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Bookshelves/General_Chemistry/Chemistry_2e_(OpenStax)/06%3A_Electronic_Structure_and_Periodic_Properties/6.04%3A_Electronic_Structure_of_Atoms_(Electron_Configurations)	Learning Objectives By the end of this section, you will be able to: Derive the predicted ground-state electron configurations of atoms Identify and explain exceptions to predicted electron configurations for atoms and ions Relate electron configurations to element classifications in the periodic table Having introduced the basics of atomic structure and quantum mechanics, we can use our understanding of quantum numbers to determine how atomic orbitals relate to one another. This allows us to determine which orbitals are occupied by electrons in each atom. The specific arrangement of electrons in orbitals of an atom determines many of the chemical properties of that atom. Orbital Energies and Atomic Structure The energy of atomic orbitals increases as the principal quantum number, n , increases. In any atom with two or more electrons, the repulsion between the electrons makes energies of subshells with different values of l differ so that the energy of the orbitals increases within a shell in the order s < p < d < f. Figure $\PageIndex{1}$: depicts how these two trends in increasing energy relate. The 1 s orbital at the bottom of the diagram is the orbital with electrons of lowest energy. The energy increases as we move up to the 2 s and then 2 p , 3 s , and 3 p orbitals, showing that the increasing n value has more influence on energy than the increasing l value for small atoms. However, this pattern does not hold for larger atoms. The 3 d orbital is higher in energy than the 4 s orbital. Such overlaps continue to occur frequently as we move up the chart. Electrons in successive atoms on the periodic table tend to fill low-energy orbitals first. Thus, many students find it confusing that, for example, the 5 p orbitals fill immediately after the 4 d , and immediately before the 6 s . The filling order is based on observed experimental results, and has been confirmed by theoretical calculations. As the principal quantum number, n , increases, the size of the orbital increases and the electrons spend more time farther from the nucleus. Thus, the attraction to the nucleus is weaker and the energy associated with the orbital is higher (less stabilized). But this is not the only effect we have to take into account. Within each shell, as the value of l increases, the electrons are less penetrating (meaning there is less electron density found close to the nucleus), in the order s > p > d > f . Electrons that are closer to the nucleus slightly repel electrons that are farther out, offsetting the more dominant electron–nucleus attractions slightly (recall that all electrons have −1 charges, but nuclei have + Z charges). This phenomenon is called shielding and will be discussed in more detail in the next section. Electrons in orbitals that experience more shielding are less stabilized and thus higher in energy. For small orbitals (1 s through 3 p ), the increase in energy due to n is more significant than the increase due to l ; however, for larger orbitals the two trends are comparable and cannot be simply predicted. We will discuss methods for remembering the observed order. The arrangement of electrons in the orbitals of an atom is called the electron configuration of the atom. We describe an electron configuration with a symbol that contains three pieces of information (Figure $\PageIndex{2}$): The number of the principal quantum shell, $n$, The letter that designates the orbital type (the subshell, $l$), and A superscript number that designates the number of electrons in that particular subshell. For example, the notation 2 p 4 (read "two–p–four") indicates four electrons in a p subshell ($l = 1$) with a principal quantum number ($n$) of 2. The notation 3 d 8 (read "three–d–eight") indicates eight electrons in the d subshell (i.e., $l = 2$) of the principal shell for which $n = 3$. The Aufbau Principle To determine the electron configuration for any particular atom, we can “build” the structures in the order of atomic numbers. Beginning with hydrogen, and continuing across the periods of the periodic table, we add one proton at a time to the nucleus and one electron to the proper subshell until we have described the electron configurations of all the elements. This procedure is called the Aufbau principle , from the German word Aufbau (“to build up”). Each added electron occupies the subshell of lowest energy available (in the order shown in Figure $\PageIndex{1}$), subject to the limitations imposed by the allowed quantum numbers according to the Pauli exclusion principle. Electrons enter higher-energy subshells only after lower-energy subshells have been filled to capacity. Figure $\PageIndex{3}$ illustrates the traditional way to remember the filling order for atomic orbitals. Since the arrangement of the periodic table is based on the electron configurations, Figure $\PageIndex{4}$ provides an alternative method for determining the electron configuration. The filling order simply begins at hydrogen and includes each subshell as you proceed in increasing Z order. For example, after filling the 3 p block up to Ar, we see the orbital will be 4s (K, Ca), followed by the 3 d orbitals. We will now construct the ground-state electron configuration and orbital diagram for a selection of atoms in the first and second periods of the periodic table. Orbital diagrams are pictorial representations of the electron configuration, showing the individual orbitals and the pairing arrangement of electrons. We start with a single hydrogen atom (atomic number 1), which consists of one proton and one electron. Referring to Figure $\PageIndex{3}$ or Figure $\PageIndex{4}$, we would expect to find the electron in the 1 s orbital. By convention, the $m_s=+\frac{1}{2}$ value is usually filled first. The electron configuration and the orbital diagram are: Following hydrogen is the noble gas helium, which has an atomic number of 2. The helium atom contains two protons and two electrons. The first electron has the same four quantum numbers as the hydrogen atom electron ($ n = 1$, $ l = 0$, $ m_ l = 0$, $m_s=+\frac{1}{2}$. The second electron also goes into the 1 s orbital and fills that orbital. The second electron has the same n, l, and ml quantum numbers, but must have the opposite spin quantum number, $m_s=-\frac{1}{2}$. This is in accord with the Pauli exclusion principle: No two electrons in the same atom can have the same set of four quantum numbers. For orbital diagrams, this means two arrows go in each box (representing two electrons in each orbital) and the arrows must point in opposite directions (representing paired spins). The electron configuration and orbital diagram of helium are: The $n = 1$ shell is completely filled in a helium atom. The next atom is the alkali metal lithium with an atomic number of 3. The first two electrons in lithium fill the 1 s orbital and have the same sets of four quantum numbers as the two electrons in helium. The remaining electron must occupy the orbital of next lowest energy, the 2 s orbital (Figure $\PageIndex{3}$ or Figure $\PageIndex{4}$). Thus, the electron configuration and orbital diagram of lithium are: An atom of the alkaline earth metal beryllium, with an atomic number of 4, contains four protons in the nucleus and four electrons surrounding the nucleus. The fourth electron fills the remaining space in the 2 s orbital. An atom of boron (atomic number 5) contains five electrons. The n = 1 shell is filled with two electrons and three electrons will occupy the n = 2 shell. Because any s subshell can contain only two electrons, the fifth electron must occupy the next energy level, which will be a 2 p orbital. There are three degenerate 2 p orbitals ( m l = −1, 0, +1) and the electron can occupy any one of these p orbitals. When drawing orbital diagrams, we include empty boxes to depict any empty orbitals in the same subshell that we are filling. Carbon (atomic number 6) has six electrons. Four of them fill the 1 s and 2 s orbitals. The remaining two electrons occupy the 2 p subshell. We now have a choice of filling one of the 2 p orbitals and pairing the electrons or of leaving the electrons unpaired in two different, but degenerate, p orbitals. The orbitals are filled as described by Hund’s rule : the lowest-energy configuration for an atom with electrons within a set of degenerate orbitals is that having the maximum number of unpaired electrons. Thus, the two electrons in the carbon 2 p orbitals have identical $n$, $l$, and $m_s$ quantum numbers and differ in their $m_l$ quantum number (in accord with the Pauli exclusion principle). The electron configuration and orbital diagram for carbon are: Nitrogen (atomic number 7) fills the 1 s and 2 s subshells and has one electron in each of the three 2 p orbitals, in accordance with Hund’s rule. These three electrons have unpaired spins. Oxygen (atomic number 8) has a pair of electrons in any one of the 2 p orbitals (the electrons have opposite spins) and a single electron in each of the other two. Fluorine (atomic number 9) has only one 2 p orbital containing an unpaired electron. All of the electrons in the noble gas neon (atomic number 10) are paired, and all of the orbitals in the n = 1 and the n = 2 shells are filled. The electron configurations and orbital diagrams of these four elements are: The alkali metal sodium (atomic number 11) has one more electron than the neon atom. This electron must go into the lowest-energy subshell available, the 3 s orbital, giving a 1 s 2 2 s 2 2 p 6 3 s 1 configuration. The electrons occupying the outermost shell orbital(s) (highest value of n ) are called valence electrons , and those occupying the inner shell orbitals are called core electrons (Figure $\PageIndex{5}$). Since the core electron shells correspond to noble gas electron configurations, we can abbreviate electron configurations by writing the noble gas that matches the core electron configuration, along with the valence electrons in a condensed format. For our sodium example, the symbol [Ne] represents core electrons, (1 s 2 2 s 2 2 p 6 ) and our abbreviated or condensed configuration is [Ne]3 s 1 . Similarly, the abbreviated configuration of lithium can be represented as [He]2 s 1 , where [He] represents the configuration of the helium atom, which is identical to that of the filled inner shell of lithium. Writing the configurations in this way emphasizes the similarity of the configurations of lithium and sodium. Both atoms, which are in the alkali metal family, have only one electron in a valence s subshell outside a filled set of inner shells. \[\begin{aligned} & Li :[ He ] 2 s^1 \\[4pt] & Na :[ Ne ] 3 s^1 \end{aligned} \nonumber \] The alkaline earth metal magnesium (atomic number 12), with its 12 electrons in a [Ne]3 s 2 configuration, is analogous to its family member beryllium, [He]2 s 2 . Both atoms have a filled s subshell outside their filled inner shells. Aluminum (atomic number 13), with 13 electrons and the electron configuration [Ne]3 s 2 3 p 1 , is analogous to its family member boron, [He]2 s 2 2 p 1 . The electron configurations of silicon (14 electrons), phosphorus (15 electrons), sulfur (16 electrons), chlorine (17 electrons), and argon (18 electrons) are analogous in the electron configurations of their outer shells to their corresponding family members carbon, nitrogen, oxygen, fluorine, and neon, respectively, except that the principal quantum number of the outer shell of the heavier elements has increased by one to n = 3. Figure $\PageIndex{6}$: shows the lowest energy, or ground-state, electron configuration for these elements as well as that for atoms of each of the known elements. When we come to the next element in the periodic table, the alkali metal potassium (atomic number 19), we might expect that we would begin to add electrons to the 3 d subshell. However, all available chemical and physical evidence indicates that potassium is like lithium and sodium, and that the next electron is not added to the 3 d level but is, instead, added to the 4 s level (Figure $\PageIndex{6}$). As discussed previously, the 3 d orbital with no radial nodes is higher in energy because it is less penetrating and more shielded from the nucleus than the 4 s , which has three radial nodes. Thus, potassium has an electron configuration of [Ar]4 s 1 . Hence, potassium corresponds to Li and Na in its valence shell configuration. The next electron is added to complete the 4 s subshell and calcium has an electron configuration of [Ar]4 s 2 . This gives calcium an outer-shell electron configuration corresponding to that of beryllium and magnesium. Beginning with the transition metal scandium (atomic number 21), additional electrons are added successively to the 3 d subshell. This subshell is filled to its capacity with 10 electrons (remember that for l = 2 [ d orbitals], there are 2 l + 1 = 5 values of m l , meaning that there are five d orbitals that have a combined capacity of 10 electrons). The 4 p subshell fills next. Note that for three series of elements, scandium (Sc) through copper (Cu), yttrium (Y) through silver (Ag), and lutetium (Lu) through gold (Au), a total of 10 d electrons are successively added to the ( n – 1) shell next to the n shell to bring that ( n – 1) shell from 8 to 18 electrons. For two series, lanthanum (La) through lutetium (Lu) and actinium (Ac) through lawrencium (Lr), 14 f electrons ( l = 3, 2 l + 1 = 7 m l values; thus, seven orbitals with a combined capacity of 14 electrons) are successively added to the ( n – 2) shell to bring that shell from 18 electrons to a total of 32 electrons. Example $\PageIndex{1}$: Quantum Numbers and Electron Configurations What is the electron configuration and orbital diagram for a phosphorus atom? What are the four quantum numbers for the last electron added? Solution The atomic number of phosphorus is 15. Thus, a phosphorus atom contains 15 electrons. The order of filling of the energy levels is 1 s , 2 s , 2 p , 3 s , 3 p , 4 s , . . . The 15 electrons of the phosphorus atom will fill up to the 3 p orbital, which will contain three electrons: The last electron added is a 3 p electron. Therefore, n = 3 and, for a p -type orbital, l = 1. The m l value could be –1, 0, or +1. The three p orbitals are degenerate, so any of these m l values is correct. For unpaired electrons, convention assigns the value of $+ \frac{1}{2}$ for the spin quantum number; thus, $m_s=+ \frac{1}{2}$ Exercise $\PageIndex{1}$ Identify the atoms from the electron configurations given: [Ar]4 s 2 3 d 5 [Kr]5 s 2 4 d 10 5 p 6 Answer Mn Xe The periodic table can be a powerful tool in predicting the electron configuration of an element. However, we do find exceptions to the order of filling of orbitals that are shown in Figure $\PageIndex{3}$: or Figure $\PageIndex{4}$. For instance, the electron configurations (shown in Figure $\PageIndex{6}$) of the transition metals chromium (Cr; atomic number 24) and copper (Cu; atomic number 29), among others, are not those we would expect. In general, such exceptions involve subshells with very similar energy, and small effects can lead to changes in the order of filling. In the case of Cr and Cu, we find that half-filled and completely filled subshells apparently represent conditions of preferred stability. This stability is such that an electron shifts from the 4 s into the 3 d orbital to gain the extra stability of a half-filled 3 d subshell (in Cr) or a filled 3 d subshell (in Cu). Other exceptions also occur. For example, niobium (Nb, atomic number 41) is predicted to have the electron configuration [Kr]5 s 2 4 d 3 . Experimentally, we observe that its ground-state electron configuration is actually [Kr]5 s 1 4 d 4 . We can rationalize this observation by saying that the electron–electron repulsions experienced by pairing the electrons in the 5 s orbital are larger than the gap in energy between the 5 s and 4 d orbitals. There is no simple method to predict the exceptions for atoms where the magnitude of the repulsions between electrons is greater than the small differences in energy between subshells. Electron Configurations and the Periodic Table As described earlier, the periodic table arranges atoms based on increasing atomic number so that elements with the same chemical properties recur periodically. When their electron configurations are added to the table (Figure $\PageIndex{6}$), we also see a periodic recurrence of similar electron configurations in the outer shells of these elements. Because they are in the outer shells of an atom, valence electrons play the most important role in chemical reactions. The outer electrons have the highest energy of the electrons in an atom and are more easily lost or shared than the core electrons. Valence electrons are also the determining factor in some physical properties of the elements. Elements in any one group (or column) have the same number of valence electrons; the alkali metals lithium and sodium each have only one valence electron, the alkaline earth metals beryllium and magnesium each have two, and the halogens fluorine and chlorine each have seven valence electrons. The similarity in chemical properties among elements of the same group occurs because they have the same number of valence electrons. It is the loss, gain, or sharing of valence electrons that defines how elements react. It is important to remember that the periodic table was developed on the basis of the chemical behavior of the elements, well before any idea of their atomic structure was available. Now we can understand why the periodic table has the arrangement it has—the arrangement puts elements whose atoms have the same number of valence electrons in the same group. This arrangement is emphasized in Figure $\PageIndex{6}$: , which shows in periodic-table form the electron configuration of the last subshell to be filled by the Aufbau principle. The colored sections of Figure $\PageIndex{6}$: show the three categories of elements classified by the orbitals being filled: main group, transition, and inner transition elements. These classifications determine which orbitals are counted in the valence shell , or highest energy level orbitals of an atom. Main group elements (sometimes called representative elements ) are those in which the last electron added enters an s or a p orbital in the outermost shell, shown in blue and red in Figure $\PageIndex{6}$. This category includes all the nonmetallic elements, as well as many metals and the metalloids. The valence electrons for main group elements are those with the highest n level. For example, gallium (Ga, atomic number 31) has the electron configuration [Ar] 4 s 2 3 d 10 4 p 1 , which contains three valence electrons (underlined). The completely filled d orbitals count as core, not valence, electrons. Transition elements or transition metals . These are metallic elements in which the last electron added enters a d orbital. The valence electrons (those added after the last noble gas configuration) in these elements include the ns and ( n – 1) d electrons. The official IUPAC definition of transition elements specifies those with partially filled d orbitals. Thus, the elements with completely filled orbitals (Zn, Cd, Hg, as well as Cu, Ag, and Au in Figure $\PageIndex{6}$) are not technically transition elements. However, the term is frequently used to refer to the entire d block (colored yellow in Figure $\PageIndex{6}$), and we will adopt this usage in this textbook. Inner transition elements are metallic elements in which the last electron added occupies an f orbital. They are shown in green in Figure $\PageIndex{6}$. The valence shells of the inner transition elements consist of the ( n – 2) f, the ( n – 1) d , and the ns subshells. There are two inner transition series: The lanthanide series: lanthanum (La) through lutetium (Lu) The actinide series: actinium (Ac) through lawrencium (Lr) Lanthanum and actinium, because of their similarities to the other members of the series, are included and used to name the series, even though they are transition metals with no f electrons. Electron Configurations of Ions Ions are formed when atoms gain or lose electrons. A cation (positively charged ion) forms when one or more electrons are removed from a parent atom. For main group elements, the electrons that were added last are the first electrons removed. For transition metals and inner transition metals, however, electrons in the s orbital are easier to remove than the d or f electrons, and so the highest ns electrons are lost, and then the ( n – 1) d or ( n – 2) f electrons are removed. An anion (negatively charged ion) forms when one or more electrons are added to a parent atom. The added electrons fill in the order predicted by the Aufbau principle. Example $\PageIndex{2}$: Predicting Electron Configurations of Ions What is the electron configuration of: Na + P 3– Al 2 + Fe 2 + Sm 3 + Solution First, write out the electron configuration for each parent atom. We have chosen to show the full, unabbreviated configurations to provide more practice for students who want it, but listing the core-abbreviated electron configurations is also acceptable. Next, determine whether an electron is gained or lost. Remember electrons are negatively charged, so ions with a positive charge have lost an electron. For main group elements, the last orbital gains or loses the electron. For transition metals, the last s orbital loses an electron before the d orbitals. Na: 1 s 2 2 s 2 2 p 6 3 s 1 . Sodium cation loses one electron, so Na + : 1 s 2 2 s 2 2 p 6 3 s 1 = Na + : 1 s 2 2 s 2 2 p 6 . P: 1 s 2 2 s 2 2 p 6 3 s 2 3 p 3 . Phosphorus trianion gains three electrons, so P 3− : 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 . Al: 1 s 2 2 s 2 2 p 6 3 s 2 3 p 1 . Aluminum dication loses two electrons Al 2 + : 1 s 2 2 s 2 2 p 6 3 s 2 3 p 1 = Al 2 + : 1 s 2 2 s 2 2 p 6 3 s 1 . Fe: 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 6 . Iron(II) loses two electrons and, since it is a transition metal, they are removed from the 4 s orbital Fe 2 + : 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 6 = 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 6 . Sm: 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 10 4 p 6 5 s 2 4 d 10 5 p 6 6 s 2 4 f 6 . Samarium trication loses three electrons. The first two will be lost from the 6 s orbital, and the final one is removed from the 4 f orbital. Sm 3+ : 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 10 4 p 6 5 s 2 4 d 10 5 p 6 6 s 2 4 f 6 = 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 10 4 p 6 5 s 2 4 d 10 5 p 6 4 f 5 . Exercise $\PageIndex{21}$ Which ion with a +2 charge has the electron configuration 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 10 4 p 6 4 d 5 ? Which ion with a +3 charge has this configuration? Answer Tc 2 + , Ru 3 +
Courses/University_of_Kansas/KU%3A_CHEM_110_GOB_Chemistry_(Sharpe_Elles)/08%3A_Energy_and_Chemical_Processes/8.3%3A_Phase_Changes	Skills to Develop The Learning Objective of this Module is to determine the heat associated with a phase change. Depending on the surrounding conditions, normal matter usually exists as one of three phases : solid, liquid, or gas. 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 $\PageIndex{1}$). Figure $\PageIndex{1}$ : Heating curve for water. As heat is added to solid water, the temperature increases until it reaches 0 ° C, the melting point. At this point, the phase change, added heat goes into changing the state from a solid to liquid. Only when this phase change is complete, the temperature can increase. Image used with permission (CC BY 3.0 Unported; Community College Consortium for Bioscience Credentials). 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 $\PageIndex{1}$ 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 joules of heat are needed to melt this ice: \[ \begin{align} \ce{heat} &= \ce{m \times ΔH_{fus}} \\[5pt] \mathrm{heat} &= \mathrm{(55.8\: \cancel{g})\left(\dfrac{79.9\: cal}{\cancel{g}}\right)=4,460\: cal} \end{align}\] Exercise $\PageIndex{1}$ 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. 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. Concept Review Exercises Explain what happens when heat flows into or out of a substance at its melting point or boiling point. How does the amount of heat required for a phase change relate to the mass of the substance? Answers The energy goes into changing the phase, not the temperature. The amount of heat is a constant per gram of substance. Key Takeaway There is an energy change associated with any phase change. Exercises How much energy is needed to melt 43.8 g of Au at its melting point of 1,064°C? How much energy is given off when 563.8 g of NaCl solidifies at its freezing point of 801°C? What mass of ice can be melted by 558 cal of energy? How much ethanol (C 2 H 5 OH) in grams can freeze at its freezing point if 1,225 cal of heat are removed? What is the heat of vaporization of a substance if 10,776 cal are required to vaporize 5.05 g? Express your final answer in joules per gram. If 1,650 cal of heat are required to vaporize a sample that has a heat of vaporization of 137 cal/g, what is the mass of the sample? What is the heat of fusion of water in calories per mole? What is the heat of vaporization of benzene (C 6 H 6 ) in calories per mole? What is the heat of vaporization of gold in calories per mole? What is the heat of fusion of iron in calories per mole? Answers 670 cal 6.98 g 8,930 J/g 1,440 cal/mol 80,600 cal/mol
Bookshelves/General_Chemistry/ChemPRIME_(Moore_et_al.)/08%3A_Properties_of_Organic_Compounds/8.19%3A_Organic_Nitrogen_Compounds	There is a tremendous variety of organic compounds which can be derived from carbon, hydrogen, and oxygen which is evident from the numerous previous sections discussing these compounds. If we include nitrogen as a possible constituent of these molecular structures, many more possibilities arise. Most of the nitrogen-containing compounds are less important commercially, however, and we will only discuss a few of them here. Amines may be derived from ammonia by replacing one, two, or all three hydrogens with alkyl groups. Some examples are The terms primary (one), secondary (two), and tertiary (three) refer to the number of hydrogens that have been replaced. Both primary and secondary amines are capable of hydrogen bonding with themselves, but tertiary amines have no hydrogens on the electronegative nitrogen atom. Amines usually have unpleasant odors, smelling “fishy“. The three methylamines listed above can all be isolated from herring brine. Amines, as well as ammonia, are produced by decomposition of nitrogen-containing compounds when a living organism dies. The methylamines are obtained commercially by condensation of methanol with ammonia over an aluminum oxide catalyst: Dimethylamine is the most important, being used in the preparation of herbicides, in rubber vulcanization, and to synthesize dimethylformamide, an important solvent. Amides are another important nitrogen containing organic compound. The key feature of an amine is a nitrogen atom bonded to a carbonyl carbon atom. Like esters , amides are formed in a condensation reaction. While esters are formed from the condensation reaction of an alcohol and a carboxylic acid, amides are formed from the condensation of an amine and a carboxylic acid: This general reaction is usually unfavorable, because the hydroxyl group acts as a bad leaving group. Organic chemists have devised methods to work around this by using certain chemicals to activate the carboxylic acid and allow for the addition of the amine. As amides are formed by condensation reactions, many important condensation polymers involve amide linkages. Nylon, for instance, is formed from the amide condensation of hexamethylenediamine and adipic acid. A second set of condensation polymers formed from amide linkages are the proteins and peptides found in your body and in all organisms. These polymers are formed from another organic nitrogen compound, the amino acid. These molecules contain both an amine group and a carboxyl group. Examples of such amino acids are glycine and lysine: Amino acids are the constituents from which proteins are made. Some, like glycine, can be synthesized in the human body, but others cannot. Lysine is an example of an essential amino acid —one which must be present in the human diet because it cannot be synthesized within the body. As mentioned, the condensation of amino acids into peptides forms amide linkages. For this reason, scientists sometimes refer to the amide backbone of a protein or peptide. A protein has a long series of amide bonds, as can be seen in the following figure showing the synthesis of a tri-peptide from three amino acids: Amino acids and proteins further discussed in the sections on enzymes and in a set of sections devoted to proteins and their chemistry in living systems. The intermolecular forces and boiling points of nitrogen-containing organic compounds may be explained according to the same principles used for oxygen-containing substances. Example $\PageIndex{1}$: Boiling Points Rationalize the following boiling points: (a) 0°C for CH 3 CH 2 CH 2 CH 3 ; (b) 11°C for CH 3 CH 2 OCH 3 ; (c) 97°C for CH 3 CH 2 CH 2 OH; and (d) 170°C for NH 2 CH 2 CH 2 OH. Solution All four molecules have very similar geometries and the same number of electrons (26 valence electrons plus 8 core electrons), and so their London forces should be about the same. Compound (a) is an alkane and is nonpolar. By contrast compound (b) is an ether and should be slightly polar. This slight polarity results in a slightly higher boiling point. Compound (c) is isomeric with compound (b) but is an alcohol. There is hydrogen bonding between molecules of (c), and its boiling point is much higher. Molecule (d) has both an amino group and a hydroxyl group, each of which can participate in hydrogen bonding. Consequently it has the highest boiling point of all.
