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Courses/Sacramento_City_College/SCC%3A_CHEM_300_-_Beginning_Chemistry/SCC%3A_CHEM_300_-_Beginning_Chemistry_(Alviar-Agnew)/07%3A_Chemical_Reactions/7.03%3A_The_Chemical_Equation | Learning Objectives Identify the reactants and products in any chemical reaction. Convert word equations into chemical equations. Use the common symbols, \(\left( s \right)\), \(\left( l \right)\), \(\left( g \right)\), \(\left( aq \right)\), and \(\rightarrow\) appropriately when writing a chemical reaction. In a chemical change, new substances are formed. In order for this to occur, the chemical bonds of the substances break, and the atoms that compose them separate and rearrange themselves into new substances with new chemical bonds. When this process occurs, we call it a chemical reaction. A chemical reaction is the process in which one or more substances are changed into one or more new substances. Reactants and Products To describe a chemical reaction, we need to indicate what substances are present at the beginning and what substances are present at the end. The substances that are present at the beginning are called reactants and the substances present at the end are called products . Sometimes when reactants are put into a reaction vessel, a reaction will take place to produce products. Reactants are the starting materials, that is, whatever we have as our initial ingredients. The products are just that—what is produced—or the result of what happens to the reactants when we put them together in the reaction vessel. If we think about baking chocolate chip cookies, our reactants would be flour, butter, sugar, vanilla, baking soda, salt, egg, and chocolate chips. What would be the products? Cookies! The reaction vessel would be our mixing bowl. \[ \underbrace{\text{Flour} + \text{Butter} + \text{Sugar} + \text{Vanilla} + \text{Baking Soda} + \text{Eggs} + \text{Chocolate Chips}}_{\text{Ingredients = Reactants}} \rightarrow \underbrace{\text{Cookies}}_{\text{Product}} \nonumber \] Writing Chemical Equations When sulfur dioxide is added to oxygen, sulfur trioxide is produced. Sulfur dioxide and oxygen, \(\ce{SO_2} + \ce{O_2}\), are reactants and sulfur trioxide, \(\ce{SO_3}\), is the product. \[ \underbrace{\ce{2 SO2(g) + O2(g) }}_{\text{Reactants}} \rightarrow \underbrace{\ce{2SO3(g)}}_{\text{Products}} \nonumber \] In chemical reactions, the reactants are found before the symbol "\(\rightarrow\)" and the products are found after the symbol "\(\rightarrow\)". The general equation for a reaction is: \[\text{Reactants } \rightarrow \text{Products} \nonumber \] There are a few special symbols that we need to know in order to "talk" in chemical shorthand. In the table below is the summary of the major symbols used in chemical equations. Table \(\PageIndex{1}\) shows a listing of symbols used in chemical equations. Symbol Description Symbol.1 Description.1 \(+\) used to separate multiple reactants or products \(\left( s \right)\) reactant or product in the solid state \(\rightarrow\) yield sign; separates reactants from products \(\left( l \right)\) reactant or product in the liquid state \(\rightleftharpoons\) replaces the yield sign for reversible reactions that reach equilibrium \(\left( g \right)\) reactant or product in the gas state \(\overset{\ce{Pt}}{\rightarrow}\) formula written above the arrow is used as a catalyst in the reaction \(\left( aq \right)\) reactant or product in an aqueous solution (dissolved in water) \(\overset{\Delta}{\rightarrow}\) triangle indicates that the reaction is being heated NaN NaN Chemists have a choice of methods for describing a chemical reaction. 1. They could draw a picture of the chemical reaction. 2. They could write a word equation for the chemical reaction: "Two molecules of hydrogen gas react with one molecule of oxygen gas to produce two molecules of water vapor." 3. They could write the equation in chemical shorthand. \[2 \ce{H_2} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{H_2O} \left( g \right) \nonumber \] In the symbolic equation, chemical formulas are used instead of chemical names for reactants and products, while symbols are used to indicate the phase of each substance. It should be apparent that the chemical shorthand method is the quickest and clearest method for writing chemical equations. We could write that an aqueous solution of calcium nitrate is added to an aqueous solution of sodium hydroxide to produce solid calcium hydroxide and an aqueous solution of sodium nitrate. Or in shorthand we could write: \[\ce{Ca(NO_3)_2} \left( aq \right) + 2 \ce{NaOH} \left( aq \right) \rightarrow \ce{Ca(OH)_2} \left( s \right) + 2 \ce{NaNO_3} \left( aq \right) \nonumber \] How much easier is that to read? Let's try it in reverse. Look at the following reaction in shorthand and write the word equation for the reaction: \[\ce{Cu} \left( s \right) + \ce{AgNO_3} \left( aq \right) \rightarrow \ce{Cu(NO_3)_2} \left( aq \right) + \ce{Ag} \left( s \right) \nonumber \] The word equation for this reaction might read something like "solid copper reacts with an aqueous solution of silver nitrate to produce a solution of copper (II) nitrate with solid silver." To turn word equations into symbolic equations, we need to follow the given steps: Identify the reactants and products. This will help you know which symbols go on each side of the arrow and where the \(+\) signs go. Write the correct formulas for all compounds. You will need to use the rules you learned in Chapter 5 (including making all ionic compounds charge balanced). Write the correct formulas for all elements. Usually this is given straight off of the periodic table. However, there are seven elements that are considered diatomic, meaning that they are always found in pairs in nature. They include those elements listed in the table. 0 1 2 3 4 5 6 7 Element Name Hydrogen Nitrogen Oxygen Fluorine Chlorine Bromine Iodine Formula \(H_2\) \(N_2\) \(O_2\) \(F_2\) \(Cl_2\) \(Br_2\) \(I_2\) Example \(\PageIndex{1}\) Transfer the following symbolic equations into word equations or word equations into symbolic equations. \(\ce{HCl} \left( aq \right) + \ce{NaOH} \left( aq \right) \rightarrow \ce{NaCl} \left( aq \right) + \ce{H_2O} \left( l \right)\) Gaseous propane, \(\ce{C_3H_8}\), burns in oxygen gas to produce gaseous carbon dioxide and liquid water. Hydrogen fluoride gas reacts with an aqueous solution of potassium carbonate to produce an aqueous solution of potassium fluoride, liquid water, and gaseous carbon dioxide. Solution a. An aqueous solution of hydrochloric acid reacts with an aqueous solution of sodium hydroxide to produce an aqueous solution of sodium chloride and liquid water. b. Reactants: propane (\(\ce{C_3H_8}\)) and oxygen (\(\ce{O_2}\)) Product: carbon dioxide (\(\ce{CO_2}\)) and water (\(\ce{H_2O}\)) \[\ce{C_3H_8} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow \ce{CO_2} \left( g \right) + \ce{H_2O} \left( l \right) \nonumber \] c. Reactants: hydrogen fluoride and potassium carbonate Products: potassium fluoride, water, and carbon dioxide \[\ce{HF} \left( g \right) + \ce{K_2CO_3} \left( aq \right) \rightarrow \ce{KF} \left( aq \right) + \ce{H_2O} \left( l \right) + \ce{CO_2} \left( g \right) \nonumber \] Exercise \(\PageIndex{1}\) Transfer the following symbolic equations into word equations or word equations into symbolic equations. Hydrogen gas reacts with nitrogen gas to produce gaseous ammonia. \(\ce{HCl} \left( aq \right) + \ce{LiOH} \left( aq \right) \rightarrow \ce{LiCl} \left( aq \right) + \ce{H_2O} \left( l \right)\) Copper metal is heated with oxygen gas to produce solid copper(II) oxide. Answer a \(H_2 (g) + N_2 (g) \rightarrow NH_3 (g)\) Answer b An aqueous solution of hydrochloric acid reacts with an aqueous solution of lithium hydroxide to produce an aqueous solution of lithium chloride and liquid water. Answer c \(Cu (s) + O_2 (g) \rightarrow CuO (s)\) Summary A chemical reaction is the process by which one or more substances are changed into one or more new substances. Chemical reactions are represented by chemical equations. Chemical equations have reactants on the left, an arrow that is read as "yields", and the products on the right. |
Courses/BridgeValley_Community_and_Technical_College/Consumer_Chemistry/07%3A_Chemical_Bonds/7.03%3A_Writing_Formulas_for_Ionic_Compounds | Learning Objectives Write the correct formula for an ionic compound. Recognize polyatomic ions in chemical formulas. Ionic compounds do not exist as molecules. In the solid state, ionic compounds are in crystal lattice containing many ions each of the cation and anion. An ionic formula, like \(\ce{NaCl}\), is an empirical formula. This formula merely indicates that sodium chloride is made of an equal number of sodium and chloride ions. Sodium sulfide, another ionic compound, has the formula \(\ce{Na_2S}\). This formula indicates that this compound is made up of twice as many sodium ions as sulfide ions. This section will teach you how to find the correct ratio of ions, so that you can write a correct formula. If you know the name of a binary ionic compound, you can write its chemical formula . Start by writing the metal ion with its charge, followed by the nonmetal ion with its charge. Because the overall compound must be electrically neutral, decide how many of each ion is needed in order for the positive and negative charges to cancel each other out. Example \(\PageIndex{1}\): Aluminum Nitride and Lithium Oxide Write the formulas for aluminum nitride and lithium oxide. Solution Unnamed: 0 Write the formula for aluminum nitride Write the formula for lithium oxide 1. Write the symbol and charge of the cation (metal) first and the anion (nonmetal) second. \(\ce{Al^{3+}} \: \: \: \: \: \ce{N^{3-}}\) \(\ce{Li^+} \: \: \: \: \: \ce{O^{2-}}\) 2. Use a multiplier to make the total charge of the cations and anions equal to each other. total charge of cations = total charge of anions 1(3+) = 1(3-) +3 = -3 total charge of cations = total charge of anions 2(1+) = 1(2-) +2 = -2 3. Use the multipliers as subscript for each ion. \(\ce{Al_1N_1}\) \(\ce{Li_2O_1}\) 4. Write the final formula. Leave out all charges and all subscripts that are 1. \(\ce{AlN}\) \(\ce{Li_2O}\) An alternative way to writing a correct formula for an ionic compound is to use the crisscross method . In this method, the numerical value of each of the ion charges is crossed over to become the subscript of the other ion. Signs of the charges are dropped. Example \(\PageIndex{2}\): The Crisscross Method for Lead (IV) Oxide Write the formula for lead (IV) oxide. Solution Crisscross Method Write the formula for lead (IV) oxide 1. Write the symbol and charge of the cation (metal) first and the anion (nonmetal) second. \(\ce{Pb^{4+}} \: \: \: \: \: \ce{O^{2-}}\) 2. Transpose only the number of the positive charge to become the subscript of the anion and the number only of the negative charge to become the subscript of the cation. NaN 3. Reduce to the lowest ratio. \(\ce{Pb_2O_4}\) 4. Write the final formula. Leave out all subscripts that are 1. \(\ce{PbO_2}\) Exercise \(\PageIndex{2}\) Write the chemical formula for an ionic compound composed of each pair of ions. the calcium ion and the oxygen ion the 2+ copper ion and the sulfur ion the 1+ copper ion and the sulfur ion Answer a: CaO Answer b: CuS Answer c: Cu 2 S Be aware that ionic compounds are empirical formulas and so must be written as the lowest ratio of the ions. Example \(\PageIndex{3}\): Sulfur Compound Write the formula for sodium combined with sulfur. Solution Crisscross Method Write the formula for sodium combined with sulfur 1. Write the symbol and charge of the cation (metal) first and the anion (nonmetal) second. \(\ce{Na^{+}} \: \: \: \: \: \ce{S^{2-}}\) 2. Transpose only the number of the positive charge to become the subscript of the anion and the number only of the negative charge to become the subscript of the cation. NaN 3. Reduce to the lowest ratio. This step is not necessary. 4. Write the final formula. Leave out all subscripts that are 1. \(\ce{Na_2S}\) Exercise \(\PageIndex{3}\) Write the formula for each ionic compound. sodium bromide lithium chloride magnesium oxide Answer a: NaBr Answer b: LiCl Answer c: MgO Polyatomic Ions Some ions consist of groups of atoms bonded together and have an overall electric charge. Because these ions contain more than one atom, they are called polyatomic ions. Polyatomic ions have characteristic formulas, names, and charges that should be memorized. For example, NO 3 − is the nitrate ion; it has one nitrogen atom and three oxygen atoms and an overall 1− charge. Table \(\PageIndex{1}\) lists the most common polyatomic ions. Name Formula ammonium ion NH4+ acetate ion C2H3O2− (also written CH3CO2−) carbonate ion CO32− chromate ion CrO42− dichromate ion Cr2O72− hydrogen carbonate ion (bicarbonate ion) HCO3− cyanide ion CN− hydroxide ion OH− nitrate ion NO3− nitrite ion NO2− permanganate ion MnO4− phosphate ion PO43− hydrogen phosphate ion HPO42− dihydrogen phosphate ion H2PO4− sulfate ion SO42− hydrogen sulfate ion (bisulfate ion) HSO4− sulfite ion SO32− The rule for constructing formulas for ionic compounds containing polyatomic ions is the same as for formulas containing monatomic (single-atom) ions: the positive and negative charges must balance. If more than one of a particular polyatomic ion is needed to balance the charge, the entire formula for the polyatomic ion must be enclosed in parentheses, and the numerical subscript is placed outside the parentheses. This is to show that the subscript applies to the entire polyatomic ion. An example is Ba(NO 3 ) 2 . Writing Formulas for Ionic Compounds Containing Polyatomic Ions Writing a formula for ionic compounds containing polyatomic ions also involves the same steps as for a binary ionic compound. Write the symbol and charge of the cation followed by the symbol and charge of the anion. Example \(\PageIndex{4}\): Calcium Nitrate Write the formula for calcium nitrate. Solution Crisscross Method Write the formula for calcium nitrate 1. Write the symbol and charge of the cation (metal) first and the anion (nonmetal) second. \(\ce{Ca^{2+}} \: \: \: \: \: \ce{NO_3^-}\) 2. Transpose only the number of the positive charge to become the subscript of the anion and the number only of the negative charge to become the subscript of the cation. The 2+ charge on Ca becomes the subscript of NO3 and the 1- charge on NO3 becomes the subscript of Ca. 3. Reduce to the lowest ratio. \(\ce{Ca_1(NO_3)_2}\) 4. Write the final formula. Leave out all subscripts that are 1. If there is only 1 of the polyatomic ion, leave off parentheses. \(\ce{Ca(NO_3)_2}\) Example \(\PageIndex{5}\) Write the chemical formula for an ionic compound composed of the potassium ion and the sulfate ion. Solution Explanation Answer Potassium ions have a charge of 1+, while sulfate ions have a charge of 2−. We will need two potassium ions to balance the charge on the sulfate ion, so the proper chemical formula is \(\ce{K2SO4}\). \(\ce{K_2SO_4}\) Exercise \(\PageIndex{5}\) Write the chemical formula for an ionic compound composed of each pair of ions. the magnesium ion and the carbonate ion the aluminum ion and the acetate ion Answer a: \(\ce{MgCO3}\) Answer b: \(\ce{Al(CH3COO)3}\) Recognizing Ionic Compounds There are two ways to recognize ionic compounds. Method 1 Compounds between metal and nonmetal elements are usually ionic. For example, \(\ce{CaBr2}\) contains a metallic element (calcium, a group 2 [or 2A] metal) and a nonmetallic element (bromine, a group 17 [or 7A] nonmetal). Therefore, it is most likely an ionic compound (in fact, it is ionic). In contrast, the compound \(\ce{NO2}\) contains two elements that are both nonmetals (nitrogen, from group 15 [or 5A] , and oxygen, from group 16 [or 6A] . It is not an ionic compound; it belongs to the category of covalent compounds discussed elsewhere. Also note that this combination of nitrogen and oxygen has no electric charge specified, so it is not the nitrite ion. Method 2 Second, if you recognize the formula of a polyatomic ion in a compound, the compound is ionic. For example, if you see the formula \(\ce{Ba(NO3)2}\), you may recognize the “\(\ce{NO3}\)” part as the nitrate ion, \(\ce{NO3^{-}}\). (Remember that the convention for writing formulas for ionic compounds is not to include the ionic charge.) This is a clue that the other part of the formula, \(\ce{Ba}\), is actually the \(\ce{Ba^{2+}}\) ion, with the 2+ charge balancing the overall 2− charge from the two nitrate ions. Thus, this compound is also ionic. Example \(\PageIndex{6}\) Identify each compound as ionic or not ionic. \(\ce{Na2O}\) \(\ce{PCl3}\) \(\ce{NH4Cl}\) \(\ce{OF2}\) Solution Explanation Answer a. Sodium is a metal, and oxygen is a nonmetal. Therefore, \(\ce{Na2O}\) is expected to be ionic via method 1. \(\ce{Na2O}\), ionic b. Both phosphorus and chlorine are nonmetals. Therefore, \(\ce{PCl3}\) is not ionic via method 1 \(\ce{PCl3}\), not ionic c. The \(\ce{NH4}\) in the formula represents the ammonium ion, \(\ce{NH4^{+}}\), which indicates that this compound is ionic via method 2 \(\ce{NH4Cl}\), ionic d. Both oxygen and fluorine are nonmetals. Therefore, \(\ce{OF2}\) is not ionic via method 1 \(\ce{OF2}\), not ionic Exercise \(\PageIndex{6}\) Identify each compound as ionic or not ionic. \(\ce{N2O}\) \(\ce{FeCl3}\) \(\ce{(NH4)3PO4}\) \(\ce{SOCl2}\) Answer a: not ionic Answer b: ionic Answer c: ionic Answer d: not ionic Summary Formulas for ionic compounds contain the symbols and number of each atom present in a compound in the lowest whole number ratio. |
Courses/University_of_California_Davis/UCD_Chem_110A%3A_Physical_Chemistry__I/UCD_Chem_110A%3A_Physical_Chemistry_I_(Larsen)/Text/08%3A_Multielectron_Atoms/8.11%3A_Hund's_Rules_Determine_the_Term_Symbols_of_the_Ground_Electronic_States | Define Hund three rules Use Hunds three rules to predict the lowest energy configuration and term symbols for multi-electron systems The Aufbau section discussed how that electrons fill the lowest energy orbitals first, and then move up to higher energy orbitals only after the lower energy orbitals are full. However, there a problem with this rule. Certainly, 1s orbitals should be filled before 2s orbitals, because the 1s orbitals have a lower value of n, and thus a lower energy. What about the three different 2p orbitals? In what order should they be filled? The answer to this question involves Hund's rule, which make a lot more sense in the context of generated term symbols that are used to combine the various \(L\) and \(S\) values represent vector additions of possible microstates. Hund’s Rules State with the largest value of \(S\) is most stable and stability decreases with decreasing \(S\). For states with same values of \(S\), the state with the largest value of \(L\) is the most stable. If states have same values of \(L\) and \(S\) then, for a subshell that is less than half filled, state with smallest \(J\) is most stable; for subshells that are more than half filled, state with largest value of \(J\) is most stable. Rank these terms associated with an electronic configuration of an atom based on energy (via Hund's rules): \(^3D\), \(^3P\), \(^3S\), \(^1D\), \(^1P\), \(^1S\) Hund's First Rule (Maximize Spin Multiplicity) According to the first rule, electrons always enter an empty orbital before they pair up. Electrons are negatively charged and, as a result, they repel each other. Electrons tend to minimize repulsion by occupying their own orbitals, rather than sharing an orbital with another electron. Furthermore, quantum-mechanical calculations have shown that the electrons in singly occupied orbitals are less effectively screened or shielded from the nucleus. There's a Coulomb repulsion between two electrons to put them in the same orbital (a spin pairing energy often discussed in Crystal Field Theory). However, there's also a quantum mechanical effect. The exchange energy (which is favorable) increases with the number of possible exchanges between electrons with the same spin and energy. In transitioning from the top state to the middle state of Figure 8.10.1 , we remove the Coulomb repulsion between electrons in the same orbital. Moreover, In transitioning from the middle state to the bottom state (most stable state predicted by Hund's first rule), we gain the exchange energy , because these two electrons are indistinguishable. Hund's Second Rule (Maximize Orbital Angular Multiplicity) What matters is the total (scalar) angular momentum, not the direction. The negative and positive signs refer only to the direction of the angular momentum, not the magnitude. The direction is furthermore arbitrary (except in, say, a magnetic or electric field). So is the spin direction, incidentally. By convention we usually draw the first electron in each orbital as "up" (positive spin). However we could just as easily draw it "down". It makes no difference - in the absence of an external EM field, the energy is the same, if only because molecules/atoms are rotating with respect to the lab frame anyway. "Up" and "down", in other words, is artificial. What matters is the relative momentum vectors of the various electrons in the system, and hence their sum total. Hund's Third Rule (Minimize less than half filled or maximize greater than half filled shells) A long time ago someone offering a reasonably simple explanation related to the fact that when the shell is more than half full, it's easier to visualize the system as an interaction between the spin and orbital momenta of holes rather than electrons, in which case the energetic stabilization term is reversed in sign. This would be because the spin angular momentum of a single hole would be opposite in sign compared to the spin angular momentum of a single electron. Taking as an example - the three p-orbitals. A situation with 1 electron and 5 electrons are functionally similar, except that one has a single electron and one has a single hole. All things being equal, the total spin angular momentum of the 1 electron system would be opposite in sign to whatever the total spin angular momentum of the 5 electron system is. So the expectations for Hund's rules would be switched. You can kind of see this if you draw out all the microstates of the 1-electron and 5-electron configurations: the everything is pretty much changed in sign in the latter case. What terms and levels can arise from an atom with the ground-state configuration of \(1s^22s^22p^6 3s^2 3p^6 4s^2 4p^1 3d^1\)? Which is the most stable (lowest in energy) state? Solution Possible states include: \(^1F_3\), \(^1D_2\), \(^1P_1\), \(^3F_4\), \(^3F_3\), \(^3F_2\), \(^3D_3\), \(^3D_2\), \(^3D_1\), \(^3P_2\), \(^3P_1\), \(^3P_0\). There are two unpaired electrons in this system from the electron configuration. Rule 1 predicts that the ground state will be a triplet with \(S=1\) so \(2S+1=3\). So the ground state is from this more narrowed list: \(^3F_4\), \(^3F_3\), \(^3F_2\), \(^3D_3\), \(^3D_2\), \(^3D_1\), \(^3P_2\), \(^3P_1\), \(^3P_0\). Rule 2 predicts a \(F\) state since that is the highest multiplicity with \(L= 3\): So the ground state is from this more narrowed list: \(^3F_4\), \(^3F_3\), \(^3F_2\) Rule 3 predicts the lowest \(J\) term since the d shell is less than half full. That is the \(J=2\) state. Therefore for this system, the atom will have a ground-state structure of \(^3F_2\) The ground configuration of a \(\ce{Ti^{2+}}\) ion is \([Ar]3d^2\). What is the term of lowest energy state? Solution Rule 1: two unpaired electrons ⇒ highest S = 1 ⇒ 2S + 1 = 3 Rule 2: two in d in parallel spin ⇒ highest L = 1 + 2 = 3 ⇒ 3F Rule 3:.L = 3, S = 1 ⇒ J = 4, 3, 2; less than half-filled ⇒ \(^3F_2\) What is the term of lowest energy state for the following atoms and ions. \(\ce{C}\): \([He]2s^2 2p^2\) \(\ce{N}\): \([He]2s^2 2p^3\) \(\ce{O}\): \([He]2s^2 2p^4\) \(\ce{Cr^{3+}}\): \([Ar]3d^3\) \(\ce{Mn^{3+}}\): \([Ar]3d^4\) \(\ce{Fe^{3+}}\): \([Ar]3d^5\) Answer \(^3P_0\), \(^4S_{3/2}\), \(^3P_2\), \(^4F_{3/2}\), \(^5D_0\), and \(^6S_{5/2}\), respectively. |
Courses/Intercollegiate_Courses/Internet_of_Science_Things/2%3A_Python_Modules/2.13%3A_Charts_and_Graphs_with_Python | Hypothes.is tag: s20iostpy13ualr Download Assignment: S20py13 Learning Objectives Students will be able to: Content Generate bar graphs Generate pie charts Generate line graphs Save your data as images Process: Import matplotlib on your Raspberry Pi Write code that that converts data in lists to charts and graphs Prior Knowledge Python concepts from previous activities Further Reading https://matplotlib.org/1.3.0/index.html Model 1: matplotlib matplotlib is a python 2D plotting library which produces publication quality figures in a variety of hardcopy formats and interactive environments across platforms. matplotlib can be used in python scripts, the python and ipython shell web application servers, and six graphical user interface toolkits. matplotlibtries to make easy things easy and hard things possible. You can generate plots, histograms, power spectra, bar charts, errorcharts, scatterplots, etc, with just a few lines of code. For more information go to http://matplotlib.org/1.3.0/index.html Task 1: For our windows PCs, we installed Python using Anaconda. It includes the matplotlib library in the install. For safe measure, update your anaconda by opening the anaconda prompt and typing conda update anaconda. Go back to Internet of Things Activity 1 (Preparing your PC for Class for a refresher if you need). Task 2: For our Raspberry Pi computers, we need install the library using apt-get. Open up a terminal window on your Raspberry Pi and complete the following commands. pi@yourhostname : ~ $ sudo apt-get update pi@yourhostname : ~ $ sudo apt-get upgrade pi@yourhostname : ~ $ sudo apt-get install libav-tools pi@yourhostname : ~ $ sudo apt-get -y install Python3-matplotlib To test if matplotlib installed correctly, open up a terminal window and type python3. When you get the python prompt (shown as >>>), type in import matplotlib. If it installed correctly, you will be given another prompt (>>>) if not, you will be given an error. Successful test for Python 3 Failed test for Python 3 pi@yourPiname -$ python3 Python 3.4.2(default, Oct 19 2014, 13:13:11) [GCC 4.9.1] on linux Type “help”, “copyright”, “credits” of “license” for more information. >>> import matplotlib >>> Note there may be a short pause before getting the next prompt. After confirming this is working, type and you will get your prompt back >>> quit() pi@yourPiname -$ pi@yourPiname -$ python3 Python 3.4.2(default, Oct 19 2014, 13:13:11) [GCC 4.9.1] on linux Type “help”, “copyright”, “credits” of “license” for more information. >>> import matplotlib Traceback (most recent call last): File “ ”, line 1, in ImportError: No module named matplotlib >>> At the prompt you can type quit() and ask your instructor for help. >>> quit() Note You can do this activity on either your PC or your Raspberry Pi. Model 1: Bar Graphs Python Program 13.1 Output import matplotlib.pyplot as plt myGraphValues = [1.0446, 0.8765, 0.7739, 0.684, 0.867] myGraphWidths = [.5, .5, .5, .5, .5] myGraphColors = ['r', '#FF6600', 'b', 'DarkViolet'] myGraphLabels = ['acetone', 'benzene', 'cyclohexane', 'heptane','toluene'] x = range(0,len(myGraphLabels)) plt.bar(x, myGraphValues, align = 'center', width=myGraphWidths,\ color=myGraphColors) plt.title('Common Organic Solvent Density') plt.ylabel('g/mL') plt.xlabel('Solvent') plt.xticks(x, myGraphLabels) plt.show() NaN Note In the first line of code, matplotlib.pyplot is imported as plt. This statement indicates that we are using the pyplot package of matplotlib, and using a shortened alias for that library. Subsequently we are using functions within the pyplot package. Critical Thinking Questions: What necessary data for the bar graph is being stored as lists? How does matplotlib know how many bars to draw on the graph? What is the function for indicating you want to generate a bar graph. What happens if you comment out the plt xticks line? 5. Graph colors can be assigned as Color symbols (‘b’, ‘g’, ‘r’, ‘c’, ‘m’, ‘y’, ‘k’, ‘w’) http://matplotlib.org/api/colors_api.html hexadecimal values for colors (e.g., ‘#FF6600’) http://www.w3schools.com/tags/ref_colorpicker.asp named colors (e.g. DarkViolet) http://matplotlib.org/examples/color/named_colors.html Using named colors, redraw the graph so that all the items are different shades of green. 6. The default value for graph width is 0.8. Change the values for myGraphWidths = [.8, .8, .8, .8. .8]. How does this alter the graph? What happens if you set myGraphWidths = [.2, .8, .5, 1, 1.2]? Each of the lists has exactly 5 elements. What happens if you add an element to one list, but not the others? In this activity, you will create a pie chart. Type in this code and run it. It will take a couple of seconds for Python to make the plot and display it, so be patient. Python Program 13.2 import matplotlib.pyplot as plt myGraphValues = [20.84, 27.54, 7.73, 36.28, 7.61] myGraphColors = ['r', '#FF6600', 'y', 'g', 'b'] myGraphLabels = ['$^{70}$Ge', '$^{72}$Ge', '$^{73}$Ge', '$^{74}$Ge', '$^{76}$Ge'] explode = (0, 0, 0, .1, 0) plt.pie(myGraphValues, colors= myGraphColors, labels = myGraphLabels, \ explode=explode, autopct='%1.1f%%', counterclock=False, \ shadow = True) plt.title('Natural Abundance of Germanium Isotopes') plt.show() A new variable called a tuple has been included named explode. Alter this line of code to determine what it does to the graph. Explain your results. myGraphLabels requires the use of superscripts in identifying the isotopes of germanium. What is the syntax for making a superscript The syntax for making a subscript is similar to a superscript except it uses a _ symbol. Change the title so that the word “Abundance” is all subscripted. Adjust each of the following arguments in the plt.pi() function. Briefly indicate what each argument affects: autopct counterclock shadow Note In creation of the pie chart, python will calculate the percentages for you. You do not need to have all the values equal to 100 if you have raw data In this activity, you will create a line graph. Type in this code and run it. It will take a couple of seconds for Python to make the plot and display it, so be patient. Python Program 13.3 import matplotlib.pyplot as plt myNO2Values = [5, 40, 16, 25, 9, 10] plt.plot(range(1,7),myNO2Values) plt.title('Average Concentration over 6 days') plt.ylabel('Concentration(ppbv)') plt.xlabel('Day of experiment') plt.show() What is the function for creating a line graph? Change line 4 in your code to read: plt.plot(range(len(myGraphValues)), myGraphValues) What is the advantage of this line of code over the previous example? What is the disadvantage of this line of code over the previous example? Add the following lines before the plt.show() function ax = plt.axes() ax.set_xlim([1,6]) ax.set_ylim([-5,50]) What effect did the second line have? What effect did the third line have After the lines you just added but before the line plt.show() add the following lines. ax.set_xticks([1,2,3,4,5,6]) ax.set_yticks([10,20,30,40,50]) What effect did the first line have? What effect did the second line have? What happens if you add the line ax.grid() Change your code to the code below. This will make our graph prettier for the moment by removing the wacky scales now that you know how to work with them and also add a second data line to the graph! Python Program 13.4 import matplotlib.pyplot as plt myNO2Values = [5, 40, 16, 25, 9, 10] myO2GraphValues = [10, 6, 22, 9, 15, 4] plt.plot(range(1,7), myNO2Values) plt.plot(range(1,7), myO2GraphValues) plt.title('Average Concentration over 6 days') plt.ylabel('Concentration(ppbv)') plt.xlabel('Day of experiment') ax = plt.axes() ax.set_xlim([1,6]) ax.set_ylim([-5,50]) ax.set_xticks([1,2,3,4,5,6]) ax.set_yticks([10,20,30,40,50]) ax.grid() plt.show() How can you add a third data line to the graph? And a fourth and so on? Change this line plt.plot(range(1,7), myNO2Values) to plt.plot(range(1,7), myNO2Values, '-oy') How does that change the appearance of the graph? Note The last argument in the plt.plot line can be altered three different ways by changing the value of the '-or'shown above. They go in the order marker, line, color. You can omit any of them. You have tons of options to replace the o with for the marker. Each one gives you a different type of marker. There are too many to list, but my favorites are Small letter o: a circle Small letter s: a square An asterisk: a star Capital letter D: a diamond Small letter d: a thin diamond You have five options for the minus sign. Each one gives you a different type of line: Nothing: no line A minus sign: solid line Two minus signs): dashed line Minus sign period: dash-dot line Colon: dotted lin You can change the line color by adding a letter corresponding to a Python color. Try it! You can add a legend to your graph by adding the command: plt.legend(['NO$_2$','O$_2$'], loc=2) Where does it place the legend? The loc argument can have values from 0 to 10. Try some of these values. Note: Saving your graphs for use elsewhere It's possible (and even likely) that you might use Python to analyze data in a course and want to produce some nice graphs and turn them in. You can save the graphs you generate with matplotlib and then paste them into a Microsoft Word (or other) document. To save the graphs, you can click the disk icon circled in red below on the raspberry pi, or green on your windowsPC. If you resize the window before clicking the save button, the resized image will be saved. This is really useful if the window that came up wasn’t quite big enough to show your entire graph. Alternatively, you can use a line of code like this. plt.savfig('MySampleImage.png', format='png') Supported file formats are png, ps, pdf, eps and svg. There is much more that you can do then we have to cover and this link provides more documentation, https://matplotlib.org/api/pyplot_api.html . |
Courses/Saint_Francis_University/CHEM_113%3A_Human_Chemistry_I_(Muino)/12%3A_Introduction_to_Organic_Chemistry_-_Alkanes/12.06%3A_Naming_Alkanes | Learning Objectives To name alkanes by the IUPAC system and write formulas for alkanes given IUPAC names As noted in previously, the number of isomers increases rapidly as the number of carbon atoms increases: there are 3 pentanes, 5 hexanes, 9 heptanes, and 18 octanes, etc. It would be difficult to assign each compound unique individual names that we could remember easily. A systematic way of naming hydrocarbons and other organic compounds has been devised by the International Union of Pure and Applied Chemistry (IUPAC). These rules, used worldwide, are known as the IUPAC System of Nomenclature. (Some of the names mentioned earlier, such as isobutane, isopentane, and neopentane, do not follow these rules and are called common names .) In the IUPAC system, a compound is named according to the number of carbons in the longest continuous chain (LCC) or parent chain and the family it belongs to. Atoms or groups attached to this carbon chain, called substituents , are then named, with their positions indicated by a numerical prefix at the beginning of the name: Prefix (substituent) – Parent (# carbons) – Suffix (family name) 2-methyl prop ane (Table \(\PageIndex{1}\)) below lists the IUPAC parent names that are used for charbon chains containing 1 to 10 carbons, along with straight-chain alkane examples for each. Notice that the suffix for each example in this table is -ane , which indicates these are members of the alk ane family. Number of Carbons Parent Chain (LCC) Name Example Alkane Name Example Condensed Structural Formula 1 meth- methane CH4 2 eth- ethane CH3CH3 3 prop- propane CH3CH2CH3 4 but- butane CH3CH2CH2CH3 5 pent- pentane CH3CH2CH2CH2CH3 6 hex- hexane CH3CH2CH2CH2CH2CH3 7 hept- heptane CH3CH2CH2CH2CH2CH2CH3 8 oct- octane CH3CH2CH2CH2CH2CH2CH2CH3 9 non- nonane CH3CH2CH2CH2CH2CH2CH2CH2CH3 10 dec- decane CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3 Atoms or groups of atoms that branch off the parent chain are called substituents . When the substituent is a carbon or group of carbons, such as –CH 3 or –CH 2 CH 3 , it is called an alkyl group . Alkyl groups are alkanes that have had one hydrogen removed to allow for binding to a main chain carbon and are named by replacing the -ane suffix of the parent hydrocarbon with -yl . For example, the –CH 3 group derived from methane (CH 4 ) results from subtracting one hydrogen atom and is called a methyl group . Removing a hydrogen from ethane, CH 3 CH 3 , gives –CH 2 CH 3 , or the ethyl group . The alkyl groups we will use most frequently are listed in Table \(\PageIndex{2}\). Alkyl groups are not independent molecules; they are parts of molecules that we consider as a unit to name compounds systematically. Parent Alkane Parent Alkane.1 Alkyl Group Alkyl Group.1 Condensed Structural Formula methane NaN methyl NaN CH3– ethane NaN ethyl NaN CH3CH2– propane NaN propyl NaN CH3CH2CH2– NaN NaN isopropyl NaN (CH3)2CH– butane NaN butyl NaN CH3CH2CH2CH2– NaN NaN sec-butyl NaN NaN NaN NaN isobutyl NaN NaN NaN NaN tert-butyl (tBu) NaN NaN Simplified IUPAC rules for naming alkanes are as follows (demonstrated in Example \(\PageIndex{1}\)). Step 1: Name the parent chain. Find the longest continuous chain, (it may not always be the most obvious chain written in one line), and name according to the number of carbon atoms it contains. Add the suffix - ane to indicate that the molecule is an alkane. Use Table \(\PageIndex{1}\) as a reference to start, but it is a good idea to commit these to memory. Step 2: Number the carbon atoms in the parent chain, giving carbons with any substituents attached the lowest number possible. These numbers are used to locate where substituents are attached to a main chain. Step 3: Name any substituents (including the location number). If the same alkyl group appears more than once, the numbers of all the carbon atoms to which it is attached are expressed. If the same group appears more than once on the same carbon atom, the number of that carbon atom is repeated as many times as the group appears. Moreover, the number of identical groups is indicated by the Greek prefixes di -, tri -, tetra -, and so on. These prefixes are not considered in determining the alphabetical order of the substituents. For example, ethyl is listed before dimethyl; the di- is simply ignored. The last alkyl group named is prefixed to the name of the parent alkane to form one word. Step 4: Write the name of the compound as a single word placing the substituent groups first (in alphabetical order), then the parent name, then the family name. Hyphens are used to separate numbers from the names of substituents; commas separate numbers from each other. When these rules are followed, every unique compound receives its own exclusive name. The rules enable us to not only name a compound from a given structure but also draw a structure from a given name. The best way to learn how to use the IUPAC system is to practice it, not just memorize the rules. It’s easier than it looks. Example \(\PageIndex{1}\) Name each compound. Solution Step 1: The LCC has five carbon atoms, and so the parent compound name is pentane. Step 2: Number the carbons in the LCC from left to right. Step 3: There is a methyl group attached to carbon #2 of the pentane chain. Step 4: The name is 2-methylpentane. Step 1: The LCC has six carbon atoms, so the parent compound is hexane. Step 2: Number the carbons in the LCC from left to right (or right to left, either way will be identical numbering). Step 3: There are two methyl groups attached to the second and fifth carbon atoms. Step 4: The name is 2,5-dimethylhexane. Step 1: The LCC has eight carbon atoms, so the parent compound is octane. Step 2: Number the carbons in the LCC from left to right to give the lower number. Step 3: There are methyl and ethyl groups, both attached to the fourth carbon atom. Step 4: The correct name is thus 4-ethyl-4-methyloctane. Exercise \(\PageIndex{1}\) Name each compound. Example \(\PageIndex{2}\) Draw the structure for each compound. 2,3-dimethylbutane 4-ethyl-2-methylheptane Solution In drawing structures, always start with the parent chain. The parent chain is butane, indicating four carbon atoms in the LCC. Then add the substituents at their proper positions. You can number the parent chain from either direction as long as you are consistent; just don’t change directions before the structure is done. The name indicates two methyl (–CH 3 ) groups, one on the second carbon atom and one on the third. Finally, fill in all the hydrogen atoms, keeping in mind that each carbon atom must have four bonds total. Adding the substituents at their proper positions gives Filling in all the hydrogen atoms gives the following condensed structural formulas (both are correct): Note that the bonds (dashes) can be shown or not; sometimes they are needed for spacing. Exercise \(\PageIndex{2}\) Draw the structure for each compound. 4-ethyloctane 3-ethyl-2-methylpentane 3,3,5-trimethylheptane |
Courses/University_of_North_Carolina_Charlotte/CHEM_2141%3A__Survey_of_Physical_Chemistry/04%3A_Entropy_and_The_Second_and_3rd_Law_of_Thermodynamics/4.06%3A_Entropy_Increases_With_Increasing_Temperature | Define entropy and its relation to energy flow. Entropy versus temperature We can put together the first and the second law for a reversible process with no other work than volume (\(PV\)) work and obtain: \[dU= δq_{rev} + δw_{rev} \nonumber \] Entropy is the dispersal of energy and is related to heat: \[δq_{rev}= TdS \nonumber \] Work is related to the change in volume: \[δw_{rev}= -PdV \nonumber \] Plugging these into our expression for \(dU\) for reversible changes: \[dU= TdS -PdV \nonumber \] We no longer have any path functions in the expression, as \(U\), \(S\) and \(V\) are all state functions . This means this expression must be an exact differential. We can generalize the expression to hold for irreversible processes, but then the expression becomes an inequality: \[dU≤ TdS - PdV \nonumber \] This equality expresses \(U\) as a function of two variables, entropy and volume: \(U(S,V)\). \(S\) and \(V\) are the natural variables of \(U\). Entropy and heat capacity At constant volume, \(dU\) becomes: \[dU=TdS \nonumber \] Recall that internal energy is related to constant volume heat capacity, \(C_V\): \[C_V=\left(\frac{dU}{dT}\right)_V \nonumber \] Combining these two expressions, we obtain: \[dS=\frac{C_V}{T}dT \nonumber \] Integrating: \[\Delta S=\int_{T_1}^{T_2}{\frac{C_V(T)}{T}dT} \nonumber \] If we know how \(C_V\) changes with temperature, we can calculate the change in entropy, \(\Delta S\). Since heat capacity is always a positive value, entropy must increase as the temperature increases. There is nothing to stop us from expressing \(U\) in other variables, e.g. \(T\) and \(V\). In fact, we can derive some interesting relationships if we do. Write \(U\) as a function of \(T\) and \(V\). Write \(U\) as a function of its natural variables. Rearrange (2) to find an expression for \(dS\). Substitute (1) into (3) and rearrange. This is the definition of \(C_V\). Write out \(S\) as a function of \(T\) and \(V\). We can also derive an expression for the change in entropy as a function of constant pressure heat capacity, \(C_P\). To start, we need to change from internal energy, \(U\), to enthalpy, \(H\): \[\begin{align*} H &= U + PV \\[4pt] dH &= dU +d(PV) \\[4pt] &= dU + PdV + VdP \end{align*} \] For reversible processes: \[\begin{align*} dH &= dU + PdV + VdP \\[4pt] &= TdS -PdV + PdV + VP \\[4pt] &= TdS + VdP\end{align*} \] The natural variables of the enthalpy are \(S\) and \(P\) (not: \(V\)). A similar derivation as above shows that the temperature change of entropy is related to the constant pressure heat capacity: \[dH=TdS+VdP \nonumber \] At constant pressure: \[dH=TdS+VdP \nonumber \] Recall that: \[C_P=\frac{dH}{dT} \nonumber \] Combining, we obtain: \[dS=\frac{C_P}{T}dT \nonumber \] Integrating: \[\Delta S=\int_{T_1}^{T_2}{\frac{C_P(T)}{T}dT} \nonumber \] This means that if we know the heat capacities as a function of temperature we can calculate how the entropy changes with temperature. Usually it is easier to obtain data under constant \(P\) conditions than for constant \(V\), so that the route with \(C_p\) is the more common one. |
Courses/University_of_Arkansas_Little_Rock/IOST_Library/Spring_2023%3A_IoST/11%3A__APIs_(week_8)/01%3A_Creating_a_Google_Service_Account | In order to programmatically upload data to a Google Cloud service your program needs to "log in", but your program is not a human being, and so you need to create a service account that allows you to run an authentication process. As of March 2023 Google uses the OAuth 2.0 authentication framework 1. Log into Google using the account you set up for this class and go to the cloud console ( console.cloud.google.com/ ) 2. Click on the Project dropdown box and choose new project 4. Give your project a name and create it. 5. Now select your project (mine is PCompBobData Go to Navigation menu, choose APIs & Services/Credentials 6. Now choose CREATE CREDENTIALS and choose Service Account ect 7. Give your account a name and create it. 8. Set yourself up as the owner 9. Complete the task |
Courses/CSU_Chico/CSU_Chico%3A_CHEM_451_-_Biochemistry_I/CHEM_451_Test/05%3A_Carbohydrates/5.1%3A_Monosaccharaides_and_Disaccharides | Learning Objectives define a sugar apply knowledge about reactions at carbonyl carbons of aldehyde and ketones in the formation of hemiacetals/hemiketals and acetals/ketals from organic chemistry to draw mechanics showing cyclization of aldoses and ketoses and (hemiacetal/hemiketal formation) and of polymerization of aldoses and ketones to form dissacharide and polysaccharide (acetal/ketals). draw open chain, Haworth (planar), and puckered (nonplanar) representations of 6 member cyclic sugars know the names of common trioses, ribose (5C) and hexoses and draw open chain, Haworth, and puckered representations of each differentiate between enantiomers, diastereomers and anomers of the hexoses explain the difference between reducing and nonreducing dissacharides draw the structure and name common sugar derivatives of the hexoses The link below is an extraordinary and free resource on glycobiology. It defines the word "glycan" as a "generic term for any sugar or assembly of sugars, in free form or attached to another molecule" and "is used interchangeably ... with saccharide or carbohydrate ." Essentials of Glycobiology, 2nd edition, is available online. Template:HideTOC |
Courses/City_College_of_San_Francisco/CCSF_Chemistry_Resources/01%3A_CHE_101_-_Introduction_to_General_Chemistry/1.02%3A_Chemical_Bonding/1.2.06%3A_Chemical_Formulas | Learning Objectives Write the correct formula for a molecule. A molecular formula is a representation of a molecule that uses chemical symbols to indicate the types of atoms followed by subscripts to show the number of atoms of each type in the molecule. (A subscript is used only when more than one atom of a given type is present.) Molecular formulas are also used as abbreviations for the names of compounds. The structural formula for a compound gives the same information as its molecular formula (the types and numbers of atoms in the molecule) but also shows how the atoms are connected in the molecule. The structural formula for methane contains symbols for one C atom and four H atoms, indicating the number of atoms in the molecule (Figure \(\PageIndex{1}\)). The lines represent bonds that hold the atoms together. (A chemical bond is an attraction between atoms or ions that holds them together in a molecule or a crystal.) We will discuss chemical bonds and see how to predict the arrangement of atoms in a molecule later. For now, simply know that the lines are an indication of how the atoms are connected in a molecule. A ball-and-stick model shows the geometric arrangement of the atoms with atomic sizes not to scale, and a space-filling model shows the relative sizes of the atoms. Although many elements consist of discrete, individual atoms, some exist as molecules made up of two or more atoms of the element chemically bonded together. For example, most samples of the elements hydrogen, oxygen, and nitrogen are composed of molecules that contain two atoms each (called diatomic molecules) and thus have the molecular formulas H 2 , O 2 , and N 2 , respectively. Other elements commonly found as diatomic molecules are fluorine (F 2 ), chlorine (Cl 2 ), bromine (Br 2 ), and iodine (I 2 ). The most common form of the element sulfur is composed of molecules that consist of eight atoms of sulfur; its molecular formula is S 8 (Figure \(\PageIndex{2}\)). It is important to note that a subscript following a symbol and a number in front of a symbol do not represent the same thing; for example, H 2 and 2H represent distinctly different species. H 2 is a molecular formula; it represents a diatomic molecule of hydrogen, consisting of two atoms of the element that are chemically bonded together. The expression 2H, on the other hand, indicates two separate hydrogen atoms that are not combined as a unit. The expression 2H 2 represents two molecules of diatomic hydrogen (Figure \(\PageIndex{3}\)). Compounds are formed when two or more elements chemically combine, resulting in the formation of bonds. For example, hydrogen and oxygen can react to form water, and sodium and chlorine can react to form table salt. We sometimes describe the composition of these compounds with an empirical formula , which indicates the types of atoms present and the simplest whole-number ratio of the number of atoms (or ions) in the compound . For example, titanium dioxide (used as pigment in white paint and in the thick, white, blocking type of sunscreen) has an empirical formula of TiO 2 . This identifies the elements titanium (Ti) and oxygen (O) as the constituents of titanium dioxide, and indicates the presence of twice as many atoms of the element oxygen as atoms of the element titanium (Figure \(\PageIndex{4}\)). As discussed previously, we can describe a compound with a molecular formula, in which the subscripts indicate the actual numbers of atoms of each element in a molecule of the compound. In many cases, the molecular formula of a substance is derived from experimental determination of both its empirical formula and its molecular mass (the sum of atomic masses for all atoms composing the molecule). For example, it can be determined experimentally that benzene contains two elements, carbon (C) and hydrogen (H), and that for every carbon atom in benzene, there is one hydrogen atom. Thus, the empirical formula is CH. An experimental determination of the molecular mass reveals that a molecule of benzene contains six carbon atoms and six hydrogen atoms, so the molecular formula for benzene is C 6 H 6 (Figure \(\PageIndex{5}\)). If we know a compound’s formula, we can easily determine the empirical formula. (This is somewhat of an academic exercise; the reverse chronology is generally followed in actual practice.) For example, the molecular formula for acetic acid, the component that gives vinegar its sharp taste, is C 2 H 4 O 2 . This formula indicates that a molecule of acetic acid (Figure \(\PageIndex{6}\)) contains two carbon atoms, four hydrogen atoms, and two oxygen atoms. The ratio of atoms is 2:4:2. Dividing by the lowest common denominator (2) gives the simplest, whole-number ratio of atoms, 1:2:1, so the empirical formula is CH 2 O. Note that a molecular formula is always a whole-number multiple of an empirical formula. Example \(\PageIndex{1}\): Empirical and Molecular Formulas Molecules of glucose (blood sugar) contain 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms. What are the molecular and empirical formulas of glucose? Solution The molecular formula is C 6 H 12 O 6 because one molecule actually contains 6 C, 12 H, and 6 O atoms. The simplest whole-number ratio of C to H to O atoms in glucose is 1:2:1, so the empirical formula is CH 2 O. Exercise \(\PageIndex{1}\) A molecule of metaldehyde (a pesticide used for snails and slugs) contains 8 carbon atoms, 16 hydrogen atoms, and 4 oxygen atoms. What are the molecular and empirical formulas of metaldehyde? Answer Molecular formula, C 8 H 16 O 4 ; empirical formula, C 2 H 4 O It is important to be aware that it may be possible for the same atoms to be arranged in different ways: Compounds with the same molecular formula may have different atom-to-atom bonding and therefore different structures. For example, could there be another compound with the same formula as acetic acid, C 2 H 4 O 2 ? And if so, what would be the structure of its molecules? If you predict that another compound with the formula C 2 H 4 O 2 could exist, then you demonstrated good chemical insight and are correct. Two C atoms, four H atoms, and two O atoms can also be arranged to form methyl formate, which is used in manufacturing, as an insecticide, and for quick-drying finishes. Methyl formate molecules have one of the oxygen atoms between the two carbon atoms, differing from the arrangement in acetic acid molecules. Acetic acid and methyl formate are examples of isomers —compounds with the same chemical formula but different molecular structures (Figure \(\PageIndex{7}\)). Note that this small difference in the arrangement of the atoms has a major effect on their respective chemical properties. You would certainly not want to use a solution of methyl formate as a substitute for a solution of acetic acid (vinegar) when you make salad dressing. Many types of isomers exist (Figure \(\PageIndex{8}\)). Acetic acid and methyl formate are structural isomers , compounds in which the molecules differ in how the atoms are connected to each other. There are also various types of spatial isomers , in which the relative orientations of the atoms in space can be different. For example, the compound carvone (found in caraway seeds, spearmint, and mandarin orange peels) consists of two isomers that are mirror images of each other. S -(+)-carvone smells like caraway, and R -(−)-carvone smells like spearmint. Summary A molecular formula uses chemical symbols and subscripts to indicate the exact numbers of different atoms in a molecule or compound. An empirical formula gives the simplest, whole-number ratio of atoms in a compound. A structural formula indicates the bonding arrangement of the atoms in the molecule. Ball-and-stick and space-filling models show the geometric arrangement of atoms in a molecule. Isomers are compounds with the same molecular formula but different arrangements of atoms. Glossary empirical formula formula showing the composition of a compound given as the simplest whole-number ratio of atoms isomers compounds with the same chemical formula but different structures molecular formula formula indicating the composition of a molecule of a compound and giving the actual number of atoms of each element in a molecule of the compound. spatial isomers compounds in which the relative orientations of the atoms in space differ structural isomer one of two substances that have the same molecular formula but different physical and chemical properties because their atoms are bonded differently structural formula shows the atoms in a molecule and how they are connected |
Courses/Maryville_College/Essential_Chemistry_for_Poisons_Potions_and_Pharmaceuticals/04%3A_Chemical_Reactions_and_Equations/4.04%3A_Ionic_Equations_-_A_Closer_Look | Learning Objectives Write ionic equations for chemical reactions between ionic compounds. Write net ionic equations for chemical reactions between ionic compounds. For single-replacement and double-replacement reactions, many of the reactions included ionic compounds—compounds between metals and nonmetals, or compounds that contained recognizable polyatomic ions. Now, we take a closer look at reactions that include ionic compounds. One important aspect about ionic compounds that differs from molecular compounds has to do with dissolution in a liquid, such as water. When molecular compounds, such as sugar, dissolve in water, the individual molecules drift apart from each other. When ionic compounds dissolve, the ions physically separate from each other . We can use a chemical equation to represent this process—for example, with NaCl: \[\ce{ NaCl(s) ->[\ce{H2O}] Na^{+}(aq) + Cl^{-}(aq)}\nonumber \] When NaCl dissolves in water, the ions separate and go their own way in solution; the ions are now written with their respective charges, and the (aq) phase label emphasizes that they are dissolved (Figure \(\PageIndex{1}\)). This process is called dissociation; we say that the ions dissociate . All ionic compounds that dissolve behave this way. This behavior was first suggested by the Swedish chemist Svante August Arrhenius [1859–1927] as part of his PhD dissertation in 1884. Interestingly, his PhD examination team had a hard time believing that ionic compounds would behave like this, so they gave Arrhenius a barely passing grade. Later, this work was cited when Arrhenius was awarded the Nobel Prize in Chemistry. Keep in mind that when the ions separate, all the ions separate. Thus, when CaCl 2 dissolves, the one Ca 2 + ion and the two Cl − ions separate from one another: \[CaCl_{2}(s)\overset{H_{2}O}{\rightarrow}Ca^{2+}(aq)+Cl^{-}(aq)+Cl^{-}(aq)\nonumber \] \[CaCl_{2}(s)\overset{H_{2}O}{\rightarrow}Ca^{2+}(aq)+2Cl^{-}(aq)\nonumber \] That is, the two chloride ions go off on their own. They do not remain as Cl 2 (that would be elemental chlorine; these are chloride ions), and they do not stick together to make Cl 2 − or Cl 2 2 − . They become dissociated ions in their own right. Polyatomic ions also retain their overall identity when they are dissolved. Example \(\PageIndex{1}\) Write the chemical equation that represents the dissociation of each ionic compound. KBr Na 2 SO 4 Solution KBr(s) → K + (aq) + Br − (aq) Not only do the two sodium ions go their own way, but the sulfate ion stays together as the sulfate ion. The dissolving equation is Na 2 SO 4 (s) → 2Na + (aq) + SO 4 2 − (aq) Exercise \(\PageIndex{1}\) Write the chemical equation that represents the dissociation of (NH 4 ) 2 S. Answer (NH 4 ) 2 S(s) → 2NH 4 + (aq) + S 2− (aq) When chemicals in solution react, the proper way of writing the chemical formulas of the dissolved ionic compounds is in terms of the dissociated ions, not the complete ionic formula. A complete ionic equation is a chemical equation in which the dissolved ionic compounds are written as separated ions. Solubility rules are very useful in determining which ionic compounds are dissolved and which are not. For example, when NaCl(aq) reacts with AgNO 3 (aq) in a double-replacement reaction to precipitate AgCl(s) and form NaNO 3 (aq), the complete ionic equation includes NaCl, AgNO 3 , and NaNO 3 written as separate ions: \[\ce{Na^{+}(aq) + Cl^{−}(aq) + Ag^{+}(aq) + NO3^{−}(aq) → AgCl(s) + Na^{+}(aq) + NO3^{−}(aq)}\nonumber \] This is more representative of what is occurring in the solution. Example \(\PageIndex{1}\) Write the complete ionic equation for each chemical reaction. KBr(aq) + AgC 2 H 3 O 2 (aq) → KC 2 H 3 O 2 (aq) + AgBr(s) MgSO 4 (aq) + Ba(NO 3 ) 2 (aq) → Mg(NO 3 ) 2 (aq) + BaSO 4 (s) Solution For any ionic compound that is aqueous, we will write the compound as separated ions. The complete ionic equation is K + (aq) + Br − (aq) + Ag + (aq) + C 2 H 3 O 2 − (aq) → K + (aq) + C 2 H 3 O 2 − (aq) + AgBr(s) The complete ionic equation is Mg 2 + (aq) + SO 4 2 − (aq) + Ba 2 + (aq) + 2NO 3 − (aq) → Mg 2 + (aq) + 2NO 3 − (aq) + BaSO 4 (s) Exercise \(\PageIndex{1}\) Write the complete ionic equation for \[\ce{CaCl2(aq) + Pb(NO3)2(aq) → Ca(NO3)2(aq) + PbCl2(s)}\nonumber \] Answer Ca 2 + (aq) + 2Cl − (aq) + Pb 2 + (aq) + 2NO 3 − (aq) → Ca 2 + (aq) + 2NO 3 − (aq) + PbCl 2 (s) You may notice that in a complete ionic equation, some ions do not change their chemical form; they stay exactly the same on the reactant and product sides of the equation. For example, in Na + (aq) + Cl − (aq) + Ag + (aq) + NO 3 − (aq) → AgCl(s) + Na + (aq) + NO 3 − (aq) the Ag + (aq) and Cl − (aq) ions become AgCl(s), but the Na + (aq) ions and the NO 3 − (aq) ions stay as Na + (aq) ions and NO 3 − (aq) ions. These two ions are examples of spectator ions—ions that do nothing in the overall course of a chemical reaction. They are present, but they do not participate in the overall chemistry. It is common to cancel spectator ions (something also done with algebraic quantities) on the opposite sides of a chemical equation: \[\cancel{Na^{+}(aq)}+Cl^{-}(aq)+Ag^{+}(aq)+\cancel{NO_{3}^{-}}(aq)\rightarrow AgCl(s)+\cancel{Na}^{+}(aq)+\cancel{NO}_{3}^{-}(aq)\nonumber \] What remains when the spectator ions are removed is called the net ionic equation , which represents the actual chemical change occurring between the ionic compounds: Cl − (aq) + Ag + (aq) → AgCl(s) It is important to reiterate that the spectator ions are still present in solution, but they do not experience any net chemical change, so they are not written in a net ionic equation. Example \(\PageIndex{1}\) Write the net ionic equation for each chemical reaction. K + (aq) + Br − (aq) + Ag + (aq) + C 2 H 3 O 2 − (aq) → K + (aq) + C 2 H 3 O 2 − (aq) + AgBr(s) Mg 2 + (aq) + SO 4 2 − (aq) + Ba 2 + (aq) + 2NO 3 − (aq) → Mg 2 + (aq) + 2NO 3 − (aq) + BaSO 4 (s) Solution In the first equation, the K + (aq) and C 2 H 3 O 2 − (aq) ions are spectator ions, so they are canceled: \[\cancel{K^{+}(aq)}+Br^{-}(aq)+Ag^{+}(aq)+\cancel{C_{2}H_{3}O_{2}^{-}(aq)}\rightarrow K^{+}(aq)+\cancel{C_{2}H_{3}O_{2}^{-}(aq)}+AgBr(s)\nonumber \] The net ionic equation is Br − (aq) + Ag + (aq) → AgBr(s) In the second equation, the Mg 2 + (aq) and NO 3 − (aq) ions are spectator ions, so they are canceled: \[\cancel{Mg^{2+}(aq)}+SO_{4}^{2-}(aq)+Ba^{2+}(aq)+\cancel{2NO_{3}^{-}(aq)}\rightarrow Mg^{2+}(aq)+\cancel{2NO_{3}^{-}(aq)}+BaSo_{4}(s)\nonumber \] The net ionic equation is SO 4 2 − (aq) + Ba 2 + (aq) → BaSO 4 (s) Exercise \(\PageIndex{1}\) Write the net ionic equation for CaCl 2 (aq) + Pb(NO 3 ) 2 (aq) → Ca(NO 3 ) 2 (aq) + PbCl 2 (s) Answer Pb 2 + (aq) + 2Cl − (aq) → PbCl 2 (s) Chemistry is Everywhere: Soluble and Insoluble Ionic Compounds The concept of solubility versus insolubility in ionic compounds is a matter of degree. Some ionic compounds are very soluble, some are only moderately soluble, and some are soluble so little that they are considered insoluble. For most ionic compounds, there is also a limit to the amount of compound that can be dissolved in a sample of water. For example, you can dissolve a maximum of 36.0 g of NaCl in 100 g of water at room temperature, but you can dissolve only 0.00019 g of AgCl in 100 g of water. We consider \(\ce{NaCl}\) soluble but \(\ce{AgCl}\) insoluble. One place where solubility is important is in the tank-type water heater found in many homes in the United States. Domestic water frequently contains small amounts of dissolved ionic compounds, including calcium carbonate (CaCO 3 ). However, CaCO 3 has the relatively unusual property of being less soluble in hot water than in cold water. So as the water heater operates by heating water, CaCO 3 can precipitate if there is enough of it in the water. This precipitate, called limescale , can also contain magnesium compounds, hydrogen carbonate compounds, and phosphate compounds. The problem is that too much limescale can impede the function of a water heater, requiring more energy to heat water to a specific temperature or even blocking water pipes into or out of the water heater, causing dysfunction. Another place where solubility versus insolubility is an issue is the Grand Canyon. We usually think of rock as insoluble. But it is actually ever so slightly soluble. This means that over a period of about two billion years, the Colorado River carved rock from the surface by slowly dissolving it, eventually generating a spectacular series of gorges and canyons. And all because of solubility! Key Takeaways Ionic compounds that dissolve separate into individual ions. Complete ionic equations show dissolved ionic solids as separated ions. Net ionic equations show only the ions and other substances that change in a chemical reaction. |
Courses/Ripon_College/CHM_321%3A_Inorganic_Chemistry/02%3A_Molecular_Symmetry_and_Group_Theory/2.05%3A_Generating_SALCs_for_Polyatomic_Molecules/2.5.05%3A_CO_(Revisted_with_Projection_Operators) | The process for finding SALCs and constructing the molecular orbital diagram for carbon dioxide was described in a previous section (Section 5.4.2) . However, we also just presented an alternative strategy, the projection operator method, to finding the shapes of SALCs in Section 5.4.4. Let's revisit carbon dioxide and demonstrate how you can use the projection operator method to find SALCs for carbon dioxide. Just as before, we can simplify the problem by approximating the \(D_{\infty h}\)point group using \(D_{2h}\). The first four steps for constructing the MO diagram would be the same as described in Section 5.4.2 . and we will pick up from that point (with Step 5) to find what the SALCs look like using the projection operator method. The projection operator method applied to \(\ce{CO2}\) Step 5.1: Label the pendent atoms. We can label the pendant atoms as \(\ce{O}_a\) and \(\ce{O}_b\), as in Figure \(\PageIndex{1}\). Step 5.2: Create an expanded character table with one pendant atom's projected position for each operation The \(D_{2h}\) character table needs no expansion because each operation is in its own class. We will arbitrarily choose to determine the new position of \(\ce{O}_a\) after each operation. \[\begin{array}{|c|cccccccc|} \hline \bf{D_{2h}} & E & C_2(z) & C_2(y) &C_2(x) & i &\sigma(xy) & \sigma(xz) & \sigma(yz)\\ \hline \bf{\text{Projection of }\ce{O}_a} & \ce{O}_a & \ce{O}_a & \ce{O}_b & \ce{O}_b & \ce{O}_b & \ce{O}_b & \ce{O}_a & \ce{O}_a \\ \hline \end{array} \nonumber \] Step 5.3: Find the contribution of each pendant atom to each SALC Create a linear combination of the projections for each or the SALCs. The irreducible representations were found in step 4 in Section 5.4.2 . Eight irreducible representations were found to be: \(2A_{g} + 2B_{1u} + B_{2g} + B_{3u} + B_{3g} + B_{2u}\). For each of the irreducible representations, multiply the projection by the respective character of the operation. \[\text{Contribution of each atom to the SALC } = \sum(\text{Projection of }H_a \times \chi) \nonumber \] The linear combination for all irreducible representations of \(D_{2h}\) are shown below. \[\text{Table }\ref{expanded2} \text{: The symmetry adapted linear combination (SALC) for each irreducible representations of \(C_3v\) are shown.} \nonumber \]\[\begin{array}{|c|cccccccc|l|} \hline \bf{D_{2h}} & E & C_2(z) & C_2(y) &C_2(x) & i &\sigma(xy) & \sigma(xz) & \sigma(yz) & \text{Linear Combination} \\ \hline \bf{\text{Projection of }\ce{O}_a} &\bf \ce{O}_a &\bf \ce{O}_a &\bf \ce{O}_b &\bf \ce{O}_b &\bf \ce{O}_b &\bf \ce{O}_b &\bf \ce{O}_a &\bf \ce{O}_a \\ A_{g} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & = \bf 4\ce{O}_a + 4\ce{O}_b \\ B_{1u} & 1 & 1 & -1 & -1 & -1 & -1 & 1 & 1 & = \bf 4\ce{O}_a - 4\ce{O}_b \\ B_{2g} & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & = \bf 0 \\ B_{3u} & 1 & -1 & -1 & 1 & -1 & 1 & 1 & -1 & = \bf 0\\ B_{3g} & 1 & -1 & -1 & 1 & 1 & -1 & -1 & 1 & = \bf 0\\ B_{2u} & 1 & -1 & 1 & -1 & -1 & 1 & -1 & 1 & = \bf 0\\ \hline \end{array} \label{expanded2} \] Notice that there are only two irreducible representations that produce SALCs, the \(A_g\) and the \(B_{1u}\). Since we found two of each (\(2A_{g} + 2B_{1u}\)) when we found the irreducible representations, we know that there will be two SALCs of \(A_g\) symmetry and two of \(B_{1u}\) symmetry; that is four total SALCs from the pendant oxygens. This is the same as what we found in Section 5.4.2 using a different method. Step 5.4: Sketch the SALCs The SALCs with \(A_g\) symmetry: Quantitatively, we can apply the normalizing factor, N , for these SALCs. Each atom contributes equally to the SALCs, thus the normalizing factor for the \(A_g\) SALC is \(N=\left(\frac{1}{\sqrt{1^2 + 1^2}}\right) = \frac{1}{\sqrt{2}}\). This tells us that each oxygen contributes \(\frac{1}{\sqrt{2}}\) to each of the normalized \(A_g\) group orbitals: \[ A_g \text{ group orbital } = \frac{1}{\sqrt{2}}\left[\psi_{O_{a}}+\psi_{O_{b}}\right] \nonumber \] The linear combination \(4\ce{O}_a + 4\ce{O}_b\) indicates that wavefunctions from each oxygen atom contribute equally to each of the two SALCs, with the same sign of the wavefunction for each. Because we know that \(A_g\) is totally symmetric (all 1's in the characters), we can assume that the SALC's will look like a symmetric arrangement of oxygen orbitals. But which orbitals contribute? We could peek at the character table to get a hint. But we'll treat this problem systematically. The SALCs with \(B_{1u}\) symmetry: The linear combination \(4\ce{O}_a - 4\ce{O}_b\) indicates that each oxygen atom contributes equally, but that they have opposite wavefunctions. Application of the normalizing factor would give us: \[ B_{1u} \text{ group orbital } = \frac{1}{\sqrt{2}}\left[\psi_{O_{a}}+\psi_{O_{b}}\right] \nonumber \] But how do we know what the SALCs look like? Again, the character table can give us some hints, but we'll use a systematic process, below. What do the SALCs look like? One way to systematically derive the SALC shapes is to perform the projection (Step 5.3) on specific groups of orbitals on the oxygen atom. For example, if we replicate Table \ref{expanded2} using just the group of oxygen 2s orbitals and the \(A_g\) and \(B_{1u}\) representations, we get the following: \[\begin{array}{|c|cccccccc|l|} \hline \bf{D_{2h}} & E & C_2(z) & C_2(y) &C_2(x) & i &\sigma(xy) & \sigma(xz) & \sigma(yz) & \text{Linear Combination} \\ \hline \bf{\text{Projection of 2s orbital of }\ce{O}_a} &\bf \ce{O}1s_a &\bf \ce{O}1s_a &\bf \ce{O}1s_b &\bf \ce{O}1s_b &\bf \ce{O}1s_b &\bf \ce{O}1s_b &\bf \ce{O}1s_a &\bf \ce{O}1s_a \\ A_{g} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & = \bf 4\ce{O}1s_a + 4\ce{O}1s_b \\ B_{1u} & 1 & 1 & -1 & -1 & -1 & -1 & 1 & 1 & = \bf 4\ce{O}1s_a - 4\ce{O}1s_b \\ \hline \end{array} \label{expanded3} \] This tells us that one of the \(A_g\) SALCs looks like two oxygen 1s orbitals with equal signs and contributions. It also tells us that one of the \(B_{1u}\) SALCs looks like two \(s\) orbitals with opposite signs and equal magnitudes. These two SALCs are shown in Figure \(\PageIndex{2}\). If we do the same for the \(2p_z\) orbitals, we need to pay attention to the orientation of the \(p\) orbital lobes. If the orbital is moved into the opposite orientation, it would get a negative sign in the projection (see below). \[\begin{array}{|c|cccccccc|l|} \hline \bf{D_{2h}} & E & C_2(z) & C_2(y) &C_2(x) & i &\sigma(xy) & \sigma(xz) & \sigma(yz) & \text{Linear Combination} \\ \hline \bf{\text{Projection of 2px orbital of }\ce{O}_a} &\bf +\ce{O}2pz_a &\bf +\ce{O}2pz_a &\bf -\ce{O}2pz_b &\bf +\ce{O}2pz_b &\bf -\ce{O}2pz_b &\bf -\ce{O}2pz_b &\bf +\ce{O}2pz_a &\bf +\ce{O}2pz_a \\ A_{g} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & = \bf 4\ce{O}2pz_a - 4\ce{O}2pz_b \\ B_{1u} & 1 & 1 & -1 & -1 & -1 & -1 & 1 & 1 & = \bf 4\ce{O}2pz_a + 4\ce{O}2pz_b \\ \hline \end{array} \label{expanded4} \] This tells us that one of the \(A_g\) SALCs is derived from two \(2p_z\) orbitals of opposite sign but equal magnitude. In other words, the two \(p\) orbitals are pointing in opposite directions, but the lobes that are facing one another are of the same sign (see Figure \(\PageIndex{3}\), left). This also tells us that one of the \(B_{1u}\) orbitals is derived from two \(p_z\) orbitals of equal magnitude and equal sign (pointing in the same directions, with lobes facing one another of opposite sign!) (see Figure \(\PageIndex{3}\), right). Now we have reached the same SALCs that we found previously in Section 5.4.2 , but by an alternate method. |
Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/07%3A_Chemical_Nomenclature/7.04%3A_Anions | What does the amount of salt in seaweed tell us? Before iodized salt was developed, some people experienced a number of developmental difficulties, including problems with thyroid gland function and intellectual disabilities. In the 1920s, we learned that these conditions could usually be treated easily with the addition of iodide anion to the diet. One easy way to increase iodide intake was to add the anion to table salt. This simple step greatly enhanced health and development. Large amounts of iodide ion are also found in seaweed such as kelp (see picture above) and saltwater fish. When a metal loses an electron, energy is needed to remove that electron. The other part of this process involves the addition of the electron to another element. The electron adds to the outer shell of the new element. Just as the loss of the electron from the metal produces a full shell, when the electron or electrons are added to the new element, it also results in a full shell. Anions Anions are negative ions that are formed when a nonmetal atom gains one or more electrons. Anions are so named because they are attracted to the anode (positive field) in an electrical field. Atoms typically gain electrons so that they will have the electron configuration of a noble gas. All the elements in Group 17 have seven valence electrons due to the outer \(ns^2 \: np^5\) configuration. Therefore, each of these elements would gain one electron and become an anion with a \(-1\) charge. Likewise, Group 16 elements form ions with a \(-2\) charge, and the Group 15 nonmetals form ions with a \(-3\) charge. Naming anions is slightly different than naming cations. The ending of the element's name is dropped and replaced with the -ide suffix. For example, \(\ce{F^-}\) is the fluoride ion, while \(\ce{O^{2-}}\) is the oxide ion. As is the case with cations, the charge on the anion is indicated by a superscript following the symbol. Common anions are listed in the table below: Anion Name Symbol and Charge fluoride \(\ce{F^-}\) chloride \(\ce{Cl^-}\) bromide \(\ce{Br^-}\) iodide \(\ce{I^-}\) oxide \(\ce{O^{2-}}\) sulfide \(\ce{S^{2-}}\) nitride \(\ce{N^{3-}}\) Uses for Anions Fluoride ion is widely used in water supplies to help prevent tooth decay. Chloride is an important component in ion balance in blood. Iodide ion is needed by the thyroid gland to make the hormone thyroxine. Summary Anions are formed by the addition of one or more electrons to the outer shell of an atom. Group 17 elements add one electron to the outer shell, group 16 elements add two electrons, and group 15 elements add three electrons. Anions are named by dropping the ending of the element's name and adding -ide . Review What is an anion? How are anions formed? Why do anions form? How are anions named? List three examples of anions with names, charges, and chemical symbols. List three ways anions are used. |
Courses/Westminster_College/CHE_180_-_Inorganic_Chemistry/08%3A_Chapter_8_-_Redox/8.1%3A_Balancing_redox_reactions | In studying redox chemistry, it is important to begin by learning to balance electrochemical reactions. Simple redox reactions (for example, H 2 + I 2 → 2 HI) can be balanced by inspection, but for more complex reactions it is helpful to have a foolproof, systematic method. The electron-ion method allows one to balance redox reactions regardless of their complexity. We illustrate this method with two examples. Example 1 - Acidic Solution I - is oxidized to IO 3 - by MnO 4 - , which is reduced to Mn 2+ . This reaction takes place in acidic solution. How can this reaction be balanced? In the electron-ion method we follow a series of ten steps: Step 1 : Write out the (unbalanced) reaction and identify the elements that are undergoing redox. MnO 4 - + I - → IO 3 - + Mn 2+ (acid) Step 2: Assign the oxidation states to each element on both sides of the equation. MnO 4 - Mn = +7 O = -2 I - I = -1 IO 3 - I = +5 O = -2 Mn 2+ Mn = +2 Step 3 : Separate the reaction into two half reactions , balancing the element undergoing redox in each. red. Mn O 4 - → Mn 2+ Manganese is being reduced ox . I - → I O 3 - Iodine is being oxidized Step 4 : Add the appropriate number of electrons to each half-reaction. Electrons are always reactants in the reduction half-reaction and products in the oxidation half reaction. red. MnO 4 - + 5 e - → Mn 2+ Mn is reduced from +7 to +2. This adds 5 electrons. ox. I - → IO 3 - + 6 e - I is oxidized from -1 to +5. This removes 6 electrons. Step 5 : Balance the charge of each half-reaction by adding hydrogen ions, H + . Since this reaction occurs in acidic solution, there are ample H + ions for charge balance. red. MnO 4 - + 5 e - + 8H + → Mn 2+ Both sides of the equation have a total 2+ charge. ox. I - → IO 3 - + 6 e - + 6 H + Both sides of the equation have a total 1- charge. Step 6 : Balance the oxygen and hydrogen atoms by adding H 2 O to the appropriate side of each half-reaction. red. MnO 4 - + 5 e - + 8 H + → Mn 2+ + 4H 2 O ox. I - + 3 H 2 O → IO 3 - + 6 e - + 6 H + Step 7 : Multiply each half-reaction by the appropriate coefficient so that there are equal numbers of electrons as reactants and products. red. (MnO 4 - + 5 e - + 8 H + → Mn 2+ + 4H 2 O) x 6 ox. (I - + 3 H 2 O → IO 3 - + 6 e - + 6 H + ) x 5 Step 8: Combine the half-reactions into an overall equation. 6 MnO 4 - + 5 I - + 30 e - + 48 H + + 15 H 2 O → 6 Mn 2+ + 5 IO 3 - + 30 e - + 30 H + + 24 H 2 O Step 9: Cancel the H + , electrons , and H 2 O : 6 MnO 4 - + 5 I - + 30 e - + 48 H + + 15 H 2 O → 6 Mn 2+ + 5 IO 3 - + 30 e - + 30 H + + 24 H 2 O The overall balanced reaction is therefore: 6 MnO 4 - (aq) + 5 I - (aq) + 18 H + (aq) → 6 Mn 2+ (aq) + 5 IO 3 - (aq) + 9 H 2 O( ℓ ) Step 10: Check your work by making sure that all elements and charges are balanced. Example 2 - Basic Solution The chromate ion ( CrO 4 2- ) oxidizes Fe(OH) 2 to Fe(OH) 3 . The CrO 4 2- ion is reduced to Cr(OH) 3 . This reaction takes place in basic solution. How can this reaction be balanced? In the electron-ion method we follow a series of ten steps: Step 1 : Write out the (unbalanced) reaction and identify the elements that are undergoing redox. CrO 4 2- + Fe(OH) 2 → Cr(OH) 3 + Fe(OH) 3 (basic) Step 2: Assign the oxidation states to each element on both sides of the equation. CrO 4 2 - Cr = +6 O = -2 Fe(OH) 2 Fe = +2 O = -2 H = +1 Cr(OH) 3 Cr = +3 O = -2 H = +1 Fe(OH) 3 Fe = +3 O = -2 H = +1 Step 3 : Separate the reaction into two half reactions , balancing the element undergoing redox in each. red. Cr O 4 - → Cr (OH) 3 Chromium is being reduced ox. Fe (OH) 2 → Fe (OH) 3 Iron i s being oxidized Step 4 : Add the appropriate number of electrons to each half-reaction. Electrons are always reactants in the reduction half-reaction and products in the oxidation half reaction. red. CrO 4 2 - + 3 e - → Cr(OH) 3 Cr is reduced from +6 to +3. This adds 3 electrons. ox. Fe(OH) 2 → Fe(OH) 3 + e - Fe is oxidized from +2 to +3. This removes 1 electron. Step 5 : Balance the charge of each half-reaction by adding hydroxide ions, OH - . Since this reaction occurs in basic solution, there are ample OH - ions for charge balance. red. CrO 4 2 - + 3 e - → Cr(OH) 3 + 5 OH - Both sides of the equation have a total 5- charge. ox. Fe(OH) 2 + OH - → Fe(OH) 3 + e - Both sides of the equation have a total 1- charge. Step 6 : Balance the oxygen and hydrogen atoms by adding H 2 O to the appropriate side of each half-reaction. red. CrO 4 2 - + 3 e - + 4 H 2 O → Cr(OH) 3 + 5 OH - ox. Fe(OH) 2 + OH - → Fe(OH) 3 + e - This half-reaction is already balanced. Step 7 : Multiply each half-reaction by the appropriate coefficient so that there are equal numbers of electrons as reactants and products. red. CrO 4 2 - + 3 e - + 4 H 2 O → Cr(OH) 3 + 5 OH - ox. (Fe(OH) 2 + OH - → Fe(OH) 3 + e - ) x 3 Step 8: Combine the half-reactions into an overall equation. CrO 4 2- + 3 Fe(OH) 2 + 3 e - + 3 OH - + 4 H 2 O → Cr(OH) 3 + 3 Fe(OH) 3 + 3 e - + 5 OH - Step 9: Cancel the O H - , electrons , and H 2 O : CrO 4 2- + 3 Fe(OH) 2 + 3 e - + 3 OH - + 4 H 2 O → Cr(OH) 3 + 3 Fe(OH) 3 + 3 e - + 5 OH - The overall balanced reaction is therefore: CrO 4 2- (aq) + 3 Fe(OH) 2 (s) + 4 H 2 O( ℓ) → Cr(OH) 3 (s) + 3 Fe(OH) 3 (s) + 2 OH - (aq) Step 10: Check your work by making sure that all elements and charges are balanced. Contributors Adapted from the Wikibook constructed by Chemistry 310 students at Penn State University . |
Courses/Duke_University/CHEM_401L%3A_Analytical_Chemistry_Lab/CHEM_401L%3A_Analytical_Chemistry_Lab_Manual/03%3A_Determining_caffeine_in_beverages_using_HPLC_(and_UV-vis)/3.04%3A_UV-vis_procedure_and_instructions_for_data_analysis | Note: You should prepare a series of standard solutions of caffeine using serial dilution. You learned how to create serial dilutions in a previous module, and in this module you should come up with your own procedure for making the diluted series of solutions. Refer to Module 2 to review the correct use of volumetric glassware and determining error from a calibration curve. The solutions you make here will also be used in the HPLC module, so plan to store your solutions in sealed vials for the next module. Materials required: Caffeine (99%. powder) Ultrapure water Volumetric glassware Caffeine control solution (provided by TA and used in this module and the next) Guidance for preparing the Standard Caffeine Samples A. Prepare a stock solution of 0.1 g/L caffeine Accurately weigh out 10.0 mg of caffeine. The caffeine can be found on the shelf near the weigh station area. Transfer the caffeine into a clean 100 mL volumetric flask. Dilute to the mark with deionized water. The stock solution will have a final concentration of 0.1 g/L. Warning There is a very small amount of caffeine powder, but it takes some time to dissolve completely. Be sure that caffeine is completely dissolved in about 50 mL of water before diluting to the volumetric mark and before proceeding to the next step. There is a sonicator in the lab that you can use to help break up small caffeine particles and aid in dissoltion. B. Prepare a serial dilution You should write out a specific plan for doing this before you get to lab (see pre-lab assignment). Carry out a serial dilution to obtain a series of standard solutions that are are approximately evenly spaced (in terms of their concentrations). The highest concentration should be about \(0.05 \pm 0.1\) g/L, and the lowest concentration solution should be approximately \(0.005 \pm 0.005\). Make enough of each solution so that you have at least 10 mL of each solution to play with. You should use ultrapure deionized water to make all solutions and dilutions. Ensure adequate mixing in each diluted solution before using it to create the next. Tips You have access to: volumetric flasks to contain of 5, 10, 25, 50, 100, 250, 500, and 1000 mL volumetric pipete to deliver volumes of 1, 2, 3, 4, 5, 10, 15, 20, 25, 50, and 100 mL automatic pipetters to deliver between 100-5000 mL When an experienced student did this lab, she used only one 25 mL volumetric flask and one 10 mL volumetric pipette to create a series of solutions. **Do not discard your solutions. Save the solutions for the next module! Procedure for analysis of caffeine with UV-vis and validating solution concentrations Start up: At the begriming of the lab period, check to confirm that the UV-vis is switched "on". The Cary 100 instruments have tungsten and deuterium lamps that must be warmed up for approximately 1 hour prior to use. The Cary 60 instrument has a xenon flash lamp, which does not require warm up. Be sure you discuss which instrument you will use with your TA, and if you are using a Cary 100, be sure it is on for 1 hour prior to use. Validate instrument: Prior to using any instrument, you should run checks to ensure the instrument is functioning properly. While the Cary UV-vis instruments run initialization procedures upon startup, there is a "Validate" program that will validate the instrument's performance. You should run a wavelength validation after the instrument has warmed up (if applicable). For Cary 100 instruments, use the deuterium (D2) lamp to validate wavelength accuracy. For the Cary 60, use the xenon (Xe) lamp to calibrate wavelength accuracy. The figure below gives a summary of the steps involved in validating the instrument. Find the "Cary WinUV" shortcut on the computer desktop and open it. The linked folder contains all the programs that operate the Cary spectrophotometer. Open the "Validate" Program. The Validate window should open. At the top left corner of the window, it should say "Validate-Online". If it says otherwise (ie "Validate-Offline, as pictured below) then it is not connected to the instrument. Wait a few moments to see if it connects. In the meantime, make sure the instrument is powered on. If it does not connect, try pressing "connect" (its where the start button is), and if that fails, get help. On the left hand side of the Validate Window, choose " Tests ". The Tests window will open. Choose "Instrument Performance Tests". There should be a list of several available tests in the "Test Order" box. If there is not, double check that the instrument is "online". Then, press the "Reset all limits to Default button" - this usually regenerated the list of available tests. It is sufficient to run one wavelength validation test. In the "Test Order" box, choose the wavelength accuracy test that uses the instrument's own lamp (either D2 or Xe). To do this, you should remove all other tests by selecting each and pressing the "remove test" button that is just to the right of the box (it looks like an exclamation point with a red "X". Do this until the test you want is the only one remaining. Press "OK" after you have specified the validation test to be run. The USP Tests box will close. Check to make sure there are no samples in the light path. Close the sample compartment. Then Press "Start" . This should initiate data collection. It will take a few minutes, and during this time you should be able to see the collected spectrum recorded on the screen. If not, get help. After the validation, copy and paste the relevant data and images from the report into your ELN. Method Settings: Open the "Scan" program on the Cary UV-vis and got to "Setup". Go to each tab and modify the method settings in the following order. Cary Tab: Set the scan range to include wavelengths of 400-200 nm (or a wider range if you prefer). Set the data interval to 1 nm and the average time to 0.1 s. Options: SBW/Energy: Use double beam mode Source: Choose either UV or UV-vis Display Options: Choose "Overlay Data" Accessories 1: This is the multi-cell, temperature-controlled sample holder. If this accessory is attached, follow the instructions below. Otherwise, uncheck the "Use cell changer" option. If the multi-cell changer is installed, select "Use cell changer". Note that the cell changer must be installed correctly and aligned to operate correctly. Check only the cell poisitions that you will use during the experiment. Select multi-zero Select multi-baseline Baseline: Select Baseline correction Reports: Select for ASCII (csv). This option saves the data as a comma delineated format that can be opened in excel or and plotted using your own computer. Auto Store: "Storage on (prompt at start)" should be selected The figure below gives a visual summary of the stesp above. After all settings have been chosen, press "OK". The instrument is ready for you to use at this point. Data collection: Depending on the sample holder and the number of samples you plan to run, the procedure below could vary slightly. Make sure to consult your instructors about any changes to this procedure. Use the Scan program to collect a baseline spectrum. You will need either one or two cuvettes containing a sample blank (ie no analyte). The cuvette should not be empty - rather it should contain the background solution of your sample, but with no analyte. (In the case here, it should be ultrapure water). Put the cuvette with a blank solution into the appropriate position(s) to baseline the instrument. Then press the "Baseline" button on the left of the Scan window. Place sample in the appropriate position(s) and press "Start". Specify the file name and folder location where the data should be saved. All data should be saved in the directory C:\Data\CHEM401L\ within a folder for your group . Make a new folder within the CHEM401L directory if one does not already exist. Your file names should follow the convention "YYYYMMDD_Initials_##". In your notebook, you should keep a log of file names and a description of what is contained in each file. Prepare a table in your notebook for recording this information. Example is below (this info is made up!): 0 1 2 3 File Location is C:\Data\CHEM401L\OurGroup File Names: 20230811_KH_## Absorbance at \(\Lambda_{max}\) Solution components Comments 20230811_KH_01 0.0001 ultrapure water This was take just to check that the blank is as expected (ie a flat line) 20230811_KH_02 0.0809 0.01 g/L caffeine in ultrapure water NaN 20230811_KH_03 NaN NaN NaN Repeat the steps above for each calibration solution and the control solution. If you use the same cuvette for each, you do not need to repeat the baseline step. However, you should ensure that your spectrum is at zero absorbance near 400 nm. If not, check with your instructors. After you have collected all data, it is helpful if you "save as" all of your data in one "bulk" .csv file. To do this, choose "file" and then "save as" and choose ".csv". Give the file a name like "YYYYMMDD_Initials_all" to indicate it has "all" your data. Open this file using excel and check that it in fact contains all the spectra that you collected. Data Transfer: Before you start cleaning up, check to be sure that you can find your data, and that they are in fact saved as .csv files. Transfer your data files to your own computer (email is fine, or use OneDrive or Box). Check following: You have YOUR data (not someone else's) You can open the files on your computer You know what is in the files and how they are structured You have a plan for how to plot the spectra and create the calibration curve from the data Treatment of Data The following should be prepared prior to moving forward with the next module. These items should be checked by your instructors, and you should plan to include them in your caffeine analysis report. (Note that the report encompasses this module and the next on HPLC) A figure containing overlaid spectra collected for the calibration curve , with wavelength of maximum absorbance indicated on the plot. This figure should include all the necessary labels and formatting that was required of previous figures/plots. Include title, labels with units, annotations, a figure number and a descriptive caption. Include a legend and choose a line type and color scheme that is legible to a reader who is colorblind. This figure could be generated using the Scan software or by plotting with MatLab or other plotting program. A plot of the calibration curve with best fit line and "the works" (ie everything that is expected of such a plot and its accoutrements - see the previous module for expectations on such a plot). Validation of the calibration curve including error analysis and comparison to a control solution. This should be used as evidence that your solutions have been prepared correctly and that the concentrations are as expected. If you cannot validate your calibration curve, use the molar extinction coefficient for caffeine reported in a reputable peer-reviewed journal to determine the actual concentrations of caffeine in your solutions, and explain why they are different than the expected concentrations. What specific wavelength(s) would be useful for monitoring caffeine in the next module? Discuss this with your instructor before leaving for the day! |
