A/Topic_D%3A_Thermochemistry/6%3A_Thermochemistry/6.02%3A_Calorimetry
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Learning Objectives Explain the technique of calorimetry Calculate and interpret heat and related properties using typical calorimetry data A calorimeter is a device used to measure the amount of heat involved in a chemical or physical process. For example, when an exothermic reaction occurs in solution in a calorimeter, the heat produced by the reaction is absorbed by the solution, which increases its temperature. When an endothermic reaction occurs, the heat required is absorbed from the thermal energy of the solution, which decreases its temperature (Figure \(\PageIndex{1}\)). The temperature change, along with the specific heat and mass of the solution, can then be used to calculate the amount of heat involved in either case. By convention, q is given a negative (-) sign when the system releases heat to the surroundings (exothermic); q is given a positive (+) sign when the system absorbs heat from the surroundings (endothermic). Scientists use well-insulated calorimeters that all but prevent the transfer of heat between the calorimeter and its environment. This enables the accurate determination of the heat involved in chemical processes, the energy content of foods, and so on. General chemistry students often use simple calorimeters constructed from polystyrene cups (Figure \(\PageIndex{2}\)). These easy-to-use “coffee cup” calorimeters allow more heat exchange with their surroundings, and therefore produce less accurate energy values. Commercial solution calorimeters are also available. Relatively inexpensive calorimeters often consist of two thin-walled cups that are nested in a way that minimizes thermal contact during use, along with an insulated cover, handheld stirrer, and simple thermometer. More expensive calorimeters used for industry and research typically have a well-insulated, fully enclosed reaction vessel, motorized stirring mechanism, and a more accurate temperature sensor (Figure \(\PageIndex{3}\)). Before we practice calorimetry problems involving chemical reactions, consider a simple example that illustrates the core idea behind calorimetry. Suppose we initially have a high-temperature substance, such as a hot piece of metal (M), and a low-temperature substance, such as cool water (W). If we place the metal in the water, heat will flow from M to W. The temperature of M will decrease, and the temperature of W will increase, until the two substances have the same temperature—that is, when they reach thermal equilibrium (Figure \(\PageIndex{4}\)). If this occurs in a calorimeter, ideally all of this heat transfer occurs between the two substances, with no heat gained or lost by either the calorimeter or the calorimeter’s surroundings. Under these ideal circumstances, the net heat change is zero: \[q_\mathrm{\,substance\: M} + q_\mathrm{\,substance\: W}=0 \nonumber\] This relationship is sometimes rearranged to more explicitly show that the heat gained by substance M is equal to the heat lost by substance W: \[q_\mathrm{\,substance\: M}=-q_\mathrm{\,substance\: W} \nonumber\] The magnitude of the heat (change) is the same for both substances, and the negative sign merely shows that q substance M and q substance W are opposite in direction of heat flow (gain or loss) but does not indicate the arithmetic sign of either q value (that is determined by whether the matter in question gains or loses heat, per definition). In the specific situation described, q substance M is a negative value and q substance W is positive, since heat is transferred from M to W. Example \(\PageIndex{1}\): Heat Transfer between Substances at Different Temperatures A hot 360-g piece of rebar (a steel rod used for reinforcing concrete) is dropped into 425 mL of water at 24.0 °C. The final temperature of the water is measured as 42.7 °C. Calculate the initial temperature of the piece of rebar. Assume the specific heat of steel is approximately the same as that for iron (Table T4), and that all heat transfer occurs between the rebar and the water (there is no heat exchange with the surroundings). Solution The