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/The_Live_Textbook_of_Physical_Chemistry_(Peverati)/18%3A_The_Schrodinger_Equation/18.02%3A_The_Time-Dependent_Schrodinger_Equation	Unfortunately, the analogy with the classical wave equation that allowed us to obtain the TISEq in the previous section cannot be extended to the time domain by considering the equation that involves the partial first derivative with respect to time. Schrödinger himself presented his time-independent equation first, and then went back and postulated the more general time-dependent equation. We are following here the same strategy and just give the time-independent variable as a postulate. The single-particle time-dependent Schrödinger equation is: \[ i\hbar\dfrac{\partial \psi(x,t)}{\partial t}=-\dfrac{\hbar^2}{2m} \dfrac{\partial^2 \psi(x,t)}{\partial x^2}+V(x)\psi(x,t) \label{19.2.1} \] where $V \in \mathbb{R}^{n}$ represents the potential energy of the system. Obviously, the time-dependent equation can be used to derive the time-independent equation. If we write the wavefunction as a product of spatial and temporal terms, $\psi(x, t) = \psi(x) f(t)$, then Equation \ref{19.2.1} becomes: \[ \psi(x) i \hbar \dfrac{df(t)}{dt} = f(t) \left[-\dfrac{\hbar^2}{2m} \dfrac{\partial^2}{\partial x^2} + V(x) \right] \psi(x), \label{19.2.2} \] which can be rearranged to: \[ \dfrac{i \hbar}{f(t)} \dfrac{df(t)}{dt} = \dfrac{1}{\psi(x)} \left[-\dfrac{\hbar^2}{2m} \dfrac{\partial^2}{\partial x^2} + V(x) \right] \psi(x). \label{19.2.3} \] Since the left-hand side of Equation \ref{19.2.3} is a function of $t$ only and the right hand side is a function of $x$ only, the two sides must equal a constant. If we tentatively designate this constant $E$ (since the right-hand side clearly must have the dimensions of energy), then we extract two ordinary differential equations, namely: \[ \dfrac{1}{f(t)} \dfrac{df(t)}{dt} = - \dfrac{i E}{\hbar} \label{19.2.4} \] and: \[ -\dfrac{\hbar^2}{2m} \dfrac{\partial^2\psi(x)}{\partial x^2} + V(x) \psi(x) = E \psi(x). \label{19.2.5} \] The latter equation is the TISEq. The former equation is easily solved to yield \[ f(t) = e^{-iEt / \hbar} \label{19.2.6} \] The solutions of Equation \ref{19.2.6}, $f(t)$, are purely oscillatory, since $f(t)$ never changes in magnitude. Thus if: \[ \psi(x, t) = \psi(x) \exp\left(\dfrac{-iEt}{\hbar}\right), \label{19.2.7} \] then the total wave function $\psi(x, t)$ differs from $\psi(x)$ only by a phase factor of constant magnitude. There are some interesting consequences of this. First of all, the quantity $\vert \psi(x, t) \vert^2$ is time independent, as we can easily show: \[ \vert \psi(x, t) \vert^2 = \psi^{}(x, t) \psi(x, t)= \psi^{}(x)\exp\left(\dfrac{iEt}{\hbar}\right)\psi(x)\exp\left(\dfrac{-iEt}{\hbar}\right)= \psi^{*}(x) \psi(x). \label{19.2.8} \] Wave functions of the form of Equation \ref{19.2.7} are called stationary states. The state $\psi(x, t)$ is “stationary,” but the particle it describes is not! Of course Equation \ref{19.2.6} represents only a particular solution to the time-dependent Schrödinger equation. The general solution is much more complicated, and the factorization of the temporal part is often not possible:$^1$ \[ \psi({\bf r}, t) = \sum_i c_i e^{-iE_it / \hbar} \psi_i({\bf r}) \nonumber \] This sections was adapted in part from Prof. C. David Sherrill’s A Brief Review of Elementary Quantum Chemistry Notes available here .
Courses/University_of_Arkansas_Little_Rock/Chem_3572%3A_Physical_Chemistry_for_Life_Sciences_(Siraj)/09%3A_Chemical_Kinetics	Chemical kinetics is the study of the speed with which a chemical reaction occurs and the factors that affect this speed. This information is especially useful for determining how a reaction occurs. 9.1: Reaction Rates In this Module, the quantitative determination of a reaction rate is demonstrated. Reaction rates can be determined over particular time intervals or at a given point in time. A rate law describes the relationship between reactant rates and reactant concentrations. Reaction rates are usually expressed as the concentration of reactant consumed or the concentration of product formed per unit time. 9.2: Reaction Order From experimental observations, scientists have established that reaction rates almost always have a power-law dependence on the concentrations of one or more of the reactants. In the following sections, we will discuss different power laws that are commonly observed in chemical reactions. 9.3: Molecularity of a Reaction If the reactions are elementary reactions, (i.e. they cannot be expressed as a series of simpler reactions), then we can directly define the rate law based on the chemical equation. 9.4: More Complex Reactions A major goal in chemical kinetics is to determine the sequence of elementary reactions, or the reaction mechanism, that comprise complex reactions. In the following sections, we will derive rate laws for complex reaction mechanisms, including reversible, parallel and consecutive reactions. 9.5: The Effect of Temperature on Reaction Rates When molecules collide, the kinetic energy of the molecules can be used to stretch, bend, and ultimately break bonds, leading to chemical reactions. If molecules move too slowly with little kinetic energy, or collide with improper orientation, they do not react and simply bounce off each other. However, if the molecules are moving fast enough with a proper collision orientation, such that the kinetic energy upon collision is greater than the minimum energy barrier, then a reaction occurs. 9.6: Potential Energy Surfaces A potential energy surface (PES) describes the potential energy of a system, especially a collection of atoms, in terms of certain parameters, normally the positions of the atoms. The surface might define the energy as a function of one or more coordinates; if there is only one coordinate, the surface is called a potential energy curve or energy profile. 9.7: Theories of Reaction Rates The macroscopic discussion of kinetics discussed in previous sections can be now expanded into a more microscopic picture in terms of molecular level properties (e..g, mass and velocities) involving two important theories: (1) collision theory and (2) transition-state theory. 9.8: Isotope Effects in Chemical Reactions The kinetic isotope effect (KIE) is a phenomenon associated with isotopically substituted molecules exhibiting different reaction rates. Isotope effects such as KIEs are invaluable tools in both physical and biological sciences and are used to aid in the understanding of reaction kinetics, mechanisms, and solvent effects. 9.9: Reactions in Solution Most of the complications of kinetics and rate processes in liquid solutions arise from the much higher density of the liquid phase. In a typical liquid solution, the solvent molecules massively outnumber the reactant solute molecules, which tend to find themselves momentarily (~10–11 sec) confined to a "hole" within the liquid. 9.10: Fast Reactions in Solution The traditional experimental methods described above all assume the possibility of following the reaction after its components have combined into a homogeneous mixture of known concentrations. But what can be done if the time required to complete the mixing process is comparable to or greater than the time needed for the reaction to run to completion? 9.11: Oscillating Reactions hus far, we have only looked at reaction systems that give rise to purely linear differential equations, however, in many instances the rate equations are nonlinear. When the differential equations are nonlinear, the behavior is considerably more complex. In particular, nonlinear equations can lead to oscillatory solutions and can also exhibit chaos. Chaotic systems are systems that are highly sensitive to small changes in the parameters of the equations or initial conditions 9.12: Chemical Kinetics (Exercises) This are exercises that to accompany the TextMap organized around Raymond Chang's Physical Chemistry for the Biosciences textbook.
Courses/Lebanon_Valley_College/CHM_311%3A_Physical_Chemistry_I_(Lebanon_Valley_College)/03%3A_Model_Systems_in_Quantum_Mechanics/3.09%3A_The_Harmonic_Oscillator_Energy_Levels	For a classical oscillator, we know exactly the position, velocity, and momentum as a function of time. The frequency of the oscillator (or normal mode) is determined by the reduced mass $\mu$ and the effective force constant $k$ of the oscillating system and does not change unless one of these quantities is changed. There are no restrictions on the energy of the oscillator, and changes in the energy of the oscillator produce changes in the amplitude of the vibrations experienced by the oscillator. For the quantum mechanical oscillator, the oscillation frequency of a given normal mode is still controlled by the mass and the force constant (or, equivalently, by the associated potential energy function). However, the energy of the oscillator is limited to certain values. The allowed quantized energy levels are equally spaced and are related to the oscillator frequencies as given by Equation $\ref{5.4.1}$ and Figure 5.4.1 . \[E_v = \left ( v + \dfrac {1}{2} \right ) \hbar \omega = \left ( v + \dfrac {1}{2} \right ) h \nu \label {5.4.1} \] with \[v = 0, 1, 2, 3, \cdots \infty \nonumber \] In a quantum mechanical oscillator, we cannot specify the position of the oscillator (the exact displacement from the equilibrium position) or its velocity as a function of time; we can only talk about the probability of the oscillator being displaced from equilibrium by a certain amount. This probability is given by \[P_{Q \rightarrow Q + dQ} = \int_{Q}^{Q + dQ} \psi ^_v (Q) \psi _v (Q) dQ \label {5.4.3} \] We can, however, calculate the average displacement and the mean square displacement of the atoms relative to their equilibrium positions. This average is just $\left \langle Q \right \rangle$, the expectation value for $Q$, and the mean square displacement is $\left \langle Q^2 \right \rangle$, the expectation value for $Q^2$. Similarly we can calculate the average momentum $\left \langle P_Q \right \rangle$, and the mean square momentum $\left \langle P^2_Q \right \rangle$, but we cannot specify the momentum as a function of time. Physically what do we expect to find for the average displacement and the average momentum? Since the potential energy function is symmetric around $Q = 0$, we expect values of $Q > 0$ to be equally as likely as $Q < 0$. The average value of $Q$ therefore should be zero. These results for the average displacement and average momentum do not mean that the harmonic oscillator is sitting still. As for the particle-in-a-box case, we can imagine the quantum mechanical harmonic oscillator as moving back and forth and therefore having an average momentum of zero. Since the lowest allowed harmonic oscillator energy, $E_0$, is $\dfrac{\hbar \omega}{2}$ and not 0, the atoms in a molecule must be moving even in the lowest vibrational energy state. This phenomenon is called the zero-point energy or the zero-point motion, and it stands in direct contrast to the classical picture of a vibrating molecule. Classically, the lowest energy available to an oscillator is zero, which means the momentum also is zero, and the oscillator is not moving. Compare the quantum mechanical harmonic oscillator to the classical harmonic oscillator at $v=1$ and $v=50$. Answer At v=1 the classical harmonic oscillator poorly predicts the results of quantum mechanical harmonic oscillator, and therefore reality. At v=1 the particle will be near the ground state and the classical model will predict the particle to spend most it's time on the outer edges when the KE goes to zero and PE is at a maximum, while the quantum model says the opposite and that the particle will be more likely to be found in the center. At v=50 the quantum model will begin to match the classical much more closely, with the particle most likely to be found at the edges. The quantum model looking more like the classical at higher quantum numbers can be referred to as the correspondence principle. Since the average values of the displacement and momentum are all zero and do not facilitate comparisons among the various normal modes and energy levels, we need to find other quantities that can be used for this purpose. We can use the root mean square deviation (see also root-mean-square displacement) (also known as the standard deviation of the displacement) and the root-mean-square momentum as measures of the uncertainty in the oscillator's position and momentum. For a molecular vibration, these quantities represent the standard deviation in the bond length and the standard deviation in the momentum of the atoms from the average values of zero, so they provide us with a measure of the relative displacement and the momentum associated with each normal mode in all its allowed energy levels. These are important quantities to determine because vibrational excitation changes the size and symmetry (or shape) of molecules. Such changes affect chemical reactivity, the absorption and emission of radiation, and the dissipation of energy in radiationless transitions. The harmonic oscillator wavefunctions form an orthonormal set; this means that all functions in the set are normalized individually \[\int \limits _{-\infty}^{\infty} \psi ^_v (x) \psi _v (x) dx = 1 \label {5.4.4} \] and are orthogonal to each other. \[\int \limits _{-\infty}^{\infty} \psi ^*_{v'} (x) \psi _v (x) dx = 0 \;\; \text {for} \;\; v' \ne v \label {5.4.5} \] The fact that a family of wavefunctions forms an orthonormal set is often helpful in simplifying complicated integrals. We will use these properties when we determine the harmonic oscillator selection rules for vibrational transitions in a molecule and calculate the absorption coefficients for the absorption of infrared radiation. Finally, we can calculate the probability that a harmonic oscillator is in the classically forbidden region. What does this tantalizing statement mean? Classically, the maximum extension of an oscillator is obtained by equating the total energy of the oscillator to the potential energy, because at the maximum extension all the energy is in the form of potential energy. If all the energy weren't in the form of potential energy at this point, the oscillator would have kinetic energy and momentum and could continue to extend further away from its rest position. Interestingly, as we show below, the wavefunctions of the quantum mechanical oscillator extend beyond the classical limit, i.e. beyond where the particle can be according to classical mechanics. The lowest allowed energy for the quantum mechanical oscillator is called the zero-point energy, $E_0 = \dfrac {\hbar \omega}{2} $. Using the classical picture described in the preceding paragraph, this total energy must equal the potential energy of the oscillator at its maximum extension. We define this classical limit of the amplitude of the oscillator displacement as $Q_0$. When we equate the zero-point energy for a particular normal mode to the potential energy of the oscillator in that normal mode, we obtain \[ \dfrac {\hbar \omega}{2} = \dfrac {k Q^2_0}{2} \label {5.4.6} \] The zero-point energy is the lowest possible energy that a quantum mechanical physical system may have. Hence, it is the energy of its ground state. Recall that $k$ is the effective force constant of the oscillator in a particular normal mode and that the frequency of the normal mode is given by Equation $\ref{5.4.1}$ which is \[\omega = \sqrt {\dfrac {k}{\mu}} \label {5.4.7} \]
Courses/Prince_Georges_Community_College/CHEM_2000%3A_Chemistry_for_Engineers_(Sinex)/Unit_6%3A_Thermo_and_Electrochemistry/Chapter_17%3A_Electrochemistry/Chapter_17.7%3A_Electrolysis	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 understand electrolysis and describe it quantitatively. In this chapter, we have described various galvanic cells in which a spontaneous chemical reaction is used to generate electrical energy. In an electrolytic cell, however, the opposite process, called electrolysis An electrochemical process in which an external voltage is applied to an electrolytic cell to drive a nonspontaneous reaction. , occurs: an external voltage is applied to drive a nonspontaneous reaction ( Figure 17.1.1 ). In this section, we look at how electrolytic cells are constructed and explore some of their many commercial applications. Note the Pattern In an electrolytic cell, an external voltage is applied to drive a nonspontaneous reaction. Electrolytic Cells If we construct an electrochemical cell in which one electrode is copper metal immersed in a 1 M Cu 2+ solution and the other electrode is cadmium metal immersed in a 1 M Cd 2+ solution and then close the circuit, the potential difference between the two compartments will be 0.74 V. The cadmium electrode will begin to dissolve (Cd is oxidized to Cd 2+ ) and is the anode, while metallic copper will be deposited on the copper electrode (Cu 2+ is reduced to Cu), which is the cathode (part (a) in Figure 17.7.1 ). The overall reaction is as follows: $ Cd\left ( s \right ) +Cu^{2+}\left ( aq \right )\rightarrow Cu\left ( s \right ) +Cd^{2+}\left ( aq \right ) \tag{17.7.1}$ This reaction is thermodynamically spntaneous as written ( $ \Delta G^{o}=nFE_{cell}^{o}= \left ( 2\; \cancel{mol \; e^{-}} \right )\left [ 96,486 \;J/\left ( \cancel{V}\cdot \cancel{mol} \right ) \right ]\left ( 0.74 \; \cancel{V} \right ) =-140 \; kJ/mol \tag{17.7.2}$ In this direction, the system is acting as a galvanic cell. Figure 17.7.1 An Applied Voltage Can Reverse the Flow of Electrons in a Galvanic Cd/Cu Cell (a) When compartments that contain a Cd electrode immersed in 1 M Cd 2+ (aq) and a Cu electrode immersed in 1 M Cu 2+ (aq) are connected to create a galvanic cell, Cd(s) is spontaneously oxidized to Cd 2+ (aq) at the anode, and Cu 2+ (aq) is spontaneously reduced to Cu(s) at the cathode. The potential of the galvanic cell is 0.74 V. (b) Applying an external potential greater than 0.74 V in the reverse direction forces electrons to flow from the Cu electrode [which is now the anode, at which metallic Cu(s) is oxidized to Cu 2+ (aq)] and into the Cd electrode [which is now the cathode, at which Cd 2+ (aq) is reduced to Cd(s)]. The anode in an electrolytic cell is positive because electrons are flowing from it, whereas the cathode is negative because electrons are flowing into it. The reverse reaction, the reduction of Cd 2+ by Cu, is thermodynamically nonspontaneous and will occur only with an input of 140 kJ. We can force the reaction to proceed in the reverse direction by applying an electrical potential greater than 0.74 V from an external power supply. The applied voltage forces electrons through the circuit in the reverse direction, converting a galvanic cell to an electrolytic cell. Thus the copper electrode is now the anode (Cu is oxidized), and the cadmium electrode is now the cathode (Cd 2+ is reduced) (part (b) in Figure 17.7.1 ). The signs of the cathode and the anode have switched to reflect the flow of electrons in the circuit. The half-reactions that occur at the cathode and the anode are as follows: $cathode:\; Cd^{2+}\left ( aq \right )+ 2e^{-} \rightarrow Cd\left ( s \right ) \;\;\; E_{cathode}^{o} = -0.40 \; V \tag{17.7.3}$ $anode:\; Cu\left ( s \right )\rightarrow Cu^{2+} \left ( aq \right ) + 2e^{-} \;\;\; E_{anode}^{o}= 0.34 \; V \tag{17.7.4}$ $overall:\; Cd^{2+}\left ( aq \right ) + Cu\left ( s \right )\rightarrow Cu^{2+} \left ( aq \right )+ Cd \left ( s \right ) \;\; E_{cell}^{o}=-0.74 \; V \tag{17.7.5}$ Because E ° cell < 0, the overall reaction—the reduction of Cd 2+ by Cu—clearly cannot occur spontaneously and proceeds only when sufficient electrical energy is applied. The differences between galvanic and electrolytic cells are summarized in Table 17.7.1 . Table 17.7.1 Comparison of Galvanic and Electrolytic Cells Property Galvanic Cell Electrolytic Cell ΔG < 0 > 0 E cell > 0 < 0 Electrode Process NaN NaN anode oxidation oxidation cathode reduction reduction Sign of Electrode NaN NaN anode − + cathode + − Electrolytic Reactions At sufficiently high temperatures, ionic solids melt to form liquids that conduct electricity extremely well due to the high concentrations of ions. If two inert electrodes are inserted into molten NaCl, for example, and an electrical potential is applied, Cl − is oxidized at the anode, and Na + is reduced at the cathode. The overall reaction is as follows: $ 2NaCl\left ( l \right ) \rightarrow Na\left ( l \right ) +Cl_{2}\left ( g \right ) \tag{17.7.6}$ This is the reverse of the formation of NaCl from its elements. The product of the reduction reaction is liquid sodium because the melting point of sodium metal is 97.8°C, well below that of NaCl (801°C). Approximately 20,000 tons of sodium metal are produced commercially in the United States each year by the electrolysis of molten NaCl in a Downs cell ( Figure 17.7.2 ). In this specialized cell, CaCl 2 (melting point = 772°C) is first added to the NaCl to lower the melting point of the mixture to about 600°C, thereby lowering operating costs. Figure 17.7.2 A Downs Cell for the Electrolysis of Molten NaCl The electrolysis of a molten mixture of NaCl and CaCl 2 results in the formation of elemental sodium and chlorine gas. Because sodium is a liquid under these conditions and liquid sodium is less dense than molten sodium chloride, the sodium floats to the top of the melt and is collected in concentric capped iron cylinders surrounding the cathode. Gaseous chlorine collects in the inverted cone over the anode. An iron screen separating the cathode and anode compartments ensures that the molten sodium and gaseous chlorine do not come into contact. Similarly, in the Hall–Heroult process used to produce aluminum commercially, a molten mixture of about 5% aluminum oxide (Al 2 O 3 ; melting point = 2054°C) and 95% cryolite (Na 3 AlF 6 ; melting point = 1012°C) is electrolyzed at about 1000°C, producing molten aluminum at the cathode and CO 2 gas at the carbon anode. The overall reaction is as follows: $ 2Al_{2}O_{3}\left ( l \right ) + 3C\left ( s \right ) \rightarrow 4Al\left ( l \right ) + 3CO_{2}\left ( g \right ) \tag{17.7.7}$ Oxide ions react with oxidized carbon at the anode, producing CO 2 (g). There are two important points to make about these two commercial processes and about the electrolysis of molten salts in general. The electrode potentials for molten salts are likely to be very different from the standard cell potentials listed in Table 17.3.1 and Standard Reduction Potentials at 25°C" , which are compiled for the reduction of the hydrated ions in aqueous solutions under standard conditions. Using a mixed salt system means there is a possibility of competition between different electrolytic reactions. When a mixture of NaCl and CaCl 2 is electrolyzed, Cl − is oxidized because it is the only anion present, but either Na + or Ca 2+ can be reduced. Conversely, in the Hall–Heroult process, only one cation is present that can be reduced (Al 3+ ), but there are three species that can be oxidized: C, O 2− , and F − . In the Hall–Heroult process, C is oxidized instead of O 2− or F − because oxygen and fluorine are more electronegative than carbon, which means that C is a weaker oxidant than either O 2 or F 2 . Similarly, in the Downs cell, we might expect electrolysis of a NaCl/CaCl 2 mixture to produce calcium rather than sodium because Na is slightly less electronegative than Ca (χ = 0.93 versus 1.00, respectively), making Na easier to oxidize and, conversely, Na + more difficult to reduce. In fact, the reduction of Na + to Na is the observed reaction. In cases where the electronegativities of two species are similar, other factors, such as the formation of complex ions, become important and may determine the outcome. Example 17.7.1 If a molten mixture of MgCl 2 and KBr is electrolyzed, what products will form at the cathode and the anode, respectively? Given: identity of salts Asked for: electrolysis products Strategy: A List all the possible reduction and oxidation products. Based on the electronegativity values shown in Figure 3.1.2 , determine which species will be reduced and which species will be oxidized. B Identify the products that will form at each electrode. Solution: A The possible reduction products are Mg and K, and the possible oxidation products are Cl 2 and Br 2 . Because Mg is more electronegative than K (χ = 1.31 versus 0.82), it is likely that Mg will be reduced rather than K. Because Cl is more electronegative than Br (3.16 versus 2.96), Cl 2 is a stronger oxidant than Br 2 . B Electrolysis will therefore produce Br 2 at the anode and Mg at the cathode. Exercise Predict the products if a molten mixture of AlBr 3 and LiF is electrolyzed. Answer: Br 2 and Al Electrolysis can also be used to drive the thermodynamically nonspontaneous decomposition of water into its constituent elements: H 2 and O 2 . However, because pure water is a very poor electrical conductor, a small amount of an ionic solute (such as H 2 SO 4 or Na 2 SO 4 ) must first be added to increase its electrical conductivity. Inserting inert electrodes into the solution and applying a voltage between them will result in the rapid evolution of bubbles of H 2 and O 2 ( Figure 17.7.3 ). The reactions that occur are as follows: $cathode:\; 2H^{+}\left ( aq \right )+ 2e^{-} \rightarrow H_{2}\left ( g \right ) \;\;\; E_{cathode}^{o} = 0 \; V \tag{17.7.8}$ $anode:\; 2H_{2}O\left ( l \right ) \rightarrow O_{2}\left ( g \right ) + 4H^{+} \left ( g \right ) + 4e^{-} \;\;\; E_{anode}^{o}= 1.23 \; V \tag{17.7.9}$ $overall:\; 2H_{2}O\left ( l \right )\rightarrow O_{2}\left ( g \right )+ 2H_{2}\left ( g \right ) \;\; E_{cell}^{o}=-1.23 \; V \tag{17.7.10}$ Figure 17.7.3 The Electrolysis of Water Applying an external potential of about 1.7–1.9 V to two inert electrodes immersed in an aqueous solution of an electrolyte such as H 2 SO 4 or Na 2 SO 4 drives the thermodynamically nonspontaneous decomposition of water into H 2 at the cathode and O 2 at the anode. For a system that contains an electrolyte such as Na 2 SO 4 , which has a negligible effect on the ionization equilibrium of liquid water, the pH of the solution will be 7.00 and [H + ] = [OH − ] = 1.0 × 10 −7 . Assuming that P(O 2 ) = P(H 2 ) = 1 atm, we can use the standard potentials and Equation 19.64 to calculate E for the overall reaction: $ E_{cell}= E_{cell}^{o}-\left ( \dfrac{0.0591 \; V}{n} \right )log\;\left ( P_{O_{2}}P_{H_{2}}^{2} \right ) \tag{17.7.11}$ $ E_{cell}= -1.23 \; V -\left ( \dfrac{0.0591 \; V}{4} \right )log\;\left ( 1 \right ) = -1.23 \; V $ Thus E cell is −1.23 V, which is the value of E ° cell if the reaction is carried out in the presence of 1 M H + rather than at pH 7.0. In practice, a voltage about 0.4–0.6 V greater than the calculated value is needed to electrolyze water. This added voltage, called an overvoltage The voltage that must be applied in electrolysis in addition to the calculated (theoretical) value to overcome factors such as a high activation energy and the formation of bubbles on a surface. , represents the additional driving force required to overcome barriers such as the large activation energy for the formation of a gas at a metal surface. Overvoltages are needed in all electrolytic processes, which explain why, for example, approximately 14 V must be applied to recharge the 12 V battery in your car. In general, any metal that does not react readily with water to produce hydrogen can be produced by the electrolytic reduction of an aqueous solution that contains the metal cation. The p -block metals and most of the transition metals are in this category, but metals in high oxidation states, which form oxoanions, cannot be reduced to the metal by simple electrolysis. Active metals, such as aluminum and those of groups 1 and 2, react so readily with water that they can be prepared only by the electrolysis of molten salts. Similarly, any nonmetallic element that does not readily oxidize water to O 2 can be prepared by the electrolytic oxidation of an aqueous solution that contains an appropriate anion. In practice, among the nonmetals, only F 2 cannot be prepared using this method. Oxoanions of nonmetals in their highest oxidation states, such as NO 3 − , SO 4 2− , PO 4 3− , are usually difficult to reduce electrochemically and usually behave like spectator ions that remain in solution during electrolysis. Note the Pattern In general, any metal that does not react readily with water to produce hydrogen can be produced by the electrolytic reduction of an aqueous solution that contains the metal cation. Electroplating In a process called electroplating A process in which a layer of a second metal is deposited on the metal electrode that acts as the cathode during electrolysis. , a layer of a second metal is deposited on the metal electrode that acts as the cathode during electrolysis. Electroplating is used to enhance the appearance of metal objects and protect them from corrosion. Examples of electroplating include the chromium layer found on many bathroom fixtures or (in earlier days) on the bumpers and hubcaps of cars, as well as the thin layer of precious metal that coats silver-plated dinnerware or jewelry. In all cases, the basic concept is the same. A schematic view of an apparatus for electroplating silverware and a photograph of a commercial electroplating cell are shown in Figure 17.7.4 . Figure 17.7.4 Electroplating (a) Electroplating uses an electrolytic cell in which the object to be plated, such as a fork, is immersed in a solution of the metal to be deposited. The object being plated acts as the cathode, on which the desired metal is deposited in a thin layer, while the anode usually consists of the metal that is being deposited (in this case, silver) that maintains the solution concentration as it dissolves. (b) In this commercial electroplating apparatus, a large number of objects can be plated simultaneously by lowering the rack into the Ag + solution and applying the correct potential. The half-reactions in electroplating a fork, for example, with silver are as follows: $cathode\left ( fork \right ):\; Ag^{+}\left ( aq \right )+ e^{-} \rightarrow Ag\left ( s \right ) \;\;\; E_{cathode}^{o} = 0.80 \; V \tag{17.7.12}$ $anode\left ( silver bar \right ):\; Ag\left ( s \right ) \rightarrow Ag^{+} \left ( aq \right ) + e^{-} \;\;\; E_{anode}^{o}= 0.80 \; V \tag{17.7.13}$ The overall reaction is the transfer of silver metal from one electrode (a silver bar acting as the anode) to another (a fork acting as the cathode). Because E ° cell = 0 V, it takes only a small applied voltage to drive the electroplating process. In practice, various other substances may be added to the plating solution to control its electrical conductivity and regulate the concentration of free metal ions, thus ensuring a smooth, even coating. Quantitative Considerations If we know the stoichiometry of an electrolysis reaction, the amount of current passed, and the length of time, we can calculate the amount of material consumed or produced in a reaction. Conversely, we can use stoichiometry to determine the combination of current and time needed to produce a given amount of material. The quantity of material that is oxidized or reduced at an electrode during an electrochemical reaction is determined by the stoichiometry of the reaction and the amount of charge that is transferred. For example, in the reaction Ag + (aq) + e − → Ag(s), 1 mol of electrons reduces 1 mol of Ag + to Ag metal. In contrast, in the reaction Cu 2+ (aq) + 2e − → Cu(s), 1 mol of electrons reduces only 0.5 mol of Cu 2+ to Cu metal. Recall that the charge on 1 mol of electrons is 1 faraday (1 F ), which is equal to 96,486 C. We can therefore calculate the number of moles of electrons transferred when a known current is passed through a cell for a given period of time. The total charge (C) transferred is the product of the current (A) and the time ( t , in seconds): $ C=A\times t \tag{17.7.14}$ The stoichiometry of the reaction and the total charge transferred enable us to calculate the amount of product formed during an electrolysis reaction or the amount of metal deposited in an electroplating process. For example, if a current of 0.60 A passes through an aqueous solution of CuSO 4 for 6.0 min, the total number of coulombs of charge that passes through the cell is as follows: $ charge=\left (0.60 \;A \right )\left ( 6.0 \; \cancel{min} \right ) \left ( 60 \; s/\cancel{min} \right ) =220 A \cdot s \tag{17.7.15}$ The number of moles of electrons transferred to Cu 2+ is therefore $ moles \; e^{-}= \dfrac{220 \; \cancel{C}}{96,486 \; \cancel{C}/mol}=2.3\times 10^{-3} \; mol \; e^{-} \tag{17.7.16}$ Because two electrons are required to reduce a single Cu 2+ ion, the total number of moles of Cu produced is half the number of moles of electrons transferred, or 1.2 × 10 −3 mol. This corresponds to 76 mg of Cu. In commercial electrorefining processes, much higher currents (greater than or equal to 50,000 A) are used, corresponding to approximately 0.5 F /s, and reaction times are on the order of 3–4 weeks. Example 17.7.2 A silver-plated spoon typically contains about 2.00 g of Ag. If 12.0 h are required to achieve the desired thickness of the Ag coating, what is the average current per spoon that must flow during the electroplating process, assuming an efficiency of 100%? Given: mass of metal, time, and efficiency Asked for: current required Strategy: A Calculate the number of moles of metal corresponding to the given mass transferred. B Write the reaction and determine the number of moles of electrons required for the electroplating process. C Use the definition of the faraday to calculate the number of coulombs required. Then convert coulombs to current in amperes. Solution: A We must first determine the number of moles of Ag corresponding to 2.00 g of Ag: $ moles \; Ag= \dfrac{2.0 \; \cancel{g}}{107.868 \; \cancel{g}/mol}=1.85\times 10^{-2} \; mol \; Ag $ B The reduction reaction is Ag + (aq) + e − → Ag(s), so 1 mol of electrons produces 1 mol of silver. C Using the definition of the faraday, $ Coulombs=\left ( 1.85\times 10^{-2} \; \cancel{mol} \right )\left ( 96,486 \; C/\cancel{mol} \right ) =1.78 \times 10^{3} \; C$ The current in amperes needed to deliver this amount of charge in 12.0 h is therefore $ amperes=\dfrac{1.78 \times 10^{3} \; C}{\left ( 12.0 \; \cancel{h} \right ) \left ( 60 \; \cancel{min} /\cancel{h} \right ) \left ( 60 \; s/\cancel{min} \right ) } $ $ \;\;\;\; =4.12\times 10^{-2} \; C/s = 4.12\times 10^{-2} \; A $ Because the electroplating process is usually much less than 100% efficient (typical values are closer to 30%), the actual current necessary is greater than 0.1 A. Exercise A typical aluminum soft-drink can weighs about 29 g. How much time is needed to produce this amount of Al(s) in the Hall–Heroult process, using a current of 15 A to reduce a molten Al 2 O 3/ Na 3 AlF 6 mixture? Answer: 5.8 h Summary In electrolysis , an external voltage is applied to drive a nonspontaneous reaction. A Downs cell is used to produce sodium metal from a mixture of salts, and the Hall–Heroult process is used to produce aluminum commercially. Electrolysis can also be used to produce H 2 and O 2 from water. In practice, an additional voltage, called an overvoltage , must be applied to overcome factors such as a large activation energy and a junction potential. Electroplating is the process by which a second metal is deposited on a metal surface, thereby enhancing an object’s appearance or providing protection from corrosion. The amount of material consumed or produced in a reaction can be calculated from the stoichiometry of an electrolysis reaction, the amount of current passed, and the duration of the electrolytic reaction. Key Takeaways In electrolysis, an external voltage is applied to drive a nonspontaneous reaction. The quantity of material oxidized or reduced can be calculated from the stoichiometry of the reaction and the amount of charge transferred. Key Equation Relationship of charge, current and time Equation 17.7.14 : C = A × t Conceptual Problems Why might an electrochemical reaction that is thermodynamically favored require an overvoltage to occur? How could you use an electrolytic cell to make quantitative comparisons of the strengths of various oxidants and reductants? Why are mixtures of molten salts, rather than a pure salt, generally used during electrolysis? Two solutions, one containing Fe(NO 3 ) 2 ·6H 2 O and the other containing the same molar concentration of Fe(NO 3 ) 3 ·6H 2 O, were electrolyzed under identical conditions. Which solution produced the most metal? Justify your answer. Numerical Problems The electrolysis of molten salts is frequently used in industry to obtain pure metals. How many grams of metal are deposited from these salts for each mole of electrons? AlCl 3 MgCl 2 FeCl 3 Electrolysis is the most direct way of recovering a metal from its ores. However, the Na + (aq)/Na(s), Mg 2+ (aq)/Mg(s), and Al 3+ (aq)/Al(s) couples all have standard electrode potentials ( E °) more negative than the reduction potential of water at pH 7.0 (−0.42 V), indicating that these metals can never be obtained by electrolysis of aqueous solutions of their salts. Why? What reaction would occur instead? What volume of chlorine gas at standard temperature and pressure is evolved when a solution of MgCl 2 is electrolyzed using a current of 12.4 A for 1.0 h? What mass of copper metal is deposited if a 5.12 A current is passed through a Cu(NO 3 ) 2 solution for 1.5 h. What mass of PbO 2 is reduced when a current of 5.0 A is withdrawn over a period of 2.0 h from a lead storage battery? Electrolysis of Cr 3+ (aq) produces Cr 2+ (aq). If you had 500 mL of a 0.15 M solution of Cr 3+ (aq), how long would it take to reduce the Cr 3+ to Cr 2+ using a 0.158 A current? Predict the products obtained at each electrode when aqueous solutions of the following are electrolyzed. AgNO 3 RbI Predict the products obtained at each electrode when aqueous solutions of the following are electrolyzed. MgBr 2 Hg(CH 3 CO 2 ) 2 Al 2 (SO 4 ) 3 Answers 5.2 L cathode: Ag(s); anode: O 2 (g); cathode: H 2 (g); anode: I 2 (s) Contributors Anonymous Modified by Joshua Halpern ( Howard University ), Scott Sinex, and Scott Johnson (PGCC)
Courses/Los_Medanos_College/Chemistry_6_and_Chemistry_7_Combined_Laboratory_Manual_(Los_Medanos_College)/02%3A_Techniques/2.01%3A_Technique_A_Laboratory_Notebook_1_2_1	Technique A : Use of Laboratory Notebook Section 1: Purpose of Technique The laboratory notebook is a permanent record of a researcher’s activities. The main purpose of maintaining a laboratory notebook is to preserve experimental plans, study design or protocol, procedures that were followed, observations, results, conclusions, and recommendations. The information that is documented must be done in a way that another scientist can replicate the study based on the information presented in the notebook or binder. Supplemental quality control activities and practices related to equipment maintenance and calibration are recorded on appropriate forms and maintained in log books as required under other laboratory Standard Operating Procedures. Unless there is a critical operation, it is generally not necessary to record supplemental data twice, though the activities should be referred to and cross-referenced in the laboratory notebook. For example, if the laboratory water supply is monitored, measured and recorded in a separate logbook, then you do not generally need to also record that monitoring result in your laboratory notebook. Section 2: How to Start Your Laboratory Notebook Requirements for a Laboratory Notebook The laboratory notebook is usually bound with sequentially numbered pages. Several styles are available. Some organizations require acid-free paper for archival purposes. Others require a carbon-copy for immediate archival at the end of the day. Pages can be ruled or graph-ruled. Some examples are provided below: For most educational settings a bound composition book is sufficient. If the pages are not sequentially numbered, then sequences can be established by diligent recording of entry dates: Laboratory notebooks should be used with either blue or black permanent ink. Many non-black or blue inks fade with time. Section 3: Components of a Laboratory Notebook Part 1: Owner’s Name , Dates of Use, and Course Title Put your name on the outside cover, inside cover, or on the first page of your notebook to identify that you own this laboratory notebook. Identify what period of time this laboratory notebook is in service. Usually that is a start date on the cover, followed by an end date when the notebook is completed. Record your course title, professor, and section number for reference. Provide an email address and contact phone number in case you leave your notebook in a classroom. Part 2: Table of Contents and Numbered Pages A Table of Contents is optional, but useful for yourself and for other people. Your Table of Contents should have the page number, title of experiment, and dates of when you started and finished your experiment. This makes it easier for others to read your laboratory notebook. Make sure to include enough pages in the front for your table of contents. After making enough space for your table of contents, if your notebook is not sequentially numbered, make sure to do so using black or blue ink. Date and initial all information collected, recorded, and calculated in your notebook at the end of each day of working in the laboratory. Part 3: Experiment Entries For each experiment include the day you started it and the title of the experiment at the beginning of the page. Document your daily plan of the experiment before starting it. Include the purpose of the experiment, background information, what techniques you will be using, what equations you are going to be using for your calculations, and what reagents and materials will you be using. Start a section for recording your observations. Write down clear, concise, and detailed descriptions of what you see. Include all that was expected and any observations you did not expect. You may have an idea of what you expect to happen if you have completed your pre-laboratory research and read the background information. Unusual observations should be recorded—these observations may lead to further experimentation. Entries should be grammatically correct, legible and honest. Include your raw data from experiment. You never know what information may be important, so write down as much as you can. You can organize data in tables. Tape or glue in any data you have printed out or recorded on a separate piece of paper. If you use a computer program, record the program name and version number if relevant. After observations, you need a data analysis section. This will include any calculations needed to properly interpret your data. Include a conclusions section of what you did well, what you could have done better, and what future scientists can do to improve the results of the experiment. Occasionally, another person may have done some parts of the experiment. Note any parts someone else may have done and have them date and initial the relevant parts of your laboratory notebook. If an experiment is too long for one page, note that the experiment continues on to the next page. If you have to skip pages because you are doing multiple experiments, include references back to the original experiment. So if Experiment 12 ends on page 23 and is continuing on page 35, a note on page 23 should read ‘continued on page 35’ and a note on page 35 should read ‘Experiment 12 (continued from page 23).’ Keeping A Legally Defensible Document All data goes in the notebook, even outliers, or “bad” data points. A failed experiment should still be documented in the laboratory notebook. Do not take out any pages or remove any data or observations you have made in your notebook. Do not skip any pages in your laboratory notebook. Cross out any unused portions of your laboratory notebook and mark it with the date and your signature. Cross out any mistakes with a single line only. Put in corrections without covering anything. Sign and date all your corrections. To correct a large section, block out with one diagonal line from corner to corner followed by a date, signature, and reason for crossing it out. Reference “Standard Operating Procedure for Use and Maintenance of Laboratory Notebooks and Project Binders”, U. S. Environmental Protection Agency Office of Pesticide Programs, Microbiology Laboratory, Environmental Science Center, Ft. Meade, MD. SOP Number: ADM-05-04; Date Revised: 06-20-17. https://www.epa.gov/sites/production.../adm-05-04.pdf (Accessed 10/29/2018). Photos from: The Book Factory, Inc., and National Brand
Bookshelves/Introductory_Chemistry/Beginning_Chemistry_(Ball)/11%3A_Solutions/11.03%3A_Quantitative_Units_of_Concentration	Learning Objective Learn to determine specific concentrations with several common units. Rather than qualitative terms (Section 11.2 - Definitions), we need quantitative ways to express the amount of solute in a solution; that is, we need specific units of concentration. In this section, we will introduce several common and useful units of concentration. Molarity (M) is defined as the number of moles of solute divided by the number of liters of solution: \[molarity \: =\: \frac{moles\: of\: solute}{liters\: of\: solution}\nonumber \] which can be simplified as \[M\: =\: \frac{mol}{L},\; or\; mol/L\nonumber \] As with any mathematical equation, if you know any two quantities, you can calculate the third, unknown, quantity. For example, suppose you have 0.500 L of solution that has 0.24 mol of NaOH dissolved in it. The concentration of the solution can be calculated as follows: \[molarity \: =\: \frac{0.24\: mol\: NaOH}{0.500L}=0.48\, M\; NaOH\nonumber \] The concentration of the solution is 0.48 M, which is spoken as "zero point forty-eight molarity" or "zero point forty-eight molar." If the quantity of the solute is given in mass units, you must convert mass units to mole units before using the definition of molarity to calculate concentration. For example, what is the molar concentration of a solution of 22.4 g of HCl dissolved in 1.56 L? First, convert the mass of solute to moles using the molar mass of HCl (36.5 g/mol): \[22.4\cancel{gHCl}\times \frac{1\: mol\: Hcl}{36.5\cancel{gHCl}}=0.614\, M\; HCl\nonumber \] Now we can use the definition of molarity to determine a concentration: \[M \: =\: \frac{0.614\: mol\: HCl}{1.56L}=0.394\, M\nonumber \] Example $\PageIndex{1}$ What is the molarity of a solution made when 32.7 g of NaOH are dissolved to make 445 mL of solution? Solution To use the definition of molarity, both quantities must be converted to the proper units. First, convert the volume units from milliliters to liters: \[445\cancel{mL}\times \frac{1\: L}{1000\cancel{mL}}=0.445\, L\nonumber \] Now we convert the amount of solute to moles, using the molar mass of NaOH, which is 40.0 g/mol: \[32.7\cancel{gNaOH}\times \frac{1\: mol\: NaOH}{40.0\cancel{gNaOH}}=0.818\, mol\: NaOH\nonumber \] Now we can use the definition of molarity to determine the molar concentration: \[M \: =\: \frac{0.818\: mol\: NaOH}{0.445L}=1.84\, M\: NaOH\nonumber \] Exercise $\PageIndex{1}$ What is the molarity of a solution made when 66.2 g of C 6 H 12 O 6 are dissolved to make 235 mL of solution? Answer 1.57 M The definition of molarity can be used to determine the amount of solute or the volume of solution, if the other information is given. Example 4 illustrates this situation. Example $\PageIndex{2}$ How many moles of solute are present in 0.108 L of a 0.887 M NaCl solution? Solution We know the volume and the molarity; we can use the definition of molarity to mathematically solve for the amount in moles. Substituting the quantities into the definition of molarity: \[0.887\, M \: =\: \frac{mol\: NaCl}{0.108L}\nonumber \] We multiply the 0.108 L over to the other side of the equation and multiply the units together; "molarity × liters" equals moles, according to the definition of molarity. So mol NaCl = (0.887 M)(0.108 L) = 0.0958 mol Exercise $\PageIndex{2}$ How many moles of solute are present in 225 mL of a 1.44 M CaCl 2 solution? Answer 0.324 mol If you need to determine volume, remember the rule that the unknown quantity must be by itself and in the numerator to determine the correct answer. Thus, rearrangement of the definition of molarity is required. Example $\PageIndex{3}$ What volume of a 2.33 M NaNO 3 solution is needed to obtain 0.222 mol of solute? Solution Using the definition of molarity, we have \[2.33\, M \: =\: \frac{0.222mol}{L}\nonumber \] To solve for the number of liters, we bring the 2.33 M over to the right into the denominator, and the number of liters over to the left in the numerator. We now have \[L \: =\: \frac{0.222mol}{2.33\, M}\nonumber \] Dividing, the volume is 0.0953 L = 95.3 mL. Exercise $\PageIndex{3}$ What volume of a 0.570 M K 2 SO 4 solution is needed to obtain 0.872 mol of solute? Answer 1.53 L A similar unit of concentration is molality ( m ), which is defined as the number of moles of solute per kilogram of solvent, not per liter of solution: \[molality\: =\: \frac{moles\: solute}{kilograms\: solvent}\nonumber \] Mathematical manipulation of molality is the same as with molarity. Another way to specify an amount is percentage composition by mass (or mass percentage , % m/m). It is defined as follows: \[\%m/m\: =\: \frac{mass\: of\: solute}{mass\: of\: entire\: sample}\times 100\%\nonumber \] It is not uncommon to see this unit used on commercial products (see Figure $\PageIndex{1}$ - Concentration in Commercial Applications). Example $\PageIndex{4}$ What is the mass percentage of Fe in a piece of metal with 87.9 g of Fe in a 113 g sample? Solution Using the definition of mass percentage, we have \[\%m/m\: =\: \frac{87.9\, g\, Fe}{113\, g\, sample}\times 100\%=77.8\%\, Fe\nonumber \] Exercise $\PageIndex{4}$ What is the mass percentage of H 2 O 2 in a solution with 1.67 g of H 2 O 2 in a 55.5 g sample? Answer 3.01% Related concentration units are parts per thousand (ppth), parts per million (ppm) and parts per billion (ppb). Parts per thousand is defined as follows: \[ppth\: =\: \frac{mass\: of\: solute}{mass\: of\: sample}\times 1000\nonumber \] There are similar definitions for parts per million and parts per billion: \[ppm\: =\: \frac{mass\: of\: solute}{mass\: of\: sample}\times 1,000,000\nonumber \] and \[ppb\: =\: \frac{mass\: of\: solute}{mass\: of\: sample}\times 1,000,000,000\nonumber \] Each unit is used for progressively lower and lower concentrations. The two masses must be expressed in the same unit of mass, so conversions may be necessary. Example $\PageIndex{5}$ If there are 0.6 g of Pb present in 277 g of solution, what is the Pb concentration in parts per thousand? Solution Use the definition of parts per thousand to determine the concentration. Substituting \[\frac{0.6g\, Pb}{277g\, solution}\times 1000=2.17\, ppth\nonumber \] Exercise $\PageIndex{5}$ If there are 0.551 mg of As in 348 g of solution, what is the As concentration in ppm? Answer 1.58 ppm As with molarity and molality, algebraic rearrangements may be necessary to answer certain questions. Example $\PageIndex{6}$ The concentration of Cl – ion in a sample of H 2 O is 15.0 ppm. What mass of Cl – ion is present in 240.0 mL of H 2 O, which has a density of 1.00 g/mL? Solution First, use the density of H 2 O to determine the mass of the sample: \[240.0\cancel{mL}\times \frac{1.00\: g}{\cancel{mL}}=240.0\, g\nonumber \] Now we can use the definition of ppm: \[15.0\, ppm\: =\: \frac{mass\: of\: solute}{240.0\: g\: solution}\times 1,000,000\nonumber \] Rearranging to solve for the mass of solute, \[mass\: solute =\: \frac{(15.0\, ppm)(240.0\: g\: solution)}{1,000,000}=0.0036g=3.6\, mg\nonumber \] Exercise $\PageIndex{6}$ The concentration of Fe 3 + ion in a sample of H 2 O is 335.0 ppm. What mass of Fe 3 + ion is present in 3,450 mL of H 2 O, which has a density of 1.00 g/mL? Answer 1.16 g For ionic solutions, we need to differentiate between the concentration of the salt versus the concentration of each individual ion. Because the ions in ionic compounds go their own way when a compound is dissolved in a solution, the resulting concentration of the ion may be different from the concentration of the complete salt. For example, if 1 M NaCl were prepared, the solution could also be described as a solution of 1 M Na + (aq) and 1 M Cl − (aq) because there is one Na + ion and one Cl − ion per formula unit of the salt. However, if the solution were 1 M CaCl 2 , there are two Cl − (aq) ions for every formula unit dissolved, so the concentration of Cl − (aq) would be 2 M, not 1 M. In addition, the total ion concentration is the sum of the individual ion concentrations. Thus for the 1 M NaCl, the total ion concentration is 2 M; for the 1 M CaCl 2 , the total ion concentration is 3 M. Key Takeaway Quantitative units of concentration include molarity, molality, mass percentage, parts per thousand (ppth), parts per million (ppm), and parts per billion (ppb).