Courses/Maryville_College/Essential_Chemistry_for_Poisons_Potions_and_Pharmaceuticals/07%3A_Air/7.05%3A_Photochemical_Smog-_Making_Haze_While_the_Sun_Shines | Learning Objectives Describe photochemical smog. List different means to address photochemical smog. Photochemical smog is a type of air pollution due to the reaction of solar radiation with airborne pollutant mixtures of nitrogen oxides (NOx) and volatile organic compounds (hydrocarbons). Smog is a byproduct of modern industrialization. Due to industry and the number of motor vehicles, this is more of a problem in large cities that have a warm, sunny and dry climate. Oxidation : Photochemical smog is also referred to as oxidizing smog. Oxidation reactions have been defined several ways. In terms of oxygen transfer, oxidation is a gain of oxygen. Oxidation can also be defined as a loss of hydrogen. The most important use of oxidation is described in terms of electron transfer. Oxidation can be described as an increase in oxidation number or loss of electrons. Oxidation numbers represents a distribution of charge. In other words, oxidation numbers represent the charge of the atom if the compound was composed of ions. Reduction : Reduction can involve the gain of hydrogen or loss of oxygen. Reduction can refer to the gain of electrons, which results in a decrease in oxidation number. Formation of Photochemical Smog The different reactions involved in the formation of photochemical smog are given below. Step 1: People begin driving in the morning, nitrogen is burned or oxidized \[N_2 + O_2 \rightarrow 2NO \nonumber \] Oxidation number of N 2 is 0. The nitrogen in NO has acquired an oxidation number of +2. Step 2: After a few hours, NO combines with O 2 , in another oxidation reaction \[2NO + O_2 \rightarrow 2NO_2 \nonumber \] The nitrogen in NO has an oxidation number of +2. The nitrogen in NO 2 has an oxidation number of +4. Step 3: Nitrogen dioxide absorbs light energy, resulting in a reduction reaction \[NO_2 \rightarrow NO + O \nonumber \] The nitrogen in NO 2 has an oxidation number of +4 and the nitrogen in NO is +2. Step 4: In sunlight, atomic oxygen combines with oxygen gas to form ozone \[O + O_2 \rightarrow O_3 \nonumber \] Step 5: Reaction is temperature and sunlight dependent \[O_3 + NO \rightleftharpoons NO_2 + O_2 \nonumber \] Alternative Reactions NO and NO 2 can also react with the hydrocarbons instead of ozone to form other volatile compounds known as PAN (peroxyacetyl nitrate) as shown in Figure . The accumulation of ozone and volatile organic compounds along with the energy from the sun forms the brown, photochemical smog seen on hot, sunny days. Panoramic view of Santiago covered by a layer of smog on May 10, 2006. The Metropolitan Region of Santiago facing the driest autumn last 28 years due to lack of rainfall, which coupled with poor air circulation, causes an increase in smog. Health Hazards Because ozone is highly reactive, it has the ability to oxidize and destroy lung tissue. Short term exposures to elevated levels of ozone (above .75 ppm) have been linked to a host of respiratory irritations including coughing, wheezing, substernal soreness, pharyngitis, and dyspnea. Prolonged exposure to the molecule has been proven to cause a permanent reduction in lung function, as well as elevate the risk of developing asthma. Sulfur dioxide is a common component of London smog. Epidemiological studies have linked short term sulfur dioxide exposure to respiratory irritations including coughing, wheezing, and pharyngitis. Other Harmful Effects of Smog Plants are harmed by exposure to nitrogen oxides, ozone, and peroxyacetyl nitrate (PAN, see above), all oxidants present in a smoggy atmosphere. PAN is the most harmful of these constituents, damaging younger plant leaves, especially. Ozone exposure causes formation of yellow spots on leaves, a condition called chlorotic stippling. Some plant species, including sword-leaf lettuce, black nightshade, quickweed, and double-fortune tomato, are extremely susceptible to damage by oxidant species in smog and are used as bioindicators of the presence of smog. Costs of crop and orchard damage by smog run into millions of dollars per year in areas prone to this kind of air pollution, such as southern California. Materials that are adversely affected by smog are generally those that are attacked by oxidants. The best example of such a material is rubber, especially natural rubber, which is attacked by ozone. Indeed, the hardening and cracking of natural rubber has been used as a test for atmospheric ozone. Visibility-reducing atmospheric aerosol particles are the most common manifestation of the harm done to atmospheric quality by smog. The smog-forming process occurs by the oxidation of organic materials in the atmosphere, and carbon-containing organic materials are the most common constituents of the aerosol particles in an atmosphere afflicted by smog. Conifer trees(pine and cypress) and citrus trees are major contributors to the organic hydrocarbons that are precursors to organic particle formation in smog. Controlling Photochemical Smog Every new vehicle sold in the United States must include a catalytic converter to reduce photochemical emissions. Catalytic converters force CO and incompletely combusted hydrocarbons to react with a metal catalyst, typically platinum, to produce CO 2 and H 2 O. Additionally, catalytic converters reduce nitrogen oxides from exhaust gases into O 2 and N 2 , eliminating the cycle of ozone formation. Many scientists have suggested that pumping gas at night could reduce photochemical ozone formation by limiting the amount of exposure VOCs have with sunlight. Preventing Smog with Green Chemistry Smog is basically a chemical problem, which would indicate that it should be amenable to chemical solutions. Indeed, the practice of green chemistry and the application of the principles of industrial ecology can help to reduce smog. This is due in large part to the fact that a basic premise of green chemistry is to avoid the generation and release of chemical species with the potential to harm the environment. The best way to prevent smog formation is to avoid the release of nitrogen oxides and organic vapors that enable smog to form. At an even more fundamental level, measures can be taken to avoid the use of technologies likely to release such substances, for example, by using alternatives to polluting automobiles for transportation. The evolution of automotive pollution control devices to reduce smog provides an example of how green chemistry can be used to reduce pollution. The first measures taken to reduce hydrocarbon and nitrogen oxide emissions from automobiles were very much command-and-control and “end-of-pipe” measures. These primitive measures implemented in the early 1970s did reduce emissions, but with a steep penalty in fuel consumption and in driving performance of vehicles. However, over the last three decades, the internal combustion automobile engine has evolved into a highly sophisticated computer-controlled machine that generally performs well, emits few air pollutants, and is highly efficient. (And it would be much more efficient if those drivers who feel that they must drive “sport utility” behemoths would switch to vehicles of a more sensible size.) This change has required an integrated approach involving reformulation of gasoline. The first major change was elimination from gasoline of tetraethyllead, an organometallic compound that poisoned automotive exhaust catalysts (and certainly was not good for people). Gasoline was also reformulated to eliminate excessively volatile hydrocarbons and unsaturated hydrocarbons (those with double bonds between carbon atoms) that are especially reactive in forming photochemical smog. An even more drastic approach to eliminating smog-forming emissions is the use of electric automobiles that do not burn gasoline. These vehicles certainly do not pollute as they are being driven, but they suffer from the probably unsolvable problem of a very limited range between charges and the need for relatively heavy batteries. However, hybrid automobiles using a small gasoline or diesel engine that provides electricity to drive electric motors propelling the automobile and to recharge relatively smaller batteries can largely remedy the emission and fuel economy problems with automobiles. The internal combustion engine on these vehicles runs only as it is needed to provide power and, in so doing, can run at a relatively uniform speed that provides maximum economy with minimum emissions. Another approach that is being used on vehicles as large as buses that have convenient and frequent access to refueling stations is the use of fuel cells that can generate electricity directly from the catalytic combination of elemental hydrogen and oxygen, producing only harmless water as a product . There are also catalytic process that can generate hydrogen from liquid fuels, such as methanol, so that vehicles carrying such a fuel can be powered by electricity generated in fuel cells. Green chemistry can be applied to devices and processes other than automobiles to reduce smog-forming emissions. This is especially true in the area of organic solvents used for parts cleaning and other industrial operations, vapors of which are often released to the atmosphere. The substitution of water with proper additives or even the use of supercritical carbon dioxide fluid can eliminate such emissions. Summary Photochemical smog is a mixture of pollutants that are formed (mostly during the hot summer months) when nitrogen oxides and volatile organic compounds (VOCs) react to sunlight, creating a brown haze above cities. Photochemical smog is formed from the reactions of natural and man-made emissions of nitrogen oxides and VOCs. Smog is a serious problem in many cities and continues to harm human health and are especially harmful for senior citizens, children, and people with heart and lung conditions such as emphysema, bronchitis, and asthma. Catalytic converters in gas powered vehicles help reduce photochemical emissions. The practice of green chemistry and the application of the principles of industrial ecology can help to reduce smog. |
Ancillary_Materials/Laboratory_Experiments/Wet_Lab_Experiments/Organic_Chemistry_Labs/Misc/Notebook | Advanced Preparation and The Notebook Before coming to the laboratory you will find that some advanced preparation can save significant time and effort in the laboratory. The advanced preparation will vary depending on the nature of the experiment, but some general rules will be applicable to all experiments. Read the experiment first and identify all of the reagents and substances that will be used and prepared during the course of the experiment. Next, collect the physical properties of these materials before coming to the laboratory. Physical properties that will be useful include melting point and boiling points (if available) of all the reagents and expected products, their densities (if liquid) and some general solubility properties. Molecular weights should be calculated for all the reagents. This information should be recorded in your notebook which will be described below. While this may seem like a lot of work, many of these same reagents will be used repeatedly. Once recorded in your notebook, this material can be copied or a reference made to the page on which this information is recorded. In addition to the physical property listed, some reference, including pagination, to the source of the information should be cited. Included below are a few valuable sources of physical property data. Copies of these sources of material can be found in the Thomas Jefferson Library and also in the laboratory. A variety of editions are available for most. "The Dictionary of Organic Compounds" Buckingham, J., editor, Chapman and Hall. "CRC Handbook of Chemistry and Physics" Weast, R. C, editor, CRC Press. "Aldrich Catalog Handbook of Fine Chemicals", Aldrich Chemical Co. "Merck Index", Windholz, M. editor, Rahway N.J. "Lange's Handbook of Chemistry", Dean, J. A., editor, McGraw-Hill. The laboratory notebook is the permanent record documenting what you did in the laboratory during a specific period of time. It should contain sufficient details and documentation that an individual with similar training could repeat your work months or years later. The laboratory notebook should contain the original data, not copies of the data. The specific requirements for entries will vary from experiment to experiment and from instructor to instructor. The following is intended to serve as a guideline of what to include in the notebook. A substantial part of the grade earned in this course will be derived from the contents of your notebook. a. Use a bound notebook. Notebooks with a spiral binding are not acceptable nor are any that allow for insertion or removal of pages. b. Write in ink. It is expected that you will make mistakes. Just draw a line or an X through the material you wish to delete. Neatness is always appreciated and, if someone else cannot read and understand what you have written, you have failed to properly record your results. c. Leave the first few pages of the notebook blank for a table of contents so that each experiment can be readily located. Pages should be numbered consecutively. You may leave some blank space on a page in order to complete an experiment but if necessary, a notation instructing the reader to where the remaining portion of the experiment is to be found, is acceptable. You may wish to include supporting documentation such as spectra that you have recorded. This documentation can be attached to the notebook with staples or tape or preferably, in a separate folder. In addition you may wish to sketch your thin layer chromatography results in your notebook since the actual plates may not survive repeated handling. d. Some notebook preparation should be completed before you come to the laboratory. What is required will vary from instructor to instructor. However, the following will apply to all. Begin with a title of the experiment and a brief sentence summarizing the purpose of the experiment and specify a reference (e.g., UMSL Chemistry 263 Manual, pp. 18-19). If the experiment involves a synthesis, include a balanced chemical equation, and physical properties of all chemicals involved. Include comments on toxicity if appropriate, any side reactions and anything out of the ordinary. Prepare an outline of the procedure and what you plan to do in the laboratory. The outline should be completed before you come to the laboratory (read Zubrick, pp. 12-24 for more detailed information) but may be on a separate sheet of paper instead of the lab notebook . This will enable you to work more efficiently. In the laboratory, you should enter directly into the notebook the masses of the reagents you used and include what you actually did and observed as you performed the experiment. Be as brief as possible. Do not simply copy the experiment from the manual to your notebook. Include a sketch of the apparatus, if one is involved. Always include the date you did the work on the outside page margin. Finally, write your observations and conclusions which is very important. When you comment on specific tests, (eg, the ferric chloride test) do not simply say the test was negative, add that it means you do not have a phenol or that you do not have salicylic acid present. It may not be possible to follow this format for each experiment you will be performing this semester. However, for those experiments that involve some synthetic component, following this format will suffice. It is important to remember that your notebook is the medium where you record what you did in the laboratory and what you observed. If you keep this in mind you will stay on the right track. Notebook Calculations The number of calculations you will need to perform in the organic laboratory is quite small but those that you will be required to perform are very important. The calculations are similar to the calculations you performed in introductory chemistry and be sure to review them if you have forgotten the details and theory behind them. What follows is a brief summary of some important definitions. The calculations associated with conversion of the starting materials to product is based on the assumption that the reaction will follow simple ideal stoichiometry. For example, in the preparation of aspirin from salicylic acid and acetic anhydride, in calculating the theoretical and actual yields, it is assumed that all of the starting material is converted to product, even though some of the starting material actually forms a polymer as a consequence of the reaction conditions and catalyst that is used. The first step in calculating yields is to determine the limiting reagent . The limiting reagent in a reaction that involves two or more reactants is simply the reagent which is present in lowest molar amount based on the stoichiometry of the reaction. This reagent will be consumed first and will limit any additional conversion to product. In the reaction of salicylic acid with acetic anhydride, the latter reagent is used in excess. There are several reasons for doing so. An important consideration whenever any reagent is used in excess, is to determine how this reagent will be separated from the product at the end of the reaction. In this case, the acetic anhydride which is not water soluble can be allowed to stand in water a while during which time it will slowly hydrolyze to acetic acid. The acetic acid which is water soluble, can be separated from the aspirin by filtration. The salicyclic acid will be the limiting reagent. As far as the calculation of the theoretical yield is concerned, we assume that every mole of salicylic acid will be converted to aspirin or acetylsalicylic acid. According to the stoichiometry of the reaction, one mole of salicyclic acid will be converted to one mole of aspirin. Therefore, from the number of moles of salicylic acid we used, we evaluate the maximum amount of aspirin that can be obtained. Multiplying the number of moles of aspirin by its molecular weight results in the theoretical yield which is usually reported in grams. In all cases, the calculation is performed similarly, regardless of how many products are formed. However, we can now evaluate the actual yield by determining how much aspirin we have actually isolated, experimentally. The % yield is simply the ratio of the actual yield divided by the theoretical yield times 100. In a multi-step synthesis, each step in the process is characterized by a limiting reagent, an actual yield and a % yield. The calculation of the overall yield of the entire process can be calculated by identifying the limiting reagent in each step. The overall yield of the process will simply be the product of the actual yield divided by the theoretical yield of the first step times the same ratio for each step in the process times 100. Thus in a two step process, if you get a 0.5 yield (50%) in the first step and the same yield in the second step, your overall yield will be 0.5 x 0.5 x 100 = 25 %. You can imagine that a multistep process operating at this efficiency is not likely to be very useful commercially. |
Courses/University_of_Georgia/CHEM_3212%3A_Physical_Chemistry_II/05%3A_Energy_and_Enthalpy/5.04%3A_Reversible_and_Irreversible_Pathways | The most common example of work in the systems discussed in this book is the work of expansion. It is also convenient to use the work of expansion to exemplify the difference between work that is done reversibly and that which is done irreversibly. The example of expansion against a constant external pressure is an example of an irreversible pathway. It does not mean that the gas cannot be re-compressed. It does, however, mean that there is a definite direction of spontaneous change at all points along the expansion. Imagine instead a case where the expansion has no spontaneous direction of change as there is no net force push the gas to seek a larger or smaller volume. The only way this is possible is if the pressure of the expanding gas is the same as the external pressure resisting the expansion at all points along the expansion. With no net force pushing the change in one direction or the other, the change is said to be reversible or to occur reversibly . The work of a reversible expansion of an ideal gas is fairly easy to calculate. If the gas expands reversibly, the external pressure (\(p_{ext}\)) can be replaced by a single value (\(p\)) which represents both the pressure of the gas and the external pressure. \[ dw = -pdV \nonumber \] or \[ w = - \int p dV \nonumber \] But now that the external pressure is not constant, \(p\) cannot be extracted from the integral. Fortunately, however, there is a simple relationship that tells us how \(p\) changes with changing \(V\) – the equation of state ! If the gas is assumed to be an ideal gas \[ w = - \int p dV -\int \left( \dfrac{nRT}{V}\right) dV \nonumber \] And if the temperature is held constant (so that the expansion follows an isothermal pathway) the nRT term can be extracted from the integral. \[ w = -nRT \int_{V_1}^{V_2} \dfrac{dV}{V} = -nRT \ln \left( \dfrac{V_2}{V_1} \right) \label{isothermal} \] Equation \ref{isothermal} is derived for ideal gases only; a van der Waal gas would result in a different version. Example \(\PageIndex{1}\): Gas Expansion What is the work done by 1.00 mol an ideal gas expanding reversibly from a volume of 22.4 L to a volume of 44.8 L at a constant temperature of 273 K? Solution Using Equation \ref{isothermal} to calculate this \[\begin{align*} w & = -(1.00 \, \cancel{mol}) \left(8.314\, \dfrac{J}{\cancel{mol}\,\cancel{ K}}\right) (273\,\cancel{K}) \ln \left( \dfrac{44.8\,L}{22.4 \,L} \right) \nonumber \\[4pt] & = -1570 \,J = 1.57 \;kJ \end{align*} \] Note : A reversible expansion will always require more work than an irreversible expansion (such as an expansion against a constant external pressure) when the final states of the two expansions are the same! The work of expansion can be depicted graphically as the area under the p-V curve depicting the expansion. Comparing examples \(\PageIndex{1}\) and \(3.1.2\), for which the initial and final volumes were the same, and the constant external pressure of the irreversible expansion was the same as the final pressure of the reversible expansion, such a graph looks as follows. The work is depicted as the shaded portion of the graph. It is clear to see that the reversible expansion (the work for which is shaded in both light and dark gray) exceeds that of the irreversible expansion (shaded in dark gray only) due to the changing pressure of the reversible expansion. In general, it will always be the case that the work generated by a reversible pathway connecting initial and final states will be the maximum work possible for the expansion. It should be noted (although it will be proven in a later chapter) that \(\Delta U\) for an isothermal reversible process involving only p-V work is 0 for an ideal gas. This is true because the internal energy, U, is a measure of a system’s capacity to convert energy into work. In order to do this, the system must somehow store that energy. The only mode in which an ideal gas can store this energy is in the translational kinetic energy of the molecules (otherwise, molecular collisions would not need to be elastic, which as you recall, was a postulate of the kinetic molecular theory!) And since the average kinetic energy is a function only of the temperature, it (and therefore \(U\)) can only change if there is a change in temperature. Hence, for any isothermal process for an ideal gas, \(\Delta U=0\). And, perhaps just as usefully, for an isothermal process involving an ideal gas, \(q = -w\), as any energy that is expended by doing work must be replaced with heat, lest the system temperature drop. Constant Volume Pathways One common pathway which processes can follow is that of constant volume. This will happen if the volume of a sample is constrained by a great enough force that it simply cannot change. It is not uncommon to encounter such conditions with gases (since they are highly compressible anyhow) and also in geological formations, where the tremendous weight of a large mountain may force any processes occurring under it to happen at constant volume. If reversible changes in which the only work that can be done is that of expansion (so-called p-V work) are considered, the following important result is obtained: \[ dU = dq + dw = dq - pdV \nonumber \] However, \(dV = 0\) since the volume is constant! As such, \(dU\) can be expressed only in terms of the heat that flows into or out of the system at constant volume \[ dU = dq_v \nonumber \] Recall that \(dq\) can be found by \[ dq = \dfrac{dq}{\partial T} dT = C\, dt \label{eq1} \] This suggests an important definition for the constant volume heat capacity (\(C_V\)) which is \[C_V \equiv \left( \dfrac{\partial U}{\partial T}\right)_V \nonumber \] When Equation \ref{eq1} is integrated the \[q = \int _{T_1}^{T_2} nC_V dt \label{isochoric} \] Example \(\PageIndex{2}\): Isochoric Pathway Consider 1.00 mol of an ideal gas with \(C_V = 3/2 R\) that undergoes a temperature change from 125 K to 255 K at a constant volume of 10.0 L. Calculate \(\Delta U\), \(q\), and \(w\) for this change. Solution Since this is a constant volume process \[w = 0 \nonumber \] Equation \ref{isochoric} is applicable for an isochoric process, \[q = \int _{T_1}^{T_2} nC_V dt \nonumber \] Assuming \(C_V\) is independent of temperature: \[\begin{align*} q & = nC_V \int _{T_1}^{T_2} dt \\[4pt] &= nC_V ( T_2-T_1) \\[4pt] & = (1.00 \, mol) \left( \dfrac{3}{2} 8.314\, \dfrac{J}{mol \, K}\right) (255\, K - 125 \,K) \\[4pt] & = 1620 \,J = 1.62\, kJ \end{align*} \] Since this a constant volume pathway, \[ \begin{align*} \Delta U & = q + \cancel{w} \\ & = 1.62 \,kJ \end{align*} \] Constant Pressure Pathways Most laboratory-based chemistry occurs at constant pressure. Specifically, it is exposed to the constant air pressure of the laboratory, glove box, or other container in which reactions are taking place. For constant pressure changes, it is convenient to define a new thermodynamic quantity called enthalpy . \[ H \equiv U+ pV \nonumber \] or \[\begin{align*} dH &\equiv dU + d(pV) \\[4pt] &= dU + pdV + Vdp \end{align*} \] For reversible changes at constant pressure (\(dp = 0\)) for which only p-V work is done \[\begin{align} dH & = dq + dw + pdV + Vdp \\[4pt] & = dq - \cancel{pdV} + \cancel{pdV} + \cancelto{0}{Vdp} \\ & = dq \label{heat} \end{align} \] And just as in the case of constant volume changes, this implies an important definition for the constant pressure heat capacity \[C_p \equiv \left( \dfrac{\partial H}{\partial T} \right)_p \nonumber \] Example \(\PageIndex{3}\): Isobaric Gas Expansion Consider 1.00 mol of an ideal gas with \(C_p = 5/2 R\) that changes temperature change from 125 K to 255 K at a constant pressure of 10.0 atm. Calculate \(\Delta U\), \(\Delta H\), \(q\), and \(w\) for this change. Solution \[q = \int_{T_1}^{T_2} nC_p dT \nonumber \] assuming \(C_p\) is independent of temperature: \[ \begin{align*} q & = nC_p \int _{T_1}^{T_2} dT \\ & = nC_p (T_2-T_1) \\ & = (1.00 \, mol) \left( \dfrac{5}{2} 8.314 \dfrac{J}{mol \, K}\right) (255\, K - 125\, K) = 2700\, J = 1.62\, kJ \end{align*} \] So via Equation \ref{heat} (specifically the integrated version of it using differences instead of differentials) \[ \Delta H = q = 1.62 \,kJ \nonumber \] \[ \begin{align*} \Delta U & = \Delta H - \Delta (pV) \\ & = \Delta H -nR\Delta T \\ & = 2700\, J - (1.00 \, mol) \left( 8.314\, \dfrac{J}{mol \, K}\right) (255\, K - 125 \,K) \\ & = 1620 \,J = 1.62\, kJ \end{align*} \] Now that \(\Delta U\) and \(q\) are determined, then work can be calculated \[\begin{align*} w & =\Delta U -q \\ & = 1.62\,kJ - 2.70\,kJ = -1.08\;kJ \end{align*} \] It makes sense that \(w\) is negative since this process is an gas expansion. Example \(\PageIndex{4}\): Isothermal Gas Expansion Calculate \(q\), \(w\), \(\Delta U\), and \(\Delta H\) for 1.00 mol of an ideal gas expanding reversibly and isothermally at 273 K from a volume of 22.4 L and a pressure of 1.00 atm to a volume of 44.8 L and a pressure of 0.500 atm. Solution Since this is an isothermal expansion, Equation\ref{isothermal} is applicable \[ \begin{align*} w & = -nRT \ln \dfrac{V_2}{V_1} \\ & = (1.00 \, mol) \left( 8.314\, \dfrac{J}{mol \, K}\right) (255\, K) \ln \left(\dfrac{44.8\,L}{22.4\,L} \right) \\ & = 1572\,J = 1.57\,kJ \\[4pt] \Delta U & = q + w \\ & = q + 1.57\,KJ \\ & = 0 \\[4pt] q &= -1.57\,kJ \end{align*} \] Since this is an isothermal expansion \[\Delta H = \Delta U + \Delta (pV) = 0 + 0 \nonumber \] where \(\Delta (pV) = 0\) due to Boyle’s Law! Adiabatic Pathways An adiabatic pathway is defined as one in which no heat is transferred (\(q = 0\)). Under these circumstances, if an ideal gas expands, it is doing work (\(w < 0\)) against the surroundings (provided the external pressure is not zero!) and as such the internal energy must drop (\(\Delta U <0 \)). And since \(\Delta U\) is negative, there must also be a decrease in the temperature (\(\Delta T < 0\)). How big will the decrease in temperature be and on what will it depend? The key to answering these questions comes in the solution to how we calculate the work done. If the adiabatic expansion is reversible and done on an ideal gas, \[dw = -pdV \nonumber \] and \[dw = nC_vdT \label{Adiabate2} \] Equating these two terms yields \[- pdV = nC_v dT \nonumber \] Using the ideal gas law for an expression for \(p\) (\(p = nRT/V\)) \[ - \dfrac{nRT}{V} dV = nC_vdT \nonumber \] And rearranging to gather the temperature terms on the right and volume terms on the left yields \[\dfrac{dV}{V} = -\dfrac{C_V}{R} \dfrac{dT}{T} \nonumber \] This expression can be integrated on the left between \(V_1\) and \(V_2\) and on the right between \(T_1\) and \(T_2\). Assuming that \(C_v/nR\) is independent of temperature over the range of integration, it can be pulled from the integrand in the term on the right. \[ \int_{V_1}^{V_2} \dfrac{dV}{V} = -\dfrac{C_V}{R} \int_{T_1}^{T_2} \dfrac{dT}{T} \nonumber \] The result is \[ \ln \left(\dfrac{V_2}{V_1} \right) = - \dfrac{C_V}{R} \ln \left( \dfrac{T_2}{T_1} \right) \nonumber \] or \[ \left(\dfrac{V_2}{V_1} \right) = \left(\dfrac{T_2}{T_1} \right)^{- \frac{C_V}{R}} \nonumber \] or \[ V_1T_1^{\frac{C_V}{R}} = V_2T_2^{\frac{C_V}{R}} \nonumber \] or \[T_1 \left(\dfrac{V_1}{V_2} \right)^{- \frac{R} {C_V}} = T_2 \label{Eq4Alternative} \] Once \(\Delta T\) is known, it is easy to calculate \(w\), \(\Delta U\) and \(\Delta H\). Example \(\PageIndex{5}\): 1.00 mol of an ideal gas (C V = 3/2 R) initially occupies 22.4 L at 273 K. The gas expands adiabatically and reversibly to a final volume of 44.8 L. Calculate \(\Delta T\), \(q\), \(w\), \(\Delta U\), and \(\Delta H\) for the expansion. Solution Since the pathway is adiabatic: \[q =0 \nonumber \] Using Equation \ref{Eq4Alternative} \[ \begin{align*} T_2 & = T_1 \left(\dfrac{V_1}{V_2} \right)^{- \frac{R} {C_V}} \\ & =(273\,K) \left( \dfrac{22.4\,L}{44.8\,L} \right)^{2/3} \\ & = 172\,K \end{align*} \] So \[\Delta T = 172\,K - 273\,K = -101\,K \nonumber \] For calculating work, we integrate Equation \ref{Adiabate2} to get \[ \begin{align*} w & = \Delta U = nC_v \Delta T \\ & = (1.00 \, mol) \left(\dfrac{3}{2} 8.314\, \dfrac{J}{mol \, K} \right) (-101\,K ) \\ & = 1.260 \,kJ \end{align*} \] \[ \begin{align*} \Delta H & = \Delta U + nR\Delta T \\ & = -1260\,J + (1.00 \, mol) \left(\dfrac{3}{2} 8.314\, \dfrac{J}{mol \, K} \right) (-101\,K ) \\ & = -2100\,J \end{align*} \] The following table shows recipes for calculating \(q\), \(w\), \(\Delta U\), and \(\Delta H\) for an ideal gas undergoing a reversible change along the specified pathway. Pathway \(q\) \(w\) \(\Delta U\) \(\Delta H\) Isothermal \(nRT \ln (V_2/V_1) \) \(-nRT \ln (V_2/V_1) \) 0 0 Isochoric \(C_V \Delta T\) 0 \(C_V \Delta T\) \(C_V \Delta T + V\Delta p\) Isobaric \(C_p \Delta T\) \(- p\Delta V\) \(C_p \Delta T - p\Delta V\) \(C_p \Delta T\) adiabatic 0 \(C_V \Delta T\) \(C_V \Delta T\) \(C_p \Delta T\) |
Courses/Mendocino_College/Introduction_to_Chemistry_(CHM_200)/02%3A_Atoms_and_Elements/2.06%3A_The_Quantum-Mechanical_Model-_Atoms_with_Orbitals | Learning Objectives Define quantum mechanics Differentiate between an orbit and an orbital. Recognize the structure of shells and subshells in an atom. Although we have discussed the general arrangement of subatomic particles in atoms, we have said little about how electrons occupy the space about the nucleus. Do they move around the nucleus at random, or do they exist in some ordered arrangement? The current model for describing the position of electrons in an atom is called the quantum mechanical model. Quantum Mechanics The study of motion of large objects such as baseballs is called mechanics, or more specifically, classical mechanics. Because of the quantum nature of the electron and other tiny particles moving at high speeds, classical mechanics is inadequate to accurately describe their motion. Quantum mechanics is the study of the motion of objects that are atomic or subatomic in size and thus demonstrate wave-particle duality. In classical mechanics, the size and mass of the objects involved effectively obscures any quantum effects, so that such objects appear to gain or lose energies in any amounts. Particles whose motion is described by quantum mechanics gain or lose energy in small pieces called quanta . One of the fundamental (and hardest to understand) principles of quantum mechanics is that the electron is both a particle and a wave. In the everyday macroscopic world of things we can see, something cannot be both. But this duality can exist in the quantum world of the submicroscopic on the atomic scale. At the heart of quantum mechanics is the idea that we cannot accurately specify the location of an electron. All we can say is that there is a probability that it exists within this certain volume of space. The scientist Erwin Schrödinger developed an equation that deals with these calculations, which we will not pursue at this time. Erwin Schrödinger. Atomic Orbitals In an atom, electrons exist in regions of space called orbitals. Quantum mechanics defines an orbital as a region of space around the nucleus where an electron has a high probability of existing, and one orbital can hold up to two electrons. There are several types of orbitals that have a specific shape and energy. Although there are many theoretical options, the atoms we encounter have electrons in 4 types of orbitals - s, p, d and f. Here are the shapes of s and p orbitals, which represent the region where we are most likely to find an electron. s Orbitals The s orbital has a spherical shape: p Orbitals For p orbitals, there are three types of orbitals that have different orientations in the atom. General Organization of Orbitals in the Atom The orbitals in an atom are organized according to their energy. Within the quantum model, Electrons in atoms can have only certain specific energies. We say that the energies of the electrons are quantized. Electrons are organized according to their energies into sets called shells (labeled by the principle quantum number, n ). Generally the higher the energy of a shell, the farther it is (on average) from the nucleus. Shells do not have specific, fixed distances from the nucleus, but an electron in a higher-energy shell will spend more time farther from the nucleus than does an electron in a lower-energy shell. Shells are further divided into subsets of electrons called subshells . The first shell has only one subshell, the second shell has two subshells, the third shell has three subshells, and so on. The subshells of each shell are labeled, in order, with the letters s , p , d , and f . Thus, the first shell has only a single s subshell (called 1 s ), the second shell has 2 s and 2 p subshells, the third shell has 3 s , 3 p , and 3 d and so forth. Shell Number of Subshells Names of Subshells 1 1 1s 2 2 2s and 2p 3 3 3s, 3p and 3d 4 4 4s, 4p, 4d and 4f Different subshells hold a different maximum number of electrons. Any s subshell can hold up to 2 electrons; p , 6; d , 10; and f , 14. Subshell Maximum Number of Electrons s 2 p 6 d 10 f 14 It is the arrangement of electrons into shells and subshells that most concerns us here, and we will focus on that in the next section. Summary Quantum mechanics involves the study of material at the atomic level. This field deals with probabilities, since we cannot definitely locate a particle. Orbitals are mathematically derived regions of space with different probabilities of having an electron. |
Courses/Ripon_College/CHM_321%3A_Inorganic_Chemistry/03%3A_Acid-Base_Chemistry_and_an_Introduction_to_Coordination_Compounds/3.06%3A_Coordination_Chemistry-_Structure_and_Isomers/3.6.04%3A_Coordination_Numbers_and_Structures | Why Do Coordination Complexes Form the Structures They Do? As with all chemical structure, coordination complexes form the structures they do so as to best stabilize the metal center and ligands through the formation of metal-ligand bonds. while avoiding destabilizing interactions like steric repulsions. The issue then is how many metal-ligand bonds should be formed and how those bonds should be arranged spatially to give the largest net stabilization possible. This question will eventually be considered in detail in connection with the nature of bonding in coordination compounds . For now, it will be helpful to think about it in terms of seven factors: The stabilizing effect of metal-ligand bond formation The driving force for complex formation is the stabilization of electrons in covalent chemical bonds. In the vast majority of cases this largely involves stabilization of the ligand lone pair is as it experiences the effective nuclear charge of the metal, although in a few involves stabilization of metal electrons by ligand nuclei (inverse ligand fields). Regardless, metal-ligand bond formation is stabilizing and classified in the way its effect is to preference the addition of ligands to the complex. Steric effects One reason coordination numbers do not increase indefinitely is that only so many ligands can fit around a metal. Exactly how many can fit depends on: Size of metal center: This is one of the more important factors and many metals tend to exhibit preferred coordination numbers, which depend on their oxidation state and size as shown in Figure \({\PageIndex{1}}\). Larger inner transition metals like the lanthanides and actinides can accommodate 9-12 sterically undemanding ligands while the smaller transition metals tend to accommodate up to six, although larger coordination numbers are more common for larger low valent metals. Thus molybdenum forms seven and eight coordinate [Mo III (CN) 7 ] 4- and [Mo IV (CN) 8 ] 4- with the sterically undemanding cyano ligand. Size and shape of ligands: As long as ligands are not excessively rigid and bulky their size is less important than the size of the metal in determining the number of ligands that coordinate. Ligands' ability to donate electrons to the metal center also tends to influence coordination number more than ligand size. However, all other things being equal for a given metal and ligand donor ability, small ligands allow higher coordination numbers while fewer bulky ligands will fit around the metal center. Ligands that are more sterically demanding in the vicinity of the metal center tend to limit the ability of other ligands to bind than those which bind through a small extended group. For example, a bulky isocyanide like tBuCN will sterically crowd the metal less than a bulky phosphine like t BuH 2 P would. For this reason the effective size of a ligand is sometimes rated in terms of either a cone angle of space they are estimated to occupy around the metal (called the Tolman cone angle, it is commonly used to evaluate phosphines' steric bulk) or the in terms of the percentage of the metal's coordination sphere the ligand occupies (called the percent buried volume, it is used to estimate the steric impact of N -heterocyclic carbenes). Repulsion between M-L bonding electrons For many complexes steric effects are neither the only effects nor the most important. Among the additional factors that should be considered are the repulsions that occur between the electrons that different ligands donate to the metal-ligand bonding. These electron-electron repulsions affect the Coordination number: When a ligand donates its electrons to a metal center to form a new metal-ligand bond, the electron density around the metal increases, raising the overall energy of the other M-L bonding electrons. This increased repulsion often limits the number of coordinated ligands. As more ligands are added the electron-electron repulsions keep increasing until the lowering of energy of the ligand electrons in the new bond is insufficient to compensate for the raising of energy of the existing M-L bonding electrons. Based on this effect alone: larger metals tend to achieve higher coordination numbers than smaller ones because the electron-electron repulsions are spread across a larger coordiantion sphere. With a given metal, more electron donating ligands have a greater tendency to form complexes with lower coordination numbers than similar neutral ones do. This is why anionic ligands (which tend to be better electron donors) tend to give lower coordination numbers than comparable neutral ligands (which tend to be weaker donors). Thus Co 2 + forms CoCl 4 2 - with chloro ligands but [Co(H 2 O) 6 ] 2+ with aqua ligands. Coordination geometry: In the Kepert model for the shapes of coordination complexes, this intraligand repulsion determines the most stable coordination geometry by causing the ligands to move as far apart form one another on the metal's coordination sphere as possible. Formally, according to the Kepert model: any of the metal's valence electrons not involved in metal-ligand bonds occupy (n-1) d orbitals and function as core electrons. As core electrons they do not influence the molecular shape electrons involved in bonding to a given ligand constitute an electron group that repels all other electron groups around the metal. all other things being equal the complex will form the geometry that minimizes the repulsions between electron groups. Notice the similarities of these postulates to those of VSEPR theory . In predicting coordination geometries in terms of electron-electron repulsions, the Kepert model is just an extension of VSEPR theory to coordination compounds. The difference between VSEPR theory and the Kepert model is that in the Kepert model only electrons involved in metal-ligand bonds count. Coordination geometries predicted by the Kepert model for coordination numbers two through nine are given in Figure \({\PageIndex{2}}\). As may be seen from the geoemtries listed in Figure \({\PageIndex{2}}\), these are just equivalent to VSEPR geometries for cases in which the number of electron groups is equal to the coordination number. The difference between optimal and suboptimal coordination geometries is greater with few ligands and becomes smaller and ligands become increasingly dispersed across the metal's coordination sphere. In complexes containing five, seven, eight, or higher coordinate metals a number of geometries that are similar in energy to the preferred geometry. These geometries, which should be regarded as accessible, are also listed in Figure \({\PageIndex{2}}\). Ligand field effects A few coordination geometries are noticeably absent from the Kepert-preferred and Kepert-accessible geometries in Figure \(\PageIndex{2}\). These include the trigonal prismatic geometries formed by compounds like W(CH 3 ) 6 and the very common square planar geometry illustrated by complexes like [PtCl 4 ] 2- and [IrClH(PPh 3 ) 2 ]. One of the reasons the Kepert model fails to predict the existence of such structures is its neglect of directional interactions involving d electrons on the metal center. Metal d electrons exert on profound influence on almost all properties of transition metal complexes, including their structures. The way in which this occurs will be explored at length in the next chapter . For now, it is enough to note that both the ligand-donated electrons surrounding a metal center and the electrons occupying particular d orbitals on that metal are oriented in specific directions relative to one another. Because of this the strength of the interactions between the ligand and metal d electrons depend on both the number of d electrons present, how strongly metal-ligand binding affects their energy, and how the ligands are arranged about the metal center. The impact of these effects, here termed ligand field effects, differ from case to case and can include distortions of the complex's geometry. For instance, an ideal octahedral coordination geometry might be tetragonally distorted by flattening or elongating it. imparting a strong preference for a non-Kepert coordination geometries. This is why, for example, 2nd and 3rd row complexes in which the metal has a d 8 electron configuration are almost always square planar. stabilizing non-Kepert geometries enough to permit complexes to adopt them in the presence of a rigid or semirigid ligand that prefers to coordinate the metal in that geometry. 3 Because of these effects square planar and trigonal prismatic geometries are also observed and the list of coordination geometries given in Figure \(\PageIndex{2}\) may be extended to that shown in Figure \(\PageIndex{3}\). Ligand geometry constraints Rigid or semirigid ligands influence the coordination geometry of metal complexes in two main ways: Bulky rigid ligands that crowd the metal center prevent other ligands from binding. Such ligands are useful for preparing low-coordinate complexes. Rigid and semirigid ligands can impose their preferred coordination geometry on a metal center. This is because these ligands energetically prefer to adopt a particular conformation when they bind a metal center. In doing so they shift the coordination geometry energy landscape toward that preferred geometry. If the shift is large enough relative to the native preference due to ligand repulsion and ligand field effects the complex will either adopt the ligand-preferred geometry or be distorted in the direction of the ligand-preferred geometry. Examples are given in Figure \(\PageIndex{4}\). The influence of ligands on coordination geometry is important in living systems in which proteins and nucleic acids can act as rigid or semirigid ligands. The ability of these ligands to distort the coordination geometries of metal atoms in ways that enable them to perform specific functions is so common that the resulting distorted geometries are termed entactic states. A particulary spectacular case of an entactic state involves the blue copper proteins azurin and plastocyanin, the structure of which is given in Figure \(\PageIndex{5}\). As may be seen from the structure in Figure \(\PageIndex{5}\), the copper in plastocyanin exibits a distorted tetrahedral coordination geometry. The protein is said to act like a medieval torture device called a rack in stretching the metal into its distorted geometry. This distortion makes its easier for the copper center to undergo facile redox reactions, enabling it to better function as an electron carrier. Crystal packing effects This effect is similar to that of ligand constraints except that in this case arise not from the structure internal to a ligand but out of the forces involves in maximizing the stabilization energy of a crystal. With lower coordination number complexes packing effects can shift the conformations of flexible ligands but only give rise to very small distortions of the overall coordination geometry. Packing effects can drive a shift in the overall coordination geometry of higher coordination number complexes, for which packing effects are significant relative to the small difference in energy between geometries. Thus while [Mo(CN) 8 ] 4- has a square antiprismatic coordination geometry in solution it exhibits a docecahedral coordination geometry in the crystals of many of its salts. Relatavistic effects on orbital energies The proximity of fast moving electrons to massive nuclei in the heavier transition elements results in relatavistic expansion and contraction of orbitals. The net results are that heavier elements tend to be smaller than expected. This effect preferences lower coordination numbers. the relative energies of orbitals shift. Orbitals which become contracted are lowered in energy while those which are expanded increase in energy, as shown for the case of gold in Figure \(\PageIndex{6A}\). The combination of smaller sizes and altered orbital energies affect coordination preferences. Relatavistic effects contribute to the greater tendency of Au(I) relative to other group 11 metals to form linear two coordinate complexes. As shown in Figure \(\PageIndex{6B}\), the relative closeness in energy of the 6 s and 5 d orbitals of gold makes mixing of these orbitals more favorable, facilitating the ability of gold to form two coordinate complexes with strong sigma bonds oriented 180° from one another. What Structures do Coordination Complexes Form? Metal complexes with coordination numbers ranging from one to 16 are known, although values greater than seven are rare for the transition metals. In this section examples of common coordination geometries will be presented in order of coordination number. Coordination Number 1 Condensed phase monocoordinate complexes are unknown for the transition metals, although the post transition metals Tl and In form monocoordinate complexes with the bulky ligands triazapentadienyl and 2,6-tris(2,4,6-triisoprophylphenyl)benzene as shown in Figure \(\PageIndex{7}\). Figure \(\PageIndex{7}\). No monocoordiante transition metal complexes are known but Tl + and In + form monocoordinate complexes with extremely bulky ligands. This work by Stephen Contakes is licensed under a Creative Commons Attribution 4.0 International License . Coordination Number 2 This coordination number is rare outside of d 10 complexes of the group 11 metals and mercury, specifically, Cu + , Ag + , Au + , and Hg 2 + . In accordance with the predictions of the Kepert model these give linear complexes. Among these: Cu + more commonly gives tetrahedral complexes but can be coaxed to give linear ones. The most prominent example is [CuCl 2 ] - , which forms when CuCl is treated with concentrated HCl under anerobic conditions. Ag + also commonly forms tetrahedral or trigonal planar complexes but can give linear ones. The most prominent example is [Ag(NH 3 ) 2 ] + , which can be formed by treating silver salts with concentrated aqueous or liquid ammonia. Au + almost always forms linear complexes but many of these formally two coordinate complexes associate as depicted in Figure \(\PageIndex{8}\). The ability of Au + to form linear complexes with cyanide is even used to selectively extract metallic gold from low grade ores. The stability of [Au(CN) 2 ] - means that the dissolution of metallic gold in aqueous cyanide is thermodynamically favorable under aerobic conditions. \[\sf{4~Au~~+~~8~CN^-~~+~~O_2~~+~~2~H_2O~~\longrightarrow~~4~[Au(CN)_2]^-~~+~~4~OH^-}\] Hg 2 + , like Au + , benefits from relativistic effects and more commonly forms two-coordinate complexes with a linear geometry. Among these is [Hg(CN) 2 ]. However, its preference for linearity is not as rigid as for Au + and so complexes with a variety of coordination geometries are known. An by means of honorary mention the mercury(I),Hg 2 2 + , forms linear complexes of the type L-Hg-Hg-L, although since Hg 2 2 + is often considered as a single unit these aren't always considered to be two coordinate complexes. Two coordinate complexes may also be formed through the use of bulky ligands that only allow for the binding of two to the metal center. Classic examples are given in Figure \(\PageIndex{9}\). Coordination Number 3 Three coordinate complexes are similar to two coordinate ones in that they are rare and, aside from the constraining influence of ligands, usually limited to d 10 metal ions such as Cu + , Ag + , Au + , Hg 2 + , and Pt 0 . As expected from the Kepert model, in the absence of constraining ligands three coordinate complexes are trigonal planar. Many of the examples involve soft ligands . They include [Cu(SPMe 3 ) 3 ] + , [Ag(PPh 3 ) 3 ] + , [Au(PPhCy 2 ) 3 ] + , HgI 3 - , and [Pt(PPh 3 ) 3 ]. Non- d 10 metal centers have successfully been constrained to adopt a three coordinate trigonal planar geometry using bulky and semirigid ligands. Examples of such three-coordinate complexes are given in Figure \(\PageIndex{10}\). Coordination Number 4 Four is one of the most common coordination numbers found in transition metal complexes. The two common four coordinate geometries are tetrahedral and square planar. Tetrahedral complexes are commonly formed by metals possessing either a d 0 or d 10 electron configuration. Monometallic examples of the former include TiCl 4 , VO 4 3 - , WS 4 2 - , MnO 4 - , CrO 4 2 - , and OsO 4 while d 10 examples are [Ni(CO) 4 ], [HgBr 4 ] 2- , [ZnCl 4 ] 2- , [CdI 4 ] 2- . Otherwise tetrahedral complexes are known but much less common. Examples usually involve good donor ligands and include [FeCl 4 ] - (d 5 ), [CoCl 4 2 - ] (d 6 ), and [NiCl 4 ] 2- (d 7 ). Second and third row transition metal centers with d 8 electron configurations like Rh + , Ir + , Pd 2 + , and Pt 2 + , Au 3 + almost exclusively exhibit square planar geometries. Beyond this, square planar geometries are often formed by Ni 2 + (d 8 ), Ni 3 + (d 7 ), and Cu 2 + (d 9 ). Examples of square planar complexes include [Cu(acac) 2 ]; [PtCl 4 ] 2- ;Wilkinson's catalyst, [RhCl(PPh 3 ) 3 ]; and Vaska's complex, trans -[Ir(CO)Cl(PPh 3 ) 2 ]. Porphyrins, phthalocyanines, and other rigid macrocycles can also impose a square planar geometry on coordinated metals. Coordination Number 5 The two common coordination geometries for five coordinate complexes are trigonal bipyramidal and square pyramidal. Five coordinate complexes are rare. Complexes that exhibit trigonal bipyramidal geometry include [Fe(CO) 5 ] and [CuCl 5 ] 3- and complexes containing tripodal ligands that preference formation of a trigonal pyramidal structure with an open coordination site, as shown in Figure \(\PageIndex{11}\). Homoleptic [Ni(CN) 5 ] 3- possesses a square pyramidal structure, although the geometry is more common for macrocyclic complexes like the iron protoporphyrin of deoxymyoglobin shown in Figure \(\PageIndex{4}\) and for complexes containing oxo and nitrido ligands, examples of which are shown in Figure \(\PageIndex{12}\). In the absence of rigid constraining ligands the relatively low energy difference between the trigonal bipyramidal and square pyramidal coordination geometries provides a mechanism for interconversion of the axial and equatorial ligands in a trigonal planar complex. For example, pentacarbonyliron(0) exhibits fluxionality involving a square pyramidal intermediate via a Berry pseudorotation mechanism, as shown in Figure \(\PageIndex{13}\) Coordination Number 6 The two common octahedral coordination geometries are octahedral and trigonal prismatic. Octahedral Of these octahedral is by far the most common mode of coordination exhibited by transition metals. Most metals can be induced to form octahedral complexes and examples exist for every d electron configuration ranging from d 0 to d 10 . Among these are the 28 metal oxidation states listed in Figure \(\PageIndex{1}\) for which octahedral is the most common coordination geometry. Because of this it is worth pointing out that octahedral complexes possessing d 4 , d 7 , or d 9 electron configurations commonly exhibit tetragonal distortions depicted in Figure \(\PageIndex{14}\). Trigonal prismatic Trigo nal prismatic coordination is related to octahedral coordination as shown in Figure \(\PageIndex{15}\). As may be seen in Figure \(\PageIndex{15}\), an octahedral coordination sphere is just a trigonal antiprism in which all edge lengths are identical. Rotation of one triangular face relative to its opposite until the two are eclipsed gives a triganal prismatic geometry. In fact, since continuation of this rotation gives another octahedral complex the trigonal prismatic geometry is an intermediate in isomerization reactions involving octahedral complexes. In contrast to octahedral coordination geometries, trigonal prismatic coordination (and distorted versions thereof) are rare and occur mostly for d 0 , d 1 , and d 2 configurations. Examples of trigonal prismatic metal centers include the d 2 Mo 4 + centers in MoS 2 , d 1 [Re(S 2 C 2 Ph 2 ) 3 ] - , and d 0 [Ta(CH 3 ) 6 ] - , of which the latter two structures are given in Figure \(\sf{\PageIndex{16}}\). Semirigid ligands like that shown in Figure \(\PageIndex{4C}\) may be used to encourage the adoption of a trigonal prismatic geometry, although once the number of d electrons present the preference for octahedral coordination is too great for a trigonal prismatic geometry to occur. Coordination Number 7 Seven coordinate complexes are rare outside of the relatively large early transition metals, lanthanides, and actinides. The three common seven coordinate geometries are pentagonal bipyramidal, monocapped octahedral, and monocapped trigonal prismatic. The latter two are often called capped octahedral, and capped trigonal prismatic, with the mono- prefix being understood. Although intraligand repulsions are smaller in the pentagonal bipyramidal coordination geometry than the capped octahedral and capped trigonal prismatic geometries the difference is small and the three structures are often close in energy. As a result the structure observed is often dependent on ligand-based constraints, crystal packing, and solvent effects that preference one geometry over the others. Heptacyano complexes are often pentagonal bipyramidal. Examples include [Mo(CN) 7 ] 3- , [W(CN) 7 ] 3- , and [Os(CN) 7 ] 3- . Seven coordinate complexes containing oxo ligands commonly are pentagonal bipyramidal with the oxo ligand(s) in the less sterically hindered axial position. Examples include [NbOF 6 ] 3- and, for the inner transition metals, [UO 2 F 5 ] 3- . Ligands that have been used to promote formation of seven-coordinate species include 15-crown-5 and 2,2':6',2'':6'',2'''-quaterpyridine and 15-crown-5. Representative complexes are given in (Figure \(\sf{\PageIndex{17}}\). Capped trigonal prismatic geometries are common for complexes of the early transition metals. Examples include [NbF 7 ] 2- ,[TaF 7 ] 2- , and [ZrF 7 ] 3- in (NH 4 ) 3 [ZrF 7 ]. Capped octahedral geometries are found in [MoMe 7 ] - , [WMe 7 ] - , and [WBr 3 (CO) 4 ], which contains three pairs of trans -Br and CO with the final CO capping the octahedron's (CO) 3 face, as shown in Figure \(\PageIndex{18}\). In seven and higher coordinate complexes ligand and crystal packing effects frequently give distorted coordination geometries. These geometries are intermediate between two or more of the idealized seven coordinate geometries, making it difficult to tell exactly which structure they are a distortion of (Figure \(\sf{\PageIndex{19}}\). Coordination Number 8 Eight coordinate complexes are rare and occurs in discrete molecules and ions only for the relatively large early transition metals, lanthanides, actinides. The three common eight coordinate geometries are square antiprismatic, dodecahedral, and bicapped trigonal prismatic. In contrast, the cubic coordination geometry is only found in ionic lattices like that of CsCl and in complexes of the inner transition metals such as Na 3 [UF 8 ]. The simplest structure is the cube, which is rare because it does not minimize interligand repulsive interactions. Square antiprismatic, dodecahedral, and bicapped trogonal prismatic coordination are more common. Of these the square antirpismatic and dodecahedral geometries can both be considered as distorted cubic structures. As shown in Figure \(\PageIndex{20}\), the square antiprism may be made by twisting one face of a cube relative to one another and the dodecahedron by folding opposing faces towards one another. As with other high-coordinate structures the energy difference between these eightfold coordination geometries is small enough that packing effects can significantly influnece the observed structure. For example, octacyanomolybdates commonly adopt a square antiprismatic coordination geometry but depending on the counterions present can give dodecahedral or bicapped trigonal prismatic complexes. Examples are given in Figure \(\sf{\PageIndex{21}}\). Coordination Number 9 Again, nine coordinate complexes typically require larger transition metals, lanthanides, and actinides. Coordination geometries are typically either tricapped trigonal prismatic or idosynchraitcally determined by the ligands. Simpleexamples include the aqua complexes [Sc(H 2 O) 9 ] 3+ , [Y(H 2 O) 9 ] 3+ , and [La(H 2 O) 9 ] 3+ as well as [TcH 9 ] 2- and [ReH 9 ] 2- . The classic example of a nine-coordinate complex, [ReH 9 ] 2- , is shown in Figure \(\PageIndex{22}\). Coordination Numbers 10-16 Coordination numbers higher than nine are extremely rare for compounds that bind in \(\kappa\) fashion (form conventional metal-ligand bonds) 14 and usually involve some combination of large metals, sterically undemanding ligands, and special ligand structures that promote higher coordination. Noteworthy examples include Twelve-coordinate [Hf(BH 4 ) 4 ], which illustrates how small multidentate ligands promote higher coordination numbers. As shown in Figure \(\PageIndex{23}\), [Hf(BH 4 ) 4 ] has a cubooctahedral structure in which BH 4 - acts as a tridentate ligand, with BH 3 units occupying triangular faces of the cubooctahedron to give a tetrahedron of BH 4 - ligands around the Hf. Twelve coordinate [Ce(NO 3 ) 6 ] 2- , in which the nitrate oxygens define an icosahedral coordination geometry as shown in Figure \(\sf{\PageIndex{24}}\). The nitrates in the structure bind the Ce center in bidentate fashion in an octahedral array. Fifteen-coordinate, [Th(H 3 BNMe 2 BH 3 ) 4 ], which also uses bridging H-B-H units that occupy little of the coordination sphere. In [Th(H 3 BNMe 2 BH 3 ) 4 ] three of the four H 3 BNMe 2 BH 3 bind in \(\kappa\) 4 fashion and one binds \(\kappa\) 3 , giving the fifteen fold coordination. 16 Sixteen-coordinate [CoB 16 ] − , which possesses the highest coordination number yet observed. Its structure is given in Figure \(\sf{\PageIndex{25}}\) The coordination geometry is an octahedral antiprism and the complex should be considered as involving a Co center in the midst of a B 16 - "molecular drum" held together by cluster bonds. References 1. Dudev, M.; Wang, J.; Dudev, T.; Lim, C., Factors Governing the Metal Coordination Number in Metal Complexes from Cambridge Structural Database Analyses. The Journal of Physical Chemistry B 2006 , 110(4) , 1889-1895. 2. Kuppuraj, G.; Dudev, M.; Lim, C., Factors Governing Metal−Ligand Distances and Coordination Geometries of Metal Complexes. The Journal of Physical Chemistry B 2009, 113 (9), 2952-2960. 3. Cremades, E.; Echeverría, J.; Alvarez, S., The Trigonal Prism in Coordination Chemistry. Chemistry – A European Journal 2010, 16 (34), 10380-10396. 4. Xiong, X.-G.; Wang, Y.-L.; Xu, C.-Q.; Qiu, Y.-H.; Wang, L.-S.; Li, J., On the gold–ligand covalency in linear [AuX 2 ] − complexes. Dalton Transactions 2015, 44 (12), 5535-5546. 5. Concepción Gimeno, M. The Chemistry of Gold in Laguna, Antonio (ed.) Modern Supramolecular Gold Chemistry: Gold-Metal Interactions and Applications . Wiley, 2008. 6. Andersen, R. A.; Faegri, K.; Green, J. C.; Haaland, A.; Lappert, M. F.; Leung, W. P.; Rypdal, K., Synthesis of bis[bis(trimethylsilyl)amido]iron(II). Structure and bonding in M[N(SiMe3)2]2 (M = manganese, iron, cobalt): two-coordinate transition-metal amides. Inorganic Chemistry 1988, 27 (10), 1782-1786. 7. Persson, I., Hydrated metal ions in aqueous solution: How regular are their structures? Pure and Applied Chemistry 2010, 8 2(10) , 1901. 8. Aramburu, J. A.; García-Fernández, P.; García-Lastra, J. M.; Moreno, M., Jahn–Teller and Non-Jahn–Teller Systems Involving CuF 6 4 – Units: Role of the Internal Electric Field in Ba2ZnF6:Cu2+ and Other Insulating Systems. The Journal of Physical Chemistry C 2017, 121(9) , 5215-5224. 9. Brown, M. D.; Levason, W.; Murray, D. C.; Popham, M. C.; Reid, G.; Webster, M., Primary and secondary coordination of crown ethers to scandium(iii). Synthesis, properties and structures of the reaction products of ScCl3(thf)3, ScCl3·6H2O and Sc(NO3)3·5H2O with crown ethers. Dalton Transactions 2003, (5), 857-865. 10. Liu, Y.; Ng, S.-M.; Lam, W. W. Y.; Yiu, S.-M.; Lau, T.-C., A Highly Reactive Seven-Coordinate Osmium(V) Oxo Complex: [OsV(O)(qpy)(pic)Cl]2+. Angewandte Chemie International Edition 2016, 55 (1), 288-291. 11. Popov, I., Jian, T., Lopez, G. et al. Cobalt-centred boron molecular drums with the highest coordination number in the CoB 16 − cluster. Nat Commun 6, 8654 (2015). https://doi.org/10.1038/ncomms9654 12. The structures are rendered from cif data reported in the following publications (A) square antiprismatic [Mo(CN) 8 ] 3- : Wen-Yan Liu, Hu Zhou, Ai-Hua Yuan, Acta Crystallographica Section E: Structure Reports Online , 2008, 64, m1151, (B) dodecahedral [Mo(CN) 8 ] 4- : B.J.Corden, J.A.Cunningham, R.Eisenberg, Inorganic Chemistry , 1970, 9, 356. 13. The structure of ReH 9 2 - is rendered from the structure reported in Abrahams, S.C.; Ginsberg, A.P.; Knox, K. Transition metal-hydrogen compounds. II. The crystal and molecular structure of potassium rhenium hydride, K 2 ReH 9 Inorganic Chemistry , 1964 , 3 , 558-567. 14. There are other complexes in which a metal may be said to interact with more than sixteen "ligand atoms" but these are not usually considered to possess a higher coordination number. For example in some \(\pi\) complexes like \(\eta\) 5 -Cp 4 U technically there are 20 C atoms fastened to the U but these complexes are better considered as 12 coordinate than twenty (since each cyclopentadienl ring is isolobal with a fac coordinated set of 3 L ligands) while the metal centers in endohedral fullerene species like La@C 60 do not interact with all sixty carbon atoms at once are so are better thought of as a metal trapped in a spacious sixty-carbon cage. 15. Zalkin, A.; Forrester, J.D.; Templeton, D.H. Crystal structure of cerium magnesium nitrate hydrate Journal of Chemical Physics , 1963 , 39 , 2881-2891. 16. Daly, S. R.; Piccoli, P. M. B.; Schultz, A. J.; Todorova, T. K.; Gagliardi, L.; Girolami, G. S., Synthesis and Properties of a Fifteen-Coordinate Complex: The Thorium Aminodiboranate [Th(H3BNMe2BH3)4]. Angewandte Chemie International Edition 2010, 49 (19), 3379-3381. 17. Popov, I., Jian, T., Lopez, G., Boldyrev, A. I.; Wang, L-S. Cobalt-centred boron molecular drums with the highest coordination number in the CoB 16 − cluster. Nat Commun 6, 8654 (2015). Contributors Stephen Contakes, Westmont College , to whom comments, corrections, and criticisms should be addressed. Consistent with the policy for original artwork made as part of this project, all unlabeled drawings of chemical structures are by Stephen Contakes and licensed under a Creative Commons Attribution 4.0 International License . |
Bookshelves/General_Chemistry/Concept_Development_Studies_in_Chemistry_(Hutchinson)/10%3A_Energetics_of_Chemical_Reactions | Foundation We begin our study of the energetics of chemical reactions with our understanding of mass relationships, determined by the stoichiometry of balanced reactions and the relative atomic masses of the elements. We will assume a conceptual understanding of energy based on the physics of mechanics, and in particular, we will assume the law of conservation of energy. In developing a molecular understanding of the reaction energetics, we will further assume our understanding of chemical bonding via valence shell electron pair sharing and molecular orbital theory. Goals The heat released or consumed in a chemical reaction is typically amongst the most easily observed and most readily appreciated consequences of the reaction. Many chemical reactions are performed routinely specifically for the purpose of utilizing the heat released by the reaction. We are interested here in an understanding of the energetics of chemical reactions. Specifically, we wish to know what factors determine whether heat is absorbed or released during a chemical reaction. With that knowledge, we seek to quantify and predict the amount of heat anticipated in a chemical reaction. We expect to find that the quantity of heat absorbed or released during a reaction is related to the bonding of the molecules involved in the reaction. Prior to answering these questions, we must first answer a few questions regarding the nature of heat. Despite our common familiarity with heat (particularly in Houston), the concept of heat is somewhat elusive to define. We recognize heat as "whatever it is that makes things hot", but this definition is too imprecise to permit measurement or any other conceptual progress. Exactly how to we define and measure heat? Observation 1: Measurement of Heat by Temperature We can define in a variety of ways a temperature scale which permits quantitative measurement of "how hot" an object is. Such scales are typically based on the expansion and contraction of materials, particularly of liquid mercury, or on variation of resistance in wires or thermocouples. Using such scales, we can easily show that heating an object causes its temperature to rise. It is important, however, to distinguish between heat and temperature. These two concepts are not one and the same. To illustrate the difference, we begin by measuring the temperature rise produced by a given amount of heat, focusing on the temperature rise in \(1000 \: \text{g}\) of water produced by burning \(1.0 \: \text{g}\) of methane gas. We discover by performing this experiment repeatedly that the temperature of this quantity of water always rises by exactly \(13.3^\text{o} \text{C}\). Therefore, the same quantity of heat must always be produced by reaction of this quantity of methane. If we burn \(1.0 \: \text{g}\) of methane to heat \(500 \: \text{g}\) of water instead, we observe a temperature rise of \(26.6^\text{o} \text{C}\). If we burn \(1.0 \: \text{g}\) of methane to heat \(1000 \: \text{g}\) of iron, we observe a temperature rise of \(123^\text{o} \text{C}\). Therefore, the temperature rise observed as a function of the quantity of material heated as well as the nature of the material heated. Consequently, \(13.3^\text{o} \text{C}\) is not an approximate measure of this quantity of heat, since we cannot say that the burning of \(1.0 \: \text{g}\) of methane "produces" \(13.3^\text{o} \text{C}\) of heat. Such a statement is clearly revealed to be nonsense, so the concepts of temperature and heat must be kept distinct. Our observations do reveal that we can relate the temperature rise produced in a substance to a fixed quantity of heat, provided that we specify the type and amount of the substance. Therefore, we define a property for each substance, called the heat capacity , which relates the temperature rise to the quantity of heat absorbed. We define \(q\) to be the quantity of heat, and \(\Delta T\) to be the temperature rise produced by this heat. The heat capacity \(C\) is defined by \[q = C \Delta T\] This equation, however, is only a definition and does not help us calculate either \(q\) or \(C\), since we know neither one. Next, however, we observe that we can also elevate the temperature of a substance mechanically , that is, by doing work on it. As simple examples, we can warm water by stirring it, or warm metal by rubbing or scraping it. (As a historical note, these observations were crucial in establishing that heat is equivalent to work in its effect on matter, demonstrating that heat is therefore a form of energy.) Although it is difficult to do, we can measure the amount of work required to elevate the temperature of \(1 \: \text{g}\) of water by \(1^\text{o} \text{C}\). We find that the amount of work required is invariably equal to \(4.184 \: \text{J}\). Consequently, adding \(4.184 \: \text{J}\) of energy to \(1 \: \text{g}\) of water must elevate the energy of the water molecules by an amount measured by \(1^\text{o} \text{C}\). By conservation of energy, the energy of the water molecules does not depend on how that energy was acquired. Therefore, the increase in energy measured by a \(1^\text{o} \text{C}\) temperature increase is the same regardless of whether the water was heated or stirred. As such, \(4.184 \: \text{J}\) must also be the amount of energy added to the water molecules when they are heated by \(1^\text{o} \text{C}\) rather than stirred. We have therefore effectively measured the heat \(q\) required to elevate the temperature of \(1 \: \text{g}\) of water by \(1^\text{o} \text{C}\). Referring back to the definition of heat capacity, we now can calculate that the heat capacity of \(1 \: \text{g}\) of water must be \(4.184 \: \frac{\text{J}}{^\text{o} \text{C}}\). The heat capacity per gram of a substance is referred to as the specific heat of the substance, usually indicated by the symbol \(c_s\). The specific heat of water is \(4.184 \: \frac{\text{J}}{^\text{o} \text{C}}\). Determining the heat capacity (or specific heat) of water is an extremely important measurement for two reasons. First, from the heat capacity of water we can determine the heat capacity of any other substance very simply. Imagine taking a hot \(5.0 \: \text{g}\) iron weight at \(100^\text{o} \text{C}\) and placing it in \(10.0 \: \text{g}\) of water at \(25^\text{o} \text{C}\). We know from experience that the iron bar will be cooled and the water will be heated until both have achieved the same temperature. This is an easy experiment to perform, and we find that the final temperature of the iron and water is \(28.8^\text{o} \text{C}\). Clearly, the temperature of the water has been raised by \(3.8^\text{o} \text{C}\). From the definition of heat capacity and the specific heat of water, we can calculate that the water must have absorbed an amount of heat \(q = \left( 10.0 \: \text{g} \right) \left( 4.184 \: \frac{\text{J}}{\text{g} ^\text{o} \text{C}} \right) \left( 3.8^\text{o} \text{C} \right) = 1.59 \: \text{J}\). By conservation of energy, this must be the amount of heat lost by the \(1 \: \text{g}\) iron weight, whose temperature was lowered by \(71.2^\text{o} \text{C}\). Again referring to the definition of heat capacity, we can calculate the specific heat of the iron bar to be \(c_s = \frac{-159 \: \text{J}}{\left( -71.2^\text{o} \text{C} \right) \left( 5.0 \: \text{g} \right)} = 0.45 \: \frac{\text{J}}{\text{g} ^\text{o} \text{C}}\). Following this procedure, we can easily produce extensive tables of heat capacities for many substances. Second, and perhaps more importantly for our purposes, we can use the known specific heat of water to measure the heat released in any chemical reaction. To analyze a previous example, we observed that the combustion of \(1.0 \: \text{g}\) of methane gas released sufficient heat to increase the temperature of \(1000 \: \text{g}\) of water by \(13.3^\text{o} \text{C}\). The heat capacity of \(1000 \: \text{g}\) of water must be \(\left( 1000 \: \text{g} \right) \left( 4.184 \: \frac{\text{J}}{\text{g} ^\text{o} \text{C}} \right) = 4184 \: \text{J}{^\text{o} \text{C}}\). Therefore, by the definition of heat capacity, elevating the temperature of \(1000 \: \text{g}\) of water by \(13.3^\text{o} \text{C}\) must require \(55,650 \: \text{J} = 55.65 \: \text{kJ}\) of heat. The method of measuring reaction energies by capturing the heat evolved in a water bath and measuring the temperature rise produced in that water bath is called calorimetry . This method is dependent on the equivalence of heat and work as transfers of energy, and on the law of conservation of energy. Following this procedure, we can straightforwardly measure the heat released or absorbed in any easily performed chemical reaction. For reactions which are difficult to initiate or which occur only under restricted conditions or which are exceedingly slow, we will require alternative methods. Observation 2: Hess' Law of Reaction Energies Hydrogen gas, which is of potential interest nationally as a clean fuel, can be generated by the reaction of carbon (coal) and water: \[\ce{C} \left( s \right) + 2 \ce{H_2O} \left( g \right) \rightarrow \ce{CO_2} \left( g \right) + 2 \ce{H_2} \left( g \right)\] Calorimetry reveals that this reaction requires the input of \(90.1 \: \text{kJ}\) of heat for every mole of \(\ce{C} \left( s \right)\) consumed. By convention, when heat is absorbed during a reaction, we consider the quantity of heat to be a positive number: in chemical terms, \(q > 0\) for an endothermic reaction. When heat is evolved, the reaction is exothermic and \(q < 0\) by convention. It is interesting to ask where this input energy goes when the reaction occurs. One way to answer this question is to consider the fact that the reaction converts one fuel, \(\ce{C} \left( s \right)\), into another, \(\ce{H_2} \left( g \right)\). To compare the energy available in each fuel, we can measure the heat evolved in the combustion of each fuel with one mole of oxygen gas. We observe that \[\ce{C} \left( s \right) + \ce{O_2} \left( g \right) \rightarrow \ce{CO_2} \left( g \right)\] produces \(393.5 \: \text{kJ}\) for one mole of carbon burned; hence \(q = -393.5 \: \text{kJ}\). The reaction \[2 \ce{H_2} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{H_2O} \left( g \right)\] produces \(483.6 \: \text{kJ}\) for two moles of hydrogen gas burned, so \(q = -483.6 \: \text{kJ}\). It is evident that more energy is available from combustion of the hydrogen fuel than from combustion of the carbon fuel, so it is not surprising that conversion of the carbon fuel to hydrogen fuel requires the input of energy. Of considerable importance is the observation that the heat input in the reaction of coal and water, \(90.1 \: \text{kJ}\) is exactly equal to the difference between the heat evolved, \(-393.5 \: \text{kJ}\), in the combustion of carbon and the heat evolved, \(-483.6 \: \text{kJ}\), in the combustion of hydrogen. This is not a coincidence: if we take the combustion of carbon and add it to the reverse of the combustion of hydrogen, we get \[\begin{align} \ce{C} \left( s \right) + \ce{O_2} \left( g \right) &\rightarrow \ce{CO_2} \left( g \right) \\ 2 \ce{H_2O} \left( g \right) &\rightarrow 2 \ce{H_2} \left( g \right) + \ce{O_2} \left( g \right) \\ \ce{C} \left( s \right) + \ce{O_2} \left( g \right) + 2 \ce{H_2O} \left( g \right) &\rightarrow \ce{CO_2} \left( g \right) + 2 \ce{H_2} \left( g \right) + \ce{O_2} \left( g \right) \end{align}\] Canceling the \(\ce{O_2} \left( g \right)\) from both sides, since it is net neither a reactant nor product. The final equation is equivalent to the reaction of carbon and water presented earlier. Thus, taking the combustion of carbon and "subtracting" the combustion of hydrogen (or more accurately, adding the reverse of the combustion of hydrogen) yields the final equation above. And, the heat of the combustion of carbon minus the heat of the combustion of hydrogen equals the heat of the reaction between carbon and water. By studying many chemical reactions in this way, we discover that this result, known as Hess' Law , is general. Hess' Law The heat of any reaction is equal to the sum of the heats of reaction for any set of reactions which in sum are equivalent to the overall reaction . (Although we have not considered the restriction, applicability of this law requires that all reactions considered proceed under similar conditions: we will consider all reactions to occur at constant pressure.) A pictorial view of Hess' Law as applied to the heat of the reaction of carbon and water is illustrative. In Figure 10.1, the reactants \(\ce{C} \left( s \right) + 2 \ce{H_2O} \left( g \right)\) are placed together in a box, representing the state of the materials involved in the reaction prior to the reaction. The products \(\ce{CO_2} \left( g \right) + 2 \ce{H_2} \left( g \right)\) are placed together in a second box representing the state of the materials involved after the reaction. The reaction arrow connecting these boxes is labeled with the heat of this reaction. Now we take these same materials and place them in a third box containing \(\ce{C} \left( s \right)\), \(\ce{O_2} \left( g \right)\), and \(2 \ce{H_2} \left( g \right)\). This box is connected to the reactant and product boxes with reaction arrows, labeled by the heats of reaction in the combustion of carbon and the combustion of hydrogen. Figure 10.1: A pictorial view of Hess' Law. This picture of Hess' Law reveals that the heat of reaction along the "path" directly connecting the reactant state to the product state is exactly equal to the total heat of reaction along the alternative "path" connecting reactants to products via the intermediate state containing \(\ce{C} \left( s \right)\), \(\ce{O_2} \left( g \right)\), and \(2 \ce{H_2} \left( g \right)\). A consequence of our observation of Hess' Law is therefore that the net heat evolved or absorbed during a reaction is independent of the path connecting the reactant to product. (This statement is again subject to our restriction that all reactions in the alternative path must occur under constant pressure conditions.) A slightly different view of Figure 10.1 results from beginning at the reactant box and following a complete circuit through the other boxes leading back to the reactant box, summing the net heats of reaction as we go. We discover that the net heat transferred (again provided that all reactions occur under constant pressure) is exactly zero. This is a statement of the conservation of energy: the energy in the reactant state does not depend upon the processes which produced that state. Therefore, we cannot extract any energy from the reactants by a process which simply recreates the reactants. Were this not the case, we could endlessly produce unlimited quantities of energy by following the circuitous path which continually reproduces the initial reactants. By this reasoning, we can define an energy function whose value for the reactants is independent of how the reactant state was prepared. Likewise, the value of this energy function in the product state is independent of how the products are prepared. We choose this function, \(H\), so that the change in the function, \(\Delta H = H_\text{products} - H_\text{reactants}\), is equal to the heat of reaction \(q\) under constant pressure conditions. \(H\), which we call the enthalpy , is a state function , since its value depends only on the state of the materials under consideration, that is, the temperature, pressure, and composition of these materials. The concept of a state function is somewhat analogous to the idea of elevation. Consider the difference in elevation between the first floor and the third floor of a building. This difference is independent of the path we choose to get from the first floor to the third floor. We can simply climb up two flights of stairs, or we can climb one flight of stairs, walk the length of the building, then walk a second flight of stairs. Or we can ride the elevator. We could even walk outside and have a crane lift us to the roof of the building, from which we can climb down to the third floor. Each path produces exactly the same elevation gain, even though the distance traveled is significantly different from one path to the next. This is simply because the elevation is a "state function". Our elevation, standing on the third floor, is independent of how we got to the third floor, and the same is true of the first floor. Since the elevation thus is a state function, the elevation gain is independent of the path. Now, the existence of an energy state function \(H\) is of considerable importance in calculating heats of reaction. Consider the prototypical reaction in Figure 10.2a, with reactants \(\ce{R}\) being converted to products \(\ce{P}\). We wish to calculate the heat absorbed or released in this reaction, which is \(\Delta H\). Since \(H\) is a state function, we can follow any path from \(\ce{R}\) to \(\ce{P}\) and calculate \(\Delta H\) along that path. In Figure 10.2b, we consider one such possible path, consisting of two reactions passing through an intermediate state containing all the atoms involved in the reaction, each in elemental form. This is a useful intermediate state since it can be used for any possible chemical reaction. For example, in Figure 10.1, the atoms involved in the reaction are \(\ce{C}\), \(\ce{H}\), and \(\ce{O}\), each of which are represented in the intermediate state in elemental form. We can see in Figure 10.2b that the \(\Delta H\) for the overall reaction is now the difference between the \(\Delta H\) in the formation of the products \(\ce{P}\) from the elements and the \(\Delta H\) in the formation of the reactants \(\ce{R}\) from the elements. a. b. Figure 10.2: Calculation of \(\Delta H\) The \(\Delta H\) values for formation of each material from the elements are thus of general utility in calculating \(\Delta H\) for any reaction of interest. We therefore define the standard formation reaction for reactant \(\ce{R}\), as \[\text{elements in standard state} \rightarrow \ce{R}\] and the heat involved in this reaction is the standard enthalpy of formation , designated by \(\Delta H_f^\text{o}\). The subscript \(f\), standing for "formation", indicates that the \(\Delta H\) is for the reaction creating the material from the elements in standard state. The superscript \(^\text{o}\) indicates that the reactions occur under constant standard pressure conditions of \(1 \: \text{atm}\). From Figure 10.2b, we see that the heat of any reaction can be calculated from \[Delta H_f^\text{o} = \Delta H_{f, \: \text{products}}^\text{o} - \Delta H_{f, \: \text{reactants}}^\text{o}\] Extensive tables of \(\Delta H_f^\text{o}\) have been compiled and published. This allows us to calculate with complete confidence the heat of reaction for any reaction of interest, even including hypothetical reactions which may be difficult to perform or impossibly slow to react. Observation 3: Bond Energies in Polyatomic Molecules The bond energy for a molecule is the energy required to separate the two bonded atoms to great distance. We recall that the total energy of the bonding electrons is lower when the two atoms are separated by the bond distance than when they are separated by a great distance. As such, the energy input required to separate the atoms elevates the energy of the electrons when the bond is broken. We can use diatomic bond energies to calculate the heat of reaction \(\Delta H\) for any reaction involving only diatomic molecules. We consider two simple examples. First, the reaction \[\ce{H_2} \left( g \right) + \ce{Br} \left( g \right) \rightarrow \ce{H} \left( g \right) + \ce{HBr} \left( g \right)\] is observed to be endothermic with heat of reaction \(70 \: \frac{\text{kJ}}{\text{mol}}\). Note that this reaction can be viewed as consisting entirely of the breaking of the \(\ce{H_2}\) bond followed by the formation of the \(\ce{HBr}\) bond. Consequently, we must input energy equal to the bond energy of \(\ce{H_2}\) \(\left( 436 \: \frac{\text{kJ}}{\text{mol}} \right)\), but in forming the \(\ce{HBr}\) bond we recover output energy equal to the bond energy of \(\ce{HBr}\) \(\left( 366 \: \frac{\text{kJ}}{\text{mol}} \right)\). Therefore the heat of the overall equation at constant pressure must be equal to the difference in these bond energies, \(70 \: \frac{\text{kJ}}{\text{mol}}\). Now we can answer the question, at least for this reaction, of where the energy "goes" during the reaction. The reason this reaction absorbs energy is that the bond which must be broken, \(\ce{H_2}\), is stronger than the bond which is formed, \(\ce{HBr}\). Note that energy is released when the \(\ce{HBr}\) bond is formed, but the amount of energy released is less than the amount of energy required to break the \(\ce{H_2}\) bond in the first place. The second example is similar: \[\ce{H_2} \left( g \right) + \ce{Br_2} \left( g \right) \rightarrow 2 \ce{HBr} \left( g \right)\] This reaction is exothermic with \(\Delta H^\text{o} = -103 \: \frac{\text{kJ}}{\text{mol}}\). In this case, we must break an \(\ce{H_2}\) bond, with energy \(436 \: \frac{\text{kJ}}{\text{mol}}\), and a \(\ce{Br_2}\) bond, with energy \(193 \: \frac{\text{kJ}}{\text{mol}}\). Since two \(\ce{HBr}\) molecules are formed, we must form two \(\ce{HBr}\) bonds, each with bond energy \(366 \: \frac{\text{kJ}}{\text{mol}}\). In total, then, breaking the bonds in the reactants requires \(629 \: \frac{\text{kJ}}{\text{mol}}\), and forming the new bonds releases \(732 \: \frac{\text{kJ}}{\text{mol}}\), for a net release of \(103 \: \frac{\text{kJ}}{\text{mol}}\). This calculation reveals that the reaction is exothermic because, although we must break one very strong bond and one weaker bond, we form two strong bonds. There are two items worth reflection in these examples. First, energy is released in a chemical reaction due to the formation of strong bonds. Breaking a bond, on the other hand, always requires the input of energy. Second, the reaction of \(\ce{H_2}\) and \(\ce{Br_2}\) does not actually proceed by the two-step process of breaking both reactant bonds, thus forming four free atoms, followed by making two new bonds. The actual process of the reaction is significantly more complicated. The details of this process are irrelevant to the energetics of the reaction, however, since, as we have shown, the heat of reaction \(\Delta H\) does not depend on the path of the reaction. This is another example of the utility of Hess' Law. We now proceed to apply this bond energy analysis to the energetics of reactions involving polyatomic molecules. A simple example is the combustion of hydrogen gas discussed previously. This is an explosive reaction, producing \(483.6 \: \text{kJ}\) per mole of oxygen. Calculating the heat of reaction from bond energies requires us to know the bond energies in \(\ce{H_2O}\). In this case, we must break not one but two bonds: \[\ce{H_2O} \left( g \right) \rightarrow 2 \ce{H} \left( g \right) + \ce{O} \left( g \right)\] The energy required to perform this reaction is measured to be \(926.0 \: \frac{\text{kJ}}{\text{mol}}\). The reaction of hydrogen and oxygen can proceed by a path in which we first break two \(\ce{H_2}\) bonds and one \(\ce{O_2}\) bond, then we follow the reverse of the decomposition of water twice: \[\begin{align} 2 \ce{H_2} \left( g \right) + \ce{O_2} \left( g \right) &\rightarrow 4 \ce{H} \left( g \right) + 2 \ce{O} \left( g \right) \\ 4 \ce{H} \left( g \right) + 2 \ce{O} \left( g \right) &\rightarrow 2 \ce{H_2O} \left( g \right) \\ 2 \ce{H_2} \left( g \right) + \ce{O_2} &\rightarrow 2 \ce{H_2O} \left( g \right) \end{align}\] Therefore, the energy of the final equation must be the energy required to break two \(\ce{H_2}\) bonds and one \(\ce{O_2}\) bond minus twice the energy of the decomposition of water. We calculate that \(\Delta H^\text{o} = 2 \times \left( 436 \: \frac{\text{kJ}}{\text{mol}} \right) + 498.3 \: \frac{\text{kJ}}{\text{mol}} - 2 \times \left( 926.9 \: \frac{\text{kJ}}{\text{mol}} \right) = -483.5 \: \frac{\text{kJ}}{\text{mol}}\). It is clear from this calculation that the formation of water is strongly exothermic because of the very large amount of energy released when two hydrogen atoms and one oxygen atom form a water molecule. It is tempting to use the heat of the decomposition of water to calculate the energy of an \(\ce{O-H}\) bond. Since breaking the two \(\ce{O-H}\) bonds in water requires \(926.9 \: \frac{\text{kJ}}{\text{mol}}\), then we might infer that breaking a single \(\ce{O-H}\) bond requires \(\frac{926.9}{2} \: \frac{\text{kJ}}{\text{mol}} = 463.5 \: \frac{\text{kJ}}{\text{mol}}\). However, the reaction \[\ce{H_2O} \left( g \right) \rightarrow \ce{OH} \left( g \right) + \ce{H} \left( g \right)\] has \(\Delta H^\text{o} = 492 \: \frac{\text{kJ}}{\text{mol}}\). Therefore, the energy required to break an \(\ce{O-H}\) bond in \(\ce{H_2O}\) is not the same as the energy required to break the \(\ce{O-H}\) bond in the \(\ce{OH}\) diatomic molecule. Stated differently, it requires more energy to break the first \(\ce{O-H}\) bond in water than is required to break the second \(\ce{O-H}\) bond. In general, we find that the energy required to break a bond between any two particular atoms depends upon the molecule those two atoms are in. Considering yet again oxygen and hydrogen, we find that the energy required to break the \(\ce{O-H}\) bond in methanol \(\left( \ce{CH_3OH} \right)\) is \(437 \: \frac{\text{kJ}}{\text{mol}}\), which differs substantially from the energy of the first \(\ce{O-H}\) bond in water. Similarly, the energy required to break a single \(\ce{C-H}\) bond in methane \(\left( \ce{CH_4} \right)\) is \(435 \: \frac{\text{kJ}}{\text{mol}}\), but the energy required to break all four \(\ce{C-H}\) bonds in methane is \(1663 \: \frac{\text{kJ}}{\text{mol}}\), which is not equal to four times the energy of one bond. As another such comparison, the energy required to break a \(\ce{C-H}\) bond is \(400 \: \frac{\text{kJ}}{\text{mol}}\) in trichloromethane \(\left( \ce{HCCl_3} \right)\), \(414 \: \frac{\text{kJ}}{\text{mol}}\) in dichloromethane \(\left( \ce{H_2CCl_2} \right)\), and \(422 \: \frac{\text{kJ}}{\text{mol}}\) in chloromethane \(\left( \ce{H_3CCl} \right)\). These observations are somewhat discouraging, since they reveal that, to use bond energies to calculate the heat of a reaction, we must first measure the bond energies for all bonds for all molecules involved in that reaction. This is almost certainly more difficult than it is desirable. On the other hand, we can note that the bond energies for similar bonds in similar molecules are close to one another. The \(\ce{C-H}\) bond energy in any one of the three chloromethanes above illustrate this quite well. We can estimate the \(\ce{C-H}\) bond energy in any one of these chloromethanes by the average \(\ce{C-H}\) bond energy in the three chloromethane molecules, which is \(412 \: \frac{\text{kJ}}{\text{mol}}\). Likewise, the average of the \(\ce{C-H}\) bond energies in methane is \(\frac{1663}{4} \: \frac{\text{kJ}}{\text{mol}} = 416 \: \frac{\text{kJ}}{\text{mol}}\) and is thus a reasonable approximation to the energy required to break a single \(\ce{C-H}\) bond in methane. By analyzing many bond energies in many molecules, we find that, in general, we can approximate the bond energy in any particular molecule by the average of the energies of similar bonds. These average bond energies can then be used to estimate the heat of a reaction without measuring all of the required bond energies. Consider for example the combustion of methane to form water and carbon dioxide: \[\ce{CH_4} \left( g \right) + 2 \ce{O_2} \left( g \right) \rightarrow \ce{CO_2} \left( g \right) + 2 \ce{H_2O} \left( g \right)\] We can estimate the heat of this reaction by using average bond energies. We must break four \(\ce{C-H}\) bonds at an energy cost of approximately \(4 \times 412 \: \frac{\text{kJ}}{\text{mol}}\) and two \(\ce{O_2}\) bonds at an energy cost of approximately \(2 \times 496 \: \frac{\text{kJ}}{\text{mol}}\). Forming the bonds in the products releases approximately \(2 \times 742 \: \frac{\text{kJ}}{\text{mol}}\) for the two \(\ce{C=O}\) double bonds and \(4 \times 463 \: \frac{\text{kJ}}{\text{mol}}\) for the \(\ce{O-H}\) bonds. Net, the heat of reaction is thus approximately \(\Delta H^\text{o} = 1648 + 992 - 1486 - 1852 = -698 \: \frac{\text{kJ}}{\text{mol}}\). This is a rather rough approximation to the actual heat of combustion of methane, \(-890 \: \frac{\text{kJ}}{\text{mol}}\). Therefore, we cannot use average bond energies to predict accurately the heat of a reaction. We can get an estimate, which may be sufficiently useful. Moreover, we can use these calculations to gain insight into the energetics of the reaction. For example, the combustion of methane is strongly exothermic, which is why methane gas (the primary component in natural gas) is an excellent fuel. From our calculation, we can see that the reaction involved breaking six bonds and forming six new bonds. The bonds formed are substantially stronger than those broken, thus accounting for the net release of energy during the reaction. Review and Discussion Questions Assume you have two samples of two different metals, X and Z. The samples are exactly the same mass. Both samples are heated to the same temperature. Then each sample is placed into separate glasses containing identical quantities of cold water, initially at identical temperatures below that of the metals. The final temperature of the water containing metal X is greater than the final temperature of the water containing metal Z. Which of the two metals has the larger heat capacity? Explain your conclusion. If each sample, initially at the same temperature, is heated with exactly \(100 \: \text{J}\) of energy, which sample has the higher final temperature? Explain how Hess' Law is a consequence of conservation of energy. Consider the reaction \[\ce{N_2O_4} \left( g \right) \rightarrow 2 \ce{NO_2} \left( g \right)\] Draw Lewis structures for each of \(\ce{N_2O_4}\) and \(\ce{NO_2}\). On the basis of these structures, predict whether the reaction is endothermic or exothermic, and explain your reasoning. Why is the bond energy of \(\ce{H_2}\) not equal to \(\Delta H_f^\text{o}\) of \(\ce{H_2}\)? For what species is the enthalpy of formation related to the bond energy of \(\ce{H_2}\)? Suggest a reason why \(\Delta H\) for the reaction \[\ce{CO_2} \left( g \right) \rightarrow \ce{CO} \left( g \right) + \ce{O} \left( g \right)\] is not equal to \(\Delta H^\text{o}\) for the reaction \[\ce{CO} \left( g \right) \rightarrow \ce{C} \left( g \right) + \ce{O} \left( g \right)\]. Determine whether the reaction is exothermic or endothermic for each of the following circumstances: The heat of combustion of the products is greater than the heat of combustion of the reactants. The enthalpy of formation of the products is greater than the enthalpy of reaction of the reactants. The total of the bond energies of the products is greater than the total of the bond energies for the reactants. |
Bookshelves/Organic_Chemistry/Catalytic_Asymmetric_Synthesis_(Punniyamurthy)/01%3A_Reactions_using_Chiral_Lewis_Acids_and_Brnsted_Acid | This module presents the recent developments in chiral Lewis acid and Brønsted acid catalysis, especially the systems having the combination of Lewis acids and Brønsted acids. This combined catalytic system has been useful in asymmetric synthesis over the past 20 years. 1.1: Brønsted Acid-Assisted Lewis Acid (BLA) Chiral Brønsted acid-assisted Lewis acids (BLAs) are efficient and versatile chiral Lewis acids for a wide range of catalytic asymmetric cycloaddition reactions. 1.2: Lewis Acid-Assisted Lewis Acid (LLA) In Lewis acid assisted chiral Lewis acids (LLAs), achiral Lewis acid is added to activate chiral Lewis acid via complex formation. The reactivity of LLA is much greater compared to that of achiral Lewis acid, and thus, the latter's presence does not affect the selectivity of the reaction. 1.3: LBA Catalysts The combination of Lewis acids and chiral Brønsted acids affords LBA catalysts. In this system, the coordination of the Lewis acids to the heteroatom of the chiral Brønsted acid results in increase the acidity of the latter. 1.4: Problems + Reference 1.5: Chiral Phosphoric Acids (PAs) The combination of Lewis acids and chiral Brønsted acids affords LBA catalysts. In this system, the coordination of the Lewis acids to the heteroatom of the chiral Brønsted acid results in increase the acidity of the latter. |