temperature of the water increases from 24.0 °C to 42.7 °C, so the water absorbs heat. That heat came from the piece of rebar, which initially was at a higher temperature. Assuming that all heat transfer was between the rebar and the water, with no heat “lost” to the surroundings, then heat given off by rebar = − heat taken in by water, or: \[q_\ce{rebar}+q_\ce{water}= 0 \nonumber\] Since we know how heat is related to other measurable quantities, we have: \[(c×m×ΔT)_\ce{rebar} + (c×m×ΔT)_\ce{water} = 0 \nonumber\] Letting f = final and i = initial, in expanded form, this becomes: \[ c_\ce{rebar}×m_\ce{rebar}×(T_\mathrm{f,rebar}−T_\mathrm{i,rebar}) + c_\ce{water}×m_\ce{water}×(T_\mathrm{f,water}−T_\mathrm{i,water}) = 0 \nonumber\] The density of water is 1.0 g/mL, so 425 mL of water = 425 g. Noting that the final temperature of both the rebar and water is 42.7 °C, substituting known values yields: \[ \mathrm{(0.449\:J/g\: °C)(360g)(42.7°C−\mathit T_\mathrm{i,rebar}) + (4.184\:J/g\: °C)(425\:g)(42.7°C−24.0°C)} = 0 \nonumber\] \[\mathrm{\mathit T_{i,rebar}=\dfrac{(4.184\:J/g\: °C)(425\:g)(42.7°C−24.0°C)}{(0.449\:J/g\: °C)(360\:g)}+42.7°C}\nonumber\] Solving this gives Ti,rebar= 248 °C, so the initial temperature of the rebar was 248 °C. Exercise \(\PageIndex{1A}\) A 248-g piece of copper is dropped into 390 mL of water at 22.6 °C. The final temperature of the water was measured as 39.9 °C. Calculate the initial temperature of the piece of copper. Assume that all heat transfer occurs between the copper and the water. Answer The initial temperature of the copper was 335.6 °C. Exercise \(\PageIndex{1B}\) A 248-g piece of copper initially at 314 °C is dropped into 390 mL of water initially at 22.6 °C. Assuming that all heat transfer occurs between the copper and the water, calculate the final temperature. Answer The final temperature (reached by both copper and water) is 38.7 °C. This method can also be used to determine other quantities, such as the specific heat of an unknown metal. Example \(\PageIndex{2}\): Identifying a Metal by Measuring Specific Heat A 59.7 g piece of metal that had been submerged in boiling water was quickly transferred into 60.0 mL of water initially at 22.0 °C. The final temperature is 28.5 °C. Use these data to determine the specific heat of the metal. Use this result to identify the metal. Solution Assuming perfect heat transfer, heat given off by metal will be absorbed by the water, or: \[q_\ce{metal} + q_\ce{water} = 0 \nonumber\] In expanded form, this is: \[c_\ce{metal}×m_\ce{metal}×(T_\mathrm{f,metal}−T_\mathrm{i, metal}) + c_\ce{water}×m_\ce{water}×(T_\mathrm{f,water}−T_\mathrm{i,water}) = 0 \nonumber\] Noting that since the metal was submerged in boiling water, its initial temperature was 100.0 °C; and that for water, 60.0 mL = 60.0 g; we have: \[\mathrm{(\mathit c_{metal})(59.7\:g)(28.5°C−100.0°C) +(4.18\:J/g\: °C)(60.0\:g)(28.5°C−22.0°C)} = 0 \nonumber\] Solving this: \[\mathrm{\mathit c_{metal}=\dfrac{−(4.184\:J/g\: °C)(60.0\:g)(6.5°C)}{(59.7\:g)(−71.5°C)}=0.38\:J/g\: °C}\nonumber\] Comparing this with values in Table T4, our experimental specific heat is closest to the value for copper (0.39 J/g °C), so we identify the metal as copper. Exercise \(\PageIndex{2}\) A 92.9-g piece of a silver/gray metal is heated to 178.0 °C, and then quickly transferred into 75.0 mL of water initially at 24.0 °C. After 5 minutes, both the metal and the water have reached the same temperature: 29.7 °C. Determine the specific heat and the identity of the metal. (Note: You should find that the specific heat is close to that of two different metals. Explain how you can confidently determine the identity of the metal). Answer \(c_{metal}= 0.13 \;J/g\; °C\) This specific heat is close to that of either gold or lead. It would be difficult to determine which metal this was based solely on the numerical values. However, the observation that the metal is silver/gray in addition to the value for the specific heat indicates that the metal is lead. When we use calorimetry to determine the heat involved in an aqueous chemical reaction, the same principles we have been discussing apply. The amount of heat absorbed by the calorimeter