Courses/University_of_North_Texas/UNT%3A_CHEM_1410_-_General_Chemistry_for_Science_Majors_I/Text/08%3A_Properties_of_Organic_Compounds/8.16%3A_Aldehydes_and_Ketones/The_Cinnamon_Trade	Cinnamon has been known for thousands of years. One of the first times it is mentioned is in the Old Testament when Moses uses cinnamon in holy oil. Using cinnamon in holy oil shows how prized this spice was in ancient times. In fact, in Greece cinnamon was used as a sacrifice to the Gods like Apollo. Although cinnamon was used in the Mediterranean, it originated in Sri Lanka. The cinnamon traveled along the spice route, and the actual origins of cinnamon were held quiet. By not allowing other traders to know the source of the spice the growers were able to keep a monopoly and control the price (at a high one of course). In the sixteenth century Portugal finally made it to Sri Lanka. In 1518 a trading fort was set up there. The Dutch fought the Portuguese and eventually won control of the cinnamon trade in 1658. The British eventually took control of the island in 1796 but by then cinnamon wasn't as highly prized. Through the spice trade cinnamon made its way out of Sri Lanka and into the cuisines and cultures of the rest of the world. In Mexico especially cinnamon is a huge part of their cuisine. In fact, Mexico is the main importer of raw cinnamon. The majority of this cinnamon is used in the processing of chocolate. Also, cinnamon in Mexico is used in desserts, hot chocolate, and many other things. The spice is used in the same kind of dishes in the United States, as well as things such as cereal. In the Middle East cinnamon is used to flavor lamb as well as soups and even as a flavor in pickling. Cinnamon's unique taste and ability to be used as a spice is the reason why it has been incorporated into cultures all around the world. The component in cinnamon that gives it it's properties is the presence of cinnamaldehyde. The essential oil that makes up cinnamon is over 90% cinnamaldehyde. Cinnamaldehyde is an organic compound that can also be classified as an aldehyde. Cinnamaldehyde is unique in that it also contains a benzene ring and a double bond, as is seen in the structure in Figure 1. Cinnamaldehyde is also used in many other foods as a flavoring. While cimmaldehyde can be synthesized, the main source of it is from cinnamon bark. More on Aldehydes and Ketones The functional group found in formaldehyde is called a carbonyl group . Two classes of compounds may be distinguished on the basis of the location of the carbonyl group. In aldehydes it is at the end of a carbon chain and has at least one hydrogen attached. In ketones the carbonyl group is attached to two carbon atoms. Some examples are cinnamaldehyde Chemical tests may be performed to determine the presence of a carbonyl group. In the video below, a solution of 2,4-dinitrophenylhydrazine is added to test tubes containing 2-propanol, an alcohol; 2-propanone (acetone), a ketone; and propionic acid a carboxylic acid. 2,4-dinitrophenylhydrazine only reacts with the carbonyl group of 2-propanone, forming an orange precipitate. This option will not work correctly. Unfortunately, your browser does not support inline frames. The reaction that occurs is: Another test can distinguish between aldehydes and ketones. Here is a video of the Silver Mirror Tollens Test for Aldehydes: This option will not work correctly. Unfortunately, your browser does not support inline frames. Tollens reagent is an aqueous solution of silver nitrate, sodium hydroxide, and a little ammonia. (In the video the ammonia comes from reaction of ammonium ions in ammonium nitrate and hydroxide ions from sodium hydroxide to form water and NH 3 .) If an aldehyde, in this case, glucose, is added to the solution, the Ag + is reduced by the aldehyde, and the aldehyde is oxidized into a carboxylic acid. This produces silver metal, which coats the flask and creates the mirror. A similar reaction does not occur for ketones, so only aldehydes produce the silver mirror. The equation for the reaction in the video is: The endings al and one signify al dehyde and ket one , respectively. The general formula for an aldehyde is , while for a ketone it is . Note that every ketone is isomeric with at least one aldehyde. Acetone, for example, has the same molecular formula (C 3 H 6 O) as propanal. Aldehyde and ketone molecules cannot hydrogen bond among themselves for the same reason that ethers cannot—they do not contain hydrogens attached to highly electronegative atoms. The carbonyl group is rather polar, however, since the difference between the electronegativities of carbon (2.5) and oxygen (3.5) is rather large, and there are usually no other dipoles in an aldehyde or ketone molecule to cancel the effect of C==O. Therefore the boiling points of aldehydes and ketones are intermediate between those of alkanes or ethers on the one hand and alcohols on the other. Acetaldehyde, CH 3 CH 2 CHO, boils at 20.8°C midway between propane (–42°C) and ethanol (78.5°C). The boiling points of propanal and acetone are compared with other organic compounds in the table of the boiling points of comparable organic compounds which shows the same trend. Of all the aldehydes and ketones, formaldehyde and acetone are of greatest commercial importance. Uses of formaldehyde have already been mentioned. Like other ketones, acetone is mainly useful as a solvent, and you may have used it for this purpose in the laboratory. Acetone and other ketones are somewhat toxic and should not be handled carelessly. From ChemPRIME: 8.15: Aldehydes and Ketones
Courses/Providence_College/CHM_331_Advanced_Analytical_Chemistry_1/11%3A_Raman_Spectroscopy/11.02%3A_Introduction_to_Lasers/11.2.02%3A_Basic_Principles/11.2.2.01%3A_Laser_Radiation_Properties	Laser Radiation Properties I Laser radiation is nearly monochromatic . Monochromatic refers to a single wavelength, or “one color” of light. Laser radiation contains a narrow band of wavelengths and can be produced closer to monochromatic than light from other sources. Laser radiation is highly directional . The radiation is produced in a beam that is spatially narrow and has low divergence relative to other light sources. Laser radiation is highly coherent , which means the waves of light emitted have a constant relative phase. The waves of light in a laser beam are thought of as in phase with one another at every point. The degree of coherence is proportional to the range of wavelengths in the light beam, or the beam’s monochromaticity. Laser radiation has both spatial and temporal coherence, characterized by the coherence length and the coherence time . Coherence Temporal coherence is the ability of light to maintain a constant phase at one point in space at two different times, separated by delay τ . Temporal coherence characterizes how well a wave can interfere with itself at two different times and increases as a source becomes more monochromatic . A coherence time ( τ cor ) and coherence length (c × τ cor , where c is the speed of light) can be calculated from the spread of wavelengths ( Δ λ ), or frequencies ( Δν ), in a beam. Expressed in terms of Δν , or “ bandwidth ”: \[\tau_{cor} = \dfrac{1}{2 \pi Δν}\] Laser Radiation Properties II Laser radiation has high brightness , a quantity defined as the power emitted per unit surface area per unit solid angle. Because laser light is emitted as a narrow beam with small divergence, the brightness of a 1 mW laser pointer, for example, is > 1,000 × ’s greater than that of the sun, which emits more than 10 25 W of radiant power a . Laser output can be continuous or pulsed . Continuous wave ( CW ) lasers are characterized by their average power , whereas peak power , energy per pulse and pulse repetition rate are figures of merit that apply to pulsed lasers. Pulse widths in the ns-ps range are employed more routinely than fs pulses, and attosecond pulses can be generated. A 10 fs pulse with only 10 mJ energy has a peak power of 10 12 W, or 1 TW! Laser Radiation Properties III The narrow range of frequencies, or wavelengths, emitted is referred to as the laser bandwidth . This output is determined by the spectral emission properties of the gain medium and the modes supported by the cavity. When the bandwidth of the gain medium is larger than the cavity mode spacing, the laser output consists of a series of narrow spectral bands (see the following figure and “Laser Radiation Properties IV” below). Cavity modes develop as a consequence of the properties of light reflection and interference. In the simplest case of a cavity formed by two flat mirrors, the allowable axial modes have wavelength λ = 2L/q , where L is the cavity length and q is an integer. The frequency spacing ( Δν ) between modes is given by Δν = c/(2L) , where c is the speed of light. Parabolic mirrors produce more complex cavity modes leading to a Gaussian beam b . Laser Radiation Properties IV Laser bandwidth frequency (Δν) and wavelength (Δλ) are related as follows: \[\Delta \lambda \approx\left(\frac{\lambda_{0}^{2}}{c}\right) \Delta v\] where λ o is the band center wavelength and c is the speed of light. A HeNe laser operating at 632.8 nm has a gain bandwidth of 1.5 GHz, or 0.002 nm. When the gain medium bandwidth is smaller than the cavity mode spacing, the laser output consists of a single mode and operates as a single frequency laser c . A HeNe laser with 20 cm cavity length has mode spacings of Δν = 750 MHz, or Δλ = .001 nm. HeNe lasers are often equipped for and use single frequency operation c , d . Mode locking e produces a fixed phase relationship between laser cavity modes and results in pulsed output. See Refs [ 2,7,8,9 ] for more details on mode locking and methods for producing ultra-short laser pulses and other aspects of single frequency laser operation. References http://www.worldoflasers.com/laserproperties.htm http://www.rp-photonics.com/resonator_modes.html http://www.rp-photonics.com/single_f...cy_lasers.html http://www.rp-photonics.com/single_f...operation.html http://www.rp-photonics.com/mode_locked_lasers.html
Courses/University_of_Connecticut/Chem_2444%3A_(Second_Semester_Organic_Chemistry)_UConn/03%3A_Conjugated_Pi-Systems_and_Aromaticity/3.08%3A_Aromatic_Heterocycles-_Pyridine_and_Pyrrole	Objectives After completing this section, you should be able to draw the structure of the common aromatic heterocycles pyridine and pyrrole. use the Hückel 4 n + 2 rule to explain the aromaticity of each of pyridine and pyrrole. draw a diagram to show the orbitals involved in forming the conjugated six‑pi‑electron systems present in aromatic heterocycles such as pyridine, pyrrole, etc. Key Terms Make certain that you can define, and use in context, the key terms below. carbocycles heterocycles Aromatic Heterocycles Many cyclic compounds have an element other than carbon present in the ring. Cyclic compounds that include one or more elements other than carbon are called a heterocycle. The most common heterocyclic compounds contain carbon along with nitrogen, oxygen, or sulfur. Because some heterocyclic compounds are aromatic, it is important to discuss how the inclusion of non-carbon atoms affects the determination of aromaticity. The reactivity and general physical properties of aromatic nitrogen heterocycles will be discussed in greater detail in Section 24.9 . Pyridine & Pyrimidine Pyridine is an example of a six-membered aromatic heterocycle and has an electronic structure similar to benzene. In the bonding picture of pyridine the five carbons and single nitrogen are all sp 2 hybridized. All six of these atoms have a p orbital perpendicular to the plane of the ring and each contains one pi electron which allows the ring to be fully conjugated. This makes the pi electron count for pyridine 6 pi electrons. In much the same fashion as benzene, the 6 pi electrons follow the 4n + 2 rule and completely fill the bonding molecular orbitals which fulfills the electronic requirement for aromaticity. The lone pair electrons on pyridine's nitrogen are contained in a sp 2 orbital that lies in the same plane as the ring and does not overlap with the p orbitals of the ring. The lone pair electrons are not part of the aromatic system and do not affect the p electron count. Pyridine's electrostatic potential map shows that the lone pair electrons, shown in a yellow/red color, are locked onto the nitrogen and not distributed around the ring. Pyrimidine is an another six-membered aromatic heterocycle that is analogous to pyridine. Pyrimidine has four carbons atoms and two nitrogen atoms that are sp 2 hybridized. Each of these atoms contributes a p orbital and a pi election allowing pyrimidine to be fully conjugated and aromatic. Both of the nitrogens in pyrimidine have lone pair electrons contained in a sp 2 orbitals and are not involved in the aromatic system. An electrostatic potential map of pyrimidine shows that neither set of lone pair electrons is distributed around the ring. Pyrrole Pyrrole is a five-membered heterocyclic ring which has 5 p orbitals and six pi electrons contributing to its aromaticity. Each carbon in pyrrole contributes one p orbital and pi electron. The nitrogen in pyrrole contributes two pi electrons by becoming sp 2 hybridized and placing its lone pair electrons into a p orbital. The electrostatic potential map of pyrrole show the nitrogen's lone pair electrons are distributed in the ring. Because the nitrogen lone pair is part of the aromatic sextet the electrons are very stable and are much less available for bonding to a proton (and if they do pick up a proton, the molecule loses aromaticity). Pyrrole is a very weak base: the conjugate acid, the pyrrolium ion, is a strong acid with a pK a of 0.4. Imidazole Imidazole is another five-membered heterocyclic ring which has 6 pi electrons and is aromatic. Both of the nitrogen atoms are sp 2 hybridized. One nitrogen is pyridine-like because it is part of a double bond and adds one pi electron to the aromatic pi ring. The other nitrogen is not part of a double bond making it pyrrole-like allowing it to contribute 2 pi electrons from it lone pair electrons. When looking at the electrostatic potential map of imidazole, the lone pair electrons on the pyrrole-like nitrogen are distributed around the ring while the lone pair electrons on the pyridine-like nitrogen are non-conjugated and locked into place. Pyrimidine and imidazole rings are particularly important in biological chemistry. Pyrimidine, for instance, is the parent ring system in cytosine, thymine, and uracil, three of the five heterocyclic amine bases found in nucleic acids. An aromatic imidazole ring is present in histidine, one of the amino acids found in proteins. Worked Example $\PageIndex{1}$ Determine if furan is aromatic. Draw an orbital diagram to support your answer. Answer For a molecule to be aromatic it must: Be cyclic, be planar, be fully conjugated , and have 4n+2 π electrons. Furan is obviously cyclic and can be assumed to be planar due to the constrains of having a five membered ring with two double bonds. To be fully conjugated each atom in the ring must have a p orbital. Although the oxygen has 4 electron groups and appears to be sp 3 hybridized, it will become sp 2 hybridized to gain the stability of conjugation. Becoming sp 2 hybridized allows the oxygen atom to put a set of lone pair electrons into a p orbital allowing it to be part of the cyclic conjugation. These two pi electrons along with the 4 pi electrons provided by the two double bonds gives furan 6 pi electrons total. This follows the 4n + 2 rule making furan aromatic. Exercises Exercise $\PageIndex{1}$ 1) Draw the orbitals of thiophene to show that it is aromatic. 2) The thiazolium ring is a five-membered sulfur containing aromatic ring system which is found in biological systems, such as thiamine diphosphate (ThDP). Describe how thiazolium ring is aromatic. Answer 1) This drawing shows thiophene has 6 electrons in the pi-orbital. 2) Similar to the last question, the drawing shows that there is only 6 electrons in the pi-system. Exercise $\PageIndex{2}$ Draw the orbitals of thiophene to show that is aromatic. Answer This drawing shows it has 6 electrons in the pi-orbital. Exercise $\PageIndex{3}$ The following ring is called a thiazolium ring. Describe how it is aromatic. Answer Similar to the last question, the drawing shows that there is only 6 electrons in the pi-system.
Courses/Oregon_Institute_of_Technology/OIT%3A_CHE_331_-_Organic_Chemistry_(Lund)/04%3A_Conformations_and_Stereochemistry/4.01%3A_Conformations_of_open-chain_organic_molecules	Before we begin our exploration of stereochemistry and chirality, we first need to consider the subject of conformational isomerism , which has to do with rotation about single bonds. We learned in section 2.1 that single bonds in organic molecules are free to rotate, due to the 'end-to-end' ( sigma ) nature of their orbital overlap. Consider the carbon-oxygen bond in ethanol, for example: with a 180 o rotation about this bond, the shape of the molecule would look quite different: Or ethane: rotation about the carbon-carbon sigma bond results in many different possible three-dimensional arrangements of the atoms. These different arrangements, resulting from sigma bond rotation, are referred to in organic chemistry as conformations . Any one specific conformation is called a conformational isomer , or conformer . In order to better visualize different conformations of a molecule, it is convenient to use a drawing convention called the Newman projection . In a Newman projection, we look lengthwise down a specific bond of interest – in this case, the carbon-carbon bond in ethane. We depict the ‘front’ atom as a dot, and the ‘back’ atom as a larger circle. Interactive mode of ethane conformationsl The six carbon-hydrogen bonds are shown as solid lines protruding from the two carbons. Note that we do not draw bonds as solid or dashed wedges in a Newman projection. Looking down the C-C bond in this way, the angle formed between a C-H bond on the front carbon and a C-H bond on the back carbon is referred to as a dihedral angle . (The dihedral angle between the hour hand and the minute hand on a clock is 0 o at noon, 90 o at 3:00, and so forth). The lowest energy conformation of ethane, shown in the figure above, is called the ‘ staggered’ conformation: all of the dihedral angles are 60 o , and the distance between the front and back C-H bonds is maximized. If we now rotate the front CH 3 group 60 ° clockwise, the molecule is in the highest energy ‘ eclipsed' conformation, where the dihedral angles are all 0 o (we stagger the bonds slightly in our Newman projection drawing so that we can see them all). The energy of the eclipsed conformation, where the electrons in the front and back C-H bonds are closer together, is approximately 12 kJ/mol higher than that of the staggered conformation. Another 60 ° rotation returns the molecule to a second staggered conformation. This process can be continued all around the 360 ° circle, with three possible eclipsed conformations and three staggered conformations, in addition to an infinite number of conformations in between these two extremes. Now let's consider butane, with its four-carbon chain. There are now three rotating carbon-carbon bonds to consider, but we will focus on the middle bond between C 2 and C 3 . Below are two representations of butane in a conformation which puts the two CH 3 groups (C 1 and C 4 ) in the eclipsed position, with the two C-C bonds at a 0 o dihedral angle. Interactive model of butane conformations If we rotate the front, (blue) carbon by 60 ° clockwise, the butane molecule is now in a staggered conformation. This is more specifically referred to as the gauche conformation of butane. Notice that although they are staggered, the two methyl groups are not as far apart as they could possibly be. A further rotation of 60 ° gives us a second eclipsed conformation (B) in which both methyl groups are lined up with hydrogen atoms. One more 60 rotation produces another staggered conformation called the anti conformation, where the two methyl groups are positioned opposite each other (a dihedral angle of 180 o ). As with ethane, the staggered conformations of butane are energy 'valleys', and the eclipsed conformations are energy 'peaks'. However, in the case of butane there are two different valleys, and two different peaks. The gauche conformation is a higher energy valley than the anti conformation due to steric strain , which is the repulsive interaction caused by the two bulky methyl groups being forced too close together. Clearly, steric strain is lower in the anti conformation. In the same way, steric strain causes the eclipsed A conformation - where the two methyl groups are as close together as they can possibly be - to be higher in energy than the two eclipsed B conformations. The diagram below summarizes the relative energies for the various eclipsed, staggered, and gauche conformations. Because the anti conformation is lowest in energy (and also simply for ease of drawing), it is conventional to draw open-chain alkanes in a 'zigzag' form, which implies anti conformation at all carbon-carbon bonds. The figure below shows, as an example, a Newman projection looking down the C 2 -C 3 bond of octane. Exercise 3.1 Using free rotation around C-C single bonds, show that (R,S) and (S,R)-tartaric acid are identical molecules. Exercise 3.2 Draw a Newman projection, looking down the C 2 -C 3 bond, of 1-butene in the conformation shown below (C 2 should be your front carbon). Solutions to exercises Online lectures from Khan Academy Newman projections part I Newman projections part II
Courses/Pasadena_City_College/Chem_2A_(Ku)_Textbook/06%3A_Molecular_Compounds_and_Structures	How do atoms make compounds? Typically they join together in such a way that they lose their identities as elements and adopt a new identity as a compound. These joins are called chemical bonds . But how do atoms join together? Ultimately, it all comes down to electrons. Before we discuss how electrons interact, we need to introduce a tool to simply illustrate electrons in an atom. 6.1: Bonding and Lewis Structures 6.1.1: Representing Valence Electrons with Dots 6.1.2: Lewis Structures of Ionic Compounds- Electrons Transferred 6.1.3: Covalent Bonds - Electrons Shared 6.1.4: Covalent Compounds - Formulas and Names 6.1.5: Naming Acids 6.1.6: Writing Lewis Structures for Covalent Compounds 6.2: Structures of Organic Molecules The term "organic" in organic chemistry originally comes from the association with living organisms. Early chemists believed that organic compounds could only be produced by living beings due to a "vital force." This changed in 1828 when Friedrich Wöhler synthesized urea, an organic compound, from inorganic materials, disproving vitalism. Today, organic chemistry is defined as the study of carbon-containing compounds, regardless of their origin. 6.2.1: Carbon - The Chemical Basis for Life 6.2.2: Hydrocarbons- Compounds Containing Only Carbon and Hydrocarbon 6.2.3: Bond-Line Formulas 6.3: Molecular Shape and Polarity 6.3.1: Predicting the Shapes of Molecules 6.3.2: Electronegativity and Polarity - Why Oil and Water Do not Mix
Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/10%3A_Liquids_and_Solids/10.07%3A_Lattice_Structures_in_Crystalline_Solids	Learning Objectives Describe the arrangement of atoms and ions in crystalline structures Compute ionic radii using unit cell dimensions Explain the use of X-ray diffraction measurements in determining crystalline structures Over 90% of naturally occurring and man-made solids are crystalline. Most solids form with a regular arrangement of their particles because the overall attractive interactions between particles are maximized, and the total intermolecular energy is minimized, when the particles pack in the most efficient manner. The regular arrangement at an atomic level is often reflected at a macroscopic level. In this module, we will explore some of the details about the structures of metallic and ionic crystalline solids, and learn how these structures are determined experimentally. The Structures of Metals We will begin our discussion of crystalline solids by considering elemental metals, which are relatively simple because each contains only one type of atom. A pure metal is a crystalline solid with metal atoms packed closely together in a repeating pattern. Some of the properties of metals in general, such as their malleability and ductility, are largely due to having identical atoms arranged in a regular pattern. The different properties of one metal compared to another partially depend on the sizes of their atoms and the specifics of their spatial arrangements. We will explore the similarities and differences of four of the most common metal crystal geometries in the sections that follow. Unit Cells of Metals The structure of a crystalline solid, whether a metal or not, is best described by considering its simplest repeating unit, which is referred to as its unit cell . The unit cell consists of lattice points that represent the locations of atoms or ions. The entire structure then consists of this unit cell repeating in three dimensions, as illustrated in Figure $\PageIndex{1}$. Let us begin our investigation of crystal lattice structure and unit cells with the most straightforward structure and the most basic unit cell. To visualize this, imagine taking a large number of identical spheres, such as tennis balls, and arranging them uniformly in a container. The simplest way to do this would be to make layers in which the spheres in one layer are directly above those in the layer below, as illustrated in Figure $\PageIndex{2}$. This arrangement is called simple cubic structure , and the unit cell is called the simple cubic unit cell or primitive cubic unit cell. In a simple cubic structure, the spheres are not packed as closely as they could be, and they only “fill” about 52% of the volume of the container. This is a relatively inefficient arrangement, and only one metal (polonium, Po) crystallizes in a simple cubic structure. As shown in Figure $\PageIndex{3}$, a solid with this type of arrangement consists of planes (or layers) in which each atom contacts only the four nearest neighbors in its layer; one atom directly above it in the layer above; and one atom directly below it in the layer below. The number of other particles that each particle in a crystalline solid contacts is known as its coordination number . For a polonium atom in a simple cubic array, the coordination number is, therefore, six. In a simple cubic lattice, the unit cell that repeats in all directions is a cube defined by the centers of eight atoms, as shown in Figure $\PageIndex{4}$. Atoms at adjacent corners of this unit cell contact each other, so the edge length of this cell is equal to two atomic radii, or one atomic diameter. A cubic unit cell contains only the parts of these atoms that are within it. Since an atom at a corner of a simple cubic unit cell is contained by a total of eight unit cells, only one-eighth of that atom is within a specific unit cell. And since each simple cubic unit cell has one atom at each of its eight “corners,” there is $8×\dfrac{1}{8}=1$ atom within one simple cubic unit cell. Example $\PageIndex{1}$: Calculating Atomic Radius and Density for Metals (Part 1) The edge length of the unit cell of alpha polonium is 336 pm. Determine the radius of a polonium atom. Determine the density of alpha polonium. Solution Alpha polonium crystallizes in a simple cubic unit cell: (a) Two adjacent Po atoms contact each other, so the edge length of this cell is equal to two Po atomic radii: $l = 2r$. Therefore, the radius of Po is \[r=\mathrm{\dfrac{l}{2}=\dfrac{336\: pm}{2}=168\: pm}\nonumber \] (b) Density is given by \[\mathrm{density=\dfrac{mass}{volume}}.