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Statistical_Mechanics/Boltzmann_Average/Proof_that___1_kT | The pressure in the state \(j\) is given by \(p_j = - (\partial E_j/\partial V)\). The average energy is \[\bar{E}=\frac{\displaystyle \sum_{J} E_{j}(N, V) e^{-\beta E_{f}(N, n)}}{\displaystyle \sum_{j} e^{-\beta E / N, n}} \label{I}\] The average pressure is \[\bar{p}=\frac{\displaystyle \sum p_{j}(N, V) e^{-\beta E_{f}(N, n)}}{\displaystyle \sum_{j} e^{-\beta E_{f}(N, n)}} \label{II}\] According the Gibbs postulate the average energy, average pressure and other average mechanical properties calculated using the partition function are equal to their thermodynamic counterparts. Note that some authors use \(\bar{E}\) and \(\bar{p}\) bar for the average quantities and elsewhere the angle bracket notation is used. These are equivalent notations. If we differentiate the expression for the average energy we can treat the denominator, \(Q\) as a function of \(V\) as well since it represents a sum over \(\exp(-\beta E_j(N,V)\). Since \(E_j\) appears both in the exponent and as a function multiplying the exponent we have \[\left(\frac{\partial E}{\partial V}\right)_{N, \beta}=\frac{\displaystyle \sum\left(\frac{\partial E_{j}}{\partial V}\right) e^{-\beta E / N, n}}{Q}-\frac{\displaystyle \sum-\beta\left(\frac{\partial E_{j}}{\partial V}\right) E_{f} e^{-\beta E / N, n}}{Q} -\frac{\displaystyle \sum_{j} E_{j}^{-\beta E_{j} / N, n} \displaystyle \sum_{j}-\beta\left(\frac{\partial E_{j}}{\partial V}\right) e^{-\beta E_{f}(N, n)}}{Q^{2}}\] Here we used the quotient rule to take the derivative. This can written compactly as \[\left(\frac{\partial E}{\partial V}\right)_{N, \beta}=-\bar{p}+\beta \overline{E p}-\beta \bar{E} \bar{p}\] We can differentiate Equation \ref{II} with respect to \(\beta\) to obtain \[\left(\frac{\partial \bar{p}}{\partial \beta}\right)_{N, V}=\bar{E} \bar{p}-\overline{E p}\] The two derivative expressions can be combined to give \[\left(\frac{\partial E}{\partial V}\right)_{N, \beta}+\beta\left(\frac{\partial \bar{p}}{\partial \beta}\right)_{N, V}=-p\] This can be compared to the thermodynamic equation of state \[\left(\frac{\partial E}{\partial V}\right)_{T}-T\left(\frac{\partial p}{\partial T}\right)_{V}=-p \label{IV}\] This can be derived as follows from \[dE = TdS – PdV.\] First take the derivative of both sides with respect to \(V\) at constant \(T\). Now, note that \[\left(\frac{\partial p}{\partial T}\right)_{V}=\left(\frac{\partial S}{\partial V}\right)_{T}\] This is known as a Maxwell relation . It is obtained from \[dA = SdT + PdV \label{III}\] From the fact that \(A\) is a state function (the Helmholtz free energy ) we know that the second cross derivatives must be equal. That is: \[\left(\frac{\partial A}{\partial V \partial T}\right)=\left(\frac{\partial A}{\partial T \partial V}\right)\] And from inspection of Equation \ref{III} we see that \[\left(\frac{\partial A}{\partial T}\right)_{V}=S\] and \[\left(\frac{\partial A}{\partial V}\right)_{T}=P\] Finally, using the relation \[T\left(\frac{\partial}{\partial T}\right)=-\frac{1}{T}\left(\frac{\partial}{\partial(1 / T)}\right)\] Showing that this is true is a little tricky. For example, we can define \(F = 1/T\). Then \[\dfrac{\partial F}{\partial T} = \dfrac{-1}{T^2}\] and \[\dfrac{\partial F}{\partial F} = 1\] So we can write \[\dfrac{\partial F}{\partial T} = \dfrac{-1}{T^2} \left(\dfrac{\partial F}{\partial F}\right)\] or \[\dfrac{\partial }{\partial T} = \dfrac{-1}{T^2} \left(\dfrac{\partial }{\partial F}\right)\] which gives Equation \ref{IV} when both sides are multiplied by \(T\). \[\left(\frac{\partial E}{\partial V}\right)_{T}+\frac{1}{T}\left(\frac{\partial p}{\partial(1 / T)}\right)_{V}=-p\] The comparison with the above equation shows that \(\beta \propto 1/T\). This proves that \(\beta = constant/T\). The constant turns out to be \(k_B\) or Boltzmann’s constant by comparison with expressions for the average energy or average pressure with known thermodynamic equations. |
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/13%3A_Molecular_Spectroscopy/13.11%3A_Time-Dependent_Perturbation_Theory | In time-independent perturbation theory the perturbation Hamiltonian is static (i.e., possesses no time dependence). Time-independent perturbation theory was presented by Erwin Schrödinger in a 1926 paper,shortly after he produced his theories in wave mechanics. Time-dependent perturbation theory, developed by Paul Dirac, studies the effect of a time-dependent perturbation V ( t ) applied to a time-independent Hamiltonian \(H_0\). Since the perturbed Hamiltonian is time-dependent, so are its energy levels and eigenstates. Thus, the goals of time-dependent perturbation theory are slightly different from time-independent perturbation theory, where one may be interested in the following quantities: The time-dependent expectation value of some observable A , for a given initial state. The time-dependent amplitudes of those quantum states that are energy eigenkets (eigenvectors) in the unperturbed system. The first quantity is important because it gives rise to the classical result of a measurement performed on a macroscopic number of copies of the perturbed system. The second quantity looks at the time-dependent probability of occupation for each eigenstate. This is particularly useful in laser physics, where one is interested in the populations of different atomic states in a gas when a time-dependent electric field is applied. We will briefly examine the method behind Dirac's formulation of time-dependent perturbation theory. Choose an energy basis | n ⟩ {\displaystyle {|n\rangle }} \(| n \rangle \) for the unperturbed system. (We drop the (0) superscripts for the eigenstates, because it is not useful to speak of energy levels and eigenstates for the perturbed system.) If the unperturbed system is in eigenstate \(|j \rangle \) at time \(t = 0\), its state at subsequent times varies only by a phase (this is the Schrödinger picture, where state vectors evolve in time and operators are constant) \[|j(t)\rangle =e^{-iE_{j}t/\hbar }|j\rangle \nonumber \] Now, introduce a time-dependent perturbing Hamiltonian \(H_1(t)\). The Hamiltonian of the perturbed system is \[H=H_{0}+H_1(t) \nonumber \] Let \(|\psi (t)\rangle \) denote the quantum state of the perturbed system at time \(t\) and obeys the time-dependent Schrödinger equation, \[H|\psi (t)\rangle =i\hbar {\dfrac {\partial }{\partial t}}|\psi (t)\rangle \nonumber \] The quantum state at each instant can be expressed as a linear combination of the complete eigenbasis of \( | n \rangle \): \[|\psi (t)\rangle =\sum _{n}c_{n}(t)e^{-iE_{n}t/\hbar }|n\rangle \nonumber \] where the \(c_n(t)\) coef ficient s are to be determined complex functions of t which we will refer to as amplitudes We have explicitly extracted the exponential phase factors \(\exp(-iE_{n}t/\hbar)\) on the right hand side. This is only a matter of convention, and may be done without loss of generality. The reason we go to this trouble is that when the system starts in the state {\displaystyle |j\rangle } \(|j\rangle \) and no perturbation is present, the amplitudes have the convenient property that, for all t, \(c_j(t) = 1\) and \(c_n(t) = 0\) if \(n \neq j\). The square of the absolute amplitude \(c_n(t)\) is the probability that the system is in state \(n\) at time \(t\), since \[|\psi (t)\rangle =\sum _{n}c_{n}(t)e^{-iE_{n}t/\hbar }|n\rangle \nonumber \] Plugging into the Schrödinger equation and using the fact that \(\partial/ \partial t\) acts by a chain rule, one obtains \[\sum _{n}\left(i\hbar {\dfrac {\partial c_{n}}{\partial t}}-c_{n}(t)V(t)\right)e^{-iE_{n}t/\hbar }|n\rangle =0~. \nonumber \] By resolving the identity in front of V , this can be reduced to a set of partial differential equations for the amplitudes, \[{\dfrac {\partial c_{n}}{\partial t}}={\dfrac {-i}{\hbar }}\sum _{k}\langle n|H_1(t)|k\rangle \,c_{k}(t)\,e^{-i(E_{k}-E_{n})t/\hbar }~. \nonumber \] The matrix elements of \(H_1\) play a similar role as in time-independent perturbation theory, being proportional to the rate at which amplitudes are shifted between states. Note, however, that the direction of the shift is modified by the exponential phase factor. Over times much longer than the energy difference \(E_k − E_n\), the phase winds around 0 several times. If the time-dependence of \(H_1\) is sufficiently slow, this may cause the state amplitudes to oscillate (e.g., such oscillations are useful for managing radiative transitions in a laser). Two-Level System Consider the two level system (i.e. \(n=1,2\)) \[ | \psi \rangle = \sum_ {n=1,2} c_n(t) | n \rangle_o \nonumber \] Solution of time-dependent perturbation for two level system: \[ i \hbar \dfrac{\partial c_1(t)}{\partial t} = c_1(t) H_{11}(t) + c_2 e^{-i \omega_o t} H_{12} (t) \nonumber \] \[ i \hbar \dfrac{\partial c_2(t)}{\partial t} = c_2(t) H_{22}(t) + c_1 e^{+i \omega_o t} H_{21} (t) \nonumber \] where the matrix elements of the permutation (in terms of the eigenstates of \(H(0)\)) are \[ \langle m | H_1(t) | n \rangle = H_{mn}(t) \nonumber \] Assume initial state is \(n=1\), and \(H_{11}=H_{22}=0\) \[ | \psi (t=0) \rangle = |1 \rangle \nonumber \] Probability of particle at \(n=2\) at time \(t\) after the perturbation is turned on (i.e., incident light): \[ c_2(t) = \dfrac{-i}{\hbar} \int_o^t e^{i \omega_o t} dt' H_{21}(t') \label{EQ1} \] where \[H_1(t) = \cos (\omega t) V(r) \nonumber \] \(V(r)\) is an amplitude of polarization vector, which we can ignore for now. If we assume incident frequency of incident light \(\omega\) is comparable to the natural frequency of oscillation from \(\omega _o\) \[\omega \approx \omega_o \nonumber \] then Equation \(\ref{EQ1}\) can be simplified to \[c_2(t) = - \dfrac{2i}{\hbar} \dfrac{\sin (\omega-\omega_o)t /2}{ (\omega-\omega_o )t} e^{i (\omega_0-\omega)t/2} H_{21} \nonumber \] Transition Probability Assume initial state is \(n=1\), and probability of transition from \(n=1\) state to \(n=2\) state is: \[ P_{12}(t) = | c_2(t) |^2 =\dfrac{4}{\hbar^2} | \dfrac{\sin (\omega-\omega_o)t /2}{ (\omega-\omega_o )t} |^2 | H_{21} |^2 \nonumber \] What does this mean? Strangely, it means that the probability of making a transition is actually oscillating sinusoidally (squared)! If you want to cause a transition, should turn off perturbation after time \(\pi / |\omega- \omega_o| \)or some odd multiple, when the system is in upper state with maximum probability. \(P_{12}(t)\) is peaked at \( \omega- \omega_o = 0\). The height of \( |H_{12}t/2 \hbar |^2\) and width of \(4\pi/t\) gets higher and narrower as time goes on. Recall this is perturbative treatment, however, and \(P_{12}(t)\) cannot get bigger than 1, so perturbation theory breaks down eventually. |
Courses/SUNY_Oneonta/Chem_221%3A_Organic_Chemistry_I_(Bennett)/3%3AStuff_to_Review_from_General_Chemistry/04%3A_Thermochemistry/4.01%3A_Prelude_to_Thermochemistry | Chemical reactions, such as those that occur when you light a match, involve changes in energy as well as matter. Societies at all levels of development could not function without the energy released by chemical reactions. In 2012, about 85% of US energy consumption came from the combustion of petroleum products, coal, wood, and garbage. We use this energy to produce electricity (38%); to transport food, raw materials, manufactured goods, and people (27%); for industrial production (21%); and to heat and power our homes and businesses (10%). 1 While these combustion reactions help us meet our essential energy needs, they are also recognized by the majority of the scientific community as a major contributor to global climate change. Useful forms of energy are also available from a variety of chemical reactions other than combustion. For example, the energy produced by the batteries in a cell phone, car, or flashlight results from chemical reactions. This chapter introduces many of the basic ideas necessary to explore the relationships between chemical changes and energy, with a focus on thermal energy. Footnotes US Energy Information Administration, Primary Energy Consumption by Source and Sector, 2012 , Total Energy [www.eia.gov] . Data derived from US Energy Information Administration, Monthly Energy Review (January 2014). |
Courses/Palomar_College/PC%3A_CHEM100_-_Fundamentals_of_Chemistry/11%3A_Redox_Reactions/11.7%3A_Combustion_Reactions | Roasting marshmallows over an open fire is a favorite past-time for campers, outdoor cook-outs, and just gathering around a fire in the backyard. The trick is to get the marshmallow a nice golden brown without catching it on fire. Too often we are not successful and we see the marshmallow burning on the stick - a combustion reaction taking place right in front of us. Combustion Reactions A combustion reaction is a reaction in which a substance reacts with oxygen gas, releasing energy in the form of light and heat. Combustion reactions must involve \(\ce{O_2}\) as one reactant. The combustion of hydrogen gas produces water vapor. \[2 \ce{H_2} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{H_2O} \left( g \right)\] Notice that this reaction also qualifies as a combination reaction. The Hindenberg was a hydrogen-filled airship that suffered an accident upon its attempted landing in New Jersey in 1937. The hydrogen immediately combusted in a huge fireball, destroying the airship and killing 36 people. The chemical reaction was a simple one: hydrogen combining with oxygen to produce water. Many combustion reactions occur with a hydrocarbon, a compound made up solely of carbon and hydrogen. The products of the combustion of hydrocarbons are carbon dioxide and water. Many hydrocarbons are used as fuel because their combustion releases very large amounts of heat energy. propane \(\left( \ce{C_3H_8} \right)\) is a gaseous hydrocarbon that is commonly used as the fuel source in gas grills. \[\ce{C_3H_8} \left( g \right) + 5 \ce{O_2} \left( g \right) \rightarrow 3 \ce{CO_2} \left( g \right) + 4 \ce{H_2O} \left( g \right)\] Example 11.6.1 Ethanol can be used as a fuel source in an alcohol lamp. The formula for ethanol is \(\ce{C_2H_5OH}\). Write the balanced equation for the combustion of ethanol. Solution Step 1: Plan the problem . Ethanol and oxygen are the reactants. As with a hydrocarbon, the products of the combustion of an alcohol are carbon dioxide and water. Step 2: Solve . Write the skeleton equation: \[\ce{C_2H_5OH} \left( l \right) + \ce{O_2} \left( g \right) \rightarrow \ce{CO_2} \left( g \right) + \ce{H_2O} \left( g \right)\] Balance the equation. \[\ce{C_2H_5OH} \left( l \right) + 3 \ce{O_2} \left( g \right) \rightarrow 2 \ce{CO_2} \left( g \right) + 3 \ce{H_2O} \left( g \right)\] Step 3: Think about your result. Combustion reactions must have oxygen as a reactant. Note that the water is produced is in the gas rather than the liquid state because of the high temperatures that accompany a combustion reaction. Summary Combustion reaction is defined and examples are given. |
Courses/Madera_Community_College/MacArthur_Chemistry_3A_v_1.2/09%3A_Attractive_Forces/9.03%3A_Phase_Transitions/9.3.03%3A_Calculations_for_Phase_Changes | Learning Objectives Calculate the e ner g y change needed for a phase change. For each phase change of a substance, there is a characteristic quantity of heat needed to perform the phase change per gram (or per mole) of material. The heat of fusion (Δ H fus ) is the amount of heat per gram (or per mole) required for a phase change that occurs at the melting point. The heat of vaporization (Δ H vap ) is the amount of heat per gram (or per mole) required for a phase change that occurs at the boiling point. If you know the total number of grams or moles of material, you can use the Δ H fus or the Δ H vap to determine the total heat being transferred for melting or solidification using these expressions: \[\text{heat} = n \times ΔH_{fus} \label{Eq1a}\] where \(n\) is the number of moles and \(ΔH_{fus}\) is expressed in energy/mole or \[\text{heat} = m \times ΔH_{fus} \label{Eq1b}\] where \(m\) is the mass in grams and \(ΔH_{fus}\) is expressed in energy/gram. For the boiling or condensation, use these expressions: \[\text{heat} = n \times ΔH_{vap} \label{Eq2a}\] where \(n\) is the number of moles) and \(ΔH_{vap}\) is expressed in energy/mole or \[\text{heat} = m \times ΔH_{vap} \label{Eq2b}\] where \(m\) is the mass in grams and \(ΔH_{vap}\) is expressed in energy/gram. Remember that a phase change depends on the direction of the heat transfer. If heat transfers in, solids become liquids, and liquids become solids at the melting and boiling points, respectively. If heat transfers out, liquids solidify, and gases condense into liquids. Some Δ H fus values are listed in Table \(\PageIndex{1}\); it is assumed that these values are for the melting point of the substance. Note that the unit of Δ H fus is kilojoules per mole, so we need to know the quantity of material to know how much energy is involved. The Δ H fus is always tabulated as a positive number. However, it can be used for both the melting and the freezing processes, minding that melting is always endothermic (so Δ H will be positive), while freezing is always exothermic (so Δ H will be negative). Substance (Melting Point) ΔHfus (kJ/mol) Water (0°C) 6.01 Aluminum (660°C) 10.70 Benzene (5.5°C) 9.95 Ethanol (−114.3°C) 5.02 Mercury (−38.8°C) 2.29 Example \(\PageIndex{1}\) What is the energy change when 45.7 g of \(\ce{H2O}\) melt at 0°C? Solution The \(ΔH_{fus}\) of \(\ce{H2O}\) is 6.01 kJ/mol. However, our quantity is given in units of grams, not moles, so the first step is to convert grams to moles using the molar mass of \(\ce{H_2O}\), which is 18.0 g/mol. Then we can use \(ΔH_{fus}\) as a conversion factor. Because the substance is melting, the process is endothermic, so the energy change will have a positive sign. \[45.7\cancel{g\: H_{2}O}\times \frac{1\cancel{mol\: H_{2}O}}{18.0\cancel{g}}\times \frac{6.01kJ}{\cancel{mol}}=15.3\,kJ \nonumber\nonumber \] Without a sign, the number is assumed to be positive. Exercise \(\PageIndex{1}\) What is the energy change when 108 g of \(\ce{C6H6}\) freeze at 5.5°C? Answer −13.8 kJ Like the solid/liquid phase change, the liquid/gas phase change involves energy. The amount of energy required to convert a liquid to a gas is called the enthalpy of vaporization (or heat of vaporization), represented as Δ H vap . Some Δ H vap values are listed in Table \(\PageIndex{2}\); it is assumed that these values are for the normal boiling point temperature of the substance, which is also given in the table. The unit for Δ H vap is also kilojoules per mole, so we need to know the quantity of material to know how much energy is involved. The Δ H vap is also always tabulated as a positive number. It can be used for both the boiling and the condensation processes as long as you keep in mind that boiling is always endothermic (so Δ H will be positive), while condensation is always exothermic (so Δ H will be negative). Substance (Normal Boiling Point) ΔHvap (kJ/mol) Water (100°C) 40.68 Bromine (59.5°C) 15.40 Benzene (80.1°C) 30.80 Ethanol (78.3°C) 38.60 Mercury (357°C) 59.23 Example \(\PageIndex{2}\) What is the energy change when 66.7 g of Br 2 (g) condense to a liquid at 59.5°C? Solution The Δ H vap of Br 2 is 15.4 kJ/mol. Even though this is a condensation process, we can still use the numerical value of Δ H vap as long as we realize that we must take energy out, so the Δ H value will be negative. To determine the magnitude of the energy change, we must first convert the amount of Br 2 to moles. Then we can use Δ H vap as a conversion factor. \[66.7\cancel{g\: Br_{2}}\times \frac{1\cancel{mol\: Br_{2}}}{159.8\cancel{g}}\times \frac{15.4kJ}{\cancel{mol}}=6.43\,kJ \nonumber\nonumber \] Because the process is exothermic, the actual value will be negative: Δ H = −6.43 kJ. Exercise \(\PageIndex{2}\) What is the energy change when 822 g of \(\ce{C2H5OH(ℓ)}\) boil at its normal boiling point of 78.3°C? Answer 689 kJ As with melting, the energy in boiling goes exclusively to changing the phase of a substance; it does not go into changing the temperature of a substance. So boiling is also an isothermal process. Only when all of a substance has boiled does any additional energy go to changing its temperature. Sometimes, a table will contain both the heat of fusion and the heat of vaporization. If you are provided such a table as a reference for performing a calculation, make sure that you are selecting the correct constant from the table. Also pay attention to the units. In the first two tables you were provided units of kJ/mol, but in the following table the units are cal/g. Use the following table to answer the questions which follow it. Substance ΔHfus (cal/g) ΔHvap (cal/g) aluminum (Al) 94.0 2602.0 gold (Au) 15.3 409.0 iron (Fe) 63.2 1504.0 water (H2O) 79.9 540.0 sodium chloride (NaCl) 123.5 691.0 ethanol (C2H5OH) 45.2 200.3 benzene (C6H6) 30.4 94.1 Example \(\PageIndex{1}\) How many cal are necessary to melt 55.8 g of ice (solid H 2 O) at 0°C? Solution We can use the relationship between heat and the heat of fusion (Eq. \(\PageIndex{1}\)b) to determine how many joules of heat are needed to melt this ice: \[ \begin{align*} \text{heat} &= m \times ΔH_{fus} \\[4pt] & = (55.8\: \cancel{g})\left(\dfrac{79.9\: cal}{\cancel{g}}\right)=4,460\: cal} \end{align*}\] Exercise \(\PageIndex{1}\) How many cal are necessary to vaporize 685 g of H 2 O at 100°C? Summary Contributions & Attributions This page was constructed from content via the following contributor(s) and edited (topically or extensively) by the LibreTexts development team to meet platform style, presentation, and quality: |
Bookshelves/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/13%3A_Amino_Acids_and_Proteins/13.02%3A_Peptides | Learning Outcomes Define polypeptide. Identify amide bond. Predict product of condensation of amino acids. Name polypeptides given the abbreviation of the amino acids. Peptide cells in our bodies have an intricate mechanism for the manufacture of proteins. Humans have to use other techniques in order to synthesize the same proteins in a lab. The chemistry of peptide synthesis is complicated. Both active groups on an amino acid can react and the amino acid sequence must be a specific one in order for the protein to function. Robert Merrifield developed the first synthetic approach for making proteins in the lab, a manual approach which was lengthy and tedious (and, he won the Nobel Prize in Chemistry in 1984 for his work). Today, however, automated systems can crank out a peptide in a very short period of time. Peptides A peptide is a combination of amino acids in which the amine group of one amino acid has undergone a reaction with the carboxyl group of another amino acid. The reaction is a condensation reaction, forming an amide group \(\left( \ce{CO-N} \right)\), shown below. A peptide bond is the amide bond that occurs between the amine nitrogen of one amino acid and the carboxyl carbon of another amino acid. The resulting molecule is called a dipeptide. Notice that the particular side chains of each amino acid are irrelevant since the \(\ce{R}\) groups are not involved in the peptide bond. The dipeptide has a free amine group on one end of the molecule (known as the \(\ce{N}\)-terminus) and a free carboxyl group on the other end (known as the \(\ce{C}\)-terminus). Each is capable of extending the chain through the formation of another peptide bond. The particular sequence of amino acids in a longer chain is called an amino acid sequence. By convention, the amino acid sequence is listed in the order such that the free amine group is on the left end of the molecule and the free carboxyl group is on the right end of the molecule. For example, suppose that a sequence of the amino acids glycine, tryptophan, and alanine is formed with the free amine group as part of the glycine and the free carboxyl group as part of the alanine. The amino acid sequence can be easily written using the abbreviations as Gly-Trp-Ala. This is a different sequence from Ala-Trp-Gly because the free amine and carboxyl groups would be on different amino acids in that case. Example \(\PageIndex{1}\) Draw the polypeptide Asp-Val-Ser. Solution 1. Identify the structures of each of the three given amino acids and draw them in the same order as given in the name. 2. Leaving the order the same, connect the amino acids to one another by forming peptide bonds. Note that the order given in the name is the same way the amino acids are connected in the molecule. The first one listed is always the \(\ce{N}\)-terminus of the polypeptide. Example \(\PageIndex{2}\) List all of the possible polypeptides that can be formed from cysteine (Cys), leucine (Leu), and arginine (Arg). Solution Although there are only three amino acids, the order in which they are bonded changes the identity, properties, and function of the resulting polypeptide. There are six possible polypeptides formed from these three amino acids. Cys-Leu-Arg Cys-Arg-Leu Leu-Cys-Arg Leu-Arg-Cys Arg-Cys-Leu Arg-Leu-Cys |
Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/05%3A_Thermochemistry/5.02%3A_The_First_Law_of_Thermodynamics | Learning Objectives To calculate changes in internal energy To study the flow of energy during a chemical reaction, we need to distinguish between a system, the small, well-defined part of the universe in which we are interested (such as a chemical reaction), and its surroundings, the rest of the universe, including the container in which the reaction is carried out (Figure \(\PageIndex{1}\)). In the discussion that follows, the mixture of chemical substances that undergoes a reaction is always the system, and the flow of heat can be from the system to the surroundings or vice versa. Three kinds of systems are important in chemistry. An open system can exchange both matter and energy with its surroundings. A pot of boiling water is an open system because a burner supplies energy in the form of heat, and matter in the form of water vapor is lost as the water boils. A closed system can exchange energy but not matter with its surroundings. The sealed pouch of a ready-made dinner that is dropped into a pot of boiling water is a closed system because thermal energy is transferred to the system from the boiling water but no matter is exchanged (unless the pouch leaks, in which case it is no longer a closed system). An isolated system exchanges neither energy nor matter with the surroundings. Energy is always exchanged between a system and its surroundings, although this process may take place very slowly. A truly isolated system does not actually exist. An insulated thermos containing hot coffee approximates an isolated system, but eventually the coffee cools as heat is transferred to the surroundings. In all cases, the amount of heat lost by a system is equal to the amount of heat gained by its surroundings and vice versa. That is, the total energy of a system plus its surroundings is constant , which must be true if energy is conserved . The state of a system is a complete description of a system at a given time, including its temperature and pressure, the amount of matter it contains, its chemical composition, and the physical state of the matter. A state function is a property of a system whose magnitude depends on only the present state of the system, not its previous history. Temperature, pressure, volume, and potential energy are all state functions. The temperature of an oven, for example, is independent of however many steps it may have taken for it to reach that temperature. Similarly, the pressure in a tire is independent of how often air is pumped into the tire for it to reach that pressure, as is the final volume of air in the tire. Heat and work, on the other hand, are not state functions because they are path dependent . For example, a car sitting on the top level of a parking garage has the same potential energy whether it was lifted by a crane, set there by a helicopter, driven up, or pushed up by a group of students (Figure \(\PageIndex{2}\)). The amount of work expended to get it there, however, can differ greatly depending on the path chosen. If the students decided to carry the car to the top of the ramp, they would perform a great deal more work than if they simply pushed the car up the ramp (unless, of course, they neglected to release the parking brake, in which case the work expended would increase substantially!). The potential energy of the car is the same, however, no matter which path they choose. Direction of Heat Flow The reaction of powdered aluminum with iron(III) oxide, known as the thermite reaction, generates an enormous amount of heat—enough, in fact, to melt steel (Figure \(\PageIndex{3}\)). The balanced chemical equation for the reaction is as follows: \[\ce{ 2Al(s) + Fe_2O_3(s) -> 2Fe(s) + Al_2O_3(s)} \label{5.2.1} \] We can also write this chemical equation as \[\ce{2Al(s) + Fe_2O_3(s) \rightarrow 2Fe(s) + Al_2O_3(s)} + \text{heat} \label{5.2.2} \] to indicate that heat is one of the products. Chemical equations in which heat is shown as either a reactant or a product are called thermochemical equations . In this reaction, the system consists of aluminum, iron, and oxygen atoms; everything else, including the container, makes up the surroundings. During the reaction, so much heat is produced that the iron liquefies. Eventually, the system cools; the iron solidifies as heat is transferred to the surroundings. A process in which heat ( q ) is transferred from a system to its surroundings is described as exothermic. By convention, \(q < 0\) for an exothermic reaction. When you hold an ice cube in your hand, heat from the surroundings (including your hand) is transferred to the system (the ice), causing the ice to melt and your hand to become cold. We can describe this process by the following thermochemical equation: \[ \ce{heat + H_2O(s) \rightarrow H_2O(l)} \label{5.2.3} \] When heat is transferred to a system from its surroundings, the process is endothermic. By convention, \(q > 0\) for an endothermic reaction. Heat is technically not a component in Chemical Reactions Technically, it is poor form to have a \(heat\) term in the chemical reaction like in Equations \(\ref{5.2.2}\) and \(\ref{5.2.3}\) since is it not a true species in the reaction. However, this is a convenient approach to represent exothermic and endothermic behavior and is commonly used by chemists. The First Law The relationship between the energy change of a system and that of its surroundings is given by the first law of thermodynamics , which states that the energy of the universe is constant. We can express this law mathematically as follows: \[U_{univ}=ΔU_{sys}+ΔU_{surr}=0 \label{5.2.4a} \] \[\Delta{U_{sys}}=−ΔU_{surr} \label{5.2.4b} \] where the subscripts univ, sys, and surr refer to the universe, the system, and the surroundings, respectively. Thus the change in energy of a system is identical in magnitude but opposite in sign to the change in energy of its surroundings. The tendency of all systems, chemical or otherwise, is to move toward the state with the lowest possible energy. An important factor that determines the outcome of a chemical reaction is the tendency of all systems, chemical or otherwise, to move toward the lowest possible overall energy state. As a brick dropped from a rooftop falls, its potential energy is converted to kinetic energy; when it reaches ground level, it has achieved a state of lower potential energy. Anyone nearby will notice that energy is transferred to the surroundings as the noise of the impact reverberates and the dust rises when the brick hits the ground. Similarly, if a spark ignites a mixture of isooctane and oxygen in an internal combustion engine, carbon dioxide and water form spontaneously, while potential energy (in the form of the relative positions of atoms in the molecules) is released to the surroundings as heat and work. The internal energy content of the \(CO_2/H_2O\) product mixture is less than that of the isooctane/ \(O_2\) reactant mixture. The two cases differ, however, in the form in which the energy is released to the surroundings. In the case of the falling brick, the energy is transferred as work done on whatever happens to be in the path of the brick; in the case of burning isooctane, the energy can be released as solely heat (if the reaction is carried out in an open container) or as a mixture of heat and work (if the reaction is carried out in the cylinder of an internal combustion engine). Because heat and work are the only two ways in which energy can be transferred between a system and its surroundings, any change in the internal energy of the system is the sum of the heat transferred (q) and the work done (w): \[ΔU_{sys} = q + w \label{5.2.5} \] Although \(q\) and \(w\) are not state functions on their own, their sum (\(ΔU_{sys}\)) is independent of the path taken and is therefore a state function. A major task for the designers of any machine that converts energy to work is to maximize the amount of work obtained and minimize the amount of energy released to the environment as heat. An example is the combustion of coal to produce electricity. Although the maximum amount of energy available from the process is fixed by the energy content of the reactants and the products, the fraction of that energy that can be used to perform useful work is not fixed. Because we focus almost exclusively on the changes in the energy of a system, we will not use “sys” as a subscript unless we need to distinguish explicitly between a system and its surroundings. Although \(q\) and \(w\) are not state functions, their sum (\(ΔU_{sys}\)) is independent of the path taken and therefore is a state function. Thus, because of the first law, we can determine \(ΔU\) for any process if we can measure both \(q\) and \(w\). Heat, \(q\), may be calculated by measuring a change in temperature of the surroundings. Work, \(w\), may come in different forms, but it too can be measured. One important form of work for chemistry is pressure-volume work done by an expanding gas. At a constant external pressure (for example, atmospheric pressure) \[w = −PΔV \label{5.2.6} \] The negative sign associated with \(PV\) work done indicates that the system loses energy when the volume increases. That is, an expanding gas does work on its surroundings, while a gas that is compressed has work done on it by the surroundings. Example \(\PageIndex{1}\) A sample of an ideal gas in the cylinder of an engine is compressed from 400 mL to 50.0 mL during the compression stroke against a constant pressure of 8.00 atm. At the same time, 140 J of energy is transferred from the gas to the surroundings as heat. What is the total change in the internal energy (ΔU) of the gas in joules? Given : initial volume, final volume, external pressure, and quantity of energy transferred as heat Asked for: total change in internal energy Strategy: Determine the sign of \(q\) to use in Equation \(\ref{5.2.5}\). From Equation \(\ref{5.2.6}\) calculate \(w\) from the values given. Substitute this value into Equation \(\ref{5.2.5}\) to calculate \(ΔU\). Solution A From Equation \(\ref{5.2.5}\), we know that ΔU = q + w. We are given the magnitude of q (140 J) and need only determine its sign. Because energy is transferred from the system (the gas) to the surroundings, q is negative by convention. B Because the gas is being compressed, we know that work is being done on the system, so \(w\) must be positive. From Equation \(\ref{5.2.5}\), \[w=-P_{\textrm{ext}}\Delta V=-8.00\textrm{ atm}(\textrm{0.0500 L} - \textrm{0.400 L})\left(\dfrac{\textrm{101.3 J}}{\mathrm{L\cdot atm}} \right)=284\textrm{ J} \nonumber \] Thus \[\begin{align*} ΔU &= q + w \\[4pt] &= −140 \,J + 284\, J \\[4pt] &= 144\, J\end{align*} \nonumber \] In this case, although work is done on the gas, increasing its internal energy, heat flows from the system to the surroundings, decreasing its internal energy by 144 J. The work done and the heat transferred can have opposite signs. Exercise \(\PageIndex{1}\) A sample of an ideal gas is allowed to expand from an initial volume of 0.200 L to a final volume of 3.50 L against a constant external pressure of 0.995 atm. At the same time, 117 J of heat is transferred from the surroundings to the gas. What is the total change in the internal energy (ΔU) of the gas in joules? Answer −216 J By convention (to chemists), both heat flow and work have a negative sign when energy is transferred from a system to its surroundings and vice versa. Summary In chemistry, the small part of the universe that we are studying is the system , and the rest of the universe is the surroundings . Open systems can exchange both matter and energy with their surroundings, closed systems can exchange energy but not matter with their surroundings, and isolated systems can exchange neither matter nor energy with their surroundings. A state function is a property of a system that depends on only its present state , not its history. A reaction or process in which heat is transferred from a system to its surroundings is exothermic . A reaction or process in which heat is transferred to a system from its surroundings is endothermic . The first law of thermodynamics states that the energy of the universe is constant. The change in the internal energy of a system is the sum of the heat transferred and the work done. The heat flow is equal to the change in the internal energy of the system plus the PV work done. When the volume of a system is constant, changes in its internal energy can be calculated by substituting the ideal gas law into the equation for ΔU. |
Courses/Smith_College/CHM_223_Chemistry_III%3A_Organic_Chemistry_(2024)/06%3A_Carbonyl_Alpha-Substitution_Reactions/6.08%3A_Alkylation_of_Enolate_Ions | Objectives After completing this section, you should be able to write a general mechanism for the attack of an enolate anion on an alkyl halide. write a reaction sequence to illustrate the preparation of carboxylic acids via the malonic ester synthesis. identify the product formed, and all the intermediates, in a given malonic ester synthesis. identify all of the compounds needed to prepare a given carboxylic acid by a malonic ester synthesis. write a detailed mechanism for each of the steps involved in a malonic ester synthesis. write a reaction sequence to illustrate the preparation of ketones through the acetoacetic ester synthesis. identify the product formed, and all the intermediates, in a given acetoacetic ester synthesis. identify all of the compounds needed to prepare a given ketone by an acetoacetic ester synthesis. write a detailed mechanism for each of the steps involved in an acetoacetic ester synthesis. identify the product or products formed when a given lactone, ester, nitrile or ketone is treated with lithium diisopropylamide followed by an alkyl halide. identify the compounds needed to prepare a given α- substituted ketone, ester, lactone or nitrile by a method involving the alkylation of an enolate anion. Key Terms Make certain that you can define, and use in context, the key terms below. alkylation malonic ester synthesis Study Notes The two syntheses discussed in this section provide routes to a wide variety of carboxylic acids and methyl ketones. You may wish to review the factors influencing S N 2 reactions ( Section 2.4 ) in conjunction with this section. You should try to memorize the structures of malonic ester and ethyl acetoacetate. The IUPAC names of these compounds are shown in the table below. Structure Common name IUPAC name NaN malonic acid propanedioic acid NaN malonic ester or diethyl malonate diethyl propanedioate NaN acetoacetic acid 3-oxobutanoic acid NaN ethyl acetoacetate or acetoacetic ester ethyl 3-oxobutanoate Alkylation of Enolates Enolates can be alkylated in the alpha position through an S N 2 reaction with alkyl halides. During this reaction an α-hydrogen is replaced with an alkyl group and a new C-C bond is formed. The limitations of S N 2 reactions still apply. This includes preferring a good primary or secondary leaving group, X = chloride, bromide, iodide, tosylate. Tertiary leaving groups cannot be used in this reaction and typically give undesired E2 elimination products. A very strong base, such as LDA, is often used because of its ability to form the enolate completely. Removal of the carbonyl starting material from the reaction mixture makes it unavailable for nucleophilic addition by the enolate. Aldehydes are usually not directly alkylated because their enolates prefer to undergo the carbonyl condensation reactions discussed later in Section 7.1 . In addition, the acidic hydrogen on carboxylic acids inhibits the formation of an enolate, and makes their direct alkylation difficult. Esters, including lactones, and symmetrical ketones readily undergo direct alkylation. However, direct alkylations, like all enolate-based reactions, can form a racemic mixture if the alkylated α- carbon produced is chiral. General Reaction Mechanism 1) Enolate formation 2) S N 2 attack Examples When an unsymmetrical ketone with two sets of non-equivalent α- hydrogens is treated with a base, two possible enolates can form. Regioselective enolate formation is possible under the proper conditions. The main determinant is whether the reaction is under kinetic control (rate) or thermodynamic control (equilibrium). Although a predominant product can be produced, a mixture of products is usually formed causing a reduction in product yield. Thermodynamic Enolates The thermodynamic enolate is formed when the more substituted α- hydrogen is removed. This leads to the more alkyl substituted, therefore the more stable, enolate to be formed. The presence of additional alkyl groups causes the formation of the thermodynamic enolate to be sterically hindered and kinetically slow, especially when a bulky base like LDA is used. Thermodynamic enolates are favored by conditions which allow for equilibration between the possible enolates. When the ketone starting material is not completely deprotonated, equilibrium between the possible enolates and the α- hydrogens of the ketone can occur. During equilibrium, interconversion between the enolates allows the lower energy of the thermodynamic enolate to dominate. Other conditions can also promote the formation of the thermodynamic enolate, such as higher reaction temperatures, or the use of a smaller less sterically hindered base such as sodium hydride (NaH). Weaker bases, such as sodium ethoxide, do not completely deprotonate the ketone starting material which also allows for enolate equilibrium to occur. Example of enolate equilibration Kinetic Enolates Kinetic enolates are favored under conditions which do not allow for equilibration between the enolates, such as the use of a strong bulky base, like LDA, in a molar equivalent to the ketone starting material. Kinetic enolates are formed when the less substituted α- hydrogen is deprotonated. Being less sterically hindered allows this α- hydrogen to be deprotonated faster even though it forms a less thermodynamically stable enolate. Using a molar equivalent of LDA completely converts the ketone starting material to an enolate, removing it from the reaction mixture and preventing equilibration between the possible enolates. Low reaction temperatures (-78 o C) prevent enolate equilibration and promote the formation of the kinetic enolate. When and enolate of an asymmetric ketone is stabilized through additional resonance forms there is no competition between possible enolates despite kinetic or thermodynamics conditions. The resonance stabilized enolate will be preferentially alkylated to the point that formation of the alkylated products of other possible enolates will be minimal. Example Malonic Ester Synthesis The malonic ester synthesis is a series of reactions which converts an alkyl halide to a carboxylic acid with two additional carbons. One important use of this synthesis pathway is that it allows for the creation of α -alkylated carboxylic acids which cannot be created by direct alkylation. The starting material of this reaction is a malonic ester: a diester derivative of malonic acid. Diethyl propanedioate, also known as diethyl malonate, is the malonic ester most commonly used in pathway. Since it is a 1,3-dicarbonyl compound, diethyl malonate has relatively acidic α -hydrogens (pK a = 12.6) and can be transformed to its enolate using sodium ethoxide as a base. Other alkoxide bases are not typically used given the possibility of a transesterification reaction. General Reaction Predicting the Product of a Malonic Ester Synthesis The product of a Malonic Ester Synthesis can be created by simply replacing the halogen on the alkyl halide with a -CH 2 CO 2 H group. Malonic ester synthesis takes place in four steps: 1) Enolate Formation Reacting diethyl malonate with sodium ethoxide (NaOEt) forms a resonance-stabilized enolate. 2) Alkylation The enolate is alkylated via an S N 2 reaction to form an monoalkylmalonic ester. 3) Ester hydrolysis and protonation After alkylation, the diester undergoes hydrolysis with sodium hydroxide to form a dicarboxylate. Subsequent protonation with acid forms a monoalkyl malonic acid. 4) Decarboxylation & Tautomerization Monoalkyl malonic acids decarboxylate when heated, forming an α -alkyl carboxylic acid and carbon dioxide (CO 2 ). Decarboxylation can only occur in compounds with a second carbonyl group two atoms away from carboxylic acid such as in malonic acids and β -keto acids. The mechanism occurs via a concerted mechanism involving a proton transfer between the carboxyl acid hydrogen and the nearby carbonyl group to form the enol of a carboxylic acid and CO 2 . The enol undergoes tautomerization to form the carboxylic acid. Example Dialkylation The presence of two α -hydrogens in malonic esters allows for a second alkylation to be performed prior to decarboxylation. This leads to dialkylated carboxylic acids. Due to the lack of stereochemical control inherent in enolate based reactions, if the two added alkyl groups are different, a racemic mixture of products will result. Examples In a variation of the dialkylation reaction - if one molar equivalent of malonic ester is reacted with one molar equivalent of a dihaloalkane and two molar equivalents of sodium ethoxide, a cyclization reaction occurs. By changing the dihaloalkane, three, four, five, and six-membered rings can be created. The Acetoacetic Ester Synthesis The acetoacetic ester synthesis is a series of reactions which converts alkyl halides into a methyl ketone with three additional carbons. This reaction creates an α -substituted methyl ketone without side-products. The starting reagent for this pathway is ethyl 3-oxobutanoate, also called ethyl acetoacetate, or acetoacetic ester. Like other 1,3-dicarbonyl compounds, ethyl acetoacetate is more acidic than ordinary esters being almost completely converted to an enolate by sodium ethoxide. General Reaction Predicting the Product of an Acetoacetic Ester Synthesis The product of a acetoacetic ester synthesis can be created by replacing halogen on the alkyl halide with a -CH 2 COCH 3 group. Reaction Steps 1) Formation of the enolate As previously described, the α -hydrogens of acetoacetic ester are rather acidic (pK a = 10.7) allowing the enolate to be easily formed when sodium ethoxide is used as a base. 2) Alkylation via an S N 2 Reaction Subsequent reaction with an alkyl halide produces a monoalkylacetoacetic ester. 3) Ester hydrolysis and decarboxylation Hydrolysis with NaOH followed by protonation produces an alkylated beta-ketoacid. β -ketoacids are easily decoboxylated to form an α -alkyl substituted methyl ketone and carbon dioxide (CO 2 ) using a similar mechanism as the malonic ester synthesis. Examples Much like the malonic ester synthesis, a second alkyl group can added before the decarboxylation step. The reaction steps of the acetoacetic ester synthesis can also be applied to other β -keto esters with acidic α -hydrogens. Because the α -hydrogens between the two carbonyls are the most acidic, they are preferentially deprotonated allowing for a single enolate to be formed. Even cyclic beta-keto esters can be alkylated and subsequently decarboxylated to give an α -alkylated cyclic ketone. Examples Direct Alkylation of Nitriles The presence of acidic α -hydrogens in nitriles gives them the ability to form an enolate equivalent which can be also be directly alkylated. General Reaction Mechanism Example Planning a Synthesis Using Enolate Alkylations When planning a synthesis that could involve enolates, the key is to recognize the functionality which can form an enolate. During retrosynthetic analysis a C-C bond is broken between the α -carbon and the β -carbon away from this functionality. It is also important to be able to identify specific groups of atoms which indicate if a malonic ester or an acetoacetic ester synthesis can be used. Having multiple C-C bonds which can be broken allows for multiple synthetic pathways. After retrosynthetically breaking the C-C bond, the fragment with the functionality will gain a hydrogen and the other fragment will gain a halogen. Sometimes the fragment with the functionality will become diethyl malonate or acetoacetic ester. Worked out example: Plan a synthesis of the following molecule using an alkylation of an enolate. Consider multiple pathways and explain which is preferable. The target molecule does not contain the appropriate fragments to utilize either the malonic ester or acetoacetic acid synthesis so direct alkylation of a ketone will likely be used. When analyzing this molecule, there are three α - β C-C bonds which could be cleaved to create a possible starting material. When looking at the possible starting materials, A and C are asymmetrical ketones and therefore can create multiple products during alkylation. B is a symmetrical ketone and should be the most likely to create the target molecule in high yield. Possible Synthesis Exercise \(\PageIndex{1}\) Propose a synthesis for each of the following molecules from this malonic ester. (a) (b) (c) Answer (a) 1) Malonic Ester, NaOEt, 2) 4-Methylbenzyl Bromide, 3) Base, 4) Acid, Heat (b) 1) Malonic Ester, NaOEt, 2) 3-bromohexane, 3) Base, 4) Acid, Eat (c) 1) Malonic Ester, NaOEt, 2) 1-Bromo-2,3,3-trimethylbutane, 3) Base, 4) Acid, Heat Exercise \(\PageIndex{2}\) Why can't you prepare tri substituted acetic acids from a malonic ester? Answer Malonic esters only contain two acid protons. Exercise \(\PageIndex{3}\) Propose a synthesis for the following molecule via a malonic ester. Answer Exercise \(\PageIndex{4}\) How might you prepare the following compounds from an alkylation reaction? (a) (b) (c) Answer (a) (b) (c) |
Ancillary_Materials/Worksheets/Worksheets%3A_General_Chemistry/Worksheets%3A_General_Chemistry_(Traditional)/Molecular_Structure_2_(Worksheet) | Name: ______________________________ Section: _____________________________ Student ID#:__________________________ Work in groups on these problems. You should try to answer the questions without referring to your textbook. If you get stuck, try asking another group for help. Q1 For each of the following, (i) draw a Lewis diagram, (ii) count the number of electron groups around the central atom and the number of bonded electron groups, (iii) draw a three-dimensional representation of the molecule, (iv) give the values of the ideal bond angles, and (v) give the name of the electron-pair and molecular geometries. Substance i ii iii iv v CH4 NaN NaN NaN NaN NaN NO3– NaN NaN NaN NaN NaN PCl5 NaN NaN NaN NaN NaN SO42– NaN NaN NaN NaN NaN Modeling can be done with physical models such as balls and sticks or electronically with computers. Molecular models are very useful for visualizing molecular shapes and conceptualizing how molecular shapes relate to physical and chemical properties such as solubility and chemical reactivity. Physical models and computer generated graphics are equally valuable visualization tools. Computer graphics are particularly good for large, complex molecules, such as proteins, where building a physical model is difficult. 2. Refer to the references cited in: ep.llnl.gov/msds/Chem120/vsepr.html Complete the following table for the molecular images found on the Web page to include: Lewis Structures, An accurate description of the shape of the molecule, Identifying the molecule as being polar or non-polar. For example, see 4A in the table. A B C 1 NaN NaN 2 NaN NaN 3 NaN NaN 4 NaN NaN Bent, Polar a) Draw all possible resonance structures for 1B and 4A. The bond lengths of the sulfur to oxygen bonds are equal (1.57 Angstroms) for every sulfur to oxygen bond in both moleclues. Briefly explain this fact despite an S-O single bond is >1.57 Angstroms and an S=O double bond is < 1.57 Angstroms. b) Calculate the formal charge for each oxygen atom in the following molecules and clearly indicate their respective charge. c) Answer the tutorial's practice problems: 3, 9 and 11 from the Purdue Web-site: http://www.chem.purdue.edu/gchelp/vsepr/ Refer to: ep.llnl.gov/msds/Chem120/hv-globalCO2.html 3. a) Prepare a table below with a list of the respective frequencies (wavenumber, cm -1 ) of the infrared (IR) absorbances and their corresponding percent transmission in the spectrum for your assigned greenhouse gas. Consult the ChemConnections, MC2/Chemlinks and NIST WebBook pages: chemistry.beloit.edu/Warming/.../infrared.html, http://webbook.nist.gov/chemistry/ b) Using IR Tutor (downloadable from: chemistry.beloit.edu/Warming/.../infrared.html), view the IR spectrum of your assigned greenhouse gas. For each of the absorbances (peaks) that you listed in question #1, add a column or row to the table with the type of molecular vibration that corresponds to each of the absorbances in the spectrum. Eg. anti-symmetrical stretching vibration. Use a separate page if necessary. Table: A molecular model kit consists of colored centers that correspond to various common geometries and connecting joints to link the centers. The connecting joints come in longer and shorter lengths to represent longer and shorter bonds. Use the shorter lengths to indicate bonds to hydrogen and the longer lengths to indicate bonds between any other two atoms. The colored centers usually follow a standard correlation with atoms: 0 1 2 3 4 5 Black C Red O White H Blue N Purple P Yellow S Silver metals Green halogens NaN NaN 4. Complete the following table by: finding an example of a molecule or ion with the given structure predicting the molecular geometry building a model estimating the bond angles. A represents a central atom, B represents a terminal atom, and E represents an unshared electron pair on the central atom. The first line is given as an example. Structure Example Molecular Geometry Bond Angles AB2 CO2 linear 180° AB2E NaN NaN NaN AB3 NaN NaN NaN AB4 NaN NaN NaN AB3E NaN NaN NaN AB2E2 NaN NaN NaN AB5 NaN NaN NaN AB6 NaN NaN NaN 5. Build models of each of the following: H 2 O, H 2 S, CO 2 . a) Consider the electronegativities of the elements and the three-dimensional molecular geometry of each of the three molecules, and match each of the following dipole moments to its corresponding molecule: 0 D, 0.95 D, 1.85 D. Indicate the electronegative and electropositive regions of each polar molecule. 6. Build a model of acetic acid, CH 3 COOH. a) Draw a Lewis diagram of the molecule. b) Identify the electron-pair and molecular geometry around each atom in the acetic acid molecule, i.e. consider each atom separately as the central atom. d) Does the molecule have only one unique shape? What factors contribute to its shape or shapes? The Dutch chemist Jacobus Van’t Hoff (1852–1911) and the French chemist Joseph Le Bel (1847–1930) reasoned that carbon with four covalent bonds must have a tetrahedral shape. This was a tremendous step forward in conceptualizing the chemistry of carbon compounds. Information and experimental methods available to them at the time were limited and sparse compared to today's scientific resources. A molecular model kit will help you to appreciate Van't Hoff's insight into carbon bonds and three dimensional shapes which are universally accepted and used today. Some molecules have the same formula, but can be separated into compounds that exhibit different physical and chemical properties. These compounds are called isomers. Since isomers have different properties, they cannot have exactly the same molecular structure. An instrument that was used to examine compounds in the late nineteenth century was a device known as a polarimeter. It worked by passing a beam of polarized light through a solution to an analyzing filter. Certain types of compounds had isomers with three dimensional shapes that would interact with the light. These compounds are referred to as being optically active. As an example, if a pair of isomers are mirror images of one another, the light of a polarimeter will rotate the beam in one direction for one of the isomers and rotate the beam by the same amount in the opposite direction for the other isomer. Van’t Hoff examined a number of carbon compounds with a polarimeter. Compounds without an isomer did not rotate light, while others with did. Consider the following data table. Compound Number of Isomers CH4 1 CH3Cl 1 CH2Cl2 1 CH2ClBr 1 CHClBrF 2 The compound CHClBrF was shown to have two isomers that were mirror images of each other. Build a molecular model of each of them. The two structures are stereoisomers. They have the same connectivity, or order of attachment among the atoms, but a different orientation of their atoms in space. Enantiomers are stereoisomers whose molecules are mirror images of each other. One enantiomer will rotate polarized light in one direction whereas the other enantiomer rotates polarized light by that same amount but in the opposite direction. What do you think will happen to a beam of polarized light if you have a 50:50 mixture of both enantiomers, which is called a racemic mixture? Consider your hands. Are they mirror images of each other? You may want to use a mirror to check. 7. Sketch three-dimensional representations of the respective CHClBrF enantiomers to show that they are mirror images. Solid lines are in the plane of the paper, dotted lines are behind the plane, and solid wedges come out from the plane. |
Courses/Pasadena_City_College/PCC_Chemistry_2A/03%3A_Atomic_Structure/3.02%3A_Nuclear_Chemistry/3.2.07%3A_Therapeutic_Radiation | Learning Objectives Distinguish between diagnostic and therapeutic radiation. Know the types of cancer treatments available. Recall the A/Z formats of a proton (particle) and gamma (ray). Compare/contrast external beam therapy and brachytherapy. Know the differences between proton and photon radiation. Realize the side effects of radiation therapy and medications available. There are many techniques used to treat cancer. Surgery can be used to remove cancerous tumors inside or on the body. With chemotherapy, ingested or injected chemicals are used to kill rapidly dividing cells (cancerous and noncancerous). For more information regarding chemotherapy, please access this link . Other cancer treatment methods include immunotherapy, stem cell replacement, hormone therapy, and targeted therapy. Radiation therapy and Chemotherapy: Two different treatment procedures Patients diagnosed with cancer might be required to do chemotherapy or radiation therapy. Sometimes, both of these methods are used for a patient. In this picture, a female patient is receiving chemotherapy through an IV. She is participating in cooling therapy while receiving her treatment. By placing her hands and feet in cooling devices, this will lower her chances of losing her finger and toenails. Cold cap therapy is also now available for chemotherapy patients. Wearing this type of device could enable a patient to keep his/her hair during chemotherapy. Current therapeutic radiation applications involve the use of gamma, x-rays, or protons. Recently, some research facilities are investigating the use of alpha and beta tagged molecules to kill cancer cells. These radioisotopes will first locate a cancer related molecule on a tumor cell. Then, the alpha or beta tagged species will inject its radiation into the tumor. Sr-89 (beta emitter) and Ra-223 (alpha emitter) have been used in clinical research trials of certain types of bone cancers. Radiation Therapy is used as a treatment to control malignant cells within cancer patients. Oncologists (specialists that deal with cancer) utilize radiation frequently to help slow or cure the spread of cancer within individuals. Radiation is specifically applied to malignant tumors in order to shrink them in size. Medical professionals, mainly radiation oncologists, administer a variety of dosages to patient, contingent to the patients current health, as well as other treatments such as chemotherapy, success of surgery, etc. External Beam Therapy (Photon and Proton Therapy) External Beam Therapy (EBT) is a method of delivering a high energy beam of radiation to the precise location of a patient's tumor. These beams can destroy cancer cells and with careful planning, NOT kill surrounding cells. The concept is to have several beams of radiation, each of which is sub-lethal, enter the body from different directions. The only place in the body where the beam would be lethal is at the point where all the beams intersect. Before the EBT process, the patient is three-dimensionally mapped using CT scans and x-rays. The patient receives small tattoos to allow the therapist to line up the beams exactly. Alignment lasers are used to precisely locate the target. The radiation beam is usually generated with a linear accelerator. The video below illustrates the basic preparation and administration of external beam therapy. Targeting Cancer - Radiation Therapy Treatment Process Photon EBT utilizes either x-ray or gamma rays. An x-ray source would require a linear accelerator to produce high energy electrons. In contrast, a gamma source incorporates a radioactive isotope (like Co-60). Keep in mind both of these technologies use ionizing radiation. As a result, cancer patients must be monitored throughout their life to ensure they do not develop other cancers, like leukemia. EBT is used to treat the following diseases as well as others: Breast cancer Colorectal cancer Head and neck cancer Lung cancer Prostate cancer The majority of radiation patients receive photon EBT. The smaller size of this machine makes this therapy an option for all sizes of hospitals and cancer treatment centers. Photon EBT equipment costs approximately three million dollars. The size and the price of this technology enables smaller facilities to keep their patients closer to home during treatment In the past ten years, another method of radiation treatment involving protons has become more available. This type of therapy requires a cyclotron to generate proton beams (recall, a proton is an ionized H-1 isotope). Unlike x-ray or gamma rays (photon therapy), protons are extremely heavy. Today's proton facilities occupy large plots of land to house (the size of a football field that is three stories high) the medical cyclotron. With a price tag of 150-200 million, these facilities are sporadically located in the United States. At this time, research facilities are working on miniaturizing proton generators. An ideal technology would reduce the cost from hundreds of millions of dollars to about twenty to thirty million per device. This would make proton therapy more available and convenient for patients. Proton Therapy versus Photon Therapy There are many advantages to choosing proton therapy over photon therapy. Unlike photon radiation, proton beams will only penetrate to the depth of the tumor and not pass through the entire body. This reduces the overall toxicity dose. In addition, fewer treatments are required for proton therapy patients than photon therapy.Unfortunately, proton therapy is more expensive than photon therapy and less common. Once approved by a facility and medical insurance, a patient may have to move temporarily to a larger city to receive treatment. Some forms of cancer have never been clinically treated with proton therapy (namely, breast cancer). Patients desiring proton therapy might not be able to receive type of therapy due to lack of research. Brachytherapy The Greek term brachy means to travel a short distance. This type of radiation involves placing ionizing pellets(seeds) or rods directly at the tumor. Photons (in the form of x-ray or gamma ray) are produced inside the body and will penetrate throughout this particular area localizing the radiation. Pellets are surgically implanted while rods can be temporarily inserted to produce radiation internally. With pellets/seeds, the patient will remain radioactive as long as these devices remain inside the body. People undergoing this type of radiotherapy need to be aware of their constant emission of radiation. Radiation that is administered through rods connected to a photon device will disperse energy immediately and not leave the patient radioactive. Brachytherapy is widely used in the treatment of cancers involving reproductive organs. Because the radiation is isolated internally, patients are less likely to experience side effects when receiving this type of treatment. Cancers that have been treated with brachytherapy are shown below: Prostate Breast Esophageal Lung Uterine Anal/Rectal Sarcomas Head and neck Radionuclide Type Half-life Cesium-131 (131Cs) Electron Capture, ε 9.7 days Cesium-137 (137Cs) β−- particles, γ-rays 30.17 years Cobalt-60 (60Co) β−- particles, γ-rays 5.26 years Iridium-192 (192Ir) γ-rays 73.8 days Iodine-125 (125I) Electron Capture, ε 59.6 days Palladium-103 (103Pd) Electron Capture, ε 17.0 days Ruthenium-106 (106Ru) β−- particles 1.02 years Radium-226 (226Ra) β−- particles 1599 years Side Effects of Radiation Therapy Patients receiving radiation therapy can experience a variety of side effects. For example, sterility could occur if reproductive organs are irradiated. Skin that has been irradiated can appear dry and feel itchy. Some patients will loose sensation in the irradiated area. Radiation can affect the production of white and red blood cells. A reduction of white blood cells results in immunity disorders. Red blood cell lose causes anemia. Gastrointestinal issues such as diarrhea and nausea are common during radiation therapy. Some patients will lose hair as well. Lastly, dry mouth and tooth decay are prevalent during radiation treatments. Medications are available to alleviate symptoms of radiation therapy. Narcotics can be prescribed to help alleviate intense pain. Prescription medications like zofran and phenergan can help with nausea. Special mouthwashes have been formulated to reduce dry mouth and cavities. Hair loss is a side effect of radiation, but only locally Radiation therapy can cause hair loss, but hair is only lost in the area being treated. For instance, radiation to your head may cause you to lose some or all the hair on your head (even eyebrows and lashes), but if you get treatment to your hip, you won’t lose the hair on your head. Access to Radiotherapy For those countries who do not have access to nuclear technologies, the International Atomic Energy Agency (IAEA) provides support and technical assistance to its member countries. When nations chose to join the IAEA, then they will gain nuclear materials and technologies for energy, medical, industrial, and agriculture purposes. Once accepted into the IAEA, a country must allow weapons inspectors to monitor any equipment that could be used to make nuclear weapons. The IAEA does not allow countries that do not have nuclear weapons to obtain them. IAEA video: Fighting Cervical Cancer: A Fair Chance for Every Woman Around the World References https://www.cancer.gov/about-cancer/treatment/types https://weillcornellgucancer.org/201...-cancer-cells/ http://www.proton-therapy.org/zapper.htm www.brachytherapy.com/ https://www.iaea.org/ |
Courses/Matanuska-Susitna_College/MatSu_College-CHEM_A104_Introduction_to_Organic_and_Biochemistry/09%3A_Solutions/9.07%3A_Le_Chatelier's_Principle | Learning Outcomes Define Le Chatelier's principle. Predict how the change in amounts of substances, temperature, or pressure will affect amounts of reactants and products present at equilibrium. Le Chatelier's Principle Chemical equilibrium was studied by the French chemist Henri Le Chatelier (1850 - 1936) and his description of how a system responds to a stress to equilibrium has become known as Le Chatelier's principle : When a chemical system that is at equilibrium is disturbed by a stress, the system will respond in order to relieve the stress. Stresses to a chemical system involve changes in the concentrations of reactants or products, changes in the temperature of the system, or changes in the pressure of the system. We will discuss each of these stresses separately. The change to the equilibrium position in every case is either a favoring of the forward reaction or a favoring of the reverse reaction. When the forward reaction is favored, the concentrations of products increase, while the concentrations of reactants decrease. When the reverse reaction is favored, the concentrations of the products decrease, while the concentrations of reactants increase. \[\begin{array}{lll} \textbf{Original Equilibrium} & \textbf{Favored Reaction} & \textbf{Result} \\ \ce{A} \rightleftharpoons \ce{B} & \text{Forward:} \: \ce{A} \rightarrow \ce{B} & \left[ \ce{A} \right] \: \text{decreases}; \: \left[ \ce{B} \right] \: \text{increases} \\ \ce{A} \rightleftharpoons \ce{B} & \text{Reverse:} \: \ce{A} \leftarrow \ce{B} & \left[ \ce{A} \right] \: \text{increases}; \: \left[ \ce{B} \right] \: \text{decreases} \end{array} \nonumber \] Effect of Concentration A change in concentration of one of the substances in an equilibrium system typically involves either the addition or the removal of one of the reactants or products. Consider the Haber-Bosch process for the industrial production of ammonia from nitrogen and hydrogen gases. \[\ce{N_2} \left( g \right) + 3 \ce{H_2} \left( g \right) \rightleftharpoons 2 \ce{NH_3} \left( g \right) \nonumber \] If the concentration of one substance in a system is increased, the system will respond by favoring the reaction that removes that substance. When more \(\ce{N_2}\) is added, the forward reaction will be favored because the forward reaction uses up \(\ce{N_2}\) and converts it to \(\ce{NH_3}\). The forward reaction speeds up temporarily as a result of the addition of a reactant. The position of equilibrium shifts as more \(\ce{NH_3}\) is produced. The concentration of \(\ce{NH_3}\) increases, while the concentrations of \(\ce{N_2}\) and \(\ce{H_2}\) decrease. After some time passes, equilibrium is reestablished with new concentrations of all three substance. As can be seen in the figure below, if more \(\ce{N_2}\) is added, a new equilibrium is achieved by the system. The new concentration of \(\ce{NH_3}\) is higher because of the favoring of the forward reaction. The new concentration of the \(\ce{H_2}\) is lower .The concentration of \(\ce{N_2}\) is higher than in the original equilibrium, but went down slightly following the addition of the \(\ce{N_2}\) that disturbed the original equilibrium. By responding in this way, the value of the equilibrium constant for the reaction, \(K_\text{eq}\), does not change as a result of the stress to the system. In other words, the amount of each substance is different but the ratio of the amount of each remains the same. If more \(\ce{NH_3}\) were added, the reverse reaction would be favored. This "favoring" of a reaction means temporarily speeding up the reaction in that direction until equilibrium is reestablished. Recall that once equilibrium is reestablished, the rates of the forward and reverse reactions are again equal. The addition of \(\ce{NH_3}\) would result in increased formation of the reactants, \(\ce{N_2}\) and \(\ce{H_2}\). An equilibrium can also be disrupted by the removal of one of the substances. If the concentration of a substance is decreased, the system will respond by favoring the reaction that replaces that substance. In the industrial Haber-Bosch process, \(\ce{NH_3}\) is removed from the equilibrium system as the reaction proceeds. As a result, the forward reaction is favored so that more \(\ce{NH_3}\) is produced. The concentrations of \(\ce{N_2}\) and \(\ce{H_2}\) decrease. Continued removal of \(\ce{NH_3}\) will eventually force the reaction to go to completion until all of the reactants are used up. If either \(\ce{N_2}\) or \(\ce{H_2}\) were removed from the equilibrium system, the reverse reaction would be favored and the concentration of \(\ce{NH_3}\) would decrease. The effect of changes in concentration on an equilibrium system according to Le Chatelier's principle is summarized in the table below. Stress Response addition of reactant forward reaction favored addition of product reverse reaction favored removal of reactant reverse reaction favored removal of product forward reaction favored Example \(\PageIndex{1}\) Given this reaction at equilibrium: \[N_{2}+3H_{2}\rightleftharpoons 2NH_{3} \nonumber \] How will it affect the reaction if the equilibrium is stressed by each change? H 2 is added. NH 3 is added. NH 3 is removed. Solution If H 2 is added, there is now more reactant, so the reaction will shift to the right (toward products) to reduce the added H 2 . If NH 3 is added, there is now more product, so the reaction will shift to the left (toward reactants) to reduce the added NH 3 . If NH 3 is removed, there is now less product, so the reaction will shift to the right (toward products) to replace the product removed. Given this reaction at equilibrium: \[CO(g)+Br_{2}(g)\rightleftharpoons COBr_{2}(g) \nonumber \] How will it affect the reaction if the equilibrium is stressed by each change? Br 2 is removed. COBr 2 is added. Answer 1. shift to the left (toward reactants) 2. shift to the left (toward reactants) Effect of Temperature Increasing or decreasing the temperature of a system at equilibrium is also a stress to the system. The equation for the Haber-Bosch process is written again below, as a thermochemical equation (i.e. it contains information about the energy gained or lost when the reaction occurs). \[\ce{N_2} \left( g \right) + 3 \ce{H_2} \left( g \right) \rightleftharpoons 2 \ce{NH_3} \left( g \right) + 91 \: \text{kJ} \nonumber \] The forward reaction is the exothermic direction: the formation of \(\ce{NH_3}\) releases heat which is why that is shown as a product. The reverse reaction is the endothermic direction: as \(\ce{NH_3}\) decomposes to \(\ce{N_2}\) and \(\ce{H_2}\), heat is absorbed. An increase in the temperature for this is like adding a product because heat is being released by the reaction. If we add a product then the reaction proceeds towards the formation of more reactants. Reducing the temperature for this system would be similar to removing a product which would favor the formation of more products. The amount of \(\ce{NH_3}\) will increase and the amount of \(\ce{N_2}\) and \(\ce{H_2}\) will decrease. For changes in concentration, the system responds in such a way that the value of the equilibrium constant, \(K_\text{eq}\) is unchanged. However, a change in temperature shifts the equilibrium and the \(K_\text{eq}\) value either increases or decreases. As discussed in the previous section, values of \(K_\text{eq}\) are dependent on the temperature. When the temperature of the system for the Haber-Bosch process is increased, the resultant shift in equilibrium towards the reactants means that the \(K_\text{eq}\) value decreases. When the temperature is decreased, the shift in equilibrium towards the products means that the \(K_\text{eq}\) value increases. Le Chatelier's principle as related to temperature changes can be illustrated easily be the reaction in which dinitrogen tetroxide is in equilibrium with nitrogen dioxide. \[\ce{N_2O_4} \left( g \right) + \text{heat} \rightleftharpoons 2 \ce{NO_2} \left( g \right) \nonumber \] Dinitrogen tetroxide \(\left( \ce{N_2O_4} \right)\) is colorless, while nitrogen dioxide \(\left( \ce{NO_2} \right)\) is dark brown in color. When \(\ce{N_2O_4}\) breaks down into \(\ce{NO_2}\), heat is absorbed (endothermic) according to the forward reaction above. Therefore, an increase in temperature (adding heat) of the system will favor the forward reaction. Conversely, a decrease in temperature (removing heat) will favor the reverse reaction. Example \(\PageIndex{2}\) Predict the effect of increasing the temperature on this equilibrium. \[PCl_{3}+Cl_{2}\rightleftharpoons PCl_{5}+60kJ \nonumber \] Solution Because energy is listed as a product, it is being produced, so the reaction is exothermic. If the temperature is increasing, a product is being added to the equilibrium, so the equilibrium shifts to minimize the addition of extra product: it shifts to the left (back toward reactants). Predict the effect of decreasing the temperature on this equilibrium. \[N_{2}O_{4}+57kJ\rightleftharpoons 2NO_{2} \nonumber \] Answer Equilibrium shifts to the left (toward reactants). Effect of Pressure Changing the pressure of an equilibrium system in which gases are involved is also a stress to the system. A change in the pressure on a liquid or a solid has a negligible effect. We will return again the equilibrium for the Haber-Bosch process. Imagine the gases are contained in a closed system in which the volume of the system is controlled by an adjustable piston as shown in the figure below. On the far left, the reaction system contains primarily \(\ce{N_2}\) and \(\ce{H_2}\), with only one molecule of \(\ce{NH_3}\) present. As the piston is pushed inwards, the pressure of the system increases according to Boyle's law. This is a stress to the equilibrium. In the middle image, the same number of molecules is now confined in a smaller space and so the pressure has increased. According to Le Chatelier's principle, the system responds in order to relieve the stress. In the image on the right, the forward reaction has been favored and more \(\ce{NH_3}\) is produced. The overall result is a decrease in the number of gas molecules in the entire system. This in turn decreases the pressure and provides a relief to the original stress of a pressure increase. An increase in pressure on an equilibrium system favors the reaction which products fewer total moles of gas. In this case, it is the forward reaction that is favored. A decrease in pressure on the above system could be achieved by pulling the piston outward, increasing the container volume. The equilibrium would respond by favoring the reverse reaction in which \(\ce{NH_3}\) decomposes to \(\ce{N_2}\) and \(\ce{H_2}\). This is because the overall number of gas molecules would increase and so would the pressure. A decrease in pressure on an equilibrium system favors the reaction which produces more total moles of gas. This is summarized in the table below. Stress Response pressure increase reaction produces fewer gas molecules pressure decrease reaction produces more gas molecules Like changes in concentration, the \(K_\text{eq}\) value for a given reaction is unchanged by a change in pressure. The amounts of each substance will change but the ratio will not. It is important to remember when analyzing the effect of a pressure change on equilibrium that only gases are affected. If a certain reaction involves liquids or solids, they should be ignored. For example, calcium carbonate decomposes according to the equilibrium reaction: \[\ce{CaCO_3} \left( s \right) \rightleftharpoons \ce{CaO} \left( s \right) + \ce{O_2} \left( g \right) \nonumber \] Oxygen is the only gas in the system. An increase in the pressure of the system slows the rate of decomposition of \(\ce{CaCO_3}\) because the reverse reaction is favored. When a system contains equal moles of gas on both sides of the equation, pressure has no effect on the equilibrium position, as in the formation of \(\ce{HCl}\) from \(\ce{H_2}\) and \(\ce{Cl_2}\). \[\ce{H_2} \left( g \right) + \ce{Cl_2} \left( g \right) \rightleftharpoons 2 \ce{HCl} \left( g \right) \nonumber \] Example \(\PageIndex{3}\) What is the effect on this equilibrium if pressure is increased? \[N_{2}(g)+3H_{2}(g)\rightleftharpoons 2NH_{3}(g) \nonumber \] Solution According to Le Chatelier's principle, if pressure is increased, then the equilibrium shifts to the side with the fewer number of moles of gas. This particular reaction shows a total of 4 mol of gas as reactants and 2 mol of gas as products, so the reaction shifts to the right (toward the products side). What is the effect on this equilibrium if pressure is decreased? \[3O_{2}(g)\rightleftharpoons 2O_{3}(g) \nonumber \] Answer Reaction shifts to the left (toward reactants). Application of Le Chatelier's Principle Oxygen transport by the blood In aerobic respiration, oxygen is transported to the cells where it is combined with glucose and metabolized to carbon dioxide, which then moves back to the lungs from which it is expelled. hemoglobin + O 2 oxyhemoglobin The partial pressure of O 2 in the air is 0.2 atm, sufficient to allow these molecules to be taken up by hemoglobin (the red pigment of blood) in which it becomes loosely bound in a complex known as oxyhemoglobin. At the ends of the capillaries which deliver the blood to the tissues, the O 2 concentration is reduced by about 50% owing to its consumption by the cells. This shifts the equilibrium to the left, releasing the oxygen so it can diffuse into the cells. Key Takeaways In a reaction at equilibrium, the introduction of more products will shift the mass balance towards more reactants, and the introduction of more reactants will lead to the formation of more products, but the ratio of Products/Reactants ( equilibrium constant ), K is unchanged. If temperature is changed, the numeric value K will change. If a reaction is exothermic (releases heat), an increase in the temperature will force the equilibrium to the left, causing the system to absorb heat and thus partially offsetting the rise in temperature. The opposite effect occurs for endothermic reactions, which are shifted to the right by rising temperature. The effect of pressure on an equilibrium is significant only for reactions which involve different numbers of moles of gases on the two sides of the equation. An increase in the total pressure will shift to the side with fewer moles of gas. A decrease in pressure will shift to the side with more moles of gas. |
Courses/DePaul_University/Thermodynamics_and_Introduction_to_Quantum_Mechanics_(Southern)/01%3A_Fundamentals_of_Thermodynamics/1.03%3A_Equations_of_State_for_Real_Gases | According to Boyle's law , the product PV is a constant at any given temperature, so a plot of PV as a function of the pressure of an ideal gas yields a horizontal straight line. This implies that any increase in the pressure of the gas is exactly counteracted by a decrease in the volume as the molecules are crowded closer together. But we know that the molecules themselves are finite objects having volumes of their own, and this must place a lower limit on the volume into which they can be squeezed. So we must reformulate the ideal gas equation of state as a relation that is true only in the limiting case of zero pressure: \[\lim_{P \rightarrow 0} PV=nRT\] So what happens when a real gas is subjected to a very high pressure? The outcome varies with both the molar mass of the gas and its temperature, but in general we can see the the effects of both repulsive and attractive intermolecular forces: Repulsive forces: As a gas is compressed, the individual molecules begin to get in each other's way, giving rise to a very strong repulsive force acts to oppose any further volume decrease. We would therefore expect the PV vs P line to curve upward at high pressures, a nd this is in fact what is observed for all gases at sufficiently high pressures. Attractive forces: At very close distances, all molecules repel each other as their electron clouds come into contact. At greater distances, however, brief statistical fluctuations in the distribution these electron clouds give rise to a universal attractive force between all molecules. The more electrons in the molecule (and thus the greater the molecular weight), the greater is this attractive force. As long as the energy of thermal motion dominates this attractive force, the substance remains in the gaseous state, but at sufficiently low temperatures the attractions dominate and the substance condenses to a liquid or solid. The universal attractive force described above is known as the dispersion , or London force. There may also be additional (and usually stronger) attractive forces related to charge imbalance in the molecule or to hydrogen bonding. These various attractive forces are often referred to collectively as van der Waals forces . A plot of PV/RT as a function of pressure is a very sensitive indicator of deviations from ideal behavior, since such a plot is just a horizontal line for an ideal gas. The two illustrations below show how these plots vary with the nature of the gas, and with temperature. Intermolecular attractions , which generally increase with molecular weight, cause the PV product to decrease as higher pressures bring the molecules closer together and thus within the range of these attractive forces; the effect is to cause the volume to decrease more rapidly than it otherwise would. The repulsive forces always eventually win out. But as the molecules begin to intrude on each others' territory, the stronger repulsive forces cause the curve to bend upward. The temperature makes a big difference! At higher temperatures, increased thermal motions overcome the effects of intermolecular attractions which normally dominate at lower pressures. So all gases behave more ideally at higher temperatures. For any gas, there is a special temperature (the Boyle temperature ) at which attractive and repulsive forces exactly balance each other at zero pressure. As you can see in this plot for methane, this some of this balance does remain as the pressure is increased. The van der Waals Equation of State How might we modify the ideal gas equation of state to take into account the effects of intermolecular interactions? The first and most well known answer to this question was offered by the Dutch scientist J.D. van der Waals (1837-1923) in 1873. The ideal gas model assumes that the gas molecules are merely points that occupy no volume; the " V " term in the equation is the volume of the container and is independent of the nature of the gas. van der Waals recognized that the molecules themselves take up space that subtracts from the volume of the container, so that the “volume of the gas” V in the ideal gas equation should be replaced by the term ( V–b ), in which b relates to the excluded volume , typically of the order of 20-100 cm 3 mol –1 . The excluded volume surrounding any molecule defines the closest possible approach of any two molecules during collision. Note that the excluded volume is greater then the volume of the molecule, its radius being half again as great as that of a spherical molecule. The other effect that van der Waals needed to correct for are the intermolecular attractive forces. These are ignored in the ideal gas model, but in real gases they exert a small cohesive force between the molecules, thus helping to hold the gas together and reducing the pressure it exerts on the walls of the container. Because this pressure depends on both the frequency and the intensity of collisions with the walls, the reduction in pressure is proportional to the square of the number of molecules per volume of space, and thus for a fixed number of molecules such as one mole, the reduction in pressure is inversely proportional to the square of the volume of the gas. The smaller the volume, the closer are the molecules and the greater will be the effect. The van der Waals equation replaces the \(P\) term in the ideal gas equation with \(P + (a / V^2)\) in which the magnitude of the constant a increases with the strength of the intermolecular attractive forces. The complete van der Waals equation of state can be written as Although most students are not required to memorize this equation, you are expected to understand it and to explain the significance of the terms it contains. You should also understand that the van der Waals constants \(a\) and \(b\) must be determined empirically for every gas. This can be done by plotting the P-V behavior of the gas and adjusting the values of a and b until the van der Waals equation results in an identical plot. The constant \(a\) is related in a simple way to the molecular radius; thus the determination of \(a\) constitutes an indirect measurement of an important microscopic quantity. Substance molar mass, g a (L2-atm mole–2) b (L mol–1) hydrogen H2 2 0.244 0.0266 helium He 4 0.034 0.0237 methane CH4 16 2.250 0.0428 water H2O 18 5.460 0.0305 nitrogen N2 28 1.390 0.0391 carbon dioxide CO2 44 3.590 0.0427 carbon tetrachloride CCl4 154 20.400 0.1383 The van der Waals equation is only one of many equations of state for real gases. More elaborate equations are required to describe the behavior of gases over wider pressure ranges. These generally take account of higher-order nonlinear attractive forces, and require the use of more empirical constants. Although we will make no use of them in this course, they are widely employed in chemical engineering work in which the behavior of gases at high pressures must be accurately predicted. |
Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/14%3A_Delocalized_Pi_Systems%3A_Investigation_by_Ultraviolet_and_Visible_Spectroscopy/14.07%3ADelocalization_among__More__than__Two__Pi__Bonds%3A__Extended_Conjugation_and__Benzene | Among the many distinctive features of benzene , its aromaticity is the major contributor to why it is so unreactive. This section will try to clarify the theory of aromaticity and why aromaticity gives unique qualities that make these conjugated alkenes inert to compounds such as Br 2 and even hydrochloric acid. It will also go into detail about the unusually large resonance energy due to the six conjugated carbons of benzene . The delocalization of the p-orbital carbons on the sp 2 hybridized carbons is what gives the aromatic qualities of benzene. This diagram shows one of the molecular orbitals containing two of the delocalized electrons, which may be found anywhere within the two "doughnuts". The other molecular orbitals are almost never drawn. Benzene, C 6 H 6 , is a planar molecule containing a ring of six carbon atoms, each with a hydrogen atom attached. The six carbon atoms form a perfectly regular hexagon. All of the carbon-carbon bonds have exactly the same lengths - somewhere between single and double bonds. There are delocalized electrons above and below the plane of the ring. The presence of the delocalized electrons makes benzene particularly stable. Benzene resists addition reactions because those reactions would involve breaking the delocalization and losing that stability. Benzene is represented by this symbol, where the circle represents the delocalized electrons, and each corner of the hexagon has a carbon atom with a hydrogen attached. Basic Structure of Benzene Because of the aromaticity of benzene, the resulting molecule is planar in shape with each C-C bond being 1.39 Å in length and each bond angle being 120°. You might ask yourselves how it's possible to have all of the bonds to be the same length if the ring is conjugated with both single (1.47 Å) and double (1.34 Å), but it is important to note that there are no distinct single or double bonds within the benzene. Rather, the delocalization of the ring makes each count as one and a half bonds between the carbons which makes sense because experimentally we find that the actual bond length is somewhere in between a single and double bond. Finally, there are a total of six p-orbital electrons that form the stabilizing electron clouds above and below the aromatic ring. Contributors Jim Clark ( Chemguide.co.uk ) |
Bookshelves/Inorganic_Chemistry/Online_Dictionary_of_Crystallography_(IUCr_Commission)/01%3A_Fundamental_Crystallography/1.59%3A_Incommensurate_magnetic_structure | An incommensurate magnetic structure is a structure in which the magnetic moments are ordered, but without periodicity that is commensurate with that of the nuclear structure of the crystal. In particular, the magnetic moments have a spin density with wave vectors that have at least one irrational component with respect to the reciprocal lattice of the atoms. Or, in the case of localized moments, the spin function S ( n + r j ) (where the j th atom has position r j in the unit cell) has Fourier components with irrational indices with respect to the reciprocal lattice of the crystal. |
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Chemical_Bonding/Molecular_Orbital_Theory/MO._Molecular_Orbitals/MO1._Introduction | At an elementary level, we can think about atoms being held together by simple electrostatic attraction. It is a fundamental principal that opposite charges attract. A positively charged ion and a negatively charged ion are held together by this force of attraction. This idea works well for ionic compounds, such as sodium chloride. It does not work well for compounds in which similar atoms are connected, such as ethane, \(C_2H_6\). Maybe the carbons are negatively charged and the hydrogens are positively charged, but what is holding the two carbons together if they have like charges? A similar problem is encountered in diatomic molecules such as \(H_2\). Is one of these hydrogen atoms negative while the other hydrogen atom is positive? In the early twentieth century, G. N. Lewis noticed a trend in the characteristics of compounds that he called "the rule of two." If you were to count up the total number of electrons in any stable compound, you would always come up with an even number -- that is, some number that is divisible by two. Perhaps, Lewis reasoned, this predominance of even numbers arises because electrons need to be in pairs. What does an element do if it has an odd number of electrons? One solution is to steal an electron from another element, or to allow one to be stolen away; these arrangements lead to ionic bonds. However, those elements not adept at stealing electrons may have a problem; they may need to share them instead. In order to share electrons, elements will have to form close associations with each other. They will become bonded together. An "ionic bond" is an electrostatic interaction between an anion and a cation. A "covalent bond" is one pair of electrons shared between two atoms. Lewis took this idea further. If you count up the valence shell electrons around each of the atoms in stable compounds, not only are there even numbers, but there are almost always the same number of electrons as there are in one of the noble gases: He, Ne, Ar, Kr (2, 8, 8, or 18). This observation is sometimes called the Lewis octet rule because so many common atoms that form stable compounds obtain 8 electrons in their outermost shell as a result. Neon is the nearest noble gas to carbon, oxygen and nitrogen, and all of these atoms adopt 8-electron configurations in stable compounds. Lewis structures illustrate how atoms can maintain these numbers of electrons by sharing with other atoms. These simple structural drawings are used to convey most of our ideas about molecular chemistry. However, additional information can often be found through quantum mechanics and a molecular orbital approach to bonding. |