is often small enough that we can neglect it (though not for highly accurate measurements, as discussed later), and the calorimeter minimizes energy exchange with the surroundings. Because energy is neither created nor destroyed during a chemical reaction, there is no overall energy change during the reaction. The heat produced or consumed in the reaction (the “system”), q reaction , plus the heat absorbed or lost by the solution (the “surroundings”), q solution , must add up to zero: \[q_\ce{reaction}+q_\ce{solution}=0\] The amount of heat produced or consumed in the reaction equals the amount of heat absorbed or lost by the solution. This concept lies at the heart of all calorimetry problems and calculations. Example \(\PageIndex{3}\): Heat Produced by an Exothermic Reaction When 50.0 mL of 0.10 M HCl( aq ) and 50.0 mL of 1.00 M NaOH( aq ), both at 22.0 °C, are added to a coffee cup calorimeter, the temperature of the mixture reaches a maximum of 28.9 °C. What is the approximate amount of heat produced by this reaction? \[\ce{HCl}(aq)+\ce{NaOH}(aq)⟶\ce{NaCl}(aq)+\ce{H2O}(l) \nonumber\] S olution To visualize what is going on, imagine that you could combine the two solutions so quickly that no reaction took place while they mixed; then after mixing, the reaction took place. At the instant of mixing, you have 100.0 mL of a mixture of HCl and NaOH at 22.0 °C. The HCl and NaOH then react until the solution temperature reaches 28.9 °C. The heat given off by the reaction is equal to that taken in by the solution. Therefore: \[q_\ce{reaction} + q_\ce{solution} = 0 \nonumber\] (It is important to remember that this relationship only holds if the calorimeter does not absorb any heat from the reaction, and there is no heat exchange between the calorimeter and its surroundings.) Next, we know that the heat absorbed by the solution depends on its specific heat, mass, and temperature change: \[q_\ce{solution}=(c×m×ΔT)_\ce{solution} \nonumber\] To proceed with this calculation, we need to make a few more reasonable assumptions or approximations. Since the solution is aqueous, we can proceed as if it were water in terms of its specific heat and mass values. The density of water is approximately 1.0 g/mL, so 100.0 mL has a mass of about 1.0 × 10 2 g (two significant figures). The specific heat of water is approximately 4.18 J/g °C, so we use that for the specific heat of the solution. Substituting these values gives: \[\mathrm{\mathit q_{solution}=(4.184\:J/g\: °C)(1.0×10^2\:g)(28.9°C−22.0°C)=2.89×10^3\:J}\nonumber\] Finally, since we are trying to find the heat of the reaction, we have: \[q_\ce{reaction}=−q_\ce{solution}=−2.89×10^3\:J \nonumber\] The negative sign indicates that the reaction is exothermic. It produces 2.89 kJ of heat. Exercise \(\PageIndex{3}\) When 100 mL of 0.200 M NaCl( aq ) and 100 mL of 0.200 M AgNO 3 ( aq ), both at 21.9 °C, are mixed in a coffee cup calorimeter, the temperature increases to 23.5 °C as solid AgCl forms. How much heat is produced by this precipitation reaction? What assumptions did you make to determine your value? Answer \(1.34 \times 10^3\; J\); assume no heat is absorbed by the calorimeter, no heat is exchanged between the calorimeter and its surroundings, and that the specific heat and mass of the solution are the same as those for water Thermochemistry of Hand Warmers When working or playing outdoors on a cold day, you might use a hand warmer to warm your hands (Figure \(\PageIndex{5}\)). A common reusable hand warmer contains a supersaturated solution of NaC 2 H 3 O 2 (sodium acetate) and a metal disc. Bending the disk creates nucleation sites around which the metastable NaC 2 H 3 O 2 quickly crystallizes (a later chapter on solutions will investigate saturation and supersaturation in more detail). The process \(\ce{NaC2H3O2}(aq)⟶\ce{NaC2H3O2}(s)\) is exothermic, and the heat produced by this process is absorbed by your hands, thereby warming them (at least for a while). If the hand warmer is reheated, the NaC 2 H 3 O 2 redissolves and can be reused. Another common hand warmer produces heat when it is ripped open, exposing iron and water in the hand warmer to oxygen in the air. One simplified version of this exothermic reaction is \[\ce{2Fe}(s)+\dfrac{3}{2}\ce{O2}(g)⟶\ce{Fe2O3}(s).