\nonumber \] The density of polonium can be found by determining the density of its unit cell (the mass contained within a unit cell divided by the volume of the unit cell). Since a Po unit cell contains one-eighth of a Po atom at each of its eight corners, a unit cell contains one Po atom. The mass of a Po unit cell can be found by: \[\mathrm{1\: Po\: unit\: cell×\dfrac{1\: Po\: atom}{1\: Po\: unit\: cell}×\dfrac{1\: mol\: Po}{6.022\times 10^{23}\:Po\: atoms}×\dfrac{208.998\:g}{1\: mol\: Po}=3.47\times 10^{−22}\:g}\nonumber \] The volume of a Po unit cell can be found by: \[V=l^3=\mathrm{(336\times 10^{−10}\:cm)^3=3.79\times 10^{−23}\:cm^3}\nonumber \] (Note that the edge length was converted from pm to cm to get the usual volume units for density.) Therefore, the density of \[\mathrm{Po=\dfrac{3.471\times 10^{−22}\:g}{3.79\times 10^{−23}\:cm^3}=9.16\: g/cm^3}\nonumber \] Exercise $\PageIndex{1}$ The edge length of the unit cell for nickel is 0.3524 nm. The density of Ni is 8.90 g/cm 3 . Does nickel crystallize in a simple cubic structure? Explain. Answer No. If Ni were simple cubic, its density would be given by: \[\mathrm{1\: Ni\: atom×\dfrac{1\: mol\: Ni}{6.022\times 10^{23}\:Ni\: atoms}×\dfrac{58.693\:g}{1\: mol\: Ni}=9.746\times 10^{−23}\:g}\nonumber \] \[V=l^3=\mathrm{(3.524\times 10^{−8}\:cm)^3=4.376\times 10^{−23}\:cm^3}\nonumber \] Then the density of Ni would be \[(\mathrm{=\dfrac{9.746\times 10^{−23}\:g}{4.376\times 10^{−23}\:cm^3}=2.23\: g/cm^3}\nonumber \] Since the actual density of Ni is not close to this, Ni does not form a simple cubic structure. Most metal crystals are one of the four major types of unit cells. For now, we will focus on the three cubic unit cells: simple cubic (which we have already seen), body-centered cubic unit cell , and face-centered cubic unit cell —all of which are illustrated in Figure $\PageIndex{5}$. (Note that there are actually seven different lattice systems, some of which have more than one type of lattice, for a total of 14 different types of unit cells. We leave the more complicated geometries for later in this module.) Some metals crystallize in an arrangement that has a cubic unit cell with atoms at all of the corners and an atom in the center, as shown in Figure $\PageIndex{6}$. This is called a body-centered cubic (BCC) solid . Atoms in the corners of a BCC unit cell do not contact each other but contact the atom in the center. A BCC unit cell contains two atoms: one-eighth of an atom at each of the eight corners ($8×\dfrac{1}{8}=1$ atom from the corners) plus one atom from the center. Any atom in this structure touches four atoms in the layer above it and four atoms in the layer below it. Thus, an atom in a BCC structure has a coordination number of eight. Atoms in BCC arrangements are much more efficiently packed than in a simple cubic structure, occupying about 68% of the total volume. Isomorphous metals with a BCC structure include K, Ba, Cr, Mo, W, and Fe at room temperature. (Elements or compounds that crystallize with the same structure are said to be isomorphous .) Many other metals, such as aluminum, copper, and lead, crystallize in an arrangement that has a cubic unit cell with atoms at all of the corners and at the centers of each face, as illustrated in Figure $\PageIndex{7}$. This arrangement is called a face-centered cubic (FCC) solid . A FCC unit cell contains four atoms: one-eighth of an atom at each of the eight corners ($8×\dfrac{1}{8}=1$ atom from the corners) and one-half of an atom on each of the six faces ($6×\dfrac{1}{2}=3$ atoms from the faces). The atoms at the corners touch the atoms in the centers of the adjacent faces along the face diagonals of the cube. Because the atoms are on identical lattice points, they have identical environments. Atoms in an FCC arrangement are packed as closely together as possible, with atoms occupying 74% of the volume. This structure is also called cubic closest packing (CCP) . In CCP , there are three repeating layers of hexagonally arranged atoms. Each atom contacts six atoms in its own layer, three in the layer above, and three in the layer below. In this arrangement, each atom touches 12 near neighbors, and therefore has a coordination number of 12. The fact that FCC and CCP arrangements are equivalent may not be immediately obvious, but why they are actually the same structure is illustrated in Figure $\PageIndex{8}$. Because closer packing maximizes the overall attractions between atoms and minimizes the total intermolecular energy, the atoms in most metals pack in this manner. We find two types of closest packing in simple metallic crystalline structures: CCP, which we have already encountered, and hexagonal closest packing (HCP) shown in Figure $\PageIndex{9}$. Both consist of repeating layers of hexagonally arranged atoms. In both types, a second layer (B) is placed on the first layer (A) so that each atom in the second layer is in contact with three atoms in the first layer. The third layer is positioned in one of two ways. In HCP , atoms in the third layer are directly above atoms in the first layer (i.e., the third layer is also type A), and the stacking consists of alternating type A and type B close-packed layers (i.e., ABABAB⋯). In CCP, atoms in the third layer are not above atoms in either of the first two layers (i.e., the third layer is type C), and the stacking consists of alternating type A, type B, and type C close-packed layers (i.e., ABCABCABC⋯). About two–thirds of all metals crystallize in closest-packed arrays with coordination numbers of 12. Metals that crystallize in an HCP structure include Cd, Co, Li, Mg, Na, and Zn, and metals that crystallize in a CCP structure include Ag, Al, Ca, Cu, Ni, Pb, and Pt. Example $\PageIndex{2}$: Calculating Atomic Radius and Density for Metals (Part 2) Calcium crystallizes in a face-centered cubic structure. The edge length of its unit cell is 558.8 pm. What is the atomic radius of Ca in this structure? Calculate the density of Ca. Solution (a) In an FCC structure, Ca atoms contact each other across the diagonal of the face, so the length of the diagonal is equal to four Ca atomic radii (d = 4 r ). Two adjacent edges and the diagonal of the face form a right triangle, with the length of each side equal to 558.8 pm and the length of the hypotenuse equal to four Ca atomic radii: \[\begin{align} a^2+a^2 &=d^2 \\[4pt] \mathrm{(558.8\:pm)^2+(558.5\:pm)^2} &=(4r)^2 \end{align} \nonumber \] Solving this gives \[r=\mathrm{\sqrt{\dfrac{(558.8\:pm)^2+(558.5\:pm)^2}{16}}}=\textrm{197.6 pmg for a Ca radius}. \nonumber \] (b) Density is given by $\mathrm{density=\dfrac{mass}{volume}}$. The density of calcium can be found by determining the density of its unit cell: for example, the mass contained within a unit cell divided by the volume of the unit cell. A face-centered Ca unit cell has one-eighth of an atom at each of the eight corners ($8 \times \dfrac{1}{8}=1$atom) and one-half of an atom on each of the six faces $6×\dfrac{1}{2}=3$ atoms), for a total of four atoms in the unit cell. The mass of the unit cell can be found by: \[\mathrm{1\: Ca\: unit\: cell×\dfrac{4\: Ca\: atoms}{1\: Ca\: unit\: cell}×\dfrac{1\: mol\: Ca}{6.022\times 10^{23}\:Ca\: atoms}×\dfrac{40.078\:g}{1\: mol\: Ca}=2.662\times 10^{−22}\:g} \nonumber \] The volume of a Ca unit cell can be found by: \[V=a^3=\mathrm{(558.8\times 10^{−10}\:cm)^3=1.745\times 10^{−22}\:cm^3} \nonumber \] (Note that the edge length was converted from pm to cm to get the usual volume units for density.) Then, the density of polonium: \[\mathrm{Po=\dfrac{2.662\times 10^{−22}\:g}{1.745\times 10^{−22}\:cm^3}=1.53\: g/cm^3} \nonumber \] Exercise $\PageIndex{2}$ Silver crystallizes in an FCC structure. The edge length of its unit cell is 409 pm. What is the atomic radius of Ag in this structure? Calculate the density of Ag. Answer a 144 pm Answer b 10.5 g/cm 3 In general, a unit cell is defined by the lengths of three axes ( a , b , and c ) and the angles ( α , β , and γ ) between them, as illustrated in Figure $\PageIndex{10}$. The axes are defined as being the lengths between points in the space lattice. Consequently, unit cell axes join points with identical environments. There are seven different lattice systems, some of which have more than one type of lattice, for a total of fourteen different unit cells, which have the shapes shown in Figure $\PageIndex{11}$. The Structures of Ionic Crystals Ionic crystals consist of two or more different kinds of ions that usually have different sizes. The packing of these ions into a crystal structure is more complex than the packing of metal atoms that are the same size. Most monatomic ions behave as charged spheres, and their attraction for ions of opposite charge is the same in every direction. Consequently, stable structures for ionic compounds result (1) when ions of one charge are surrounded by as many ions as possible of the opposite charge and (2) when the cations and anions are in contact with each other. Structures are determined by two principal factors: the relative sizes of the ions and the ratio of the numbers of positive and negative ions in the compound. In simple ionic structures, we usually find the anions, which are normally larger than the cations, arranged in a closest-packed array. (As seen previously, additional electrons attracted to the same nucleus make anions larger and fewer electrons attracted to the same nucleus make cations smaller when compared to the atoms from which they are formed.) The smaller cations commonly occupy one of two types of holes (or interstices) remaining between the anions. The smaller of the holes is found between three anions in one plane and one anion in an adjacent plane. The four anions surrounding this hole are arranged at the corners of a tetrahedron, so the hole is called a tetrahedral hole . The larger type of hole is found at the center of six anions (three in one layer and three in an adjacent layer) located at the corners of an octahedron; this is called an octahedral hole . Figure $\PageIndex{12}$ illustrates both of these types of holes. Depending on the relative sizes of the cations and anions, the cations of an ionic compound may occupy tetrahedral or octahedral holes, as illustrated in Figure $\PageIndex{13}$. Relatively small cations occupy tetrahedral holes, and larger cations occupy octahedral holes. If the cations are too large to fit into the octahedral holes, the anions may adopt a more open structure, such as a simple cubic array. The larger cations can then occupy the larger cubic holes made possible by the more open spacing. There are two tetrahedral holes for each anion in either an HCP or CCP array of anions. A compound that crystallizes in a closest-packed array of anions with cations in the tetrahedral holes can have a maximum cation:anion ratio of 2:1; all of the tetrahedral holes are filled at this ratio. Examples include Li 2 O, Na 2 O, Li 2 S, and Na 2 S. Compounds with a ratio of less than 2:1 may also crystallize in a closest-packed array of anions with cations in the tetrahedral holes, if the ionic sizes fit. In these compounds, however, some of the tetrahedral holes remain vacant. Example $\PageIndex{3}$: Occupancy of Tetrahedral Holes Zinc sulfide is an important industrial source of zinc and is also used as a white pigment in paint. Zinc sulfide crystallizes with zinc ions occupying one-half of the tetrahedral holes in a closest-packed array of sulfide ions. What is the formula of zinc sulfide? Solution Because there are two tetrahedral holes per anion (sulfide ion) and one-half of these holes are occupied by zinc ions, there must be $\dfrac{1}{2}×2$, or 1, zinc ion per sulfide ion. Thus, the formula is ZnS. Exercise $\PageIndex{3}$: Lithium selenide Lithium selenide can be described as a closest-packed array of selenide ions with lithium ions in all of the tetrahedral holes. What it the formula of lithium selenide? Answer $\ce{Li2Se}$ The ratio of octahedral holes to anions in either an HCP or CCP structure is 1:1. Thus, compounds with cations in octahedral holes in a closest-packed array of anions can have a maximum cation:anion ratio of 1:1. In NiO, MnS, NaCl, and KH , for example, all of the octahedral holes are filled. Ratios of less than 1:1 are observed when some of the octahedral holes remain empty. Example $\PageIndex{4}$: Stoichiometry of Ionic Compounds Sapphire Aluminum oxide crystallizes with aluminum ions in two-thirds of the octahedral holes in a closest-packed array of oxide ions. What is the formula of aluminum oxide? Solution Because there is one octahedral hole per anion (oxide ion) and only two-thirds of these holes are occupied, the ratio of aluminum to oxygen must be $\dfrac{2}{3}$:1, which would give $\mathrm{Al_{2/3}O}$. The simplest whole number ratio is 2:3, so the formula is Al 2 O 3 . Exercise $\PageIndex{4}$ The white pigment titanium oxide crystallizes with titanium ions in one-half of the octahedral holes in a closest-packed array of oxide ions. What is the formula of titanium oxide? Answer $\ce{TiO2}$ In a simple cubic array of anions, there is one cubic hole that can be occupied by a cation for each anion in the array. In CsCl, and in other compounds with the same structure, all of the cubic holes are occupied. Half of the cubic holes are occupied in SrH 2 , UO 2 , SrCl 2 , and CaF 2 . Different types of ionic compounds often crystallize in the same structure when the relative sizes of their ions and their stoichiometries (the two principal features that determine structure) are similar. Unit Cells of Ionic Compounds Many ionic compounds crystallize with cubic unit cells, and we will use these compounds to describe the general features of ionic structures. When an ionic compound is composed of cations and anions of similar size in a 1:1 ratio, it typically forms a simple cubic structure. Cesium chloride, CsCl, (Figure $\PageIndex{14}$) is an example of this, with Cs + and Cl − having radii of 174 pm and 181 pm, respectively. We can think of this as chloride ions forming a simple cubic unit cell, with a cesium ion in the center; or as cesium ions forming a unit cell with a chloride ion in the center; or as simple cubic unit cells formed by Cs + ions overlapping unit cells formed by Cl − ions. Cesium ions and chloride ions touch along the body diagonals of the unit cells. One cesium ion and one chloride ion are present per unit cell, giving the l:l stoichiometry required by the formula for cesium chloride. Note that there is no lattice point in the center of the cell, and CsCl is not a BCC structure because a cesium ion is not identical to a chloride ion. We have said that the location of lattice points is arbitrary. This is illustrated by an alternate description of the CsCl structure in which the lattice points are located in the centers of the cesium ions. In this description, the cesium ions are located on the lattice points at the corners of the cell, and the chloride ion is located at the center of the cell. The two unit cells are different, but they describe identical structures. When an ionic compound is composed of a 1:1 ratio of cations and anions that differ significantly in size, it typically crystallizes with an FCC unit cell, like that shown in Figure $\PageIndex{15}$. Sodium chloride, NaCl, is an example of this, with Na + and Cl − having radii of 102 pm and 181 pm, respectively. We can think of this as chloride ions forming an FCC cell, with sodium ions located in the octahedral holes in the middle of the cell edges and in the center of the cell. The sodium and chloride ions touch each other along the cell edges. The unit cell contains four sodium ions and four chloride ions, giving the 1:1 stoichiometry required by the formula, NaCl. The cubic form of zinc sulfide, zinc blende, also crystallizes in an FCC unit cell, as illustrated in Figure $\PageIndex{16}$. This structure contains sulfide ions on the lattice points of an FCC lattice. (The arrangement of sulfide ions is identical to the arrangement of chloride ions in sodium chloride.) The radius of a zinc ion is only about 40% of the radius of a sulfide ion, so these small Zn 2 + ions are located in alternating tetrahedral holes, that is, in one half of the tetrahedral holes. There are four zinc ions and four sulfide ions in the unit cell, giving the empirical formula ZnS. A calcium fluoride unit cell, like that shown in Figure $\PageIndex{17}$, is also an FCC unit cell, but in this case, the cations are located on the lattice points; equivalent calcium ions are located on the lattice points of an FCC lattice. All of the tetrahedral sites in the FCC array of calcium ions are occupied by fluoride ions. There are four calcium ions and eight fluoride ions in a unit cell, giving a calcium:fluorine ratio of 1:2, as required by the chemical formula, CaF 2 . Close examination of Figure $\PageIndex{17}$ will reveal a simple cubic array of fluoride ions with calcium ions in one half of the cubic holes. The structure cannot be described in terms of a space lattice of points on the fluoride ions because the fluoride ions do not all have identical environments. The orientation of the four calcium ions about the fluoride ions differs. Calculation of Ionic Radii If we know the edge length of a unit cell of an ionic compound and the position of the ions in the cell, we can calculate ionic radii for the ions in the compound if we make assumptions about individual ionic shapes and contacts. Example $\PageIndex{5}$: Calculation of Ionic Radii The edge length of the unit cell of LiCl (NaCl-like structure, FCC) is 0.514 nm or 5.14 Å. Assuming that the lithium ion is small enough so that the chloride ions are in contact, calculate the ionic radius for the chloride ion. Note: The length unit angstrom, Å, is often used to represent atomic-scale dimensions and is equivalent to 10 −10 m. Solution On the face of a LiCl unit cell, chloride ions contact each other across the diagonal of the face: Drawing a right triangle on the face of the unit cell, we see that the length of the diagonal is equal to four chloride radii (one radius from each corner chloride and one diameter—which equals two radii—from the chloride ion in the center of the face), so $d = 4r$. From the Pythagorean theorem , we have: \[a^2+a^2=d^2 \nonumber \] which yields: \[\mathrm{(0.514\:nm)^2+(0.514\:nm)^2}=(4r)^2=16r^2 \nonumber \] Solving this gives: \[r=\mathrm{\sqrt{\dfrac{(0.514\:nm)^2+(0.514\:nm)^2}{16}}=0.182\: nm\:(1.82\: Å)\:for\: a\: Cl^−\: radius.} \nonumber \] Exercise $\PageIndex{6}$ The edge length of the unit cell of KCl (NaCl-like structure, FCC) is 6.28 Å. Assuming anion-cation contact along the cell edge, calculate the radius of the potassium ion. The radius of the chloride ion is 1.82 Å. Answer The radius of the potassium ion is 1.33 Å. It is important to realize that values for ionic radii calculated from the edge lengths of unit cells depend on numerous assumptions, such as a perfect spherical shape for ions, which are approximations at best. Hence, such calculated values are themselves approximate and comparisons cannot be pushed too far. Nevertheless, this method has proved useful for calculating ionic radii from experimental measurements such as X-ray crystallographic determinations. X-Ray Crystallography The size of the unit cell and the arrangement of atoms in a crystal may be determined from measurements of the diffraction of X-rays by the crystal, termed X-ray crystallography . Diffraction is the change in the direction of travel experienced by an electromagnetic wave when it encounters a physical barrier whose dimensions are comparable to those of the wavelength of the light. X-rays are electromagnetic radiation with wavelengths about as long as the distance between neighboring atoms in crystals (on the order of a few Å). When a beam of monochromatic X-rays strikes a crystal, its rays are scattered in all directions by the atoms within the crystal. When scattered waves traveling in the same direction encounter one another, they undergo interference , a process by which the waves combine to yield either an increase or a decrease in amplitude (intensity) depending upon the extent to which the combining waves’ maxima are separated (Figure $\PageIndex{18}$). When X-rays of a certain wavelength, λ , are scattered by atoms in adjacent crystal planes separated by a distance, d , they may undergo constructive interference when the difference between the distances traveled by the two waves prior to their combination is an integer factor, n , of the wavelength. This condition is satisfied when the angle of the diffracted beam, θ , is related to the wavelength and interatomic distance by the equation: \[nλ=2d\sin \theta \label{Eq1} \] This relation is known as the Bragg equation in honor of W. H. Bragg , the English physicist who first explained this phenomenon. Figure $\PageIndex{18}$ illustrates two examples of diffracted waves from the same two crystal planes. The figure on the left depicts waves diffracted at the Bragg angle, resulting in constructive interference, while that on the right shows diffraction and a different angle that does not satisfy the Bragg condition, resulting in destructive interference. An X-ray diffractometer, such as the one illustrated in Figure $\PageIndex{20}$, may be used to measure the angles at which X-rays are diffracted when interacting with a crystal as described earlier. From such measurements, the Bragg equation may be used to compute distances between atoms as demonstrated in the following example exercise. Example $\PageIndex{6}$: Using the Bragg Equation In a diffractometer, X-rays with a wavelength of 0.1315 nm were used to produce a diffraction pattern for copper. The first order diffraction ( n = 1) occurred at an angle θ = 25.25°. Determine the spacing between the diffracting planes in copper. Solution The distance between the planes is found by solving the Bragg equation (Equation $\ref{Eq1}$) for d . This gives \[d=\dfrac{nλ}{2\sinθ}=\mathrm{\dfrac{1(0.1315\:nm)}{2\sin(25.25°)}=0.154\: nm}\nonumber \] Exercise $\PageIndex{6}$ A crystal with spacing between planes equal to 0.394 nm diffracts X-rays with a wavelength of 0.147 nm. What is the angle for the first order diffraction? Answer 21.9° X-ray Crystallographer Rosalind Franklin The discovery of the structure of DNA in 1953 by Francis Crick and James Watson is one of the great achievements in the history of science. They were awarded the 1962 Nobel Prize in Physiology or Medicine, along with Maurice Wilkins , who provided experimental proof of DNA’s structure. British chemist Rosalind Franklin made invaluable contributions to this monumental achievement through her work in measuring X-ray diffraction images of DNA. Early in her career, Franklin’s research on the structure of coals proved helpful to the British war effort. After shifting her focus to biological systems in the early 1950s, Franklin and doctoral student Raymond Gosling discovered that DNA consists of two forms: a long, thin fiber formed when wet (type “B”) and a short, wide fiber formed when dried (type “A”). Her X-ray diffraction images of DNA provided the crucial information that allowed Watson and Crick to confirm that DNA forms a double helix, and to determine details of its size and structure. Franklin also conducted pioneering research on viruses and the RNA that contains their genetic information, uncovering new information that radically changed the body of knowledge in the field. After developing ovarian cancer, Franklin continued to work until her death in 1958 at age 37. Among many posthumous recognitions of her work, the Chicago Medical School of Finch University of Health Sciences changed its name to the Rosalind Franklin University of Medicine and Science in 2004, and adopted an image of her famous X-ray diffraction image of DNA as its official university logo. Key Concepts and Summary The structures of crystalline metals and simple ionic compounds can be described in terms of packing of spheres. Metal atoms can pack in hexagonal closest-packed structures, cubic closest-packed structures, body-centered structures, and simple cubic structures. The anions in simple ionic structures commonly adopt one of these structures, and the cations occupy the spaces remaining between the anions. Small cations usually occupy tetrahedral holes in a closest-packed array of anions. Larger cations usually occupy octahedral holes. Still larger cations can occupy cubic holes in a simple cubic array of anions. The structure of a solid can be described by indicating the size and shape of a unit cell and the contents of the cell. The type of structure and dimensions of the unit cell can be determined by X-ray diffraction measurements. Glossary body-centered cubic (BCC) solid crystalline structure that has a cubic unit cell with lattice points at the corners and in the center of the cell body-centered cubic unit cell simplest repeating unit of a body-centered cubic crystal; it is a cube containing lattice points at each corner and in the center of the cube Bragg equation equation that relates the angles at which X-rays are diffracted by the atoms within a crystal coordination number number of atoms closest to any given atom in a crystal or to the central metal atom in a complex cubic closest packing (CCP) crystalline structure in which planes of closely packed atoms or ions are stacked as a series of three alternating layers of different relative orientations (ABC) diffraction redirection of electromagnetic radiation that occurs when it encounters a physical barrier of appropriate dimensions face-centered cubic (FCC) solid crystalline structure consisting of a cubic unit cell with lattice points on the corners and in the center of each face face-centered cubic unit cell simplest repeating unit of a face-centered cubic crystal; it is a cube containing lattice points at each corner and in the center of each face hexagonal closest packing (HCP) crystalline structure in which close packed layers of atoms or ions are stacked as a series of two alternating layers of different relative orientations (AB) hole (also, interstice) space between atoms within a crystal isomorphous possessing the same crystalline structure octahedral hole open space in a crystal at the center of six particles located at the corners of an octahedron simple cubic unit cell (also, primitive cubic unit cell) unit cell in the simple cubic structure simple cubic structure crystalline structure with a cubic unit cell with lattice points only at the corners space lattice all points within a crystal that have identical environments tetrahedral hole tetrahedral space formed by four atoms or ions in a crystal unit cell smallest portion of a space lattice that is repeated in three dimensions to form the entire lattice X-ray crystallography experimental technique for determining distances between atoms in a crystal by measuring the angles at which X-rays are diffracted when passing through the crystal Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected] ).