Courses/Saint_Marys_College_Notre_Dame_IN/CHEM_118_(Under_Construction)/CHEM_118_Textbook/09%3A_Acids_Bases_and_Buffers/9.03%3A_Strong_and_Weak_Acids_and_Acid_Ionization_Constant_%5C(%5Cleft(_K_%5Ctext%7Ba%7D_%5Cright)%5C) | The etching of glass is a slow process that can produce beautiful artwork. Traditionally, the glass has been treated with dilute hydrofluoric acid which gradually dissolves the glass under it. Parts of the piece that should not be etched are covered with wax or some other non-reactive material. In more recent times, compounds such as ammonium bifluoride have been used. Whichever chemical is employed, the artist must be very careful not to get any on their skin. Strong and Weak Acids and Acid Ionization Constant Acids are classified as either strong or weak, based on their ionization in water. A strong acid is an acid which is completely ionized in an aqueous solution. Hydrogen chloride \(\left( \ce{HCl} \right)\) ionizes completely into hydrogen ions and chloride ions in water. \[\ce{HCl} \left( g \right) \rightarrow \ce{H^+} \left( aq \right) + \ce{Cl^-} \left( aq \right)\nonumber \] A weak acid is an acid that ionizes only slightly in an aqueous solution. Acetic acid (found in vinegar) is a very common weak acid. Its ionization is shown below. \[\ce{CH_3COOH} \left( aq \right) \rightleftharpoons \ce{H^+} \left( aq \right) + \ce{CH_3COO^-} \left( aq \right)\nonumber \] The ionization of acetic acid is incomplete, and so the equation is shown with a double arrow. The extent of ionization of weak acids varies, but is generally less than \(10\%\). A \(0.10 \: \text{M}\) solution of acetic acid is only about \(1.3\%\) ionized, meaning that the equilibrium strongly favors the reactants. Weak acids, like strong acids, ionize to yield the \(\ce{H^+}\) ion and a conjugate base. Because \(\ce{HCl}\) is a strong acid, its conjugate base \(\left( \ce{Cl^-} \right)\) is extremely weak. The chloride ion is incapable of accepting the \(\ce{H^+}\) ion and becoming \(\ce{HCl}\) again. In general, the stronger the acid, the weaker its conjugate base. Likewise, the weaker the acid, the stronger its conjugate base. Acid Conjugate Base Strong Acids NaN \(\ce{HCl}\) (hydrochloric acid) (strongest) \(\ce{Cl^-}\) (chloride ion) (weakest) \(\ce{H_2SO_4}\) (sulfuric acid) \(\ce{HSO_4^-}\) (hydrogen sulfate ion) \(\ce{HNO_3}\) (nitric acid) \(\ce{NO_3^-}\) (nitrate ion) Weak Acids NaN \(\ce{H_3PO_4}\) (phosphoric acid) \(\ce{H_2PO_4^-}\) (dihydrogen phosphate ion) \(\ce{CH_3COOH}\) (acetic acid) \(\ce{CH_3COO^-}\) (acetate ion) \(\ce{H_2CO_3}\) (carbonic acid) \(\ce{HCO_3^-}\) (hydrogen carbonate ion) \(\ce{HCN}\) (hydrocyanic acid) (weakest) \(\ce{CN^-}\) (cyanide ion) (strongest) Strong acids are \(100\%\) ionized in solution. Weak acids are only slightly ionized. Phosphoric acid is stronger than acetic acid, and so is ionized to a greater extent. Acetic acid is stronger than carbonic acid, and so on. The Acid Ionization Constant, \(K_\text{a}\) The ionization for a general weak acid, \(\ce{HA}\), can be written as follows: \[\ce{HA} \left( aq \right) \rightarrow \ce{H^+} \left( aq \right) + \ce{A^-} \left( aq \right)\nonumber \] 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 \(1.3 \times 10^{-2}\) Oxalic acid \(\ce{H_2C_2O_4} \rightleftharpoons \ce{H^+} + \ce{HC_2O_4^-}\) \(\ce{HC_2O_4} \rightleftharpoons \ce{H^+} + \ce{C_2O_4^{2-}}\) \(6.5 \times 10^{-2}\) \(6.1 \times 10^{-5}\) Phosphoric acid \(\ce{H_3PO_4} \rightleftharpoons \ce{H^+} + \ce{H_2PO_4^-}\) \(\ce{H_2PO_4^-} \rightleftharpoons \ce{H^+} + \ce{HPO_4^{2--}}\) \(\ce{HPO_4^{2-}} \rightleftharpoons \ce{H^+} + \ce{PO_4^{3-}}\) \(7.5 \times 10^{-3}\) \(6.2 \times 10^{-8}\) \(4.8 \times 10^{-13}\) 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 \(\ce{H_2CO_3} \rightleftharpoons \ce{H^+} + \ce{HCO_3^-}\) \(\ce{HCO_3^-} \rightleftharpoons \ce{H^+} + \ce{CO_3^{2-}}\) \(4.2 \times 10^{-7}\) \(4.8 \times 10^{-11}\) Hydrocyanic acid \(\ce{HCN} \rightleftharpoons \ce{H^+} + \ce{CN^-}\) \(4.9 \times 10^{-10}\) Summary A strong acid is an acid which is completely ionized in an aqueous solution. A weak acid is an acid that ionizes only slightly in an aqueous solution. The acid ionization constant \(\left( K_\text{a} \right)\) is defined. |
Bookshelves/General_Chemistry/Chemistry_2e_(OpenStax)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.11%3A_Occurrence_Preparation_and_Properties_of_Halogens | Learning Objectives By the end of this section, you will be able to: Describe the preparation, properties, and uses of halogens Describe the properties, preparation, and uses of halogen compounds The elements in group 17 are the halogens. These are the elements fluorine, chlorine, bromine, iodine, and astatine. These elements are too reactive to occur freely in nature, but their compounds are widely distributed. Chlorides are the most abundant; although fluorides, bromides, and iodides are less common, they are reasonably available. In this section, we will examine the occurrence, preparation, and properties of halogens. Next, we will examine halogen compounds with the representative metals followed by an examination of the interhalogens. This section will conclude with some applications of halogens. Occurrence and Preparation All of the halogens occur in seawater as halide ions. The concentration of the chloride ion is 0.54 M ; that of the other halides is less than 10 –4 M . Fluoride also occurs in minerals such as CaF 2 , Ca(PO 4 ) 3 F, and Na 3 AlF 6 . Chloride also occurs in the Great Salt Lake and the Dead Sea, and in extensive salt beds that contain NaCl, KCl, or MgCl 2 . Part of the chlorine in your body is present as hydrochloric acid, which is a component of stomach acid. Bromine compounds occur in the Dead Sea and underground brines. Iodine compounds are found in small quantities in Chile saltpeter, underground brines, and sea kelp. Iodine is essential to the function of the thyroid gland. The best sources of halogens (except iodine) are halide salts. It is possible to oxidize the halide ions to free diatomic halogen molecules by various methods, depending on the ease of oxidation of the halide ion. Fluoride is the most difficult to oxidize, whereas iodide is the easiest. The major method for preparing fluorine is electrolytic oxidation. The most common electrolysis procedure is to use a molten mixture of potassium hydrogen fluoride, KHF 2 , and anhydrous hydrogen fluoride. Electrolysis causes HF to decompose, forming fluorine gas at the anode and hydrogen at the cathode. It is necessary to keep the two gases separated to prevent their explosive recombination to reform hydrogen fluoride. Most commercial chlorine comes from the electrolysis of the chloride ion in aqueous solutions of sodium chloride; this is the chlor-alkali process discussed previously. Chlorine is also a product of the electrolytic production of metals such as sodium, calcium, and magnesium from their fused chlorides. It is also possible to prepare chlorine by the chemical oxidation of the chloride ion in acid solution with strong oxidizing agents such as manganese dioxide (MnO 2 ) or sodium dichromate (Na 2 Cr 2 O 7 ). The reaction with manganese dioxide is: \[\ce{MnO_2(s) + 2 Cl^{-}(aq) + 4 H3O^{+}(aq) \longrightarrow Mn^{2+}(aq) + Cl_2(g) + 6 H2O(l)} \nonumber \] The commercial preparation of bromine involves the oxidation of bromide ion by chlorine: \[\ce{2 Br^{-}(aq) + Cl_2(g) \longrightarrow Br_2(l) + 2 Cl^{-}(aq)} \nonumber \] Chlorine is a stronger oxidizing agent than bromine. This method is important for the production of essentially all domestic bromine. Some iodine comes from the oxidation of iodine chloride, ICl, or iodic acid, HlO 3 . The commercial preparation of iodine utilizes the reduction of sodium iodate, NaIO 3 , an impurity in deposits of Chile saltpeter, with sodium hydrogen sulfite: \[\ce{2 IO_3^{-}(aq) + 5 HSO_3^{-}(aq) \longrightarrow 3 HSO_4^{-}(aq) + 2 SO_4^{2-}(aq) + H2O(l) + I_2(s)} \nonumber \] Properties of the Halogens Fluorine is a pale yellow gas, chlorine is a greenish-yellow gas, bromine is a deep reddish-brown liquid, and iodine is a grayish-black crystalline solid. Liquid bromine has a high vapor pressure, and the reddish vapor is readily visible in Figure \(\PageIndex{1}\). Iodine crystals have a noticeable vapor pressure. When gently heated, these crystals sublime and form a beautiful deep violet vapor. Bromine is only slightly soluble in water, but it is miscible in all proportions in less polar (or nonpolar) solvents such as chloroform, carbon tetrachloride, and carbon disulfide, forming solutions that vary from yellow to reddish-brown, depending on the concentration. Iodine is soluble in chloroform, carbon tetrachloride, carbon disulfide, and many hydrocarbons, giving violet solutions of I 2 molecules. Iodine dissolves only slightly in water, giving brown solutions. It is quite soluble in aqueous solutions of iodides, with which it forms brown solutions. These brown solutions result because iodine molecules have empty valence d orbitals and can act as weak Lewis acids towards the iodide ion. The equation for the reversible reaction of iodine (Lewis acid) with the iodide ion (Lewis base) to form triiodide ion, is: \[\ce{I_2(s) + I^{-}(aq) \longrightarrow I_3^{-}(aq)} \nonumber \] The easier it is to oxidize the halide ion, the more difficult it is for the halogen to act as an oxidizing agent. Fluorine generally oxidizes an element to its highest oxidation state, whereas the heavier halogens may not. For example, when excess fluorine reacts with sulfur, SF 6 forms. Chlorine gives SCl 2 and bromine, S 2 Br 2 . Iodine does not react with sulfur. Fluorine is the most powerful oxidizing agent of the known elements. It spontaneously oxidizes most other elements; therefore, the reverse reaction, the oxidation of fluorides, is very difficult to accomplish. Fluorine reacts directly and forms binary fluorides with all of the elements except the lighter noble gases (He, Ne, and Ar). Fluorine is such a strong oxidizing agent that many substances ignite on contact with it. Drops of water inflame in fluorine and form O 2 , OF 2 , H 2 O 2 , O 3 , and HF. Wood and asbestos ignite and burn in fluorine gas. Most hot metals burn vigorously in fluorine. However, it is possible to handle fluorine in copper, iron, or nickel containers because an adherent film of the fluoride salt passivates their surfaces. Fluorine is the only element that reacts directly with the noble gas xenon. Although it is a strong oxidizing agent, chlorine is less active than fluorine. Mixing chlorine and hydrogen in the dark makes the reaction between them to be imperceptibly slow. Exposure of the mixture to light causes the two to react explosively. Chlorine is also less active towards metals than fluorine, and oxidation reactions usually require higher temperatures. Molten sodium ignites in chlorine. Chlorine attacks most nonmetals (C, N 2 , and O 2 are notable exceptions), forming covalent molecular compounds. Chlorine generally reacts with compounds that contain only carbon and hydrogen (hydrocarbons) by adding to multiple bonds or by substitution. In cold water, chlorine undergoes a disproportionation reaction: \[\ce{Cl_2(aq) + 2 H2O(l) \longrightarrow HOCl(aq) + H3O^{+}(aq) + Cl^{-}(aq)} \nonumber \] Half the chlorine atoms oxidize to the 1+ oxidation state (hypochlorous acid), and the other half reduce to the 1− oxidation state (chloride ion). This disproportionation is incomplete, so chlorine water is an equilibrium mixture of chlorine molecules, hypochlorous acid molecules, hydronium ions, and chloride ions. When exposed to light, this solution undergoes a photochemical decomposition: \[\ce{2 HOCl(aq) + 2 H2O(l) ->[\text{sunlight}] 2 H3O^{+}(aq) + 2 Cl^{-}(aq) + O2(g)} \nonumber \] The nonmetal chlorine is more electronegative than any other element except fluorine, oxygen, and nitrogen. In general, very electronegative elements are good oxidizing agents; therefore, we would expect elemental chlorine to oxidize all of the other elements except for these three (and the nonreactive noble gases). Its oxidizing property, in fact, is responsible for its principal use. For example, phosphorus(V) chloride, an important intermediate in the preparation of insecticides and chemical weapons, is manufactured by oxidizing the phosphorus with chlorine: \[\ce{P_4(s) + 10 Cl_2(g) \longrightarrow 4 PCl_5(l)} \nonumber \] A great deal of chlorine is also used to oxidize, and thus to destroy, organic or biological materials in water purification and in bleaching. The chemical properties of bromine are similar to those of chlorine, although bromine is the weaker oxidizing agent and its reactivity is less than that of chlorine. Iodine is the least reactive of the halogens. It is the weakest oxidizing agent, and the iodide ion is the most easily oxidized halide ion. Iodine reacts with metals, but heating is often required. It does not oxidize other halide ions. Compared with the other halogens, iodine reacts only slightly with water. Traces of iodine in water react with a mixture of starch and iodide ion, forming a deep blue color. This reaction is a very sensitive test for the presence of iodine in water. Halides of the Representative Metals Thousands of salts of the representative metals have been prepared. The binary halides are an important subclass of salts. A salt is an ionic compound composed of cations and anions, other than hydroxide or oxide ions. In general, it is possible to prepare these salts from the metals or from oxides, hydroxides, or carbonates. We will illustrate the general types of reactions for preparing salts through reactions used to prepare binary halides. The binary compounds of a metal with the halogens are the halides . Most binary halides are ionic. However, mercury, the elements of group 13 with oxidation states of 3+, tin(IV), and lead(IV) form covalent binary halides. The direct reaction of a metal and a halogen produce the halide of the metal. Examples of these oxidation-reduction reactions include: \[\begin{align*} \ce{Cd(s) + Cl_2(g) & \longrightarrow CdCl_2(s)} \\[4pt][4pt] \ce{2 Ga(l) + 3 Br_2(l) & \longrightarrow 2 GaBr_3(s)} \end{align*} \] Link to Learning Reactions of the alkali metals with elemental halogens are very exothermic and often quite violent. Under controlled conditions, they provide exciting demonstrations for budding students of chemistry. You can view the initial heating of the sodium that removes the coating of sodium hydroxide, sodium peroxide, and residual mineral oil to expose the reactive surface. The reaction with chlorine gas then proceeds very nicely. If a metal can exhibit two oxidation states, it may be necessary to control the stoichiometry in order to obtain the halide with the lower oxidation state. For example, preparation of tin(II) chloride requires a 1:1 ratio of Sn to Cl 2 , whereas preparation of tin(IV) chloride requires a 1:2 ratio: \[\begin{align*} \ce{Sn(s) + Cl_2(g) &\longrightarrow SnCl_2(s) }\\[4pt][4pt] \ce{Sn(s) + 2 Cl_2(g) &\longrightarrow SnCl_4(l)} \end{align*} \] The active representative metals—those that are easier to oxidize than hydrogen—react with gaseous hydrogen halides to produce metal halides and hydrogen. The reaction of zinc with hydrogen fluoride is: \[\ce{Zn(s) + 2 HF(g) \longrightarrow ZnF_2(s) + H_2(g)} \nonumber \] The active representative metals also react with solutions of hydrogen halides to form hydrogen and solutions of the corresponding halides. Examples of such reactions include: \[\begin{align*} \ce{Cd(s) + 2 HBr(aq) &\longrightarrow CdBr_2(aq) + H_2(g)} \\[4pt][4pt] \ce{Sn(s) + 2 HI(aq) &\longrightarrow SnI_2(aq) + H_2(g)} \end{align*} \] Hydroxides, carbonates, and some oxides react with solutions of the hydrogen halides to form solutions of halide salts. It is possible to prepare additional salts by the reaction of these hydroxides, carbonates, and oxides with aqueous solution of other acids: \[\begin{align*} \ce{CaCo_3(s) + 2 HCl(aq) &\longrightarrow CaCl_2(aq) + CO_2(g) + H2O(l)} \\[4pt][4pt] \ce{TlOH(aq) + HF(aq) &\longrightarrow TlF(aq) + H2O(l)} \end{align*} \] A few halides and many of the other salts of the representative metals are insoluble. It is possible to prepare these soluble salts by metathesis reactions that occur when solutions of soluble salts are mixed (see Figure \(\PageIndex{2}\)). Metathesis reactions are examined in the chapter on the stoichiometry of chemical reactions. Several halides occur in large quantities in nature. The ocean and underground brines contain many halides. For example, magnesium chloride in the ocean is the source of magnesium ions used in the production of magnesium. Large underground deposits of sodium chloride, like the salt mine shown in Figure \(\PageIndex{3}\), occur in many parts of the world. These deposits serve as the source of sodium and chlorine in almost all other compounds containing these elements. The chlor-alkali process is one example. Interhalogens Compounds formed from two or more different halogens are interhalogens . Interhalogen molecules consist of one atom of the heavier halogen bonded by single bonds to an odd number of atoms of the lighter halogen. The structures of IF 3 , IF 5 , and IF 7 are illustrated in Figure \(\PageIndex{4}\). Formulas for other interhalogens, each of which comes from the reaction of the respective halogens, are in Table \(\PageIndex{1}\). Note from Table \(\PageIndex{1}\) that fluorine is able to oxidize iodine to its maximum oxidation state, 7+, whereas bromine and chlorine, which are more difficult to oxidize, achieve only the 5+-oxidation state. A 7+-oxidation state is the limit for the halogens. Because smaller halogens are grouped about a larger one, the maximum number of smaller atoms possible increases as the radius of the larger atom increases. Many of these compounds are unstable, and most are extremely reactive. The interhalogens react like their component halides; halogen fluorides, for example, are stronger oxidizing agents than are halogen chlorides. The ionic polyhalides of the alkali metals, such as KI 3 , KICl 2 , KICl 4 , CsIBr 2 , and CsBrCl 2 , which contain an anion composed of at least three halogen atoms, are closely related to the interhalogens. As seen previously, the formation of the polyhalide \(\ce{I3^{-}}\) anion is responsible for the solubility of iodine in aqueous solutions containing an iodide ion. YX YX3 YX5 YX7 ClF(g) ClF3(g) ClF5(g) NaN BrF(g) BrF3(l) BrF5(l) NaN BrCl(g) NaN NaN NaN IF(s) IF3(s) IF5(l) IF7(g) ICl(l) ICl3(s) NaN NaN IBr(s) NaN NaN NaN Applications The fluoride ion and fluorine compounds have many important uses. Compounds of carbon, hydrogen, and fluorine are replacing Freons (compounds of carbon, chlorine, and fluorine) as refrigerants. Teflon is a polymer composed of –CF 2 CF 2 – units. Fluoride ion is added to water supplies and to some toothpastes as SnF 2 or NaF to fight tooth decay. Fluoride partially converts teeth from Ca 5 (PO 4 ) 3 (OH) into Ca 5 (PO 4 ) 3 F. Chlorine is important to bleach wood pulp and cotton cloth. The chlorine reacts with water to form hypochlorous acid, which oxidizes colored substances to colorless ones. Large quantities of chlorine are important in chlorinating hydrocarbons (replacing hydrogen with chlorine) to produce compounds such as tetrachloride (CCl 4 ), chloroform (CHCl 3 ), and ethyl chloride (C 2 H 5 Cl), and in the production of polyvinyl chloride (PVC) and other polymers. Chlorine is also important to kill the bacteria in community water supplies. Bromine is important in the production of certain dyes, and sodium and potassium bromides are used as sedatives. At one time, light-sensitive silver bromide was a component of photographic film. Iodine in alcohol solution with potassium iodide is an antiseptic (tincture of iodine). Iodide salts are essential for the proper functioning of the thyroid gland; an iodine deficiency may lead to the development of a goiter. Iodized table salt contains 0.023% potassium iodide. Silver iodide is useful in the seeding of clouds to induce rain; it was important in the production of photographic film and iodoform, CHI 3 , is an antiseptic. |
Courses/Sacramento_City_College/SCC%3A_CHEM_300_-_Beginning_Chemistry/SCC%3A_CHEM_300_-_Beginning_Chemistry_(Alviar-Agnew)/13%3A_Solutions/13.08%3A_Solution_Stoichiometry | Learning Objectives Determine amounts of reactants or products in aqueous solutions. As we learned previously, double replacement reactions involve the reaction between ionic compounds in solution and, in the course of the reaction, the ions in the two reacting compounds are “switched” (they replace each other). Because these reactions occur in aqueous solution, we can use the concept of molarity to directly calculate the number of moles of reactants or products that will be formed, and therefore their amounts (i.e. volume of solutions or mass of precipitates). As an example, lead (II) nitrate and sodium chloride react to form sodium nitrate and the insoluble compound, lead (II) chloride. \[\ce{Pb(NO_3)_2 (aq) + 2 NaCl (aq) → PbCl_2 (s) + 2 NaNO3 (aq)} \label{EQ1} \] In the reaction shown above, if we mixed 0.123 L of a 1.00 M solution of \(\ce{NaCl}\) with 1.50 M solution of \(\ce{Pb(NO3)2}\), we could calculate the volume of \(\ce{Pb(NO3)2}\) solution needed to completely precipitate the \(\ce{Pb^{2+}}\) ions. The molar concentration can also be expressed as the following: \[1.00\,M \,\ce{NaCl} = \dfrac{1.00 \; mol \; \ce{NaCl}}{1 \; L \; \ce{NaCl} \; \text{solution}} \nonumber \] and \[1.50\,M\, \ce{Pb(NO3)2} = \dfrac{1.50 \; mol \; \ce{Pb(NO_3)_2}}{1 \; L \; \ce{Pb(NO_3)_2} \text{solution}} \nonumber \] First, we must examine the reaction stoichiometry in the balanced reaction (Equation \ref{EQ1}). In this reaction, one mole of \(\ce{Pb(NO3)2}\) reacts with two moles of \(\ce{NaCl}\) to give one mole of \(\ce{PbCl2}\) precipitate. Thus, the concept map utilizing the stoichiometric ratios is: so the volume of lead (II) nitrate that reacted is calculated as: \[0.123 \; L \; \ce{NaCl} \, \text{solution} \times \dfrac{1.00 \; mol \; \ce{NaCl}}{1 \; L \; \ce{NaCl} \; \text{solution}} \times \dfrac{1 \; mol \; \ce{Pb(NO_3)_2}}{2 \; mol \; \ce{NaCl}} \times \dfrac{1 \; L \; \ce{Pb(NO_3)_2} \; \text{solution}}{1.5 \; mol \; \ce{Pb(NO_3)_2}} = 0.041 \; \ce{Pb(NO_3)_2}\,L \; \text{solution} \nonumber \] This volume makes intuitive sense for two reasons: (1) the number of moles of \(\ce{Pb(NO3)2}\) required is half of the number of moles of \(\ce{NaCl}\), based off of the stoichiometry in the balanced reaction (Equation \ref{EQ1}); (2) the concentration of \(\ce{Pb(NO3)2}\) solution is 50% greater than the \(\ce{NaCl}\) solution, so less volume is needed. Example \(\PageIndex{1}\) What volume (in L) of 0.500 M sodium sulfate will react with 275 mL of 0.250 M barium chloride to completely precipitate all \(\ce{Ba^{2+}}\) in the solution? Solution Steps for Problem Solving Example \(\PageIndex{1}\) Identify the "given" information and what the problem is asking you to "find." Given: 275 mL BaCl2 0.250 M \(\ce{BaCl2}\) or \(\displaystyle \dfrac{0.250\; mol BaCl_2}{1\; L\; BaCl_2\; solution}\) 0.500 M \(\ce{Na2SO4}\) or \(\displaystyle \dfrac{0.500\; mol Na_2SO_4}{1\; L\; Na_2SO_4\; solution}\) Find: Volume \(\ce{Na2SO4}\) solution. Set up and balance the chemical equation. \(\ce{Na2SO4(aq) + BaCl2(aq) -> BaSO4(s)} + \underline{2} \ce{NaCl (aq)}\) An insoluble product is formed after the reaction. List other known quantities. 1 mol of Na2SO4 to 1 mol BaCl2 1000 mL = 1 L Prepare a concept map and use the proper conversion factor. NaN Cancel units and calculate. \(\displaystyle 275\cancel{mL \; BaCl_2 \; solution}\times \dfrac{1\cancel{L}}{1000\cancel{mL}}\times \dfrac{0.250 \cancel{mol \; BaCl_2}}{1 \cancel{L\;BaCl_2 \; solution}}\times \dfrac{1 \cancel{mol \; Na_2SO_4}}{1 \cancel{mol \; BaCl_2}}\times \dfrac{1\; L \; Na_2SO_4 \; solution}{0.500 \cancel{mol Na_2SO_4}}\) = 0.1375 L sodium sulfate Think about your result. The lesser amount (almost half) of sodium sulfate is to be expected as it is more concentrated than barium chloride. Also, the units are correct. Exercise \(\PageIndex{1}\) What volume of 0.250 M lithium hydroxide will completely react with 0.500 L of 0.250 M of sulfuric acid solution? Answer 0.250 L \(\ce{LiOH}\) solution |
Courses/Williams_School/Chemistry_I/01%3A_Essential_Ideas_of_Chemistry/1.06%3A_Mathematical_Treatment_of_Measurement_Results | Learning Objectives Explain the factor-label method to mathematical calculations involving quantities Describe how to use the factor-label to carry out unit conversions for a given property and computations involving two or more properties Convert between the three main temperature units: Fahrenheit, Celsius, and Kelvin It is often the case that a quantity of interest may not be easy (or even possible) to measure directly but instead must be calculated from other directly measured properties and appropriate mathematical relationships. For example, consider measuring the average speed of an athlete running sprints. This is typically accomplished by measuring the time required for the athlete to run from the starting line to the finish line, and the distance between these two lines, and then computing speed from the equation that relates these three properties: \[\mathrm{speed=\dfrac{distance}{time}} \nonumber \] An Olympic-quality sprinter can run 100 m in approximately 10 s, corresponding to an average speed of \[\mathrm{\dfrac{100\: m}{10\: s}=10\: m/s} \nonumber \] Note that this simple arithmetic involves dividing the numbers of each measured quantity to yield the number of the computed quantity (100/10 = 10) and likewise dividing the units of each measured quantity to yield the unit of the computed quantity (m/s = m/s). Now, consider using this same relation to predict the time required for a person running at this speed to travel a distance of 25 m. The same relation between the three properties is used, but in this case, the two quantities provided are a speed (10 m/s) and a distance (25 m). To yield the sought property, time, the equation must be rearranged appropriately: \[\mathrm{time=\dfrac{distance}{speed}} \nonumber \] The time can then be computed as: \[\mathrm{\dfrac{25\: m}{10\: m/s}=2.5\: s} \nonumber \] Again, arithmetic on the numbers (25/10 = 2.5) was accompanied by the same arithmetic on the units (m/m/s = s) to yield the number and unit of the result, 2.5 s. Note that, just as for numbers, when a unit is divided by an identical unit (in this case, m/m), the result is “1”—or, as commonly phrased, the units “cancel.” These calculations are examples of a versatile mathematical approach known as dimensional analysis (or the factor-label method ). Dimensional analysis is based on this premise: the units of quantities must be subjected to the same mathematical operations as their associated numbers . This method can be applied to computations ranging from simple unit conversions to more complex, multi-step calculations involving several different quantities. Conversion Factors and Dimensional Analysis A ratio of two equivalent quantities expressed with different measurement units can be used as a unit conversion factor . For example, the lengths of 2.54 cm and 1 in. are equivalent (by definition), and so a unit conversion factor may be derived from the ratio, \[\mathrm{\dfrac{2.54\: cm}{1\: in.}\:(2.54\: cm=1\: in.)\: or\: 2.54\:\dfrac{cm}{in.}} \nonumber \] Several other commonly used conversion factors are given in Table \(\PageIndex{1}\). Length Volume Mass 1 m = 1.0936 yd 1 L = 1.0567 qt 1 kg = 2.2046 lb 1 in. = 2.54 cm (exact) 1 qt = 0.94635 L 1 lb = 453.59 g 1 km = 0.62137 mi 1 ft3 = 28.317 L 1 (avoirdupois) oz = 28.349 g 1 mi = 1609.3 m 1 tbsp = 14.787 mL 1 (troy) oz = 31.103 g When we multiply a quantity (such as distance given in inches) by an appropriate unit conversion factor, we convert the quantity to an equivalent value with different units (such as distance in centimeters). For example, a basketball player’s vertical jump of 34 inches can be converted to centimeters by: \[\mathrm{34\: \cancel{in.} \times \dfrac{2.54\: cm}{1\:\cancel{in.}}=86\: cm} \nonumber \] Since this simple arithmetic involves quantities , the premise of dimensional analysis requires that we multiply both numbers and units . The numbers of these two quantities are multiplied to yield the number of the product quantity, 86, whereas the units are multiplied to yield \[\mathrm{\dfrac{in.\times cm}{in.}}. \nonumber \] Just as for numbers, a ratio of identical units is also numerically equal to one, \[\mathrm{\dfrac{in.}{in.}=1} \nonumber \] and the unit product thus simplifies to cm . (When identical units divide to yield a factor of 1, they are said to “cancel.”) Using dimensional analysis, we can determine that a unit conversion factor has been set up correctly by checking to confirm that the original unit will cancel, and the result will contain the sought (converted) unit. Example \(\PageIndex{1}\): Using a Unit Conversion Factor The mass of a competition Frisbee is 125 g. Convert its mass to ounces using the unit conversion factor derived from the relationship 1 oz = 28.349 g (Table \(\PageIndex{1}\)). Solution If we have the conversion factor, we can determine the mass in kilograms using an equation similar the one used for converting length from inches to centimeters. \[x\:\mathrm{oz=125\: g\times unit\: conversion\: factor}\nonumber \] We write the unit conversion factor in its two forms: \[\mathrm{\dfrac{1\: oz}{28.349\: g}\:and\:\dfrac{28.349\: g}{1\: oz}}\nonumber \] The correct unit conversion factor is the ratio that cancels the units of grams and leaves ounces. \[\begin{align*} x\:\ce{oz}&=\mathrm{125\:\cancel{g}\times \dfrac{1\: oz}{28.349\:\cancel{g}}}\\ &=\mathrm{\left(\dfrac{125}{28.349}\right)\:oz}\\ &=\mathrm{4.41\: oz\: (three\: significant\: figures)} \end{align*} \nonumber \] Exercise \(\PageIndex{1}\) Convert a volume of 9.345 qt to liters. Answer 8.844 L Beyond simple unit conversions, the factor-label method can be used to solve more complex problems involving computations. Regardless of the details, the basic approach is the same—all the factors involved in the calculation must be appropriately oriented to insure that their labels (units) will appropriately cancel and/or combine to yield the desired unit in the result. This is why it is referred to as the factor-label method. As your study of chemistry continues, you will encounter many opportunities to apply this approach. Example \(\PageIndex{2}\): Computing Quantities from Measurement Results What is the density of common antifreeze in units of g/mL? A 4.00-qt sample of the antifreeze weighs 9.26 lb. Solution Since \(\mathrm{density=\dfrac{mass}{volume}}\), we need to divide the mass in grams by the volume in milliliters. In general: the number of units of B = the number of units of A \(\times\) unit conversion factor. The necessary conversion factors are given in Table 1.7.1: 1 lb = 453.59 g; 1 L = 1.0567 qt; 1 L = 1,000 mL. We can convert mass from pounds to grams in one step: \[\mathrm{9.26\:\cancel{lb}\times \dfrac{453.59\: g}{1\:\cancel{lb}}=4.20\times 10^3\:g}\nonumber \] We need to use two steps to convert volume from quarts to milliliters. Convert quarts to liters. \[\mathrm{4.00\:\cancel{qt}\times\dfrac{1\: L}{1.0567\:\cancel{qt}}=3.78\: L}\nonumber \] Convert liters to milliliters. \[\mathrm{3.78\:\cancel{L}\times\dfrac{1000\: mL}{1\:\cancel{L}}=3.78\times10^3\:mL}\nonumber \] Then, \[\mathrm{density=\dfrac{4.20\times10^3\:g}{3.78\times10^3\:mL}=1.11\: g/mL}\nonumber \] Alternatively, the calculation could be set up in a way that uses three unit conversion factors sequentially as follows: \[\mathrm{\dfrac{9.26\:\cancel{lb}}{4.00\:\cancel{qt}}\times\dfrac{453.59\: g}{1\:\cancel{lb}}\times\dfrac{1.0567\:\cancel{qt}}{1\:\cancel{L}}\times\dfrac{1\:\cancel{L}}{1000\: mL}=1.11\: g/mL}\nonumber \] Exercise \(\PageIndex{2}\) What is the volume in liters of 1.000 oz, given that 1 L = 1.0567 qt and 1 qt = 32 oz (exactly)? Answer \(\mathrm{2.956\times10^{-2}\:L}\) Example \(\PageIndex{3}\): Computing Quantities from Measurement Results While being driven from Philadelphia to Atlanta, a distance of about 1250 km, a 2014 Lamborghini Aventador Roadster uses 213 L gasoline. What (average) fuel economy, in miles per gallon, did the Roadster get during this trip? If gasoline costs $3.80 per gallon, what was the fuel cost for this trip? Solution (a) We first convert distance from kilometers to miles: \[\mathrm{1250\: km\times\dfrac{0.62137\: mi}{1\: km}=777\: mi}\nonumber \] and then convert volume from liters to gallons: \[\mathrm{213\:\cancel{L}\times\dfrac{1.0567\:\cancel{qt}}{1\:\cancel{L}}\times\dfrac{1\: gal}{4\:\cancel{qt}}=56.3\: gal}\nonumber \] Then, \[\mathrm{(average)\: mileage=\dfrac{777\: mi}{56.3\: gal}=13.8\: miles/gallon=13.8\: mpg}\nonumber \] Alternatively, the calculation could be set up in a way that uses all the conversion factors sequentially, as follows: \[\mathrm{\dfrac{1250\:\cancel{km}}{213\:\cancel{L}}\times\dfrac{0.62137\: mi}{1\:\cancel{km}}\times\dfrac{1\:\cancel{L}}{1.0567\:\cancel{qt}}\times\dfrac{4\:\cancel{qt}}{1\: gal}=13.8\: mpg}\nonumber \] (b) Using the previously calculated volume in gallons, we find: \[\mathrm{56.3\: gal\times\dfrac{$3.80}{1\: gal}=$214}\nonumber \] Exercise \(\PageIndex{3}\) A Toyota Prius Hybrid uses 59.7 L gasoline to drive from San Francisco to Seattle, a distance of 1300 km (two significant digits). What (average) fuel economy, in miles per gallon, did the Prius get during this trip? If gasoline costs $3.90 per gallon, what was the fuel cost for this trip? Answer a 51 mpg Answer b $62 Converting Between Units: https://youtu.be/wSaOh48k8Wg Conversion of Temperature Units We use the word temperature to refer to the hotness or coldness of a substance. One way we measure a change in temperature is to use the fact that most substances expand when their temperature increases and contract when their temperature decreases. The mercury or alcohol in a common glass thermometer changes its volume as the temperature changes. Because the volume of the liquid changes more than the volume of the glass, we can see the liquid expand when it gets warmer and contract when it gets cooler. To mark a scale on a thermometer, we need a set of reference values: Two of the most commonly used are the freezing and boiling temperatures of water at a specified atmospheric pressure. On the Celsius scale, 0 °C is defined as the freezing temperature of water and 100 °C as the boiling temperature of water. The space between the two temperatures is divided into 100 equal intervals, which we call degrees. On the Fahrenheit scale, the freezing point of water is defined as 32 °F and the boiling temperature as 212 °F. The space between these two points on a Fahrenheit thermometer is divided into 180 equal parts (degrees). Defining the Celsius and Fahrenheit temperature scales as described in the previous paragraph results in a slightly more complex relationship between temperature values on these two scales than for different units of measure for other properties. Most measurement units for a given property are directly proportional to one another (y = mx). Using familiar length units as one example: \[\mathrm{length\: in\: feet=\left(\dfrac{1\: ft}{12\: in.}\right)\times length\: in\: inches} \nonumber \] where y = length in feet, x = length in inches, and the proportionality constant, m, is the conversion factor. The Celsius and Fahrenheit temperature scales, however, do not share a common zero point, and so the relationship between these two scales is a linear one rather than a proportional one (\(y = mx + b\)). Consequently, converting a temperature from one of these scales into the other requires more than simple multiplication by a conversion factor, m, it also must take into account differences in the scales’ zero points (\(b\)). The linear equation relating Celsius and Fahrenheit temperatures is easily derived from the two temperatures used to define each scale. Representing the Celsius temperature as \(x\) and the Fahrenheit temperature as \(y\), the slope, \(m\), is computed to be: \[\begin{align*} m &=\dfrac{\Delta y}{\Delta x} \\[4pt] &= \mathrm{\dfrac{212\: ^\circ F - 32\: ^\circ F}{100\: ^\circ C-0\: ^\circ C}} \\[4pt] &= \mathrm{\dfrac{180\: ^\circ F}{100\: ^\circ C}} \\[4pt] &= \mathrm{\dfrac{9\: ^\circ F}{5\: ^\circ C} }\end{align*} \nonumber \] The y-intercept of the equation, b , is then calculated using either of the equivalent temperature pairs, (100 °C, 212 °F) or (0 °C, 32 °F), as: \[\begin{align*} b&=y-mx \\[4pt] &= \mathrm{32\:^\circ F-\dfrac{9\:^\circ F}{5\:^\circ C}\times0\:^\circ C} \\[4pt] &= \mathrm{32\:^\circ F} \end{align*} \nonumber \] The equation relating the temperature scales is then: \[\mathrm{\mathit{T}_{^\circ F}=\left(\dfrac{9\:^\circ F}{5\:^\circ C}\times \mathit{T}_{^\circ C}\right)+32\:^\circ C} \nonumber \] An abbreviated form of this equation that omits the measurement units is: \[\mathrm{\mathit{T}_{^\circ F}=\dfrac{9}{5}\times \mathit{T}_{^\circ C}+32} \nonumber \] Rearrangement of this equation yields the form useful for converting from Fahrenheit to Celsius: \[\mathrm{\mathit{T}_{^\circ C}=\dfrac{5}{9}(\mathit{T}_{^\circ F}+32)} \nonumber \] As mentioned earlier in this chapter, the SI unit of temperature is the kelvin (K). Unlike the Celsius and Fahrenheit scales, the kelvin scale is an absolute temperature scale in which 0 (zero) K corresponds to the lowest temperature that can theoretically be achieved. The early 19th-century discovery of the relationship between a gas's volume and temperature suggested that the volume of a gas would be zero at −273.15 °C. In 1848, British physicist William Thompson, who later adopted the title of Lord Kelvin, proposed an absolute temperature scale based on this concept (further treatment of this topic is provided in this text’s chapter on gases). The freezing temperature of water on this scale is 273.15 K and its boiling temperature 373.15 K. Notice the numerical difference in these two reference temperatures is 100, the same as for the Celsius scale, and so the linear relation between these two temperature scales will exhibit a slope of \(\mathrm{1\:\dfrac{K}{^\circ\:C}}\). Following the same approach, the equations for converting between the kelvin and Celsius temperature scales are derived to be: \[T_{\ce K}=T_{\mathrm{^\circ C}}+273.15 \nonumber \] \[T_\mathrm{^\circ C}=T_{\ce K}-273.15 \nonumber \] The 273.15 in these equations has been determined experimentally, so it is not exact. Figure \(\PageIndex{1}\) shows the relationship among the three temperature scales. Recall that we do not use the degree sign with temperatures on the kelvin scale. Although the kelvin (absolute) temperature scale is the official SI temperature scale, Celsius is commonly used in many scientific contexts and is the scale of choice for nonscience contexts in almost all areas of the world. Very few countries (the U.S. and its territories, the Bahamas, Belize, Cayman Islands, and Palau) still use Fahrenheit for weather, medicine, and cooking. Example \(\PageIndex{4}\): Conversion from Celsius Normal body temperature has been commonly accepted as 37.0 °C (although it varies depending on time of day and method of measurement, as well as among individuals). What is this temperature on the kelvin scale and on the Fahrenheit scale? Solution \[\mathrm{K= {^\circ C}+273.15=37.0+273.2=310.2\: K}\nonumber \] \[\mathrm{^\circ F=\dfrac{9}{5}\:{^\circ C}+32.0=\left(\dfrac{9}{5}\times 37.0\right)+32.0=66.6+32.0=98.6\: ^\circ F}\nonumber \] Exercise \(\PageIndex{4}\) Convert 80.92 °C to K and °F. Answer 354.07 K, 177.7 °F Example \(\PageIndex{5}\): Conversion from Fahrenheit Baking a ready-made pizza calls for an oven temperature of 450 °F. If you are in Europe, and your oven thermometer uses the Celsius scale, what is the setting? What is the kelvin temperature? Solution \[\mathrm{^\circ C=\dfrac{5}{9}(^\circ F-32)=\dfrac{5}{9}(450-32)=\dfrac{5}{9}\times 418=232 ^\circ C\rightarrow set\: oven\: to\: 230 ^\circ C}\hspace{20px}\textrm{(two significant figures)}\nonumber \] \[\mathrm{K={^\circ C}+273.15=230+273=503\: K\rightarrow 5.0\times 10^2\,K\hspace{20px}(two\: significant\: figures)}\nonumber \] Exercise \(\PageIndex{5}\) Convert 50 °F to °C and K. Answer 10 °C, 280 K Units of Temperature: https://youtu.be/DTPo0HDMz3o Summary Measurements are made using a variety of units. It is often useful or necessary to convert a measured quantity from one unit into another. These conversions are accomplished using unit conversion factors, which are derived by simple applications of a mathematical approach called the factor-label method or dimensional analysis. This strategy is also employed to calculate sought quantities using measured quantities and appropriate mathematical relations. Key Equations \(T_\mathrm{^\circ C}=\dfrac{5}{9}\times T_\mathrm{^\circ F}-32\) \(T_\mathrm{^\circ F}=\dfrac{9}{5}\times T_\mathrm{^\circ C}+32\) \(T_\ce{K}={^\circ \ce C}+273.15\) \(T_\mathrm{^\circ C}=\ce K-273.15\) Exercise \(\PageIndex{1}\) Use the table earlier in the chapter to help you complete the following unit conversions. Go to two decimal places with your answers. 15 mL to tbsp 9 kg to lbs 5 km to mi 200 g to lbs Answer 1.01 tbsp. 19.84 lbs 3.11 mi 0.44 lbs Glossary dimensional analysis (also, factor-label method) versatile mathematical approach that can be applied to computations ranging from simple unit conversions to more complex, multi-step calculations involving several different quantities Fahrenheit unit of temperature; water freezes at 32 °F and boils at 212 °F on this scale unit conversion factor ratio of equivalent quantities expressed with different units; used to convert from one unit to a different unit |
Courses/Heartland_Community_College/CHEM_120%3A_Fundamentals_of_Chemistry/03%3A_Molecules_and_Compounds/3.07%3A_Ionic_Bonding%3A__Writing_Chemical_Formulas_and_Chemical_Names | Learning Objectives Write the chemical names of ionic compounds containing main group elements. In the previous section, the process for writing the chemical formula of an ionic compound containing main group elements was presented and applied. The chemical name of a compound is derived based on the information included in its chemical formula. A chemical name should uniquely correspond to a single chemical formula. In other words, no two chemical formulas should share a common chemical name. The procedure for writing the chemical name of an ionic compound containing main group elements is explained in the following paragraphs. Naming Ionic Compounds Containing Main Group Elements The chemical name of an ionic compound is based solely on the identities of the ions that it contains. Specifically, the names of the ions are modified by removing the word "ion" from each, and the remaining terms are written in the order in which they appear in the corresponding ionic chemical formula. Since the subscripts in an ionic chemical formula are the result of achieving charge-balance between the compound's constituent ions, referencing subscripts in an ionic chemical name is considered redundant. Therefore, ionic compounds do not include any numerical prefixes. For example, consider K 3 N, which is the chemical formula for the ionic compound that is formed when nitrogen and potassium bond with one another. These elements bond with one another as ions, not as neutral atoms. Therefore, more accurately, K 3 N is the chemical formula for the ionic compound that is formed when the nitride ion , N –3 , the anion formed upon the ionization of nitrogen , and the potassium ion , K +1 , the cation formed when potassium ionizes, bond with one another. Recall that the suffix of an anion is "-ide," as a verbal indicator of its negative charge. When naming an ionic compound, the word "ion" is removed from each of these terms, as no charges are explicitly-written in an ionic chemical formula. Each constituent particle , such as N –3 and K +1 , is charged and, consequently, has a name that includes the word "ion." However, an ionic compound , such as K 3 N , is a net-neutral species, due to the charge-balance achieved between these particles. Therefore, the term "ion" should not be incorporated into the chemical name of an ionic compound. In this example, " nitride ion " is shortened to " nitride ," and " potassium ion " simply becomes " potassium ." Finally, since the cation is symbolized before the anion in an ionic chemical formula, the cation term appears first in the chemical name of an ionic compound. Therefore, in this example, the term " potassium " is written before " nitride ." As the subscripts in an ionic chemical formula are not referenced in an ionic chemical name, the result of combining these terms, " potassium nitride ," is the chemically-correct name for K 3 N . Example \(\PageIndex{1}\) Write the chemical name of Al 2 S 3 , the ionic compound that is formed when aluminum and sulfur bond with one another. Solution More accurately, Al 2 S 3 is the chemical formula for the ionic compound that is formed when the aluminum ion , Al +3 , the cation formed when aluminum ionizes, and the sulfide ion , S –2 , the anion formed upon the ionization of sulfur , bond with one another. When naming an ionic compound, the word "ion" is removed from each of these terms, as no charges are explicitly-written in an ionic chemical formula. As a result, " sulfide ion " is shortened to " sulfide ," and " aluminum ion " simply becomes " aluminum ." Finally, since the cation is symbolized before the anion in an ionic chemical formula, the cation term appears first in the chemical name of an ionic compound. Therefore, the term " aluminum " is written before " sulfide ." As the subscripts in an ionic chemical formula are not referenced in an ionic chemical name, the result of combining these terms, " aluminum sulfide ," is the chemically-correct name for Al 2 S 3 . Exercise \(\PageIndex{1}\) Write the chemical name that corresponds to each of the following chemical formulas. CaO, the ionic compound that is formed when calcium and oxygen bond with one another. MgCl 2 , the ionic compound that is formed when chlorine and magnesium bond with one another. Answer a More accurately, Ca O is the chemical formula for the ionic compound that is formed when the calcium ion , Ca +2 , the cation formed when calcium ionizes) and the oxide ion , O –2 , the anion formed upon the ionization of oxygen , bond with one another. When naming an ionic compound, the word "ion" is removed from each of these terms, as no charges are explicitly-written in an ionic chemical formula. As a result, " oxide ion " is shortened to " oxide ," and " calcium ion " simply becomes " calcium ." Finally, since the cation is symbolized before the anion in an ionic chemical formula, the cation term appears first in the chemical name of an ionic compound. Therefore, the term " calcium " is written before " oxide ." As the subscripts in an ionic chemical formula are not referenced in an ionic chemical name, the result of combining these terms, " calcium oxide ," is the chemically-correct name for Ca O . Answer b More accurately, Mg Cl 2 is the chemical formula for the ionic compound that is formed when the chloride ion , Cl –1 , and the magnesium ion , Mg +2 , bond with one another. The word "ion" is removed from each of these terms, as no charges are explicitly-written in an ionic chemical formula. As a result, " chloride ion " is shortened to " chloride ," and " magnesium ion " simply becomes " magnesium ." Finally, the cation term appears first in the chemical name of an ionic compound, so the term " magnesium " is written before " chloride ." As the subscripts in an ionic chemical formula are not referenced in an ionic chemical name, " magnesium chloride ," is the chemically-correct name for Mg Cl 2 . |
Courses/Ursinus_College/CHEM322%3A_Inorganic_Chemistry/06%3A_Solid_State_Chemistry | 6.1: Solid State Structures 6.1.1: Cubic Lattices and Close Packing 6.1.2: Ionic Radii and Radius Ratios 6.2: Crystalline Solids 6.2.1: Types of Crystalline Solids 6.2.2: Alloys and Intermetallics 6.2.3: The Imperfect Solid State 6.3: X-Ray Crystallography of Solids 6.3.1: Miller Indices (hkl) 6.3.2: X-rays and X-ray Diffraction 6.3.3: Powder X-ray Diffraction 6.4: Energetics of Ionic Solids 6.4.1: Lattice Enthlapies of Ionic Solids 6.4.2: Born Haber Cycles 6.4.3: Lattice Energies and Solubility 6.4.4: Theoretical Lattice Energy Calculations 6.4.5: Kapustinskii Equation 6.5: Band Theory and Conductivity 6.5.1: Bonding in Metals and Semicondoctors 6.5.2: The Fermi Level 6.5.3: Semiconductors- Band Gaps, Colors, Conductivity and Doping 6.5.4: Periodic Trends- Metals, Semiconductors, and Insulators 6.5.5: Semiconductor p-n Junctions 6.5.6: Diodes, LEDs and Solar Cells 6.5.7: Superconductors |
Courses/Chabot_College/Introduction_to_General_Organic_and_Biochemistry/06%3A_Ionic_and_Molecular_Compounds/6.E%3A_Ionic_Bonding_and_Simple_Ionic_Compounds_(Exercises_I) | These are homework exercises to accompany Chapter 3 of the Ball et al. " The Basics of GOB Chemistry " Textmap. 3.1: Two Types of Bonding Concept Review Exercises What is the octet rule? How are ionic bonds formed? Answers The octet rule is the concept that atoms tend to have eight electrons in their valence electron shell. Ionic bonds are formed by the attraction between oppositely charged ions. Exercises Why is an ionic compound unlikely to consist of two positively charged ions? Why is an ionic compound unlikely to consist of two negatively charged ions? A calcium atom has two valence electrons. Do you think it will lose two electrons or gain six electrons to obtain an octet in its outermost electron shell? An aluminum atom has three valence electrons. Do you think it will lose three electrons or gain five electrons to obtain an octet in its outermost electron shell? A selenium atom has six valence electrons. Do you think it will lose six electrons or gain two electrons to obtain an octet in its outermost electron shell? An iodine atom has seven valence electrons. Do you think it will lose seven electrons or gain one electron to obtain an octet in its outermost electron shell? Answers Positive charges repel each other, so an ionic compound is not likely between two positively charged ions. 3. It is more likely to lose two electrons. 5. It is more likely to gain two electrons. 3.2: Ions Concept Review Exercises What are the two types of ions? Use Lewis diagrams to illustrate the formation of an ionic compound from a potassium atom and an iodine atom. When the following atoms become ions, what charges do they acquire? Li S Ca F Answers Cations have positive charges, and anions have negative charges. 1+ 2− 2+ 1− Key Takeaways Ions can be positively charged or negatively charged. A Lewis diagram is used to show how electrons are transferred to make ions and ionic compounds. Exercises Identify each as a cation, an anion, or neither. H + Cl − O 2 Ba 2 + CH 4 CS 2 Identify each as a cation, an anion, or neither. NH 3 Br − H − Hg 2 + CCl 4 SO 3 Write the electron configuration for each ion. Li + Mg 2 + F − S 2− Write the electron configuration for each ion. Na + Be 2 + Cl − O 2− Draw Lewis diagrams for the ions listed in Exercise 3. Also include Lewis diagrams for the respective neutral atoms as a comparison. Draw Lewis diagrams for the ions listed in Exercise 4. Also include Lewis diagrams for the respective neutral atoms as a comparison. Using Lewis diagrams, show the electron transfer for the formation of LiF. Using Lewis diagrams, show the electron transfer for the formation of MgO. Using Lewis diagrams, show the electron transfer for the formation of Li 2 O. Using Lewis diagrams, show the electron transfer for the formation of CaF 2 . What characteristic charge do atoms in the first column of the periodic table have when they become ions? What characteristic charge do atoms in the second column of the periodic table have when they become ions? What characteristic charge do atoms in the third-to-last column of the periodic table have when they become ions? What characteristic charge do atoms in the next-to-last column of the periodic table have when they become ions? Answers 1. cation anion neither cation neither neither 3. 1 s 2 1 s 2 2 s 2 2 p 6 1 s 2 2 s 2 2 p 6 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 5. a. b. c. d. 1+ 2− 3.3: Formulas for Ionic Compounds Concept Review Exercises What information is contained in the formula of an ionic compound? Why do the chemical formulas for some ionic compounds contain subscripts, while others do not? Write the chemical formula for the ionic compound formed by each pair of ions. Mg 2 + and I − Na + and O 2− Answers the ratio of each kind of ion in the compound Sometimes more than one ion is needed to balance the charge on the other ion in an ionic compound. MgI 2 Na 2 O Key Takeaways Proper chemical formulas for ionic compounds balance the total positive charge with the total negative charge. Groups of atoms with an overall charge, called polyatomic ions, also exist. Exercises Write the chemical formula for the ionic compound formed by each pair of ions. Na + and Br − Mg 2 + and Br − Mg 2 + and S 2− Write the chemical formula for the ionic compound formed by each pair of ions. K + and Cl − Mg 2 + and Cl − Mg 2 + and Se 2 − Write the chemical formula for the ionic compound formed by each pair of ions. Na + and N 3− Mg 2 + and N 3− Al 3 + and S 2− Write the chemical formula for the ionic compound formed by each pair of ions. Li + and N 3− Mg 2 + and P 3− Li + and P 3− Write the chemical formula for the ionic compound formed by each pair of ions. Fe 3 + and Br − Fe 2 + and Br − Au 3 + and S 2− Au + and S 2− Write the chemical formula for the ionic compound formed by each pair of ions. Cr 3 + and O 2− Cr 2 + and O 2− Pb 2 + and Cl − Pb 4 + and Cl − Write the chemical formula for the ionic compound formed by each pair of ions. Cr 3 + and NO 3 − Fe 2 + and PO 4 3 − Ca 2 + and CrO 4 2 − Al 3 + and OH − Write the chemical formula for the ionic compound formed by each pair of ions. NH 4 + and NO 3 − H + and Cr 2 O 7 2 − Cu + and CO 3 2 − Na + and HCO 3 − For each pair of elements, determine the charge for their ions and write the proper formula for the resulting ionic compound between them. Ba and S Cs and I For each pair of elements, determine the charge for their ions and write the proper formula for the resulting ionic compound between them. K and S Sc and Br Which compounds would you predict to be ionic? Li 2 O (NH 4 ) 2 O CO 2 FeSO 3 C 6 H 6 C 2 H 6 O Which compounds would you predict to be ionic? Ba(OH) 2 CH 2 O NH 2 CONH 2 (NH 4 ) 2 CrO 4 C 8 H 18 NH 3 Answers NaBr MgBr 2 MgS Na 3 N Mg 3 N 2 Al 2 S 3 FeBr 3 FeBr 2 Au 2 S 3 Au 2 S Cr(NO 3 ) 3 Fe 3 (PO 4 ) 2 CaCrO 4 Al(OH) 3 Ba 2 + , S 2− , BaS Cs + , I − , CsI ionic ionic not ionic ionic not ionic not ionic 3.4: Ionic Nomenclature Concept Review Exercises Briefly describe the process for naming an ionic compound. In what order do the names of ions appear in the names of ionic compounds? Which ionic compounds can be named using two different systems? Give an example. Answers Name the cation and then the anion but don’t use numerical prefixes. the cation name followed by the anion name Ionic compounds in which the cation can have more than one possible charge have two naming systems. FeCl 3 is either iron(III) chloride or ferric chloride (answers will vary). Key Takeaways Each ionic compound has its own unique name that comes from the names of the ions. Exercises Name each ion. Ra 2 + P 3− H 2 PO 4 − Sn 4 + Name each ion. Cs + As 3 − HSO 4 − Sn 2 + Name the ionic compound formed by each pair of ions. Na + and Br − Mg 2 + and Br − Mg 2 + and S 2− Name the ionic compound formed by each pair of ions. K + and Cl − Mg 2 + and Cl − Mg 2 + and Se 2 − Name the ionic compound formed by each pair of ions. Na + and N 3− Mg 2 + and N 3− Al 3 + and S 2− Name the ionic compound formed by each pair of ions. Li + and N 3− Mg 2 + and P 3− Li + and P 3− Name the ionic compound formed by each pair of ions. Use both the Stock and common systems, where appropriate. Fe 3 + and Br − Fe 2 + and Br − Au 3 + and S 2− Au + and S 2− Name the ionic compound formed by each pair of ions. Use both the Stock and common systems, where appropriate. Cr 3 + and O 2− Cr 2 + and O 2− Pb 2 + and Cl − Pb 4 + and Cl − Name the ionic compound formed by each pair of ions. Use both the Stock and common systems, where appropriate. Cr 3 + and NO 3 − Fe 2 + and PO 4 3 − Ca 2 + and CrO 4 2 − Al 3 + and OH − Name the ionic compound formed by each pair of ions. Use both the Stock and common systems, where appropriate. NH 4 + and NO 3 − H + and Cr 2 O 7 2 − Cu + and CO 3 2 − Na + and HCO 3 − Give two names for each compound. Al(HSO 4 ) 3 Mg(HSO 4 ) 2 Give two names for each compound. Co(HCO 3 ) 2 LiHCO 3 Answers the radium ion the phosphide ion the dihydrogen phosphate ion the tin(IV) ion or the stannic ion sodium bromide magnesium bromide magnesium sulfide sodium nitride magnesium nitride aluminum sulfide iron(III) bromide or ferric bromide iron(II) bromide or ferrous bromide gold(III) sulfide or auric sulfide gold(I) sulfide or aurous sulfide chromium(III) nitrate or chromic nitrate iron(II) phosphate or ferrous phosphate calcium chromate aluminum hydroxide aluminum hydrogen sulfate or aluminum bisulfate magnesium hydrogen sulfate or magnesium bisulfate 3.5: Formula Mass Concept Review Exercises What is the relationship between atomic mass and formula mass? How are subscripts used to determine a formula mass when more than one polyatomic ion is present in a chemical formula? Answers The formula mass is the sum of the atomic masses of the atoms in the formula. The subscript is distributed throughout the parentheses to determine the total number of atoms in the formula. Key Takeaways Formula masses of ionic compounds can be determined from the masses of the atoms in their formulas. Exercises What is the formula mass for the ionic compound formed by each pair of ions? Na + and Br − Mg 2 + and Br − Mg 2 + and S 2− What is the formula mass for the ionic compound formed by each pair of ions? K + and Cl − Mg 2 + and Cl − Mg 2 + and Se 2 − What is the formula mass for the ionic compound formed by each pair of ions? Na + and N 3− Mg 2 + and N 3− Al 3 + and S 2− What is the formula mass for the ionic compound formed by each pair of ions? Li + and N 3− Mg 2 + and P 3− Li + and P 3− What is the formula mass for each compound? FeBr 3 FeBr 2 Au 2 S 3 Au 2 S What is the formula mass for each compound? Cr 2 O 3 CrO PbCl 2 PbCl 4 What is the formula mass for each compound? Cr(NO 3 ) 3 Fe 3 (PO 4 ) 2 CaCrO 4 Al(OH) 3 What is the formula mass for each compound? NH 4 NO 3 H 2 Cr 2 O 7 Cu 2 CO 3 NaHCO 3 What is the formula mass for each compound? Al(HSO 4 ) 3 Mg(HSO 4 ) 2 What is the formula mass for each compound? Co(HCO 3 ) 2 LiHCO 3 Answers 102.90 amu 184.11 amu 56.38 amu 83.00 amu 100.93 amu 150.17 amu 295.50 amu 215.60 amu 490.30 amu 426.10 amu 238.00 amu 357.49 amu 156.08 amu 78.01 amu 318.22 amu 218.47 amu |
Courses/Rio_Hondo/Chemistry_110%3A_An_Introduction_to_General_Organic_and_Biological_Chemistry_(Garg)/13%3A_Carbohydrates/13.01%3A_Prelude_to_Carbohydrates | In the United States, 17.9 million people have been diagnosed with diabetes, and experts estimate that at least another 5.7 million people have the disease but have not been diagnosed. In 2006, diabetes was the seventh leading cause of death, listed on 72,507 death certificates. Moreover, it was a contributing factor in over 200,000 deaths in which the cause was listed as something else, such as heart or kidney disease. People with diabetes are impaired in their ability to metabolize glucose, a sugar needed by the body for energy; as a result, excessive quantities of glucose accumulate in the blood and the urine. The characteristic symptoms of diabetes are weight loss, constant hunger, extreme thirst, and frequent urination (the kidneys excrete large amounts of water in an attempt to remove the excess sugar from the blood). An important diagnostic test for diabetes is the oral glucose tolerance test, which measures the level of glucose in blood plasma. A first measurement is made after a fast of at least 8 h, followed by another measurement 2 h after the person drinks a flavored solution of 75 g of glucose dissolved in water. At the second measurement, the glucose plasma level should be no higher than 139 mg/dL. Individuals with a value between 140 and 199 mg/dL are diagnosed with prediabetes, while those with a value of 200 mg/dL or above are diagnosed with diabetes. Following a diagnosis of diabetes a person will need to monitor his or her blood glucose levels daily (or more often) using a glucose meter. |
Courses/Grand_Rapids_Community_College/CHM_120%3A_Survey_of_General_Chemistry_(Crandell)/10%3A_Thermochemistry_and_Thermodynamics/10.04%3A_Thermodynamics/10.4.01%3A_Spontaneity/10.4.1.01%3A_Entropy | Learning Objectives By the end of this section, you will be able to: Define entropy Explain the relationship between entropy and the number of microstates Predict the sign of the entropy change for chemical and physical processes In 1824, at the age of 28, Nicolas Léonard Sadi Carnot (Figure \(\PageIndex{1}\)) published the results of an extensive study regarding the efficiency of steam heat engines. A later review of Carnot’s findings by Rudolf Clausius introduced a new thermodynamic property that relates the spontaneous heat flow accompanying a process to the temperature at which the process takes place. This new property was expressed as the ratio of the reversible heat ( q rev ) and the kelvin temperature ( T ). In thermodynamics, a reversible process is one that takes place at such a slow rate that it is always at equilibrium and its direction can be changed (it can be “reversed”) by an infinitesimally small change in some condition. Note that the idea of a reversible process is a formalism required to support the development of various thermodynamic concepts; no real processes are truly reversible, rather they are classified as irreversible . Similar to other thermodynamic properties, this new quantity is a state function, so its change depends only upon the initial and final states of a system. In 1865, Clausius named this property en tropy (\(S\)) and defined its change for any process as the following: \[\Delta S=\frac{q_{ rev }}{T} \nonumber \] The entropy change for a real, irreversible process is then equal to that for the theoretical reversible process that involves the same initial and final states. Entropy and Microstates Following the work of Carnot and Clausius, Ludwig Boltzmann developed a molecular-scale statistical model that related the entropy of a system to the number of microstates ( W ) possible for the system. A microstate is a specific configuration of all the locations and energies of the atoms or molecules that make up a system. The relation between a system’s entropy and the number of possible microstates is \[S=k \ln W \nonumber \] where k is the Boltzmann constant, 1.38 10 −23 J/K. As for other state functions, the change in entropy for a process is the difference between its final (\(S_f\)) and initial (\(S_i\)) values: \[\Delta S=S_{ f }-S_{ i }=k \ln W_{ f }-k \ln W_{ i }=k \ln \frac{W_{ f }}{W_{ i }} \nonumber \] For processes involving an increase in the number of microstates, W f > W i , the entropy of the system increases and Δ S > 0. Conversely, processes that reduce the number of microstates, W f < W i , yield a decrease in system entropy, Δ S < 0. This molecular-scale interpretation of entropy provides a link to the probability that a process will occur as illustrated in the next paragraphs. Consider the general case of a system comprised of N particles distributed among n boxes. The number of microstates possible for such a system is n N . For example, distributing four particles among two boxes will result in 2 4 = 16 different microstates as illustrated in Figure \(\PageIndex{2}\). Microstates with equivalent particle arrangements (not considering individual particle identities) are grouped together and are called distributions . The probability that a system will exist with its components in a given distribution is proportional to the number of microstates within the distribution. Since entropy increases logarithmically with the number of microstates, the most probable distribution is therefore the one of greatest entropy . For this system, the most probable configuration is one of the six microstates associated with distribution (c) where the particles are evenly distributed between the boxes, that is, a configuration of two particles in each box. The probability of finding the system in this configuration is \(\frac{6}{16}\) or \(\frac{3}{8}\) The least probable configuration of the system is one in which all four particles are in one box, corresponding to distributions (a) and (e), each with a probability of The probability of finding all particles in only one box (either the left box or right box) is then or \[\left(\frac{1}{16}+\frac{1}{16}\right)=\frac{2}{16} \text { or } \frac{1}{8} \nonumber \] As you add more particles to the system, the number of possible microstates increases exponentially (2 N ). A macroscopic (laboratory-sized) system would typically consist of moles of particles ( N ~ 10 23 ), and the corresponding number of microstates would be staggeringly huge. Regardless of the number of particles in the system, however, the distributions in which roughly equal numbers of particles are found in each box are always the most probable configurations. This matter dispersal model of entropy is often described qualitatively in terms of the disorder of the system. By this description, microstates in which all the particles are in a single box are the most ordered, thus possessing the least entropy. Microstates in which the particles are more evenly distributed among the boxes are more disordered, possessing greater entropy. The previous description of an ideal gas expanding into a vacuum (Figure 16.4) is a macroscopic example of this particle-in-a-box model. For this system, the most probable distribution is confirmed to be the one in which the matter is most uniformly dispersed or distributed between the two flasks. Initially, the gas molecules are confined to just one of the two flasks. Opening the valve between the flasks increases the volume available to the gas molecules and, correspondingly, the number of microstates possible for the system. Since W f > W i , the expansion process involves an increase in entropy (Δ S > 0) and is spontaneous. A similar approach may be used to describe the spontaneous flow of heat. Consider a system consisting of two objects, each containing two particles, and two units of thermal energy (represented as “*”) in Figure \(\PageIndex{3}\). The hot object is comprised of particles A and B and initially contains both energy units. The cold object is comprised of particles C and D , which initially has no energy units. Distribution (a) shows the three microstates possible for the initial state of the system, with both units of energy contained within the hot object. If one of the two energy units is transferred, the result is distribution (b) consisting of four microstates. If both energy units are transferred, the result is distribution (c) consisting of three microstates. Thus, we may describe this system by a total of ten microstates. The probability that the heat does not flow when the two objects are brought into contact, that is, that the system remains in distribution (a), is \(\frac{3}{10}\). More likely is the flow of heat to yield one of the other two distribution, the combined probability being \(\frac{7}{10}\). The most likely result is the flow of heat to yield the uniform dispersal of energy represented by distribution (b), the probability of this configuration being \(\frac{4}{10}\). This supports the common observation that placing hot and cold objects in contact results in spontaneous heat flow that ultimately equalizes the objects’ temperatures. And, again, this spontaneous process is also characterized by an increase in system entropy. Example \(\PageIndex{1}\): Determination of Δ S Calculate the change in entropy for the process depicted below. Solution The initial number of microstates is one, the final six: \[\Delta S=k \ln \frac{W_{ c }}{W_{ a }}=1.38 \times 10^{-23} J / K \times \ln \frac{6}{1}=2.47 \times 10^{-23} J / K \nonumber \] The sign of this result is consistent with expectation; since there are more microstates possible for the final state than for the initial state, the change in entropy should be positive. Exercise \(\PageIndex{1}\) Consider the system shown in Figure \(\PageIndex{3}\). What is the change in entropy for the process where all the energy is transferred from the hot object ( AB ) to the cold object ( CD )? Answer 0 J/K Predicting the Sign of Δ S The relationships between entropy, microstates, and matter/energy dispersal described previously allow us to make generalizations regarding the relative entropies of substances and to predict the sign of entropy changes for chemical and physical processes. Consider the phase changes illustrated in Figure \(\PageIndex{4}\). In the solid phase, the atoms or molecules are restricted to nearly fixed positions with respect to each other and are capable of only modest oscillations about these positions. With essentially fixed locations for the system’s component particles, the number of microstates is relatively small. In the liquid phase, the atoms or molecules are free to move over and around each other, though they remain in relatively close proximity to one another. This increased freedom of motion results in a greater variation in possible particle locations, so the number of microstates is correspondingly greater than for the solid. As a result, S liquid > S solid and the process of converting a substance from solid to liquid (melting) is characterized by an increase in entropy, Δ S > 0. By the same logic, the reciprocal process (freezing) exhibits a decrease in entropy, Δ S < 0. Now consider the gaseous phase, in which a given number of atoms or molecules occupy a much greater volume than in the liquid phase. Each atom or molecule can be found in many more locations, corresponding to a much greater number of microstates. Consequently, for any substance, S gas > S liquid > S solid , and the processes of vaporization and sublimation likewise involve increases in entropy, Δ S > 0. Likewise, the reciprocal phase transitions, condensation and deposition, involve decreases in entropy, Δ S < 0. According to kinetic-molecular theory, the temperature of a substance is proportional to the average kinetic energy of its particles. Raising the temperature of a substance will result in more extensive vibrations of the particles in solids and more rapid translations of the particles in liquids and gases. At higher temperatures, the distribution of kinetic energies among the atoms or molecules of the substance is also broader (more dispersed) than at lower temperatures. Thus, the entropy for any substance increases with temperature (Figure \(\PageIndex{5}\)). Link to Learning Try this simulator with interactive visualization of the dependence of particle location and freedom of motion on physical state and temperature. The entropy of a substance is influenced by the structure of the particles (atoms or molecules) that comprise the substance. With regard to atomic substances, heavier atoms possess greater entropy at a given temperature than lighter atoms, which is a consequence of the relation between a particle’s mass and the spacing of quantized translational energy levels (a topic beyond the scope of this text). For molecules, greater numbers of atoms increase the number of ways in which the molecules can vibrate and thus the number of possible microstates and the entropy of the system. Finally, variations in the types of particles affects the entropy of a system. Compared to a pure substance, in which all particles are identical, the entropy of a mixture of two or more different particle types is greater. This is because of the additional orientations and interactions that are possible in a system comprised of nonidentical components. For example, when a solid dissolves in a liquid, the particles of the solid experience both a greater freedom of motion and additional interactions with the solvent particles. This corresponds to a more uniform dispersal of matter and energy and a greater number of microstates. The process of dissolution therefore involves an increase in entropy, Δ S > 0. Considering the various factors that affect entropy allows us to make informed predictions of the sign of Δ S for various chemical and physical processes as illustrated in Example \(\PageIndex{2}\). Example \(\PageIndex{2}\): Predicting the Sign of ∆ S Predict the sign of the entropy change for the following processes. Indicate the reason for each of your predictions. One mole liquid water at room temperature one mole liquid water at 50 °C \(Ag^{+}(aq)+ Cl^{-}(aq) \longrightarrow AgCl (s)\) \(C_6 H_6(l)+\frac{15}{2} O_2(g) \longrightarrow 6 CO_2(g)+3 H_2O(l)) \(NH_3(s) \longrightarrow NH_3(l)\) Solution positive, temperature increases negative, reduction in the number of ions (particles) in solution, decreased dispersal of matter negative, net decrease in the amount of gaseous species positive, phase transition from solid to liquid, net increase in dispersal of matter Exercise \(\PageIndex{1}\) Predict the sign of the entropy change for the following processes. Give a reason for your prediction. \(NaNO_3(s) \longrightarrow Na^{+}(aq)+ NO_3{ }^{-}(aq)\) the freezing of liquid water \(CO_2(s) \longrightarrow CO_2(g)\) \(CaCO_3(s) \longrightarrow CaO (s)+ CO_2(g)\) Answer (a) Positive; The solid dissolves to give an increase of mobile ions in solution. (b) Negative; The liquid becomes a more ordered solid. (c) Positive; The relatively ordered solid becomes a gas. (d) Positive; There is a net increase in the amount of gaseous species. |
Courses/South_Puget_Sound_Community_College/CHEM_110%3A_Chemical_Concepts/02%3A_Units_and_Measurements/2.07%3A_Conversion_between_Units_with_Conversion_Factor_(Single-step_Conversion_Problems) | Learning Objectives To convert a value reported in one unit to a corresponding value in a different unit using conversion factors. During your studies of chemistry (and physics also), you will note that mathematical equations are used in many different applications. Many of these equations have a number of different variables with which you will need to work. You should also note that these equations will often require you to use measurements with their units. Algebra skills become very important here! Converting Between Units with Conversion Factors A conversion factor is a factor used to convert one unit of measurement into another. A simple conversion factor can convert meters into centimeters, or a more complex one can convert miles per hour into meters per second. Since most calculations require measurements to be in certain units, you will find many uses for conversion factors. Always remember that a conversion factor has to represent a fact; this fact can either be simple or more complex. For instance, you already know that 12 eggs equal 1 dozen. A more complex fact is that the speed of light is \(1.86 \times 10^5\) miles/\(\text{sec}\). Either one of these can be used as a conversion factor depending on what type of calculation you are working with (Table \(\PageIndex{1}\)). English Units Metric Units Quantity 1 ounce (oz) 28.35 grams (g) *mass 1 fluid once (oz) 29.6 mL volume 2.205 pounds (lb) 1 kilogram (kg) *mass 1 inch (in) 2.54 centimeters (cm) length 0.6214 miles (mi) 1 kilometer (km) length 1 quarter (qt) 0.95 liters (L) volume *Pounds and ounces are technically units of force, not mass, but this fact is often ignored by the non-scientific community. Of course, there are other ratios which are not listed in Table \(\PageIndex{1}\). They may include: Ratios embedded in the text of the problem (using words such as per or in each , or using symbols such as / or %). Conversions in the metric system, as covered earlier in this chapter. Common knowledge ratios (such as 60 seconds \(=\) 1 minute). If you learned the SI units and prefixes described, then you know that 1 cm is 1/100th of a meter. \[ 1\; \rm{cm} = \dfrac{1}{100} \; \rm{m} = 10^{-2}\rm{m} \nonumber \] or \[100\; \rm{cm} = 1\; \rm{m} \nonumber \] Suppose we divide both sides of the equation by \(1 \text{m}\) (both the number and the unit): \[\mathrm{\dfrac{100\:cm}{1\:m}=\dfrac{1\:m}{1\:m}} \nonumber \] As long as we perform the same operation on both sides of the equals sign, the expression remains an equality. Look at the right side of the equation; it now has the same quantity in the numerator (the top) as it has in the denominator (the bottom). Any fraction that has the same quantity in the numerator and the denominator has a value of 1: \[ \dfrac{ \text{100 cm}}{\text{1 m}} = \dfrac{ \text{1000 mm}}{\text{1 m}}= \dfrac{ 1\times 10^6 \mu \text{m}}{\text{1 m}}= 1 \nonumber \] We know that 100 cm is 1 m, so we have the same quantity on the top and the bottom of our fraction, although it is expressed in different units. Performing Dimensional Analysis Dimensional analysis is amongst the most valuable tools that physical scientists use. Simply put, it is the conversion between an amount in one unit to the corresponding amount in a desired unit using various conversion factors. This is valuable because certain measurements are more accurate or easier to find than others. The use of units in a calculation to ensure that we obtain the final proper units is called dimensional analysis . Here is a simple example. How many centimeters are there in 3.55 m? Perhaps you can determine the answer in your head. If there are 100 cm in every meter, then 3.55 m equals 355 cm. To solve the problem more formally with a conversion factor, we first write the quantity we are given, 3.55 m. Then we multiply this quantity by a conversion factor, which is the same as multiplying it by 1. We can write 1 as \(\mathrm{\dfrac{100\:cm}{1\:m}}\) and multiply: \[ 3.55 \; \rm{m} \times \dfrac{100 \; \rm{cm}}{1\; \rm{m}} \nonumber \] The 3.55 m can be thought of as a fraction with a 1 in the denominator. Because m, the abbreviation for meters, occurs in both the numerator and the denominator of our expression, they cancel out: \[\dfrac{3.55 \; \cancel{\rm{m}}}{ 1} \times \dfrac{100 \; \rm{cm}}{1 \; \cancel{\rm{m}}} \nonumber \] The final step is to perform the calculation that remains once the units have been canceled: \[ \dfrac{3.55}{1} \times \dfrac{100 \; \rm{cm}}{1} = 355 \; \rm{cm} \nonumber \] In the final answer, we omit the 1 in the denominator. Thus, by a more formal procedure, we find that 3.55 m equals 355 cm. A generalized description of this process is as follows: quantity (in old units) × conversion factor = quantity (in new units) You may be wondering why we use a seemingly complicated procedure for a straightforward conversion. In later studies, the conversion problems you encounter will not always be so simple . If you master the technique of applying conversion factors, you will be able to solve a large variety of problems. In the previous example, we used the fraction \(\dfrac{100 \; \rm{cm}}{1 \; \rm{m}}\) as a conversion factor. Does the conversion factor \(\dfrac{1 \; \rm m}{100 \; \rm{cm}}\) also equal 1? Yes, it does; it has the same quantity in the numerator as in the denominator (except that they are expressed in different units). Why did we not use that conversion factor? If we had used the second conversion factor, the original unit would not have canceled, and the result would have been meaningless. Here is what we would have gotten: \[ 3.55 \; \rm{m} \times \dfrac{1\; \rm{m}}{100 \; \rm{cm}} = 0.0355 \dfrac{\rm{m}^2}{\rm{cm}} \nonumber \] For the answer to be meaningful, we have to construct the conversion factor in a form that causes the original unit to cancel out . Figure \(\PageIndex{1}\) shows a concept map for constructing a proper conversion. General Steps in Performing Dimensional Analysis Identify the " given " information in the problem. Look for a number with units to start this problem with. What is the problem asking you to " find "? In other words, what unit will your answer have? Use ratios and conversion factors to cancel out the units that aren't part of your answer, and leave you with units that are part of your answer. When your units cancel out correctly, you are ready to do the math . You are multiplying fractions, so you multiply the top numbers and divide by the bottom numbers in the fractions. Significant Figures in Conversions How do conversion factors affect the determination of significant figures? Numbers in conversion factors based on prefix changes, such as kilograms to grams, are not considered in the determination of significant figures in a calculation because the numbers in such conversion factors are exact. Exact numbers are defined or counted numbers, not measured numbers, and can be considered as having an infinite number of significant figures. (In other words, 1 kg is exactly 1,000 g, by the definition of kilo-.) Counted numbers are also exact. If there are 16 students in a classroom, the number 16 is exact. In contrast, conversion factors that come from measurements (such as density, as we will see shortly) or that are approximations have a limited number of significant figures and should be considered in determining the significant figures of the final answer. Example \(\PageIndex{1}\) Empty DataFrame Columns: [(Unnamed: 0_level_0, Steps for Problem Solving, Identify the "given" information and what the problem is asking you to "find.", List other known quantities., Prepare a concept map and use the proper conversion factor., Cancel units and calculate., Think about your result.), (Example \(\PageIndex{1}\), The average volume of blood in an adult male is 4.7 L. What is this volume in milliliters?, Given: 4.7 L Find: mL, \(1\, mL = 10^{-3} L \), Unnamed: 1_level_4, \( 4.7 \cancel{\rm{L}} \times \dfrac{1 \; \rm{mL}}{10^{-3}\; \cancel{\rm{L}}} = 4,700\; \rm{mL}\) or \( 4.7 \cancel{\rm{L}} \times \dfrac{1,000 \; \rm{mL}}{1\; \cancel{\rm{L}}} = 4,700\; \rm{mL}\) or 4.7 x 103 2SF, not ambiguous, The amount in mL should be 1000 times larger than the given amount in L.), (Example \(\PageIndex{2}\), A hummingbird can flap its wings once in 18 ms. How many seconds are in 18 ms?, Given: 18 ms Find: s, \(1 \,ms = 10^{-3} s \), Unnamed: 2_level_4, \( 18 \; \cancel{\rm{ms}} \times \dfrac{10^{-3}\; \rm{s}}{1 \; \cancel{\rm{ms}}} = 0.018\; \rm{s}\) or \( 18 \; \cancel{\rm{ms}} \times \dfrac{1\; \rm{s}}{1,000 \; \cancel{\rm{ms}}} = 0.018\; \rm{s}\), The amount in s should be 1/1000 the given amount in ms.)] Index: [] Exercise \(\PageIndex{1}\) Perform each conversion. 101,000 ns to seconds 32.08 kg to grams 1.53 grams to cg Answer a: \(1.01000 x 10^{-4} s \) Answer b: \(3.208 x 10^{4} g \) Answer c: \(1.53 x 10^{2} cg \) Video: Converting Units with Conversion Factors by Tyler Dewitt Summary Conversion factors are used to convert one unit of measurement into another. Dimensional analysis (unit conversions) involves the use of conversion factors that will cancel unwanted units and produce the appropriate units. |
Bookshelves/General_Chemistry/CLUE%3A_Chemistry_Life_the_Universe_and_Everything/02%3A_Electrons_and_Orbitals/2.8%3A_In-Text_References | This may (or may not) be helpful: http://www.cv.nrao.edu/course/astr53.../LarmorRad.pdf ↵ http://phet.colorado.edu/en/simulati...e-interference ↵ Link to “Dr. Quantum” double slit experiment: http://www.youtube.com/watch?v=DfPeprQ7oGc ↵ In fact, there is lots of light within your eyeball, even in the dark, due to black body radiation. You do not see it because it is not energetic enough to activate your photosensing cells. See: http://blogs.discovermagazine.com/co...ose-your-eyes/ ↵ http://www.physorg.com/news76249412.html ↵ \(h= 6.626068 \times 10^{-34} \mathrm{~m}^{2} \mathrm{kg/s}\) (or joule-seconds, where a joule is the kinetic energy of a \(2 \mathrm{~kg}\) mass moving at a velocity of \(1 \mathrm{~m/s}\)) ↵ This is known as the Rayleigh-Jeans law. ↵ http://phet.colorado.edu/simulations...lectric_Effect ↵ One type of semi-exception is illustrated by what are known as two- and multi-photon microscopes, in which two lower energy photons hit a molecule at almost the same moment, allowing their energies to be combined; see http://en.Wikipedia.org/wiki/Two-pho...ion_microscopy . ↵ For a more complex explanation, see: http://www.coffeeshopphysics.com/art...y_of_rainbows/ ↵ Bohr model applet particle and wave views: http://www.walter-fendt.de/ph11e/bohrh.htm ↵ Although the resting mass of a photon is zero, a moving photon does has an effective mass because it has energy. ↵ Good reference: http://ww2010.atmos.uiuc.edu/(Gl)/gu...asics/wvl.rxml ↵ see https://www.youtube.com/watch?v=6SxzfZ8bRO4 and https://www.youtube.com/watch?v=1920gi3swe4 ↵ These experiments are carried out using atoms in the gas phase in order to simplify the measurement. ↵ There are a number different ways of defining the size of an atom, and in fact the size depends on the atom’s chemical environment (for example, whether it is bonded to another atom or not). In fact, we can only measure the positions of atomic nuclei, and it is impossible to see where the electron cloud actually ends; remember that orbitals are defined as the surface within which there is a 90% probability of finding an electron. Therefore, we often use the van der Waals radius, which is half the distance between the nuclei of two adjacent unbonded atoms. ↵ For more information see: http://winter.group.shef.ac.uk/orbit.../1s/index.html http://www.uark.edu/misc/julio/orbitals/index.html ↵ This is called the Pauli exclusion principle, which states that no two electrons may occupy the same quantum state; that is, no two electrons can have the same value for all four quantum numbers. ↵ We should note that this model for calculating the effective nuclear charge is just that – a model. It provides us with an easy way to predict the relative attractions between the nuclei and electrons, but there are of course more accurate ways of calculating the attraction which take into account the fact that the nuclei is only partial shielded by the core electrons. ↵ This is often called Hund’s rule. Just as passengers on a bus do not sit together until they have to, neither do electrons. ↵ |
Courses/Oregon_Institute_of_Technology/OIT_(Lund)%3A_Organic_Chemistry_with_a_Biological_Emphasis_(Soderberg)/11%3A_Nucleophilic_Acyl_Substitution_Reactions/11.04%3A_The_Relative_Reactivity_of_Carboxylic_Acid_Derivatives | In carboxylic acid derivatives, the partial positive charge on the carbonyl carbon is stabilized by electron donation from nonbonding electrons on the adjacent heteroatom, which has the effect of decreasing electrophilicity. Among the carboxylic acid derivatives, carboxylate groups are the least reactive towards nucleophilic acyl substitution, followed by amides, then carboxylic esters and carboxylic acids, thioesters, and finally acyl phosphates, which are the most reactive among the biologically relevant acyl groups. Acid anhydrides and acid chlorides are laboratory reagents that are analogous to thioesters and acyl phosphates, in the sense that they too are highly reactive carboxylic acid derivatives. Section 11.8 near the end of this chapters includes information about the chemistry of these two reagents. Relative reactivity of carboxylic acid derivatives: The reactivity trend of the carboxylic acid derivatives can be understood by evaluating the basicity of the leaving group (acyl X group) - remember from section 8.4 that weaker bases are better leaving groups. A thioester is more reactive than an ester, for example, because a thiolate (RS-) is a weaker base and better leaving group than an alcoxide (\(RO\)-). Recall from chapter 7 that the \(pK_a\) of a thiol is about 10, while the \(pK_a\) of an alcohol is 15 or higher: a stronger conjugate acid means a weaker conjugate base. In general, if the incoming nucleophile is a weaker base than the ‘acyl X’ group that is already there, it will also be the better leaving group, and thus the first nucleophilic step will simply reverse itself and we’ll get the starting materials back: In general, acyl substitution reactions convert higher energy carboxylic acid derivatives into derivatives of lower energy. Thioesters, for example, are often converted directly into carboxylic esters in biochemical reactions, but not the other way around. To go 'uphill' - from a carboxylate to a thioester, for example, requires the 'coupling' of the uphill reaction to an energetically favorable reaction. We will see how this works in the next section. |
Courses/University_of_California_Davis/Chem_205%3A_Symmetry_Spectroscopy_and_Structure/02%3A_Electronic_Spectroscopy | Electron spectroscopy is an analytical technique to study the electronic structure and its dynamics in atoms and molecules. In general an excitation source such as x-rays, electrons or synchrotron radiation will eject an electron from an inner-shell orbital of an atom. 2.1: Transition Integrals 2.2: Vibronic Transitions 2.3: Broadening Mechanisms Spectrum lines are not infinitesimally narrow; they have a finite width. A graph of radiance or intensity per unit wavelength (or frequency) versus wavelength (or frequency) is the line profile. There are several causes of line broadening, some internal to the atom, others external, and each produces its characteristic profile. Analysis of the exact shape of a line profile may give us information about the physical conditions, such as temperature and pressure, in a stellar atmosphere. 2.4: The Fate of Electronic Transitions 2.5: Electronic State and Transitions 2.6: Introduction to Symmetry 2.7: The Carbonyl Group 2.8: Symmetry and Formaldehyde 2.9: Configuration Interaction The best energies obtained at the Hartree-Fock level are still not accurate, because they use an average potential for the electron-electron interactions. Configuration interaction (CI) methods help to overcome this limitation. Because electrons interact and repel each other, their motion in atoms is correlated. When one electron is close to the nucleus, the other tends to be far away. When one is on one side, the other tends to be on the other side. 2.10: Measures of Transition Amplitudes 2.11: Term Symbols Term symbols are a shorthand method used to describe the energy, angular momentum, and spin multiplicity of an atom in any particular state. From a spectroscopic perspective, we need to know the values for the various types of angular momenta. 2.12: Absorption Spectrum of Formaldehyde 2.13: Assignment of Bands Based on Solvent Effects 2.14: Solvent Effect of Fluorescence 2.15: Breaking Symmetries 2.16: Charge Transfer Bands 2.17: Conjugation Length |