\] Salt in the hand warmer catalyzes the reaction, so it produces heat more rapidly; cellulose, vermiculite, and activated carbon help distribute the heat evenly. Other types of hand warmers use lighter fluid (a platinum catalyst helps lighter fluid oxidize exothermically), charcoal (charcoal oxidizes in a special case), or electrical units that produce heat by passing an electrical current from a battery through resistive wires. Example \(\PageIndex{4}\): Heat Flow in an Instant Ice Pack When solid ammonium nitrate dissolves in water, the solution becomes cold. This is the basis for an “instant ice pack” (Figure \(\PageIndex{5}\)). When 3.21 g of solid NH 4 NO 3 dissolves in 50.0 g of water at 24.9 °C in a calorimeter, the temperature decreases to 20.3 °C. Calculate the value of q for this reaction and explain the meaning of its arithmetic sign. State any assumptions that you made. Solution We assume that the calorimeter prevents heat transfer between the solution and its external environment (including the calorimeter itself), in which case: \[q_\ce{rxn} + q_\ce{soln} = 0 \nonumber\] with “rxn” and “soln” used as shorthand for “reaction” and “solution,” respectively. Assuming also that the specific heat of the solution is the same as that for water, we have: \[\begin{align*} &q_\ce{rxn}=−q_\ce{soln}=−(c×m×ΔT)_\ce{soln}\\ &=\mathrm{−[(4.184J/g\: °C)×(53.2\:g)×(20.3°C−24.9°C)]}\\ &=\mathrm{−[(4.184J/g\: °C)×(53.2\:g)×(−4.6°C)]}\\ &+\mathrm{1.0×10^3\:J=+1.0\:kJ} \end{align*}\] The positive sign for q indicates that the dissolution is an endothermic process. Exercise \(\PageIndex{4}\) When a 3.00-g sample of KCl was added to 3.00 × 10 2 g of water in a coffee cup calorimeter, the temperature decreased by 1.05 °C. How much heat is involved in the dissolution of the KCl? What assumptions did you make? Answer 1.33 kJ; assume that the calorimeter prevents heat transfer between the solution and its external environment (including the calorimeter itself) and that the specific heat of the solution is the same as that for water. If the amount of heat absorbed by a calorimeter is too large to neglect or if we require more accurate results, then we must take into account the heat absorbed both by the solution and by the calorimeter. The calorimeters described are designed to operate at constant (atmospheric) pressure and are convenient to measure heat flow accompanying processes that occur in solution. A different type of calorimeter that operates at constant volume, colloquially known as a bomb calorimeter , is used to measure the energy produced by reactions that yield large amounts of heat and gaseous products, such as combustion reactions. (The term “bomb” comes from the observation that these reactions can be vigorous enough to resemble explosions that would damage other calorimeters.) This type of calorimeter consists of a robust steel container (the “bomb”) that contains the reactants and is itself submerged in water (Figure \(\PageIndex{6}\)). The sample is placed in the bomb, which is then filled with oxygen at high pressure. A small electrical spark is used to ignite the sample. The energy produced by the reaction is trapped in the steel bomb and the surrounding water. The temperature increase is measured and, along with the known heat capacity of the calorimeter, is used to calculate the energy produced by the reaction. Bomb calorimeters require calibration to determine the heat capacity of the calorimeter and ensure accurate results. The calibration is accomplished using a reaction with a known q , such as a measured quantity of benzoic acid ignited by a spark from a nickel fuse wire that is weighed before and after the reaction. The temperature change produced by the known reaction is used to determine the heat capacity of the calorimeter. The calibration is generally performed each time before the calorimeter is used to gather research data. Video \(\PageIndex{1}\): Video of view how a bomb calorimeter is prepared for action. Example \(\PageIndex{5}\): Bomb Calorimetry When 3.12 g of glucose, C 6 H 12 O 6 , is burned in a bomb