Courses/Chabot_College/Chem_12A%3A_Organic_Chemistry_Fall_2022/04%3A_Acids_and_Bases/4.02%3A_Brnsted-Lowry_Acids_and_Bases_(Review)	learning objective recognize acids and bases The Brønsted-Lowry Theory of Acids and Bases In 1923, Danish chemist Johannes Brønsted and English chemist Thomas Lowry independently proposed new definitions for acids and bases, ones that focus on proton transfer. A Brønsted-Lowry acid is any species that can donate a proton (H + ) to another molecule. A Brønsted-Lowry base is any species that can accept a proton from another molecule. In short, a Brønsted-Lowry acid is a proton donor (PD) , while a Brønsted-Lowry base is a proton acceptor (PA) . A Brønsted-Lowry acid is a proton donor, while a Brønsted-Lowry base is a proton acceptor. Let us use the reaction of ammonia in water to demonstrate the Brønsted-Lowry definitions of an acid and a base. Ammonia and water molecules are reactants, while the ammonium ion and the hydroxide ion are products: \[\ce{NH3(aq) + H2O (ℓ) <=> NH^{+}4(aq) + OH^{−}(aq) }\label{Eq1}\] What has happened in this reaction is that the original water molecule has donated a hydrogen ion to the original ammonia molecule, which in turn has accepted the hydrogen ion. We can illustrate this as follows: Because the water molecule donates a hydrogen ion to the ammonia, it is the Brønsted-Lowry acid, while the ammonia molecule—which accepts the hydrogen ion—is the Brønsted-Lowry base. Thus, ammonia acts as a base in both the Arrhenius sense and the Brønsted-Lowry sense. Is an Arrhenius acid like hydrochloric acid still an acid in the Brønsted-Lowry sense? Yes, but it requires us to understand what really happens when HCl is dissolved in water. Recall that the hydrogen atom is a single proton surrounded by a single electron. To make the hydrogen ion , we remove the electron, leaving a bare proton. Do we really have bare protons floating around in aqueous solution? No, we do not. What really happens is that the H + ion attaches itself to H 2 O to make H 3 O + , which is called the hydronium ion . For most purposes, H + and H 3 O + represent the same species, but writing H 3 O + instead of H + shows that we understand that there are no bare protons floating around in solution. Rather, these protons are actually attached to solvent molecules. The Hydronium IOn A proton in aqueous solution may be surrounded by more than one water molecule, leading to formulas like $\ce{H5O2^{+}}$ or $\ce{H9O4^{+}}$ rather than $\ce{H3O^{+}}$. It is simpler, however, to use $\ce{H3O^{+}}$ to represent the hydronium ion. With this in mind, how do we define HCl as an acid in the Brønsted-Lowry sense? Consider what happens when HCl is dissolved in H 2 O: \[\ce{HCl(g) + H_2O (ℓ) \rightarrow H_3O^{+}(aq) + Cl^{−}(aq) }\label{Eq2}\] We can depict this process using Lewis electron dot diagrams: Now we see that a hydrogen ion is transferred from the HCl molecule to the H 2 O molecule to make chloride ions and hydronium ions. As the hydrogen ion donor, HCl acts as a Brønsted-Lowry acid; as a hydrogen ion acceptor, H 2 O is a Brønsted-Lowry base. So HCl is an acid not just in the Arrhenius sense but also in the Brønsted-Lowry sense. Moreover, by the Brønsted-Lowry definitions, H 2 O is a base in the formation of aqueous HCl. So the Brønsted-Lowry definitions of an acid and a base classify the dissolving of HCl in water as a reaction between an acid and a base—although the Arrhenius definition would not have labeled H 2 O a base in this circumstance. A Brønsted-Lowry acid is a proton (hydrogen ion) donor. A Brønsted-Lowry base is a proton (hydrogen ion) acceptor. All Arrhenius acids and bases are Brønsted-Lowry acids and bases as well. However, not all Brønsted-Lowry acids and bases are Arrhenius acids and bases. Example $\PageIndex{1}$ Aniline (C 6 H 5 NH 2 ) is slightly soluble in water. It has a nitrogen atom that can accept a hydrogen ion from a water molecule just like the nitrogen atom in ammonia does. Write the chemical equation for this reaction and identify the Brønsted-Lowry acid and base. Solution C 6 H 5 NH 2 and H 2 O are the reactants. When C 6 H 5 NH 2 accepts a proton from H 2 O, it gains an extra H and a positive charge and leaves an OH − ion behind. The reaction is as follows: \[\ce{C6H5NH2(aq) + H2O(ℓ) <=> C6H5NH3^{+}(aq) + OH^{−}(aq)} \nonumber\] Because C 6 H 5 NH 2 accepts a proton, it is the Brønsted-Lowry base. The H 2 O molecule, because it donates a proton, is the Brønsted-Lowry acid. Exercise $\PageIndex{1}$ Identify the Brønsted-Lowry acid and the Brønsted-Lowry base in this chemical equation. \[\ce{H2PO4^{-} + H_2O <=> HPO4^{2-} + H3O^{+}}\] Answer: Brønsted-Lowry acid: H 2 PO 4 - ; Brønsted-Lowry base: H 2 O Exercise $\PageIndex{2}$ Which of the following compounds is a Bronsted-Lowry base? HCl HPO 4 2 - H 3 PO 4 NH 4 + CH 3 NH 3 + Answer: A Brønsted-Lowry Base is a proton acceptor, which means it will take in an H + . This eliminates $\ce{HCl}$, $\ce{H3PO4}$ , $\ce{NH4^{+}}$ and $\ce{CH_3NH_3^{+}}$ because they are Bronsted-Lowry acids. They all give away protons. In the case of $\ce{HPO4^{2-}}$, consider the following equation: \[\ce{HPO4^{2-} (aq) + H2O (l) \rightarrow PO4^{3-} (aq) + H3O^{+}(aq) } \nonumber\] Here, it is clear that HPO 4 2 - is the acid since it donates a proton to water to make H 3 O + and PO 4 3 - . Now consider the following equation: \[ \ce{ HPO4^{2-}(aq) + H2O(l) \rightarrow H2PO4^{-} + OH^{-}(aq)} \nonumber\] In this case, HPO 4 2 - is the base since it accepts a proton from water to form H 2 PO 4 - and OH - . Thus, HPO 4 2 - is an acid and base together, making it amphoteric. Since HPO 4 2 - is the only compound from the options that can act as a base, the answer is (b) HPO 4 2- . Conjugate Acid-Base Pair In reality, all acid-base reactions involve the transfer of protons between acids and bases. For example, consider the acid-base reaction that takes place when ammonia is dissolved in water. A water molecule (functioning as an acid) transfers a proton to an ammonia molecule (functioning as a base), yielding the conjugate base of water, $\ce{OH^-}$, and the conjugate acid of ammonia, $\ce{NH4+}$: In the reaction of ammonia with water to give ammonium ions and hydroxide ions, ammonia acts as a base by accepting a proton from a water molecule, which in this case means that water is acting as an acid. In the reverse reaction, an ammonium ion acts as an acid by donating a proton to a hydroxide ion, and the hydroxide ion acts as a base. The conjugate acid–base pairs for this reaction are $NH_4^+/NH_3$ and $H_2O/OH^−$. The strongest acids are at the bottom left, and the strongest bases are at the top right. The conjugate base of a strong acid is a very weak base, and, conversely, the conjugate acid of a strong base is a very weak acid. Example $\PageIndex{2}$ Identify the conjugate acid-base pairs in this equilibrium. \[\ce{CH3CO2H + H2O <=> H3O^{+} + CH3CO2^{-}} \nonumber\] Solution Similarly, in the reaction of acetic acid with water, acetic acid donates a proton to water, which acts as the base. In the reverse reaction, $H_3O^+$ is the acid that donates a proton to the acetate ion, which acts as the base. Once again, we have two conjugate acid–base pairs: the parent acid and its conjugate base ($CH_3CO_2H/CH_3CO_2^−$) and the parent base and its conjugate acid ($H_3O^+/H_2O$). Example $\PageIndex{3}$ Identify the conjugate acid-base pairs in this equilibrium. \[(CH_{3})_{3}N + H_{2}O\rightleftharpoons (CH_{3})_{3}NH^{+} + OH^{-} \nonumber\] Solution One pair is H 2 O and OH − , where H 2 O has one more H + and is the conjugate acid, while OH − has one less H + and is the conjugate base. The other pair consists of (CH 3 ) 3 N and (CH 3 ) 3 NH + , where (CH 3 ) 3 NH + is the conjugate acid (it has an additional proton) and (CH 3 ) 3 N is the conjugate base. Exercise $\PageIndex{3}$ Identify the conjugate acid-base pairs in this equilibrium. \[\ce{NH2^{-} + H2O\rightleftharpoons NH3 + OH^{-}} \nonumber\] Answer: H 2 O (acid) and OH − (base); NH 2 − (base) and NH 3 (acid)
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Courses/Fullerton_College/Introductory_Chemistry_for_Allied_Health_(Chan)/14%3A_Lipids/14.E%3A_Exercises	Additional Exercises The melting point of elaidic acid is 52°C. What trend is observed when comparing the melting points of elaidic acid, oleic acid, and stearic acid? Explain. Would you expect the melting point of palmitelaidic acid to be lower or higher than that of elaidic acid? Explain. Examine the labels on two brands of margarine and two brands of shortening and list the oils used in the various brands. Draw a typical lecithin molecule that incorporates glycerol, palmitic acid, oleic acid, phosphoric acid, and choline. Circle all the ester bonds. In cerebrosides, is the linkage between the fatty acid and sphingosine an amide bond or an ester bond? Justify your answer. Serine is an amino acid that has the following structure. Draw the structure for a phosphatidylserine that contains a palmitic acid and a palmitoleic acid unit. Explain whether each compound would be expected to diffuse through the lipid bilayer of a cell membrane. potassium chloride CH 3 CH 2 CH 2 CH 2 CH 2 CH 3 fructose Identify the role of each steroid hormone in the body. progesterone aldosterone testosterone cortisol How does the structure of cholic acid differ from that of cholesterol? Which compound would you expect to be more polar? Why? What fatty acid is the precursor for the prostaglandins? Identify three biological effects of prostaglandins. Why is it important to determine the ratio of LDLs to HDLs, rather than just the concentration of serum cholesterol? Answers Stearic acid has the highest melting point, followed by elaidic acid, and then oleic acid with the lowest melting point. Elaidic acid is a trans fatty acid, and the carbon chains can pack together almost as tightly as those of the saturated stearic acid. Oleic acid is a cis fatty acid, and the bend in the hydrocarbon chain keeps these carbon chains from packing as closely together; fewer interactions lead to a much lower melting point. The melting point of palmitelaidic acid should be lower than that of elaidic acid because it has a shorter carbon chain (16, as compared to 18 for elaidic acid). The shorter the carbon chain, the lower the melting point due to a decrease in intermolecular interactions. regulates the menstrual cycle and maintains pregnancy regulates salt metabolism by stimulating the kidneys to retain sodium and excrete potassium stimulates and maintains male sex characteristics stimulates the conversion of proteins to carbohydrates arachidonic acid induce smooth muscle contraction, lower blood pressure, and contribute to the inflammatory response
Bookshelves/Environmental_Chemistry/Environmental_Toxicology_(van_Gestel_et_al.)/06%3A_Risk_Assessment_and_Regulation/6.05%3A_Regulatory_Frameworks	6.5. Regulatory Frameworks Regulatory frameworks Authors : Charles Bodar and Joop de Knecht Reviewers : Kees van Gestel Learning objectives: You should be able to explain how the potential environmental risks of chemicals are legally being controlled in the EU and beyond mention the different regulatory bodies involved in the regulation of different categories of chemicals explain the purpose of the Classification, Labelling and Packaging (CLP) approach and its difference with the risk assessment of chemicals Keywords: chemicals, environmental regulations, hazard, risk Introduction There is no single, overarching global regulatory framework to manage the risks of all chemicals. Instead, different regulations or directives have been developed for different categories of chemicals. These categories are typically related to the usage of the chemicals. Important categories are industrial chemicals (solvents, plasticizers, etc.), plant protection products, biocides and human and veterinary drugs. Some chemicals may belong to more than one category. Zinc, for example, is used in the building industry, but it also has biocidal applications (antifouling agent) and zinc oxide is used as a veterinary drug. In the European Union, each chemical category is subject to specific regulations or directives providing the legal conditions and requirements to guarantee a safe production and use of chemicals. A key element of all legal frameworks is the requirement that sufficient data on a chemical should be made available. Valid data on production and identity (e.g. chemical structure), use volumes, emissions, environmental fate properties and the (eco)toxicity of a chemical are the essential building blocks for a sound assessment and management of environmental risks. Rules for the minimum data set that should be provided by the actors involved (e.g. producers or importers) are laid down in various regulatory frameworks. With this data, both hazard and risk assessments can be carried out according to specified technical guidelines. The outcome of the assessment is then used for risk management, which is focused on minimizing any risk by taking measures, ranging from requests for additional data to restrictions on the particular use or a full-scale ban of a chemical. REACH REACH is a regulation of the European Union, adopted to improve the protection of human health and the environment from the risks that can be posed by chemicals, while enhancing the competitiveness of the EU chemicals industry. REACH stands for Registration, Evaluation, Authorisation and Restriction of Chemicals. The REACH regulation entered into force on 1st June 2007 to streamline and improve the former legislative frameworks on new and existing chemical substances. It replaced approximately forty community regulations and directives by one single regulation. REACH establishes procedures for collecting and assessing information on the properties, hazards and risks of substances. REACH applies to a very broad spectrum of chemicals: from industrial to household applications, and very much more. It requires that EU manufacturers and importers register their chemical substances if produced or imported in annual amounts of > 1 tonne, unless the substance is exempted from registration under REACH. At quantities of > 10 tonnes, the manufacturers, importers, and down-stream users are responsible to show that their substances do not adversely affect human health or the environment. The amount of standard information required to show safe use depends on the quantity of the substance that is manufactured or imported. Before testing on vertebrate animals like fish and mammals, the use of alternative methods must be considered. The European Chemical Agency ( ECHA ) coordinates and facilitates the REACH program. For production volumes above 10 tonnes per year, industry has to prepare a risk assessment, taking into account all risk management measures envisaged, and document this in a chemical safety assessment (CSA). A CSA should include an exposure assessment, hazard or dose-response assessment, and a risk characterization showing risks ratios below 1.0, i.e. safe use (see sections on REACH Human and REACH Eco ). Classification, Labelling and Packaging (CLP) The EU CLP regulation requires manufacturers, importers or downstream users of substances or mixtures to classify, label and package their hazardous chemicals appropriately before placing them on the market. When relevant information (e.g. ecotoxicity data) on a substance or mixture meets the classification criteria in the CLP regulation, the hazards of a substance or mixture are identified by assigning a certain hazard class and category. An important CLP hazard class is ' Hazardous to the aquatic environment' , which is divided into categories based on the criteria, for example, Category Acute 1, representing the most (acute) toxic chemicals (LC50/EC50 ≤ 1 mg/L). CLP also sets detailed criteria for the labelling elements, such as the well-known pictograms (Figure 1). Plant protection products regulation Plant protection products (PPPs) are pesti cides that are mainly used to keep crops healthy and prevent them from being damaged by disease and infestation. They include among others herbicides, fungicides, insecticides, acaricides, plant growth regulators and repellents (see section on Crop Protection Products ). PPPs fall under the EU Regulation (EC) No 1107/2009 which determines that PPPs cannot be placed on the market or used without prior authorization. The European Food and Safety Authority (EFSA) coordinates the EU regulation on PPPs. Biocides regulation The distinction between biocides and PPP is not always straightforward, but as a general rule of thumb the PPP regulation applies to substances used by farmers for crop protection while the biocides regulation covers all other pesticide applications. Different applications of the same active ingredient, one as a PPP and the other as a biocide, may thus fall under different regulations. Biocides are used to protect humans, animals, materials or articles against harmful organisms like pests or bacteria, by the action of the active substances contained in the biocidal product. Examples of biocides are antifouling agents, preservatives and disinfectants. According to the EU Biocidal Products Regulation ( BPR ), all biocidal products require an authorization before they can be placed on the market, and the active substances contained in the biocidal product must be previously approved. The European Chemical Agency (ECHA) coordinates and facilitates the BPR. More or less similar to other legislations, for biocides the environmental risk assessment is mainly performed by comparing compartmental concentrations (PEC) with the concentration below which unacceptable effects on organisms will most likely not occur (PNEC). Veterinary and human pharmaceuticals regulation Since 2006, EU law requires an environmental risk assessment (ERA) for all new applications for a marketing authorization of human and veterinary pharmaceuticals. For both products, guidance documents have been developed for conducting an ERA based on two phases. The first phase estimates the exposure of the environment to the drug substance. Based on an action limit the assessment may be terminated. In the second phase, information about the fate and effects in the environment is obtained and assessed. For conducting an ERA a base set, including ecotoxicity data, is required. For veterinary medicines, the ERA is part of a risk-benefit analysis, in which the positive therapeutic effects are weighed against any environment risks, whereas for human medicines the environmental concerns are excluded from the risk-benefit analysis. The European Medicines Agency (EMA) is responsible for the scientific evaluation, supervision and safety monitoring of medicines in the EU. Harmonization of testing Testing chemicals is an important aspect of risk assessment, e.g. testing for toxicity, for degradation or for a physicochemical property like the K ow (see Chapter 3 ). The outcome of a test may vary depending on the conditions, e.g. temperature, test medium or light conditions. For this reason there is an incentive to standardize the test conditions and to harmonize the testing procedures between agencies and countries. This would also avoid duplication of testing, and leading to a more efficient and effective testing system. The Organization for Economic Co-operation and Development (OECD) assists its member governments in developing and implementing high-quality chemical management policies and instruments. One of the key activities to achieve this goal is the development of harmonized guidelines to test and assess the risks of chemicals leading to a system of mutual acceptance of chemical safety data among OECD countries. The OECD also developed Principles of Good Laboratory Practice (GLP) to ensure that studies are of sufficient quality and rigor and are verifiable. The OECD also facilitates the development of new tools to obtain more safety information and maintain quality while reducing costs, time and animal testing, such as the OECD QSAR toolbox . 6.5. Question 1 What are the major categories of chemicals for which regulatory frameworks are present controlling the environmental risks of these chemicals? 6.5. Question 2 Is a chemical producer in the EU allowed to put a new chemical on the market without a registration or authorization? 6.5. Question 3 Is the CLP regulation based on the hazard, exposure and/or risk of chemicals? 6.5. Question 4 The amount of minimum information that is required under REACH for making a proper hazard and risk assessment is dependent on what? And what do you think is the rationale behind it? 6.5. Question 5 Name three European agencies or authorities that are coordinating important regulatory frameworks in the EU? 6.5.1. REACH human Authors : Theo Vermeire Reviewers : Tim Bowmer Learning objective: You should be able to: outline how human risk assessment of chemicals is performed under REACH; explain the regulatory function of human risk assessment in REACH. Keywords : REACH, chemical safety assessment, human, RCR, DNEL, DMEL Human risk assessment under REACH The REACH Regulation aims to ensure a high level of protection of human health and the environment, including the promotion of alternative methods for assessment of hazards of substances, as well as the free circulation of substances on the internal market while enhancing competitiveness and innovation. Risk assessment under REACH aims to realize such a level of protection for humans that the likelihood of adverse effects occurring is low, taking into account the nature of the potentially exposed population (including sensitive groups) and the severity of the effect(s). Industry therefore has to prepare a risk assessment (in REACH terminology: chemical safety assessment, CSA) for all relevant stages in the life cycle of the chemical, taking into account all risk management measures envisaged, and document this in the chemical safety report (CSR). Risk characterization in the context of a CSA is the estimation of the likelihood that adverse effect levels occur due to actual or predicted exposure to a chemical. The human populations considered, or protection goals, are workers, consumers and humans exposed via the environment. In risk characterization, exposure levels are compared to reference levels to yield "risk characterization ratios" (RCRs) for each protection goal. RCRs are derived for all endpoints (e.g. skin and eye irritation, sensitization, repeated dose toxicity) and time scales. It should be noted that these RCRs have to be derived for all stages in the life-cycle of a compound. Environmental exposure assessment for humans Humans can be exposed through the environment directly via inhalation of indoor and ambient air, soil ingestion and dermal contact, and indirectly via food products and drinking water (Figure 1). REACH does not consider direct exposure via soil. In the REACH exposure scenario, assessment of human exposure through the environment can be divided into three steps: Determination of the concentrations in intake media (air, soil, food, drinking water); Determination of the total daily intake of these media; Combining concentrations in the media with total daily intake (and, if necessary, using a factor for bioavailability through the route of uptake concerned). A fourth step may be the consideration of aggregated exposure taking into account exposure to the same substance in consumer products and at the workplace. Moreover, there may be similar substances, acting via the same mechanism of action, that may have to be considered in the exposure assessment, for instance, as a worst case, by applying the concept of dose or concentration addition. The section on Environmental realistic scenarios (PECs) - Human explains the concept of exposure scenarios and how concentrations in environmental compartments are derived. Hazard identification and dose-response assessment The aim of hazard identification is to classify chemicals and to select key data for the dose-response assessment to derive a safe reference level, which in REACH terminology is called the DNEL (Derived No Effect Level) or DMEL (Derived Minimal Effect Level). For human end-points, a distinction is made between substances considered to have a threshold for toxicity and those without a threshold. For threshold substances, a No-Observed-Adverse Effect Level (NOAEL) or Lowest-Observed-Adverse-Effect Level (LOAEL) is derived, typically from toxicity studies with laboratory animals such as rats and mice. Alternatively a Benchmark Dose (BMD) can be derived by fitting a dose-response model to all observations. These toxicity values are then extrapolated to a DNEL using assessment factors to correct for uncertainty and variability. The most frequently used assessment factors are those for interspecies differences and those for intraspecies variability (see section on Setting safe standards ). Additionally, factors can be applied to account for remaining uncertainties such as those due to a poor database. For substances considered to exert their effect by a non-threshold mode of action, especially mutagenicity and carcinogenicity, it is generally assumed, as a default assumption, that even at very low levels of exposure residual risks cannot be excluded. That said, recent progress has been made on establishing scientific, 'health-based' thresholds for some genotoxic carcinogens. For non-threshold genotoxic carcinogens it is recommended to derive a DMEL, if the available data allow. A DMEL is a cancer risk value considered to be of very low concern, e.g. a 1 in a million tumour risk after lifetime exposure to the chemical and using a conservative linear dose-response model. There is as yet no EU-wide consensus on acceptable levels of cancer risk. Risk characterization Safe use of substances is demonstrated when: • RCRs are below one, both at local and regional level. For threshold substances, the RCR is the ratio of the estimated exposure (concentration or dose) and the DNEL; for non-threshold substances the DMEL is used. • The likelihood and severity of an event such as an explosion occurring due to the physicochemical properties of the substance as determined in the hazard assessment is negligible. A risk characterization needs to be carried out for each exposure scenario (see Section on Environmental realistic scenarios (PECs) - Human ) and human population. The assessment consists of a comparison of the exposure of each human population known to be or likely to be exposed with the appropriate DNELs or DMELs and an assessment of the likelihood and severity of an event occurring due to the physicochemical properties of the substance. 0 1 2 Example of a deterministic assessment (Vermeire et al., 2001) Exposure assessment Based on an emission estimation for processing of dibutylphthalate (DBP) as a softener in plastics, the concentrations in environmental compartments were estimated. Based on modelling as schematically presented in Figure 1, the total human dose was determined to be 93 ug.kg bw-1. PEC-air 2.4 µg.m-3 PEC-surface water 2.8 µg.l-1 PEC- grassland soil 0.15 mg.kg-1 PEC-porewater agric. soil 3.2 µg.l-1 PEC-porewater grassl. soil 1.4 µg.l-1 PEC-groundwater 3.2 µg.l-1 Total Human Dose 93 µg.kgbw-1.d-1 Effects assessment The total dose should be compared to a DNEL for humans. DBP is not considered a genotoxic carcinogen but is toxic to reproduction and therefore the risk assessment is based on endpoints assumed to have a threshold for toxicity. The lowest NOAEL of DBP was observed in a two-generation reproduction test in rats and at the lowest dose-level in the diet (52 mg.kgbw-1.d-1 for males and 80 mg.kgbw-1.d-1 for females) a reduced number of live pups per litter and decreased pup weights were seen in the absence of maternal toxicity. The lowest dose level of 52 mg.kgbw-1.d-1 was chosen as the NOAEL. The DNEL was derived by the application of an overall assessment factor of 1000, accounting for interspecies differences, human variability and uncertainties due to a non-chronic exposure period. Risk characterisation The deterministic estimate of the RCR would be based on the deterministic exposure estimate of 0.093 mg.kgbw-1.d-1 and the deterministic DNEL of 0.052 mg.kgbw-1.d-1. The deterministic RCR would then be 1.8, based on the NOAEL. Since this is higher than one, this assessment indicates a concern, requiring a refinement of the assessment or risk management measures. NaN NaN PEC-air 2.40 µg.m-3 PEC-surface water 2.80 µg.l-1 PEC- grassland soil 0.15 mg.kg-1 PEC-porewater agric. soil 3.20 µg.l-1 PEC-porewater grassl. soil 1.40 µg.l-1 PEC-groundwater 3.20 µg.l-1 Total Human Dose 93.00 µg.kgbw-1.d-1 Additional reading Van Leeuwen C.J., Vermeire T.G. (Eds.) (2007) Risk assessment of chemicals: an introduction. Springer, Dordrecht, The Netherlands, ISBN 978-1-4020-6102-8 (e-book), https://doi.org/10.1007/978-1-4020-6102-8 . Vermeire, T., Jager, T., Janssen, G., Bos, P., Pieters, M. (2001) A probabilistic human health risk assessment for environmental exposure to dibutylphthalate. Journal of Human and Ecological Risk Assessment 7, 1663-1679. 