Courses/Tennessee_State_University/Inorganic_Chemistry_(CHEM_5000_4200)/01%3A_Map-_Inorganic_Chemistry-I_(LibreTexts)/03%3A_Simple_Bonding_Theory/3.04%3A_Hydrogen_Bonding | A hydrogen bond is an intermolecular force (IMF) that forms a special type of dipole-dipole attraction when a hydrogen atom bonded to a strongly electronegative atom exists in the vicinity of another electronegative atom with a lone pair of electrons. Intermolecular forces (IMFs) occur between molecules. Other examples include ordinary dipole-dipole interactions and dispersion forces. Hydrogen bonds are are generally stronger than ordinary dipole-dipole and dispersion forces, but weaker than true covalent and ionic bonds. The evidence for hydrogen bonding Many elements form compounds with hydrogen. If you plot the boiling points of the compounds of the group 14 elements with hydrogen, you find that the boiling points increase as you go down the group. The increase in boiling point happens because the molecules are getting larger with more electrons, and so van der Waals dispersion forces become greater. If you repeat this exercise with the compounds of the elements in groups 15 , 16, and 17 with hydrogen, something odd happens. Although the same reasoning applies for group 4 of the periodic table, the boiling point of the compound of hydrogen with the first element in each group is abnormally high. In the cases of \(NH_3\), \(H_2O\) and \(HF\) there must be some additional intermolecular forces of attraction, requiring significantly more heat energy to break the IMFs. These relatively powerful intermolecular forces are described as hydrogen bonds. Origin of Hydrogen Bonding The molecules capable of hydrogen bonding include the following: Notice that in each of these molecules: The hydrogen is attached directly to a highly electronegative atoms, causing the hydrogen to acquire a highly positive charge. Each of the highly electronegative atoms attains a high negative charge and has at least one "active" lone pair. Lone pairs at the 2-level have electrons contained in a relatively small volume of space, resulting in a high negative charge density. Lone pairs at higher levels are more diffuse and, resulting in a lower charge density and lower affinity for positive charge. If you are not familiar with electronegativity , you should follow this link before you go on. Consider two water molecules coming close together. The \(\delta^+\) hydrogen is so strongly attracted to the lone pair that it is almost as if you were beginning to form a co-ordinate (dative covalent) bond . It doesn't go that far, but the attraction is significantly stronger than an ordinary dipole-dipole interaction . Hydrogen bonds have about a tenth of the strength of an average covalent bond, and are constantly broken and reformed in liquid water. If you liken the covalent bond between the oxygen and hydrogen to a stable marriage, the hydrogen bond has "just good friends" status. Water is an ideal example of hydrogen bonding. Notice that each water molecule can potentially form four hydrogen bonds with surrounding water molecules: two with the hydrogen atoms and two with the with the oxygen atoms. There are exactly the right numbers of \(\delta^+\) hydrogens and lone pairs for every one of them to be involved in hydrogen bonding. This is why the boiling point of water is higher than that of ammonia or hydrogen fluoride. In the case of ammonia, the amount of hydrogen bonding is limited by the fact that each nitrogen only has one lone pair. In a group of ammonia molecules, there are not enough lone pairs to go around to satisfy all the hydrogens. In hydrogen fluoride, the problem is a shortage of hydrogens. In water, two hydrogen bonds and two lone pairs allow formation of hydrogen bond interactions in a lattice of water molecules. Water is thus considered an ideal hydrogen bonded system. More complex examples of hydrogen bonding The hydration of negative ions When an ionic substance dissolves in water, water molecules cluster around the separated ions. This process is called hydration . Water frequently attaches to positive ions by co-ordinate (dative covalent) bonds. It bonds to negative ions using hydrogen bonds. If you are interested in the bonding in hydrated positive ions, you could follow this link to co-ordinate (dative covalent) bonding . The diagram shows the potential hydrogen bonds formed with a chloride ion, Cl-. Although the lone pairs in the chloride ion are at the 3-level and would not normally be active enough to form hydrogen bonds, they are made more attractive by the full negative charge on the chlorine in this case. However complicated the negative ion, there will always be lone pairs that the hydrogen atoms from the water molecules can hydrogen bond to. Hydrogen bonding in alcohols An alcohol is an organic molecule containing an -OH group. Any molecule which has a hydrogen atom attached directly to an oxygen or a nitrogen is capable of hydrogen bonding. Hydrogen bonds also occur when hydrogen is bonded to fluorine, but the HF group does not appear in other molecules. Molecules with hydrogen bonds will always have higher boiling points than similarly sized molecules which don't have an -O-H or an -N-H group. The hydrogen bonding makes the molecules "stickier," such that more heat (energy) is required to separate them. This phenomenon can be used to analyze boiling point of different molecules, defined as the temperature at which a phase change from liquid to gas occurs. Ethanol, \(\ce{CH3CH2-O-H}\), and methoxymethane, \(\ce{CH3-O-CH3}\), both have the same molecular formula, \(\ce{C2H6O}\). They have the same number of electrons, and a similar length. The van der Waals attractions (both dispersion forces and dipole-dipole attractions) in each will be similar. However, ethanol has a hydrogen atom attached directly to an oxygen; here the oxygen still has two lone pairs like a water molecule. Hydrogen bonding can occur between ethanol molecules, although not as effectively as in water. The hydrogen bonding is limited by the fact that there is only one hydrogen in each ethanol molecule with sufficient + charge. In methoxymethane, the lone pairs on the oxygen are still there, but the hydrogens are not sufficiently + for hydrogen bonds to form. Except in some rather unusual cases, the hydrogen atom has to be attached directly to the very electronegative element for hydrogen bonding to occur. The boiling points of ethanol and methoxymethane show the dramatic effect that the hydrogen bonding has on the stickiness of the ethanol molecules: 0 1 ethanol (with hydrogen bonding) 78.5°C methoxymethane (without hydrogen bonding) -24.8°C The hydrogen bonding in the ethanol has lifted its boiling point about 100°C. It is important to realize that hydrogen bonding exists in addition to van der Waals attractions. For example, all the following molecules contain the same number of electrons, and the first two have similar chain lengths. The higher boiling point of the butan-1-ol is due to the additional hydrogen bonding. Comparing the two alcohols (containing -OH groups), both boiling points are high because of the additional hydrogen bonding; however, the values are not the same. The boiling point of the 2-methylpropan-1-ol isn't as high as the butan-1-ol because the branching in the molecule makes the van der Waals attractions less effective than in the longer butan-1-ol. Hydrogen bonding in organic molecules containing nitrogen Hydrogen bonding also occurs in organic molecules containing N-H groups; recall the hydrogen bonds that occur with ammonia. Examples range from simple molecules like CH 3 NH 2 (methylamine) to large molecules like proteins and DNA. The two strands of the famous double helix in DNA are held together by hydrogen bonds between hydrogen atoms attached to nitrogen on one strand, and lone pairs on another nitrogen or an oxygen on the other one. Donors and Acceptors In order for a hydrogen bond to occur there must be both a hydrogen donor and an acceptor present. The donor in a hydrogen bond is usually a strongly electronegative atom such as N, O, or F that is covalently bonded to a hydrogen bond. The hydrogen acceptor is an electronegative atom of a neighboring molecule or ion that contains a lone pair that participates in the hydrogen bond. Why does a hydrogen bond occur? Since the hydrogen donor (N, O, or F) is strongly electronegative, it pulls the covalently bonded electron pair closer to its nucleus, and away from the hydrogen atom. The hydrogen atom is then left with a partial positive charge, creating a dipole-dipole attraction between the hydrogen atom bonded to the donor and the lone electron pair of the acceptor. This results in a hydrogen bond.(see Interactions Between Molecules With Permanent Dipoles) Types of hydrogen bonds Although hydrogen bonds are well-known as a type of IMF, these bonds can also occur within a single molecule, between two identical molecules, or between two dissimilar molecules. Intramolecular hydrogen bonds Intramolecular hydrogen bonds are those which occur within one single molecule. This occurs when two functional groups of a molecule can form hydrogen bonds with each other. In order for this to happen, both a hydrogen donor a hydrogen acceptor must be present within one molecule, and they must be within close proximity of each other in the molecule. For example, intramolecular hydrogen bonding occurs in ethylene glycol (C 2 H 4 (OH) 2 ) between its two hydroxyl groups due to the molecular geometry. Intermolecular hydrogen bonds Intermolecular hydrogen bonds occur between separate molecules in a substance. They can occur between any number of like or unlike molecules as long as hydrogen donors and acceptors are present in positions where they can interact with one another. For example, intermolecular hydrogen bonds can occur between NH 3 molecules alone, between H 2 O molecules alone, or between NH 3 and H 2 O molecules. Properties and effects of hydrogen bonds On Boiling Point When we consider the boiling points of molecules, we usually expect molecules with larger molar masses to have higher normal boiling points than molecules with smaller molar masses. This, without taking hydrogen bonds into account, is due to greater dispersion forces (see Interactions Between Nonpolar Molecules). Larger molecules have more space for electron distribution and thus more possibilities for an instantaneous dipole moment. However, when we consider the table below, we see that this is not always the case. Compound Molar Mass Normal Boiling Point \(H_2O\) 18 g/mol 373 K \(HF\) 20 g/mol 292.5 K \(NH_3\) 17 g/mol 239.8 K \(H_2S\) 34 g/mol 212.9 K \(HCl\) 36.4 g/mol 197.9 K \(PH_3\) 34 g/mol 185.2 K We see that H 2 O, HF, and NH 3 each have higher boiling points than the same compound formed between hydrogen and the next element moving down its respective group, indicating that the former have greater intermolecular forces. This is because H 2 O, HF, and NH 3 all exhibit hydrogen bonding, whereas the others do not. Furthermore, \(H_2O\) has a smaller molar mass than HF but partakes in more hydrogen bonds per molecule, so its boiling point is higher. On Viscosity The same effect that is seen on boiling point as a result of hydrogen bonding can also be observed in the viscosity of certain substances. Substances capable of forming hydrogen bonds tend to have a higher viscosity than those that do not form hydrogen bonds. Generally, substances that have the possibility for multiple hydrogen bonds exhibit even higher viscosities. Factors preventing Hydrogen bonding Electronegativity Hydrogen bonding cannot occur without significant electronegativity differences between hydrogen and the atom it is bonded to. Thus, we see molecules such as PH 3 , which do not participate in hydrogen bonding. PH 3 exhibits a trigonal pyramidal molecular geometry like that of ammonia, but unlike NH 3 it cannot hydrogen bond. This is due to the similarity in the electronegativities of phosphorous and hydrogen. Both atoms have an electronegativity of 2.1, and thus, there is no dipole moment. This prevents the hydrogen atom from acquiring the partial positive charge needed to hydrogen bond with the lone electron pair in another molecule. (see Polarizability ) Atom Size The size of donors and acceptors can also affect the ability to hydrogen bond. This can account for the relatively low ability of Cl to form hydrogen bonds. When the radii of two atoms differ greatly or are large, their nuclei cannot achieve close proximity when they interact, resulting in a weak interaction. Hydrogen Bonding in Nature Hydrogen bonding plays a crucial role in many biological processes and can account for many natural phenomena such as the Unusual properties of Water . In addition to being present in water, hydrogen bonding is also important in the water transport system of plants, secondary and tertiary protein structure, and DNA base pairing. Plants The cohesion-adhesion theory of transport in vascular plants uses hydrogen bonding to explain many key components of water movement through the plant's xylem and other vessels. Within a vessel, water molecules hydrogen bond not only to each other, but also to the cellulose chain that comprises the wall of plant cells. Since the vessel is relatively small, the attraction of the water to the cellulose wall creates a sort of capillary tube that allows for capillary action . This mechanism allows plants to pull water up into their roots. Furthermore, hydrogen bonding can create a long chain of water molecules, which can overcome the force of gravity and travel up to the high altitudes of leaves. Proteins Hydrogen bonding is present abundantly in the secondary structure of proteins , and also sparingly in tertiary conformation. The secondary structure of a protein involves interactions (mainly hydrogen bonds) between neighboring polypeptide backbones which contain nitrogen-hydrogen bonded pairs and oxygen atoms. Since both N and O are strongly electronegative, the hydrogen atoms bonded to nitrogen in one polypeptide backbone can hydrogen bond to the oxygen atoms in another chain and vice-versa. Though they are relatively weak, these bonds offer substantial stability to secondary protein structure because they repeat many times and work collectively. In tertiary protein structure, interactions are primarily between functional R groups of a polypeptide chain; one such interaction is called a hydrophobic interaction. These interactions occur because of hydrogen bonding between water molecules around the hydrophobe that further reinforces protein conformation. References Brown, et al. Chemistry:The Central Science. 11th ed. Upper Saddle River, New Jersey: Pearson/Prentice Hall, 2008. Chang, Raymond. General Chemistry:The Essential Concepts. 3rd ed. New York: Mcgraw Hill, 2003 Petrucci, et al. General Chemistry: Principles & Modern Applications. 9th ed. Upper Saddle River, New Jersey: Pearson/Prentice Hall, 2007. |
Courses/Nassau_Community_College/Organic_Chemistry_I_and_II/09%3A_Alkynes/9.03%3A_Reactions_of_Alkynes_-_Addition_of_HX_and_X%E2%82%82 | Learning Objective predict the products and specify the reagents for the Electrophilic Addition Reactions (EARs) of alkynes with HX and X 2 Addition by Electrophilic Reagents Since the most common chemical transformation of a carbon-carbon double bond is an addition reaction, we might expect the same to be true for carbon-carbon triple bonds. Indeed, most of the alkene addition reactions also take place with alkynes with similar regio- and stereoselectivity. When the addition reactions of electrophilic reagents, such as strong Brønsted acids and halogens, to alkynes are studied we find a curious paradox. The reactions are even more exothermic than the additions to alkenes, and yet the rate of addition to alkynes is slower by a factor of 100 to 1000 than addition to equivalently substituted alkenes. The reaction of one equivalent of bromine with 1-penten-4-yne, for example, gave 4,5-dibromo-1-pentyne as the chief product. HC≡C-CH 2 -CH=CH 2 + Br 2 → HC≡C-CH 2 -CH Br CH 2 Br Although these electrophilic additions to alkynes are sluggish, they do take place and generally display Markovnikov Rule regioselectivity and anti-stereoselectivity. One problem, of course, is that the products of these additions are themselves substituted alkenes and can therefore undergo further addition. Because of their high electronegativity, halogen substituents on a double bond act to reduce its nucleophilicity, and thereby decrease the rate of electrophilic addition reactions. Consequently, there is a delicate balance as to whether the product of an initial addition to an alkyne will suffer further addition to a saturated product. Although the initial alkene products can often be isolated and identified, they are commonly present in mixtures of products and may not be obtained in high yield. The following reactions illustrate many of these features. In the last example, 1,2-diodoethene does not suffer further addition inasmuch as vicinal-diiodoalkanes are relatively unstable. As a rule, electrophilic addition reactions to alkenes and alkynes proceed by initial formation of a pi-complex , in which the electrophile accepts electrons from and becomes weakly bonded to the multiple bond. Such complexes are formed reversibly and may then reorganize to a reactive intermediate in a slower, rate-determining step. Reactions with alkynes are more sensitive to solvent changes and catalytic influences than are equivalent alkenes. Why are the reactions of alkynes with electrophilic reagents more sluggish than the corresponding reactions of alkenes? After all, addition reactions to alkynes are generally more exothermic than additions to alkenes, and there would seem to be a higher π-electron density about the triple bond ( two π-bonds versus one ). Two factors are significant in explaining this apparent paradox. First, although there are more π-electrons associated with the triple bond, the sp-hybridized carbons exert a strong attraction for these π-electrons, which are consequently bound more tightly to the functional group than are the π-electrons of a double bond. This is seen in the ionization potentials of ethylene and acetylene. 0 1 2 3 Acetylene HC≡CH + Energy → [HC≡CH •(+) + e(–) NaN ΔH = +264 kcal/mole Ethylene H2C=CH2 + Energy → [H2C=CH2] •(+) + e(–) NaN ΔH = +244 kcal/mole Ethane H3C–CH3 + Energy → [H3C–CH3] •(+) + e(–) NaN ΔH = +296 kcal/mole As defined by the preceding equations, an ionization potential is the minimum energy required to remove an electron from a molecule of a compound. Since pi-electrons are less tightly held than sigma-electrons, we expect the ionization potentials of ethylene and acetylene to be lower than that of ethane, as is the case. Gas-phase proton affinities show the same order, with ethylene being more basic than acetylene, and ethane being less basic than either. Since the initial interaction between an electrophile and an alkene or alkyne is the formation of a pi-complex, in which the electrophile accepts electrons from and becomes weakly bonded to the multiple bond, the relatively slower reactions of alkynes becomes understandable. A second factor is presumed to be the stability of the carbocation intermediate generated by sigma-bonding of a proton or other electrophile to one of the triple bond carbon atoms. This intermediate has its positive charge localized on an unsaturated carbon, and such vinyl cations are less stable than their saturated analogs. Indeed, we can modify our earlier ordering of carbocation stability to include these vinyl cations in the manner shown below. It is possible that vinyl cations stabilized by conjugation with an aryl substituent are intermediates in HX addition to alkynes of the type Ar-C≡C-R, but such intermediates are not formed in all alkyne addition reactions. 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Carbocation Stability CH3(+) ≈ RCH=CH(+) < RCH2(+) ≈ RCH=CR(+) < R2CH(+) ≈ CH2=CH-CH2(+) < C6H5CH2(+) ≈ R3C(+) Methyl 1°-Vinyl 1° 2°-Vinyl 2° 1°-Allyl 1°-Benzyl 3° NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN Carbocation Stability CH3(+) ≈ RCH=CH(+) < RCH2(+) ≈ RCH=CR(+) < R2CH(+) ≈ CH2=CH-CH2(+) < C6H5CH2(+) ≈ R3C(+) Carbocation Stability Methyl NaN 1°-Vinyl NaN 1° NaN 2°-Vinyl NaN 2° NaN 1°-Allyl NaN 1°-Benzyl NaN 3° Application of the Hammond postulate indicates that the activation energy for the generation of a vinyl cation intermediate would be higher than that for a lower energy intermediate. This is illustrated for alkenes versus alkynes by the following energy diagrams. Despite these differences, electrophilic additions to alkynes have emerged as exceptionally useful synthetic transforms. Addition of Hydrogen Halide to an Alkyne Summary: Reactivity order of hydrogen halides : HI > HB r> HCl > HF. Follows Markovnikov’s rule: Hydrogen adds to the carbon with the greatest number of hydrogens, the halogen adds to the carbon with fewest hydrogens. Protination occurs on the more stable carbocation. With the addition of HX, haloalkenes form. With the addition of excess HX, you get anti addition forming a geminal dihaloalkane. Addition of a HX to an Internal Alkyne As shown in Figure 2 below, the \(\pi\) electrons react with the hydrogen of the HBr and because the alkyne carbons are equivalent it does not matter which carbon adds the hydrogen. Once the hydrogen is covalently bonded to one of the carbons, the bromide will react with the carbocation intermediate to form a vinyl halide as shown in the example of forming 2-bromobutene from 2-butyne reacting with HBr. The reaction below assumes a 1:1 mole ratio of the alkyne and HBr. Now, what happens if there is excess HBr? Addition due to excess HX yields a geminal dihaloalkane Here, the electrophilic addition proceeds with the same steps used to achieve the product in Addition of a HX to an Internal Alkyne. The \(\pi\) electrons react with the hydrogen (shown in blue) adding it to the carbon on the left because the lone pair electrons of the bromine can help stabilize the carbocation intermediate that reacts with the bromide ions to form a geminal dihalide. Addition of HX to Terminal Alkyne For terminal alkynes, the carbon atoms sharing the triple bond are not equivalent. The addition of HX to terminal alkynes occurs in a Markovnikov-manner in which the halide attaches to the most substituted carbon. The pi electrons react with the hydrogen and it bonds to the terminal carbon. The bromide reacts with the resulting carbocation intermediate to form the vinyl halide. The overall reaction and mechanism are shown below. Addition due to excess HBr present Similar to the addition of excess HBr to internal alkynes, both halides will add to the same carbon to form a geminal dihalide. HBr Addition With Radical Mechanism Most hydrogen halide reactions with terminal alkynes occur in a Markovnikov-manner in which the halide attaches to the most substituted carbon since it is the most positively polarized. However, there are two specific reactions among alkynes where anti-Markovnikov reactions take place: the radical addition of HBr and Hydroboration Oxidation reactions. For alkynes, an anti-Markovnikov addition takes place for terminal alkynes. The Br of the Hydrogen Bromide (H-Br) attaches to the less substituted 1-carbon of the terminal alkyne shown below in an anti-Markovnikov manner while the Hydrogen proton attaches to the second carbon. As mentioned above, the first carbon is the less substituted carbon since it has fewer bonds attached to carbons and other substituents. The H-Br reagent must also be reacted with heat or some other radicial initiator such as a peroxide in order for this reaction to proceed in this manner. This presence of the radical or heat leads to the anti-Markovnikov addition since it produces the most stable reaction. The product of a terminal alkyne that is reacted with a peroxide (or light) and H-Br is a 1-bromoalkene. Regioselectivity : The Bromine can attach in a syn or anti manner which means the resulting alkene can be both cis and tran s . Syn addition is when both Hydrogens attach to the same face or side of the double bond (i.e. cis ) while the anti addition is when they attach on opposite sides of the bond ( trans ). Halogenation of Alkynes The additon of X 2 to alkynes is analogous to the addition of X2 to alkenes. The halogen molecule becomes polarized by the approach of the nucleophilic alkyne. The pi electrons of the alkyne react with the bromine to form a carbon-bromine bond and cyclic halonium ion with halide as the leaving group. The formation of the cyclic halonium ion requires anti-addition of the nucleophilic halide to produce a vicinal dihalide alkene as shown in the reaction below. Exercise Draw the structure, and give the IUPAC name, of the product formed in each of the reactions listed below. $\ce{\sf{CH3-C#C-CH3->[\displaystyle{\textrm{1 equiv}}][\displaystyle{\textrm{HCl}}]}}$ $\ce{\sf{CH3-C#C-CH3->[\displaystyle{\textrm{excess}}][\displaystyle{\textrm{HCl}}]}}$ $\ce{\sf{CH3-C#C-CH3->[\displaystyle{\textrm{1 equiv}}][\displaystyle{\textrm{Br}_2}]}}$ $\ce{\sf{CH3-C#C-CH3->[\displaystyle{\textrm{excess}}][\displaystyle{\textrm{Br}_2}]}}$ $\ce{\sf{CH3CH2-C#C-H->[\displaystyle{\textrm{1 equiv}}][\displaystyle{\textrm{HCl}}]}}$ $\ce{\sf{CH3CH2-C#C-H->[\displaystyle{\textrm{excess}}][\displaystyle{\textrm{HCl}}]}}$ $\ce{\sf{CH3CH2CH2-C#C-H->[\displaystyle{\textrm{1 equiv}}][\displaystyle{\textrm{Br}_2}]}}$ $\ce{\sf{CH3CH2CH2-C#C-H->[\displaystyle{\textrm{excess}}][\displaystyle{\textrm{Br}_2}]}}$ Answer ( Z )-2-chloro-2-butene 2,2-dichlorobutane ( E )-2,3-dibromo-2-butene 2,2,3,3-tetrabromobutane 2-chloro-1-butene 2,2-dichlorobutane ( E )-1,2-dibromo-1-pentene 1,1,2,2-tetrabromopentane |
Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/14%3A_Developing_a_Standard_Method/14.06%3A_Additional_Resources | The following set of experiments provide practical examples of the optimization of experimental conditions. Examples include simplex optimization, factorial designs for developing empirical models of response surfaces, and fitting experimental data to theoretical models of the response surface. Amenta, D. S.; Lamb, C. E.; Leary, J. J. “Simplex Optimization of Yield of sec -Butylbenzene in a Friedel-Crafts Alkylation,” J. Chem. Educ. 1979 , 56 , 557–558. Gozálvez, J. M.; García-Diaz, J. C. “Mixture Design Experiments Applied to the Formulation of Colo- rant Solutions,” J. Chem. Educ. 2006 , 83 , 647–650. Harvey, D. T.; Byerly, S.; Bowman, A.; Tomlin, J. “Optimization of HPLC and GC Separations Using Response Surfaces,” J. Chem. Educ. 1991 , 68 , 162–168. Krawcyzk, T.; Shupska, R.; Baj, S. “Applications of Chemiluminescence in the Teaching of Experimental Design,” J. Chem. Educ. 2015 , 92 , 317–321. Leggett, D. L. “Instrumental Simplex Optimization,” J. Chem. Educ. 1983 , 60 , 707–710. Oles, P. J. “Fractional Factorial Experimental Design as a Teaching Tool for Quantitative Analysis,” J. Chem. Educ. 1998 , 75 , 357–359. Palasota, J. A.; Deming, S.N. “Central Composite Experimental Design,” J. Chem. Educ. 1992 , 69 , 560–561. Sangsila, S.; Labinaz, G.; Poland, J. S.; vanLoon, G. W. “An Experiment on Sequential Simplex Optimization of an Atomic Absorption Analysis Procedure,” J. Chem. Educ. 1989 , 66 , 351–353. Santos-Delgado, M. J.; Larrea-Tarruella, L. “A Didactic Experience of Statistical Analysis for the De- termination of Glycine in a Nonaqueous Medium using ANOVA and a Computer Program,” J. Chem. Educ. 2004 , 81 , 97–99. Shavers, C. L.; Parsons, M. L.; Deming, S. N. “Simplex Optimization of Chemical Systems,” J. Chem Educ. 1979 , 56 , 307–309. Stieg, S. “A Low-Noise Simplex Optimization Experiment,” J. Chem. Educ. 1986 , 63 , 547–548. Stolzberg, R. J. “Screening and Sequential Experimentation: Simulations and Flame Atomic Absorption Spectrometry Experiments,” J. Chem. Educ. 1997 , 74 , 216–220. Van Ryswyk, H.; Van Hecke, G. R. “Attaining Optimal Conditions,” J. Chem. Educ. 1991 , 66 , 878– 882. The following texts and articles provide an excellent discussion of optimization methods based on searching algorithms and mathematical modeling use factorial designs, including a discussion of the relevant calculations. A few of these sources discuss other types of experimental designs. Analytical Methods Committee “Experimental design and optimization (1): an introduction to some basic concepts,” AMCTB 24, 2006. Analytical Methods Committee “Experimental design and optimization (2): handling uncontrolled factors,” AMCTB 26, 2006. Analytical Methods Committee “Experimental design and optimization (3): some fractional factorial designs,” AMCTB 36, 2009. Analytical Methods Committee “Experimental design and optimisation (4): Plackett–Burman de- signs,” AMCTB 55, 2013. Bayne, C. K.; Rubin, I. B. Practical Experimental Designs and Optimization Methods for Chemists , VCH Publishers: Deerfield Beach, FL; 1986. Bezerra, M. A.; Santelli, R. E.; Oliveira, E. P.; Villar, L. S.; Escaleira, L. A. “Response surface methodology (RSM) as a tool for optimization in analytical chemistry,” Talanta 2008 , 76 , 965–977. Box, G. E. P. “Statistical Design in the Study of Analytical Methods,” Analyst 1952 , 77 , 879–891. Deming, S. N.; Morgan, S. L. Experimental Design: A Chemometric Approach , Elsevier: Amsterdam, 1987. Ferreira, S. L. C.; dos Santos, W. N. L.; Quintella, C. M.; Neto, B. B.; Bosque-Sendra, J. M. “Doehlert Matrix: A Chemometric Tool for Analytical Chemistry—Review,” Talanta 2004 , 63 , 1061–1067. Ferreira, S. L. C.; Bruns, R. E.; Ferreira, H. S.; Matos, G. D.; David, J. M.; Brandão, G. C.; da Silva, E. G. P.; Portugal, L. A.; dos Reis, P. S.; Souza, A. S.; dos Santos, W. N. L. “Box-Behnken Design: An Alternative for the Optimization of Analytical Methods,” Anal. Chim. Acta 2007 , 597 , 179–186. Gonzalez, A. G. “Two Level Factorial Experimental Designs Based on Multiple Linear Regression Models: A Tutorial Digest Illustrated by Case Studies,” Anal. Chim. Acta 1998 , 360 , 227–241. Goupy, J. “What Kind of Experimental Design for Finding and Checking Robustness of Analytical Methods?” Anal. Chim. Acta 2005 , 544 , 184–190. Hendrix, C. D. “What Every Technologist Should Know About Experimental Design,” Chemtech 1979 , 9 , 167–174. Hendrix, C. D. “Through the Response Surface with Test Tube and Pipe Wrench,” Chemtech 1980 , 10 , 488–497. Leardi, R. “Experimental Design: A Tutorial,” Anal. Chim. Acta 2009 , 652 , 161–172. Liang, Y. “Comparison of Optimization Methods,” Chromatography Review 1985 , 12(2) , 6–9. Morgan, E. Chemometrics: Experimental Design , John Wiley and Sons: Chichester, 1991. Walters, F. H.; Morgan, S. L.; Parker, L. P., Jr.; Deming, S. N. Sequential Simplex Optimization , CRC Press: Boca Raton, FL, 1991. The following texts provide additional information about ANOVA calculations, including discussions of two-way analysis of variance. Graham, R. C. Data Analysis for the Chemical Sciences , VCH Publishers: New York, 1993. Miller, J. C.; Miller, J. N. Statistics for Analytical Chemistry , Ellis Horwood Limited: Chichester, 1988. The following resources provide additional information on the validation of analytical methods. Gonzalez, A. G.; Herrador, M. A. “A Practical Guide to Analytical Method Validation, Including Measurement Uncertainty and Accuracy Profiles,” Trends Anal. Chem. 2007 , 26 , 227–238. Thompson, M.; Ellison, S. L. R.; Wood, R. “Harmonized Guidelines for Single-Laboratory Validation of Analytical Methods,” Pure Appl. Chem. 2002 , 74 , 835–855. |
Courses/Brevard_College/CHE_104%3A_Principles_of_Chemistry_II/04%3A_Thermochemistry_and_Thermodynamics/4.03%3A_Exothermic_and_Endothermic_Processes | Units of Heat Heat flow is measured in one of two common units: the calorie and the joule. The joule \(\left( \text{J} \right)\) is the SI unit of energy. The calorie is familiar because it is commonly used when referring to the amount of energy contained within food. A calorie \(\left( \text{cal} \right)\) is the quantity of heat required to raise the temperature of 1 gram of water by \(1^\text{o} \text{C}\). For example, raising the temperature of \(100 \: \text{g}\) of water from \(20^\text{o} \text{C}\) to \(22^\text{o} \text{C}\) would require \(100 \times 2 = 200 \: \text{cal}\). Calories contained within food are actually kilocalories \(\left( \text{kcal} \right)\). In other words, if a certain snack contains 85 food calories, it actually contains \(85 \: \text{kcal}\) or \(85,000 \: \text{cal}\). In order to make the distinction, the dietary calorie is written with a capital C. \[1 \: \text{kilocalorie} = 1 \: \text{Calorie} = 1000 \: \text{calories}\] To say that the snack "contains" 85 Calories means that \(85 \: \text{kcal}\) of energy are released when that snack is processed by the human body. Heat changes in chemical reactions are typically measured in joules rather than calories. The conversion between a joule and a calorie is shown below. \[1 \: \text{J} = 0.2390 \: \text{cal or} \: 1 \: \text{cal} = 4.184 \: \text{J}\] We can calculate the amount of heat released in kilojoules when a 400 Calorie hamburger is digested: \[400. \: \text{Cal} = 400. \: \text{kcal} \times \frac{4.184 \: \text{kJ}}{1 \: \text{kcal}} = 1.67 \times 10^3 \: \text{kJ}\] Summary Common units for heat include Joules (J) and calories (cal); 1 cal = 1.184 J 1 food Calorie (Cal) = 1000 cal |
Courses/Oregon_Institute_of_Technology/OIT%3A_CHE_331_-_Organic_Chemistry_(Lund)/08%3A_Intro_to_Spectroscopy_-_UV-Vis%2C_IR%2C_MS/8.00%3A_Prelude_to_Structure_Determination_I | (Photo credit: https://www.flickr.com/photos/vamapaull/ ) Introduction: a foiled forgery I n the end, it was a 'funky yellow color' that led to the demise of Charles Heller's not-so-illustrious career in the world of collectable art. William Aiken Walker was a 19th-century 'genre' painter, known for his small scenes of sharecroppers working the fields in the post-Civil War south. For much of his career, he traveled extensively, throughout the southern states but also to New York City and even as far as Cuba. He earned a decent living wherever he went by setting up shop on the sidewalk and selling his paintings to tourists, usually for a few dollars each. While he never became a household name in the art world, he was prolific and popular, and his paintings are today considered collectible, often selling for upwards of ten thousand dollars. In August, 1994, Robert Hicklin, an art gallery owner in Charleston, South Carolina, was appraising a Walker painting brought to him by another South Carolina art dealer named Rick Simons. Hicklin's years of experience with Walker paintings told him that something just wasn't right with this one - he was particularly bothered by one of the pigments used, which he later described in a story in the Maine Antique Digest as a 'funky yellow color'. Reluctantly, he told Simons that it almost certainly was a fake. Hoping that Hicklin was wrong, Simons decided to submit his painting to other experts for analysis, and eventually it ended up in the laboratory of James Martin, whose company Orion Analytical specializes in forensic materials analysis. Using a technique called infrared spectroscopy, Martin was able to positively identify the suspicious yellow pigment as an organic compound called 'pigment yellow 3'. As it turns out, Pigment Yellow 3 had not become available in the United States until many years after William Aiken Walker died. Simons had purchased his painting from a man named Robert Heller for $9,500. When Heller approached him again to offer several more Walker paintings for sale, Simons contacted the FBI. A few days later, with FBI agents listening in, Simons agreed to buy two more Walker paintings. When he received them, they were promptly analyzed and found to be fake. Heller, who turned out to be a convicted felon, was arrested and eventually imprisoned. (For more details on this story, see Chemical and Engineering News, Sept 10, 2007, p. 28 ). ***** In the first three chapters of this text, we have focused our efforts on learning about the structure of organic compounds. Now that we know what organic molecules look like, we can begin to address, in the next two chapters, the question of how we get this knowledge in the first place. How are chemists able to draw with confidence the bonding arrangements in organic molecules, even simple ones such as acetone or ethanol? How was James Martin at Orion Analytical able to identify the chemical structure of the pigment compound responsible for the 'funky yellow color' in the forged William Aiken Walker painting? This chapter is devoted to three very important techniques used by chemists to learn about the structures of organic molecules. First, we will learn how mass spectrometry can provide us with information about the mass of a molecule as well as the mass of fragments into which the molecule has been broken. Then, we will begin our investigation of molecular spectroscopy, which is the study of how electromagnetic radiation at different wavelengths interacts in different ways with molecules - and how these interactions can be quantified, analyzed, and interpreted to gain information about molecular structure. After a brief overview of the properties of light and the elements of a molecular spectroscopy experiment, we will move to a discussion of infrared (IR) spectroscopy, the key technique used in the detection of the Walker forgery, and a way to learn about functional groups present in an organic compound. Then, we will consider ultraviolet-visible (UV-vis) spectroscopy, with which chemists gain information about conjugated pi -bonding systems in organic molecules. Among other applications, we will see how information from UV-vis spectroscopy can be used to measure the concentration of biomolecules compounds in solution. Looking ahead, Chapter 5 will be devoted to nuclear magnetic resonance (NMR) spectroscopy, where we use ultra-strong magnets and radio frequency radiation to learn about the electronic environment of individual atoms in a molecule and use this information to determine the atom-to-atom bonding arrangement. For most organic chemists, NMR is one of the most powerful analytical tools available in terms of the wealth of detailed information it can provide about the structure of a molecule. The structure determination techniques we will be studying in this chapter and the next primarily attempt to address the following questions about an organic molecule: Chapter 4: Mass spectrometry (MS): What is the atomic weight of the molecule and its common fragments? Infrared (IR) spectroscopy: what functional groups does the molecule contain? Ultraviolet-visible (UV-Vis) spectroscopy: What is the nature of conjugated pi-bonding systems in the molecule? Chapter 5: Nuclear magnetic resonance spectroscopy (NMR): What is the overall bonding framework of the molecule? |
Courses/Riverland_Community_College/CHEM_1121%3A_General_Organic_and_Biochemistry/08%3A_States_of_Matter_and_the_Gas_Laws/8.01%3A_Phase_Changes | Learning Objectives Determine the heat associated with a phase change. Matter can exist in one of several different states, including a gas, liquid, or solid state. The amount of energy in molecules of matter determines the state of matter . A gas is a state of matter in which atoms or molecules have enough energy to move freely. The molecules come into contact with one another only when they randomly collide. A liquid is a state of matter in which atoms or molecules are constantly in contact but have enough energy to keep changing positions relative to one another. A solid is a state of matter in which atoms or molecules do not have enough energy to move. They are constantly in contact and in fixed positions relative to one another. The following are the changes of state: Solid → Liquid Melting or fusion Liquid → Gas Vaporization Liquid → Solid Freezing Gas → Liquid Condensation Solid → Gas Sublimation If heat is added to a substance, such as in melting, vaporization, and sublimation, the process is endothermic . In this instance, heat is increasing the speed of the molecules causing them move faster (examples: solid to liquid; liquid to gas; solid to gas). If heat is removed from a substance, such as in freezing and condensation, then the process is exothermic . In this instance, heat is decreasing the speed of the molecules causing them move slower (examples: liquid to solid; gas to liquid). These changes release heat to the surroundings. The amount of heat needed to change a sample from solid to liquid would be the same to reverse from liquid to solid. The only difference is the direction of heat transfer. Example \(\PageIndex{1}\) Label each of the following processes as endothermic or exothermic. water boiling ice forming on a pond Solution endothermic - you must put a pan of water on the stove and give it heat in order to get water to boil. Because you are adding heat/energy, the reaction is endothermic. exothermic - think of ice forming in your freezer instead. You put water into the freezer, which takes heat out of the water, to get it to freeze. Because heat is being pulled out of the water, it is exothermic. Heat is leaving. Label each of the following processes as endothermic or exothermic. water vapor condensing gold melting Answer a. exothermic b. endothermic A phase change is a physical process in which a substance goes from one phase to another. Usually the change occurs when adding or removing heat at a particular temperature, known as the melting point or the boiling point of the substance. The melting point is the temperature at which the substance goes from a solid to a liquid (or from a liquid to a solid). The boiling point is the temperature at which a substance goes from a liquid to a gas (or from a gas to a liquid). The nature of the phase change depends on the direction of the heat transfer. Heat going into a substance changes it from a solid to a liquid or a liquid to a gas. Removing heat from a substance changes a gas to a liquid or a liquid to a solid. Two key points are worth emphasizing. First, at a substance’s melting point or boiling point, two phases can exist simultaneously. Take water (H 2 O) as an example. On the Celsius scale, H 2 O has a melting point of 0°C and a boiling point of 100°C. At 0°C, both the solid and liquid phases of H 2 O can coexist. However, if heat is added, some of the solid H 2 O will melt and turn into liquid H 2 O. If heat is removed, the opposite happens: some of the liquid H 2 O turns into solid H 2 O. A similar process can occur at 100°C: adding heat increases the amount of gaseous H 2 O, while removing heat increases the amount of liquid H 2 O (Figure \(\PageIndex{1}\)). Water is a good substance to use as an example because many people are already familiar with it. Other substances have melting points and boiling points as well. Second, as shown in Figure \(\PageIndex{1}\), the temperature of a substance does not change as the substance goes from one phase to another . In other words, phase changes are isothermal (isothermal means “constant temperature”). Again, consider H 2 O as an example. Solid water (ice) can exist at 0°C. If heat is added to ice at 0°C, some of the solid changes phase to make liquid, which is also at 0°C. Remember, the solid and liquid phases of H 2 O can coexist at 0°C. Only after all of the solid has melted into liquid does the addition of heat change the temperature of the substance. For each phase change of a substance, there is a characteristic quantity of heat needed to perform the phase change per gram (or per mole) of material. The heat of fusion (Δ H fus ) is the amount of heat per gram (or per mole) required for a phase change that occurs at the melting point. The heat of vaporization (Δ H vap ) is the amount of heat per gram (or per mole) required for a phase change that occurs at the boiling point. If you know the total number of grams or moles of material, you can use the Δ H fus or the Δ H vap to determine the total heat being transferred for melting or solidification using these expressions: \[\text{heat} = n \times ΔH_{fus} \label{Eq1a} \] wher e \(n\) is th e number of moles and \(ΔH_{fus}\) is expressed in energy/mole or \[\text{heat} = m \times ΔH_{fus} \label{Eq1b} \] where \(m\) is the mass in grams and \(ΔH_{fus}\) is expressed in energy/gram. For the boiling or condensation, use these expressions: \[\text{heat} = n \times ΔH_{vap} \label{Eq2a} \] wher e \(n\) is the number of moles) and \(ΔH_{vap}\) is expressed in energy/mole or \[\text{heat} = m \times ΔH_{vap} \label{Eq2b} \] wh ere \(m\) i s the mass in grams and \(ΔH_{vap}\) is expressed in energy/gram. Remember that a phase change depends on the direction of the heat transfer. If heat transfers in, solids become liquids, and liquids become solids at the melting and boiling points, respectively. If heat transfers out, liquids solidify, and gases condense into liquids. At these points, there are no changes in temperature as reflected in the above equations. Example \(\PageIndex{2}\) How much heat is necessary to melt 55.8 g of ice (solid H 2 O) at 0°C? The heat of fusion of H 2 O is 79.9 cal/g. Solution We can use the relationship between heat and the heat of fusion (Equation \(\PageIndex{1}\)) to determine how many cal of heat are needed to melt this ice: \[ \begin{align*} \ce{heat} &= \ce{m \times ΔH_{fus}} \\[4pt] \mathrm{heat} &= \mathrm{(55.8\: \cancel{g})\left(\dfrac{79.9\: cal}{\cancel{g}}\right)=4,460\: cal} \end{align*} \nonumber \] Exercise \(\PageIndex{2}\) How much heat is necessary to vaporize 685 g of H 2 O at 100°C? The heat of vaporization of H 2 O is 540 cal/g. Answer \[ \begin{align*} \ce{heat} &= \ce{m \times ΔH_{vap}} \\[4pt] \mathrm{heat} &= \mathrm{(685\: \cancel{g})\left(\dfrac{540\: cal}{\cancel{g}}\right)=370,000\: cal} \end{align*} \nonumber \] Table \(\PageIndex{1}\) lists the heats of fusion and vaporization for some common substances. Note the units on these quantities; when you use these values in problem solving, make sure that the other variables in your calculation are expressed in units consistent with the units in the specific heats or the heats of fusion and vaporization. Substance ΔHfus (cal/g) ΔHvap (cal/g) aluminum (Al) 94.0 2602.0 gold (Au) 15.3 409.0 iron (Fe) 63.2 1504.0 water (H2O) 79.9 540.0 sodium chloride (NaCl) 123.5 691.0 ethanol (C2H5OH) 45.2 200.3 benzene (C6H6) 30.4 94.1 Sublimation There is also a phase change where a solid goes directly to a gas: \[\text{solid} \rightarrow \text{gas} \label{Eq3} \] This phase change is called sublimation . Each substance has a characteristic heat of sublimation associated with this process. For example, the heat of sublimation (Δ H sub ) of H 2 O is 620 cal/g. We encounter sublimation in several ways. You may already be familiar with dry ice, which is simply solid carbon dioxide (CO 2 ). At −78.5°C (−109°F), solid carbon dioxide sublimes, changing directly from the solid phase to the gas phase: \[\mathrm{CO_2(s) \xrightarrow{-78.5^\circ C} CO_2(g)} \label{Eq4} \] Solid carbon dioxide is called dry ice because it does not pass through the liquid phase. Instead, it does directly to the gas phase. (Carbon dioxide can exist as liquid but only under high pressure.) Dry ice has many practical uses, including the long-term preservation of medical samples. Even at temperatures below 0°C, solid H 2 O will slowly sublime. For example, a thin layer of snow or frost on the ground may slowly disappear as the solid H 2 O sublimes, even though the outside temperature may be below the freezing point of water. Similarly, ice cubes in a freezer may get smaller over time. Although frozen, the solid water slowly sublimes, redepositing on the colder cooling elements of the freezer, which necessitates periodic defrosting (frost-free freezers minimize this redeposition). Lowering the temperature in a freezer will reduce the need to defrost as often. Under similar circumstances, water will also sublime from frozen foods (e.g., meats or vegetables), giving them an unattractive, mottled appearance called freezer burn. It is not really a “burn,” and the food has not necessarily gone bad, although it looks unappetizing. Freezer burn can be minimized by lowering a freezer’s temperature and by wrapping foods tightly so water does not have any space to sublime into. Key Takeaway There is an energy change associated with any phase change. |
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.41%3A_Entanglement_Reveals_a_Conflict_Between_Local_Realism_and_Quantum_Theory | A tensor algebra approach is used to demonstrate the challenge to the local realistic position of reality that quantum mechanical entanglement creates. The example is drawn from Chapter 3 of David Z Albert's text, Quantum Mechanics and Experience . A quon (any entity that exhibits both wave and particle aspects in the peculiar quantum manner - Nick Herbert, Quantum Reality , page 64) has a variety of properties each of which can take on two values. For example, it has the property of hardness and can be either hard or soft . It also has the property of color and can be either black or white . In the matrix formulation of quantum mechanics these states are represented by the following vectors. \[ \begin{matrix} \text{Hard} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} & \text{Soft} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} & \text{Black} = \begin{pmatrix} \frac{1}{ \sqrt{2}} \\ \frac{1}{ \sqrt{2}} \end{pmatrix} & \text{White} = \begin{pmatrix} \frac{1}{ \sqrt{2}} \\ \frac{-1}{ \sqrt{2}} \end{pmatrix} \end{matrix} \nonumber \] Hard and Soft represent an orthonormal basis in the two-dimensional Hardness vector space. \[ \begin{matrix} \text{Hard}^T \text{Hard} = 1 & \text{Soft}^T \text{Soft} = 1 & \text{Hard}^T \text{Hard} = 0 \end{matrix} \nonumber \] Likewise Black and White are an orthonormal in the two-dimensional Color vector space. \[ \begin{matrix} \text{Black}^T \text{Black} = 1 & \text{White}^T \text{White} = 1 & \text{Black}^T \text{White} = 0 \end{matrix} \nonumber \] The relationship between the two bases is reflected in the following projection calculations. \[ \begin{matrix} \text{Hard}^T \text{Black} = 0.707 & \text{Hard}^T \text{White} = 0.707 & \text{Soft}^T \text{Black} = 0.707 & \text{Soft}^T \text{White} = -0.707 & \frac{1}{ \sqrt{2}} = 0.707 \end{matrix} \nonumber \] Clearly Black and White can be written as superpositions of Hard and Soft, and vice versa. \[ \begin{matrix} \frac{1}{ \sqrt{2}} \text{(Hard + Soft)} = \begin{pmatrix} 0.707 \\ 0.707 \end{pmatrix} & \frac{1}{ \sqrt{2}} \text{(Hard - Soft)} = \begin{pmatrix} 0.707 \\ -0.707 \end{pmatrix} \\ \frac{1}{ \sqrt{2}} \text{(Black + White)} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} & \frac{1}{ \sqrt{2}} \text{(Black - White)} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \end{matrix} \nonumber \] Hard, Soft, Black and White are measurable properties and the vectors representing them are eigenstates of the Hardness and Color operators with eigenvalues +/- 1. Operators \[ \begin{matrix} \text{Hardness} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} & \text{Color} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} & I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \end{matrix} \nonumber \] \[ \begin{matrix} \text{Eigenvalue +1} & \text{Eigenvalue -1} \\ \text{Hardness Hard} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} & \text{Hardness Soft} = \begin{pmatrix} 0 \\ -1 \end{pmatrix} \\ \text{Color Black} = \begin{pmatrix} 0.707 \\ 0.707 \end{pmatrix} & \text{Color White} = \begin{pmatrix} -0.707 \\ 0.707 \end{pmatrix} \end{matrix} \nonumber \] Hard and Soft are not eigenfunctions of the Color operator, and Black and White are not eigenfunctions of the Hardness operator . \[ \begin{matrix} \text{Hardness Black} = \begin{pmatrix} 0.707 \\ -0.707 \end{pmatrix} & \text{Hardness White} = \begin{pmatrix} 0.707 \\ 0.707 \end{pmatrix} \\ \text{Color Hard} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} & \text{Color Soft} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \end{matrix} \nonumber \] As the Hardness-Color commutator shows, the Hardness and Color operators do not commute. They represent incompatible observables; observables that cannot simultaneously have well-defined values. \[ \text{Hardness Color} - \text{Color Hardness} = \begin{pmatrix} 0 & 2 \\ -2 & 0 \end{pmatrix} \nonumber \] We now proceed with an analysis of the implications of the following two-quon entangled state, expressed in tensor format. A pair of quons is prepared in the following "singlet" state; one is hard and one is soft. (The Appendix shows how to set this state up using Mathcad.) \[ | \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ | \text{Hard} \rangle_1 | \text{Soft} \rangle_2 - | \text{Soft} \rangle_1 | \text{Hard} \rangle_2 \right] = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \\ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \\ 1 \end{pmatrix} - \begin{pmatrix} 0 \\ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \\ 0 \end{pmatrix} \right] \nonumber \] \[ \Psi = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \\ 1 \\ -1 \\ 0 \end{pmatrix} \nonumber \] Given |Ψ> the expectation value for measuring Hardness on the first quon is 0. The same is true for the second quon. In other words, it is equally likely for either quon to be Hard or Soft . ( Kronecker is Mathcad's command for tensor multiplication of matrices. See the Appendix for more detail.) \[ \begin{matrix} \Psi^T \text{kronecker(Hardness, I)} \Psi = 0 & \Psi^T \text{kronecker(I, Hardness)} \Psi = 0 \end{matrix} \nonumber \] However, if one quon is found to be Hard by measurement, the second will be measured Soft , and vice versa. In other words, there is perfect anti-correlation between the joint measurement of this property on the two quons. \[ \Psi^T \text{kronecker(Hardness, Hardness)} \Psi = -1 \nonumber \] Given |Ψ> the expectation value for measuring Color on the first quon is 0. The same is true for the second quon. In other words, it is equally likely for either quon to be Black or White . \[ \begin{matrix} \Psi^T \text{kronecker(Color, I)} \Psi = 0 & \Psi^T \text{kronecker(I, Color)} \Psi = 0 \end{matrix} \nonumber \] However, if one quon is found to be Black by measurement, the second will be measured White and vice versa. In other words, there is perfect anti-correlation between the joint measurement of this property on the two quons. \[ \Psi^T \text{kronecker(Color, Color)} \Psi = -1 \nonumber \] Furthermore, as the following calculations show, there is no correlation between the measurement outcomes on Color and Hardness . \[ \begin{matrix} \Psi^T \text{kronecker(Hardness, Color)} \Psi = 0 & \Psi^T \text{kronecker(Color, Hardness)} \Psi = 0 \end{matrix} \nonumber \] As the foundation for their belief in local realism, Einstein, Podolsky and Rosen (EPR) defined the concept of element of reality in their famous 1935 Physical Review paper, "If, without in any way disturbing a system, we can predict with certainty (i.e. with probability equal to unity) the value of a physical quantity, then there exists an element of reality corresponding to this physical quantity." It would seem from the above results, namely these, \[ \begin{matrix} \Psi^T \text{kronecker(Color, Color)} \Psi = -1 & \Psi^T \text{kronecker(Hardness, Hardness)} \Psi = -1 \end{matrix} \nonumber \] that according to EPR both hardness and color are elements of reality . If the hardness of quon 1 is measured and found to be soft, we know without measurement (given the reliability of quantum mechanical predictions) that quon 2 is hard. Likewise, if the color of quon 2 is measured and found to be white, we know without measurement that quon 1 is black. On the basis of these calculations, the realist constructs the following table which assigns well-defined hardness and color states to both quons and is consistent with all the quantum calculations. \[ \begin{pmatrix} \text{Quon 1} & \text{Quon 2} & \text{HardnessHardness} & \text{ColorColor} & \text{HardnessColor} \\ \text{HB} & \text{SW} & -1 & -1 & -1 \\ \text{HW} & \text{SB} & -1 & -1 & 1 \\ \text{SB} & \text{HW} & -1 & -1 & 1 \\ \text{SW} & \text{HB} & -1 & -1 & -1 \\ \text{Realist} & \text{AverageValue} & -1 & -1 &0 \\ \text{Quantum} & \text{AverageValue} & -1 & -1 & 0 \end{pmatrix} \nonumber \] The problem with this interpretation is that it has previously been shown that the Hardness and Color operators do not commute, meaning that they represent incompatible observables. Incompatible observables cannot be known (determined) simultaneously. A contradiction between the EPR reality criterion and quantum mechanics has thus been shown to exist. Appendix Tensor multiplication is used to construct the initial states using Mathcad commands submatrix , kronecker , and augment . \[ \Psi = \frac{1}{ \sqrt{2}} \begin{bmatrix} \text{submatrix} \left[ \text{kronecker} \left[ \text{augment} \left[ \begin{pmatrix} 1 \\ 0 \end{pmatrix},~ \begin{pmatrix} 0 \\ 0 \end{pmatrix} \right],~ \text{augment} \left[ \begin{pmatrix} 0 \\ 1 \end{pmatrix},~ \begin{pmatrix} 0 \\ 0 \end{pmatrix} \right] \right], ~1,~4,~1,~1 \right] ~... \\ +- \text{submatrix} \left[ \text{kronecker} \left[ \text{augment} \left[ \begin{pmatrix} 0 \\ 1 \end{pmatrix},~ \begin{pmatrix} 0 \\ 0 \end{pmatrix} \right],~ \text{augment} \left[ \begin{pmatrix} 1 \\ 0 \end{pmatrix},~ \begin{pmatrix} 0 \\ 0 \end{pmatrix} \right] \right], ~1,~4,~1,~1 \right] \end{bmatrix} \nonumber \] Kronecker is the Mathcad command that carries out the tensor multiplication of matrices. For example, consider the tensor multiplication of the Hardness and Color matrix operators. \[ Hardness \otimes Color = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \otimes \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} & 0 \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \\ 0 & \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} & -1 \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & -1 & 0 \end{pmatrix} \nonumber \] \[ \begin{matrix} \text{kronecker(Hardness, Color)} = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & -1 & 0 \end{pmatrix} & \text{kronecker} \begin{bmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix},~ \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \end{bmatrix} = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & -1 & 0 \end{pmatrix} \end{matrix} \nonumber \] |
Courses/University_of_Illinois_Springfield/CHE_124%3A_General_Chemistry_for_the_Health_Professions_(Morsch_and_Andrews)/05%3A_Introduction_to_Chemical_Reactions/5.1%3A_The_Law_of_Conservation_of_Matter | Skills to Develop Correctly define a law as it pertains to science. State the law of conservation of matter. In science, a law is a general statement that explains a large number of observations. Before being accepted, a law must be verified many times under many conditions. Laws are therefore considered the highest form of scientific knowledge and are generally thought to be inviolable. Scientific laws form the core of scientific knowledge. One scientific law that provides the foundation for understanding in chemistry is the law of conservation of matter. It states that in any given system that is closed to the transfer of matter (in and out), the amount of matter in the system stays constant. A concise way of expressing this law is to say that the amount of matter in a system is conserved . With the development of more precise ideas on elements, compounds and mixtures, scientists began to investigate how and why substances react. French chemist A. Lavoisier laid the foundation to the scientific investigation of matter by describing that substances react by following certain laws. These laws are called the laws of chemical combination. These eventually formed the basis of Dalton's Atomic Theory of Matter. Law of Conservation of Mass According to this law, during any physical or chemical change, the total mass of the products remains equal to the total mass of the reactants. \[ \overbrace{\underbrace{\ce{HgO (s)}}_{\text{100 g}}}^{\text{Mercuric oxide}} \rightarrow \underbrace{\overbrace{\ce{Hg (l) }}^{\text{Mercury}}}_{\text{92.6 g}} + \underbrace{\overbrace{\ce{O2 (g)}}^{\text{Oxygen}}}_{\text{7.4 g}} \] The law of conservation of mass is also known as the "law of indestructibility of matter." Example \(\PageIndex{1}\) If heating 10 grams of \(\ce{CaCO3}\) produces 4.4 g of \(\ce{CO2}\) and 5.6 g of \(\ce{CaO}\), show that these observations are in agreement with the law of conservation of mass. A sample of calcium carbonate (CaCO3). Image used with permission (Public Domain; Walkerma ). SOLUTION Mass of the reactants: \(10 \,g\) Mass of the products: \(4.4 \,g+ 5.6\, g = 10\, g\). Because the mass of the reactants is equal to the mass of the products, the observations are in agreement with the law of conservation of mass. That does this mean for chemistry? In any chemical change, one or more initial substances change into a different substance or substances. Both the initial and final substances are composed of atoms because all matter is composed of atoms. According to the law of conservation of matter, matter is neither created nor destroyed, so we must have the same number and type of atoms after the chemical change as were present before the chemical change. Before looking at explicit examples of the law of conservation of matter, we need to examine the method chemists use to represent chemical changes. Exercise \(\PageIndex{1}\) What is the law of conservation of matter? How does the law of conservation of matter apply to chemistry? Answer a The law of conservation of matter states that in any given system that is closed to the transfer of matter, the amount of matter in the system stays constant Answer b The law of conservation of matter says that in chemical reactions, the total mass of the products must equal the total mass of the reactants. Summary The amount of matter in a closed system is conserved. Contributors Binod Shrestha (University of Lorraine) Exercises Express the law of conservation of matter in your own words. Explain why the concept of conservation of matter is considered a scientific law. Answer Matter may not be created or destroyed. |
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/The_Live_Textbook_of_Physical_Chemistry_(Peverati)/18%3A_The_Schrodinger_Equation/18.01%3A_The_Time-Independent_Schrodinger_Equation | We can start the derivation of the single-particle time-independent Schrödinger equation (TISEq) from the equation that describes the motion of a wave in classical mechanics: \[ \psi(x,t)=\exp[i(kx-\omega t)], \label{19.1.1} \] where \(x\) is the position, \(t\) is time, \(k=\dfrac{2\pi}{\lambda}\) is the wave vector, and \(\omega=2\pi\nu\) is the angular frequency of the wave. If we are not concerned with the time evolution, we can consider uniquely the derivatives of Equation \ref{19.1.1} with respect to the location, which are: \[ \begin{aligned} \dfrac{\partial \psi}{\partial x} &=ik\exp[i(kx-\omega t)] = ik\psi, \\ \dfrac{\partial^2 \psi}{\partial x^2} &=i^2k^2\exp[i(kx-\omega t)] = -k^2\psi, \end{aligned} \label{19.1.2} \] where we have used the fact that \(i^2=-1\). Assuming that particles behaves as wave—as proven by de Broglie’s we can now use the first of de Broglie’s equation, Equation 17.5.4 , we can replace \(k=\dfrac{p}{\hbar}\) to obtain: \[ \dfrac{\partial^2 \psi}{\partial x^2} = -\dfrac{p^2\psi}{\hbar^2}, \label{19.1.3} \] which can be rearranged to: \[ p^2 \psi = -\hbar^2 \dfrac{\partial^2 \psi}{\partial x^2}. \label{19.1.4} \] The total energy associated with a wave moving in space is simply the sum of its kinetic and potential energies: \[ E = \dfrac {p^{2}}{2m} + V(x), \label{19.1.5} \] from which we can obtain: \[ p^2 = 2m[E - V(x)], \label{19.1.6} \] which we can then replace into Equation \ref{19.1.4} to obtain: \[ 2m[E-V(x)]\psi = - \hbar^2 \dfrac{\partial^2 \psi}{\partial x^2}, \label{19.1.7} \] which can then be rearranged to the famous time-independent Schrödinger equation (TISEq) : \[ - \dfrac{\hbar^2}{2m} \dfrac{\partial^2 \psi}{\partial x^2} + V(x) \psi = E\psi, \label{19.1.8} \] A two-body problem can also be treated by this equation if the mass \(m\) is replaced with a reduced mass \(\mu = \dfrac{m_1 m_2}{m_1+m_2}\). |
Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/2_p-Block_Elements/Group_17%3A_The_Halogens/0Group_17%3A_Physical_Properties_of_the_Halogens | Some chemical and physical properties of the halogens are summarized in Table \(\PageIndex{1}\). It can be seen that there is a regular increase in many of the properties of the halogens proceeding down group 17 from fluorine to iodine. This includes their melting points, boiling points, the intensity of their color, the radius of the corresponding halide ion, and the density of the element. On the other hand, there is a regular decrease in the first ionization energy as we go down this group. As a result, there is a regular increase in the ability to form high oxidation states and a decrease in the oxidizing strength of the halogens from fluorine to iodine. Property F Cl Br I Atomic number, Z 9 17 35 53 Ground state electronic configuration [He]2s2 2p5 [Ne]3s2 3p5 [Ar]3d10 4s2 4p5 [Kr]4d10 5s2 5p5 color pale yellow gas yellow-green gas red-brown liquid blue-black solid Density of liquids at various temperatures, /kg m-3 1.51 (85 °K) 1.66 (203 °K) 3.19 (273 °K) 3.96 (393 °K) Melting point, /K 53.53 171.6 265.8 386.85 Boiling point, /K 85.01 239.18 331.93 457.5 Enthalpy of atomization, ΔaH° (298K) / kJ mol-1 79.08 121.8 111.7 106.7 Standard enthalpy of fusion of X2, ΔfusH°(mp) / kJ mol-1 0.51 6.4 10.57 15.52 Standard enthalpy of vaporization of X2, ΔvapH°(bp) / kJ mol-1 6.62 20.41 29.96 41.57 First ionization energy, IE1 / kJ mol-1 1681 1251.1 1139.9 1008.4 ΔEAH1°(298K) / kJ mol-1 -333 -348 -324 -295 ΔhydH°(X-,g) / kJ mol-1 -504 -361 -330 -285 ΔhydS°(X-,g) / JK-1 mol-1 -150 -90 -70 -50 ΔhydG°(X-,g) / kJ mol-1 -459 -334 -309 -270 Standard redox potential, E°(X2 /2X-) /V 2.87 1.36 1.09 0.54 Covalent radius, rcov = ½ X-X bond length /pm 72 100 114.2 133.3 Ionic radius, rion for X- /pm 133 181 196 220 van der Waals radius, rv /pm 135 180 195 215 X-X(g)bond energy /kJ mol-1 159 243 193 151 H-X(g)bond energy /kJ mol-1 562 431 366 299 C-X(g)bond energy /kJ mol-1 484 338 276 238 Pauling electronegativity, χP 3.98 3.16 2.96 2.66 Color The origin of the color of the halogens stems from the excitation between the highest occupied π* molecular orbital and the lowest unoccupied σ* molecular orbital. The energy gap between the HOMO and LUMO decreases according to F 2 > Cl 2 > Br 2 > I 2 . The amount of energy required for excitation depends upon the size of the atom. Fluorine is the smallest element in the group and the force of attraction between the nucleus and the outer electrons is very large. As a result, it requires a large excitation energy and absorbs violet light (high energy) and so appears pale yellow. On the other hand, iodine needs significantly less excitation energy and absorbs yellow light of low energy. Thus it appears dark violet. Using similar arguments, it is possible to explain the greenish yellow color of chlorine and the reddish brown color of bromine. Figure \(\PageIndex{1}\): Molecular orbital diagram for fluorine. The halogens show a variety of colors when dissolved in different solvents. Solutions of iodine can be bright violet in CCl 4 , pink or reddish brown in aromatic hydrocarbons, and deep brown in alcohols, for example. This variety can be explained by weak donor-acceptor interaction and complex formation. The presence of charge-transfer bands further supports this since they are thought to be derived from interaction with the HOMO σ u * orbital. The X-ray structure of some of these complexes have been obtained, and often the intense color can be used for characterization and determination such as the bright blue color of iodine in the presence of starch. In the case of the solid formed between dibromine and benzene, the structure is shown below and a new charge transfer band occurs at 292 nm. The Br-Br bond length is essentially unchanged from that of dibromine (228 pm). Figure \(\PageIndex{2}\): Structure of dibromine and benzene complex In a study of the reaction of dibromine with substituted phosphines in diethyl ether, all but one showed a tetrahedral arrangement where one bromine was linked to the phosphorus.[3] \[R_3P + Br_2 (Et_2O, N_2/r.t.) \rightarrow R_3PBr_2 \label{1} \] The X-ray study of the triethylphosphine was interpreted as [Et 3 PBr]Br, where the Br-Br separation was 330 pm. This is considerably longer than the 228 pm found above and was taken to mean that the compound was ionic. In the case of the tri(perfluorophenyl)phosphine, however, the structure showed both bromines linked to give a trigonal bipyramid arrangement with D 3 symmetry. Why (C 6 F 5 ) 3 PBr 2 was the only R 3 PBr 2 compound that adopted trigonal bipyramidal geometry was reasoned to be due to the very low basicity of the parent tertiary phosphine. Melting and Boiling Points Intermolecular forces are the attractive forces between molecules without which all substances would be gases. The various types of these interactions span large differences in energy and for the halogens and interhalogens are generally quite small. The dispersion forces involved in these cases are called London forces (after Fritz Wolfgang London, 1900-1954). They are derived from momentary oscillations of electron charge in atoms and hence are present between all particles (atoms, ions and molecules). The ease with which the electron cloud of an atom can be distorted to become asymmetric is termed the molecule's polarizability . The greater the number of electrons an atom has, the farther the outer electrons will be from the nucleus, and the greater the chance for them to shift positions within the molecule. This means that larger nonpolar molecules tend to have stronger London dispersion forces. This is evident when considering the diatomic elements in group 17, the Halogens. All of these diatomic elements are nonpolar, covalently bonded molecules. Descending the group, fluorine and chlorine are gases, bromine is a liquid, and iodine is a solid. For nonpolar molecules, the farther you go down the group, the stronger the London dispersion forces. To picture how this occurs, compare the situation 1) where the electrons are evenly distributed and then consider 2) an instantaneous dipole that would arise from an uneven distribution of electrons on one side of the nucleus. When two molecules are close together, the instantaneous dipole of one molecule can induce a dipole in the second molecule. This results in synchronized motion of the electrons and an attraction between them. 3) Multiply this effect over numerous molecules and the overall result is that the attraction keeps these molecules together, and for diiodine is sufficient to make this a solid. Figure \(\PageIndex{3}\): On average the electron cloud for molecules can be considered to be spherical in shape. When two non-polar molecules approach, attractions or repulsions between the electrons and nuclei can lead to distortions in their electron clouds (i.e., dipoles are induced). When more molecules interact, these induced dipoles lead to intermolecular attraction. The changes seen in the variation of MP and BP for the dihalogens and binary interhalogens can be attributed to the increase in the London dispersion forces of attraction between the molecules. In general they increase with increasing atomic number. Figure 4: Redox Properties The most characteristic chemical feature of the halogens is their oxidizing strength. Fluorine has the strongest oxidizing ability, so that a simple chemical preparation is almost impossible and it must be prepared by electrolysis. Note that since fluorine reacts explosively with water, oxidizing it to dioxygen, finding reaction conditions for any reaction can be difficult. When fluorine is combined with other elements they generally exhibit high oxidation states. Chlorine is the next strongest oxidizing agent, but it can be prepared by chemical oxidation. Most elements react directly with chlorine, bromine and iodine, with decreasing reactivity going down the group, but often the reaction must be activated by heat or UV light. [2] The energy changes in redox process are: Enthalpy of atomization, Δ EA H 1 , Δ hyd H°(X - ,g) The redox potential, E°, X 2 /2X - , measures a free-energy change, usually dominated by the ΔH term. The values in the Table below show that there is a decrease in oxidizing strength proceeding down the group (2.87, 1.36, 1.09, 0.54 V). This can be explained by comparing the steps shown above. 1) atomization of the dihalide is the energy required to break the molecule into atoms: \[½ X_{2(g)} \rightarrow X_{(g)} \label{2} \] Note that only F 2 and Cl 2 are gases in their natural states, so the energies associated with atomization of Br 2 and I 2 require converting the liquid or solid to gas first. 2) Δ EA H 1 is the energy liberated when the atom is converted into a negative ion and is related to the Electron Affinity \[X_{(g)} + e^- \rightarrow X^-_{(g)} \label{3} \] Addition of an electron to the small F atom is accompanied by larger e - /e - repulsion than is found for the larger Cl, Br or I atoms. This would suggest that the process for F should be less exothermic than for Cl and not fit the trend that shows a general decrease going down the group. 3) Δ hyd H°(X - ,g) is the energy liberated upon the hydration of the ion, the Hydration energy. \[X^-_{(g)} + H_2O \rightarrow X^-_{(aq)} \label{4} \] The overall reaction is then: \[½ X_{2(g)} \rightarrow X^-_{(aq)} \label{5} \] Table \(\PageIndex{2}\) Halogen atomization energy (kJ mol-1) ΔEAH1 (kJ mol-1) hydration enthalpy (kJ mol-1) overall (kJ mol-1) F 79.08 -333 -504 -758 Cl 121.80 -348 -361 -587 Br 111.70 -324 -330 -542 I 106.70 -295 -285 -473 This shows a very negative energy change for the fluoride compared to the others in the group. This comes about because of two main factors: the high hydration energy and the low atomization energy. For F 2 2) is less than for Cl 2 , but since the energy needed to break the F-F bond is also less and the hydration more, the total energy drop is much greater. In spite of their lower atomization energies, Br 2 and I 2 are weaker oxidizing agents than Cl 2 and this is due to their smaller Δ EA H 1 and smaller Δ hyd H°. It can be seen that the Δ EA H 1 value for fluorine is in between those for chlorine and bromine and so this value alone does not provide a good explanation for the observed variation. Each of the halogens is able to oxidize any of the heavier halogens situated below it in the group. They can oxidize hydrogen and nonmetals such as: \[X_2 + H_{2(g)} \rightarrow 2HX_{(g)} \label{6} \] In water, the halogens disproportionate according to: \[X_2 + H_2O_{(l)} \rightarrow HX_{(aq)} + HXO_{(aq)} \label{7} \] where \(X=Cl, Br, I\). When base is added then the reaction goes to completion forming hypohalites, or at higher temperatures, halates; for example, heating dichlorine: \[3Cl_{2(g)} + 6OH^-_{(aq)} \rightarrow ClO^-_{3(aq)} + 5Cl^-_{(aq)} + 3H_2O(l) \label{8} \] First Ionization Energies The trend seen for the complete removal of an electron from the gaseous halogen atoms is that fluorine has the highest IE 1 and iodine the lowest. To overcome the attractive force of the nucleus means that energy is required, so the Ionization Energies are all positive. The variation with size can be explained, since as the size increases it take less energy to remove an electron. This inverse relationship is seen for all the groups, not just group 17. As the distance from the nucleus to the outermost electrons increases, the attraction decreases so that those electrons are easier to remove. The high value of IE 1 for fluorine is such that it does not exhibit any positive oxidation states, whereas Cl, Br and I can exist in oxidation states as high as 7. Oxidation states Fluorine is the most electronegative element in the periodic table and exists in all its compounds in either the -1 or 0 oxidation state. Chlorine, bromine, and iodine, however, can be found in a range of oxidation states including: +1, +3, +5, and +7, as shown below. Oxidation States Examples -1 CaF2, HCl, NaBr, AgI 0 F2, Cl2, Br2, I2 1 HClO, ClF 3 HClO2, ClF3 5 HClO3, BrF5, [BrF6]-, IF5 7 HClO4, BrF6+, IF7, [IF8]- In general, odd-numbered groups (like group 17) form odd-numbered oxidation states, and this can be explained since all stable molecules contain paired electrons (free radicals are obviously much more reactive). When covalent bonds are formed or broken, two electrons are involved, so the oxidation state changes by 2. When difluorine reacts with diiodine, initially iodine monofluoride is formed. \[I_2 + F_2 \rightarrow 2IF \nonumber \] Adding a second difluorine uses two more iodine valence electrons to form two more bonds: \[2IF + F_2 \rightarrow IF_3 \nonumber \] References 1. "Inorganic Chemistry" - C. Housecroft and A.G. Sharpe, Prentice Hall, 3rd Ed., Dec 2007, ISBN13: 978-0131755536, ISBN10: 0131755536, Chapter 17. 2. "Chemistry. The Molecular Nature of Matter and Change" - M.S. Silberberg, McGraw Hill Higher Education, 4th Ed., 2006, ISBN13: 978-0072558203, Chapters 8, 12 and 14. 3. Stephen M. Godfrey, Charles A. McAuliffe, Imran Mushtaq, Robin G. Pritchard and Joanne M. Sheffield, J. Chem. Soc., Dalton Trans., 1998, 3815-3818 |
Courses/Modesto_Junior_College/Chemistry_142%3A_Pre-General_Chemistry_(Maki)/01%3A_Text_(Maki)/1.06%3A_Chemical_Compounds | 1.6.1: Ions Ions can be positively charged or negatively charged. A Lewis diagram is used to show how electrons are transferred to make ions and ionic compounds. 1.6.2: Formulas for Ionic Compounds Proper chemical formulas for ionic compounds balance the total positive charge with the total negative charge. Groups of atoms with an overall charge, called polyatomic ions, also exist. 1.6.3: Ionic Nomenclature Each ionic compound has its own unique name that comes from the names of the ions. After learning a few more details about the names of individual ions, you will be a step away from knowing how to name ionic compounds. This section begins the formal study of nomenclature, the systematic naming of chemical compounds. 1.6.3.1: Practice Naming 1.6.4: Covalent Compounds - Formulas and Names The chemical formula of a simple covalent compound can be determined from its name. The name of a simple covalent compound can be determined from its chemical formula. 1.6.5: Acids An acid is a compound of the H+ ion dissolved in water. Acids have their own naming system. Acids have certain chemical properties that distinguish them from other compounds. 1.6.5.1: More Practice Naming 1.6.6: Lewis Structures of Ionic Compounds- Electrons Transferred The tendency to form species that have eight electrons in the valence shell is called the octet rule. The attraction of oppositely charged ions caused by electron transfer is called an ionic bond. The strength of ionic bonding depends on the magnitude of the charges and the sizes of the ions. 1.6.6.1: Covalent Lewis Structures- Electrons Shared 1.6.6.2: Writing Lewis Structures for Covalent Compounds 1.6.6.3: Practice Lewis Structures 1.6.7: Predicting the Shapes of Molecules The approximate shape of a molecule can be predicted from the number of electron groups and the number of surrounding atoms. 1.6.7.1: Practice Shape 1.6.8: Electronegativity and Polarity - Why Oil and Water Don’t Mix Covalent bonds can be nonpolar or polar, depending on the electronegativities of the atoms involved. Covalent bonds can be broken if energy is added to a molecule. The formation of covalent bonds is accompanied by energy given off. Covalent bond energies can be used to estimate the enthalpy changes of chemical reactions. 1.6.8.1: Practice Polarity |
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/DeVoes_Thermodynamics_and_Chemistry/14%3A_Galvanic_Cells/14.03%3A_Molar_Reaction_Quantities_of_the_Cell_Reaction | \( \newcommand{\tx}[1]{\text{#1}} % text in math mode\) \( \newcommand{\subs}[1]{_{\text{#1}}} % subscript text\) \( \newcommand{\sups}[1]{^{\text{#1}}} % superscript text\) \( \newcommand{\st}{^\circ} % standard state symbol\) \( \newcommand{\id}{^{\text{id}}} % ideal\) \( \newcommand{\rf}{^{\text{ref}}} % reference state\) \( \newcommand{\units}[1]{\mbox{$\thinspace$#1}}\) \( \newcommand{\K}{\units{K}} % kelvins\) \( \newcommand{\degC}{^\circ\text{C}} % degrees Celsius\) \( \newcommand{\br}{\units{bar}} % bar (\bar is already defined)\) \( \newcommand{\Pa}{\units{Pa}}\) \( \newcommand{\mol}{\units{mol}} % mole\) \( \newcommand{\V}{\units{V}} % volts\) \( \newcommand{\timesten}[1]{\mbox{$\,\times\,10^{#1}$}}\) \( \newcommand{\per}{^{-1}} % minus one power\) \( \newcommand{\m}{_{\text{m}}} % subscript m for molar quantity\) \( \newcommand{\CVm}{C_{V,\text{m}}} % molar heat capacity at const.V\) \( \newcommand{\Cpm}{C_{p,\text{m}}} % molar heat capacity at const.p\) \( \newcommand{\kT}{\kappa_T} % isothermal compressibility\) \( \newcommand{\A}{_{\text{A}}} % subscript A for solvent or state A\) \( \newcommand{\B}{_{\text{B}}} % subscript B for solute or state B\) \( \newcommand{\bd}{_{\text{b}}} % subscript b for boundary or boiling point\) \( \newcommand{\C}{_{\text{C}}} % subscript C\) \( \newcommand{\f}{_{\text{f}}} % subscript f for freezing point\) \( \newcommand{\mA}{_{\text{m},\text{A}}} % subscript m,A (m=molar)\) \( \newcommand{\mB}{_{\text{m},\text{B}}} % subscript m,B (m=molar)\) \( \newcommand{\mi}{_{\text{m},i}} % subscript m,i (m=molar)\) \( \newcommand{\fA}{_{\text{f},\text{A}}} % subscript f,A (for fr. pt.)\) \( \newcommand{\fB}{_{\text{f},\text{B}}} % subscript f,B (for fr. pt.)\) \( \newcommand{\xbB}{_{x,\text{B}}} % x basis, B\) \( \newcommand{\xbC}{_{x,\text{C}}} % x basis, C\) \( \newcommand{\cbB}{_{c,\text{B}}} % c basis, B\) \( \newcommand{\mbB}{_{m,\text{B}}} % m basis, B\) \( \newcommand{\kHi}{k_{\text{H},i}} % Henry's law constant, x basis, i\) \( \newcommand{\kHB}{k_{\text{H,B}}} % Henry's law constant, x basis, B\) \( \newcommand{\arrow}{\,\rightarrow\,} % right arrow with extra spaces\) \( \newcommand{\arrows}{\,\rightleftharpoons\,} % double arrows with extra spaces\) \( \newcommand{\ra}{\rightarrow} % right arrow (can be used in text mode)\) \( \newcommand{\eq}{\subs{eq}} % equilibrium state\) \( \newcommand{\onehalf}{\textstyle\frac{1}{2}\D} % small 1/2 for display equation\) \( \newcommand{\sys}{\subs{sys}} % system property\) \( \newcommand{\sur}{\sups{sur}} % surroundings\) \( \renewcommand{\in}{\sups{int}} % internal\) \( \newcommand{\lab}{\subs{lab}} % lab frame\) \( \newcommand{\cm}{\subs{cm}} % center of mass\) \( \newcommand{\rev}{\subs{rev}} % reversible\) \( \newcommand{\irr}{\subs{irr}} % irreversible\) \( \newcommand{\fric}{\subs{fric}} % friction\) \( \newcommand{\diss}{\subs{diss}} % dissipation\) \( \newcommand{\el}{\subs{el}} % electrical\) \( \newcommand{\cell}{\subs{cell}} % cell\) \( \newcommand{\As}{A\subs{s}} % surface area\) \( \newcommand{\E}{^\mathsf{E}} % excess quantity (superscript)\) \( \newcommand{\allni}{\{n_i \}} % set of all n_i\) \( \newcommand{\sol}{\hspace{-.1em}\tx{(sol)}}\) \( \newcommand{\solmB}{\tx{(sol,$\,$$m\B$)}}\) \( \newcommand{\dil}{\tx{(dil)}}\) \( \newcommand{\sln}{\tx{(sln)}}\) \( \newcommand{\mix}{\tx{(mix)}}\) \( \newcommand{\rxn}{\tx{(rxn)}}\) \( \newcommand{\expt}{\tx{(expt)}}\) \( \newcommand{\solid}{\tx{(s)}}\) \( \newcommand{\liquid}{\tx{(l)}}\) \( \newcommand{\gas}{\tx{(g)}}\) \( \newcommand{\pha}{\alpha} % phase alpha\) \( \newcommand{\phb}{\beta} % phase beta\) \( \newcommand{\phg}{\gamma} % phase gamma\) \( \newcommand{\aph}{^{\alpha}} % alpha phase superscript\) \( \newcommand{\bph}{^{\beta}} % beta phase superscript\) \( \newcommand{\gph}{^{\gamma}} % gamma phase superscript\) \( \newcommand{\aphp}{^{\alpha'}} % alpha prime phase superscript\) \( \newcommand{\bphp}{^{\beta'}} % beta prime phase superscript\) \( \newcommand{\gphp}{^{\gamma'}} % gamma prime phase superscript\) \( \newcommand{\apht}{\small\aph} % alpha phase tiny superscript\) \( \newcommand{\bpht}{\small\bph} % beta phase tiny superscript\) \( \newcommand{\gpht}{\small\gph} % gamma phase tiny superscript\) \( \newcommand{\upOmega}{\Omega}\) \( \newcommand{\dif}{\mathop{}\!\mathrm{d}} % roman d in math mode, preceded by space\) \( \newcommand{\Dif}{\mathop{}\!\mathrm{D}} % roman D in math mode, preceded by space\) \( \newcommand{\df}{\dif\hspace{0.05em} f} % df\) \(\newcommand{\dBar}{\mathop{}\!\mathrm{d}\hspace-.3em\raise1.05ex{\Rule{.8ex}{.125ex}{0ex}}} % inexact differential \) \( \newcommand{\dq}{\dBar q} % heat differential\) \( \newcommand{\dw}{\dBar w} % work differential\) \( \newcommand{\dQ}{\dBar Q} % infinitesimal charge\) \( \newcommand{\dx}{\dif\hspace{0.05em} x} % dx\) \( \newcommand{\dt}{\dif\hspace{0.05em} t} % dt\) \( \newcommand{\difp}{\dif\hspace{0.05em} p} % dp\) \( \newcommand{\Del}{\Delta}\) \( \newcommand{\Delsub}[1]{\Delta_{\text{#1}}}\) \( \newcommand{\pd}[3]{(\partial #1 / \partial #2 )_{#3}} % \pd{}{}{} - partial derivative, one line\) \( \newcommand{\Pd}[3]{\left( \dfrac {\partial #1} {\partial #2}\right)_{#3}} % Pd{}{}{} - Partial derivative, built-up\) \( \newcommand{\bpd}[3]{[ \partial #1 / \partial #2 ]_{#3}}\) \( \newcommand{\bPd}[3]{\left[ \dfrac {\partial #1} {\partial #2}\right]_{#3}}\) \( \newcommand{\dotprod}{\small\bullet}\) \( \newcommand{\fug}{f} % fugacity\) \( \newcommand{\g}{\gamma} % solute activity coefficient, or gamma in general\) \( \newcommand{\G}{\varGamma} % activity coefficient of a reference state (pressure factor)\) \( \newcommand{\ecp}{\widetilde{\mu}} % electrochemical or total potential\) \( \newcommand{\Eeq}{E\subs{cell, eq}} % equilibrium cell potential\) \( \newcommand{\Ej}{E\subs{j}} % liquid junction potential\) \( \newcommand{\mue}{\mu\subs{e}} % electron chemical potential\) \( \newcommand{\defn}{\,\stackrel{\mathrm{def}}{=}\,} % "equal by definition" symbol\) \( \newcommand{\D}{\displaystyle} % for a line in built-up\) \( \newcommand{\s}{\smash[b]} % use in equations with conditions of validity\) \( \newcommand{\cond}[1]{\\[-2.5pt]{}\tag*{#1}}\) \( \newcommand{\nextcond}[1]{\\[-5pt]{}\tag*{#1}}\) \( \newcommand{\R}{8.3145\units{J$\,$K$\per\,$mol$\per$}} % gas constant value\) \( \newcommand{\Rsix}{8.31447\units{J$\,$K$\per\,$mol$\per$}} % gas constant value - 6 sig figs\) \( \newcommand{\jn}{\hspace3pt\lower.3ex{\Rule{.6pt}{2ex}{0ex}}\hspace3pt} \) \( \newcommand{\ljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}} \hspace3pt} \) \( \newcommand{\lljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace1.4pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace3pt} \) This e-book will denote the molar reaction Gibbs energy of a cell reaction by \(\Delsub{r}G\cell\). This notation distinguishes it from the molar reaction Gibbs energy \(\Delsub{r}G\) of the direct reaction, which may have a different value because in the cell the chemical potential of an ionic species is affected by the electric potential of its phase. \(\Delsub{r}G\cell\) is defined by \begin{equation} \Delsub{r}G\cell \defn \sum_i\nu_i\mu_i \tag{14.3.1} \end{equation} where the sum is over the reactants and products of the cell reaction. \(\Delsub{r}G\cell\) is also equal to the partial derivative \(\pd{G\cell}{\xi}{T,p}\), where \(\xi\) is the advancement of the cell reaction. 14.3.1 Relation between \(\Delsub{r}G\cell\) and \(\Eeq\) When a galvanic cell is in a zero-current equilibrium state, both electrode reactions are at reaction equilibrium. In the electrode reaction at the left electrode, electrons are a product with stoichiometric number equal to \(z\). At the right electrode, electrons are a reactant with stoichiometric number equal to \(-z\). We can write the conditions for electrode reaction equilibria as follows: \begin{equation} \tx{At the left electrode} \quad \sum_i\nu_i\mu_i + z\mue(\tx{LE}) = 0 \tag{14.3.2} \end{equation} \begin{equation} \tx{At the right electrode} \quad \sum_j\nu_j\mu_j - z\mue(\tx{RE}) = 0 \tag{14.3.3} \end{equation} In these equations, the sum over \(i\) is for the chemical species (excluding electrons) of the electrode reaction at the left electrode, and the sum over \(j\) is for the chemical species of the electrode reaction at the right electrode. \(\mue(\tx{LE})\) is the chemical potential of electrons in the electron conductor of the left electrode, and \(\mue(\tx{RE})\) is the chemical potential of electrons in the electron conductor of the right electrode. Adding Eqs. 14.3.2 and 14.3.3, we obtain \begin{equation} \sum_i\nu_i\mu_i + \sum_j\nu_j\mu_j + z[ \mue(\tx{LE})-\mue(\tx{RE}) ] = 0 \tag{14.3.4} \end{equation} The first two terms on the left side of Eq. 14.3.4 are sums over all the reactants and products of the cell reaction. From Eq. 14.3.1, we recognize the sum of these terms as the molar reaction Gibbs energy of the cell reaction: \begin{equation} \sum_i\nu_i\mu_i + \sum_j\nu_j\mu_j = \Delsub{r}G\cell \tag{14.3.5} \end{equation} Substituting from Eq. 14.3.5 into Eq. 14.3.4 and solving for \(\Delsub{r}G\cell\), we obtain \begin{equation} \Delsub{r}G\cell = -z[ \mue(\tx{LE})-\mue(\tx{RE}) ] \tag{14.3.6} \end{equation} In a zero-current equilibrium state, there is electron transfer equilibrium between the left electron conductor and the left terminal, and between the right electron conductor and the right terminal: \(\mue(\tx{LE})=\mue(\tx{LT})\) and \(\mue(\tx{RE})=\mue(\tx{RT})\), where \(\mue(\tx{LT})\) and \(\mue(\tx{RT})\) are the chemical potentials of electrons in the left terminal and right terminal, respectively. Thus we can rewrite Eq. 14.3.6 as \begin{equation} \Delsub{r}G\cell = -z[ \mue(\tx{LT})-\mue(\tx{RT}) ] \tag{14.3.7} \end{equation} Making substitutions from Eq. 14.2.2 for \(\mue(\tx{LT})\) and \(\mue(\tx{RT})\), and recognizing that \(\mue(0)\) is the same in both terminals because they have the same composition, we obtain \begin{equation} \begin{split} \Delsub{r}G\cell & = -zF(\phi\subs{R}-\phi\subs{L}) \cr & = -zF\Eeq \end{split} \tag{14.3.8} \end{equation} We can see from Eq. 14.3.1 that the value of \(\Delsub{r}G\cell\) has nothing to do with the composition of the terminals. The relations of Eq. 14.3.8 were derived for a cell with both terminals made of the same metal. We can make the following deductions for such a cell: Equation 14.3.8 can be derived by a different route. According to Eq. 5.8.6, reversible electrical work at constant \(T\) and \(p\) is equal to the Gibbs energy change: \(\dw\subs{el, rev}=\dif G\cell\). Making the substitution \(\dw\subs{el, rev}=\Eeq\dQ\sys\) (from Eq. 3.8.8), with \(\dQ\sys\) set equal to \(-zF\dif\xi\) (Eq. 14.1.1), followed by division by \(\dif\xi\), gives \(-zF\Eeq = \pd{G\cell}{\xi}{T,p}\), or \(\Delsub{r}G\cell = -zF\Eeq\). Strictly speaking, this derivation applies only to a cell without a liquid junction. In a cell with a liquid junction, the electric current is carried across the junction by different ions depending on the direction of the current, and the cell is therefore not reversible. 14.3.2 Relation between \(\Delsub{r}G\cell\) and \(\Delsub{r}G\) Now imagine a reaction vessel that has the same temperature and pressure as the galvanic cell, and contains the same reactants and products at the same activities as in the cell. This reaction vessel, unlike the cell, is not part of an electrical circuit. In it, the reactants and products are in direct contact with one another, so there is no constraint preventing a spontaneous direct reaction. For example, the reaction vessel corresponding to the zinc–copper cell of Fig. 14.2 would have zinc and copper strips in contact with a solution of both ZnSO\(_4\) and CuSO\(_4\). Another example is the slow direct reaction in a cell without liquid junction described in Sec. 14.2.1. Let the reaction equation of the direct reaction be written with the same stoichiometric numbers \(\nu_i\) as in the reaction equation for the cell reaction. The direct reaction in the reaction vessel is described by this equation or its reverse, depending on which direction is spontaneous for the given activities. The question now arises whether the molar reaction Gibbs energy \(\Delsub{r}G\cell\) of the cell reaction is equal to the molar reaction Gibbs energy \(\Delsub{r}G\) of the direct reaction. Both \(\Delsub{r}G\cell\) and \(\Delsub{r}G\) are defined by the sum \(\sum_i\!\nu_i\mu_i\). Both reactions have the same values of \(\nu_i\), but the values of \(\mu_i\) for charged species are in general different in the two systems because the electric potentials are different. Consider first a cell without a liquid junction. This kind of cell has a single electrolyte solution, and all of the reactant and product ions of the cell reaction are in this solution phase. The same solution phase is present in the reaction vessel during the direct reaction. When all ions are in the same phase, the value of \(\sum_i\!\nu_i\mu_i\) is independent of the electric potentials of any of the phases (see the comment following Eq. 11.8.4), so that the molar reaction Gibbs energies are the same for the cell reaction and the direct reaction: \begin{gather} \s{ \Delsub{r}G\cell = \Delsub{r}G } \tag{14.3.9} \cond{(no liquid junction)} \end{gather} Next, consider a cell with two electrolyte solutions separated by a liquid junction. For the molar reaction Gibbs energy of the cell reaction, we write \begin{equation} \Delsub{r}G\cell = \sum_i\nu_i\mu_i(\phi_i) + \sum_j\nu_j\mu_j(\phi_j) \tag{14.3.10} \end{equation} The sums here include all of the reactants and products appearing in the cell reaction, those with index \(i\) being at the left electrode and those with index \(j\) at the right electrode. Let the solution at the left electrode be phase \(\pha\) and the solution at the right electrode be phase \(\phb\). Then making the substitution \(\mu_i(\phi)=\mu_i(0)+z_iF\phi\) (Eq. 10.1.6) gives us \begin{equation} \Delsub{r}G\cell = \sum_i\nu_i\mu_i(0) + \sum_j\nu_j\mu_j(0) + \sum_i\nu_i z_i F\phi\aph + \sum_j\nu_j z_j F\phi\bph \tag{14.3.11} \end{equation} The sum of the first two terms on the right side of Eq. 14.3.11 is the molar reaction Gibbs energy of a reaction in which the reactants and products are in phases of zero electric potential. According to the comment following Eq. 11.8.4, the molar reaction Gibbs energy would be the same if the ions were in a single phase of any electric potential. Consequently the sum \(\sum_i\!\nu_i\mu_i(0){+}\sum_j\!\nu_j\mu_j(0)\) is equal to \(\Delsub{r}G\) for the direct reaction. The conservation of charge during advancement of the electrode reactions at the left electrode and the right electrode is expressed by \(\sum_i\!\nu_i z_i - z = 0\) and \(\sum_j\!\nu_j z_j + z = 0\), respectively. Equation 14.3.11 becomes \begin{gather} \s{ \Delsub{r}G\cell = \Delsub{r}G - zF\Ej } \tag{14.3.12} \cond{(cell with liquid junction)} \end{gather} where \(\Ej = \phi\bph-\phi\aph\) is the liquid junction potential. Finally, in Eqs. 14.3.9 and 14.3.12 we replace \(\Delsub{r}G\cell\) by \(-zF\Eeq\) (Eq. 14.3.8) and solve for \(\Eeq\): \begin{gather} \s{ \Eeq = -\frac{\Delsub{r}G}{zF} } \tag{14.3.13} \cond{(cell without liquid junction)} \end{gather} \begin{gather} \s{ \Eeq = -\frac{\Delsub{r}G}{zF} + \Ej } \tag{14.3.14} \cond{(cell with liquid junction)} \end{gather} \(\Eeq\) can be measured with great precision. If a reaction can be carried out in a galvanic cell without liquid junction, Eq. 14.3.13 provides a way to evaluate \(\Delsub{r}G\) under given conditions. If the reaction can only be carried out in a cell with a liquid junction, Eq. 14.3.14 can be used for this purpose provided that the liquid junction potential \(\Ej\) can be assumed to be negligible or can be estimated from theory. Note that the cell has reaction equilibrium only if \(\Delsub{r}G\) is zero. The cell has thermal, mechanical, and transfer equilibrium when the electric current is zero and the cell potential is the zero-current cell potential \(\Eeq\). Equations 14.3.13 and 14.3.14 show that in order for the cell to also have reaction equilibrium, \(\Eeq\) must equal the liquid junction potential if there is a liquid junction, or be zero otherwise. These are the conditions of an exhausted, “dead” cell that can no longer do electrical work. 14.3.3 Standard molar reaction quantities Consider a hypothetical galvanic cell in which each reactant and product of the cell reaction is in its standard state at unit activity, and in which a liquid junction if present has a negligible liquid junction potential. The equilibrium cell potential of this cell is called the standard cell potential of the cell reaction, \(\Eeq\st\). An experimental procedure for evaluating \(\Eeq\st\) will be described in Sec. 14.5. In this hypothetical cell, \(\Delsub{r}G\cell\) is equal to the standard molar reaction Gibbs energy \(\Delsub{r}G\st\). From Eq. 14.3.13, or Eq. 14.3.14 with \(\Ej\) assumed equal to zero, we have \begin{equation} \Delsub{r}G\st=-zF\Eeq\st \tag{14.3.15} \end{equation} \(\Delsub{r}G\st\) is the molar reaction Gibbs energy when each reactant and product is at unit activity and, if it is an ion, is in a phase of zero electric potential. Since \(\Delsub{r}G\st\) is equal to \(-RT\ln K\) (Eq. 11.8.10), we can write \begin{equation} \ln K = \frac{zF}{RT}\Eeq\st \tag{14.3.16} \end{equation} Equation 14.3.16 allows us to evaluate the thermodynamic equilibrium constant \(K\) of the cell reaction by a noncalorimetric method. Consider for example the cell \[ \ce{Ag} \jn \ce{Ag+}\tx{(aq)} \lljn \ce{Cl-}\tx{(aq)} \jn \ce{AgCl}\tx{(s)} \jn \ce{Ag} \] in which the pair of dashed vertical bars indicates a liquid junction of negligible liquid junction potential. The electrode reactions are \begin{align*} & \tx{Ag(s)} \arrow \tx{Ag\(^+\)(aq)} + \tx{e\(^-\)} \cr & \tx{AgCl(s)} + \tx{e\(^-\)} \arrow \tx{Ag(s)} + \tx{Cl\(^-\)(aq)} \end{align*} and the cell reaction is \begin{equation*} \ce{AgCl}\tx{(s)} \arrow \ce{Ag+}\tx{(aq)} + \ce{Cl-}\tx{(aq)} \end{equation*} The equilibrium constant of this reaction is the solubility product \(K\subs{s}\) of silver chloride (Sec. 12.5.5). At \(298.15\K\), the standard cell potential is found to be \(\Eeq\st=-0.5770\V\). We can use this value in Eq. 14.3.16 to evaluate \(K\subs{s}\) at \(298.15\K\) (see Prob. 14.5). Equation 14.3.16 also allows us to evaluate the standard molar reaction enthalpy by substitution in Eq. 12.1.13: \begin{gather} \s{ \begin{split} \Delsub{r}H\st & = RT^2\frac{\dif\ln K}{\dif T}\cr & = zF\left(T\frac{\dif \Eeq\st}{\dif T}-\Eeq\st \right) \end{split} } \tag{14.3.17} \cond{(no solute standard states} \nextcond{based on concentration)} \end{gather} Finally, by combining Eqs. 14.3.15 and 14.3.17 with \(\Delsub{r}G\st=\Delsub{r}H\st-T\Delsub{r}S\st\), we obtain an expression for the standard molar reaction entropy: \begin{gather} \s{ \Delsub{r}S\st = zF\frac{\dif \Eeq\st}{\dif T} } \tag{14.3.18} \cond{(no solute standard states} \nextcond{based on concentration)} \end{gather} Because \(G\), \(H\), and \(S\) are state functions, the thermodynamic equilibrium constant and the molar reaction quantities evaluated from \(\Eeq\st\) and \(\dif\Eeq\st/\dif T\) are the same quantities as those for the reaction when it takes place in a reaction vessel instead of in a galvanic cell. However, the heats at constant \(T\) and \(p\) are not the same (Sec. 11.3.1). During a reversible cell reaction, \(\dif S\) must equal \(\dq/T\), and \(\dq/\dif\xi\) is therefore equal to \(T\Delsub{r}S\st\) during a cell reaction taking place reversibly under standard state conditions at constant \(T\) and \(p\). |
Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Named_Reactions/Eschweiler-Clarke_reaction | Welcome to the Chemistry Library. This Living Library is a principal hub of the LibreTexts project , which is a multi-institutional collaborative venture to develop the next generation of open-access texts to improve postsecondary education at all levels of higher learning. The LibreTexts approach is highly collaborative where an Open Access textbook environment is under constant revision by students, faculty, and outside experts to supplant conventional paper-based books. Campus Bookshelves Bookshelves Learning Objects |
Courses/can/CHEM_231%3A_Organic_Chemistry_I_Textbook/06%3A_An_Overview_of_Organic_Reactions/6.01%3A_Why_This_Chapter | All chemical reactions, whether they take place in the laboratory or in living organisms, follow the same “rules.” Reactions in living organisms often look more complex than laboratory reactions because of the size of the biomolecules and the involvement of biological catalysts called enzymes, but the principles governing all chemical reactions are the same. To understand both organic and biological chemistry, it’s necessary to know not just what occurs but also why and how chemical reactions take place. In this chapter, we’ll start with an overview of the fundamental kinds of organic reactions, we’ll see why reactions occur, and we’ll see how reactions can be described. Once this background is out of the way, we’ll then be ready to begin studying the details of organic chemistry in future chapters. When first approached, organic chemistry might seem overwhelming. It’s not so much that any one part is difficult to understand, it’s that there are so many parts: tens of millions of compounds, dozens of functional groups, and an apparently endless number of reactions. With study, though, it becomes evident that there are only a few fundamental ideas that underlie all organic reactions. Far from being a collection of isolated facts, organic chemistry is a beautifully logical subject that is unified by a few broad themes. When these themes are understood, learning organic chemistry becomes much easier and memorization is minimized. The aim of this book is to describe the themes and clarify the patterns that unify organic chemistry in future chapters. |