calorimeter, the temperature of the calorimeter increases from 23.8 °C to 35.6 °C. The calorimeter contains 775 g of water, and the bomb itself has a heat capacity of 893 J/°C. How much heat was produced by the combustion of the glucose sample? Solution The combustion produces heat that is primarily absorbed by the water and the bomb. (The amounts of heat absorbed by the reaction products and the unreacted excess oxygen are relatively small and dealing with them is beyond the scope of this text. We will neglect them in our calculations.) The heat produced by the reaction is absorbed by the water and the bomb: \(\begin{align*} &q_\ce{rxn}=−(q_\ce{water}+q_\ce{bomb})\\ &=\mathrm{−[(4.184\:J/g\: °C)×(775\:g)×(35.6°C−23.8°C)+893\:J/°C×(35.6°C−23.8°C)]}\\ &=\mathrm{−(38,300\:J+10,500\:J)}\\ &=\mathrm{−48,800\: J=−48.8\: kJ} \end{align*}\) This reaction released 48.7 kJ of heat when 3.12 g of glucose was burned. Exercise \(\PageIndex{5}\) When 0.963 g of benzene, C 6 H 6 , is burned in a bomb calorimeter, the temperature of the calorimeter increases by 8.39 °C. The bomb has a heat capacity of 784 J/°C and is submerged in 925 mL of water. How much heat was produced by the combustion of the glucose sample? Answer 39.0 kJ Since the first one was constructed in 1899, 35 calorimeters have been built to measure the heat produced by a living person. 1 These whole-body calorimeters of various designs are large enough to hold an individual human being. More recently, whole-room calorimeters allow for relatively normal activities to be performed, and these calorimeters generate data that more closely reflect the real world. These calorimeters are used to measure the metabolism of individuals under different environmental conditions, different dietary regimes, and with different health conditions, such as diabetes. In humans, metabolism is typically measured in Calories per day. A nutritional calorie (Calorie) is the energy unit used to quantify the amount of energy derived from the metabolism of foods; one Calorie is equal to 1000 calories (1 kcal), the amount of energy needed to heat 1 kg of water by 1 °C. Measuring Nutritional Calories In your day-to-day life, you may be more familiar with energy being given in Calories, or nutritional calories, which are used to quantify the amount of energy in foods. One calorie (cal) = exactly 4.184 joules, and one Calorie (note the capitalization) = 1000 cal, or 1 kcal. (This is approximately the amount of energy needed to heat 1 kg of water by 1 °C.) The macronutrients in food are proteins, carbohydrates, and fats or oils. Proteins provide about 4 Calories per gram, carbohydrates also provide about 4 Calories per gram, and fats and oils provide about 9 Calories/g. Nutritional labels on food packages show the caloric content of one serving of the food, as well as the breakdown into Calories from each of the three macronutrients (Figure \(\PageIndex{7}\)). For the example shown in (b), the total energy per 228-g portion is calculated by: \[\mathrm{(5\:g\: protein×4\:Calories/g)+(31\:g\: carb×4\:Calories/g)+(12\:g\: fat×9\:Calories/g)=252\:Calories} \label{5.3.X}\] So, you can use food labels to count your Calories. But where do the values come from? And how accurate are they? The caloric content of foods can be determined by using bomb calorimetry; that is, by burning the food and measuring the energy it contains. A sample of food is weighed, mixed in a blender, freeze-dried, ground into powder, and formed into a pellet. The pellet is burned inside a bomb calorimeter, and the measured temperature change is converted into energy per gram of food. Today, the caloric content on food labels is derived using a method called the Atwater system that uses the average caloric content of the different chemical constituents of food, protein, carbohydrate, and fats. The average amounts are those given in the equation and are derived from the various results given by bomb calorimetry of whole foods. The carbohydrate amount is discounted a certain amount for the fiber content, which is indigestible carbohydrate. To determine the energy content of a food, the quantities of carbohydrate, protein, and fat are each multiplied