6.5.1. Question 1 Uncertainty happens! It is inherent to risk assessment. Where, in your view, are the greatest sources of uncertainty in the process of risk assessment? 6.5.1. Question 2 Are there risks identified in the example for humans indirectly exposed via the environment? To what extent are these potential or realistic risks by asking yourself the following questions: • Do you have relevant toxicological and exposure data ? • Are these fixed values or not? • How relevant or adverse are the toxicological effects observed? • Were appropriate assessment factors used? What would you recommend as a strategy to reduce the identified risks sufficiently? 6.5.2. REACH environment Author : Joop de Knecht Reviewers : Watze de Wolf Keywords : REACH, European chemicals regulation Introduction REACH establishes procedures for collecting and assessing information on the properties, hazards and risks of substances. At quantities of > 10 tonnes, the manufacturers, importers, and down-stream users must show that their substances do not adversely affect human health or the environment for the uses and operational conditions registered. The amount of standard information required to show safe use depends on the quantity of the substance that is manufactured or imported. This section explains how risks to the environment are assessed in REACH. Data requirements As a minimum requirement, all substances manufactured or imported in quantities of 1 tonne or more need to be tested in acute toxicity tests on Daphnia and algae, while also information should be provided on biodegradability (Table 1). Physical-chemical properties relevant for environmental fate assessment that have to be provided at this tonnage level are water solubility, vapour pressure and octanol-water partition coefficient. At 10 tonnes or more, this should be supplemented with an acute toxicity test on fish and an activated sludge respiration inhibition test. At this tonnage level, also an adsorption/desorption screening and a hydrolysis test should be performed. If the chemical safety assessment, performed at 100 tonnes or more in case a substance is classified based on hazard information , indicates the need to investigate further the effects on aquatic organisms, the chronic toxicity on these aquatic species should be determined. If the substance has a high potential for bioaccumulation (for instance a log K ow > 3), also the bioaccumulation in aquatic species should be determined. The registrant should also determine the acute toxicity to terrestrial species or, in the absence of these data, consider the equilibrium partitioning method (EPM) to assess the hazard to soil organisms. To further investigate the fate of the substance in surface water, sediment and soil, simulation tests on its degradation should be conducted and when needed further information on the adsorption/desorption should be provided. At 1000 tonnes or more, chronic tests on terrestrial and sediment-living species should be conducted if further refinement of the safety assessment is needed. Before testing vertebrate animals like fish and mammals, the use of alternative methods and all other options must be considered to comply with the regulations regarding (the reduction of) animal testing. Table 1 Required ecotoxicological and environmental fate information as defined in REACH 0 1 1-10 t/y Acute Aquatic toxicity (invertebrates, algae) Ready biodegradability 10-100 t/y Acute Aquatic toxicity (fish) Activated sludge respiration, inhibition Hydrolysis as a function of pH Adsorption/ desorption screening test 100-1000 t/y Chronic Aquatic toxicity (invertebrates, fish) Bioaccumulation Surface water, soil and sediment simulation (degradation) test Acute terrestrial toxicity Further information on adsorption/desorption ≥ 1000 t/y Further fate and behaviour in the environment of the substance and/or degradation products Chronic terrestrial toxicity Sediment toxicity Avian toxicity Safety assessment For substances that are classified based on hazard information the registrant should assess the environmental safety of a substance, by comparing the predicted environmental concentration (PEC) with the Predicted No Effect Concentration (PNEC), resulting in a Risk Characterisation Ratio (RCR=PEC/PNEC). The use of the substance is considered to be safe when the RCR <1. Chapter 16 of the ECHA guidance offers methods to estimate the PEC based on the tonnage, use and operational conditions, standardised through a set of use descriptors, particularly the Environmental Release Categories (ERCs). These ERCs are linked to conservative default release factors to be used as a starting point for a first tier environmental exposure assessment. When substances are emitted via waste water, the physical-chemical and fate properties of the chemical substance are then used to predict its behaviour in the Wastewater Treatment Plant (WWTP). Subsequently, the release of treated wastewater is used to estimate the concentration in fresh and marine surface water. The concentration in sediment is estimated from the PEC in water and experimental or estimated sediment-water partitioning coefficient (Kp sed ). Soil concentrations are estimated from deposition from air and the application of sludge from an WWTP. The guidance offers default values for all relevant parameters, thus a generic local PEC can be calculated and considered applicable to all local emissions in Europe, although the default values can be adapted to specific conditions if justified. The local risk for wide-dispersive uses (e.g. from consumers or small, non- industrial companies) is estimated for a default WWTP serving 10,000 inhabitants. In addition, a regional assessment is conducted for a standard area, a region represented by a typical densely populated EU area located in Western Europe (i.e. about 20 million inhabitants, distributed in a 200 x 200 km 2 area). For calculating the regional PECs, a multi-media fate-modelling approach is used (e.g. the SimpleBox model; see Section on Multicompartment fate modelling ). All releases to each environmental compartment for each use, assumed to constitute a constant and continuous flux, are summed and averaged over the year and steady-state concentrations in the environmental compartments are calculated. The regional concentrations are used as background concentrations in the calculation of the local concentrations. The PNEC is calculated using the lowest toxicity value and an assessment factor (AF) related to the amount of information (see section Setting safe standards or chapter 10 of the REACH guidance . If only the minimum set of aquatic acute toxicity data is available, i.e. LC50s or EC50s for algae, daphnia and fish, a default value of 1000 is used. When one, two or three or more long-term tests are available, a default AF of 100, 50 and 10 is applied to No Observed Effect Concentrations (NOECs), respectively. The idea behind lowering the AF when more data become available is that the amount of uncertainty around the PNEC is being reduced. In the absence of ecotoxicological data for soil and/or sediment-dwelling organisms, the PNEC soil and/or PNEC sed may be provisionally calculated using the EPM. This method uses the PNEC water for aquatic organisms and the suspended matter/water partitioning coefficient as inputs. For substances with a log K ow >5 (or with a corresponding log Kp value), the PEC/PNEC ratio resulting from the EPM is increased by a factor of 10 to take into account possible uptake through the ingestion of sediment. If the PEC/PNEC is greater than 1 a sediment test must be conducted. If one, two or three long-term No Observed Effect Concentrations (NOECs) from sediment invertebrate species representing different living and feeding conditions are available, the PNEC can be derived using default AFs of 100, 50 and 10, respectively. For data rich chemicals, the PNEC can be derived using Species Sensitivity Distributions (SSD) or other higher-tier approaches. 6.5.2. Question 1 Who is responsible within the EU that industrial chemicals do not pose a risk for the environment? 6.5.2. Question 2 In which circumstances will an environmental risk be identified? 6.5.2. Question 3 Describe how the ecotoxicogical safety levels (PNECs) for the aquatic environment are derived depending on the ecotoxicological information available? 6.5.3. Pesticides (EFSA) under review 6.5.4. Environmental Risk Assessment of Pharmaceuticals in Europe Author : Gerd Maack Reviewers : Ad Ragas, Julia Fabrega, Rhys Whomsley Learning objectives: You should be able to Explain the philosophy and objective of the environmental risk assessment of pharmaceuticals. mention the key aspects of the tiered approach of the assessment identify the exposure routes for human and veterinary medicinal products and should know the respective consequences in the assessment Keywords: Human pharmaceuticals, veterinary pharmaceuticals, environmental impact, tiered approach Introduction Pharmaceuticals are a crucial element of modern medicine and confer significant benefits to society. About 4,000 active pharmaceutical ingredients are being administered worldwide in prescription medicines, over-the-counter medicines, and veterinary medicines. They are designed to be efficacious and stable, as they need to pass different barriers i.e. skin, the gastrointestinal system (GIT), or even the blood-brain barrier before reaching the target cells. Each target system has a different pH and different lipophilicity and the GIT is in addition colonised with specific bacteria, specialized to digest, dissolve and disintegrate organic molecules. As a consequence of this stability, most of the pharmaceutical ingredients are stable in the environment as well and could cause effects on non-target organisms. The active ingredients comprise a variety of synthetic chemicals produced by pharmaceutical companies in both the industrialized and the developing world at a rate of 100,000 tons per year. While pharmaceuticals are stringently regulated in terms of efficacy and safety for patients, as well as for target animal safety, user and consumer safety, the potential effects on non-target organisms and environmental effects are regulated comparably weakly. The authorisation procedure requires an environmental risk assessment (ERA) to be submitted by the applicants for each new human and veterinary medicinal product. The assessment encompasses the fate and behaviour of the active ingredient in the environment and its ecotoxicity based on a catalogue of standardised test guidelines. In the case of veterinary pharmaceuticals, constraints to reduce risk and thus ensure safe usage can be stipulated in most cases. In the case of human pharmaceuticals, it is far more difficult to ensure risk reduction through restriction of the drug's use due to practical and ethical reasons. Because of their unique benefits, a restriction is not reasonable. This is reflected in the legal framework, as a potential effect on the environment is not included in the final benefit risk assessment for a marketing authorisation. Exposure pathways Human pharmaceuticals Human pharmaceuticals enter the environment mainly via surface waters through sewage systems and sewage treatment plants. The main exposure pathways are excretion and non-appropriate disposal. Typically only a fraction of the medicinal product taken is metabolised by the patients, meaning that the main share of the active ingredient is excreted unchanged into the wastewater system. Furthermore, sometimes the metabolites themselves are pharmacologically active. Yet, no wastewater treatment plant is able to degrade all active ingredients. So medicinal products are commonly found in surface water, to some extent in ground water, and sometimes even in drinking water. However, the concentrations in drinking water are orders of magnitude lower than therapeutic concentrations. An additional exposure pathway for human pharmaceuticals is the spreading of sewage sludge on soil, if the sludge is used as fertilizer on farmland. See for more details the Link " The Drugs We Wash Away: Pharmaceuticals, Drinking Water and the Environment ". Veterinary pharmaceuticals Veterinary pharmaceuticals on the other hand enter the environment mainly via soil, either indirectly, if the slurry and manure from mass livestock production is spread onto agricultural land as fertiliser, or directly from pasture animals. Moreover, pasture animals might additionally excrete directly into surface water. Pharmaceuticals can also enter the environment via the detour of manure used in biogas plants. Assessment schemes Despite the differences mentioned above, the general scheme of the environmental risk assessment of human and veterinary pharmaceuticals is similar. Both assessments start with an exposure assessment. Only if specific trigger values are reached an in-depth assessment of fate, behaviour and effects of the active ingredient is necessary. Environmental risk assessment of human pharmaceuticals In Europe, an ERA for human pharmaceuticals has to be conducted according to the Guideline on Environmental Risk Assessment of Medicinal Products for Human Use (EMA 2006). This ERA consists of two phases. Phase I is a pre-screening for estimating the exposure in surface water, and if this Predicted Environmental Exposure Concentration (PEC) does not reach the action limit of 0.01 µg/L, in most cases, the ERA can stop. In case this action limit is reached or exceeded, a base set of aquatic toxicology, and fate and behaviour data need to be supplied in phase II Tier A. A risk assessment, comparing the PEC with the Predicted No Effect Concentration (PNEC), needs to be conducted. If in this step a risk is still identified for a specific compartment, a substance and compartment-specific refinement and risk assessment in Phase II Tier B needs to be conducted (Figure 2). Phase I: Estimation of Exposure In Phase I, the PEC calculation is restricted to the aquatic compartment. The estimation should be based on the drug substance only, irrespective of its route of administration, pharmaceutical form, metabolism and excretion. The initial calculation of the PEC in surface water assumes: The predicted amount used per capita per year is evenly distributed over the year and throughout the geographic area (Dose ai ); A fraction of the overall market penetration (F pen ), in other words 'how many people will take the medicinal product? Normally a default value of 1% is used; The sewage system is the drug's main route of entry into surface water. The following formula is used to estimate the PEC in surface water: PEC surfacewater = mg/l DOSE ai = Maximum daily dose consumed per capita [mg . inh -1. d -1 ] F pen = Fraction of market penetration (= 1% by default) WASTE inhab = Amount of wastewater per inhabitant per day (= 200 l by default) DILUTION = Dilution Factor (= 10 by default) Three factors of this formula, i.e. F pen , Waste inhab and the Dilution Factor, are default values, meaning that the PEC surfacewater in Phase I entirely depends on the dose of the active ingredient. The Fpen can be refined by providing reasonably justified market penetration data, e.g. based on published epidemiological data. If the PEC surfacewater value is equal to or above 0.01 μg/l (mean dose ≥ 2 mg cap -1 d -1 ), a Phase II environmental fate and effect analysis should be performed. Otherwise, the ERA can stop. However, in some cases, the action limit may not be applicable. For instance, medicinal substances with a log K ow > 4.5 are potential PBT candidates and should be screened for persistence (P), bioaccumulation potential (B), and toxicity (T) independently of the PEC value. Furthermore, some substances may affect vertebrates or lower animals at concentrations lower than 0.01 μg/L. These substances should always enter Phase II and a tailored risk assessment strategy should be followed which addresses the specific mechanism of action of the substance. This is often true for e.g. hormone active substances (see section on Endocrine disruption ). The required tests in a Phase II assessment (see below) need to cover the most sensitive life stage, and the most sensitive endpoint needs to be assessed. This means for instance that for substances affecting reproduction, the organism needs to be exposed to the substance during gonad development and the reproductive output needs to be assessed. Phase II: Environmental Fate and Effects Analysis A Phase II assessment is conducted by evaluating the PEC/PNEC ratio based on a base set of data and the predicted environmental concentration from Tier A. If a potential environmental impact is indicated, further testing might be needed to refine PEC and PNEC values in Tier B. Under certain circumstances, effects on sediment-dwelling organisms and terrestrial environmental fate and effects analysis are also required. Experimental studies should follow standard test protocols, e.g. OECD guidelines . It is not acceptable to use QSAR estimation, modelling and extrapolation from e.g. a substance with a similar mode of action and molecular structure (read across). This is in clear contrast to other regulations like e.g. REACH. Human pharmaceuticals are used all year round without any major fluctuations and peaks. The only exemption are substances used against cold and influenza. These substances have a clear peak in the consumption in autumn and winter times. In developed countries in Europe and North America, antibiotics display a similar peak as they are prescribed to support the substances used against viral infections. The guideline reflects this exposure scenario and asks explicitly for long-term effect tests for all three trophic levels: algae, aquatic invertebrates and vertebrates (i.e., fish). In order to assess the physio chemical fate, amongst other tests the sorption behaviour and fate in a water/sediment system should be determined. If, after refinement, the possibility of environmental risks cannot be excluded, precautionary and safety measures may consist of: An indication of potential risks presented by the medicinal product for the environment. Product labelling, Summary Product Characteristics (SPC), Package Leaflet (PL) for patient use, product storage and disposal. Labelling should generally aim at minimising the quantity discharged into the environment by appropriate mitigation measures Environmental risk assessment of veterinary pharmaceuticals In the EU, the Environmental Risk Assessment (ERA) is conducted for all veterinary medicinal products. The structure of an ERA for Veterinary Medicinal Products (VMPs) is quite similar to the ERA for Human Medicinal Products. It is also tier based and starts with an exposure assessment in Phase I. Here, the potential for environmental exposure is assessed based on the intended use of the product. It is assumed that products with limited environmental exposure will have negligible environmental effects and thus can stop in Phase I. Some VMPs that might otherwise stop in Phase I as a result of their low environmental exposure, may require additional hazard information to address particular concerns associated with their intrinsic properties and use. This approach is comparable to the assessment of Human Pharmaceutical Products, see above. Phase I: Estimation of Environmental Exposure For the exposure assessment, a decision tree was developed (Figure 3). The decision tree consists of a number of questions, and the answers of the individual questions will conclude in the extent of the environmental exposure of the product. The goal is to determine if environmental exposure is sufficiently significant to consider if data on hazard properties are needed for characterizing a risk. Products with a low environmental exposure are considered not to pose a risk to the environment and hence these products do not need further assessment. However, if the outcome of Phase I assessment is that the use of the product leads to significant environmental exposure, then additional environmental fate and effect data are required. Examples for products with a low environmental exposure are, among others are products for companion animals only and products that result in a Predicted Environmental Concentration in soil (PECsoil) of less than 100 µg/kg, based on a worst-case estimation. Phase II: Environmental Fate and Effects Analysis A Phase II assessment is necessary if either the trigger of 100 µg/kg in the terrestrial branch or the trigger of 1 µg/L in the aquatic branch is reached. It is also necessary, if the substance is a parasiticide for food producing animals. A Phase II is also required for substances that would in principle stop in Phase I, but there are indications that an environmental risk at very low concentrations is likely due to their hazardous profile (e.g., endocrine active medicinal products). This is comparable to the assessment for Human Pharmaceutical Products. For Veterinary Pharmaceutical Products also the Phase II assessment is sub-divided into several Tiers, see Figure 4. For Tier A, a base set of studies assessing the physical-chemical properties, the environmental fate, and effects of the active ingredient is necessary. For Tier A, acute effect tests are suggested, assuming a more peak like exposure scenario due to e.g. applying manure and dung on fields and meadows, in contrast to the permanent exposure of human pharmaceuticals. If for a specific trophic level, e.g. dung fauna or algae, a risk is identified (PEC/PNEC ≥1) (see Introduction to Chapter 6), long-term tests for this level have to be conducted in Tier B. For the trophic levels, without an identified risk, the assessment can stop. If the risk still applies with these long-term studies, a further refinement with field studies in Tier C can be conducted. Here a co-operation with a competent authority is strongly recommended, as these tests are tailored, reflected by the individual design of these field studies. In addition, and independent of this, risk mitigation measures can be imposed to reduce the exposure concentration (PEC). These can be, beside others, that animals must remain stabled for a certain amount of time after the treatment, to ensure that the concentration of active ingredient in excreta is low enough to avoid adverse effects on dung fauna and their predators. Alternatively, the treated animals are denied access to water as the active ingredient has harmful effects on aquatic organisms. Conclusion The Environmental Risk Assessment of Human and Veterinary Medicinal Products is a straightforward, tiered-based process with the possibility to exit at several steps in the assessment procedure. Depending on the dose, the physico-chemical properties, and the anticipated use, this can be quite early in the procedure. On the other hand, for very potent substances with specific modes of action the guidelines are flexible enough to allow specific assessments covering these modes of action. The ERA guideline for human medicinal products entered into application 2006 and many data gaps exist for products approved prior to 2006. Although there is a legal requirement for an ERA dossier for all marketing authorisation applications, new applications for pharmaceuticals on the market before 2006 are only required to submit ERA data under certain circumstances (e.g. significant increase in usage). Even for some of the blockbusters, like Ibuprofen, Diclofenac, and Metformin, full information on fate, behaviour and effects on non-target organisms is currently lacking. Furthermore, systematic post-authorisation monitoring and evaluation of potential unintended ecotoxicological effects does not exist. The market authorisation for pharmaceuticals does not expire, in contrast to e.g. an authorisation of pesticides, which needs to be renewed every 10 years. For Veterinary Medicinal Products, an in-depth ERA is necessary for food producing animals only. An ERA for non-food animals can stop with question 3 in Phase I (Figure 3) as it is considered that the use of products for companion animals leads to negligible environmental concentrations, which might not be necessarily the case. Here, the guideline does not reflect the state of the art of scientific and regulatory knowledge. For example, the market authorisation, as a pesticide or biocide, has been withdrawn or strongly restricted for some potent insecticides like imidacloprid and fipronil which both are authorised for use in companion animals. Further Reading Pharmaceuticals in the Environment: https://www.umweltbundesamt.de/en/publikationen/pharmaceuticals-in-the-environment-the-global Recommendations for reducing micro-pollutants in waters: https://www.umweltbundesamt.de/publikationen/recommendations-for-reducing-micropollutants-in 6.5.4. Question 1 Why are pharmaceuticals a problem for non-target organisms and for the environment? 6.5.4. Question 2 How do Human pharmaceuticals enter the environment? 6.5.4. Question 3 How do Veterinary pharmaceuticals enter the environment? 6.5.4. Question 4 What is the general scheme of the environmental risk assessment of human and veterinary pharmaceuticals? 6.5.4. Question 5 Why are long-term tests needed for the assessment of human pharmaceuticals, in contrast to the assessment of veterinary pharmaceuticals? 6.5.5. Water Framework Directive under review 6.5.6. Policy on soil and groundwater regulation Author: Frank Swartjes Reviewers : Kees van Gestel, Ad Ragas, Dietmar Müller-Grabherr Learning objectives : You should be able to explain how different countries regulate soil contamination issues list some differences between different policy systems on soil and groundwater regulations describe how risk assessment procedures are implemented in policy Keywords : Policy on soil contamination, Water Framework Directive, screening values comparison, Thematic Soil Strategy, Common Forum History As a bomb hit, soil contamination came onto the political agenda in the United States and in Europe through a number of disasters in the late 1970s and early 1980s. Starting point was the 1978 Love Canal disaster in upper New York State, USA, in which a school and a number of residences had been built on a former landfill for chemical waste disposal with thousands of tonnes of dangerous chemical wastes, and became a national media event. In Europe in 1979, the residential site of Lekkerkerk in the Netherlands became an infamous national event. Again, a residential area had been built on a former waste dump, which included chemical waste from the painting industry, and with channels and ditches that had been filled in with chemical waste-containing materials. Since these events, soil contamination-related policies emerged one after the other in different countries in the world. Crucial elements of these policies were a benchmark date for a ban on bringing pollutants in or on the soil ('prevention'), including a strict policy, e.g. duty of care, for contaminations that are caused after the benchmark date, financial liability for polluting activities, tools for assessing the quality of soil and groundwater, and management solutions (remediation technologies and facilities for disposal). Evolution in soil policies Objectives in soil policies often show evolution over time and changes go along with developing new concepts and approaches for implementing policies. In general, soil policies often develop from a maximum risk control until a functional approach. The corresponding tools for implementation usually develop from a set of screening values towards a systemic use of frameworks, enabling sound environmental protection while improving the cost-benefit-balance. Consequently, soil policy implementation usually goes through different stages. In general terms, four different stages can be distinguished, i.e., related to maximum risk control, to the use of screening values, to the use of frameworks and based on a functional approach. Maximum risk control follows the precaution principle and is a stringent way of assessing and managing contamination by trying to avoid any risk. Procedures based on screening values allow for a distinction in polluted and non-polluted sites for which the former, the polluted sites, require some kind of intervention. The scientific underpinning of the earliest generations of screening values was limited and expert judgement played an important role. Later, more sophisticated screening values emerged, based on risk assessment. This resulted in screening values for individual contaminants within the contaminant groups metals and metalloids, other inorganic contaminants (e.g., cyanides), polycyclic aromatic hydrocarbons (PAHs), monocyclic aromatic hydrocarbons (including BETX (benzene, toluene, xylene)), persistent organic pollutants (including PCBs and dioxins), volatile organic contaminants (including trichloroethylene, tetrachloroethylene, 1,1,1-trichloroethane, and vinyl chloride), petroleum hydrocarbons and, in a few countries only, asbestos. For some contaminants such as PAHs, sum-screening values for groups were derived in several countries, based on toxicity equivalents. In a procedure based on frameworks , often the same screening values generally act as a trigger for further, more detailed site-specific investigations in one or two additional assessment steps. In the functional approach , soil and groundwater must be suited for the land use it relates to (e.g., agricultural or residential land) and the functions (e.g., drinking water abstraction, irrigation water) it performs. Some countries skip the maximum risk control and sometimes also the screening values stages and adopt a framework and/or a functional approach. European collaboration and legislation In Europe, collaboration was strengthened by concerted actions such as CARACAS (concerted action on risk assessment for contaminated sites in the European Union; 1996 - 1998) and CLARINET (Contaminated Land Rehabilitation Network for Environmental Technologies; 1998 - 2001). These concerted actions were followed up by fruitful international networks that are still are active today. These are the Common Forum , which is a network of contaminated land policy makers, regulators and technical advisors from Environment Authorities in European Union member states and European Free Trade Association countries, and NICOLE (Network for Industrially Co-ordinated Sustainable Land Management in Europe), which is a leading forum on industrially co-ordinated sustainable land management in Europe. NICOLE is promoting co-operation between industry, academia and service providers on the development and application of sustainable technologies. In 2000, the EU Water Framework Directive (WFD; Directive 2000/60/EC ) was adopted by the European Commission, followed by the Groundwater Directive ( Directive 2006/118/EC ) in 2006 (European parliament and the council of the European Union, 2019b). The environmental objectives are defined by the WFD. Moreover, 'good chemical status' and the 'no deterioration clause' account for groundwater bodies. 