Courses/DePaul_University/Physical_Chemistry_for_Biological_Sciences/16%3A_Chemical_Kinetics_I_-_Rate_Laws/16.03%3A_First-Order_Reactions_Show_an_Exponential_Decay_of_Reactant_Concentration_with_Time | A first order rate law would take the form \[ \dfrac{d[A]}{dt} = k[A] \nonumber \] Again, separating the variables by placing all of the concentration terms on the left and all of the time terms on the right yields \[ \dfrac{d[A]}{[A]} =-k\,dt \nonumber \] This expression is also easily integrated as before \[ \int_{[A]=0}^{[A]} \dfrac{d[A]}{[A]} =-k \int_{t=0}^{t=t}\,dt \nonumber \] Noting that \[ \dfrac{dx}{x} = d (\ln x) \nonumber \] The form of the integrated rate law becomes \[ \ln [A] - \ln [A]_o = kt \nonumber \] or \[ \ln [A] = \ln [A]_o - kt \label{In1} \] This form implies that a plot of the natural logarithm of the concentration is a linear function of the time. And so a plot of ln[A] as a function of time should produce a linear plot, the slope of which is -k, and the intercept of which is ln[A] 0 . Example \(\PageIndex{1}\): Consider the following kinetic data. Use a graph to demonstrate that the data are consistent with first order kinetics. Also, if the data are first order, determine the value of the rate constant for the reaction. 0 1 2 3 4 5 6 7 8 9 Time (s) 0.000 10.000 20.000 50.000 100.000 150.000 200.000 250.000 300.00 [A] (M) 0.873 0.752 0.648 0.414 0.196 0.093 0.044 0.021 0.01 Solution The plot looks as follows: From this plot, it can be seen that the rate constant is 0.0149 s -1 . The concentration at time \(t = 0\) can also be inferred from the intercept. It should also be noted that the integrated rate law (Equation \ref{In1}) can be expressed in exponential form: \[ [A] = [A]_o e^{-kt} \nonumber \] Because of this functional form, 1 st order kinetics are sometimes referred to as exponential decay kinetics. Many processes, including radioactive decay of nuclides follow this type of rate law. |
Courses/Fullerton_College/Introductory_Biochemistry/10%3A_Amino_Acids_Proteins_and_Enzymes/10.11%3A_Enzyme_Inhibition | 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. Concept Review Exercises What are the characteristics of an irreversible inhibitor? In what ways does a competitive inhibitor differ from a noncompetitive inhibitor? Answers It inactivates an enzyme by bonding covalently to a particular group at the active site. A competitive inhibitor structurally resembles the substrate for a given enzyme and 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 and can bind to either the free enzyme or the enzyme-substrate complex. Exercises What amino acid is present in the active site of all enzymes that are irreversibly inhibited by nerve gases such as DIFP? Oxaloacetate, (-OOCCH2COCOO-), inhibits succinate dehydrogenase. Would you expect oxaloacetate to be a competitive or noncompetitive inhibitor? Explain. Answer serine 2. One would expect oxaloacetate to be a competitive inhibitor, since it is similar in structure to succinate and thus, may be able to fit in succinate dehydrogenase's active site. |
Courses/Sacramento_City_College/SCC%3A_Chem_309_-_General_Organic_and_Biochemistry_(Bennett)/Text/05%3A_Covalent_Bonding_and_Simple_Molecular_Compounds/5.6%3A_Intermolecular_Forces | The physical properties of melting point, boiling point, vapor pressure, evaporation, viscosity, surface tension, and solubility are related to the strength of attractive forces between molecules. These attractive forces are called Intermolecular Forces. The amount of "stick togetherness" is important in the interpretation of the various properties listed above. Introduction There are four types of intermolecular forces. Most of the intermolecular forces are identical to bonding between atoms in a single molecule. Intermolecular forces just extend the thinking to forces between molecules and follows the patterns already set by the bonding within molecules. Ionic Forces The forces holding ions together in ionic solids are electrostatic forces. Opposite charges attract each other. These are the strongest intermolecular forces. Ionic forces hold many ions in a crystal lattice structure. Dipole Forces Polar covalent molecules are sometimes described as "dipoles", meaning that the molecule has two "poles". One end (pole) of the molecule has a partial positive charge while the other end has a partial negative charge. The molecules will orientate themselves so that the opposite charges attract principle operates effectively. For example, hydrochloric acid (\(HCl\)) is a polar molecule with the partial positive charge on the hydrogen and the partial negative charge on the chlorine. A network of partial + and - charges attract molecules to each other. Hydrogen Bonding The hydrogen bond is really a special case of dipole forces. A hydrogen bond is the attractive force between the hydrogen attached to an electronegative atom of one molecule and an electronegative atom of a different molecule. Usually the electronegative atom is oxygen, nitrogen, or fluorine. In other words - The hydrogen on one molecule attached to O or N that is attracted to an O or N of a different molecule. In the graphic below, the hydrogen is partially positive and attracted to the partially negative charge on the oxygen or nitrogen. Because oxygen has two lone pairs, two different hydrogen bonds can be made to each oxygen. This is a very specific bond as indicated. Some combinations that are not hydrogen bonds include: hydrogen to another hydrogen or hydrogen to a carbon. London Dispersion or Induced Dipole or Van der Waals Forces Forces between essentially non-polar molecules are the weakest of all intermolecular forces. "Temporary dipoles" are formed by the shifting of electron clouds within molecules. These temporary dipoles attract or repel the electron clouds of nearby non-polar molecules. The temporary dipoles may exist for only a fraction of a second but a force of attraction also exist for that fraction of time. The strength of induced dipole forces depends on how easily electron clouds can be distorted. Large atoms or molecules with many electrons far removed from the nucleus are more easily distorted. Contributors Charles Ophardt, Professor Emeritus, Elmhurst College Virtual Chembook |
Bookshelves/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/06%3A_Chemical_Bonding_I-_Drawing_Lewis_Structures_and_Determining_Molecular_Shapes/6.05%3A_Exceptions_to_the_Octet_Rule-_Odd-Electron_Species_Incomplete_Octets_and_Expanded_Octets | Learning Objectives To assign a Lewis dot symbol to elements not having an octet of electrons in their compounds. Three cases can be constructed that do not follow the octet rule, and as such, they are known as the exceptions to the octet rule. Following the Octet Rule for Lewis Dot Structures leads to the most accurate depictions of stable molecular and atomic structures and because of this we always want to use the octet rule when drawing Lewis Dot Structures. However, it is hard to imagine that one rule could be followed by all molecules. There is always an exception, and in this case, three exceptions: When there are an odd number of valence electrons When there are too few valence electrons When there are too many valence electrons Exception 1: Species with Odd Numbers of Electrons The first exception to the Octet Rule is when there are an odd number of valence electrons. An example of this would be Nitrogen (II) Oxide also called nitric oxide (\(\ce{NO}\). Nitrogen has 5 valence electrons while Oxygen has 6. The total would be 11 valence electrons to be used. The Octet Rule for this molecule is fulfilled in the above example, however that is with 10 valence electrons. The last one does not know where to go. The lone electron is called an unpaired electron. But where should the unpaired electron go? The unpaired electron is usually placed in the Lewis Dot Structure so that each element in the structure will have the lowest formal charge possible. The formal charge is the perceived charge on an individual atom in a molecule when atoms do not contribute equal numbers of electrons to the bonds they participate in . No formal charge at all is the most ideal situation. An example of a stable molecule with an odd number of valence electrons would be nitric oxide. nitric oxide has 11 valence electrons. If you need more information about formal charges, see Lewis Structures. If we were to imagine nitric oxide had ten valence electrons we would come up with the Lewis Structure (Figure \(\PageIndex{1}\)): Let's look at the formal charges of Figure \(\PageIndex{2}\) based on this Lewis structure. Nitrogen normally has five valence electrons. In Figure \(\PageIndex{1}\), it has two lone pair electrons and it participates in two bonds (a double bond) with oxygen. This results in nitrogen having a formal charge of +1. Oxygen normally has six valence electrons. In Figure \(\PageIndex{1}\), oxygen has four lone pair electrons and it participates in two bonds with nitrogen. Oxygen therefore has a formal charge of 0. The overall molecule here has a formal charge of +1 (+1 for nitrogen, 0 for oxygen. +1 + 0 = +1). However, if we add the eleventh electron to nitrogen (because we want the molecule to have the lowest total formal charge), it will bring both the nitrogen and the molecule's overall charges to zero, the most ideal formal charge situation. That is exactly what is done to get the correct Lewis structure for nitric oxide (Figure \(\PageIndex{2}\)): Free Radicals There are actually very few stable molecules with odd numbers of electrons that exist, since that unpaired electron is willing to react with other unpaired electrons. Most odd electron species are highly reactive, which we call Free Radicals. Because of their instability, free radicals bond to atoms in which they can take an electron from in order to become stable, making them very chemically reactive. Radicals are found as both reactants and products, but generally react to form more stable molecules as soon as they can. In order to emphasize the existence of the unpaired electron, radicals are denoted with a dot in front of their chemical symbol as with \(\cdot OH\), the hydroxyl radical. An example of a radical you may by familiar with already is the gaseous chlorine atom, denoted \(\cdot Cl\). Interestingly, an odd Number of Valence Electrons will result in the molecule being paramagnetic. Exception 2: Incomplete Octets The second exception to the Octet Rule is when there are too few valence electrons that results in an incomplete Octet. There are even more occasions where the octet rule does not give the most correct depiction of a molecule or ion. This is also the case with incomplete octets. Species with incomplete octets are pretty rare and generally are only found in some beryllium, aluminum, and boron compounds including the boron hydrides. Let's take a look at one such hydride, BH 3 (Borane). If one were to make a Lewis structure for BH 3 following the basic strategies for drawing Lewis structures, one would probably come up with this structure (Figure \(\PageIndex{2}\)): The problem with this structure is that boron has an incomplete octet; it only has six electrons around it. Hydrogen atoms can naturally only have only 2 electrons in their outermost shell (their version of an octet), and as such there are no spare electrons to form a double bond with boron. One might surmise that the failure of this structure to form complete octets must mean that this bond should be ionic instead of covalent. However, boron has an electronegativity that is very similar to hydrogen, meaning there is likely very little ionic character in the hydrogen to boron bonds, and as such this Lewis structure, though it does not fulfill the octet rule, is likely the best structure possible for depicting BH 3 with Lewis theory. One of the things that may account for BH 3 's incomplete octet is that it is commonly a transitory species, formed temporarily in reactions that involve multiple steps. Let's take a look at another incomplete octet situation dealing with boron, BF 3 (Boron trifluorine). Like with BH 3 , the initial drawing of a Lewis structure of BF 3 will form a structure where boron has only six electrons around it (Figure \(\PageIndex{4}\)). If you look Figure \(\PageIndex{4}\), you can see that the fluorine atoms possess extra lone pairs that they can use to make additional bonds with boron, and you might think that all you have to do is make one lone pair into a bond and the structure will be correct. If we add one double bond between boron and one of the fluorines we get the following Lewis Structure (Figure \(\PageIndex{5}\)): Each fluorine has eight electrons, and the boron atom has eight as well! Each atom has a perfect octet, right? Not so fast. We must examine the formal charges of this structure. The fluorine that shares a double bond with boron has six electrons around it (four from its two lone pairs of electrons and one each from its two bonds with boron). This is one less electron than the number of valence electrons it would have naturally (Group Seven elements have seven valence electrons), so it has a formal charge of +1. The two flourines that share single bonds with boron have seven electrons around them (six from their three lone pairs and one from their single bonds with boron). This is the same amount as the number of valence electrons they would have on their own, so they both have a formal charge of zero. Finally, boron has four electrons around it (one from each of its four bonds shared with fluorine). This is one more electron than the number of valence electrons that boron would have on its own, and as such boron has a formal charge of -1. This structure is supported by the fact that the experimentally determined bond length of the boron to fluorine bonds in BF 3 is less than what would be typical for a single bond (see Bond Order and Lengths). However, this structure contradicts one of the major rules of formal charges: Negative formal charges are supposed to be found on the more electronegative atom(s) in a bond, but in the structure depicted in Figure \(\PageIndex{5}\), a positive formal charge is found on fluorine, which not only is the most electronegative element in the structure, but the most electronegative element in the entire periodic table (\(\chi=4.0\)). Boron on the other hand, with the much lower electronegativity of 2.0, has the negative formal charge in this structure. This formal charge-electronegativity disagreement makes this double-bonded structure impossible. However the large electronegativity difference here, as opposed to in BH 3 , signifies significant polar bonds between boron and fluorine, which means there is a high ionic character to this molecule. This suggests the possibility of a semi-ionic structure such as seen in Figure \(\PageIndex{6}\): None of these three structures is the "correct" structure in this instance. The most "correct" structure is most likely a resonance of all three structures: the one with the incomplete octet (Figure \(\PageIndex{4}\)), the one with the double bond (Figure \(\PageIndex{5}\)), and the one with the ionic bond (Figure \(\PageIndex{6}\)). The most contributing structure is probably the incomplete octet structure (due to Figure \(\PageIndex{5}\) being basically impossible and Figure \(\PageIndex{6}\) not matching up with the behavior and properties of BF 3 ). As you can see even when other possibilities exist, incomplete octets may best portray a molecular structure. As a side note, it is important to note that BF 3 frequently bonds with a F - ion in order to form BF 4 - rather than staying as BF 3 . This structure completes boron's octet and it is more common in nature. This exemplifies the fact that incomplete octets are rare, and other configurations are typically more favorable, including bonding with additional ions as in the case of BF 3 . Example \(\PageIndex{1}\): \(NF_3\) Draw the Lewis structure for boron trifluoride (BF 3 ). Solution 1. Add electrons (3*7) + 3 = 24 2. Draw connectivities: 3. Add octets to outer atoms: 4. Add extra electrons (24-24=0) to central atom: 5. Does central electron have octet? NO. It has 6 electrons Add a multiple bond (double bond) to see if central atom can achieve an octet: 6. The central Boron now has an octet (there would be three resonance Lewis structures) However... In this structure with a double bond the fluorine atom is sharing extra electrons with the boron. The fluorine would have a '+' partial charge, and the boron a '-' partial charge, this is inconsistent with the electronegativities of fluorine and boron. Thus, the structure of BF 3 , with single bonds, and 6 valence electrons around the central boron is the most likely structure BF 3 reacts strongly with compounds which have an unshared pair of electrons which can be used to form a bond with the boron: Exception 3: Expanded Valence Shells More common than incomplete octets are expanded octets where the central atom in a Lewis structure has more than eight electrons in its valence shell. In expanded octets, the central atom can have ten electrons, or even twelve. Molecules with expanded octets involve highly electronegative terminal atoms, and a nonmetal central atom found in the third period or below , which those terminal atoms bond to. For example, \(PCl_5\) is a legitimate compound (whereas \(NCl_5\)) is not: Expanded valence shells are observed only for elements in period 3 (i.e. n=3) and beyond The 'octet' rule is based upon available n s and n p orbitals for valence electrons (2 electrons in the s orbitals, and 6 in the p orbitals). Beginning with the n=3 principle quantum number, the d orbitals become available ( l =2). The orbital diagram for the valence shell of phosphorous is: Hence, the third period elements occasionally exceed the octet rule by using their empty d orbitals to accommodate additional electrons. Size is also an important consideration: The larger the central atom, the larger the number of electrons which can surround it Expanded valence shells occur most often when the central atom is bonded to small electronegative atoms, such as F, Cl and O. There is currently much scientific exploration and inquiry into the reason why expanded valence shells are found. The top area of interest is figuring out where the extra pair(s) of electrons are found. Many chemists think that there is not a very large energy difference between the 3p and 3d orbitals, and as such it is plausible for extra electrons to easily fill the 3d orbital when an expanded octet is more favorable than having a complete octet. This matter is still under hot debate, however and there is even debate as to what makes an expanded octet more favorable than a configuration that follows the octet rule. One of the situations where expanded octet structures are treated as more favorable than Lewis structures that follow the octet rule is when the formal charges in the expanded octet structure are smaller than in a structure that adheres to the octet rule, or when there are less formal charges in the expanded octet than in the structure a structure that adheres to the octet rule. Example \(\PageIndex{2}\): The \(SO_4^{-2}\) ion Such is the case for the sulfate ion, SO 4 -2 . A strict adherence to the octet rule forms the following Lewis structure: If we look at the formal charges on this molecule, we can see that all of the oxygen atoms have seven electrons around them (six from the three lone pairs and one from the bond with sulfur). This is one more electron than the number of valence electrons then they would have normally, and as such each of the oxygens in this structure has a formal charge of -1. Sulfur has four electrons around it in this structure (one from each of its four bonds) which is two electrons more than the number of valence electrons it would have normally, and as such it carries a formal charge of +2. If instead we made a structure for the sulfate ion with an expanded octet, it would look like this: Looking at the formal charges for this structure, the sulfur ion has six electrons around it (one from each of its bonds). This is the same amount as the number of valence electrons it would have naturally. This leaves sulfur with a formal charge of zero. The two oxygens that have double bonds to sulfur have six electrons each around them (four from the two lone pairs and one each from the two bonds with sulfur). This is the same amount of electrons as the number of valence electrons that oxygen atoms have on their own, and as such both of these oxygen atoms have a formal charge of zero. The two oxygens with the single bonds to sulfur have seven electrons around them in this structure (six from the three lone pairs and one from the bond to sulfur). That is one electron more than the number of valence electrons that oxygen would have on its own, and as such those two oxygens carry a formal charge of -1. Remember that with formal charges, the goal is to keep the formal charges (or the difference between the formal charges of each atom) as small as possible. The number of and values of the formal charges on this structure (-1 and 0 (difference of 1) in Figure \(\PageIndex{12}\), as opposed to +2 and -1 (difference of 3) in Figure \(\PageIndex{12}\)) is significantly lower than on the structure that follows the octet rule, and as such an expanded octet is plausible, and even preferred to a normal octet, in this case. Example \(\PageIndex{3}\): The \(ICl_4^-\) Ion Draw the Lewis structure for \(ICl_4^-\) ion. Solution 1. Count up the valence electrons: 7+(4*7)+1 = 36 electrons 2. Draw the connectivities: 3. Add octet of electrons to outer atoms: 4. Add extra electrons (36-32= 4 ) to central atom: 5. The ICl 4 - ion thus has 12 valence electrons around the central Iodine (in the 5 d orbitals) Expanded Lewis structures are also plausible depictions of molecules when experimentally determined bond lengths suggest partial double bond characters even when single bonds would already fully fill the octet of the central atom. Despite the cases for expanded octets, as mentioned for incomplete octets, it is important to keep in mind that, in general, the octet rule applies. Summary Following the Octet Rule for Lewis Dot Structures leads to the most accurate depictions of stable molecular and atomic structures and because of this we always want to use the octet rule when drawing Lewis Dot Structures. There are three exceptions: (1) When there are an odd number of valence electrons, (2) When there are too few valence electrons, and (3) when there are too many valence electrons References Petrucci, Ralph H.; Harwood, William S.; Herring, F. G.; Madura, Jeffrey D. General Chemistry: Principles & Modern Applications . 9th Ed. New Jersey. Pearson Education, Inc. 2007. Moore, John W.; Stanitski, Conrad L. ; Jurs, Peter C. Chemistry; The Molecular Science . 2nd Ed. 2004. |
Courses/University_of_Kentucky/CHE_103%3A_Chemistry_for_Allied_Health_(Soult)/13%3A_Amino_Acids_and_Proteins/13.03%3A_Protein_Structure | Learning Outcomes Describe the four levels of protein structure. Identify the two types of secondary structure in proteins. Describe the interactive forces in each level of protein structure. Distinguish between globular and fibrous proteins. Define denaturation of proteins. Identify ways in which proteins are denatured. Hemoglobin is a complex protein which has a quaternary structure and contains iron. There are four subunits in the hemoglobin molecule - two alpha subunits and two beta subunits. Each subunit contains one iron ion, whose oxidation state changes from \(+2\) to \(+3\) and back again, depending upon the environment around the iron. When oxygen binds to the iron, the three-dimensional shape of the molecule changes. Upon release of the oxygen to the cells, the shape changes again. With hemoglobin of normal structure, this shift in conformation does not present any problems. However, individuals with hemoglobin S do experience serious complications. This hemoglobin has one amino acid in the two beta chains that is different from the amino acid at that point in the primary structure of normal hemoglobin. The result of this one structural change is aggregation of the individual protein molecules when oxygen is released. Adjacent hemoglobin molecules come in contact with one another and clump up, causing the red cells to deform and break. This abnormality, known as sickle cell, is genetic in nature. A person may inherit the gene from one parent and have sickle cell trait (only some of the hemoglobin is hemoglobin S), which is usually not life-threatening. Inheriting the gene from both parents will result in sickle cell disease, a very serious condition. Proteins A polypeptide is a sequence of amino acids between ten and one hundred in length. A protein is a peptide that is greater than one hundred amino acids in length. Proteins are very prevalent in living organisms. Hair, skin, nails, muscles, and the hemoglobin in red blood cells are some of the important parts of your body that are made of different proteins. The wide array of chemical and physiological properties of proteins is a function of their amino acid sequences. Since proteins generally consist of one hundred or more amino acids, the number of amino acid sequences that are possible is virtually limitless. The three-dimensional structure of a protein is very critical to its function. This structure can be broken down into four levels. The primary structure is the amino acid sequence of the protein. The amino acid sequence of a given protein is unique and defines the function of that protein. Peptide bonds form the connections between the amino acids. The secondary structure is a highly irregular sub-structure of the protein. The two most common types of protein secondary structure are the alpha helix (see figure below) and the beta sheet (see figure below). An alpha helix consists of amino acids that adopt a spiral shape. A beta pleated sheet (like a fan-folded paper) is alternating rows of amino acids that line up in a side-by-side fashion. In both cases, the secondary structures are stabilized by extensive hydrogen bonding between the side chains. The interaction of the various side chains in the amino acid, specifically the hydrogen bonding, leads to the adoption of a particular secondary structure. The hydrogen bonding occurs between amino acids that are close to each other in the primary structure. The tertiary structure is the overall three-dimensional structure of the protein. A typical protein consists of several sections of a specific secondary structure (alpha helix or beta sheet) along with other areas in which a more random structure occurs. These areas combine to produce the tertiary structure. The tertiary structure is stabilized by forces similar to the intermolecular forces previously seen between molecules. These attractive forces include London dispersion forces (hydrophobic), hydrogen bonding, dipole-dipole forces, ion-dipole interactions, ionic salt bridges, and disulfide bonds (see figure below). Some protein molecules consist of multiple protein subunits. The quaternary structure of a protein refers to the specific interaction and orientation of the subunits of that protein. The quaternary structure is a result of the same types of interactions as seen in tertiary structure but between different subunits. Quaternary structures can have different numbers of subunits. For example, hemoglobin contains four subunits while insulin contains two subunits. Hemoglobin is a very large protein found in red blood cells and whose function is to bind and carry oxygen throughout the bloodstream. Hemoglobin consists of four subunits - two \(\alpha\) subunits (yellow) and two \(\beta\) subunits (gray) - which then come together in a specific and defined way through interactions of the side chains (see figure below). Hemoglobin also contains four iron atoms, located in the middle of each of the four subunits. The iron atoms are part of a structure called a porphyrin, shown in red in the figure. Some proteins consist of only one subunit and thus do not have a quaternary structure. The figure below diagrams the interaction of the four levels of protein structure. Globular and Fibrous Proteins Once proteins form and have developed all levels of their structure, they can be classified as either fibrous or globular. These classifications give the basic shape of the entire protein molecule. While many proteins are globular proteins (see figure below), keratin proteins are fibrous (see figure below) and make up the hair, nails, and the outer layer of skin. Denaturation of Proteins The highly organized structures of proteins are truly masterworks of chemical architecture. But highly organized structures tend to have a certain delicacy, and this is true of proteins. Denaturation is the term used for any change in the three-dimensional structure of a protein that renders it incapable of performing its assigned function. A denatured protein cannot do its job because there is a change in the secondary, tertiary, or quaternary structure (see figure below). A wide variety of reagents and conditions, such as heat, organic compounds, pH changes, and heavy metal ions can cause protein denaturation. Anyone who has fried an egg has observed denaturation. The clear egg white turns opaque as the albumin denatures and coagulates. The primary structures of proteins are quite sturdy. In general, fairly vigorous conditions are needed to hydrolyze peptide bonds. At the secondary through quaternary levels, however, proteins are quite vulnerable to attack, though they vary in their vulnerability to denaturation. The delicately folded globular proteins are much easier to denature than are the tough, fibrous proteins of hair and skin. There are a variety of ways to denature proteins including those below. Heat above \(50^\text{o} \text{C}\) Strong acids Strong bases Ionic compounds (i.e. \(\ce{NaCl}\)) Reducing agents Detergents Heavy metal ions Agitation If you have ever had a hair permanent or chemically straightened your hair, the process involved the denaturation of proteins. The reducing agent (usually an ammonium compound) breaks the disulfide bonds in the hair. The hair is then curled or straightened which aligns the amino acids in a different pattern. An oxidizing agent is applied and the disulfide bonds reform between different amino acids. The change is permanent for the hair that you have at the time but new hair growing in will have the structure of the original proteins and your hair is back to its normal state. |
Courses/Lumen_Learning/Book%3A_United_States_History%3A_Reconstruction_to_the_Present_(Lumen)/07%3A_Module_5-_An_Age_of_Progressive_Reform/7.07%3A_Video-_The_Progressive_Era | This video teaches you about the Progressive Era in the United States. In the late 19th and early 20th century in America, there was a sense that things could be improved upon. A sense that reforms should be enacted. A sense that progress should be made. As a result, we got the Progressive Era, which has very little to do with automobile insurance, but a little to do with automobiles. All this overlapped with the Gilded Age, and is a little confusing, but here we have it. Basically, people were trying to solve some of the social problems that came with the benefits of industrial capitalism. To oversimplify, there was a competition between the corporations’ desire to keep wages low and workers’ desire to have a decent life. Improving food safety, reducing child labor, and unions were all on the agenda in the Progressive Era. While progress was being made, and people were becoming more free, these gains were not equally distributed. Jim Crow laws were put in place in the south, and immigrant rights were restricted as well. So once again on Crash Course, things aren’t so simple. https://youtu.be/i0Q4zPR4G7M?t=1s All rights reserved content The Progressive Era: Crash Course US History #27. Provided by : Crash Course. Located at : https://youtu.be/i0Q4zPR4G7M?t=1s . License : All Rights Reserved . License Terms : Standard YouTube License |
Bookshelves/General_Chemistry/Chem1_(Lower)/13%3A_Acid-Base_Equilibria/13.05%3A_Acid_Base_Titration | Learning Objectives Make sure you thoroughly understand the following essential concepts: Sketch out a plot representing the titration of a strong monoprotic acid by a strong base, or of a strong base titrated by a strong acid. Identify the equivalence point and explain its significance. On the plot referred to above, draw a similar plot that would correspond to the same system at a higher or lower concentration . Sketch out a plot representing the titration of a weak monoprotic acid by a strong base, or of a weak base titrated by a strong acid. Identify the equivalence point and half-equivalence points. Explain what an acid-base indicator is, and how it works. When solutions of some polyprotic acids are titrated with strong base, not all of the equivalence points can be observed. Explain the most common reasons for this. Calculate the molarity of a monoprotic acid HA whose titration endpoint occurs after V ml of strong base of a given concentration has been added. The objective of an acid-base titration is to determine \(C_a\), the nominal concentration of acid in the solution. In its simplest form, titration is carried out by measuring the volume of the solution of strong base required to complete the reaction \[\ce{H_nA + n OH- → n A- + n H_2O} \label{0-1}\] in which \(n\) is the number of replaceable hydrogens in the acid. The point at which this reaction is just complete is known as the equivalence point . This is to be distinguished from the end point , which is the value we observe experimentally. A replaceable hydrogen atom (sometimes called an "acidic" hydrogen) is one that can be donated to a strong base — that is, to an OH – ion. Thus in acetic acid HCOO H , only the hydrogen in the carboxyl group is considered "replaceable". What we actually measure, of course, is the volume of titrant delivered by the burette. Learning to properly control the stopcock at the bottom of the burette usually requires some instruction and practice, as does the reading of the volume. For highly precise work, the concentration of the titrant itself must be determined in a separate experiment known as "standardization". Understanding Titration Curves A plot showing the pH of the solution as a function of the quantity of base added is known as a titration curve . These plots can be constructed by plotting the pH as a function of either the volume of base added, or the equivalent fraction \(ƒ\) which is simply the number of moles of base added per mole of acid present in the solution. In most of the titration curves illustrated in this section, we plot pH as a function of \(ƒ\). It's worth taking some time to thoroughly familiarize yourself with the general form of a titration curve such as the one shown below, in which a weak acid HA is titrated with a strong base, typically sodium hydroxide. The equivalence point occurs at the pH at which the equivalent fraction ƒ of base added is unity. At this point, the reaction \[HA + OH^– → AB^– + H_2O \label{1-1}\] is stoichiometrically complete; a solution initially containing n moles of a monoprotic acid HA will now be identical to one containing the same number of moles of the conjugate base A – . At the half-equivalence point ƒ = 0.5, the concentrations of the conjugate species are identical: [HA] = [A – ]. This, of course corresponds to a buffer solution (hence the relatively flat part of the curve) whose pH is the same as pK a . As base is added beyond ƒ = 1, the pH begins to level off, suggesting that another buffered system has come into play. In this case it involves the solvent (water) and hydroxide ion: {H 2 O} ≈ {OH-}. A similar effect is seen at the low-pH side of the curve when a strong acid is titrated, as in the plot for the titration of HCl below. In this case, the buffering is due to {H 3 O + ) ≈ {H2O}. How can this be? Surely, the concentration of OH – , even when the pH approaches 14, cannot be anything like that of [H 2 O] which will be about 55.5 M in most solutions! This fine point (along with the mention of H 2 O/OH – buffering) is rarely mentioned in elementary courses because the theory behind it involves some rather esoteric elements of solution thermodynamics. However, in case you are curious, note that the curly brackets in {H 2 O} ≈ {OH – } denote activities, not concentrations. And by convention the activity of a pure liquid (H 2 O in this case) is unity. At a pH of around 12, pOH = 2, [OH – ] = .01. At this rather high ion concentration, {OH – } will be somewhat smaller than this, but the two activities will be similar enough to produce the buffering effect we observe. The pH of the solution at its equivalence point will be 7 if we are titrating a strong acid with strong base, as in HCl + NaOH → H 2 O + NaCl. However, if the acid is weak, as in the above plot, the solution will be alkaline. This pH can be calculated from C b and K b in a manner exactly analogous to that used for calculating the pH of a solution of a weak acid in water. It is important to understand that the equivalent fraction ƒ of base that must be added to reach the equivalence point is independent of the strength of the acid and of its concentration in the solution. The whole utility of titration as a means of quantitative analysis rests on this independence; we are in all cases measuring only the total number of moles of “acidic” hydrogens in the sample undergoing titration. Acid and base strengths determine the shape of the curve Although the strength of an acid has no effect on the location of the equivalence point, it does affect the shape of the titration curve and can be estimated on a plot of the curve. The weaker the acid being titrated, the higher the initial pH (at ƒ=0), and the smaller will be the vertical height of the plot near the equivalence point. As we shall see later, this can make it difficult to locate the equivalence point if the acid is extremely weak. Estimating the acid strength As shown in plot above, the pK a of a weak acid can be estimated by noting the pH that corresponds to the half-titration point ƒ = 0.5. Recalling that the pH is controlled by the ratio of conjugate species concentrations \[pH = pK_a + \log \dfrac{[A^]}{[HA]} \label{1-2}\] it will be apparent that this equation reduces to pH = pK a when the titration is half complete (that is, when [HA] = [A – ]), the pH of the solution will be identical to the pK a of the acid. This equation does not work for strong acids owing to the strong buffering that occurs at the very low pH at which ƒ = 0.5. As indicated here, the buffering has nothing to do with the acid HCl itself (which does not exist as such in water), but rather with its dissociation products H 3 O + and OH – , "the strongest acid and base that can exist in water." Monoprotic titration curves The following two principles govern the detailed shape of a titration curve: The stronger the acid or base, the greater will be the slope of the curve near the equivalence point; The weaker the acid or base, the greater will the deviation of the pH from neutrality at the equivalence point. It is important to understand the reasons for these two relations. The second is the simplest to explain. Titration of an acid HA with a base such as NaOH results in a solution of NaA; that is, a solution of the conjugate base A – . Being a base, it will react with water to yield an excess of hydroxide ions, leaving a slightly alkaline solution. Titration of a weak base with an acid will have the opposite effect. The extent of the jump in the pH at the equivalence point is determined by a combination of factors. In the case of a weak acid, for example, the initial pH is likely to be higher, so the titration curve starts higher. Further, the weaker the acid, the stronger will be its conjugate base, so the higher will be the pH at the equivalence point. These two factors raise the bottom part of the titration curve. The upper extent of the curve is of course limited by the concentration and strength of the titrant. These principles are clearly evident in the above plots for the titrations of acids and bases having various strengths. Notice the blue curves that represent the titration of pure water (a very weak acid) with strong acid or base. Monoprotic titration curve gallery When both the titrant and sample are "strong", we get long vertical plots at ƒ = 1. Adding even half a drop of titrant can take us across the equivalence point! When one of the reactants is weak, the pH changes rapidly at first until buffering sets in. ← In (C), the onset of H 2 O/OH- buffering near ƒ=1 makes the equivalence point more difficult to locate. "Weak/weak" titrations tend to be problematic as the buffered regions move closer to ƒ=1. The equivalence point pH of 7 in these examples reflects the near-equality of pK a and pK b of the reactants. Dealing with very weak acids It can be difficult to reliably detect the equivalence point in the titration of boric acid ( pK a = 9.3) or of other similarly weak acids from the shape of the titration curve*. *An interesting student laboratory experiment that employs an auxiliary reagent (mannitol) to make boric acid stronger and thus more readily titratable was described in J. Chem Ed . 2012, 89, 767-770. The problem here is that aqueous solutions are buffered against pH change at very low and very high pH ranges. An extreme example occurs in the titration of pure water with a strong acid or base. At these extremes of pH the concentrations of H 3 O + and of OH – are sufficiently great that a competing buffer system (either H 3 O + /H 2 O or H 2 O/OH – , depending on whether the solution is highly acidic or highly alkaline) comes into play. Why we usually use a "strong" titrant The above plots clearly show that the most easily-detectable equivalence points occur when an acid with is titrated with a strong base such as sodium hydroxide (or a base is titrated with a strong acid.) In practice, many of the titrations carried out in research, industry, and clinical practice involve mixtures of more than one acid. Examples include natural waters, physiological fluids, fruit juices, wine making, brewing, and industrial effluents. For titrating these kinds of samples, the use of anything other than a strong titrant presents the possibility that the titrant may be weaker than one or more of the "stronger" components in the sample, in which case it would be incapable of titrating these components to completion. In terms of proton-free energies, the proton source (the acidic titrant) would be unable to deliver an equivalent quantity of protons to the (stronger) component of the mixture. Polyprotic acids There will be as many equivalence points as there are replaceable hydrogens in an acid. Thus in the extremely important carbonate system , equivalence points are seen at both ƒ=1 and ƒ=2: In general, there are two requirements for a clearly discernible jump in the pH to occur in a polyprotic titration: The successive K a 's must differ by several orders of magnitude; The pH of the equivalence point must not be very high or very low. Separation of successive equilibrium constants The effect of the first point is seen by comparing the titration curves of two diprotic acids, sulfurous and succinic. The appearance of only one equivalence point in the latter is a consequence of the closeness of the first and second acid dissociation constants. The pK a 's of sulfurous acid (below, left) are sufficiently far apart that its titration curve can be regarded as the superposition of those for two independent monoprotic acids having the corresponding K a 's. This reflects the fact that the two acidic –OH groups are connected to the same central atom, so that the local negative charge that remains when HSO 3 – is formed acts to suppress the second dissociation step. *It can be shown that in the limit of large n , the ratio of K 1 /K 2 for a symmetrical dicarboxylic acid HOOC-(CH 2 ) n - COOH converges to a value of 4. In succinic acid , the two –COOH groups are physically more separated and thus tend to dissociate independently*. Inspection of the species distribution curves for succinic acid (above, right) reveals that the fraction of the ampholyte HA can never exceed 75 percent. That is, there is no pH at which the reaction H 2 A → HA – + H + can be said to be “complete” while at the same time the second step HA – → A 2– + H + has occurred to only a negligible extent. Thus the rise in the pH that would normally be expected as HA is produced will be prevented by consumption of OH – in the second step which will be well underway at that point; only when all steps are completed and hydroxide ion is no longer being consumed will the pH rise. Two other examples of polyprotic acids whose titration curves do not reveal all of the equivalence points are sulfuric and phosphoric acids. Owing to the leveling effect, the apparent K a1 of H 2 SO 4 is so close to K a2 = 0.01 that the effect is the same as in succinic acid, so only the second equivalence point is detected. In phosphoric acid , the third equivalence point (for HPO 4 2 – ) is obscured by H 2 O-OH – buffering as explained previously. Detection of the equivalence point Whether or not the equivalence point is revealed by a distinct "break" in the titration curve, it will correspond to a unique hydrogen ion concentration which can be calculated in advance. There are many ways of determining the equivalence point of an acid-base titration. Indicators Don't overshoot the equiv point! The traditional method of detecting the equivalence point has been to employ an indicator dye, which is a second acid-base system in which the protonated and deprotonated forms differ in color, and whose pK a is close to the pH expected at the equivalence point. If the acid being titrated is not a strong one, it is important to keep the indicator concentration as low as possible in order to prevent its own consumption of OH – from distorting the titration curve. The observed color change of an indicator does not take place sharply, but occurs over a range of about 1.5 to 2 pH units. Indicators are therefore only useful in the titration of acids and bases that are sufficiently strong to show a definite break in the titration curve. Some plants contain coloring agents that can act as natural pH indicators. These include cabbage (shown), beets, and hydrangea flowers. For a strong acid - strong base titration, almost any indicator can be used, although phenolphthalein is most commonly employed. For titrations involving weak acids or bases, as in the acid titration of sodium carbonate solution shown here, the indicator should have a pK close to that of the substance being titrated. When titrating a polyprotic acid or base, multiple indicators are required if more than one equivalence point is to be seen. The pK a s of phenolphthalein and methyl orange are 9.3 and 3.7, respectively. Potentiometry: Use a pH meter The pH meter detects the voltage produced when the H + ions in the solution displace Na + ions from a thin glass membrane that is dipped into the solution. A more modern way of finding an equivalence point is to follow the titration by means of a pH meter. Because it involves measuring the electrical potential difference between two electrodes, this method is known as potentiometry . Until around 1980, pH meters were too expensive for regular use in student laboratories, but this has changed; potentiometry is now the standard tool for determining equivalence points. Improved plotting methods Plotting the pH after each volume increment of titrant has been added can yield a titration curve as detailed as desired, but there are better ways of locating the equivalence point. The most common of these is to take the first or second derivatives of the plot: d (pH)/d V or d 2 (pH)/d V 2 (of course, for finite increments of pH and volume, these terms would be expressed as Δ(pH)/Δ V and Δ 2 (pH)/Δ V 2 .) A second-derivative curve locates the inflection point by finding where the rate at which the pH changes is zero. The differential plot , showing rate-of-change of pH against titrant volume, locates the inflection point which is also the equivalence point In a standard plot of pH-vs-volume of titrant added, the inflection point is located visually as half-way along the steepest part of the curve. The idealized plots shown above are unlikely to be seen in practice. When the titration is carried out manually, the titrant is added in increments, so even the simple titration curve must be constructed from points subject to uncertainties in volume measurement and pH (especially if the latter is visually estimated by color change of an indicator.) If this data is then converted to differential form, these uncertainties add a certain amount of "noise" to the data. A second-derivative plot uses pH readings on both sides of the equivalence point, making it easier to locate in the presence of noise. Locating the equivalence point depends very strongly on correct reading of only one or two pH readings near the top of the plot. A simple curve, plotted from a small number of pH readings, will not always unambiguously locate the equivalence point. The "noise" in differential plots can usually be minimized by keeping the titrant and analyte concentrations above 10 –3 M . Some other ways of following a titration Monitoring the pH by means of an indicator or by potentiometry as described above are the standard ways of detecting the equivalence point of a titration. However, we have already seen that in certain cases involving polyprotic acids or bases, some of the equivalence points are obscured by their close proximity to others, or by the buffering that occurs near the extremes of the pH range. Similar problems can arise when the solution to be titrated contains several different acids, as often happens when fluids connected with industrial processes must be monitored. Take the temperature! Acid-base neutralization reactions HA + B → A – + BH + are always exothermic; when protons fall from their level in the acid to that in the base, most of the free energy drop gets released as heat. If the acid and base are both strong (i.e., totally dissociated), the enthalpy of neutralization for the reaction \[\ce{H3O+ + OH- → 2 H2O}\] is -68 kJ mol –1 . See this Wikipedia page for more on thermometric titrations, including many examples. Note also the video on this topic in the "Videos" section near the end of this page. Thermometric titrations are not limited to acid-base determinations; they can also be used to follow precipitation-, complex formation-, and oxidation-reduction reactions. They can be used with polyprotic acids or bases, and with mixtures containing more than one acid or base. They are able to follow acid-base titrations that must be carried out in non-aqueous solvents, where other titration methods are not possible. A typical thermometric titration curve consists of two branches, beginning with a steep rise in temperature as the titrant being added reacts with the analyte, liberating heat. Once the equivalence point is reached, the rise quickly diminishes as heat production stops. Then, as the mixture begins to cool, the plot assumes a negative slope. Although a rough indication of the equivalence point can be estimated by extrapolating the linear parts of the curve (blue dashed lines), the differential methods described above are generally preferred. Follow the electrolytic conductance Acids and bases are electrolytes , meaning that their solutions conduct electric current. The conductivity of such solutions depends on the concentrations of the ions, and to a lesser extent, on the nature of the particular ions. Any chemical reaction in which there is a change in the total quantity of ions in the solution can usually be followed by monitoring the conductance. Acid-base titrations fall into this category. Consider, for example, the titration of hydrochloric acid with sodium hydroxide. This can be described by the equation \[\ce{H^{+} + Cl^{–} + Na^{+} + OH^{–} → H2O + Na^{+} + Cl^{–}} \label{2-1}\] which shows that two of the four species of ions being combined disappear at the equivalence point. During the course of the titration, the conductance of the solution falls as H + and Cl – ions are consumed. At the equivalence point the conductance passes through a minimum, and then rises as continued addition of titrant adds more Na + and OH – ions to the solution. Each kind of ion makes its own contribution to the solution conductivity. If we could observe the contribution of each ion separately, see that the slopes for H + and OH – are much greater. This reflects the much greater conductivities of these ions owing to their uniquely rapid movement through the solution by hopping across water molecules. However, because the conductances of individual ions cannot be observed directly, conductance measurements always register the total conductances of all ions in the solution. The change in conductance that is actually observed during the titration of HCl by sodium hydroxide is the sum of the ionic conductances shown above. For most ordinary acid-base titrations, conductimetry rarely offers any special advantage over regular volumetric analysis using indicators or potentiometry. This is especially true if the acid being titrated is weak; if the pK a is much below 2, the rising salt line (Na + when titrating with NaOH) will overwhelm the fall in the contribution the small amount of H + makes to the conductance, thus preventing any minimum in the total conductance curve from being seen. However, in some special cases such as those illustrated below, conductimetry is the only method capable of yielding useful results. These examples illustrate two unique capabilities of conductimetric titrations: (left) Titration of a mixture of two acids and (right) Titration of a strong polyprotic acid → Because pure H 2 SO 2 is a neutral molecule, it makes no contribution to the conductance, which rises to a maximum at the equivalence point. Automating the process In four years of college lab sessions, many Chemistry majors will likely carry out fewer than a dozen titrations. However, in the real world, time is money, and long gone are the days when technicians were employed full time just to titrate multiple samples in such enterprises as breweries, food processing (such as blending of canned orange juice), clinical labs, and biochemical research. Titration calculations A titration is carried out by adding a sufficient volume \Vo of the titrant solution to a known volume \Vt of the solution being titrated. This addition continues until the end point is reached. The end point is our experimental approximation of the equivalence point at which the acid-base reaction is stoichiometrically complete (ƒ = 1). The quantity we actually measure at the end point is the volume V_ep of titrant delivered to the solution undergoing titration. The solution being titrated is often referred to as the analyte (the substance being "analyzed") or, less commonly, as the titrand . We shall employ the latter term in what follows. In a simple titration of a monoprotic acid HA by a base B, the equivalence point corresponds to completion of the reaction \[\ce{HA + B → A^{–} + BH^{+} }\label{3-1}\] where equimolar quantities of HA and B have been combined; that is, M HA = M B (where M represents the number of moles.) Recall that the number of moles is given by the product of the volume and concentration: L × mol L –1 = mol, mL x mMol ml –1 = mmol. Because we are measuring the volume of titrant rather than the number of moles, we need to use its concentration to link the two quantities. So if we are titrating the base B with acid HA, the end point is reached when a volume V of HA has been added. The number of moles of HA we have added at the end point is given by the product of its volume and concentration \[M_{HA} = V_{HA} \times C_{HA} \label{3-2}\] And because the reaction \ref{3-1} is now complete, M HA = M B ; thus, at the end point of any monoprotic titration, \[V_{HA} C_{HA} = V_B C_B \label{3-3}\] Equation \ref{3-3} is i mportant! In any titration, both the volume and the concentration of the titrant are known, so the unknown concentration is easily calculated. In titrations carried out in the laboratory, the titrant is delivered by a burette that is usually calibrated in milliliters, so it is more convenient to express M HA in millimoles and C HA in millimoles/mL (mMol ml –1 ); note that the latter is numerically the same as moles/L . Example \(\PageIndex{1}\) 50.0 mL of 0.100 M hydrochloric acid is titrated with 0.200 M sodium hydroxide. What volume of NaOH solution will have been added at the equivalence point? Solution First, find the number of moles of HCl in the titrand: \[\begin{align*} M_{\ce{HCl}} &= C_{\ce{HCl}} \times V_{\ce{HCl}} \\[4pt] &= (0.100\, mMol\, mL^{–1}) \times (50.0\, mL) \\[4pt] &= 5.0 \,mMol\,\ce{HCl} \end{align*}\] This same number of moles of NaOH must be delivered by the burette in order to reach the equivalence point ( i.e. , M NaOH = 5.0 mMol.) The volume of NaOH solution M NaOH required is V NaOH = ( M NaOH / C NaOH ) = (5.0 mMol) / (0.200 mMol/mL) = 25 mL |
Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/12%3A_Solids_and_Modern_Materials/12.02%3A_Materials_for_Structure/12.2.01%3A_Concrete | The most popular building material in the world today, concrete was first developed and exploited by the ancient Romans, who used it to create monumental public spaces such as aqueducts , churches, baths, and the Colosseum. Although concrete is a durable material with many building applications, the design of Roman concrete structures reinforced Roman ideas about social status and imperial power. This chapter explores the rich history of concrete and its legacy in the modern world, touching upon the role of concrete in ancient Rome, today’s technical advances in concrete construction, and concrete’s environmental drawbacks. The chapter also examines how concrete construction is shaped by societal ideals today, just as it was by societal ideas in ancient Rome. Pourable, moldable, durable, waterproof, and relatively easy and inexpensive to manufacture, concrete is the world’s most popular building material. We live, work, and play on and in buildings and roads constructed from it. Architects exploit its properties to create artistic tours de force as well as utilitarian monuments Origin of Concrete As far back as the sixth millennium (6000–5000 BCE), the ancient Mesopotamians knew that heating calcium carbonate, a substance occurring naturally in limestone rocks, creates a new substance, known today as quicklime, in a process described chemically as CACO₃ + heat(1000° C) =CO₂ + CaO. This chemical reaction releases carbon dioxide into the atmosphere (more on this later). The resulting material, when mixed with water, bonds to other surfaces. The early residents of Çatalhöyük, an ancient city in modern-day Turkey used this substance to coat their walls, providing a surface for painted decoration. Modern Concrete Despite more highly mechanized manufacturing techniques, the process of creating concrete still relies on the basic chemical reaction exploited by the Romans, the reaction that releases carbon dioxide, a major pollutant, into the atmosphere. The world’s yearly production of concrete has been rising steadily and today is over four billion tons. Concrete production is responsible for nearly eight percent of the anthropogenic (human-made) greenhouse gases released into the air. By some estimates, the external climate and health damages caused by concrete production amount to approximately 74 percent of the value of the industry itself, putting the external costs of concrete higher than natural gas and oil and only slightly less than coal. As even more concrete is produced, the amount of carbon dioxide will also rise unless we develop smarter, greener methods of production. Concrete manufacturers recognize this problem, but any solution will have to be cost effective for worldwide adoption to take place. The first step in making concrete is to produce cement as discussed in the video below. An additional cost of manufacturing concrete is the need for sand as a component of the finished product. Today, sand is becoming an increasingly rare and sought-after commodity. As the Romans knew, only sand worn by water (river or ocean sand), not sand that has been exposed to the elements (desert sand), is suitable. While there are problems are connected with new construction, older concrete structures pose other issues. Most modern concrete is reinforced by metal bars that lead to eventual cracking as the metal expands and contracts. Can we recycle ruined concrete buildings? How can we stabilize and repair buildings? Engineers are working to develop new technologies that can sense imminent structural issues before a bridge or building collapses. To prevent damage in new construction, engineers developed Smartcrete, a form of concrete that can repair itself. New methods of concrete construction such as Ductal, which requires no metal, and the use of cloth as a framing material are also potential answers to this problem. Although some striking modern architectural monuments have been built from concrete—the Guggenheim Museum in New York, for example—many consider modern concrete stark and ugly because of the many utilitarian buildings constructed from it. Because, unlike the Romans, we can make concrete with a smooth surface, it is not necessary for us to cover it with other materials. New concrete technologies continue to emerge. Among the possibilities is concrete laid by robots. Acknowledging the close link between buildings and social structure allows us to wonder if there might be hidden costs to this technology. What types of workforce changes would occur if machines took over this aspect of building construction? Would using robots free humans to do other things or would it merely eliminate a large category of jobs? Future of Concrete Building construction remains the primary use of concrete today. Might there not be other uses for such a versatile material beyond architecture? Architects, engineers, and others are beginning to address this question. For example, kitchen designers are using concrete for countertops, taking advantage of its durability, cost effectiveness, modern appearance, and ability to resist water. People are even considering how ancient Roman concrete might be used to address sea level rise associated with global warming. To encourage thinking about a common material in a different light, the American Society for Civil Engineering sponsors an annual concrete canoe contest. This challenge forces students to broaden their ideas about possible applications for this common material. Engineering students from across the US attempt to build and race a concrete canoe. They are judged not only on the results of the race, but also on their design concept. Concrete is certainly not the first material that comes to mind when thinking about canoes, although it is waterproof. But a concrete canoe suggests that if we think beyond the limits imposed on the use of concrete by our societal worldview and historical traditions, we may be able to find newer and more effective ways to use this versatile material. Contributed By Excerpted from a longer piece by Mary Ann Eaverly, University of Florida |
Courses/Saint_Marys_College_Notre_Dame_IN/CHEM_431%3A_Inorganic_Chemistry_(Haas)/CHEM_431_Readings/09%3A_Acid-Base_and_Donor-Acceptor_Chemistry/9.03%3A_Brnsted-Lowry_Concept | The Brønsted-Lowry acid-base concept The Brønsted-Lowry acid-base concept overcomes the Arrhenius system's inability to describe reactions that take place outside of aqueous solution by moving the focus away from the solution and onto the acid and base themselves. It does this by redefining acid-base reactivity as involving the transfer of a hydrogen ion, \(H^+\), between an acid and a base. Specifically, a Brønsted acid is a substance that loses an \(H^+\) ion by donating it to a base. This means that a Brønsted base is defined as a substance which accepts \(H^+\) from an acid when it reacts. Because the Brønsted-Lowry concept is concerned with \(H^+\) ion transfer rather than the creation of a particular chemical species, it is able to handle a diverse array of acid-base concepts. In fact, from the viewpoint of the Brønsted-Lowry concept, Arrhenius acids and bases are just a special case involving hydrogen ion donation and acceptance involving water. Arrhenius acids donate \(H^+\) ion to water, which acts as a Brønsted base in accepting it to give \(H_3O^+\): \[\label{ 6.3.1} \] Similarly, Arrhenius bases act as Brønsted bases in accepting a hydrogen ion from the Brønsted acid water: \[ \nonumber \] In this way it can be seen that Arrhenius acids and bases are defined in terms of their causing hydrogen ions to be donated to and abstracted from water, respectively, while Brønsted acids and bases are defined in terms of their ability to donate and accept hydrogen ions to and from anything. Becasue the Brønsted-Lowry concept can handle any sort of hydrogen ion transfer it readily accommodates many reactions that Arrhenius theory cannot, including those that take place outside of water, such as the reaction between gaseous hydrochloric acid and ammonia: \[ \nonumber \] The classification of acids as strong or weak usually refers to their ability to donate or abstract hydrogen ions to or from water to give \(H_3O^+\) and \(OH^-\), respectively, i.e., their Arrhenius acidity and basicity. However, acids and bases may be classified as strong and weak under the Brønsted-Lowry definition based on whether they completely transfer or accept hydrogen ions; it is just that in this case it is important to specify the conditions under which a given acid or base acts strong or weak. For example, acetic acid acts as a weak acid in water but is a strong acid in triethylamine, since in the latter case it completely transfers a hydrogen ion to triethylamine to give triethylammonium acetate. Alternatively, the acidity or basicity of a compound may be specified using a thermodynamic scale like the Hammett acidity . Conjugate Acids and Bases By redefining acids and bases in terms of hydrogen ion donation and acceptance, the Brønsted-Lowry system makes it easy to recognize that when an acid loses its hydrogen ion it becomes a substance that is capable of receiving it back again, namely, a base. Consider, for example, the base dissociation of ammonia in water. When ammonia acts as a Brønsted base and receives a hydrogen ion from water, ammonium ion and hydroxide are formed: \[ \nonumber \] The ammonium ion is itself a weak acid that can undergo dissociation: \[ \nonumber \] In this case ammonia and ammonium ion are acid-base conjugates. In general acids and bases that differ by a single ionizable hydrogen ion are said to be conjugates of one another. The strengths of conjugates vary reciprocally with one another, so the stronger the acid the weaker the base and vice versa. For example, in water, acetic acid acts as a weak Brønsted acid: \[ \nonumber \] and acetic acid's conjugate base, acetate, acts as a weak Brønsted base. \[ \nonumber \] However, in liquid ammonia acetic acid acts as a strong Brønsted acid: \[ \nonumber \] while its conjugate base, acetate, is neutral. \[ \nonumber \] The reciprocal relationship between the strengths of acids and their conjugate bases has several consequences: Under conditions when an acid or base acts as a weak acid or base its conjugate acts as weak as well. Conversely, when an acid or base acts as a strong acid or base its conjugate acts as a neutral species. When a Brønsted acid and base react with one another, the equilibrium favors formation of the weakest acid-base pair. That is why the acid-base reaction between acetic acid and ammonia in liquid ammonia proceeded to give the weak acid ammonium ion and neutral acetate. This consequence is particularly important for understanding the behavior of acids and bases in nonaqueous solvents, as illustrated by the following example. Example \(\PageIndex{1}\) Can a solution of lithium diisopropylamide in heptane be used to form lithium cyclopentadienide? The \(pK_a\) of cyclopentadiene and diisopropylamine are ~15 and 40, respectively, and the proposed reaction is as follows: Solution: Since cyclopentadiene is a stronger acid than diisopropylamine (the stronger the acid the lower the \(pK_a\)) the equilibrium will favor protonation of the diisopropylamine by cyclopentadiene. Consequently addition of a heptane solution of lithium diisopropylamide to monomeric cyclopentadiene should give lithium cyclopentadienide. |
Bookshelves/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/21%3A_d-Block_Metal_Chemistry_-_The_First_Row_Metals/21.11%3A_Group_10_-_Nickel/21.11D%3A_Nickel(I) | Welcome to the Chemistry Library. This Living Library is a principal hub of the LibreTexts project , which is a multi-institutional collaborative venture to develop the next generation of open-access texts to improve postsecondary education at all levels of higher learning. The LibreTexts approach is highly collaborative where an Open Access textbook environment is under constant revision by students, faculty, and outside experts to supplant conventional paper-based books. Campus Bookshelves Bookshelves Learning Objects |
Bookshelves/Biological_Chemistry/An_Introduction_to_Medicinal_Chemistry_and_Molecular_Recognition_(de_Araujo_et_al.)/01%3A_Chapters/1.02%3A_How_are_Drugs_Discovered | “The best way to discover a new drug is to start with an old one.” – James Black 2.1 Natural Products and Natural Product Analogs Drug discovery is not always carried out at a molecular level. Herbs, berries, roots, barks, and other natural products have been go-to ailments for humans since antiquity. These natural medicines were often ‘prescribed’ without any prior knowledge of their active constituents. Over time, multiple ancient populations accumulated their own pharmacopeia for natural products , several of which have served as inspiration for the isolation and/or synthesis of more innovative and potent compounds. These remedies derived from plants, animals, or microorganisms have foundational properties that transformed the landscape of modern medicine. Figure 2.1 highlights currently employed drugs that were derived from natural sources along with a selection of case studies below. 2.1.1 Aspirin The discovery of aspirin and its growth to one of the most widely used over-the-counter analgesic (“pain-relieving”) drugs, spans multiple millennia, and involves many landmark developments in isolation and synthetic chemistry. Different civilizations independently converged on the use of willow trees, particularly the white willow ( Salix alba ) and meadowsweet ( Filipendula ulmaria ), for fever-reducing and pain-relieving properties. The earliest documented use can be found in the Ebers Papyrus, an ancient Egyptian medical text (3000 BC) although Reverend Edward Stone (1763) is often credited with the first scientific study in demonstrating the therapeutic properties . In 1828, Professor Joseph Buncher (Germany) was able to extract the active ingredient (which he named salicin), and it was later refined to salicylic acid by French chemist Charles Gerhardt (1853). Dr. Felix Hoffman (a German chemist at Bayer) was able to acetylate the compound, which alleviated several adverse side effects associated with salicin and salicylic acid, resulting in the compound acetylsalicylic acid (Figure 2.2). This landmark discovery represents the first synthetic drug from pharma, which galvanized a new era in therapeutics. Bayer registered the drug under the name Aspirin in 1899, with the namesake as a derivatization from yet another natural species of meadowsweet ( Spirea ulmaria ). 2.1.2 Morphine and Codeine The powerful analgesic properties of the poppy plant ( Papaver somniferum ) were also known by multiple civilizations, with archaeological evidence of human use as far back as 5000 BC in the Mediterranean region. The seed capsules of poppy plants contain a milky substance called latex that contains opium. Opium is comprised of a mixture of chemical compounds called alkaloids (a loosely defined subset of naturally occurring organic compounds containing at least one nitrogen). In 1804, the German pharmacist, Friedrich Sertürner, isolated morphine (Figure 2.3) from opium (with the name emerging from Morpheus, the Greek god of dreams). Similarly, French chemist Pierre-Jean Robiquet was able to isolate codeine (parent compound of morphine) as another component from opium in 1832. Codeine was named after the Greek word “kodeia” referring to poppy head. Other euphoria-inducing drugs can also be synthesized from opium including heroin and oxycodone, although these are all tightly controlled substances due to their heavily addictive properties. 2.1.3 Penicillin In 1928, Scottish physician, Alexander Fleming, was conducting research on the growth properties of the bacteria, Staphylococcus aureus . He returned from a two-week vacation and discovered that his cultures were contaminated by mold, but also that the bacteria could not grow in close proximity to the fungi on the culture plate. The fungi was determined to be Penicillium notatum. After culturing it, Fleming confirmed that the growing broth harboured an anti-bacterial substance that was effective on Gram-positive pathogens but not on Gram-negative bacteria. In 1929, he re-named the broth, penicillin, in place of the more colloquial name, “mold juice”. The chemical structure was determined and purified by another team (Howard Florey and Edward Chain) in 1940, followed by the first treatment in patients in 1942. (Figure 2.4) Its uptake into the clinics was largely accelerated during World War II with its success in tackling bacterial infections and the trio of scientists awarded the Nobel Prize in Physiology or Medicine in 1945. 2.2 Modern Rational Drug Design One of the common themes evident in drug discovery via natural products is that it is grounded in astute observation of often-serendipitous circumstances, followed by empirical trial-and-error methods, and ultimately isolation of the key active ingredient. Over the early 20 th century, this framework for drug discovery transitioned to a more step-wise hypothesis-driven approach with the goal of systematically developing, synthesizing, and exploring new (non-natural) composition-of-matter. This reform in therapeutic methodology was accelerated by the creation of the US FDA in 1938, followed by substantial administrative changes in the Kefauver-Harris Drug Amendments of 1962, as a result of the thalidomide drug scandals. The US FDA oversees public health in the US including the approval of new drugs and medical devices. The Canadian counterpart is referred to as Health Canada, which was instituted in 1919 in response to the Spanish Flu outbreak, and later reformed in 1993. Both Health Canada and the US FDA collaborate closely (via the Regulatory Cooperation Council Joint Forward Plan) to harmonize several drug-related policies, although they are separate entities and can/will uphold different guidelines. 2.3 Typical Stages of the Drug Discovery Pipeline For a medicinal chemist, the stages of the drug discovery pipeline are often represented in a linear, forward-moving strategy, although the overall process is highly iterative and can involve multiple cycles of optimization and re-visiting of each stage (Figure 2.5). 2.3.1 Stage 1: Target Selection In the first stage of drug discovery, a target or specific disease is selected. A target refers to a biomolecule where pharmacological or genetic intervention would (directly or indirectly) disrupt a critical pathological biochemical process. For example, >95% of chronic myeloid leukemia (CML) cases result from a random genetic mutation that creates the fusion protein, BCR-ABL kinase. BCR-ABL kinase is therefore a target in CML, and is often treated with the drug, imatinib (or one of its analogues). (Figure 2.6) One disease can have more than one target, and similarly, one target may be relevant in more than one disease. The rationale for selecting a target is usually supported by an array of genetic, biological, and physiological data that suggest it will be efficacious to knock-down while not compromising overall safety to the patient. This is generally referred to as target validation . For example, disrupting the enzyme Protein Kinase A (PKA) is not a feasible option in the rare fibrolamellar hepatocellular (FLC), despite the fact that its variants have been shown to drive this cancer. This is because of the importance of PKA in the heart for maintaining appropriate cardiac functioning. However, there are many cases where background biological information is not readily available, or the target is not well-studied. In these cases, a greater importance is placed on pre-clinical investigations and safety trials. 2.3.2 Stage 2: Hit Identification and Hit-to-Lead / Lead Discovery Following selection of the disease/target, the next stage involves identifying a chemical compound that can interact with the target. Usually, a series of compounds are evaluated in an experiment that involves the desired target. This experiment, called an assay , can involve a variety of tools to quantify the interaction between the compound and biomolecular target; for example, monitoring the interaction (binding event) via nuclear magnetic resonance (NMR), quantifying a change in the target activity via fluorescence, or observing cell death via imaging. The types of assays will vary, depending on both the target and the nature/origin of the compounds being tested, and are often custom-designed for each drug discovery program. The output of the assay will usually provide an initial group of compounds that can engage with the target, which are referred to as hit compounds . The activity of these hit compounds can vary substantially, but they serve as general starting points to create additional generations of molecules with better activity. The goal of subsequent organic synthesis campaigns is to improve the hit compounds and create/select a lead compound . A lead compound is a compound that: Interacts with the target to achieve the desired biological activity. Is amenable to synthetic modifications. Can reach the target once administered. Distinguishing and selecting lead compounds will be discussed later in this text. However, an initial hit molecule differs from a final lead molecule in that it usually has lower biological activity, and has not been validated to reach the target upon administration. As such, these two properties are classical outcomes of hit-to-lead or lead discovery campaigns, where the goal is to chemically synthesize new molecules that build and optimize a hit compound into a lead molecule. 2.3.3 Stage 3: Lead Optimization Although identifying a lead molecule is a monumental step forward in the drug discovery pipeline, it is often chemically distant from the final drug candidate. This is usually because the lead molecule has specific properties that can limit its efficacy. For example, the structure-activity relationship (SAR) optimization rounds from the previous hit-to-lead studies may have identified specific functional groups required for target binding (biological activity), but the oral absorption of the compound may not be sufficient, as a result of these chemical groups. Therefore, at the lead optimization stage, the goal is often to preserve the structural elements that maintain potency of the lead compound, while optimizing a new molecule with improved physical properties for bioavailability. It is conceivable that potency of the newly developed candidate molecules may be reduced, as other parameters are optimized. This is a balancing act to fine-tune the optimal overall properties between potency, efficacy, permeability, and stability. For example, Bayer (in partnership with Onyx Pharmaceuticals) initially screened ~200,000 compounds to explore the target Raf1 kinase, which led to the initial hit molecule 1 (Figure 2.7) with a half-maximal inhibitory concentration (IC 50 ) of 17 µM. The urea and phenyl group were identified to be critical for potency and another ~1000 bis-aryl urea analogs were generated that led to an isoxazole derivative as the lead compound ( 2 ) (Figure 2.7). The compound underwent lead optimization to the final development candidate BAY 43-9006 (sorafenib) with a potent affinity for Raf-1 (IC 50 = 6 nM). Crystallographic studies (which were determined retrospectively) revealed that the urea moiety was crucial in the SAR, due to the formation of two critical H-bonds (with the backbone aspartic acid residue of the DFG loop and the glutamate side chain of the αC helix of the target protein – structural elements that will be discussed in Section 2.5.1). Additional key features from BAY-43-9006 that were identified include the 4-pyridyl ring that mimics the natural adenine scaffold and the chlorobenzene ring which interacts with a hydrophobic pocket behind the orthosteric ATP binding site. Although the lack of structural information did not allow for visualization of these effects at the time of compound synthesis, these features improved the binding capacity of the drug which was represented by the lower IC 50 values and indicated that they were important in the lead optimization process. 2.3.4 Stage 4: Preclinical / Clinical Development & Regulatory Approval The focus of this resource is on medicinal chemistry, and following completion of lead optimization and pre-clinical trials, the compound is generally optimized and there is a diminished motivation for synthesizing newer compounds. However, the subsequent regulatory steps are briefly summarized below. Following lead optimization, the top developmental candidate needs to be evaluated in more complex models. This usually involves a mammalian model such as a mouse or rat that has been modified to express a phenotype representing the disease state. For example, exploring cancer tumors can involve engrafting specific cancer cells into a mouse and monitoring the tumor volume upon compound treatment (as a measure of cancer-killing potential). For neurological diseases, this could involve genetically introducing known mutations and monitoring specific exercises such as novel object recognition time tests (as a measure of memory function). Generally, these studies are benchmarked to the standard-of-care which is the current drug regimen used in clinical settings, and this provides an understanding of the degree to which the candidate molecule can improve outcomes. Although it is not an absolute requirement, usually two preclinical models are required to advance to human trials. Clinical trials are an expensive and complex endeavour with the goal of demonstrating that the drug candidate is safe and efficacious. Biotech companies looking to evaluate their drug candidates in humans (in Canada) require approval from Health Canada. The classical route involves Phase I trials, which are predominantly focussed on safety and pharmacokinetics of the drug candidate in healthy individuals. However, for rare diseases, the compound may be administered to a population with the disease of interest in Phase I. Following positive results, a Phase II clinical trial would be initiated with a cohort of patients from the disease population. Phase III clinical trials would have an even larger disease population for improved sampling statistics. Depending on the type of dosing performed, the clinical trial may also be labelled with an a or b suffix. Clinical trials cost millions of dollars and require years to prepare clinical centres and teams, recruit patients, administer the drugs, and analyze the results. On top of this, the success rate of a drug to advance through all three clinical trials with positive data is ~7%. Overall, it is estimated that advancing a compound through the drug discovery pipeline may take 15-20 years and cost 1-2 billion dollars. Although this may seem daunting, understanding why drugs fail at the late stage, helps front-load these issues into the medicinal chemistry thought process at the beginning of the drug discovery pipeline. The results of many clinical case-studies have revealed benefits and challenges of different functional groups and the goal of medicinal chemistry is to build on these learnings to avoid attrition of drugs at late clinical stages. 2.4 Rational Drug Design: The “Magic Bullet” Concept Paul Ehrlich, a German physician, pioneered a number of concepts that took a foothold in the early applications of medicinal chemistry. One of his primary contributions was in the field of immunity, for which he was awarded the Nobel Prize in Physiology or Medicine in 1908. He developed a “side-chain theory” which purported that specific chemical structures could elicit a response by immune cells in the blood, similar to the protein-ligand “lock-and-key” biochemistry model. He also adopted the concept of a “magic bullet” (a concept from German stories of a bullet that locks on to a target and cannot miss once fired), proposing that it would be possible to develop a compound that mimics these properties and specifically engage invading foreign entities with a high degree of specificity. Ehrlich pursued his idea of a magic bullet, generating libraries of compounds based on the toxic drug atoxyl for the treatment of syphilis, which was a major public health threat in Europe at the time. His goal was to identify a compound that would selectively kill the responsible bacterium, Treponema pallidum , without harming healthy cells. Eventually, his team developed compound 606 or arsphenamine which was marketed to Hoechst AG under the name salvarsan in 1910. Although this compound was a substantial improvement over the standard-of-care (mercury-containing compounds), it was highly unstable in air, which could lead to multiple and serious adverse side effects. Another 300 derivatives later, his team developed a more degradation-resistant drug, which was called neosalvarsan (Figure 2.8). Although these compounds were superseded by the discovery of penicillin in the 1940s, the approach adopted by Ehrlich foreshadowed the iterative drug discovery pipeline to optimize new selective compounds. 2.5 Rational Drug Design: Understanding the Target Understanding the target will shape the drug discovery program and alter the overall medicinal chemistry strategy. If considering just the human genome, there are approximately 20,000 genes that encode potential targets. About 15% of the genome can be pharmacologically altered with today’s collection of drugs. This means that there are potentially thousands of targets without drugs (although not all of these proteins are disease-relevant or would be significant effectors of disease if targeted by a drug). Additionally, there are a wealth of non-human proteins, such as pathogenic biomolecules from bacteria or viruses that cannot be accessed by current drugs. Although there are many different pathways, and therefore many protein targets that require pharmacologic intervention, current targets are asymmetrically skewed toward certain privileged protein families. For example, nearly two-thirds of all drugs on the market, target proteins from either the kinase or G-protein-coupled receptor (GPCR) superfamilies. Part of the reason why these protein classes are over-represented in drug discovery is because they exist at the top of large biochemical cascades, and therefore blockading their action can yield powerful responses. GPCRs and nuclear receptors have many different types of drugs that act on the same target (i.e. there are many selective drugs for these targets), whereas for kinases the reverse trend is true – there are many different kinases that are targeted by a smaller number of drugs, which indicates many more “off-targets”. Additionally, GPCRs are located on the cell surface, which reduces the requirement of a drug to cross an additional membrane. Importantly, kinases have been heavily studied both in structure and function which helps guide the drug discovery process. Designing drugs based on protein structure represents a critical step in the rational development of drugs based on protein-ligand interactions. Although the “lock-and-key” model of protein-ligand binding has been updated to involve “induced-fit” models that account for the fluidity of the protein and ligand conformations, it still provides a useful classical analogy for understanding the goals of designing new drugs for protein targets. This model has been applied to understand the different conformations of kinases and how drugs can engage these targets. 2.5.1 Kinases Protein kinases are enzymes that are responsible for transferring a phosphate group from ATP to another protein with a hydroxy-side chain (e.g. tyrosine). This post-translational modification introduces a di-anionic charge which is unique to a phosphorylated residue (and not found on any naturally incorporated amino acid). As such the consequences of phosphorylation can result in drastically altered protein conformations or interaction partners. There are two main types of kinases, Serine/Threonine kinases (~385 proteins) and Tyrosine Kinases (90 proteins) which specifically phosphorylate their namesake residue. Phosphorylation via Ser/Thr kinases is usually associated with large-scale conformational changes whereas as action by Tyr kinases can result in protein localization changes. There are also rarer kinases that are capable of phosphorylating non-conventional residues such as Histidine, Arginine, and Lysine. The kinase structure is highly conserved and the secondary structural elements are well-defined. The active kinase domain contains two lobes, a small N-lobe (largely comprised anti-parallel β-sheets) and a large C-lobe (largely α-helical) connected by a hinge region. The ATP binding site is sandwiched at the interface of these lobes and the adenine ring forms H-bonds with the hinge region. Kinases cycle between an inactive (open) conformation and an active (closed) conformation. The open conformation enables ATP / ADP to access the active site hinge region, whereas the closed conformation facilitates formation of the functionally active cleft. Both the N- and C- lobes have key structural features associated with each state. The N-lobe contains a large helix (labelled an αC-helix) that forms the roof of the active site and will either rotate away (open/inactive) or towards (closed/active) the orthosteric site. The C-lobe contains a conserved loop (referred to as an Activation or DFG loop) that begins with residues Asp, Phe, and Gly and ends with Ala, Pro, and Glu. The Asp side chains bind Mg 2 + which helps coordinate the phosphate of ATP. In the active conformation, the Asp faces into the pocket, whereas the inactive conformation has the Asp extruding out of the pocket. (These conformational switches also have the Phe entering and blocking the adenine binding pocket). Kinases represent the archetypical drug binding sites – they contain a well-defined binding pocket that is buried, hydrophobic, with affinity for a known ligand (ATP). In fact, the binding pockets of kinases and the inhibitors that have emerged are so extensively studied that they are stratified across 6 types: Type I inhibitors bind directly at the orthosteric (ATP binding) site and leverage conserved catalytic residues in their interactions. These inhibitors are ATP-competitive and have excellent shape complementary by exploiting the rigid nature of the active site. For example, crizotinib and dasatinib are Type I kinase inhibitors. Type II inhibitors bind at the ATP binding site but in the DFG-out (inactive) conformation. In this conformation, a new hydrophobic site is exposed between the C-helix of the small lobe and the DFG motif. The example of sorafentib, which is discussed above, is a Type II inhibitor. Type III inhibitors bind deeper in the ATP binding site, in a hydrophobic pocket behind the active site. These inhibitors have substantially improved selectivity, since this hydrophobic pocket is kinase-specific. Moreover, these inhibitors do not have a heterocyclic group that normally mimics the adenine ring. However, this pocket is often transient and occupancy does not guarantee inhibition. Type IV inhibitors bind at an allosteric site, completely distinct from the ATP site. Type V inhibitors are bivalent inhibitors with chemical moieties that engage with structural elements on both lobes of the kinase. Type VI inhibitors are covalent inhibitors that will form a covalent bond with a residue (usually cysteine) on the kinase. An important aspect for inhibitors is that over time, cells can become resistant to the effects of inhibitors. This can emerge from mutations in the target that reduce binding affinity or capacity of the inhibitor. In kinases, this can often occur with a gatekeeper residue , which is a key residue that is located close to the hinge region and guards access to the pockets behind the adenine ring. In BCR-ABL1 (CML) the T315I gatekeeper mutation results in steric hindrance that impedes binding of imatinib. In EGRR the T790M gatekeeper mutation induces resistance by increasing affinity for ATP. New drugs need to be designed that can overcome these mutations by avoiding the gatekeeper, allosterically engaging the target, or harnessing a separate biochemical pathway altogether. For example, the drug ponatinib bypasses the gatekeeper mutation via an ethynyl group, whose linear structure sterically evades the bulky T315I mutation and can continue to block BCR-ABL activity. |