by the average Calories per gram for each and the products summed to obtain the total energy. Summary Calorimetry is used to measure the amount of thermal energy transferred in a chemical or physical process. This requires careful measurement of the temperature change that occurs during the process and the masses of the system and surroundings. These measured quantities are then used to compute the amount of heat produced or consumed in the process using known mathematical relations. Calorimeters are designed to minimize energy exchange between the system being studied and its surroundings. They range from simple coffee cup calorimeters used by introductory chemistry students to sophisticated bomb calorimeters used to determine the energy content of food. Footnotes 1 Francis D. Reardon et al. “The Snellen human calorimeter revisited, re-engineered and upgraded: Design and performance characteristics.” Medical and Biological Engineering and Computing 8 (2006)721–28, http://link.springer.com/article/10....517-006-0086-5 . Glossary bomb calorimeter device designed to measure the energy change for processes occurring under conditions of constant volume; commonly used for reactions involving solid and gaseous reactants or products calorimeter device used to measure the amount of heat absorbed or released in a chemical or physical process calorimetry process of measuring the amount of heat involved in a chemical or physical process nutritional calorie (Calorie) unit used for quantifying energy provided by digestion of foods, defined as 1000 cal or 1 kcal surroundings all matter other than the system being studied system portion of matter undergoing a chemical or physical change being studied
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Courses/University_of_North_Texas/UNT%3A_CHEM_1410_-_General_Chemistry_for_Science_Majors_I/Text/02%3A_Atoms_Molecules_and_Chemical_Reactions/2.07%3A_The_Amount_of_Substance%3A_Moles
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According to the atomic theory , atoms are the units of chemical reactions. The formula HgBr 2 indicates that each molecule of this substance contains one mercury and two bromine atoms. Therefore, if we ask how much bromine is required to make a given quantity of mercury (II) bromide, the answer is two bromine atoms for each mercury atom or two bromine atoms per molecule. In other words, how much substance we have depends in a very important way on how many atoms or molecules are present. So far, we've dealt with mass ratios. Is there a way to change masses of atoms into numbers of atoms, so it is easy to see how much of one element will react with another, just by looking at the number of atoms that are needed? As we see below, there seems to be no fundamental connection between the number of atoms or molecules in the chemical equations, and typical measures of "how much": \(1 \text{ Hg } (l)\) + \( 1 \text{Br}_{2} (l)\) \(\rightarrow\) \(1 \text{ HgBr}_{2} (s)\) 1 atom NaN 1 molecule NaN 1 molecule 1.00 g NaN 0.797 g NaN 1.797 g 1.00 ml NaN 3.47 ml NaN 0.30 ml "How much?" in the above sense of the quantity of atoms or molecules present is not the same thing as "how much" in terms of volume or mass. It takes 3.47 cm 3 Br 2 ( l ) to react with a 1 cm 3 sample of Hg( l ). That same 1 cm 3 Hg( l ) would weigh 13.59 g, but only 10.83 g Br 2 ( l ) would be needed to react with it. In terms of volume, more bromine than mercury is needed, while in terms of mass, less bromine than mercury is required. In the atomic sense, however, there are exactly twice as many bromine atoms as mercury atoms and twice as much bromine as mercury. Luckily, the International System of Measurements (IUPAC) has a measure of amount that reflects the number of atoms present, and it is called the mole. For perspective, 1 mole of salt cubes (like those seen below) would form a cube 27 miles square . For additional perspective, it would take the fastest marathoner in the world just about 2 hours to run the length of one side. A mole is a huge number... Find out more about it on the next page! Image credits go to Tony L. Wong on Flickr.
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