Courses/Lumen_Learning/Book%3A_US_History_I_(OS_Collection)_(Lumen)/14%3A_Cotton_is_King%3A_The_Antebellum_South%2C_1800-1860/14.5%3A_Wealth_and_Culture_in_the_South | Learning Objectives By the end of this section, you will be able to: Assess the distribution of wealth in the antebellum South Describe the southern culture of honor Identify the main proslavery arguments in the years prior to the Civil War During the antebellum years, wealthy southern planters formed an elite master class that wielded most of the economic and political power of the region. They created their own standards of gentility and honor, defining ideals of southern white manhood and womanhood and shaping the culture of the South. To defend the system of forced labor on which their economic survival and genteel lifestyles depended, elite southerners developed several proslavery arguments that they levied at those who would see the institution dismantled. SLAVERY AND THE WHITE CLASS STRUCTURE The South prospered, but its wealth was very unequally distributed. Upward social mobility did not exist for the millions of slaves who produced a good portion of the nation’s wealth, while poor southern whites envisioned a day when they might rise enough in the world to own slaves of their own. Because of the cotton boom, there were more millionaires per capita in the Mississippi River Valley by 1860 than anywhere else in the United States. However, in that same year, only 3 percent of whites owned more than fifty slaves, and two-thirds of white households in the South did not own any slaves at all. Distribution of wealth in the South became less democratic over time; fewer whites owned slaves in 1860 than in 1840. As the wealth of the antebellum South increased, it also became more unequally distributed, and an ever-smaller percentage of slaveholders held a substantial number of slaves. At the top of southern white society stood the planter elite, which comprised two groups. In the Upper South, an aristocratic gentry, generation upon generation of whom had grown up with slavery, held a privileged place. In the Deep South, an elite group of slaveholders gained new wealth from cotton. Some members of this group hailed from established families in the eastern states (Virginia and the Carolinas), while others came from humbler backgrounds. South Carolinian Nathaniel Heyward, a wealthy rice planter and member of the aristocratic gentry, came from an established family and sat atop the pyramid of southern slaveholders. He amassed an enormous estate; in 1850, he owned more than eighteen hundred slaves. When he died in 1851, he left an estate worth more than $2 million (approximately $63 million in 2014 dollars). As cotton production increased, new wealth flowed to the cotton planters. These planters became the staunchest defenders of slavery, and as their wealth grew, they gained considerable political power. One member of the planter elite was Edward Lloyd V, who came from an established and wealthy family of Talbot County, Maryland. Lloyd had inherited his position rather than rising to it through his own labors. His hundreds of slaves formed a crucial part of his wealth. Like many of the planter elite, Lloyd’s plantation was a masterpiece of elegant architecture and gardens. The grand house of Edward Lloyd V advertised the status and wealth of its owner. In its heyday, the Lloyd family’s plantation boasted holdings of forty-two thousand acres and one thousand slaves. One of the slaves on Lloyd’s plantation was Frederick Douglass, who escaped in 1838 and became an abolitionist leader, writer, statesman, and orator in the North. In his autobiography, Douglass described the plantation’s elaborate gardens and racehorses, but also its underfed and brutalized slave population. Lloyd provided employment opportunities to other whites in Talbot County, many of whom served as slave traders and the “slave breakers” entrusted with beating and overworking unruly slaves into submission. Like other members of the planter elite, Lloyd himself served in a variety of local and national political offices. He was governor of Maryland from 1809 to 1811, a member of the House of Representatives from 1807 to 1809, and a senator from 1819 to 1826. As a representative and a senator, Lloyd defended slavery as the foundation of the American economy. Wealthy plantation owners like Lloyd came close to forming an American ruling class in the years before the Civil War. They helped shape foreign and domestic policy with one goal in view: to expand the power and reach of the cotton kingdom of the South. Socially, they cultivated a refined manner and believed whites, especially members of their class, should not perform manual labor. Rather, they created an identity for themselves based on a world of leisure in which horse racing and entertainment mattered greatly, and where the enslavement of others was the bedrock of civilization. In this painting by Felix Octavius Carr Darley, a yeoman farmer carrying a scythe follows his livestock down the road. Below the wealthy planters were the yeoman farmers, or small landowners. Below yeomen were poor, landless whites, who made up the majority of whites in the South. These landless white men dreamed of owning land and slaves and served as slave overseers, drivers, and traders in the southern economy. In fact, owning land and slaves provided one of the only opportunities for upward social and economic mobility. In the South, living the American dream meant possessing slaves, producing cotton, and owning land. Despite this unequal distribution of wealth, non-slaveholding whites shared with white planters a common set of values, most notably a belief in white supremacy. Whites, whether rich or poor, were bound together by racism. Slavery defused class tensions among them, because no matter how poor they were, white southerners had race in common with the mighty plantation owners. Non-slaveholders accepted the rule of the planters as defenders of their shared interest in maintaining a racial hierarchy. Significantly, all whites were also bound together by the constant, prevailing fear of slave uprisings. D. R. Hundley on the Southern Yeoman D. R. Hundley was a well-educated planter, lawyer, and banker from Alabama. Something of an amateur sociologist, he argued against the common northern assumption that the South was made up exclusively of two tiers of white residents: the very wealthy planter class and the very poor landless whites. In his 1860 book, Social Relations in Our Southern States , Hundley describes what he calls the “Southern Yeomen,” a social group he insists is roughly equivalent to the middle-class farmers of the North. But you have no Yeomen in the South, my dear Sir? Beg your pardon, our dear Sir, but we have—hosts of them. I thought you had only poor White Trash? Yes, we dare say as much—and that the moon is made of green cheese! . . . Know, then, that the Poor Whites of the South constitute a separate class to themselves; the Southern Yeomen are as distinct from them as the Southern Gentleman is from the Cotton Snob. Certainly the Southern Yeomen are nearly always poor, at least so far as this world’s goods are to be taken into account. As a general thing they own no slaves; and even in case they do, the wealthiest of them rarely possess more than from ten to fifteen. . . . The Southern Yeoman much resembles in his speech, religious opinions, household arrangements, indoor sports, and family traditions, the middle class farmers of the Northern States. He is fully as intelligent as the latter, and is on the whole much better versed in the lore of politics and the provisions of our Federal and State Constitutions. . . . [A]lthough not as a class pecuniarily interested in slave property, the Southern Yeomanry are almost unanimously pro-slavery in sentiment. Nor do we see how any honest, thoughtful person can reasonably find fault with them on this account. —D. R. Hundley, Social Relations in Our Southern States , 1860 What elements of social relations in the South is Hundley attempting to emphasize for his readers? In what respects might his position as an educated and wealthy planter influence his understanding of social relations in the South? Because race bound all whites together as members of the master race, non-slaveholding whites took part in civil duties. They served on juries and voted. They also engaged in the daily rounds of maintaining slavery by serving on neighborhood patrols to ensure that slaves did not escape and that rebellions did not occur. The practical consequence of such activities was that the institution of slavery, and its perpetuation, became a source of commonality among different economic and social tiers that otherwise were separated by a gulf of difference. Southern planters exerted a powerful influence on the federal government. Seven of the first eleven presidents owned slaves, and more than half of the Supreme Court justices who served on the court from its inception to the Civil War came from slaveholding states. However, southern white yeoman farmers generally did not support an active federal government. They were suspicious of the state bank and supported President Jackson’s dismantling of the Second Bank of the United States. They also did not support taxes to create internal improvements such as canals and railroads; to them, government involvement in the economic life of the nation disrupted what they perceived as the natural workings of the economy. They also feared a strong national government might tamper with slavery. Planters operated within a larger capitalist society, but the labor system they used to produce goods—that is, slavery—was similar to systems that existed before capitalism, such as feudalism and serfdom. Under capitalism, free workers are paid for their labor (by owners of capital) to produce commodities; the money from the sale of the goods is used to pay for the work performed. As slaves did not reap any earnings from their forced labor, some economic historians consider the antebellum plantation system a “pre-capitalist” system. HONOR IN THE SOUTH “The Modern Tribunal and Arbiter of Men’s Differences,” an illustration that appeared on the cover of The Mascot, a newspaper published in nineteenth-century New Orleans, reveals the importance of dueling in southern culture; it shows men bowing before an altar on which are laid a pistol and knife. A complicated code of honor among privileged white southerners, dictating the beliefs and behavior of “gentlemen” and “ladies,” developed in the antebellum years. Maintaining appearances and reputation was supremely important. It can be argued that, as in many societies, the concept of honor in the antebellum South had much to do with control over dependents, whether slaves, wives, or relatives. Defending their honor and ensuring that they received proper respect became preoccupations of whites in the slaveholding South. To question another man’s assertions was to call his honor and reputation into question. Insults in the form of words or behavior, such as calling someone a coward, could trigger a rupture that might well end on the dueling ground. Dueling had largely disappeared in the antebellum North by the early nineteenth century, but it remained an important part of the southern code of honor through the Civil War years. Southern white men, especially those of high social status, settled their differences with duels, before which antagonists usually attempted reconciliation, often through the exchange of letters addressing the alleged insult. If the challenger was not satisfied by the exchange, a duel would often result. The dispute between South Carolina’s James Hammond and his erstwhile friend (and brother-in-law) Wade Hampton II illustrates the southern culture of honor and the place of the duel in that culture. A strong friendship bound Hammond and Hampton together. Both stood at the top of South Carolina’s society as successful, married plantation owners involved in state politics. Prior to his election as governor of the state in 1842, Hammond became sexually involved with each of Hampton’s four teenage daughters, who were his nieces by marriage. “[A]ll of them rushing on every occasion into my arms,” Hammond confided in his private diary, “covering me with kisses, lolling on my lap, pressing their bodies almost into mine . . . and permitting my hands to stray unchecked.” Hampton found out about these dalliances, and in keeping with the code of honor, could have demanded a duel with Hammond. However, Hampton instead tried to use the liaisons to destroy his former friend politically. This effort proved disastrous for Hampton, because it represented a violation of the southern code of honor. “As matters now stand,” Hammond wrote, “he [Hampton] is a convicted dastard who, not having nerve to redress his own wrongs, put forward bullies to do it for him. . . . To challenge me [to a duel] would be to throw himself upon my mercy for he knows I am not bound to meet him [for a duel].” Because Hampton’s behavior marked him as a man who lacked honor, Hammond was no longer bound to meet Hampton in a duel even if Hampton were to demand one. Hammond’s reputation, though tarnished, remained high in the esteem of South Carolinians, and the governor went on to serve as a U.S. senator from 1857 to 1860. As for the four Hampton daughters, they never married; their names were disgraced, not only by the whispered-about scandal but by their father’s actions in response to it; and no man of honor in South Carolina would stoop so low as to marry them. GENDER AND THE SOUTHERN HOUSEHOLD The antebellum South was an especially male-dominated society. Far more than in the North, southern men, particularly wealthy planters, were patriarchs and sovereigns of their own household. Among the white members of the household, labor and daily ritual conformed to rigid gender delineations. Men represented their household in the larger world of politics, business, and war. Within the family, the patriarchal male was the ultimate authority. White women were relegated to the household and lived under the thumb and protection of the male patriarch. The ideal southern lady conformed to her prescribed gender role, a role that was largely domestic and subservient. While responsibilities and experiences varied across different social tiers, women’s subordinate state in relation to the male patriarch remained the same. Writers in the antebellum period were fond of celebrating the image of the ideal southern woman. One such writer, Thomas Roderick Dew, president of Virginia’s College of William and Mary in the mid-nineteenth century, wrote approvingly of the virtue of southern women, a virtue he concluded derived from their natural weakness, piety, grace, and modesty. In his Dissertation on the Characteristic Differences Between the Sexes , he writes that southern women derive their power not by leading armies to combat, or of enabling her to bring into more formidable action the physical power which nature has conferred on her. No! It is but the better to perfect all those feminine graces, all those fascinating attributes, which render her the center of attraction, and which delight and charm all those who breathe the atmosphere in which she moves; and, in the language of Mr. Burke, would make ten thousand swords leap from their scabbards to avenge the insult that might be offered to her. By her very meekness and beauty does she subdue all around her. Such popular idealizations of elite southern white women, however, are difficult to reconcile with their lived experience: in their own words, these women frequently described the trauma of childbirth, the loss of children, and the loneliness of the plantation. This cover illustration from Harper’s Weekly in 1861 shows the ideal of southern womanhood. Louisa Cheves McCord’s “Woman’s Progress” Louisa Cheves McCord was born in Charleston, South Carolina, in 1810. A child of some privilege in the South, she received an excellent education and became a prolific writer. As the excerpt from her poem “Woman’s Progress” indicates, some southern women also contributed to the idealization of southern white womanhood. Sweet Sister! stoop not thou to be a man! Man has his place as woman hers; and she As made to comfort, minister and help; Moulded for gentler duties, ill fulfils His jarring destinies. Her mission is To labour and to pray; to help, to heal, To soothe, to bear; patient, with smiles, to suffer; And with self-abnegation noble lose Her private interest in the dearer weal Of those she loves and lives for. Call not this— (The all-fulfilling of her destiny; She the world’s soothing mother)—call it not, With scorn and mocking sneer, a drudgery. The ribald tongue profanes Heaven’s holiest things, But holy still they are. The lowliest tasks Are sanctified in nobly acting them. Christ washed the apostles’ feet, not thus cast shame Upon the God-like in him. Woman lives Man’s constant prophet. If her life be true And based upon the instincts of her being, She is a living sermon of that truth Which ever through her gentle actions speaks, That life is given to labour and to love.—Louisa Susanna Cheves McCord, “Woman’s Progress,” 1853 What womanly virtues does Louisa Cheves McCord emphasize? How might her social status, as an educated southern woman of great privilege, influence her understanding of gender relations in the South? For slaveholding whites, the male-dominated household operated to protect gendered divisions and prevalent gender norms; for slave women, however, the same system exposed them to brutality and frequent sexual domination. The demands on the labor of slave women made it impossible for them to perform the role of domestic caretaker that was so idealized by southern men. That slaveholders put them out into the fields, where they frequently performed work traditionally thought of as male, reflected little the ideal image of gentleness and delicacy reserved for white women. Nor did the slave woman’s role as daughter, wife, or mother garner any patriarchal protection. Each of these roles and the relationships they defined was subject to the prerogative of a master, who could freely violate enslaved women’s persons, sell off their children, or separate them from their families. DEFENDING SLAVERY John C. Calhoun, shown here in a ca. 1845 portrait by George Alexander Healy, defended states’ rights, especially the right of the southern states to protect slavery from a hostile northern majority. With the rise of democracy during the Jacksonian era in the 1830s, slaveholders worried about the power of the majority. If political power went to a majority that was hostile to slavery, the South—and the honor of white southerners—would be imperiled. White southerners keen on preserving the institution of slavery bristled at what they perceived to be northern attempts to deprive them of their livelihood. Powerful southerners like South Carolinian John C. Calhoun highlighted laws like the Tariff of 1828 as evidence of the North’s desire to destroy the southern economy and, by extension, its culture. Such a tariff, he and others concluded, would disproportionately harm the South, which relied heavily on imports, and benefit the North, which would receive protections for its manufacturing centers. The tariff appeared to open the door for other federal initiatives, including the abolition of slavery. Because of this perceived threat to southern society, Calhoun argued that states could nullify federal laws. This belief illustrated the importance of the states’ rights argument to the southern states. It also showed slaveholders’ willingness to unite against the federal government when they believed it acted unjustly against their interests. As the nation expanded in the 1830s and 1840s, the writings of abolitionists—a small but vocal group of northerners committed to ending slavery—reached a larger national audience. White southerners responded by putting forth arguments in defense of slavery, their way of life, and their honor. Calhoun became a leading political theorist defending slavery and the rights of the South, which he saw as containing an increasingly embattled minority. He advanced the idea of a concurrent majority, a majority of a separate region (that would otherwise be in the minority of the nation) with the power to veto or disallow legislation put forward by a hostile majority. Calhoun’s idea of the concurrent majority found full expression in his 1850 essay “Disquisition on Government.” In this treatise, he wrote about government as a necessary means to ensure the preservation of society, since society existed to “preserve and protect our race.” If government grew hostile to society, then a concurrent majority had to take action, including forming a new government. “Disquisition on Government” advanced a profoundly anti-democratic argument. It illustrates southern leaders’ intense suspicion of democratic majorities and their ability to effect legislation that would challenge southern interests. Go to the Internet Archive to read John C. Calhoun’s “Disquisition on Government.” Why do you think he proposed the creation of a concurrent majority? White southerners reacted strongly to abolitionists’ attacks on slavery. In making their defense of slavery, they critiqued wage labor in the North. They argued that the Industrial Revolution had brought about a new type of slavery—wage slavery—and that this form of “slavery” was far worse than the slave labor used on southern plantations. Defenders of the institution also lashed out directly at abolitionists such as William Lloyd Garrison for daring to call into question their way of life. Indeed, Virginians cited Garrison as the instigator of Nat Turner’s 1831 rebellion. The Virginian George Fitzhugh contributed to the defense of slavery with his book Sociology for the South, or the Failure of Free Society (1854). Fitzhugh argued that laissez-faire capitalism, as celebrated by Adam Smith, benefited only the quick-witted and intelligent, leaving the ignorant at a huge disadvantage. Slaveholders, he argued, took care of the ignorant—in Fitzhugh’s argument, the slaves of the South. Southerners provided slaves with care from birth to death, he asserted; this offered a stark contrast to the wage slavery of the North, where workers were at the mercy of economic forces beyond their control. Fitzhugh’s ideas exemplified southern notions of paternalism. George Fitzhugh’s Defense of Slavery George Fitzhugh, a southern writer of social treatises, was a staunch supporter of slavery, not as a necessary evil but as what he argued was a necessary good, a way to take care of slaves and keep them from being a burden on society. He published Sociology for the South, or the Failure of Free Society in 1854, in which he laid out what he believed to be the benefits of slavery to both the slaves and society as a whole. According to Fitzhugh: [I]t is clear the Athenian democracy would not suit a negro nation, nor will the government of mere law suffice for the individual negro. He is but a grown up child and must be governed as a child . . . The master occupies towards him the place of parent or guardian. . . . The negro is improvident; will not lay up in summer for the wants of winter; will not accumulate in youth for the exigencies of age. He would become an insufferable burden to society. Society has the right to prevent this, and can only do so by subjecting him to domestic slavery. In the last place, the negro race is inferior to the white race, and living in their midst, they would be far outstripped or outwitted in the chase of free competition. . . . Our negroes are not only better off as to physical comfort than free laborers, but their moral condition is better. What arguments does Fitzhugh use to promote slavery? What basic premise underlies his ideas? Can you think of a modern parallel to Fitzhugh’s argument? The North also produced defenders of slavery, including Louis Agassiz, a Harvard professor of zoology and geology. Agassiz helped to popularize polygenism, the idea that different human races came from separate origins. According to this formulation, no single human family origin existed, and blacks made up a race wholly separate from the white race. Agassiz’s notion gained widespread popularity in the 1850s with the 1854 publication of George Gliddon and Josiah Nott’s Types of Mankind and other books. The theory of polygenism codified racism, giving the notion of black inferiority the lofty mantle of science. One popular advocate of the idea posited that blacks occupied a place in evolution between the Greeks and chimpanzees. This 1857 illustration by an advocate of polygenism indicates that the “Negro” occupies a place between the Greeks and chimpanzees. What does this image reveal about the methods of those who advocated polygenism? Section Summary Although a small white elite owned the vast majority of slaves in the South, and most other whites could only aspire to slaveholders’ wealth and status, slavery shaped the social life of all white southerners in profound ways. Southern culture valued a behavioral code in which men’s honor, based on the domination of others and the protection of southern white womanhood, stood as the highest good. Slavery also decreased class tensions, binding whites together on the basis of race despite their inequalities of wealth. Several defenses of slavery were prevalent in the antebellum era, including Calhoun’s argument that the South’s “concurrent majority” could overrule federal legislation deemed hostile to southern interests; the notion that slaveholders’ care of their chattel made slaves better off than wage workers in the North; and the profoundly racist ideas underlying polygenism. A Open Assessments element has been excluded from this version of the text. You can view it online here: http://pb.libretexts.org/ios/?p=254 Review Question How did defenders of slavery use the concept of paternalism to structure their ideas? Answer to Review Question Defenders of slavery, such as George Fitzhugh, argued that only the clever and the bright could truly benefit within a laissez-faire economy. Premising their argument on the notion that slaves were, by nature, intellectually inferior and less able to compete, such defenders maintained that slaves were better off in the care of paternalistic masters. While northern workers found themselves trapped in wage slavery, they argued, southern slaves’ needs—for food, clothing, and shelter, among other things—were met by their masters’ paternal benevolence. Glossary concurrent majority a majority of a separate region (that would otherwise be in the minority of the nation) with the power to veto or disallow legislation put forward by a hostile majority polygenism the idea that blacks and whites come from different origins CC licensed content, Shared previously US History. Authored by : P. Scott Corbett, Volker Janssen, John M. Lund, Todd Pfannestiel, Paul Vickery, and Sylvie Waskiewicz. Provided by : OpenStax College. Located at : http://openstaxcollege.org/textbooks/us-history . License : CC BY: Attribution . License Terms : Download for free at http://cnx.org/content/col11740/latest/ |
Courses/Modesto_Junior_College/Chemistry_143%3A_Introductory_College_Chemistry_(Brzezinski)/CHEM_143%3A_Text_(Brzezinski)/09%3A_Solutions/9.02%3A_Solutions_of_Solids_Dissolved_in_Water-_How_to_Make_Rock_Candy | Learning Objectives Define electrolytes and non electrolytes Explain why solutions form. Discuss the idea of water as the "universal solvent". Explain how water molecules attract ionic solids when they dissolve in water. We have learned that solutions can be formed in a variety of combinations using solids, liquids, and gases. We also know that solutions have constant composition, and that this composition can be varied up to a point to maintain the homogeneous nature of the solution. But how exactly do solutions form? Why is it that oil and water will not form a solution, and yet vinegar and water will? Why could we dissolve table salt in water, but not in vegetable oil? The reasons why solutions will form will be explored in this section, along with a discussion of why water is used most frequently to dissolve substances of various types. Solubility and Saturation Table salt \(\left( \ce{NaCl} \right)\) readily dissolves in water. In most cases, only a certain maximum amount of solute can be dissolved in a given amount of solvent. This maximum amount is specified as the solubility of the solute. It is usually expressed in terms of the amount of solute that can dissolve in 100 g of the solvent at a given temperature. Table \(\PageIndex{1}\) lists the solubilities of some simple ionic compounds. These solubilities vary widely. NaCl can dissolve up to 31.6 g per 100 g of H 2 O, while AgCl can dissolve only 0.00019 g per 100 g of H 2 O. Solute Solubility (g per 100 g of H2O at 25°C) AgCl 0.00019 CaCO3 0.00060 KBr 70.70000 NaCl 36.10000 NaNO3 94.60000 When the maximum amount of solute has been dissolved in a given amount of solvent, we say that the solution is saturated with solute. When less than the maximum amount of solute is dissolved in a given amount of solute, the solution is unsaturated . These terms are also qualitative terms because each solute has its own solubility. A solution of 0.00019 g of AgCl per 100 g of H 2 O may be saturated, but with so little solute dissolved, it is also rather dilute. A solution of 36.1 g of NaCl in 100 g of H 2 O is also saturated, but rather concentrated. In some circumstances, it is possible to dissolve more than the maximum amount of a solute in a solution. Usually, this happens by heating the solvent, dissolving more solute than would normally dissolve at regular temperatures, and letting the solution cool down slowly and carefully. Such solutions are called supersaturated solutions and are not stable; given an opportunity (such as dropping a crystal of solute in the solution), the excess solute will precipitate from the solution.The figure below illustrates the above process and shows the distinction between unsaturated and saturated. How can you tell if a solution is saturated or unsaturated? If more solute is added and it does not dissolve, then the original solution was saturated. If the added solute dissolves, then the original solution was unsaturated. A solution that has been allowed to reach equilibrium, but which has extra undissolved solute at the bottom of the container, must be saturated. Electrolyte Solutions: Dissolved Ionic Solids 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 (good conductor) . If only a relatively small fraction of the dissolved substance undergoes the ion-producing process, the substance is a weak electrolyte (does not conduct electricity as well) . 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 \(\PageIndex{1}\)). 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. Example \(\PageIndex{1}\): Identifying Ionic Compounds Which compound(s) will dissolve in solution to separate into ions? \(\ce{LiF}\) \(\ce{P_2F_5}\) \(\ce{C_2H_5OH}\) Solution \(\ce{LiF}\) will separate into ions when dissolved in solution, because it is an ionic compound. \(\ce{P_2F_5}\) and \(\ce{C_2H_5OH}\) are both covalent and will stay as molecules in a solution. Exercise \(\PageIndex{1}\) Which compounds will dissolve in solution to separate into ions? C 6 H 12 O 11 , glucose CCl 4 CaCl 2 AgNO 3 Answer c & d How Temperature Influences Solubility The solubility of a substance is the amount of that substance that is required to form a saturated solution in a given amount of solvent at a specified temperature. Solubility is often measured as the grams of solute per \(100 \: \text{g}\) of solvent. The solubility of sodium chloride in water is \(36.0 \: \text{g}\) per \(100 \: \text{g}\) water at \(20^\text{o} \text{C}\). The temperature must be specified because solubility varies with temperature. For gases, the pressure must also be specified. Solubility is specific for a particular solvent. We will consider solubility of material in water as solvent. The solubility of the majority of solid substances increases as the temperature increases. However, the effect is difficult to predict and varies widely from one solute to another. The temperature dependence of solubility can be visualized with the help of a solubility curve , a graph of the solubility vs. temperature (Figure \(\PageIndex{4}\)). Notice how the temperature dependence of \(\ce{NaCl}\) is fairly flat, meaning that an increase in temperature has relatively little effect on the solubility of \(\ce{NaCl}\). The curve for \(\ce{KNO_3}\), on the other hand, is very steep and so an increase in temperature dramatically increases the solubility of \(\ce{KNO_3}\). Several substances—\(\ce{HCl}\), \(\ce{NH_3}\), and \(\ce{SO_2}\)—have solubility that decreases as temperature increases. They are all gases at standard pressure. When a solvent with a gas dissolved in it is heated, the kinetic energy of both the solvent and solute increase. As the kinetic energy of the gaseous solute increases, its molecules have a greater tendency to escape the attraction of the solvent molecules and return to the gas phase. Therefore, the solubility of a gas decreases as the temperature increases. Solubility curves can be used to determine if a given solution is saturated or unsaturated. Suppose that \(80 \: \text{g}\) of \(\ce{KNO_3}\) is added to \(100 \: \text{g}\) of water at \(30^\text{o} \text{C}\). According to the solubility curve, approximately \(48 \: \text{g}\) of \(\ce{KNO_3}\) will dissolve at \(30^\text{o} \text{C}\). This means that the solution will be saturated since \(48 \: \text{g}\) is less than \(80 \: \text{g}\). We can also determine that there will be \(80 - 48 = 32 \: \text{g}\) of undissolved \(\ce{KNO_3}\) remaining at the bottom of the container. Now suppose that this saturated solution is heated to \(60^\text{o} \text{C}\). According to the curve, the solubility of \(\ce{KNO_3}\) at \(60^\text{o} \text{C}\) is about \(107 \: \text{g}\). Now the solution is unsaturated since it contains only the original \(80 \: \text{g}\) of dissolved solute. Now suppose the solution is cooled all the way down to \(0^\text{o} \text{C}\). The solubility at \(0^\text{o} \text{C}\) is about \(14 \: \text{g}\), meaning that \(80 - 14 = 66 \: \text{g}\) of the \(\ce{KNO_3}\) will re-crystallize. Summary Solubility is the specific amount of solute that can dissolve in a given amount of solvent. Saturated and unsaturated solutions are defined. Ionic compounds dissolve in polar solvents, especially water. This occurs when the positive cation from the ionic solid is attracted to the negative end of the water molecule (oxygen) and the negative anion of the ionic solid is attracted to the positive end of the water molecule (hydrogen). Water is considered the universal solvent since it can dissolve both ionic and polar solutes, as well as some nonpolar solutes (in very limited amounts). The solubility of a solid in water increases with an increase in temperature. Vocabulary Miscible - Liquids that have the ability to dissolve in each other. Immiscible - Liquids that do not have the ability to dissolve in each other. Electrostatic attraction - The attraction of oppositely charged particles. |
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Topics_in_Thermodynamics_of_Solutions_and_Liquid_Mixtures/01%3A_Modules/1.10%3A_Gibbs_Energies/1.10.29%3A_Gibbs_Energies-_Binary_Liquid_Mixtures-_Excess_Thermodynamic_Variables | The properties of binary mixtures are complicated. As a point of reference the excess molar properties of two non-aqueous binary liquid mixtures are often discussed. The mixtures are (A) trichloromethane + propanone, and (B) tetrachloromethane + methanol. A snapshot of the thermodynamic properties of a given binary mixture (at fixed \(\mathrm{T}\) and \(\mathrm{p}\)) is provided by combined plots of \({\mathrm{G}_{\mathrm{m}}}^{\mathrm{E}}\), \({\mathrm{H}_{\mathrm{m}}}^{\mathrm{E}}\) and \(\mathrm{T} \, \mathrm{S}_{\mathrm{m}}^{\mathrm{E}}\) as a function of mixture composition [1-5]. In effect the starting point is the Gibbs energy leading to first, second, third and fourth derivatives [6]. At this stage we make some sweeping (and dangerous) generalizations. For most binary aqueous mixtures, \({\mathrm{G}_{\mathrm{m}}}^{\mathrm{E}}\) is a smooth function of water m mole fraction \(x_{1}\), with an extremum near \(x_{1} = 0.5\). Rarely for a given mixture does the sign of \({\mathrm{G}_{\mathrm{m}}}^{\mathrm{E}}\) change across the mole faction range m although this feature is not unknown; e.g. water + 1,1,1,3,3,3- hexafluropropan-2-ol mixtures at \(298.15 \mathrm{~K}\) [7] but contrast water + 2,2,2- trifluorethanol mixtures [8] where at \(298.2 \mathrm{~K} {\mathrm{~G}_{\mathrm{m}}}^{\mathrm{E}}\) is positive across the m whole mole fraction range. However a change in sign of \({\mathrm{H}_{\mathrm{m}}}^{\mathrm{E}}\) and \(\mathrm{T} \, {\mathrm{S}_{\mathrm{m}}}^{\mathrm{E}}\) and \({\mathrm{V}_{\mathrm{m}}}^{\mathrm{E}}\) with change in mole fraction composition is quite common. For mixture A, \({\mathrm{G}_{\mathrm{m}}}^{\mathrm{E}}\) is negative indicating that \(\Delta_{\text{mix}}\mathrm{G}_{\mathrm{m}}\) is more negative than in the case of an ideal binary liquid mixture. In the case of Mixture A, the negative \({\mathrm{G}_{\mathrm{m}}}^{\mathrm{E}}\) is linked with a marked exothermic mixing; \({\mathrm{H}_{\mathrm{m}}}^{\mathrm{E}} < 0\). The latter is attributed to strong inter-component hydrogen bonding. For both mixtures A and B the signs of \({\mathrm{G}_{\mathrm{m}}}^{\mathrm{E}}\) and \({\mathrm{H}_{\mathrm{m}}}^{\mathrm{E}}\) are the same. This feature is characteristic of binary non-aqueous liquid mixtures where in most instances, \(\left|\mathrm{H}_{\mathrm{m}}^{\mathrm{E}}\right|>\left|\mathrm{T} \, \mathrm{S}_{\mathrm{mm}}^{\mathrm{E}}\right|\). \({\mathrm{G}_{\mathrm{m}}}^{\mathrm{E}}\) and \({\mathrm{H}_{\mathrm{m}}}^{\mathrm{E}}\) are both positive for mixture B. Here the pattern is understood in terms of disruption of methanol-methanol hydrogen bonding ( i.e. intracomponent interaction) by the second component. Again we note that a positive \({\mathrm{G}_{\mathrm{m}}}^{\mathrm{E}}\) means that the tendency for \(\Delta_{\text{mix}}\mathrm{G}_{\mathrm{m}}\) to be negative (cf. ideal mixtures) is opposed. Through a series of mixtures with increasing \({\mathrm{G}_{\mathrm{m}}}^{\mathrm{E}}\), a stage is reached where the magnitude of \({\mathrm{G}_{\mathrm{m}}}^{\mathrm{E}}\) is such that phase separation occurs [1]. For m many binary non-aqueous binary liquid mixtures the phase diagram for liquid miscibility has an upper critical solution temperature UCST. In other words only at high temperatures is the liquid mixture miscible in all proportions. Often binary aqueous mixtures are used as solvents for the following reason. The solubilities of salts in water(l) are high because ‘water is a polar solvent’ but the solubilities of apolar solutes are low. However the solubilities of apolar substances in organic solvents (e.g. ethanol) are high. If the chemical reaction being studied involves both polar and apolar solutes, judicious choice of the composition of a binary aqueous mixture leads to a solvent where the solubilities of both polar and apolar solutes are high. Nevertheless the task of accounting for the properties of binary aqueous mixtures is awesome. For this reason the classification introduced by Franks [9] has considerable merit. A distinction is drawn between Typically Aqueous and Typically Non-Aqueous Binary Aqueous Mixtures, based on the the thermodynamic excess functions, \({\mathrm{G}_{\mathrm{m}}}^{\mathrm{E}}\), \({\mathrm{H}_{\mathrm{m}}}^{\mathrm{E}}\) and \(\mathrm{T} \, {\mathrm{S}_{\mathrm{m}}}^{\mathrm{E}}\). Davis has explored how the properties of many binary aqueous mixtures can be subdivided on the basis of the ranges of mole fraction compositions [10]. Footnotes [1] J. S. Rowlinson and F. L. Swinton, Liquids and Liquid Mixtures, Butterworths, London, 3rd. edn., 1982. [2] K. N. Marsh, Annu. Rep. Prog. Chem., Sect. C, Phys. Chem.,1984, 81 , 209-245; Pure Appl. Chem.,1983, 55 ,467. [3] G. Scatchard. Chem. Rev.,1931, 8 ,321. [4] G. Scatchard, Chem. Rev.,1940, 44 ,7. [5] With respect to compressibilities; G. Douheret, C. Moreau and A. Viallard, Fluid Phase Equilib., 1985, 22 ,289. [6] Y. Koga, J.Phys.Chem.,1996, 100 ,5172; Y. Koga, K. Nishikawa and P. Westh, J. Phys. Chem.A,2004, 108 ,3873. [7] M. J. Blandamer, J. Burgess, A. Cooney, H. J. Cowles, I. M. Horn, K. J. Martin, K. W. Morcom and P. Warrick, J. Chem. Soc. Faraday Trans.,1990, 86 ,2209. A. Kivinen, J. Murto and A. Viit, Suomen Kemist.,Sect. B,1967, 40 ,298. [8] R. Jadot and M. Fraiha, J. Chem. Eng. Data, 1988, 33 ,237. [9] F. Franks, in Hydrogen –Bonded Solvent Systems, ed. A. K. Covington and P. Jones, Taylor and Francis, London, 1968, pp.31-47. [10] M. I. Davis, Thermochim Acta, 1984,77,421; 1985, 90 ,313; 1987, 120 ,299. M. I. Davis, M. C. Molina and G. Douheret.1988, 131 , 153. G. Douheret, A. Pal and M. I. Davis, J. Chem. Thermodyn., 1990, 22 , 99. G. Douheret, A. Pal, H. Hoiland, O. Anowi and M. I. Davis, J. Chem. Thermodyn., 1991, 23 ,569. G. Douheret, A. H. Roux, M. I. Davis, M. E. Hernandez, H. Hoiland and E. Hogseth, J. Solution Chem.,1993, 22 ,1041. G. Douheret, C. Moreau and A. Viallard, Fluid Phase Equilib., 1985, 22 , 277, 289. |
Courses/Oregon_Institute_of_Technology/OIT%3A_CHE_333_-_Organic_Chemistry_III_(Lund)/New_Page/6%3A_Overview_of_Organic_Reactivity/6.4%3A_Comparing_Biological_Reactions_to_Laboratory_Reactions | The focus of this book is on organic chemistry in a biological context. At various points in our investigation of organic reactivity, however, we will also be considering some non-biological, laboratory counterparts of reactions that occur in living cells. The reason for this is two-fold: first of all, it is often instructive to compare and contrast similar reactions taking place in very different environments, and sometimes the similarities are quite striking. Secondly, even those who intend to pursue a career in the life or health sciences can benefit from some exposure to the kind of challenges that professional organic or medicinal chemists work on: if you are working as a biologist for a pharmaceutical company for example, you will be better able to appreciate the contributions of your chemist colleagues if you are able to make the intellectual connection between the reactions they are running in flasks and the those that are taking place in the cells you are studying. Below we briefly outline the differences between laboratory and biological reactions: Catalysts : The vast majority of biological organic reactions are catalyzed by enzymes. While chemists synthesizing molecules in the laboratory sometimes make use of enzyme-catalyzed reactions, it is much more common to use non-biological catalysts (often containing transition metals), acids or bases as catalysts, or no catalyst at all. Solvent : Biological organic reactions occur in the aqueous environment of the cell. In the laboratory, organic reactions can be run in a wide variety of solvents, ranging from the very nonpolar (such as hexane) to the very polar, such as methanol,water, or even ionic liquids. Most commonly, though, laboratory reactions are run in relatively non-polar solvents such as diethyl ether or dichloromethane. Reactant mixture : The aqueous environment of a cell is an extremely complex mixture of thousands of different biomolecules in solution at low concentrations (usually nanomolar to millimolar), whereas the components of a laboratory reaction have usually been purified, and are present in much higher concentrations. Temperature : Biological reactions take place within a narrow temperature range specific to the organism: a little too cold and the enzymes catalyzing the reactions are 'frozen', a little too hot and the enzymes will come unfolded, or 'denature'. Laboratory reactions can be run at a variety of temperatures, sometimes at room temperature, sometimes at the boiling point of the solvent, and sometimes at very low temperatures (such as when a reaction flask is immersed in a dry ice-acetone bath). pH : Biological reactions take place in aqueous solution buffered to a specific pH: about pH 7 for most living things. Accordingly, highly acidic or basic species are unlikely to be reactants or intermediates in a biological reaction mechanism. Laboratory reactions are often carried out in the presence of strong acids or bases. |
Courses/American_River_College/CHEM_305%3A_Introduction_to_Chemistry_(Zumalt)/01%3A_Unit_1/1.05%3A_Scientific_Notation_-_Writing_Large_and_Small_Numbers | Learning Objectives Express a large number or a small number in scientific notation. Carry out arithmetical operations and express the final answer in scientific notation Chemists often work with numbers that are exceedingly large or small. For example, entering the mass in grams of a hydrogen atom into a calculator would require a display with at least 24 decimal places. A system called scientific notation avoids much of the tedium and awkwardness of manipulating numbers with large or small magnitudes. In scientific notation, these numbers are expressed in the form \[ N \times 10^n \nonumber \] where N is greater than or equal to 1 and less than 10 (1 ≤ N < 10), and n is a positive or negative integer (10 0 = 1). The number 10 is called the base because it is this number that is raised to the power \(n\). Although a base number may have values other than 10, the base number in scientific notation is always 10. A simple way to convert numbers to scientific notation is to move the decimal point as many places to the left or right as needed to give a number from 1 to 10 (N). The magnitude of n is then determined as follows: If the decimal point is moved to the left n places, n is positive. If the decimal point is moved to the right n places, n is negative. Another way to remember this is to recognize that as the number N decreases in magnitude, the exponent increases and vice versa. The application of this rule is illustrated in Example \(\PageIndex{1}\). Example \(\PageIndex{1}\): Expressing Numbers in Scientific Notation Convert each number to scientific notation. 637.8 0.0479 7.86 12,378 0.00032 61.06700 2002.080 0.01020 Solution Unnamed: 0 Explanation Answer a To convert 637.8 to a number from 1 to 10, we move the decimal point two places to the left: 637.8 Because the decimal point was moved two places to the left, n = 2. \(6.378 \times 10^2\) b To convert 0.0479 to a number from 1 to 10, we move the decimal point two places to the right: 0.0479 Because the decimal point was moved two places to the right, n = −2. \(4.79 \times 10^{−2}\) c This is usually expressed simply as 7.86. (Recall that 100 = 1.) \(7.86 \times 10^0\) d Because the decimal point was moved four places to the left, n = 4. \(1.2378 \times 10^4\) e Because the decimal point was moved four places to the right, n = −4. \(3.2 \times 10^{−4}\) f Because the decimal point was moved one place to the left, n = 1. \(6.106700 \times 10^1\) g Because the decimal point was moved three places to the left, n = 3. \(2.002080 \times 10^3\) h Because the decimal point was moved two places to the right, n = -2. \(1.020 \times 10^{−2}\) Addition and Subtraction Before numbers expressed in scientific notation can be added or subtracted, they must be converted to a form in which all the exponents have the same value. The appropriate operation is then carried out on the values of N. Example \(\PageIndex{2}\) illustrates how to do this. Example \(\PageIndex{2}\): Expressing Sums and Differences in Scientific Notation Carry out the appropriate operation and then express the answer in scientific notation. \( (1.36 \times 10^2) + (4.73 \times 10^3) \nonumber\) \((6.923 \times 10^{−3}) − (8.756 \times 10^{−4}) \nonumber\) Solution Unnamed: 0 Explanation Answer a Both exponents must have the same value, so these numbers are converted to either \((1.36 \times 10^2) + (47.3 \times 10^2) = \) \((1.36 + 47.3) \times 10^2 = 48.66 × 10^2\) or \((0.136 \times 10^3) + (4.73 \times 10^3) =\) \((0.136 + 4.73) \times 10^3) = 4.87 \times 10^3\). Choosing either alternative gives the same answer, reported to two decimal places. In converting 48.66 × 102 to scientific notation, \(n\) has become more positive by 1 because the value of \(N\) has decreased. \(4.87 \times 10^3\) b Converting the exponents to the same value gives either \((6.923 \times 10^{-3}) − (0.8756 \times 10^{-3})= \) \((6.923 − 0.8756) \times 10^{−3}\) or \((69.23 \times 10^{-4}) − (8.756 \times 10^{-4}) =\) \( (69.23 − 8.756) \times 10^{−4} = 60.474 \times 10^{−4}\). In converting 60.474 × 10-4 to scientific notation, \(n\) has become more positive by 1 because the value of \(N\) has decreased. \(6.047 \times 10^{−3}\) Multiplication and Division When multiplying numbers expressed in scientific notation, we multiply the values of \(N\) and add together the values of \(n\). Conversely, when dividing, we divide \(N\) in the dividend (the number being divided) by \(N\) in the divisor (the number by which we are dividing) and then subtract n in the divisor from n in the dividend. In contrast to addition and subtraction, the exponents do not have to be the same in multiplication and division. Examples of problems involving multiplication and division are shown in Example \(\PageIndex{3}\). Example \(\PageIndex{3}\): Expressing Products and Quotients in Scientific Notation Perform the appropriate operation and express your answer in scientific notation. \( (6.022 \times 10^{23})(6.42 \times 10^{−2}) \nonumber\) \( \dfrac{ 1.67 \times 10^{-24} }{ 9.12 \times 10 ^{-28} } \nonumber \) \( \dfrac{ (6.63 \times 10^{−34})(6.0 \times 10) }{ 8.52 \times 10^{−2}} \nonumber \) Solution Unnamed: 0 Explanation Unnamed: 2 a In multiplication, we add the exponents: \((6.022 \times 10^{23})(6.42 \times 10^{−2})\)\(= (6.022)(6.42) \times 10^{[23 + (−2)]} = 38.7 \times 10^{21} \nonumber \) In converting \(38.7 \times 10^{21}\) to scientific notation, \(n\) has become more positive by 1 because the value of \(N\) has decreased. \(3.87 \times 10^{22}\) b In division, we subtract the exponents: \({1.67 \times 10^{−24} \over 9.12 \times 10^{−28}} = \) \({1.67 \over 9.12} \times 10^{[−24 − (−28)]} = 0.183 \times 10^4 \) In converting \(0.183 \times 10^4\) to scientific notation, \(n\) has become more negative by 1 because the value of \(N\) has increased. \( 1.83 \times 10^3\) c This problem has both multiplication and division: \( {(6.63 \times 10^{−34})(6.0 \times 10) \over (8.52 \times 10^{−2})} = \) \({39.78 \over 8.52} \times 10^{[−34 + 1 − (−2)]} \nonumber \) \( 4.7\times 10^{-31}\) |
Bookshelves/Biological_Chemistry/Concepts_in_Biophysical_Chemistry_(Tokmakoff)/06%3A_Dynamics_and_Kinetics/22%3A_Biophysical_Reaction_Dynamics/22.01%3A_Concepts_and_Definitions | Time-dependent problems in molecular biophysics: How do molecular systems change? How does a molecular system change its microscopic configuration? How are molecules transported? How does a system sample its thermodynamically accessible states? Two types of descriptions of time-dependent processes: Kinetics : Describes the rates of interconversion between states. This is typically measured by most experiments. It does not directly explain how processes happen, but it can be used to predict the time-dependent behavior of populations from a proposed mechanism. Dynamics : A description of the time-evolving molecular structures involved in a process, with the objective of gaining insight into mechanism. At a molecular level, this information is typically more readily available from dynamical simulations of a model than from experiments. There is no single way to describe biophysical kinetics and dynamics, so we will survey a few approaches. The emphasis here will be on the description and analysis of time-dependent phenomena, and not on the experimental or computational methods used to obtain the data. Two common classes of problems: Barrier crossing or activated processes : For a solution phase process, evolution between two or more states separated by a barrier whose energy is \( \gg k_BT\). A description of “rare events” when the system rapidly jumps between states. Includes chemical reactions described by transition-state theory. \(\rightarrow\) We’ll look at two state problems. Diffusion processes : Transport in the absence of significant enthalpic barriers. Many small barriers on the scale of \(k_BT\) lead to “friction”, rapid randomization of momenta, and thereby diffusion. Now let’s start with some basic definitions of terms we will use often: Coordinates Refers to many types of variables that are used to describe the structure or configuration of a system. For instance, this may refer to the positions of atoms in a MD simulation as a function of time {r N ,t}, or these Cartesian variables might be transformed onto a set of internal coordinates (such as bond lengths, bond angles, and torsion angles), or these positions may be projected onto a different collective coordinate. Unlike our simple lattice models, the transformation from atomic to collective coordinate is complex when the objective is to calculate a partition function, since the atomic degrees of freedom are all correlated. Collective coordinate A coordinate that reflects a sum/projection over multiple internal variables—from a high-dimensional space to a lower one. Example: Solvent coordinate in electron transfer. In polar solvation, the position of the electron is governed by the stabilization by the configuration of solvent dipoles. An effective collective coordinate could be the difference in electrostatic potential between the donor and acceptor sites: \(q ~ Φ_A‒Φ_D\). Example: RMSD variation of structure with coordinates from a reference state. \[ R M S D=\sqrt{\frac{1}{n} \sum_{i=1}^{n}\left(\mathbf{r}_{i}-\mathbf{r}_{i}^{0}\right)^{2}} \] where \(r\) is the position of an atom in an n atom molecule. Sometimes the term “order parameter” gets used to describe a collective coordinate. This term originated in the description of changes of symmetry at phase transitions, and is a more specific term than order parameter. While order parameters are collective variables, collective variables are not necessarily order parameters. Reaction coordinate An internal variable that describes the forward progress of a reaction or process. Typically an abstract quantity, and not a simple configurational or geometrical coordinate. In making a connection to molecular structure, often the optimal reaction coordinate is not known or cannot be described, and so we talk about a “good reaction coordinate” as a collective variable that is a good approximate description of the progress of the reaction. Energy Landscape A structure is characterized by an energy of formation. There are many forms of energy that we will use, including free energy (G, A), internal energy or enthalpy (E, H), interaction potential (U, V), ... so we will have to be careful to define the energy for a problem. Most of the time, though, we are interested in free energy. The energy landscape is used to express the relative stability of different states, the position and magnitude of barriers between states, and possible configurational entropy of certain states. It is closely related to the free energy of the system, and is often used synonymously with the potential of mean force. The energy landscape expresses how the energy of a system (typically, but it is not limited to, free energy) depends on one or more coordinates of the system. It is often used as a free energy analog of a potential energy surface. For many-particle systems, they can be presented as a reduced dimensional surface by projecting onto one or a few degrees of freedom of interest, by integrating over the remaining degrees of freedom. “Energy landscapes” represent the free energy (or rather the negative of the logarithm of the probability) along a particular coordinate. Let’s remind ourselves of some definitions. The free energy of the system is calculated from . \[ A = -k_BT \ln{Z} \] where Z is the partition function. The free energy is a number that reflects the thermally weighted number of microstates available to the system. The free energy determines the relative probability of occupying two states of the system: \[ \dfrac{P_A}{P_B} = e^{-(A_A-A_B)/k_BT} \] The energy landscape is most closely related to a potential of mean force \[ F(x) = -k_BT\ln{P(x)} \] P(x) is the probability density that reflects the probability for observing the system at a position x. As such it is equivalent to decomposing the free energy as a function of the coordinate x. Whereas the partition function is evaluated by integrating a Boltzmann weighting over all degrees of freedom, P(x) is obtained by integrating over all degrees of freedom except x. States We will use the term “state” in the thermodynamic sense: a distinguishable minimum or basin on free energy surface. States refer to a region of phase–space where you persist long compared to thermal fluctuations. The regions where there is a high probability of observing the system. One state is distinguished from another kinetically by a time-scale separation. The rate of evolving within a state is faster than the rate of transition between states. Configuration Can refer to a distinct microstate or a structure that has been averaged over a local energy basin. You average over configurations (integrate over q) to get states (macrostates). Transition State The transition state or transition–state ensemble, often labeled ‡, refers to those barrier configurations that have equal probability of making a transition forward or backward. It’s not really a “state” by our definition, but a barrier or saddle point along a reaction coordinate. |
Courses/UW-Whitewater/Chem_260%3A_Inorganic_Chemistry_(Girard)/08%3A_Solids/8.01%3A_The_Solid_State_of_Matter | Learning Objectives Define and describe the bonding and properties of ionic, molecular, metallic, and covalent network crystalline solids Describe the main types of crystalline solids: ionic solids, metallic solids, covalent network solids, and molecular solids Explain the ways in which crystal defects can occur in a solid When most liquids are cooled, they eventually freeze and form crystalline solids , solids in which the atoms, ions, or molecules are arranged in a definite repeating pattern. It is also possible for a liquid to freeze before its molecules become arranged in an orderly pattern. The resulting materials are called amorphous solids or noncrystalline solids (or, sometimes, glasses). The particles of such solids lack an ordered internal structure and are randomly arranged (Figure \(\PageIndex{1}\)). Metals and ionic compounds typically form ordered, crystalline solids. Substances that consist of large molecules, or a mixture of molecules whose movements are more restricted, often form amorphous solids. For examples, candle waxes are amorphous solids composed of large hydrocarbon molecules. Some substances, such as boron oxide (Figure \(\PageIndex{2}\)), can form either crystalline or amorphous solids, depending on the conditions under which it is produced. Also, amorphous solids may undergo a transition to the crystalline state under appropriate conditions. Crystalline solids are generally classified according the nature of the forces that hold its particles together. These forces are primarily responsible for the physical properties exhibited by the bulk solids. The following sections provide descriptions of the major types of crystalline solids: ionic, metallic, covalent network, and molecular. Ionic Solids Ionic solids , such as sodium chloride and nickel oxide, are composed of positive and negative ions that are held together by electrostatic attractions, which can be quite strong (Figure \(\PageIndex{3}\)). Many ionic crystals also have high melting points. This is due to the very strong attractions between the ions—in ionic compounds, the attractions between full charges are (much) larger than those between the partial charges in polar molecular compounds. This will be looked at in more detail in a later discussion of lattice energies. Although they are hard, they also tend to be brittle, and they shatter rather than bend. Ionic solids do not conduct electricity; however, they do conduct when molten or dissolved because their ions are free to move. Many simple compounds formed by the reaction of a metallic element with a nonmetallic element are ionic. Metallic Solids Metallic solids such as crystals of copper, aluminum, and iron are formed by metal atoms Figure \(\PageIndex{4}\). The structure of metallic crystals is often described as a uniform distribution of atomic nuclei within a “sea” of delocalized electrons. The atoms within such a metallic solid are held together by a unique force known as metallic bonding that gives rise to many useful and varied bulk properties. All exhibit high thermal and electrical conductivity, metallic luster, and malleability. Many are very hard and quite strong. Because of their malleability (the ability to deform under pressure or hammering), they do not shatter and, therefore, make useful construction materials. The melting points of the metals vary widely. Mercury is a liquid at room temperature, and the alkali metals melt below 200 °C. Several post-transition metals also have low melting points, whereas the transition metals melt at temperatures above 1000 °C. These differences reflect differences in strengths of metallic bonding among the metals. Covalent Network Solids Covalent network solids include crystals of diamond, silicon, some other nonmetals, and some covalent compounds such as silicon dioxide (sand) and silicon carbide (carborundum, the abrasive on sandpaper). Many minerals have networks of covalent bonds. The atoms in these solids are held together by a network of covalent bonds, as shown in Figure \(\PageIndex{5}\). To break or to melt a covalent network solid, covalent bonds must be broken. Because covalent bonds are relatively strong, covalent network solids are typically characterized by hardness, strength, and high melting points. For example, diamond is one of the hardest substances known and melts above 3500 °C. Molecular Solids Molecular solids , such as ice, sucrose (table sugar), and iodine, as shown in Figure \(\PageIndex{6}\), are composed of neutral molecules. The strengths of the attractive forces between the units present in different crystals vary widely, as indicated by the melting points of the crystals. Small symmetrical molecules (nonpolar molecules), such as H 2 , N 2 , O 2 , and F 2 , have weak attractive forces and form molecular solids with very low melting points (below −200 °C). Substances consisting of larger, nonpolar molecules have larger attractive forces and melt at higher temperatures. Molecular solids composed of molecules with permanent dipole moments (polar molecules) melt at still higher temperatures. Examples include ice (melting point, 0 °C) and table sugar (melting point, 185 °C). Properties of Solids A crystalline solid, like those listed in Table \(\PageIndex{1}\) has a precise melting temperature because each atom or molecule of the same type is held in place with the same forces or energy. Thus, the attractions between the units that make up the crystal all have the same strength and all require the same amount of energy to be broken. The gradual softening of an amorphous material differs dramatically from the distinct melting of a crystalline solid. This results from the structural nonequivalence of the molecules in the amorphous solid. Some forces are weaker than others, and when an amorphous material is heated, the weakest intermolecular attractions break first. As the temperature is increased further, the stronger attractions are broken. Thus amorphous materials soften over a range of temperatures. Type of Solid Type of Particles Type of Attractions Properties Examples ionic ions ionic bonds hard, brittle, conducts electricity as a liquid but not as a solid, high to very high melting points NaCl, Al2O3 metallic atoms of electropositive elements metallic bonds shiny, malleable, ductile, conducts heat and electricity well, variable hardness and melting temperature Cu, Fe, Ti, Pb, U covalent network atoms of electronegative elements covalent bonds very hard, not conductive, very high melting points C (diamond), SiO2, SiC molecular molecules (or atoms) IMFs variable hardness, variable brittleness, not conductive, low melting points H2O, CO2, I2, C12H22O11 Graphene: Material of the Future Carbon is an essential element in our world. The unique properties of carbon atoms allow the existence of carbon-based life forms such as ourselves. Carbon forms a huge variety of substances that we use on a daily basis, including those shown in Figure \(\PageIndex{7}\). You may be familiar with diamond and graphite, the two most common allotropes of carbon. (Allotropes are different structural forms of the same element.) Diamond is one of the hardest-known substances, whereas graphite is soft enough to be used as pencil lead. These very different properties stem from the different arrangements of the carbon atoms in the different allotropes. You may be less familiar with a recently discovered form of carbon: graphene. Graphene was first isolated in 2004 by using tape to peel off thinner and thinner layers from graphite. It is essentially a single sheet (one atom thick) of graphite. Graphene, illustrated in Figure \(\PageIndex{8}\), is not only strong and lightweight, but it is also an excellent conductor of electricity and heat. These properties may prove very useful in a wide range of applications, such as vastly improved computer chips and circuits, better batteries and solar cells, and stronger and lighter structural materials. The 2010 Nobel Prize in Physics was awarded to Andre Geim and Konstantin Novoselov for their pioneering work with graphene. Crystal Defects In a crystalline solid, the atoms, ions, or molecules are arranged in a definite repeating pattern, but occasional defects may occur in the pattern. Several types of defects are known, as illustrated in Figure \(\PageIndex{9}\). Vacancies are defects that occur when positions that should contain atoms or ions are vacant. Less commonly, some atoms or ions in a crystal may occupy positions, called interstitial sites , located between the regular positions for atoms. Other distortions are found in impure crystals, as, for example, when the cations, anions, or molecules of the impurity are too large to fit into the regular positions without distorting the structure. Trace amounts of impurities are sometimes added to a crystal (a process known as doping) in order to create defects in the structure that yield desirable changes in its properties. For example, silicon crystals are doped with varying amounts of different elements to yield suitable electrical properties for their use in the manufacture of semiconductors and computer chips. Summary Some substances form crystalline solids consisting of particles in a very organized structure; others form amorphous (noncrystalline) solids with an internal structure that is not ordered. The main types of crystalline solids are ionic solids, metallic solids, covalent network solids, and molecular solids. The properties of the different kinds of crystalline solids are due to the types of particles of which they consist, the arrangements of the particles, and the strengths of the attractions between them. Because their particles experience identical attractions, crystalline solids have distinct melting temperatures; the particles in amorphous solids experience a range of interactions, so they soften gradually and melt over a range of temperatures. Some crystalline solids have defects in the definite repeating pattern of their particles. These defects (which include vacancies, atoms or ions not in the regular positions, and impurities) change physical properties such as electrical conductivity, which is exploited in the silicon crystals used to manufacture computer chips. Glossary amorphous solid (also, noncrystalline solid) solid in which the particles lack an ordered internal structure covalent network solid solid whose particles are held together by covalent bonds crystalline solid solid in which the particles are arranged in a definite repeating pattern interstitial sites spaces between the regular particle positions in any array of atoms or ions ionic solid solid composed of positive and negative ions held together by strong electrostatic attractions metallic solid solid composed of metal atoms molecular solid solid composed of neutral molecules held together by intermolecular forces of attraction vacancy defect that occurs when a position that should contain an atom or ion is vacant |
Courses/Tennessee_State_University/Inorganic_Chemistry_(CHEM_5000_4200)/01%3A_Map-_Inorganic_Chemistry-I_(LibreTexts)/02%3A_Atomic_Structure/2.04%3A_Problems_(do_we_want_this_here) | Example Exercises The following series of problems reviews general understanding of the aforementioned material. Exercises 1. Based on the periodic trends for ionization energy, which element has the highest ionization energy? 2. Which has a larger atomic radius: nitrogen or oxygen? 3. Which element is more electronegative, sulfur (S) or selenium (Se)? 4. Why is the electronegativity value of most noble gases zero? 5. Rewrite the following list in order of decreasing electron affinity: fluorine (F), phosphorous (P), sulfur (S), boron (B). 6. Which of these elements has a smaller atomic radius than sulfur (S): O, Cl, Ca, Li Answer 1 Helium (He). Answer 2 Atomic radius increases from right to left on the periodic table. Therefore, nitrogen is larger than oxygen. Answer 3 Sulfur (S). Note that sulfur and selenium share the same column. Electronegativity increases up a column. This indicates that sulfur is more electronegative than selenium. Answer 4 Because of their full valence electron shell, the noble gases are extremely stable and do not readily lose or gain electrons. Answer 5 Fluorine (F)>Sulfur (S)>Phosphorous (P)>Boron (B). Explanation: Electron affinity generally increases from left to right and from bottom to top. Answer 6 Oxygen (O) is the only element in the list with a smaller atomic radius than S. Periodic trends indicate that atomic radius increases down a group (from top to bottom) and from left to right across a period. |
Courses/Providence_College/Organic_Chemistry_II/13%3A_Conjugation/13.01%3A_Allyl_System | We have previously described the molecular orbitals of ethylene, which contains only a single \(π\) bond. When we have consecutive double bonds (\(π\) systems), the molecular orbital description actually changes dramatically and we say that the system is conjugated . Conjugated systems are present in anything that has resonance delocalization, but it also has implications for simple neutral systems. Electrons are delocalized if \(π\) bonds are conjugated. The molecules below have electrons that can spread out over multiple atoms. Thus, our molecular orbital description must account for this. Allyl System Let’s take a look at the simplest system that has conjugation – where electrons can be delocalized over three atoms. This is known as the allyl system . Notice that there is never any way to place the –, +, or * on the central carbon atom! Why? Let’s construct the molecular orbitals for the allyl group’s \(π\) system. 1. n atomic orbitals combine to form n molecular orbitals 2. lowest energy molecular orbital has zero nodal planes 3. any increase in energy level results in an additional nodal plane If we consider just the allyl anion, what you might notice is that in the HOMO, the coefficients are equal at the ends of the system (if symmetrical). The electron density is found at the ends, NEVER at the central carbon atom because there is a node there! This should tell you that if the allyl anion is reacting as a nucleophile, the electrons it will use will be the HOMO electrons, and the only atoms that have electron density in the HOMO are the ends! What this means is that the molecular orbitals will tell us where all of the “chemistry” (reactivity) will occur. It is the same for the allyl cation and the allyl radical and is predicted by the molecular orbitals. All chemistry must occur in \(π\) 2 . These are all examples of symmetrical systems – what about asymmetric systems? Well, the only thing that changes is the size of the coefficients in the molecular orbital diagram. Consider the asymmetric system below – which carbon would you expect to have more electron density in \(π\) 2 ? In other words, which end of the \(π\) system is more reactive? It turns out that the resonance form that contains the negative charge on the more highly substituted carbon atom is more reactive because of the extra \(σ\)-donation into the \(π\) system. This raises the coefficient on this carbon atom and makes it more reactive (better orbital overlap with an electrophile). What about a masked allyl system? In N , N -dimethylformamide (DMF) , the electrons on nitrogen can delocalize into the \(π\) system of the carbonyl to create a dipolar resonance form with the negative charge on oxygen (making oxygen more basic). Because this delocalization occurs similarly to a simple allyl system, we say that the nitrogen atom is sp 2 hybridized and NOT sp 3 hybridized, and the lone pairs on nitrogen are in an atomic p orbital that can align with the \(π\) C-O bond. This means that there is partial double bond character in the C-N bond and the C-O bond. What effect does this have? Well, it means that in the dipolar resonance form, the methyl groups are in different magnetic environments since there is not rotation about the C-N bond. Thus, there are two peaks for the methyl groups in the 13 C NMR, not one. |
Courses/South_Puget_Sound_Community_College/Chem_121%3A_Introduction_to_Chemistry/02%3A_Chapter_2_-_Measurements/2.01%3A_Units_in_Measurements/2.1.03%3A_Measurement_and_Scale_-_The_Mole_Concept/2.1.3.01%3A_Avogadro's_Number_and_the_Mole | Learning Objectives Make sure you thoroughly understand the following essential ideas: Define Avogadro's number and explain why it is important to know. Define the mole . Be able to calculate the number of moles in a given mass of a substance, or the mass corresponding to a given number of moles. Define molecular weight , formula weight , and molar mass ; explain how the latter differs from the first two. Be able to find the number of atoms or molecules in a given weight of a substance. Find the molar volume of a solid or liquid, given its density and molar mass. Explain how the molar volume of a metallic solid can lead to an estimate of atomic diameter. The chemical changes we observe always involve discrete numbers of atoms that rearrange themselves into new configurations. These numbers are HUGE— far too large in magnitude for us to count or even visualize, but they are still numbers , and we need to have a way to deal with them. We also need a bridge between these numbers, which we are unable to measure directly, and the weights of substances, which we do measure and observe. The mole concept provides this bridge, and is central to all of quantitative chemistry. Counting Atoms: Avogadro's Number Owing to their tiny size, atoms and molecules cannot be counted by direct observation. But much as we do when "counting" beans in a jar, we can estimate the number of particles in a sample of an element or compound if we have some idea of the volume occupied by each particle and the volume of the container. Once this has been done, we know the number of formula units (to use the most general term for any combination of atoms we wish to define) in any arbitrary weight of the substance. The number will of course depend both on the formula of the substance and on the weight of the sample. However, if we consider a weight of substance that is the same as its formula (molecular) weight expressed in grams, we have only one number to know: Avogadro's number . Avogadro's number Avogadro's number is known to ten significant digits: \[N_A = 6.022141527 \times 10^{23}.\] However, you only need to know it to three significant figures: \[N_A \approx 6.02 \times 10^{23}. \label{3.2.1}\] So \(6.02 \times 10^{23}\) of what ? Well, of anything you like: apples, stars in the sky, burritos. However, the only practical use for \(N_A\) is to have a more convenient way of expressing the huge numbers of the tiny particles such as atoms or molecules that we deal with in chemistry. Avogadro's number is a collective number , just like a dozen. Students can think of \(6.02 \times 10^{23}\) as the "chemist's dozen". Before getting into the use of Avogadro's number in problems, take a moment to convince yourself of the reasoning embodied in the following examples. Example \(\PageIndex{1}\): Mass ratio from atomic weights The atomic weights of oxygen and carbon are 16.0 and 12.0 atomic mass units (\(u\)), respectively. How much heavier is the oxygen atom in relation to carbon? Solution Atomic weights represent the relative masses of different kinds of atoms. This means that the atom of oxygen has a mass that is \[\dfrac{16\, \cancel{u}}{12\, \cancel{u}} = \dfrac{4}{3} ≈ 1.33 \nonumber\] as great as the mass of a carbon atom. Example \(\PageIndex{2}\): Mass of a single atom The absolute mass of a carbon atom is 12.0 unified atomic mass units (\(u\)). How many grams will a single oxygen atom weigh? Solution The absolute mass of a carbon atom is 12.0 \(u\) or \[12\,\cancel{u} \times \dfrac{1.6605 \times 10^{–24}\, g}{1 \,\cancel{u}} = 1.99 \times 10^{–23} \, g \text{ (per carbon atom)} \nonumber\] The mass of the oxygen atom will be 4/3 greater (from Example \(\PageIndex{1}\)): \[ \left( \dfrac{4}{3} \right) 1.99 \times 10^{–23} \, g = 2.66 \times 10^{–23} \, g \text{ (per oxygen atom)} \nonumber\] Alternatively we can do the calculation directly like with carbon: \[16\,\cancel{u} \times \dfrac{1.6605 \times 10^{–24}\, g}{1 \,\cancel{u}} = 2.66 \times 10^{–23} \, g \text{ (per oxygen atom)} \nonumber\] Example \(\PageIndex{3}\): Relative masses from atomic weights Suppose that we have \(N\) carbon atoms, where \(N\) is a number large enough to give us a pile of carbon atoms whose mass is 12.0 grams. How much would the same number, \(N\), of oxygen atoms weigh? Solution We use the results from Example \(\PageIndex{1}\) again. The collection of \(N\) oxygen atoms would have a mass of \[\dfrac{4}{3} \times 12\, g = 16.0\, g. \nonumber\] Exercise \(\PageIndex{1}\) What is the numerical value of \(N\) in Example \(\PageIndex{3}\)? Answer Using the results of Examples \(\PageIndex{2}\) and \(\PageIndex{3}\). \[N \times 1.99 \times 10^{–23} \, g \text{ (per carbon atom)} = 12\, g \nonumber\] or \[N = \dfrac{12\, \cancel{g}}{1.99 \times 10^{–23} \, \cancel{g} \text{ (per carbon atom)}} = 6.03 \times 10^{23} \text{atoms} \nonumber \] There are a lot of atoms in 12 g of carbon. Things to understand about Avogadro's number It is a number , just as is "dozen", and thus is dimensionless . It is a huge number, far greater in magnitude than we can visualize Its practical use is limited to counting tiny things like atoms, molecules, "formula units", electrons, or photons. The value of N A can be known only to the precision that the number of atoms in a measurable weight of a substance can be estimated. Because large numbers of atoms cannot be counted directly, a variety of ingenious indirect measurements have been made involving such things as Brownian motion and X-ray scattering . The current value was determined by measuring the distances between the atoms of silicon in an ultrapure crystal of this element that was shaped into a perfect sphere. (The measurement was made by X-ray scattering.) When combined with the measured mass of this sphere, it yields Avogadro's number. However, there are two problems with this: The silicon sphere is an artifact, rather than being something that occurs in nature, and thus may not be perfectly reproducible. The standard of mass, the kilogram, is not precisely known, and its value appears to be changing. For these reasons, there are proposals to revise the definitions of both N A and the kilogram. Moles and their Uses The mole (abbreviated mol) is the the SI measure of quantity of a "chemical entity" , which can be an atom, molecule, formula unit, electron or photon. One mole of anything is just Avogadro's number of that something. Or, if you think like a lawyer, you might prefer the official SI definition: Definition: The Mole The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon 12 Avogadro's number (Equation \ref{3.2.1}) like any pure number, is dimensionless. However, it also defines the mole, so we can also express N A as 6.02 × 10 23 mol –1 ; in this form, it is properly known as Avogadro's constant . This construction emphasizes the role of Avogadro's number as a conversion factor between number of moles and number of "entities". Example \(\PageIndex{4}\): number of moles in N particles How many moles of nickel atoms are there in 80 nickel atoms? Solution \[\dfrac{80 \;atoms}{6.02 \times 10^{23} \; atoms\; mol^{-1}} = 1.33 \times 10^{-22} mol \nonumber\] Is this answer reasonable? Yes, because 80 is an extremely small fraction of \(N_A\). Molar Mass The atomic weight, molecular weight, or formula weight of one mole of the fundamental units (atoms, molecules, or groups of atoms that correspond to the formula of a pure substance) is the ratio of its mass to 1/12 the mass of one mole of C 12 atoms, and being a ratio, is dimensionless. But at the same time, this molar mass (as many now prefer to call it) is also the observable mass of one mole ( N A ) of the substance, so we frequently emphasize this by stating it explicitly as so many grams (or kilograms) per mole: g mol –1 . It is important always to bear in mind that the mole is a number and not a mass . But each individual particle has a mass of its own, so a mole of any specific substance will always correspond to a certain mass of that substance. Example \(\PageIndex{5}\): Boron content of borax Borax is the common name of sodium tetraborate, \(\ce{Na2B4O7}\). how many moles of boron are present in 20.0 g of borax? how many grams of boron are present in 20.0 g of borax? Solution The formula weight of \(\ce{Na2B4O7}\) so the molecular weight is: \[(2 \times 23.0) + (4 \times 10.8) + (7 \times 16.0) = 201.2 \nonumber\] 20 g of borax contains (20.0 g) ÷ (201 g mol –1 ) = 0.10 mol of borax, and thus 0.40 mol of B. 0.40 mol of boron has a mass of (0.40 mol) × (10.8 g mol –1 ) = 4.3 g . Example \(\PageIndex{6}\): Magnesium in chlorophyll The plant photosynthetic pigment chlorophyll contains 2.68 percent magnesium by weight. How many atoms of Mg will there be in 1.00 g of chlorophyll? Solution Each gram of chlorophyll contains 0.0268 g of Mg, atomic weight 24.3. Number of moles in this weight of Mg: (.0268 g) / (24.2 g mol –1 ) = 0.00110 mol Number of atoms: (0.00110 mol) × (6.02E23 mol –1 ) = \(6.64 \times 10^{20}\) Is this answer reasonable? (Always be suspicious of huge-number answers!) Yes, because we would expect to have huge numbers of atoms in any observable quantity of a substance. Molar Volume This is the volume occupied by one mole of a pure substance. Molar volume depends on the density of a substance and, like density, varies with temperature owing to thermal expansion, and also with the pressure. For solids and liquids, these variables ordinarily have little practical effect, so the values quoted for 1 atm pressure and 25°C are generally useful over a fairly wide range of conditions. This is definitely not the case with gases, whose molar volumes must be calculated for a specific temperature and pressure. Example \(\PageIndex{7}\): Molar Volume of a Liquid Methanol, CH 3 OH, is a liquid having a density of 0.79 g per milliliter. Calculate the molar volume of methanol. Solution The molar volume will be the volume occupied by one molar mass (32 g) of the liquid. Expressing the density in liters instead of mL, we have \[V_M = \dfrac{32\; g\; mol^{–1}}{790\; g\; L^{–1}}= 0.0405 \;L \;mol^{–1} \nonumber\] The molar volume of a metallic element allows one to estimate the size of the atom. The idea is to mentally divide a piece of the metal into as many little cubic boxes as there are atoms, and then calculate the length of each box. Assuming that an atom sits in the center of each box and that each atom is in direct contact with its six neighbors (two along each dimension), this gives the diameter of the atom. The manner in which atoms pack together in actual metallic crystals is usually more complicated than this and it varies from metal to metal, so this calculation only provides an approximate value. Example \(\PageIndex{8}\): Radius of a Strontium Atom The density of metallic strontium is 2.60 g cm –3 . Use this value to estimate the radius of the atom of Sr, whose atomic weight is 87.6. Solution The molar volume of Sr is: \[\dfrac{87.6 \; g \; mol^{-1}}{2.60\; g\; cm^{-3}} = 33.7\; cm^3\; mol^{–1}\] The volume of each "box" is" \[\dfrac{33.7\; cm^3 mol^{–1}} {6.02 \times 10^{23}\; mol^{–1}} = 5.48 \times 10^{-23}\; cm^3\] The side length of each box will be the cube root of this value, \(3.79 \times 10^{–8}\; cm\). The atomic radius will be half this value, or \[1.9 \times 10^{–8}\; cm = 1.9 \times 10^{–10}\; m = 190 pm\] Note : Your calculator probably has no cube root button, but you are expected to be able to find cube roots; you can usually use the x y button with y =0.333. You should also be able estimate the magnitude of this value for checking. The easiest way is to express the number so that the exponent is a multiple of 3. Take \(54 \times 10^{-24}\), for example. Since 3 3 =27 and 4 3 = 64, you know that the cube root of 55 will be between 3 and 4, so the cube root should be a bit less than 4 × 10 –8 . So how good is our atomic radius? Standard tables give the atomic radius of strontium is in the range 192-220 pm. |
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page) Unknown License : 0% (0 page) Unknown License : 0% (0 page) Unknown License : 0% (0 page) Unknown License : 0% (0 page) Unknown License : 0% (0 page) Unknown License : 0% (0 page) Unknown License : 0% (0 page) Unknown License : 0% (0 page) Unknown License : 0% (0 page) Unknown License : 0% (0 page) Unknown License : 0% (0 page) Unknown License : 0% (0 page) By Page MacArthur Chemistry 3A v 1.2 — Undeclared Front Matter — Unknown License TitlePage — Unknown License InfoPage — Unknown License Table of Contents — Undeclared For the Instructor — Unknown License For the Student — Unknown License How Science Works — Unknown License 0.1: Hypothesis, Theories, and Laws — Undeclared 0.2: The Scientific Method - How Chemists Think — CC BY-NC-SA 4.0 1: Basics of Measurement — Unknown License 1.1: Taking Measurements- Numbers and Units — Unknown License 1.1.1: Scientific Notation - Writing Large and Small Numbers — Unknown License 1.1.2: The Basic Units of Measurement — Unknown License 1.2: 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Courses/can/CHEM_231%3A_Organic_Chemistry_I_Textbook/01%3A_Structure_and_Bonding/1.11%3A_Hybridization_of_Nitrogen_Oxygen_Phosphorus_and_Sulfur | The valence-bond concept of orbital hybridization described in the previous four sections is not limited to carbon. Covalent bonds formed by other elements can also be described using hybrid orbitals. Look, for instance, at the nitrogen atom in methylamine (CH 3 NH 2 ), an organic derivative of ammonia (NH 3 ) and the substance responsible for the odor of rotting fish. The experimentally measured H–N–H bond angle in methylamine is 107.1°, and the C–N–H bond angle is 110.3°, both of which are close to the 109.5° tetrahedral angle found in methane. We therefore assume that nitrogen forms four sp 3 -hybridized orbitals, just as carbon does. One of the four sp 3 orbitals is occupied by two nonbonding electrons (a lone pair), and the other three hybrid orbitals have one electron each. Overlap of these three half-filled nitrogen orbitals with half-filled orbitals from other atoms (C or H) gives methylamine. Note that the unshared lone pair of electrons in the fourth sp 3 hybrid orbital of nitrogen occupies as much space as an N–H bond does and is very important to the chemistry of methylamine and other nitrogen-containing organic molecules. Like the carbon atom in methane and the nitrogen atom in methylamine, the oxygen atom in methanol (methyl alcohol) and many other organic molecules can be described as sp 3 -hybridized. The C–O–H bond angle in methanol is 108.5°, very close to the 109.5° tetrahedral angle. Two of the four sp 3 hybrid orbitals on oxygen are occupied by nonbonding electron lone pairs, and two are used to form bonds. In the periodic table, phosphorus and sulfur are the third-row analogs of nitrogen and oxygen, and the bonding in both can be described using hybrid orbitals. Because of their positions in the third row, however, both phosphorus and sulfur can expand their outer-shell octets and form more than the typical number of covalent bonds. Phosphorus, for instance, often forms five covalent bonds, and sulfur often forms four. Phosphorus is most commonly encountered in biological molecules in compounds called organophosphates , which contain a phosphorus atom bonded to four oxygens, with one of the oxygens also bonded to carbon. Methyl phosphate, CH 3 OPO 3 2 − , is the simplest example. The O–P–O bond angle in such compounds is typically in the range 110° to 112°, implying sp 3 hybridization for phosphorus orbitals. Sulfur is most commonly encountered in biological molecules either in compounds called thiols, which have a sulfur atom bonded to one hydrogen and one carbon, C–S–H or in sulfides, which have a sulfur atom bonded to two carbons, C–S–C. Produced by some bacteria, methanethiol (CH 3 SH) is the simplest example of a thiol, and dimethyl sulfide, H 3 C–S–CH 3 , is the simplest example of a sulfide. Both can be described by approximate sp 3 hybridization around sulfur, although both have significant deviation from the 109.5° tetrahedral angle. Exercise \(\PageIndex{1}\) Identify all nonbonding lone pairs of electrons in the following molecules, and tell what geometry you expect for each of the indicated atoms. The oxygen atom in dimethyl ether, CH 3 –O–CH 3 The nitrogen atom in trimethylamine, The phosphorus atom in phosphine, PH 3 The sulfur atom in the amino acid methionine, Answer O has 2 lone pairs and is sp3-hybridized. N has 1 lone pair and is sp3-hybridized. P has 1 lone pair and is sp3-hybridized. S has 2 lone pairs and is sp 3 -hybridized. |
Courses/Indiana_Tech/EWC%3A_CHEM_2300_-_Introductory_Organic_(Budhi)/9%3A_Energy_Metabolism/9.04%3A_Overview_of_Stage_II_of_Catabolism | Learning Objectives To describe the role of acetyl-CoA in metabolism. A metabolic pathway is a series of biochemical reactions by which an organism converts a given reactant to a specific end product. There are specific metabolic pathways—which are different for carbohydrates, triglycerides, and proteins—that break down the products of stage I of catabolism (monosaccharides, fatty acids, and amino acids) to produce a common end product, acetyl-coenzyme A (acetyl-CoA) in stage II of catabolism. Acetyl-CoA is shown in Figure \(\PageIndex{1}\). The acetyl unit, derived (as we will see) from the breakdown of carbohydrates, lipids, and proteins, is attached to coenzyme A, making the acetyl unit more reactive. Acetyl-CoA is used in a myriad of biochemical pathways. For example, it may be used as the starting material for the biosynthesis of lipids (such as triglycerides, phospholipids, or cholesterol and other steroids). Most importantly for energy generation, it may enter the citric acid cycle and be oxidized to produce energy, if energy is needed and oxygen is available. The various fates or uses of acetyl-CoA are summarized in Figure \(\PageIndex{1}\). Glycolysis Glycolysis is the catabolic process in which glucose is converted into pyruvate via ten enzymatic steps. There are three regulatory steps, each of which is highly regulated that are separated into two phases: the "priming phase" because it requires an input of energy in the form of 2 ATPs per glucose molecule and the "pay off phase" because energy is released in the form of 4 ATPs, 2 per glyceraldehyde molecule. The end result of Glycolysis is two new pyruvate molecules which can then be fed into the Citric Acid cycle (also known as the Kreb's Cycle ) if oxygen is present, or can be reduced to lactate or ethanol in the absence of of oxygen using a process known as fermentation . Glycolysis occurs within almost all living cells and is the primary source of Acetyl-CoA, which is the molecule responsible for the majority of energy output under aerobic conditions. The structures of Glycolysis intermediates can be found in Figure \(\PageIndex{3}\). Phase 1: The "Priming Step" The first phase of Glycolysis requires an input of energy in the form of ATP (adenosine triphosphate). alpha-D- Glucose is phosphorolated at the 6 carbon by ATP via the enzyme Hexokinase (Class: Transferase) to yield alpha-D-Glucose-6-phosphate (G-6-P). This is a regulatory step which is negatively regulated by the presence of glucose-6-phosphate. alpha-D-Glucose-6-phosphate is then converted into D- Fructose -6-phosphate (F-6-P) by Phosphoglucoisomerase (Class: Isomerase) D-Fructose-6-phosphate is once again phosphorolated this time at the 1 carbon position by ATP via the enzyme Phosphofructokinase (Class: Transferase) to yield D-Fructose-1,6-bisphosphate (FBP). This is the committed step of glycolysis because of its large \(\Delta G\) value. D-Fructose-1,6-bisphosphate is then cleaved into two, three carbon molecules; Dihydroxyacetone phosphate (DHAP) and D-Glyceraldehyde-3-phosphate (G-3-P) by the enzyme Fructose bisphosphate aldolase (Class: Lyase) Because the next portion of Glycolysis requires the molecule D-Glyceraldehyde-3-phosphate to continue Dihydroxyacetone phosphate is converted into D-Glyceraldehyde-3-phosphate by the enzyme Triose phosphate isomerase (Class: Isomerase) Phase 2: The "Pay Off Step" The second phase of Glycolysis where 4 molecules of ATP are produced per molecule of glucose. Enzymes appear in red: D-Glyceraldehyde-3-phosphate is phosphorolated at the 1 carbon by the enzyme Glyceraldehyde-3-phosphate dehodrogenase to yield the high energy molecule 1,3-Bisphosphoglycerate (BPG) ADP is then phosphorolated at the expense of 1,3-Bisphosphoglycerate by the enzyme Phosphoglycerate kinase (Class: Transferase) to yield ATP and 3-Phosphoglycerate (3-PG) 3-Phosphoglycerate is then converted into 2-Phosphoglycerate by Phosphoglycerate mutase in preparation to yield another high energy molecule 2-Phosphoglycerate is then converted to phosphoenolpyruvate (PEP) by Enolase. H 2 O, potassium, and magnesium are all released as a result. ADP is once again phosphorolated, this time at the expense of PEP by the enzyme pyruvate kinase to yield another molecule of ATP and and pyruvate. This step is regulated by the energy in the cell. The higher the energy of the cell the more inhibited pyruvate kinase becomes. Indicators of high energy levels within the cell are high concentrations of ATP, Acetyl-CoA, Alanine, and cAMP. Because Glucose is split to yield two molecules of D-Glyceraldehyde-3-phosphate, each step in the "Pay Off" phase occurs twice per molecule of glucose. Beta-Oxidation The best source of energy for eukaryotic organisms are fats. Glucose offers a ratio 6.3 moles of ATP per carbon while saturated fatty acids offer 8.1 ATP per carbon. Also the complete oxidation of fats yields enormous amounts of water for those organisms that do not have adequate access to drinkable water. Camels and killer whales are good example of this, they obtain their water requirements from the complete oxidation of fats. There are four distinct stages in the oxidation of fatty acids. Fatty acid degradation takes place within the mitochondria and requires the help of several different enzymes. In order for fatty acids to enter the mitochondria the assistance of two carrier proteins is required, Carnitine acyltransferase I and II. It is also interesting to note the similarities between the four steps of beta-oxidation and the later four steps of the TCA cycle . Entry into Beta-oxidation Most fats stored in eukaryotic organisms are stored as triglycerides as seen below. In order to enter into beta-oxidation bonds must be broken usually with the use of a Lipase. The end result of these broken bonds are a glycerol molecule and three fatty acids in the case of triglycerides. Other lipids are capable of being degraded as well. Activation Step Once the triglycerides are broken down into glycerol and fatty acids they must be activated before they can enter into the mitochondria and proceed on with beta-oxidation. This is done by Acyl-CoA synthetase to yield fatty acyl-CoA. After the fatty acid has been acylated it is now ready to enter into the mitochondria. There are two carrier proteins (Carnitine acyltransferase I and II), one located on the outer membrane and one on the inner membrane of the mitochondria. Both are required for entry of the Acyl-CoA into the mitochondria. Once inside the mitochondria the fatty acyl-CoA can enter into beta-oxidation. Oxidation Step A fatty acyl-CoA is oxidized by Acyl-CoA dehydrogenase to yield a trans alkene. This is done with the aid of an [FAD] prosthetic group. Hydration Step The trans alkene is then hydrated with the help of Enoyl-CoA hydratase Oxidation Step The alcohol of the hydroxyacly-CoA is then oxidized by NAD + to a carbonyl with the help of Hydroxyacyl-CoA dehydrogenase. NAD + is used to oxidize the alcohol rather then [FAD] because NAD + is capable of the alcohol while [FAD] is not. Cleavage Finally acetyl-CoA is cleaved off with the help of Thiolase to yield an Acyl-CoA that is two carbons shorter than before. The cleaved acetyl-CoA can then enter into the TCA and ETC because it is already within the mitochondria. Summary Acetyl-CoA is formed from the breakdown of carbohydrates, lipids, and proteins. It is used in many biochemical pathways. References Garrett, H., Reginald and Charles Grisham. Biochemistry. Boston: Twayne Publishers, 2008. Raven, Peter. Biology. Boston: Twayne Publishers, 2005. |
Courses/Saint_Francis_University/CHEM_113%3A_Human_Chemistry_I_(Muino)/12%3A_Introduction_to_Organic_Chemistry_-_Alkanes/12.02%3A_Families_of_Organic_Molecules_-_Functional_Groups | Learning Objectives Identify and describe functional groups in organic molecules. Organic molecules can be classified into families based on structural similarities. Within a family, molecules have similar physical behavior and often have predictable chemical reactivity. The structural components differentiating different organic families involve specific arrangements of atoms or bonds, called functional groups . If you understand the behavior of a particular functional group, you can describe the general properties of that class of compounds. The simplest organic compounds are in the alkane family and contain only carbon–carbon and carbon–hydrogen single bonds but do not have any specific functional group. Hydrocarbons containing at least one carbon–carbon double bond, (denoted C=C), are in the alkene family. Alkynes have at least one carbon–carbon triple bond (C≡C). Both carbon–carbon double bonds and triple bonds chemically react in specific ways that differ from reactions of alkanes and each other, making these specific functional groups. In the next few chapters, we will learn more about additional functional groups that are made up of atoms or groups of atoms attached to hydrocarbons. Being able to recognize different functional groups will help to understand and describe common medications and biomolecules such as amino acids, carbohydrates, and fats. Table \(\PageIndex{1}\) and Figure \(\PageIndex{1}\) below list several of the functional groups to become familiar with as you learn about organic chemistry. Family Name Functional Group Structure Simple Example Structure Simple Example Name Name Suffix alkane none CH3CH2CH3 propane -ane alkene NaN H2C=CH2 ethene (ethylene) -ene alkyne NaN HC≡CH ethyne (acetylene) -yne aromatic NaN NaN benzene none alkyl halide (X = F, Cl, Br, I) CH3CH2Cl chloroethane none alcohol NaN CH3CH2OH ethanol -ol ether NaN CH3CH2–O–CH2CH3 diethyl ether none* amine NaN CH3CH2NH2 ethylamine -amine aldehyde NaN NaN ethanal -al ketone NaN NaN propanone (acetone) -one carboxylic acid NaN NaN ethanoic acid (acetic acid) -oic acid anhydride NaN NaN acetic anhydride none ester NaN NaN methyl ethanoate (methyl acetate) -ate amide NaN NaN acetamide -amide thiol NaN CH3CH2SH ethanethiol -thiol disulfide NaN CH3S–SCH3 dimethyl disulfide none sulfide NaN CH3CH2SCH3 ethyl methyl sulfide none Atoms and bonds in red indicate the functional group. Bonds not specified are attached to R groups (carbons and hydrogens). *Ethers do not have a suffix in their common name; all ethers end with the word ether. Atoms and bonds in red indicate the functional group. Bonds not specified are attached to R groups (carbons and hydrogens). *Ethers do not have a suffix in their common name; all ethers end with the word ether. Atoms and bonds in red indicate the functional group. Bonds not specified are attached to R groups (carbons and hydrogens). *Ethers do not have a suffix in their common name; all ethers end with the word ether. Atoms and bonds in red indicate the functional group. Bonds not specified are attached to R groups (carbons and hydrogens). *Ethers do not have a suffix in their common name; all ethers end with the word ether. Atoms and bonds in red indicate the functional group. Bonds not specified are attached to R groups (carbons and hydrogens). *Ethers do not have a suffix in their common name; all ethers end with the word ether. Figure \(\PageIndex{1}\): Functional groups in organic chemistry. (CC BY-NC-ND, CompoundChem.com). |
Courses/CSU_Chico/CSU_Chico%3A_CHEM_451_-_Biochemistry_I/CHEM_451_Test/04%3A_Protein_Structure/4.4%3A_Protein_Folding_-_in_Vivo_and_in_Vitro/D5._Multiple_Conformations_from_The_Same_Sequence | 1. Silent Single nucleotide polymorphisms (SNPs): For some amino acids, multiple triplet nucleotide sequences (codons) in the coding regions of a gene for a protein lead to the incorporation of the same amino acid in the protein sequence. Hence two proteins identical in amino acid sequence might have slightly different nucleotide sequences in the gene that encodes them. Such single nucleotide polymorphisms (SNPs) in coding regions were thought to have no effect on the tertiary structure and biological function of a protein if the single nucleotide variation did not lead to the insertion of a different amino acid into the growing peptide chain (i.e the codons were synonymous and the mutations presumably silent with no effect). Recently single nucleotide polymorphisms (SNPs) in the gene for the product of the MDR1 (multidrug resistance 1) gene, P-glycoprotein, was shown to result in a protein with different substrate specificity and inhibitor interactions, and hence a different 3D structure. One possible explanation for this observation is a difference in the rate of translation of the mRNA for this membrane protein. Different rates might lead to different intra- and intermolecular associations, which could lead to different final 3D structures as the protein cotranslationally folds and inserts into the membrane. This would especially be true if two possible structures where close enough in free energy but separated by a significant activation energy barrier, precluding simple conformational rearrangement of one conformation to another. 2. Metamorphic Proteins: In addition to prion proteins, it appears that many proteins can adopt more than one conformation under the same set of conditions. In contrast to prion proteins, however, in which the formation of the beta-structure variant is irreversible since the conformational change is associated with aggregation, many proteins can change conformations reversibly. Often, these changes do not appear to be associated only with binding interactions that trigger the change. Murzin has described proteins that change conformations on change of pH (viral glycoproteins), redox state (chloride channel), disulfide isomerization (lysozyme), and bound ligand (RNA polymerase as it initiates and then elongates the growing RNA polymer). He cites two proteins that appear to changes state without external signals. These include Mad2, in which the two conformers share extensive similarity, and Ltn10 (lymphotactin), in which they don't. One form of lymphotactin (Ltn 10) binds to similar lymphokine receptors, while the other (Ltn 40) binds to heparin. Folding kinetics may play a part in these examples as well, as proteins capable of folding to two conformers independently and quickly might prevent misfolding and aggregation that might occur if they had to completely unfold first before a conformational transition. Both Mad2 and Ltn10 alter conformation through transient formations of dimers, which facilitate conformational changes without widespread unfolding. Mutations in Ltn10 can cause the protein to adopt the Ltn40 conformation, Hence primordial "metamorphic" proteins could, by simple mutation, produce new protein functionalities. 3. Intrinsically Disordered Proteins (IDPs): Many examples of proteins that are partially or completely disordered but still retain biological function have been found. At first glance this might appear to be unexpected, since how could such a protein bind its natural ligand with specificity and selectivity to express its function? Of course one could postulate ligand binding would induce conformational changes necessary for function (such as catalysis) in an extreme example of an induced fit of a ligand compared to a "lock-and-key" fit. Decades ago, Linus Pauling predicted that antibodies, proteins that recognize foreign molecules (antigens), would bind loosely to the antigen, followed by a conformational change to form a more complementary and tighter fit. This was the easiest way to allow for a finite number of possible protein antibodies to bind a seemingly endless number of possible foreign molecules. This is indeed one method in which antibodies can recognize foreign antigens. Antibodies that bind to antigen with high affinity and hence high specificity more likely bind through a lock and key fit. (Pauling, however, didn't know that the genes that encode the proteins chains in antibodies are differentially spliced and subjected to enhanced mutational rates which allow the generation of incredible antibody diversity from a limited set of genes.) It's been estimated that over half of all native proteins have regions (greater than 30 amino acids) that are disordered, and upwards of 20% of proteins are completely disordered. Regions of disorder are enriched in polar and charged side chains which follows since these might expected to assume many available conformations in aqueous solutions compared to sequences enriched in hydrophobic side chains, which would probably collapse into a compact core stabilized by the hydrophobic effect. Mutations in the disordered regions tend to preserve the disordered region, suggesting that the disordered region is advantageous for "future" function. In addition, mutations that cause a noncoding sequence to produce a coding one invariably produce disordered protein sequences. Disordered proteins tend to have regulatory properties and bind multiple ligands, in comparison to ordered one, which are involved in highly specific ligand binding necessary for catalysis and transport. The intracellular concentration of disordered proteins has also been shown to be lower than ordered proteins, possibly to prevent occurrences of inappropriate binding interactions mediated through hydrophobic interactions, for example. Processes to accomplish this include more rapid mRNA and protein degradation and slower translation of mRNA for disordered proteins. For a similar reason, misfolded proteins are targeted for degradation as well. Figure A below shows the mean net charge vs the mean hydrophobicity for 275 folded and 91 natively unfolded proteins. Figure B shows the relative amino acid composition of globular (ordered) proteins compared to regions of disorder greater than 10 amino acids in disordered proteins. The two different grey bars were obtained with two different versions of the software used to analyze the proteins. Again the graph shows an enrichment of hydrophilic amino acids in disordered proteins. Figure: Characteristics of Intrinsically Disordered Proteins from open access journal: Dunker, A. et al. BMC Genomics 2008, 9 ( Suppl 2):S1 doi:10.1186/1471-2164-9-S2-S1 Many experimental methods can be used to detect disordered regions in proteins. Such regions are not resolved well in X-Ray crystal structures (have high B factors). NMR solution structures would show multiple, and differing conformations. CD spectroscopy likewise would show ill-defined secondary structure. In addition solution measurements of size (light scattering, centrifugation) would show larger size distributions for a given protein. What types of proteins contain disorder? The above experimental and new computational methods have been developed to classify proteins as to their degree of disorder. There appears to be more IDPs in eukaryotes than in archea and prokaryotes. Many IDPS are involved in cell signaling processes (when external molecules signal cells to respond by proliferating, differentiating, dying, etc). Most appear to reside in the nucleus. The largest percentage of known IDPs bind to other proteins and also to DNA. These results suggest that IEPs are essential to protein function and probably confer significant advantages to eukaryotic cells as multiple functions can be elicited from the interaction of a single IEP (derived from a single gene) with different protein binding partners. This would greatly extend the effective genome size in humans, for examples, from around 25,000 with specified function, to many more. This doesn't even take into account the increase functionalities derived from post-translational chemical modifications. We will discuss intrinsically disordered proteins further in Chapter 5. What is clear from recent finding is that protein structure is fluid and complex and our simple notions and words to denote proteins as either native or denatured are misguided and constrain our ideas about how protein structure elicits biological function. For example, what does the word "native" mean, if proteins exist in multiple states in vivo and in vitro simultaneously? Dunker et al (2001) have coined the concept "Protein Trinity" to move past the notion that a single protein folds to a single state which elicits a single function. Rather each of the states in the "trinity", the ordered, collapsed (or molten globule) and extended (random coil) coexist in the cell. Hence all can be considered "native" and all contribute to the function of the cell. A single IDP could bind to many different protein partners, each producing different final structures and functions. IDPs would also be more accessible and hence susceptible to proteolysis, which would lead to a simple mechanism to control their concentrations, an important way to regulate their biological activity. Their propensity to post-translational chemical modification would likewise lead to new types of biological regulation. Figure: The Protein Trinity: Ordered, Collapsed and Extended States These ideas have profound ramifications for our understanding of the expression of cellular phenotype. In addition, a whole new world of drug target is available by finding drugs that modulate the transitions between ordered, collapsed and extended protein states. Likewise, side effects of drugs might be understood by investigating drug effects of these transitions in IDPs not initially targeted. PONDR - Predictor of Naturally Occurring Disorder Database of Protein Disorder 4. Catalysis by Molten Globule: A recent example (Bemporad) that a bacterial acylphosphatase has catalytic activity as a molten globule further questions our notions of structure and enzyme activity. In this example, substrate interaction did not induce global conformational changes in the protein. Molecular dynamics simulations showed that many partially disordered conformations of the protein are present, and the disorder involved the active site. However, parts of the protein are more ordered and form a "scaffold" which keeps the catalytic and substrate binding amino acids near enough that binding could engender conformational rearrangements at the active side and subsequent catalytic activity. |
Courses/University_of_Arkansas_Little_Rock/Chem_1402%3A_General_Chemistry_1_(Belford)/Text/9%3A_Orbital_Hybridization_and_Molecular_Orbitals/9.1%3A_Orbitals_and_Theories_of_Chemical_Bonding | Introduction In Chapter 7 we described atomic orbitals for the hydrogen-like atom, based on the wavefunction of an electron in a spherical harmonic orbital around a positive nuclei and then we extended it to the multi-electron atom. In each of these cases we were describing an isolated atom. In a molecule there is more than one nucleus, and so the wavefunction will be different. There are two basic approaches we will take to describing molecular orbitals, the valence bond approach and the molecular orbital approach. We will also identify two different types of molecular orbitals, \(\sigma\) and \(\pi\). \(\sigma\) orbitals: These are orbitals that have electron density along the internuclear axis, and the sigma bonding orbitals from valence bond theory account for the VSEPR orientations. \(\pi\) orbitals: These orbitals result from the overlap of p orbitals and have a node along the internuclear axis. They are involved in multiple bonds. So a double bond is 1 \(\sigma\) and 1 \(\pi\), while a triple bond is a \(\sigma\) and 2 \(\pi\)s. The above two images show the types of overlap involving atomic s and p orbitals. But in molecular systems these are often not appropriate to describe molecular orbitals, and so we are going to look at two different approaches to molecular orbitals, Valence Bond Theory and Molecular Orbital Theory Valence Bond Theory Valence Bond (VB) Theory: In section 9.2 we cover Valence bond theory, where we treat the molecular orbitals that result in the VSEPR orientations as being the result of the hybridization of atomic orbitals (s,p,d,...), that is, atomic orbitals mix to form molecular orbitals. There are several things to recognize for two orbitals to "mix". First, the number of molecular orbitals produced equals the number of atomic orbitals mixed. Second, the atomic orbitals being mixed need to be of similar energy. Third, the resulting molecular orbitals acquire the geometry, nodal features and orientation of the orbitals that are mixed. So if you mix an s orbital with a p, it will have a shape that is half S and half P in characteristics. Valence Bond Theory can account for VSEPR geometries, but fails on some more advanced topics like some magnetic properties, or the concept of antibonding orbitals. To understand these we need to take a more complicated approach, molecular orbital theory. But there is a lot of value to valence bond theory and chemists use it all the time. Molecular Orbital Theory Molecular Orbital (MO) Theory: Molecular orbital theory is the direct quantum mechanical solution to the multi-nuclei wave function. These result in two types of wave functions, bonding and antibonding orbitals. In section 9.3 we will look at this for the simplest case, the homonuclear and heteronuclear diatomic. We will also look at magnetic properties like paramagnetism and show how MO theory can account for these types of properties in molecules like diatomic oxygen, when valence theory does not. |
Courses/Chabot_College/Introduction_to_General_Organic_and_Biochemistry/03%3A_Matter_and_Energy | 3.1: What is Matter? Matter is anything that has mass and volume (takes up space). For most common objects that we deal with every day, it is fairly simple to demonstrate that they have mass and take up space. You might be able to imagine, however, the difficulty for people several hundred years ago to demonstrate that air has mass and volume. Air (and all other gases) are invisible to the eye, have very small masses compared to equal amounts of solids and liquids, and are quite easy to compress (change volume). 3.2: Classifying Matter According to Its State—Solid, Liquid, and Gas Three states of matter exist—solid, liquid, and gas. Solids have a definite shape and volume. Liquids have a definite volume, but take the shape of the container. Gases have no definite shape or volume. 3.3: Classifying Matter According to Its Composition One useful way of organizing our understanding of matter is to think of a hierarchy that extends down from the most general and complex, to the simplest and most fundamental. Matter can be classified into two broad categories: pure substances and mixtures. A pure substance is a form of matter that has a consistent composition and properties that are constant throughout the sample. A material composed of two or more substances is a mixture. 3.4: Differences in Matter- Physical and Chemical Properties A physical property is a characteristic of a substance that can be observed or measured without changing the identity of the substance. Physical properties include color, density, hardness, melting points, and boiling points. A chemical property describes the ability of a substance to undergo a specific chemical change. 3.5: Changes in Matter - Physical and Chemical Changes Change is happening all around us all of the time. Just as chemists have classified elements and compounds, they have also classified types of changes. Changes are either classified as physical or chemical changes. Chemists learn a lot about the nature of matter by studying the changes that matter can undergo. Chemists make a distinction between two different types of changes that they study—physical changes and chemical changes. 3.6: Energy When we speak of using energy, we are really referring to transferring energy from one place to another. Although energy is used in many kinds of different situations, all of these uses rely on energy being transferred in one of two ways—energy can be transferred as heat or as work. 3.7: Energy and Chemical and Physical Change Phase changes involve changes in energy. All chemical reactions involve changes in energy. This may be a change in heat, electricity, light, or other forms of energy. Reactions that absorb energy are endothermic. Reactions that release energy are exothermic. 3.8: Energy and Heat Capacity Calculations Heat is a familiar manifestation of transferring energy. When we touch a hot object, energy flows from the hot object into our fingers, and we perceive that incoming energy as the object being “hot.” Conversely, when we hold an ice cube in our palms, energy flows from our hand into the ice cube, and we perceive that loss of energy as “cold.” In both cases, the temperature of the object is different from the temperature of our hand. 3.9: Temperature - Random Motion of Molecules and Atoms Three different scales are commonly used to measure temperature: Fahrenheit (expressed as °F), Celsius (°C), and Kelvin (K). 3.E: Matter and Energy (Exercises) |
Courses/University_of_Florida/CHM2047%3A_One-Semester_General_Chemistry_(Kleiman)/13%3A_Solutions/13.01%3A_Solids_Liquids_and_Phase_Transitions/13.1.02%3A_Intermolecular_Forces_-_Origins_in_Molecular_Structure | Learning Objectives To describe the intermolecular forces in liquids. The properties of liquids are intermediate between those of gases and solids, but are more similar to solids. In contrast to intra molecular forces, such as the covalent bonds that hold atoms together in molecules and polyatomic ions, inter molecular forces hold molecules together in a liquid or solid. Intermolecular forces are generally much weaker than covalent bonds. For example, it requires 927 kJ to overcome the intramolecular forces and break both O–H bonds in 1 mol of water, but it takes only about 41 kJ to overcome the intermolecular attractions and convert 1 mol of liquid water to water vapor at 100°C. (Despite this seemingly low value, the intermolecular forces in liquid water are among the strongest such forces known!) Given the large difference in the strengths of intra- and intermolecular forces, changes between the solid, liquid, and gaseous states almost invariably occur for molecular substances without breaking covalent bonds . The properties of liquids are intermediate between those of gases and solids but are more similar to solids. Intermolecular forces determine bulk properties such as the melting points of solids and the boiling points of liquids. Liquids boil when the molecules have enough thermal energy to overcome the intermolecular attractive forces that hold them together, thereby forming bubbles of vapor within the liquid. Similarly, solids melt when the molecules acquire enough thermal energy to overcome the intermolecular forces that lock them into place in the solid. Coulombic Interactions Through various experiments, Charles Augustin de Coulomb found a way to explain the interactions between charged particles, which in turn helped to explain where the stabilities and instabilities of various particles come from. While the entities that hold atoms together within a molecule can be attributed to bonds, the forces that create these bonds can be explained by Coulomb Forces. Thus, the physical basis behind the bonding of two atoms can be explained. Coulomb’s findings indicate that like charges repel each other and unlike charges attract one another. Thus electrons, which are negatively charged, repel each other but attract protons. Likewise, protons repel each other. Each atom is made up of a nucleus in the center, which consists of a number of protons and neutrons, depending upon the element in question. Surrounding the nucleus are electrons that float around the nucleus in what can be thought of as a cloud. As two atoms approach one another, the protons of one atom attract the electrons of the other atom. Similarly, the protons of the other atom attract the electrons of the first atom. As a result, the simultaneous attraction of the components from one atom to another create a bond. This interaction can be summarized mathematically and is known as Coulombic forces : \[ F = k \dfrac{q_{1}q_{2}}{r^{2}} \label{C}\] In this mathematical representation of Coulomb's observations, \(F\) is the electrical force acting between two atoms with \(q_1\) and \(q_2\) representing the magnitude of the charges of each atom, \(r\) is the distance between the two atoms. and \(k\) is a constant. From Equation \ref{C}, the electrostatic force between two charges is inversely proportional to the square of the distance separating the two atoms. Ion–Ion Interactions The interactions between ions (ion - ion interactions or charge-charge interactions) are the easiest to understand since such interactions are just a simple application of Coulombic forces (Equation \ref{C}). This specific interaction operates over relatively long distances in the gas phase and is responsible for the attraction of opposite charge ions and the repulsion of like charged ions. Coulombic forces are also involved in all forms of chemical bonding; when they act between separate charged particles they are especially strong. Thus the energy required to pull a mole of \(\ce{Na^{+}}\) and \(\ce{F^{–}}\) ions apart in the sodium fluoride crystal is greater than that needed to break the a covalent bonds of a mole of \(\ce{H2}\). The effects of ion-ion attraction are seen most directly in salts such as \(\ce{NaF}\) and \(\ce{NaCl}\) that consist of oppositely-charged ions arranged in inter-penetrating crystal lattices. According to Coulomb's Law the force between two charged particles is given by \[ \underbrace{F= \dfrac{q_1q_2}{4\pi\epsilon_0 r^2}}_{\text{ion-ion Force}} \label{7.2.1}\] Instead of using SI units, chemists often prefer to express atomic-scale distances in picometers and charges as electron charge (±1, ±2, etc.) Using these units, the proportionality constant \(1/4\pi\epsilon\) works out to \(2.31 \times 10^{16}\; J\; pm\). The sign of \(F\) determines whether the force will be attractive (–) or repulsive (+); notice that the latter is the case whenever the two q 's have the same sign. Two oppositely-charged particles flying about in a vacuum will be attracted toward each other, and the force becomes stronger and stronger as they approach until eventually they will stick together and a considerable amount of energy will be required to separate them. They form an ion-pair , a new particle which has a positively-charged area and a negatively-charged area. There are fairly strong interactions between these ion pairs and free ions, so that these clusters tend to grow, and they will eventually fall out of the gas phase as a liquid or solid (depending on the temperature). Equation \ref{7.2.1} is an example of an inverse square law ; the force falls off as the square of the distance. A similar law governs the manner in which the illumination falls off as you move away from a point light source; recall this the next time you walk away from a street light at night, and you will have some feeling for what an inverse square law means. The stronger the attractive force acting between two particles, the greater the amount of work required to separate them. Work represents a flow of energy, so the foregoing statement is another way of saying that when two particles move in response to a force, their potential energy is lowered. This work is found by integrating the negative of the force function with respect to distance over the distance moved. Thus the energy that must be supplied in order to completely separate two oppositely-charged particles initially at a distance \(r_0\) is given by \[ w= - \int _{r_o} ^{\infty} \dfrac{q_1q_2}{4\pi\epsilon_0 r^2}dr = - \dfrac{q_1q_2}{4\pi\epsilon_0 r_o} \label{7.2.2}\] hence, the potential (\(V_{ion-ion}\)) responsible for the ion-ion force is \[ \underbrace{V_{ion-ion} = \dfrac{q_1q_2}{4\pi\epsilon_0 r} }_{\text{ion-ion potential}} \label{7.2.3}\] Example \(\PageIndex{1}\) When sodium chloride is melted, some of the ion pairs vaporize and form neutral \(\ce{NaCl}\) dimers. How much energy would be released when one mole of \(\ce{Na^{+}}\) and \(\ce{Cl^{–}}\) ions are brought together to generate dimers in this way? The bondlength of \(\ce{NaCl}\) is 237 pm. Solution The interactions involved in forming \(\ce{NaCl}\) dimers is the ion-ion forces with a potential energy given by Equation \ref{7.2.3}. However, this is the energy of interaction for one pair of \(\ce{Na^{+}}\) and \(\ce{Cl^{–}}\) ion and needs to be scaled by a mole. So the energy released will be \[\begin{align*}E &= N_a V(\ce{NaCl}) \\[4pt] &= N_a\dfrac{q_1q_2}{4\pi\epsilon_0 r} \end{align*}\] The \(r\) in this equation is the distance between the two ions, which is the bondlength of 237 pm (\(237 \times 10^{-12}m\)). \[\begin{align*}E &= (6.022 \times 10^{23} ) \underbrace{(8.987 \times 10^9 N m^2/C^2 )}_{1/4\pi\epsilon_o} \dfrac{(+1.6 \times 10^{-19}C) (-1.6 \times 10^{-19}C) }{ 237 \times 10^{-12} m} \\[4pt] &= –584 \;kJ/mol \end{align*}\] This is not the energy needed to separate one mole of NaCl since that is a lattice and has more than pairwise interactions and require addressing the geometric orientation of the lattice (see Madelung Constants for more details). Intermolecular forces are electrostatic in nature; that is, they arise from the electrostatic interaction between positively and negatively charged species. Like covalent and ionic bonds, intermolecular interactions are the sum of both attractive and repulsive components. Because electrostatic interactions fall off rapidly with increasing distance between molecules, intermolecular interactions are most important for solids and liquids, where the molecules are close together. These interactions become important for gases only at very high pressures, where they are responsible for the observed deviations from the ideal gas law at high pressures (i.e., real gases). Ion-Dipole Interactions A dipole that is close to a positive or negative ion will orient itself so that the end whose partial charge is opposite to the ion charge will point toward the ion. This kind of interaction is very important in aqueous solutions of ionic substances; H 2 O is a highly polar molecule, so that in a solution of sodium chloride, for example, the Na + ions will be enveloped by a shell of water molecules with their oxygen-ends pointing toward these ions, while H 2 O molecules surrounding the Cl – ions will have their hydrogen ends directed inward. As a consequence of ion-dipole interactions, all ionic species in aqueous solution are hydrated; this is what is denoted by the suffix in formulas such as K + (aq), etc. The strength of ion-dipole attraction depends on the magnitude of the dipole moment and on the charge density of the ion. This latter quantity is just the charge of the ion divided by its volume. Owing to their smaller sizes, positive ions tend to have larger charge densities than negative ions, and they should be more strongly hydrated in aqueous solution. The hydrogen ion, being nothing more than a bare proton of extremely small volume, has the highest charge density of any ion; it is for this reason that it exists entirely in its hydrated form H 3 O + in water. Since there is now both attractive and repulsive interactions and they both get weaker as the ion and dipole distance increases while also approaching each other in strength, the net ion-dipole is an inverse square relationship as shown in Equation \ref{11.2.2}. \[ \underbrace{ E\: \propto \: \dfrac{-|q_1|\mu_2}{r^2}}_{\text{ion-dipole potential}} \label{11.2.2}\] \(r\) is the distance of separation. \(q\) is the charge of the ion ( only the magnitude of the charge is shown here.) \(\mu\) is the permanent dipole moment of the polar molecule. \[ \vec{\mu} = q' \, \vec{d} \label{11.2.3}\] where \(q'\) is the partial charge of each end of the dipole and \(d\) is the separation between the charges within the dipole Ion-Ion vs. Ion-Dipole potentials There are several differences between ion-ion potential (Equation \ref{7.2.3}) and the ion-dipole potential (Equation \ref{11.2.2}) interactions. First, the potential of ion/dipole interactions are negative and net interaction will always be attractive, since the attraction of the opposite dipole to the ion will make it closer than the dipole with the like charge. By using the absolute value of the charge of the ion, and placing a negative sign in front of the equation, this results in a lowering of the potential energy (\(\mu\) is positive). Second, the potential drops off quicker in Equation \ref{11.2.2}, where it is an inverse square relationship to the radius (\(1/r^2\)), while a simple charge-charge interaction (Equation \ref{7.2.3}) has a linear inverse relationship (\(1/r\)). This means the ion-dipole are a shorter range interaction and diminish more rapidly the father the polar molecule is from the ion. Third, note that the units of the two equations are the same, as \(\mu\) has the units of charge X distance. However, the distance in \(\mu\) is the distance between the dipoles of the polar molecule, while the distance denoted by the "r" is the distance between the ion and the dipole. It needs to be understood that the molecules in a solution are rotating and vibrating and actual systems are quite complicated (Figure \(\PageIndex{4}\)). What is important to realize is that these interactions are Coulombic in nature and how the mathematical equations describe this in terms of the magnitude of the charges and their distances from each other. In this course we will not be calculating dipole moments or the magnitudes of them, but understanding how to read the equations, and developing qualitative understandings that allow us to predict trends. Dipole–Dipole Interactions Polar covalent bonds behave as if the bonded atoms have localized fractional charges that are equal but opposite (i.e., the two bonded atoms generate a dipole ). If the structure of a molecule is such that the individual bond dipoles do not cancel one another, then the molecule has a net dipole moment. Molecules with net dipole moments tend to align themselves so that the positive end of one dipole is near the negative end of another and vice versa, as shown in part (a) in Figure \(\PageIndex{1}\). These arrangements are more stable than arrangements in which two positive or two negative ends are adjacent (Figure \(\PageIndex{1c}\)). Hence dipole–dipole interactions , such as those in Figure \(\PageIndex{5b}\), are attractive intermolecular interactions , whereas those in Figure \(\PageIndex{5d}\) are repulsive intermolecular interactions . Because molecules in a liquid move freely and continuously, molecules always experience both attractive and repulsive dipole–dipole interactions simultaneously, as shown in Figure \(\PageIndex{2}\). On average, however, the attractive interactions dominate. Because each end of a dipole possesses only a fraction of the charge of an electron, dipole–dipole interactions are substantially weaker than the interactions between two ions, each of which has a charge of at least ±1, or between a dipole and an ion, in which one of the species has at least a full positive or negative charge. In addition, the attractive interaction between dipoles falls off much more rapidly with increasing distance than do the ion–ion interactions. Recall that the attractive energy between two ions is proportional to 1/ r , where r is the distance between the ions. Doubling the distance ( r → 2 r ) decreases the attractive energy by one-half. In contrast, the energy of the interaction of two dipoles is proportional to 1/ r 6 , so doubling the distance between the dipoles decreases the strength of the interaction by 2 6 , or 64-fold: \[V=-\dfrac{2\mu_{A}^2\mu_{B}^2}{3(4\pi\epsilon_{0})^2r^6}\dfrac{1}{k_{B}T} \label{5}\] Thus a substance such as HCl, which is partially held together by dipole–dipole interactions, is a gas at room temperature and 1 atm pressure, whereas NaCl, which is held together by interionic interactions, is a high-melting-point solid. Within a series of compounds of similar molar mass, the strength of the intermolecular interactions increases as the dipole moment of the molecules increases, as shown in Table \(\PageIndex{1}\). Using what we learned about predicting relative bond polarities from the electronegativities of the bonded atoms, we can make educated guesses about the relative boiling points of similar molecules. Compound Molar Mass (g/mol) Dipole Moment (D) Boiling Point (K) C3H6 (cyclopropane) 42 0.0 240 CH3OCH3 (dimethyl ether) 46 1.3 248 CH3CN (acetonitrile) 41 3.9 355 The attractive energy between two ions is proportional to 1/ r , whereas the attractive energy between two dipoles is proportional to 1/ r 6 . Example \(\PageIndex{2}\) Arrange ethyl methyl ether (\(\ce{CH3OCH2CH3}\)), 2-methylpropane [isobutane, \(\ce{(CH3)2CHCH3}\)], and acetone (\(\ce{CH3COCH3}\)) in order of increasing boiling points. Their structures are as follows: Given: compounds Asked for: order of increasing boiling points Strategy: Compare the molar masses and the polarities of the compounds. Compounds with higher molar masses and that are polar will have the highest boiling points. Solution: The three compounds have essentially the same molar mass (58–60 g/mol), so we must look at differences in polarity to predict the strength of the intermolecular dipole–dipole interactions and thus the boiling points of the compounds. The first compound, 2-methylpropane, contains only C–H bonds, which are not very polar because C and H have similar electronegativities. It should therefore have a very small (but nonzero) dipole moment and a very low boiling point. Ethyl methyl ether has a structure similar to H 2 O; it contains two polar C–O single bonds oriented at about a 109° angle to each other, in addition to relatively nonpolar C–H bonds. As a result, the C–O bond dipoles partially reinforce one another and generate a significant dipole moment that should give a moderately high boiling point. Acetone contains a polar C=O double bond oriented at about 120° to two methyl groups with nonpolar C–H bonds. The C–O bond dipole therefore corresponds to the molecular dipole, which should result in both a rather large dipole moment and a high boiling point. Thus we predict the following order of boiling points: 2-methylpropane < ethyl methyl ether < acetone. This result is in good agreement with the actual data: 2-methylpropane, boiling point = −11.7°C, and the dipole moment (μ) = 0.13 D; methyl ethyl ether, boiling point = 7.4°C and μ = 1.17 D; acetone, boiling point = 56.1°C and μ = 2.88 D. Exercise \(\PageIndex{1}\) Arrange carbon tetrafluoride (CF 4 ), ethyl methyl sulfide (CH 3 SC 2 H 5 ), dimethyl sulfoxide [(CH 3 ) 2 S=O], and 2-methylbutane [isopentane, (CH 3 ) 2 CHCH 2 CH 3 ] in order of decreasing boiling points. Answer dimethyl sulfoxide (boiling point = 189.9°C) > ethyl methyl sulfide (boiling point = 67°C) > 2-methylbutane (boiling point = 27.8°C) > carbon tetrafluoride (boiling point = −128°C) London Dispersion Forces Thus far we have considered only interactions between polar molecules, but other factors must be considered to explain why many nonpolar molecules, such as bromine, benzene, and hexane, are liquids at room temperature, and others, such as iodine and naphthalene, are solids. Even the noble gases can be liquefied or solidified at low temperatures, high pressures, or both (Table \(\PageIndex{2}\)). What kind of attractive forces can exist between nonpolar molecules or atoms? This question was answered by Fritz London (1900–1954), a German physicist who later worked in the United States. In 1930, London proposed that temporary fluctuations in the electron distributions within atoms and nonpolar molecules could result in the formation of short-lived instantaneous dipole moments , which produce attractive forces called London dispersion forces between otherwise nonpolar substances. Substance Molar Mass (g/mol) Melting Point (°C) Boiling Point (°C) Ar 40 −189.4 −185.9 Xe 131 −111.8 −108.1 N2 28 −210 −195.8 O2 32 −218.8 −183.0 F2 38 −219.7 −188.1 I2 254 113.7 184.4 CH4 16 −182.5 −161.5 Consider a pair of adjacent He atoms, for example. On average, the two electrons in each He atom are uniformly distributed around the nucleus. Because the electrons are in constant motion, however, their distribution in one atom is likely to be asymmetrical at any given instant, resulting in an instantaneous dipole moment. As shown in part (a) in Figure \(\PageIndex{3}\), the instantaneous dipole moment on one atom can interact with the electrons in an adjacent atom, pulling them toward the positive end of the instantaneous dipole or repelling them from the negative end. The net effect is that the first atom causes the temporary formation of a dipole, called an induced dipole , in the second. Interactions between these temporary dipoles cause atoms to be attracted to one another. These attractive interactions are weak and fall off rapidly with increasing distance. London was able to show with quantum mechanics that the attractive energy between molecules due to temporary dipole–induced dipole interactions falls off as 1/ r 6 . Doubling the distance therefore decreases the attractive energy by 2 6 , or 64-fold. Instantaneous dipole–induced dipole interactions between nonpolar molecules can produce intermolecular attractions just as they produce interatomic attractions in monatomic substances like Xe. This effect, illustrated for two H 2 molecules in part (b) in Figure \(\PageIndex{3}\), tends to become more pronounced as atomic and molecular masses increase (Table \(\PageIndex{2}\)). For example, Xe boils at −108.1°C, whereas He boils at −269°C. The reason for this trend is that the strength of London dispersion forces is related to the ease with which the electron distribution in a given atom can be perturbed. In small atoms such as He, the two 1 s electrons are held close to the nucleus in a very small volume, and electron–electron repulsions are strong enough to prevent significant asymmetry in their distribution. In larger atoms such as Xe, however, the outer electrons are much less strongly attracted to the nucleus because of filled intervening shells. As a result, it is relatively easy to temporarily deform the electron distribution to generate an instantaneous or induced dipole. The ease of deformation of the electron distribution in an atom or molecule is called its polarizability . Because the electron distribution is more easily perturbed in large, heavy species than in small, light species, we say that heavier substances tend to be much more polarizable than lighter ones. For similar substances, London dispersion forces get stronger with increasing molecular size. The polarizability of a substance also determines how it interacts with ions and species that possess permanent dipoles. Thus London dispersion forces are responsible for the general trend toward higher boiling points with increased molecular mass and greater surface area in a homologous series of compounds, such as the alkanes (part (a) in Figure \(\PageIndex{4}\)). The strengths of London dispersion forces also depend significantly on molecular shape because shape determines how much of one molecule can interact with its neighboring molecules at any given time. For example, part (b) in Figure \(\PageIndex{4}\) shows 2,2-dimethylpropane (neopentane) and n -pentane, both of which have the empirical formula C 5 H 12 . Neopentane is almost spherical, with a small surface area for intermolecular interactions, whereas n -pentane has an extended conformation that enables it to come into close contact with other n -pentane molecules. As a result, the boiling point of neopentane (9.5°C) is more than 25°C lower than the boiling point of n -pentane (36.1°C). All molecules, whether polar or nonpolar, are attracted to one another by London dispersion forces in addition to any other attractive forces that may be present. In general, however, dipole–dipole interactions in small polar molecules are significantly stronger than London dispersion forces, so the former predominate. Example \(\PageIndex{2}\) Arrange n -butane, propane, 2-methylpropane [isobutene, (CH 3 ) 2 CHCH 3 ], and n -pentane in order of increasing boiling points. Given: compounds Asked for: order of increasing boiling points Strategy: Determine the intermolecular forces in the compounds and then arrange the compounds according to the strength of those forces. The substance with the weakest forces will have the lowest boiling point. Solution: The four compounds are alkanes and nonpolar, so London dispersion forces are the only important intermolecular forces. These forces are generally stronger with increasing molecular mass, so propane should have the lowest boiling point and n -pentane should have the highest, with the two butane isomers falling in between. Of the two butane isomers, 2-methylpropane is more compact, and n -butane has the more extended shape. Consequently, we expect intermolecular interactions for n -butane to be stronger due to its larger surface area, resulting in a higher boiling point. The overall order is thus as follows, with actual boiling points in parentheses: propane (−42.1°C) < 2-methylpropane (−11.7°C) < n -butane (−0.5°C) < n -pentane (36.1°C). Exercise \(\PageIndex{2}\) Arrange GeH 4 , SiCl 4 , SiH 4 , CH 4 , and GeCl 4 in order of decreasing boiling points. Answer GeCl 4 (87°C) > SiCl 4 (57.6°C) > GeH 4 (−88.5°C) > SiH 4 (−111.8°C) > CH 4 (−161°C) Summary Intermolecular forces are electrostatic in nature and include van der Waals forces and hydrogen bonds. Molecules in liquids are held to other molecules by intermolecular interactions, which are weaker than the intramolecular interactions that hold the atoms together within molecules and polyatomic ions. Transitions between the solid and liquid or the liquid and gas phases are due to changes in intermolecular interactions but do not affect intramolecular interactions. The three major types of intermolecular interactions are dipole–dipole interactions, London dispersion forces (these two are often referred to collectively as van der Waals forces ), and hydrogen bonds. Dipole–dipole interactions arise from the electrostatic interactions of the positive and negative ends of molecules with permanent dipole moments; their strength is proportional to the magnitude of the dipole moment and to 1/ r 6 , where r is the distance between dipoles. London dispersion forces are due to the formation of instantaneous dipole moments in polar or nonpolar molecules as a result of short-lived fluctuations of electron charge distribution, which in turn cause the temporary formation of an induced dipole in adjacent molecules. Like dipole–dipole interactions, their energy falls off as 1/ r 6 . Larger atoms tend to be more polarizable than smaller ones because their outer electrons are less tightly bound and are therefore more easily perturbed. |
Courses/CSU_Chico/CSU_Chico%3A_CHEM_451_-_Biochemistry_I/CHEM_451_Test/04%3A_Protein_Structure/4.9%3A_Steroidogenic_Acute_Regulatory_Protein | This page contains assessment/exam questions using data, figures, and graphs from research journals such as the Journal of Biological Chemistry which allow their use, or from journals such as from PLOS that are completely open access The papers and topics chosen were selected to assess student understanding of the American Society for Biochemistry and Molecular Biology (ASBMB) foundational concepts and learning objectives as well as MCAT2015 foundational concepts and objectives. These two sets of standards broadly overlap. Both ASBMB and the MCAT2015 strongly emphasis scientific inquiry and reasoning skills, which are perhaps best assessed by open-ended questions derived from the literature in which students must employ higher level Bloom skills of application and analysis. These questions can also be used by students who seek more opportunities to practice interpreting research literature results. The ability to apply, analyze, and evaluate information and concepts are at the heart of scientific inquiry and reasoning skills which are central to the new ASBMB and MCAT2015 competency standards. The questions in this learning module are designed to assess these competencies using open-ended responses instead of multiple-choice questions. Answers can be found at the link at the bottom of this page. Research Paper: A pH-dependent Molten Globule Transition Is Required for Activity of the Steroidogenic Acute Regulatory Protein, StAR . Bo Y. Baker, Dustin C. Yaworsky, and Walter L. Miller J. Biol. Chem., Vol. 280, Issue 50, 41753-41760, December 16, 2005 Steroids synthesis requires cholesterol, which get converted to pregnenolone by an enzyme in the inner mitochondrial membrane (IM). A regulatory protein, which we will call SR , is found in the outer mitochondrial membrane (OM) and is involved in cholesterol movement from the OM to the IM. Mutations in SR can cause congenital lipoid adrenal hyperplasia, a potentially lethal disease. The protein is synthesized in the cytoplasm as a 37,000 MW pre-protein, whose N terminal "leader sequence" targets SR to the mitochondria. Upon translocation of the protein to the IM, the leader is cleaved by proteolysis and is removed from the final active protein. The C-terminus must be critical for it's function since mutations in this region causes the hyperplasia discussed above. Models of SR suggest that it is similar to related protein that have a β-barrel structure with a hydrophobic binding pocket which can bind cholesterol. If the first 62 N-terminal amino acids are removed from the precursor ("preprotein"), the protein ( N-62 SR ) retains activity. When added to LUVs with a lipid composition similar to the OM, only the C-terminal alpha helical end of the protein is protected from added proteases, suggesting that that end inserts into the OM, effectively tethering the protein to the OM. Using spectral analysis and proteolysis of the protein, it appears that the protein may adopt a molten globule state at both low pH or when bound to a LUV. A conformation change appears to be induced when the protein interacts with protonated head groups of the LUV. The C-terminal helix is more protected from proteolysis at pH 4.0 compared to 6.5. Modeling studies based on x-ray structures of homologous proteins show that the access channel to the hydrophobic binding site in the protein is not large enough to pass cholesterol. These results show that the protein must change conformations to elicit its biological activity. The experiments described below were done to determine if a conformational change occurs in the protein on interaction with lipids, and whether a molten globule state of the protein is formed. A model structure of N-62 SR is shown below in Figure 1A. It has a tightly folded N-terminal domain and a more loosely folded C-terminal domain. A molecular dynamics simulation of the protein under 3 different protonation states was done in a bath of 9631 water molecules, in order to simulate possible conformational changes in the protein. The first charge state was defined to be neutral, simulating pH 7.0. Next they changed the protonation state of amino acids to reflect expected charges under acidic (pH 4) condition, which might promote a change to the molten globule state. In a last trial, they left the majority of the protein under neutral pH condition, but changed just the charge state of the C-terminal helix to acidic (pH 4.0) conditions. FIGURE 1. Modeling and molecular dynamics of N-62 SR. A , ribbon diagram of the model of N-62 SR; the salt bridges between Asp (D)106 and Arg (R) 272 and between Glu (E)136 and Arg (R) 274 are shown as ball-and-stick representations ( white , carbon; blue , nitrogen; red , oxygen). B , total energy levels of MD simulations as a function of time. Black , neutral; red , acidic ; green , acidified C-helix . C , root mean square deviation ( RMSD ) of -carbons during MD of the initial N-62 SR structure. D , images of Asp-106-Arg-272 after 2800 ps of MD under neutral ( top ) and acidified C-helix conditions ( bottom ). The carboxyl carbon (Cγ) of Arg-106 in the ω1 loop ( left ) is bonded to two oxygen atoms, (Oδ1 and Oδ2), and the guanidino carbon (C ) of Arg-272 in the C-helix ( right ) is covalently bonded to two nitrogen atoms (N 2 and N 1). Under neutral conditions, Arg-106 forms two hydrogen bonds to Arg-272 (Oδ1-Nω2 and Oδ2-Nω1), but under acidified C-helix conditions, the ω1 loop and C-helix movement and bonding are disrupted. 1. Which amino acids would be protonated on the shift from neutral to acidic (pH 4.0) conditions. Explain why the authors ran a simulation with just the C-terminal helix in the acidic protonation state. 2. Compare the energies of the neutral and acidified strucutures (1B) and the rigidity of the two structures in 1D. Which one is more compact? Flexible? Higher in energy? Which would be more consistent with a molten globule conformation? Do these structures support the notion that the protein might interact with the membrane through its C-terminal helix? 3. Snapshots of molecular dynamics simulations were taken every 100 picoseconds, and the distance between the Cγ of Asp (D)106 in the ω loop and the guanidino group of Arg (R) 272 on the C-terminal alpha helix was determined and represented graphically below in Figure 2C (top panel) below (same color coding as before). The bottom panel in 2C shows the distance between Asp (D)136 and Arg (R) 274. FIGURE 2. Movement of the StAR molecule during molecular dynamics. A , C superposition of conformations from MD snapshots every 100 ps over the 3-ns simulations at neutral ( left panel ), acidic ( middle panel ), and acidified C-helix ( right panel ) conditions. The protein is dynamic, particularly in the distal ends and the loop regions, but the overall structure is well maintained at neutral and acidic pHs. At low pH, conformation transitions are particularly prominent at the Ω 1 region. B , interatomic distances (in �) between Arg-272 and Asp-106 as a function of time during molecular dynamics simulations. The carboxyl carbon (Cγ) of Arg-106 in the Ω1 loop is bonded to two oxygen atoms, termed Oδ1 and Oδ2, and the guanidino carbon (Cζ ) of Arg-272 in the C-helix is covalently bonded to two nitrogen atoms, termed Nω1 and Nω2. The distances are plotted between these atoms and shown as follows: in red ,Oδ1/Nω2; green ,Oδ1/Nω1; black ,Oδ2/Nω1; and yellow ,Oδ2/Nω2. Under neutral conditions, the Oδ1/Nω2 and Oδ2/Nω1 distances are 2.8 � 0.1 �, favoring hydrogen bonding; pairings of Oδ1/Nω1 and Oδ2/Nω2 are less likely. Upper panel , neutral; middle panel , acidic; lower panel , acidified C-helix. C , interatomic distances during MD. Top , distance (in �) between Cζ in Arg-272 and Cγ in Asp-106. Bottom , distance (in �) between Cζ in Arg-274 and Cδ in Glu-136. Black , neutral; red , acidic; green , acidified C-helix. Referring to the location of the amino acids structure shown in Fig 1A, and the data in Fig 2C, which part of the protein is likely to change to promote access of cholesterol to sterol binding pocket in the protein? In an attempt to run an additional control for their "in silico" molecular dynamics experiment, the authors found two pairs of amino acids with the amino acids in each pair being approximately 4 angstroms from each other in the C helix and adjacent loop. The two pairs were Ser 100/Ser 261 and Asp 106/Ala 268. They changed them to Cys and created "in silico" the following mutation pairs (in single letter code): S100C/S261C and D106C/A268C. Note: DA mutant places the C268 closer to the end of the C helix than in the SS mutant, where the C is position 261 4. Why did they believe that changes these amino acids to Cys would not dramatically change the folding of the protein? 5. Energy minimization results showed that the Cα backbone of the C-terminal helix or Ω loop did not change. What would the likely effect be on the local structure in the vicinity of the pair? How would the molecular dynamics simulation in the top panel of 2C change? The mutants described above where made using site-specific mutagenesis, and SDS-polyacrylamide gels run. 6. Why did the mutants (DA) and SS migrated further on the gel than the wild type (WT) unmutated proteins? 7. When the mutant proteins were treated first with diothiothreitol (DTT) before electrophoreis, they comigrated with the WT proteins. When the mutant proteins were treated with diamide (oxidizes sulfhydryls to disulfides), their migration was not affected. The structures of DTT and diamide are shown below. Explain the gel results. FIGURE 3. Characterization of the disulfide mutants. A , model of N-62 SR showing the locations of the SS mutant replacing Ser-100 and Ser-261 and the DA mutant replacing Asp-106 and Ala-268 with Cys. Both mutants are shown in a single molecule, but each was prepared separately. Note that the perspective is from "behind" the molecule, as compared with Figure 1 A . B , SDS-PAGE gel of the SS ( right ) and DA ( left ) mutants . Lane designations in each gel are as follows: M , molecular weight standards; WT , wild type N-62 SR; SS and DA , untreated mutant proteins; diamide, mutant plus diamide; DTT , mutant plus DTT. 8. The amino acid sequence, as derived from the DNA sequence, shows that the WT N-60 protein has 3 Cys and the two mutants (DA and SS mutated as above to contain two Cys) each have 5. The amount of free Cys in each protein was determined using Ellman's reagent. Draw a reaction to show how a free Cys (represented as RSH below) reacts with Ellmans' reagent. What instrument would you use to measure the reaction and quantitate the levels of Cys? Reaction mechanism : 9. Computer modeling of the WT N60 SR shows no disulfides and two accessible Cys. Compare the number of Cys residues determine by Ellman's for each protein in the presence and absence of sodium dodecyl sulfate. 0 1 2 Protein # Cys - SDS # Cys + SDS WT 2 3 DA 2 3 SS 2 3 Why? 10. A comparison of the CD spectra and the fluorescence emission spectra of the WT protein and two mutants, SS and DA were made. Why was this done? FIGURE 6. Spectroscopic analysis. Top , far UV-CD spectra; bottom , fluorescence spectra of wild type ( black ), SS mutant ( blue ), and DA mutant ( red ) of N-62 SR. Next CD measurements were made on the WT, SS and DA mutants at various pH to s FIGURE 7. A , far-UV CD spectra of the wild type ( left ), SS mutant ( middle ), and DA mutant ( right ) of N-62 StAR. Each protein was assessed at pH 2 ( solid red line ), pH 2.5 ( dotted red line ), pH 3.0 ( solid green line ), pH 3.5 ( dotted green line ), pH 4.0 ( solid blue line ), pH 4.5 ( dotted blue line ), pH 6.0 ( solid tan line ), and pH 7.4 ( black line ). B , predicted -helical content of wild-type ( black ), SS ( blue ), and DA N-62 SR ( red ). 11. Compare the spectra in Fig 7A of the mutants, especially the DA mutant) to the WT. Also note the data in 7B. What is the overall effect on the change in the structure of the protein within in the mutants? Is all the data consistent with the idea that formation of the molten globule might play a role in the activity of the protein? 12. Based on your conclusions and the data, rank the following proteins as to their cholesterol binding activity: WT, SS mutant, DA mutant. Explain. . Answers: Literature Learning Module: Proteins 1 |
Courses/Prince_Georges_Community_College/CHEM_2000%3A_Chemistry_for_Engineers_(Sinex)/Unit_4%3A_Nomenclature_and_Reactions/Chapter_10%3A_Nomenclature/Chapter_10.3%3A_Naming_Ionic_Compounds | 0 1 2 NaN Prince George's Community College General Chemistry for Engineering CHM 2000 NaN Unit I: Atoms Unit II: Molecules Unit III: States of Matter Unit IV: Reactions Unit V: Kinetics & Equilibrium Unit VI: Thermo & Electrochemistry Unit VII: Nuclear Chemistry Unit I: Atoms Unit II: Molecules Unit III: States of Matter Unit IV: Reactions Unit V: Kinetics & Equilibrium Unit VI: Thermo & Electrochemistry Unit VII: Nuclear Chemistry Unit I: Atoms Unit II: Molecules Unit III: States of Matter Unit IV: Reactions Unit V: Kinetics & Equilibrium Unit VI: Thermo & Electrochemistry Unit VII: Nuclear Chemistry Learning Objective To name ionic compounds. The empirical and molecular formulas discussed in the preceding section are precise and highly informative, but they have some disadvantages. First, they are inconvenient for routine verbal communication. For example, saying “C-A-three-P-O-four-two” for Ca 3 (PO 4 ) 2 is much more difficult than saying “calcium phosphate.” In addition, many compounds, especially organic compounds, have the same empirical and molecular formulas but different arrangements of atoms, which result in very different chemical and physical properties. In such cases, it is necessary for the compounds to have different names that distinguish among the possible arrangements. Many compounds, particularly those that have been known for a relatively long time, have more than one name: a common name (sometimes more than one) and a systematic name, which is the name assigned by adhering to specific rules. Like the names of most elements, the common names of chemical compounds generally have historical origins, although they often appear to be unrelated to the compounds of interest. For example, the systematic name for KNO 3 is potassium nitrate, but its common name is saltpeter. In this text, we use a systematic nomenclature to assign meaningful names to the millions of known substances. Unfortunately, some chemicals that are widely used in commerce and industry are still known almost exclusively by their common names; in such cases, you must be familiar with the common name as well as the systematic one. The objective of this and the next two sections is to teach you to write the formula for a simple inorganic compound from its name—and vice versa—and introduce you to some of the more frequently encountered common names. We begin with binary ionic compounds , which contain only two elements. The procedure for naming such compounds is outlined in Figure 10.3.1 and uses the following steps: Figure 10.3.1 Naming an Ionic Compound Place the ions in their proper order: cation and then anion. Name the cation. Metals that form only one cation. As noted in Section 3.3 , these metals are usually in groups 1–3, 12, and 13. The name of the cation of a metal that forms only one cation is the same as the name of the metal (with the word ion added if the cation is by itself). For example, Na + is the sodium ion, Ca 2 + is the calcium ion, and Al 3 + is the aluminum ion. Metals that form more than one cation. As shown in Figure 10.3.2, many metals can form more than one cation. This behavior is observed for most transition metals, many actinides, and the heaviest elements of groups 13–15. We have discussed the reasons for this. In particular, transition metals tend to lose electrons in the ns orbitals before they lose electrons in the (n-1)d orbitals. In such cases, the positive charge on the metal is indicated by a roman numeral in parentheses immediately following the name of the metal. Thus Cu + is copper(I) (read as “copper one”), Fe 2 + is iron(II), Fe 3 + is iron(III), Sn 2 + is tin(II), and Sn 4 + is tin(IV). An older system of nomenclature for such cations is still widely used, however. The name of the cation with the higher charge is formed from the root of the element’s Latin name with the suffix - ic attached, and the name of the cation with the lower charge has the same root with the suffix - ous . The names of Fe 3 + , Fe 2 + , Sn 4 + , and Sn 2 + are therefore ferric, ferrous, stannic, and stannous, respectively. Even though this text uses the systematic names with roman numerals, you should be able to recognize these common names because they are still often used. For example, on the label of your dentist’s fluoride rinse, the compound chemists call tin(II) fluoride is usually listed as stannous fluoride. Some examples of metals that form more than one cation are in Table 10.3.1 along with the names of the ions. Note that the simple Hg + cation does not occur in chemical compounds. Instead, all compounds of mercury(I) contain a dimeric cation, Hg 2 2 + , in which the two Hg atoms are bonded together. Table 10.3.1 Common Cations of Metals That Form More Than One Ion Cation Systematic Name Common Name Cation.1 Systematic Name.1 Common Name.1 Cr2+ chromium(II) chromous Cu2+ copper(II) cupric Cr3+ chromium(III) chromic Cu+ copper(I) cuprous Mn2+ manganese(II) manganous* Hg2+ mercury(II) mercuric Mn3+ manganese(III) manganic* Hg22+ mercury(I) mercurous† Fe2+ iron(II) ferrous Sn4+ tin(IV) stannic Fe3+ iron(III) ferric Sn2+ tin(II) stannous Co2+ cobalt(II) cobaltous* Pb4+ lead(IV) plumbic* Co3+ cobalt(III) cobaltic* Pb2+ lead(II) plumbous* * Not widely used. * Not widely used. * Not widely used. * Not widely used. * Not widely used. * Not widely used. †The isolated mercury(I) ion exists only as the gaseous ion. †The isolated mercury(I) ion exists only as the gaseous ion. †The isolated mercury(I) ion exists only as the gaseous ion. †The isolated mercury(I) ion exists only as the gaseous ion. †The isolated mercury(I) ion exists only as the gaseous ion. †The isolated mercury(I) ion exists only as the gaseous ion. Polyatomic cations. The names of the common polyatomic cations that are relatively important in ionic compounds (such as, the ammonium ion) are in Table 10.2.1 Name the anion. Monatomic anions. Monatomic anions are named by adding the suffix - ide to the root of the name of the parent element; thus, Cl − is chloride, O 2− is oxide, P 3− is phosphide, N 3− is nitride (also called azide), and C 4− is carbide. Because the charges on these ions can be predicted from their position in the periodic table, it is not necessary to specify the charge in the name. Examples of monatomic anions are in Table 10.1.2 . Polyatomic anions. Polyatomic anions typically have common names that you must learn; some examples are in Table 10.2.1 . Polyatomic anions that contain a single metal or nonmetal atom plus one or more oxygen atoms are called oxoanions (or oxyanions) . In cases where only two oxoanions are known for an element, the name of the oxoanion with more oxygen atoms ends in - ate , and the name of the oxoanion with fewer oxygen atoms ends in - ite . For example, NO 3− is nitrate and NO 2− is nitrite. The halogens and some of the transition metals form more extensive series of oxoanions with as many as four members. In the names of these oxoanions, the prefix per - is used to identify the oxoanion with the most oxygen (so that ClO 4 − is perchlorate and ClO 3 − is chlorate), and the prefix hypo - is used to identify the anion with the fewest oxygen (ClO 2 − is chlorite and ClO − is hypochlorite). The relationship between the names of oxoanions and the number of oxygen atoms present is diagrammed in Figure 10.3.3 . Differentiating the oxoanions in such a series is no trivial matter. For example, the hypochlorite ion is the active ingredient in laundry bleach and swimming pool disinfectant, but compounds that contain the perchlorate ion can explode if they come into contact with organic substances. Write the name of the compound as the name of the cation followed by the name of the anion. It is not necessary to indicate the number of cations or anions present per formula unit in the name of an ionic compound because this information is implied by the charges on the ions. You must consider the charge of the ions when writing the formula for an ionic compound from its name, however. Because the charge on the chloride ion is −1 and the charge on the calcium ion is +2, for example, consistent with their positions in the periodic table, simple arithmetic tells you that calcium chloride must contain twice as many chloride ions as calcium ions to maintain electrical neutrality. Thus the formula is CaCl 2 . Similarly, calcium phosphate must be Ca 3 (PO 4 ) 2 because the cation and the anion have charges of +2 and −3, respectively. The best way to learn how to name ionic compounds is to work through a few examples, referring to Figure 10.3.1 , Table 10.1.2 , Table 10.2.1 , and Table 10.3.1 as needed. Figure 10.3.2 Metals That Form More Than One Cation and Their Locations in the Periodic Table With only a few exceptions, these metals are usually transition metals or actinides. Figure 10.3.3 The Relationship between the Names of Oxoanions and the Number of Oxygen Atoms Present Note the Pattern Cations are always named before anions. Most transition metals, many actinides, and the heaviest elements of groups 13–15 can form more than one cation. Example 10.3.1 Write the systematic name (and the common name if applicable) for each ionic compound. LiCl MgSO 4 (NH 4 ) 3 PO 4 Cu 2 O Given: empirical formula Asked for: name Strategy: A If only one charge is possible for the cation, give its name, consulting Table 10.1.2 or Table 10.2.1 if necessary. If the cation can have more than one charge ( Table 10.3.1 ), specify the charge using roman numerals. B If the anion does not contain oxygen, name it according to step 3a, using Table 10.1.2 and Table 10.3.1 if necessary. For polyatomic anions that contain oxygen, use Table 10.2.1 and the appropriate prefix and suffix listed in step 3b. C Beginning with the cation, write the name of the compound. Solution: A B Lithium is in group 1, so we know that it forms only the Li + cation, which is the lithium ion. Similarly, chlorine is in group 7, so it forms the Cl − anion, which is the chloride ion. C Because we begin with the name of the cation, the name of this compound is lithium chloride, which is used medically as an antidepressant drug. A B The cation is the magnesium ion, and the anion, which contains oxygen, is sulfate. C Because we list the cation first, the name of this compound is magnesium sulfate. A hydrated form of magnesium sulfate (MgSO 4 ·7H 2 O) is sold in drugstores as Epsom salts, a harsh but effective laxative. A B The cation is the ammonium ion (from Table 10.2.1 ), and the anion is phosphate. C The compound is therefore ammonium phosphate, which is widely used as a fertilizer. It is not necessary to specify that the formula unit contains three ammonium ions because three are required to balance the negative charge on phosphate. A B The cation is a transition metal that often forms more than one cation ( Table 10.3.1 ). We must therefore specify the positive charge on the cation in the name: copper(I) or, according to the older system, cuprous. The anion is oxide. C The name of this compound is copper(I) oxide or, in the older system, cuprous oxide. Copper(I) oxide is used as a red glaze on ceramics and in antifouling paints to prevent organisms from growing on the bottoms of boats. Cu 2 O . The bottom of a boat is protected with a red antifouling paint containing copper(I) oxide, Cu 2 O. Exercise Write the systematic name (and the common name if applicable) for each ionic compound. CuCl 2 MgCO 3 FePO 4 Answer: copper(II) chloride (or cupric chloride) magnesium carbonate iron(III) phosphate (or ferric phosphate) Example 10.3.2 Write the formula for each compound. calcium dihydrogen phosphate aluminum sulfate chromium(III) oxide Given: systematic name Asked for: formula Strategy: A Identify the cation and its charge using the location of the element in the periodic table and Table 10.1.2 Table 10.2.1 , and Table 10.3.1 . If the cation is derived from a metal that can form cations with different charges, use the appropriate roman numeral or suffix to indicate its charge. B Identify the anion using Table 10.1.2 and Table 10.2.1 Beginning with the cation, write the compound’s formula and then determine the number of cations and anions needed to achieve electrical neutrality. Solution: A Calcium is in group 2, so it forms only the Ca 2 + ion. B Dihydrogen phosphate is the \(H_2PO_4^−\) ion ( Table 10.2.1 ). Two \(H_2PO_4^−\) ions are needed to balance the positive charge on Ca 2 + , to give Ca(H 2 PO 4 ) 2 . A hydrate of calcium dihydrogen phosphate, Ca(H 2 PO 4 ) 2 ·H 2 O, is the active ingredient in baking powder. A Aluminum, near the top of group 13 in the periodic table, forms only one cation, Al 3 + ( Figure 10.3.2 ). B Sulfate is SO 4 2− ( Table 10.2.1 ). To balance the electrical charges, we need two \(Al^{3+}\) cations and three \(SO_4^{2−}\) anions, giving Al 2 (SO 4 ) 3 . Aluminum sulfate is used to tan leather and purify drinking water. A Because chromium is a transition metal, it can form cations with different charges. The roman numeral tells us that the positive charge in this case is +3, so the cation is \(Cr^{3+}\). B Oxide is \(O^{2−}\). Thus two cations (\(Cr^{3+}\)) and three anions (\(O^{2−}\)) are required to give an electrically neutral compound, \(Cr_2O_3\). This compound is a common green pigment that has many uses, including camouflage coatings. Chromium(III) oxide (\(Cr_2O_3\)) is a common pigment in dark green paints, such as camouflage paint. Figure used with permission from Wikipedia Exercise Write the formula for each compound. barium chloride sodium carbonate iron(III) hydroxide Answer: BaCl 2 Na 2 CO 3 Fe(OH) 3 Summary Ionic compounds are named according to systematic procedures, although common names are widely used. Systematic nomenclature enables us to write the structure of any compound from its name and vice versa. Ionic compounds are named by writing the cation first, followed by the anion. If a metal can form cations with more than one charge, the charge is indicated by roman numerals in parentheses following the name of the metal. Oxoanions are polyatomic anions that contain a single metal or nonmetal atom and one or more oxygen atoms. Key Takeaway There is a systematic method used to name ionic compounds. Conceptual Problems Name each cation. K + Al 3 + NH 4 + Mg 2 + Li + Name each anion. Br − CO 3 2− S 2− NO 3 − HCO 2 − F − ClO − C 2 O 4 2− Name each anion. PO 4 3− Cl − SO 3 2− CH 3 CO 2 − HSO 4 − ClO 4 − NO 2 − O 2− Name each anion. SO 4 2− CN − Cr 2 O 7 2− N 3− OH − I − O 2 2− Name each compound. MgBr 2 NH 4 CN CaO KClO 3 K 3 PO 4 NH 4 NO 2 NaN 3 Name each compound. NaNO 3 Cu 3 (PO 4 ) 2 NaOH Li 4 C CaF 2 NH 4 Br MgCO 3 Name each compound. RbBr Mn 2 (SO 4 ) 3 NaClO (NH 4 ) 2 SO 4 NaBr KIO 3 Na 2 CrO 4 Name each compound. NH 4 ClO 4 SnCl 4 Fe(OH) 2 Na 2 O MgCl 2 K 2 SO 4 RaCl 2 Name each compound. KCN LiOH CaCl 2 NiSO 4 NH 4 ClO 2 LiClO 4 La(CN) 3 Answer rubidium bromide manganese(III) sulfate sodium hypochlorite ammonium sulfate sodium bromide potassium iodate sodium chromate Numerical Problems For each ionic compound, name the cation and the anion and give the charge on each ion. BeO Pb(OH) 2 BaS Na 2 Cr 2 O 7 ZnSO 4 KClO NaH 2 PO 4 For each ionic compound, name the cation and the anion and give the charge on each ion. Zn(NO 3 ) 2 CoS BeCO 3 Na 2 SO 4 K 2 C 2 O 4 NaCN FeCl 2 Write the formula for each compound. magnesium carbonate aluminum sulfate potassium phosphate lead(IV) oxide silicon nitride sodium hypochlorite titanium(IV) chloride disodium ammonium phosphate Write the formula for each compound. lead(II) nitrate ammonium phosphate silver sulfide barium sulfate cesium iodide sodium bicarbonate potassium dichromate sodium hypochlorite Write the formula for each compound. zinc cyanide silver chromate lead(II) iodide benzene copper(II) perchlorate Write the formula for each compound. calcium fluoride sodium nitrate iron(III) oxide copper(II) acetate sodium nitrite Write the formula for each compound. sodium hydroxide calcium cyanide magnesium phosphate sodium sulfate nickel(II) bromide calcium chlorite titanium(IV) bromide Write the formula for each compound. sodium chlorite potassium nitrite sodium nitride (also called sodium azide) calcium phosphide tin(II) chloride calcium hydrogen phosphate iron(II) chloride dihydrate Write the formula for each compound. potassium carbonate chromium(III) sulfite cobalt(II) phosphate magnesium hypochlorite nickel(II) nitrate hexahydrate Contributors Anonymous Modified by Joshua Halpern, Scott Sinex and Scott Johnson |
Courses/Lumen_Learning/Book%3A_English_Composition_I-3_(Lumen)/28%3A_Grammar%3A_Sentence_Structure/27.1%3A_Outcome%3A_Punctuation | Critique the use of common punctuation marks. Now that we’ve learned about the different types of words, it’s time to learn punctuation. These little marks can often be the cause of a lot of heartaches and headaches. Errors in punctuation can often have unintended meanings. For example consider the difference the comma makes in these two sentences: Let’s eat, Grandpa. Let’s eat Grandpa. However, punctuation doesn’t exist simply to cause problems; in fact, it was created to help communication. These marks were invented to guide readers through passages—to let them know how and where words relate to each other. When you learn the rules of punctuation, you equip yourself with an extensive toolset so you can better craft language to communicate the exact message you want. As we mentioned at the beginning of this module, different style guides have slightly different rules for grammar. This is especially true when it comes to punctuation. This outcome will cover the MLA rules for punctuation, but we’ll also make note of rules from other styles when they’re significantly different. What You Will Learn to Do Critique the use of end punctuation: periods, question marks, exclamation marks Critique the use of commas Critique the use of semicolons and colons Critique the use of hyphens and dashes Critique the use of apostrophes and quotation marks Critique the use of brackets, parentheses, and ellipses CC licensed content, Original Outcome: Punctuation. Provided by : Lumen Learning. License : CC BY: Attribution |
Courses/Grinnell_College/CHM_363%3A_Physical_Chemistry_1_(Grinnell_College)/05%3A_Boltzmann/5.05%3A_Partition_Functions_can_be_Decomposed_into_Partition_Functions_of_Each_Degree_of_Freedom | From the previous sections, the partition function for a system of \(N\) indistinguishable and independent molecules is: \[ Q(N,V,\beta) = \dfrac{\sum_i{e^{-\beta E_i}}}{N!} \label{ID1} \] And the average energy of the system is: \[ \langle E \rangle = kT^2 \left(\dfrac{\partial \ln{Q}}{\partial T}\right) \label{ID2} \] We can combine these two equations to obtain: \[ \begin{split} \langle E \rangle &= kT^2 \left(\dfrac{\partial \ln{Q}}{\partial T}\right)_{N,V} \\ &= NkT^2 \left(\dfrac{\partial \ln{q}}{\partial T}\right)_V \\ &= N\sum_i{\epsilon_i \dfrac{e^{-\epsilon_i/kT}}{q(V,T)}} \end{split} \label{ID3} \] The average energy is equal to: \[ \langle E \rangle = N \langle \epsilon \rangle \label{aveE} \] where \(\langle \epsilon \rangle\) is the average energy of a single particle. If we compare Equation \(\ref{ID2}\) with Equation \(\ref{ID2}\), we can see: \[ \langle \epsilon \rangle = \sum_i{\epsilon_i \dfrac{e^{-\epsilon_i/kT}}{q(V,T)}} \nonumber \] The probability that a particle is in state \(i\), \(\pi_i\), is given by: \[ \langle \epsilon \rangle = \dfrac{e^{-\epsilon_i/kT}}{q(V,T)} = \dfrac{e^{-\epsilon_i/kT}}{\sum_i{e^{-\epsilon_i/kT}}} \nonumber \] The energy of a particle is a sum of the energy of each degree of freedom for that particle. In the case of a molecule, the energy is: \[ \epsilon = \epsilon_\text{trans} + \epsilon_\text{rot} + \epsilon_\text{vib} + \epsilon_\text{elec} \nonumber \] The molecular partition function is the product of the degree of freedom partition functions: \[ q(V,T) = q_\text{trans} q_\text{rot} q_\text{vib} q_\text{elec} \nonumber \] The partition function for each degree of freedom follows the same is related to the Boltzmann distribution. For example, the vibrational partition function is: \[ q_\text{vib} = \sum_i{e^{-\epsilon_i/kT}} \nonumber \] The average energy of each degree of freedom follows the same pattern as before. For example, the average vibrational energy is: \[ \langle \epsilon_\text{vib} \rangle = kT^2\dfrac{\partial \ln{q_\text{vib}}}{\partial t} = -\dfrac{\partial \ln{q_\text{vib}}}{\partial \beta} \nonumber \] |
Courses/Oregon_Institute_of_Technology/OIT%3A_CHE_333_-_Organic_Chemistry_III_(Lund)/New_Page/7%3A_Acid-base_Reactions | 7.0: Prelude to Acid-base Reactions The glass flask sitting on a bench in Dr. Barry Marshall's lab in Perth, Western Australia, contained about thirty milliliters of a distinctly unappetizing murky, stinking yellowish liquid. A few days earlier he had poured a nutrient broth into the flask, then dropped in a small piece of tissue sample taken from the stomach of a patient suffering from chronic gastritis. 7.1: Overview of Acid-Base Reactions We’ll begin our discussion of acid-base chemistry with a couple of essential definitions. The first of these was proposed in 1923 by the Danish chemist Johannes Brønsted and the English chemist Thomas Lowry, and has come to be known as the Brønsted-Lowry definition of acidity and basicity. 7.2: The Acidity Constant You are no doubt aware that some acids are stronger than others. The relative acidity of different compounds or functional groups – in other words, their relative capacity to donate a proton to a common base under identical conditions – is quantified by a number called the acidity constant, abbreviated K 7.3: Structural Effects on Acidity and Basicity Now that we know how to quantify the strength of an acid or base, our next job is to gain an understanding of the fundamental reasons behind why one compound is more acidic or more basic than another. This is a big step: we are, for the first time, taking our knowledge of organic structure and applying it to a question of organic reactivity. 7.4: Acid-base Properties of Phenols Resonance effects involving aromatic structures can have a dramatic influence on acidity and basicity. Notice, for example, the difference in acidity between phenol and cyclohexanol. 7.5: Acid-base properties of nitrogen-containing functional groups Many of the acid-base reactions we will see throughout our study of biological organic chemistry involve functional groups which contain nitrogen. In general, a nitrogen atom with three bonds and a lone pair of electrons can potentially act as a proton-acceptor (a base) - but basicity is reduced if the lone pair electrons are stabilized somehow. 7.6: Carbon Acids So far, we have limited our discussion of acidity and basicity to heteroatom acids, where the acidic proton is bound to an oxygen, nitrogen, sulfur, or halogen. However, carbon acids - in which the proton to be donated is bonded to a carbon atom - play an integral role in biochemistry. 7.7: Polyprotic Acids Polyprotic acids are capable of donating more than one proton. The most important polyprotic acid group from a biological standpoint is triprotic phosphoric acid. Because phosphoric acid has three acidic protons, it also has three pKa values. 7.8: Effects of enzyme microenvironment on acidity and basicity Virtually all biochemical reactions take place inside the active site pocket of an enzyme, rather than free in aqueous solution. The microenvironment inside an enzyme's active site can often be very different from the environment outside in the aqueous solvent. Consider, for example, the side chain carboxylate on an aspartate residue in an enzyme. 7.E: Acid-base Reactions (Exercises) 7.S: Acid-base Reactions (Summary) |
Bookshelves/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/06%3A_Lipids/6.03%3A_Glycerolipids | Learning Objectives Define glycerolipids and triglycerides and understand their structures. Understand the difference in composition and properties of fats and oils and their primary role as energy storage molecules. Understand the chemical properties of fats and oils, including hydrolysis, saponification, and partial or complete hydrogenation. What is a glycerolipid? Glycerolipids have two components: propane-1,2,3-triol, also called glycerol, and one, two, or three fatty acids. Recall that a condensation reaction between an alcohol and a carboxylic acid forms an ester by eliminating a water molecule. The following example shows the reaction of glycerol with three molecules of stearic acid, creating a triester. A mono-, di-, or tri-ester of glycerol is called glycerolipid. The name of a glycerolipid begins with glyceryl and is followed by carboxylate. For example, the triester of glycerol shown above is glyceryl tristearate, which is commonly known as tristearin. Glycerolipids that are triesters of glycerol with three fatty acids are called triacylglycerol or triglycerides . For example, glyceryl tristearate shown above is a triglyceride. Triglycerides may be esters of three molecules of the same fatty acids, e.g., tristearin. Still, often the triglycerides found in nature are esters of two or three different fatty acids. Fatty acids usually found in fats and vegetable oils include lauric, myristic, palmitic, stearic, oleic, linoleic, and linolenic acids. An example of a mixed triglyceride of stearic, oleic, and palmitic acid is shown below. Fats and oils Fats and oils are triglycerides used as energy storage molecules in animals and plants. Energy storage is essential for hibernating animals that live in icy environments. They have plenty of food available during summer but no food and below-freezing temperatures in winter. They eat plants, seeds, and nuts and accumulate fat that becomes their energy source when hibernating in winter. For example, the polar bear shown on the right can accumulate up to 14 kg of fat per week in summer when the food is plenty and uses it as an energy source when the animal hibernates for 4 to 7 months in winter. Although glycogen is a quick energy source, fats and oils are more energy dense for two reasons: first, fats and oils are in a more reduced form than glycogen and release more energy per mass upon oxidation, and second, glycogen is hydrophilic and absorbs water, increasing its mass. Physical properties of fats and oils Pure fats and oils are colorless, odorless, and tasteless. The color, taste, and odor associated with butter -an animal fat, and olive oil -a vegetable oil, is due to a small number of other substances mixed with the triglycerides. Fats come from animal sources, such as meat, whole milk, lard, butter, and cheese. Vegetable oils come from plant sources, e.g., soybeans, peanuts, sunflowers, etc. Fats are usually solid at room temperature because they have a higher proportion of long-chain unsaturated fatty acids with higher melting points, e.g., lard and butter shown in Figure \(\PageIndex{1}\). Vegetable oils are usually liquid at room temperature because they contain more unsaturated or short-chain saturated fatty acids. For example, canola oil, safflower oil, flaxseed oil, sunflower oil, corn oil, olive oil, soyabean oil, peanut oil, and cottonseed oil are vegetable oils that are liquid at room temperature and have a higher proportion of unsaturated fatty acids as shown in Figure \(\PageIndex{2}\). Palm and coconut oils are vegetable oils with a higher balance of saturated fatty acids. Still, they are liquid at room temperature because they have short-chain saturated fatty acids. For example, coconut mainly comprises lauric acid (12:0), a short-chain fatty acid. Chemical properties of fats and oils Three important reactions of triglycerides are hydrolysis by water in the presence of acid or lipase enzyme, saponification, i.e., base-promoted hydrolysis, and hydrogenation of double bonds in the fatty acids. Hydrolysis Easters are hydrolyzed (split) by water in the presence of an acid catalyst into alcohol and carboxylic acid. The same reaction happens with triglycerides, i.e., fats and vegetable oils—for example, tripalmitin hydrolyzes into glycerol and three palmitic acid molecules, as shown below. Lipase enzymes do the same reaction during the digestion of triglycerides, e.g., trilaurin is hydrolyzed by lipase to glycerol and three molecules of lauric acid, as shown in the reaction below. The digestion of lipids starts in the mouth with lingual lipases secreted by glands in the tongue and continues in the stomach with lingual and gastric lipases. Glycerol in the hydrolysis product is soluble in water, but fatty acids are not. Fatty acids are emulsified in the stomach and later mixed with bile and pancreatic juice to continue the process of digestion. Saponification Saponification is base-promoted hydrolysis of triglycerides that produces glycerol and salts of the fatty acids, a soap, as shown below for the case of saponification of tripalmitin. Different varieties of soaps are shown in the figure on the right. Sodium salts of fatty acids make hard soaps, while potassium salts make soft soaps. Three of the most common soaps are sodium stearate, oleate, and linoleate. Salts of saturated fatty acids make rigid, and those of polyunsaturated fatty acids make soft soap. Perfumes are added for scent and dyes for color, sand is added in scouring soap, and the air is blown into the soap to make it float on water. Hydrogenation Alkenes add \(\ce{H2}\) and covert to alkane in the presence of \(\ce{Ni}\), \(\ce{Pt}\), or \(\ce{Pd}\) catalyst. The same reaction happens with the \(\ce{C=C}\) in unsaturated fatty acids found in triglycerides. The conversion of unsaturated fatty acids to saturated fatty acids increases the melting point of triglycerides. Therefore, vegetable oils with more unsaturated fatty acids become semisolid or solid after partial or complete hydrogenation, as illustrated by the reaction in Figure \(\PageIndex{3}\). Hydrogenation of vegetable oils is a commercial process to convert vegetable oils to semisolid products like margarine and shortening. The driving force for the industrial process of hydrogenation of vegetable oils is that margarine or shortenings are used as cheaper alternatives to butter with a longer shelf-life. Complete hydrogenation of vegetable oils converts them into hard margarine. Often partial hydrogenation is performed as the partially hydrogenated product is semisolid and mimics butter better than hard shortening. Dyes and flavors are mixed with it to mimic butter's color, taste, and odor. Unwanted trans-fatty acids are by-products of the hydrogenation of vegetable oils. Like all other processed food products, hydrogenated vegetable oil products are associated with health problems. Partial hydrogenation converts some of the cis -\(\ce{C=C}\) bonds to trans -\(\ce{C=C}\) bonds, as illustrated in Figure \(\PageIndex{3}\). Trans -\(\ce{C=C}\) bonds are rare in some fatty acids. Research reports indicate that trans-fatty acids affect blood cholesterol like unsaturated fatty acids. Research reports also indicated that trans-fatty acids increase the level of low-density lipids (LDL), considered bad cholesterol, and decrease the level of high-density lipids (HDL), regarded as good cholesterol. Health organizations force the manufacturers to show the trans-fatty acid contents on the product labels. They are trying to educate the consumers about trans-fatty acids, as shown in Figure \(\PageIndex{4}\) from U.S. Food and Drug Administration. |
Bookshelves/Organic_Chemistry/Organic_Chemistry_(OpenStax)/08%3A_Alkenes_-_Reactions_and_Synthesis/8.14%3A_Chemistry_MattersTerpenes-_Naturally_Occurring_Alkenes | Ever since its discovery in Persia around 1000 A.D., it has been known that steam distillation , the distillation of plant materials together with water, produces a fragrant mixture of liquids called essential oils. The resulting oils have long been used as medicines, spices, and perfumes, and their investigation played a major role in the emergence of organic chemistry as a science during the 19th century. Chemically, plant essential oils consist largely of mixtures of compounds called terpenoids—small organic molecules with an immense diversity of structure. More than 60,000 different terpenoids are known. Some are open-chain molecules, and others contain rings; some are hydrocarbons, and others contain oxygen. Hydrocarbon terpenoids, in particular, are known as terpenes, and all contain double bonds. For example: Regardless of their apparent structural differences, all terpenoids are related. According to a formalism called the isoprene rule, they can be thought of as arising from head-to-tail joining of 5-carbon isoprene units (2-methyl-1,3-butadiene). Carbon 1 is the head of the isoprene unit, and carbon 4 is the tail. For example, myrcene contains two isoprene units joined head to tail, forming an 8-carbon chain with two 1-carbon branches. α -Pinene similarly contains two isoprene units assembled into a more complex cyclic structure, and humulene contains three isoprene units. See if you can identify the isoprene units in α -pinene, humulene, and β -santalene. Terpenes (and terpenoids) are further classified according to the number of 5-carbon units they contain. Thus, monoterpenes are 10-carbon substances derived from two isoprene units, sesquiterpenes are 15-carbon molecules derived from three isoprene units, diterpenes are 20-carbon substances derived from four isoprene units, and so on. Monoterpenes and sesquiterpenes are found primarily in plants, but the higher terpenoids occur in both plants and animals, and many have important biological roles. The triterpenoid lanosterol, for instance, is the biological precursor from which all steroid hormones are made. Isoprene itself is not the true biological precursor of terpenoids. Nature instead uses two “isoprene equivalents”—isopentenyl diphosphate and dimethylallyl diphosphate—which are themselves made by two different routes depending on the organism. Lanosterol, in particular, is biosynthesized from acetic acid by a complex pathway that has been worked out in great detail. We’ll look at the subject more closely in Sections 27.6 and 27.8 . |
Courses/Rutgers_University/General_Chemistry/Chapter_7._Chemical_Reactions_and_Chemical_Quantities/7.0%3A_Prelude_to_Stoichiometry | Solid-fuel rockets are a central feature in the world’s space exploration programs, including the new Space Launch System being developed by the National Aeronautics and Space Administration (NASA) to replace the retired Space Shuttle fleet (Figure \(\PageIndex{1}\)). The engines of these rockets rely on carefully prepared solid mixtures of chemicals combined in precisely measured amounts. Igniting the mixture initiates a vigorous chemical reaction that rapidly generates large amounts of gaseous products. These gases are ejected from the rocket engine through its nozzle, providing the thrust needed to propel heavy payloads into space. Both the nature of this chemical reaction and the relationships between the amounts of the substances being consumed and produced by the reaction are critically important considerations that determine the success of the technology. This chapter will describe how to symbolize chemical reactions using chemical equations, how to classify some common chemical reactions by identifying patterns of reactivity, and how to determine the quantitative relations between the amounts of substances involved in chemical reactions—that is, the reaction stoichiometry . |
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