'Prevent and limit' as an objective aims to control direct or indirect contaminant inputs to groundwater, and distinguishes for 'preventing hazardous substances' to enter groundwater as well as 'limiting other non-hazardous substances'. Moreover, the European Commission adopted a Soil Thematic Strategy , with soil contamination being one out of the seven identified threats. A proposal for a Soil Framework Directive, launched in 2006, with the objective to protect soils across the EU, was formally withdrawn in 2014 because of a lack of support from some countries. Policies in the world Today, most countries in Europe and North America, Australia and New Zealand, and several countries in Asia and Middle and South America, have regulations on soil and groundwater contamination. The policies, however, differ substantially in stage, extent and format. Some policies only cover prevention, e.g., blocking or controlling the inputs of chemicals onto the soil surface and in groundwater bodies. Other policies cover prevention, risk based quality assessment and risk management procedures and include elaborated technical tools, which enable a decent and uniform approach. In particular the larger countries such as the USA, Germany and Spain, policies differ between states or provinces within the country. And even in countries with a policy on the federal level, the responsibilities for different steps in the soil contamination chain are very different for the different layers of authorities (at the national, regional and municipal level). In Figure 1 the European countries are shown that have a procedure based on frameworks (as described above), including risk-based screening values. It is difficult, if not impossible, to summarise all policies on soil and groundwater protection worldwide. Alternatively, some general aspects of these policies are given here. A fair first basic element in nearly all soil and groundwater policies, relating to prevention of contamination, is the declaration of a formal point in time after which polluting soil and groundwater is considered an illegal act. For soil and groundwater quality assessment and management, most policies follow the risk-based land management approach as the ultimate form of the functional approach described above. Central in this approach are the risks for specific targets that need to be protected up to a specified level. Different protection targets are considered. Not surprisingly, 'human health' is the primary protection target that is adopted in nearly all countries with soil and groundwater regulations. Moreover, the ecosystem is an important protection target for soil, while for groundwater the ecosystem as a protection target is under discussion. Another interesting general characteristic of mature soil and groundwater policies is the function-specific approach. The basic principle of this approach is that land must be suited for its purpose. As a consequence, the appraisal of a contaminated site in a residential area, for instance, follows a much more stringent concept than that of an industrial site. Risk assessment tools Risk assessment tools often form the technical backbone of policies. Since the late 1980s risk assessment procedures for soil and groundwater quality appraisal were developed. In the late 1980s the exposure model CalTOX was developed by the Californian Department of Toxic Substances Control in the USA, a few years later the CSOIL model in the Netherlands (Van den Berg, 1991/ 1994/ 1995). In Figure 2, the flow chart of the Dutch CSOIL exposure model is given as an example. Three elements are recognized in CSOIL, like in most exposure models: (1) contaminant distribution over the soil compartments; (2) contaminant transfer from (the different compartments of) the soil into contact media; and (3) direct and indirect exposure to humans. The major exposure pathways are exposure through soil ingestion, crop consumption and inhalation of indoor vapours (Elert et al., 2011). Today, several exposure models exist (see Figure 3 for some 'national' European exposure models). However, these exposure models may give quite different exposure estimates for the same exposure scenario (Swartjes, 2007). Moreover, procedures were developed for ecological risk assessment, including the Species Sensitivity Distributions (see section on SSDs ), based on empirical relations between concentration in soil or groundwater and the percentage of species or ecological processes that experience adverse effects (PAF: potentially Affected Fraction). For site specific risk assessment, the TRIAD approach was developed, based on three lines of evidence, i.e., chemically-based, toxicity-based and using data from ecological field surveys (see section on the TRIAD approach ). In the framework of the HERACLES network, another attempt was made to summarizing different EU policies on polluted soil and groundwater. A strong plea was made for harmonisation of risk assessment tools (Swartjes et al., 2009). The authors also described a procedure for harmonization based on the development of a toolbox with standardized and flexible risk assessment tools. Flexible tools are meant to cover national or regional differences in cultural, climatic and geological (e.g., soil type, depth of the groundwater table) conditions. It is generally acknowledged, however, that policy decisions should be taken on the national level. In 2007, an analysis of the differences of soil and groundwater screening values and of the underlying regulatory frameworks, human health and ecological risk assessment procedures (Carlon, 2007) was launched. Although screening values are difficult to compare, since frameworks and objectives of screening values differ significantly, a general conclusion can be drawn for e.g. the screening values at the potentially unacceptable risk level (often used as 'action' values , i.e. values that trigger further research or intervention when exceeded). For the 20 metals, most soil screening values (from 13 countries or regions) show between a factor of 10 and 100 difference between the lowest and highest values. For the 23 organic pollutants considered, most soil screening values (from 15 countries or regions) differ by a factor of between 100 and 1000, but for some organic pollutants these screening values differ by more than four orders of magnitude. These conclusions are merely relevant from a policy viewpoint. Technically, these conclusions are less relevant, since, the screenings values are derived from a combination of different protection targets and tools and based on different policy decisions. Differences in screening values are explained by differences in geographical and biological and socio-cultural factors in different countries and regions, different national regulatory and policy decisions and variability in scientific/ technical tools. References Carlon, C. (Ed.) (2007). Derivation methods of soil screening values in Europe. A review and evaluation of national procedures towards harmonisation, JRC Scientific and Technical report EUR 22805 EN. Elert, M., Bonnard, R., Jones, C., Schoof, R.A., Swartjes, F.A. (2011). Human Exposure Pathways. Chapter 11 in: Swartjes, F.A. (Ed.), Dealing with Contaminated Sites. From theory towards practical application. Springer Publishers, Dordrecht. Swartjes, F.A. (2007). Insight into the variation in calculated human exposure to soil contaminants using seven different European models. Integrated Environmental Assessment and Management 3, 322-332. Swartjes, F.A., D'Allesandro, M., Cornelis, Ch., Wcislo, E., Müller, D., Hazebrouck, B., Jones, C., Nathanail, C.P. (2009). Towards consistency in risk assessment tools for contaminated sites management in the EU . The HERACLES strategy from the end of 2009 onwards. National Institute for Public Health and the Environment (RIVM), Bilthoven, The Netherlands, RIVM Report 711701091. Van den Berg, R. (1991/1994/1995). Exposure of humans to soil contamination. A quantitative and qualitative analyses towards proposals for human toxicological C‑quality standards (revised version of the 1991/ 1994 reports). National Institute for Public Health and the Environment (RIVM), Bilthoven, The Netherlands, RIVM-report no. 725201011. Further reading Swartjes, F.A. (Ed.) (2011). Dealing with Contaminated Sites. From theory towards practical application . Springer Publishers, Dordrecht. Rodríguez-Eugenio, N., McLaughlin, M., Pennock, D. (2018). Soil Pollution: a hidden reality . Rome, FAO. 6.5.6. Question 1 What is the logical first step in policy on soil and groundwater protection, related to ' prevention' ? 6.5.6. Question 2 What are the most frequently used protection targets on policies on soil and groundwater in the world? 6.5.6. Question 3 What role should screening values play in sophisticated risk assessment procedures? 6.5.6. Question 4 What is the ideal approach when developing a new soil and groundwater policy? 6.5.6. Question 5 Regarding the harmonization of risk assessment tools: why are flexible risk assessment tools necessary? 6.5.7. Drinking water in preparation
Courses/South_Puget_Sound_Community_College	CHEM 110: Chemical Concepts Surveys chemical concepts for non-science majors: This OER text was created by the faculty of South Puget Sound Community College, Open Education and Equitable and Inclusive Pedagogy projects 2024. Front Matter 1: The Chemical World 2: Units and Measurements 3: Matter 4: Atoms and Periodic Table 5: Chemical Bonds, Ionic Bonds and Ionic Compounds 6: Covalent Bonds, Covalent Compounds 7: Chemical Reactions 8: Atomic Mass, Reaction Stoichiometry 9: Molecular Polarity, Intermolecular Forces and Solubility 10: Radiation, Electromagnetic Waves, Biological Effects of UV Radiation 11: Climate and The Environment Back Matter Chem 121: Introduction to Chemistry Introductory Chemistry for Allied Health: This OER text was created by the faculty of South Puget Sound Community College, Exceptional Faculty Grant 2022. Front Matter 1: Chapter 1 - Matter 2: Chapter 2 - Measurements 3: Chapter 3 - Atomic Structure 4: Chapter 4 - Ionic Bonding 5: Chapter 5A - Covalent Bonding 6: Chapter 5B - Intermolecular Forces and Molecular Models 7: Chapter 6 - Chemical Reactions and Stoichiometry 8: Chapter 7 - Solutions 9: Chapter 8A - Acids, bases and pH 10: Chapter 8B - Acid-Base Buffers 11: Chapter 9 - Gases 12: Chapter 10 - Nuclear Chemistry Back Matter
Courses/San_Diego_Miramar_College/Chemistry_201%3A_General_Chemistry_II_(Garces)/01%3A_Liquids_Solids_and_Intermolecular_Forces	Template:HideTOC 1.1: Structure Determines Properties If you go far enough out in space, for instance, onto the International Space Station, gravity becomes negligible, and the laws of physics act differently than here on Earth. Just how might water act in a place of zero gravity? 1.2: Solids, Liquids, and Gases- A Molecular Comparison The state of a substance depends on the balance between the kinetic energy of the individual particles (molecules or atoms) and the intermolecular forces. The kinetic energy keeps the molecules apart and moving around, and is a function of the temperature of the substance and the intermolecular forces try to draw the particles together. 1.3: Intermolecular Forces- The Forces that Hold Condensed Phases Together Molecules in liquids are held to other molecules by intermolecular interactions, which are weaker than the intramolecular interactions that hold molecules and polyatomic ions together. 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. 1.4: Intermolecular Forces in Action- Surface Tension, Viscosity, and Capillary Action Surface tension, capillary action, and viscosity are unique properties of liquids that depend on the nature of intermolecular interactions. Surface tension is the energy required to increase the surface area of a liquid. Surfactants are molecules that reduce the surface tension of polar liquids like water. Capillary action is the phenomenon in which liquids rise up into a narrow tube called a capillary. The viscosity of a liquid is its resistance to flow. 1.5: Vaporization and Vapor Pressure Because the molecules of a liquid are in constant motion and possess a wide range of kinetic energies, at any moment some fraction of them has enough energy to escape from the surface of the liquid to enter the gas or vapor phase. This process, called vaporization or evaporation, generates a vapor pressure above the liquid. Molecules in the gas phase can collide with the liquid surface and reenter the liquid via condensation. Eventually, a steady state or dynamic equilibrium is reached. 1.6: Sublimation and Fusion The heat energy which a solid absorbs when it melts is called the enthalpy of fusion or heat of fusion and is usually quoted on a molar basis. (The word fusion means the same thing as “melting.”) 1.7: Heating Curve for Water Freezing, condensation, and deposition, which are the reverse of fusion, sublimation, and vaporization—are exothermic. Thus heat pumps that use refrigerants are essentially air-conditioners running in reverse. Heat from the environment is used to vaporize the refrigerant, which is then condensed to a liquid in coils within a house to provide heat. The energy changes that occur during phase changes can be quantified by using a heating or cooling curve. 1.8: Water - An Extraordinary Substance Water is an unusual compound with unique physical properties. As a result, its the compound of life. Yet, its the most abundant compound in the biosphere of Earth. These properties are related to its electronic structure, bonding, and chemistry. However, due to its affinity for a variety of substances, ordinary water contains other substances. Few of us has used, seen or tested pure water, based on which we discuss its chemistry.
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Topics_in_Thermodynamics_of_Solutions_and_Liquid_Mixtures/01%3A_Modules/1.12%3A_Expansions/1.12.02%3A_Heat_Capacity-_Isobaric-_Partial_Molar-_Solution	Equilibrium isobaric heat capacities of solutions can be treated for most purposes as extensive variables. Thus for an aqueous solution prepared using $\mathrm{n}_{1}$ moles of solvent (water) and $\mathrm{n}_{j}$ moles of a simple solute $j$ the isobaric (equilibrium) heat capacity of the solution $\mathrm{C}_{\mathrm{p}}(\mathrm{aq})$ can be related to the composition of the solution using equations (a) and (b) [1]. \[\mathrm{C}_{\mathrm{p}}(\mathrm{aq} ; \mathrm{A}=0)=\mathrm{n}_{1} \, \mathrm{C}_{\mathrm{p} 1}(\mathrm{aq})+\mathrm{n}_{\mathrm{j}} \, \mathrm{C}_{\mathrm{pj}}(\mathrm{aq}) \nonumber \] where \[\mathrm{C}_{\mathrm{pl}}(\mathrm{aq})=\left(\frac{\partial \mathrm{C}_{\mathrm{p}}(\mathrm{aq} ; \mathrm{A}=0)}{\partial \mathrm{n}_{1}}\right)_{\mathrm{T}, \mathrm{p}, \mathrm{n}(\mathrm{j})} \quad \text { and } \mathrm{C}_{\mathrm{pj}}(\mathrm{aq})=\left(\frac{\partial \mathrm{C}_{\mathrm{p}}(\mathrm{aq} ; \mathrm{A}=0)}{\partial \mathrm{n}_{\mathrm{j}}}\right)_{\mathrm{T}, \mathrm{p}, \mathrm{n}(\mathrm{l})} \nonumber \] Similar equations are encountered in a discussion of the partial molar enthalpies but with reference to these properties we develop a number of strategies because it is not possible to determine the enthalpy of a solution. In the present case the outlook is much more favourable because it is possible to measure isobaric heat capacities of solutions [1-3]. The fact that we can measure the temperature dependence of the equilibrium enthalpy of a solution but not the actual enthalpy is an interesting philosophical point. Nevertheless it is informative to develop the analysis starting from equations relating partial molar enthalpies and compositions of solutions. A given aqueous solution is prepared using $\mathrm{n}_{1}$ moles of solvent (water) and $\mathrm{n}_{j}$ moles of solute. The partial molar enthalpies are related to the composition of the solution by the following equations. Thus \[\mathrm{H}_{1}(\mathrm{aq})=\mathrm{H}_{1}^{}(\ell)+\mathrm{R} \, \mathrm{T}^{2} \, \mathrm{M}_{1} \, \mathrm{m}_{\mathrm{j}} \,(\partial \phi / \partial \mathrm{T})_{\mathrm{p}} \nonumber \] and \[\mathrm{H}_{\mathrm{j}}(\mathrm{aq})=\mathrm{H}_{\mathrm{j}}^{\infty}(\mathrm{aq})-\mathrm{R} \, \mathrm{T}^{2} \,\left[\partial \ln \left(\gamma_{\mathrm{j}}\right) / \partial \mathrm{T}\right)_{\mathrm{p}} \nonumber \] At all $\mathrm{T}$ and $\mathrm{p}$, \[\operatorname{limit}\left(m_{j} \rightarrow 0\right) \phi=1 \text { and } \gamma_{j}=1 \nonumber \] By definition, \[\mathrm{C}_{\mathrm{p} 1}(\mathrm{aq})=\left(\frac{\partial \mathrm{C}_{\mathrm{p}}}{\partial \mathrm{n}_{1}}\right)_{\mathrm{T}, \mathrm{p}, \mathrm{n}(\mathrm{j})}=\left(\frac{\partial \mathrm{H}_{1}}{\partial \mathrm{T}}\right)_{\mathrm{p}}=\left(\frac{\partial^{2} \mathrm{H}}{\partial \mathrm{n}_{1} \, \partial \mathrm{T}}\right)_{\mathrm{p}, \mathrm{n}(\mathrm{j})} \nonumber \] And, \[\mathrm{C}_{\mathrm{pj}}(\mathrm{aq})=\left(\frac{\partial \mathrm{C}_{\mathrm{p}}}{\partial \mathrm{n}_{\mathrm{j}}}\right)_{\mathrm{T}, \mathrm{p}, \mathrm{n}(1)}=\left(\frac{\partial \mathrm{H}_{\mathrm{j}}}{\partial \mathrm{T}}\right)_{\mathrm{p}}=\left(\frac{\partial^{2} \mathrm{H}}{\partial \mathrm{n}_{\mathrm{j}} \, \partial \mathrm{T}}\right)_{\mathrm{p}, \mathrm{n}(1)} \nonumber \] The latter two equations trace the story from the enthalpy of the solution to partial molar isobaric heat capacities. Using equation (f) in conjunction with equation (c) we obtain an equation for dependence for $\mathrm{C}_{\mathrm{p}1}(\mathrm{aq})$ on molality $\mathrm{m}_{j}$. \[\mathrm{C}_{\mathrm{pl} 1}(\mathrm{aq})=\mathrm{C}_{\mathrm{pl} 1}^{}(\ell)+2 \, \mathrm{R} \, \mathrm{T} \, \mathrm{M}_{1} \, \mathrm{m}_{\mathrm{j}} \,(\partial \phi / \partial \mathrm{T})_{\mathrm{p}}+\mathrm{R} \, \mathrm{T}^{2} \, \mathrm{M}_{1} \, \mathrm{m}_{\mathrm{j}} \,\left(\partial^{2} \phi / \partial \mathrm{T}^{2}\right)_{\mathrm{p}} \nonumber \] Similarly using equations (d) and (g), we obtain an equation relating $\mathrm{C}_{\mathrm{p}j}(\mathrm{aq})$ and molality $\mathrm{m}_{j}$; the origin of the two minus signs is the Gibbs - Helmholtz Equation. \[\mathrm{C}_{\mathrm{pj}}(\mathrm{aq})=\mathrm{C}_{\mathrm{pj}}^{\infty}(\mathrm{aq})-2 \, \mathrm{R} \, \mathrm{T} \,\left[\partial \ln \left(\gamma_{\mathrm{j}}\right) / \partial \mathrm{T}\right]_{\mathrm{p}}-\mathrm{R} \, \mathrm{T}^{2} \,\left[\partial^{2} \ln \left(\gamma_{\mathrm{j}}\right) / \partial \mathrm{T}^{2}\right]_{\mathrm{p}} \nonumber \] We consider a solution prepared using $1 \mathrm{~kg}$ of water and $\mathrm{m}_{j}$ moles of solute $j$ [4]. \[\mathrm{C}_{\mathrm{p}}\left(\mathrm{aq} ; \mathrm{w}_{1}=1 \mathrm{~kg}\right)=\left(1 / \mathrm{M}_{1}\right) \, \mathrm{C}_{\mathrm{pl}}(\mathrm{aq})+\mathrm{m}_{\mathrm{j}} \, \mathrm{C}_{\mathrm{pj}} \text { (aq) } \nonumber \] If the thermodynamic properties of the solution are ideal, from the definitions of both practical osmotic coefficient $\phi$ and activity coefficient $\gamma_{j}$, the last two terms in equations (h) and (i) are zero. \[\mathrm{C}_{\mathrm{p}}\left(\mathrm{aq} ; \mathrm{id} ; \mathrm{w}_{1}=1 \mathrm{~kg}\right)=\left(1 / \mathrm{M}_{1}\right) \, \mathrm{C}_{\mathrm{p} 1}^{}(\ell)+\mathrm{m}_{\mathrm{j}} \, \mathrm{C}_{\mathrm{pj}}^{\infty}(\mathrm{aq}) \nonumber \] This is an interesting equation because two experiments yield $\mathrm{C}_{\mathrm{p}}(\mathrm{aq} ; \mathrm{id})$ and $\mathrm{C}_{\mathrm{p} 1}^{}(\ell)$. Hence, granted the ideal conditions, we obtain an estimate of $\mathrm{C}_{\mathrm{pj}}^{\infty}(\mathrm{aq})$, limiting isobaric heat capacity of solute $j$ in solution. Unfortunately the assumption concerning ideal properties of a solution is often unrealistic. Nevertheless equation (k) offers a reference against which we can examine the properties of real solutions. Footnotes [1] An important technological development was the design of the Picker flow calorimeter; P. Picker, P.-A. Leduc, P. R. Philip and J. E. Desnoyers, J. Chem. Thermodyn.,1971, 3 ,631. [2] For details of calibration of the Picker calorimeter; D. E.White and R. H. Ward, J. Solution Chem.,1982, 11 ,223. [3] For extension to measurement of thermal expansion coefficients; J. F. Alary, M. N. Simard, J. Dumont and C. Jolicoeur, J. Solution Chem.,1982, 11 ,755. [4] $\left[\mathrm{J} \mathrm{K}^{-1} \mathrm{~kg}^{-1}\right]=\left[\mathrm{kg} \, \mathrm{mol}^{-1}\right]^{-1} \,\left[\mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\right]+\left[\operatorname{mol~kg}{ }^{-1}\right] \,\left[\mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\right]$
Courses/Nassau_Community_College/CHE200_-_Introduction_to_Organic_Chemistry_(Resch)/01%3A_Introduction_to_Organic_Structure_and_Bonding_I/1.04%3A_Structures_of_some_important_biomolecules	Because we are focusing in this textbook on biologically relevant organic chemistry, we will frequently be alluding to important classes of biological molecules such as lipids, carbohydrates, proteins, and nucleic acids (DNA and RNA). Now is a good time to go through a quick overview of what these molecules look like. These are large, complex molecules and there is a lot of information here: you are not expected to memorize these structures or even, at this point, to fully understand everything presented in this section. For now, just read through the section and get what you can out of it, and work on recognizing the fundamental things you have just learned: common bonding patterns, formal charges, functional groups, and so forth. Later, you can come back to this section for review when these biomolecules are referred to in different contexts throughout the remainder of the book. Many of the biomolecules that we will be talking in this section are polymers . To understand what a polymer is, simply picture a long chain made by connecting lots of individual beads, each of which is equipped with two hooks. In chemical terminology, each bead is a monomer compound, the hooks are linking groups , and the whole chain is a polymer. Although lipids can be described as biopolymers, we will use the monomer-linking group terminology in particular when we talk about carbohydrates, protein, and nucleic acids. Click on the links below to learn about each of these important types of biological molecules: Lipids Carbohydrates Amino acids and proteins Nucleic acids
Courses/University_of_Arkansas_Little_Rock/IOST_Library/07%3A_Electronics_Book/01%3A_Electric_Fundamentals/07%3A_Multimeter	The multimeter is an essential tool for debugging electric circuits. Parts of a Multimeter Measuring Voltage The voltmeter is placed in parallel with the circuit component you wish to measure the voltage drop Plug the ground probe into the common port and the hot probe into the voltage port SparkFun YouTube Video at 2:40 minutes showing how to use multimeter to test for voltage. Measuring Resistance To measure resistance you disconnect the device you are measuring from everything else (make sure to disconnect the power), and then connect the probes across the device. The multimeter will then send a small current through the circuit and measure the resistance. Note the different ranges have different SI prefixes. Reading of 1 indicates the resistance is beyond the range of current setting and you need to increase the range. Reading 0f 0.01 indicates the resistance is too small and you need to decrease the range. Note, if you are measuring across a polar device like a diode or LED you must send the current in the correct direction (switch the direction of the probes). SparkFun YouTube Video at 4:59 minutes showing how to use multimeter to test for resistance Measuring Current To measure current you need to set the multimeter in series with the circuit you wish to measure. Most multimeters have a seperate port for measuring current. Note, you may need an oscilloscope to see fluctuating current due to capacitors, inductors and other electrical components. SparkFun YouTube Video at 6.32 minutes showing how to use multimeter to test for current Applications of Multimeters Testing a Battery 0:59 min YouTube by AdaFruit Industries: Test Batteries with a Multimeter - Collin's Lab Notes Testing a Fuse 0:57 min YouTube by AdaFruit Industries: "Blown a Fuse?-Collin's Lab Notes" Resources Sceince Buddies: How to Use a Multimeter SparkFun: How to Use a Multimeter
Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(LibreTexts)/02%3A_Measurement_and_Problem_Solving/2.01%3A_Taking_Measurements	Learning Objectives Express quantities properly, using a number and a unit. A coffee maker’s instructions tell you to fill the coffee pot with 4 cups of water and to use 3 scoops of coffee. When you follow these instructions, you are measuring. When you visit a doctor’s office, a nurse checks your temperature, height, weight, and perhaps blood pressure (Figure $\PageIndex{1}$); the nurse is also measuring. Chemists measure the properties of matter and express these measurements as quantities. A quantity is an amount of something and consists of a number and a unit. The number tells us how many (or how much), and the unit tells us what the scale of measurement is. For example, when a distance is reported as “5 kilometers,” we know that the quantity has been expressed in units of kilometers and that the number of kilometers is 5. If you ask a friend how far they walk from home to school, and the friend answers “12” without specifying a unit, you do not know whether your friend walks 12 kilometers, 12 miles, 12 furlongs, or 12 yards. Both a number and a unit must be included to express a quantity properly. To understand chemistry, we need a clear understanding of the units chemists work with and the rules they follow for expressing numbers. The next two sections examine the rules for expressing numbers. Example $\PageIndex{1}$ Identify the number and the unit in each quantity. one dozen eggs 2.54 centimeters a box of pencils 88 meters per second Solution The number is one, and the unit is a dozen eggs. The number is 2.54, and the unit is centimeter. The number 1 is implied because the quantity is only a box. The unit is box of pencils. The number is 88, and the unit is meters per second. Note that in this case the unit is actually a combination of two units: meters and seconds. Key Take Away Identify a quantity properly with a number and a unit.
Courses/Saint_Francis_University/CHEM_113%3A_Human_Chemistry_I_(Muino)/04%3A_Molecular_Compounds/4.06%3A_Molecular_Formulas_and_Lewis_Structures	Learning Objectives Understand the different ways to represent molecules. There are many "universal languages" in the world. Musicians of every culture recognize music embodied in a series of notes on a staff. This passage from a Bach cello suite could be played by any trained musician from any country, because there is an agreement as to what the symbols on the page mean. In the same way, molecules are represented using symbols and a language that all chemists agree upon. Molecular Formulas A molecular formula is a chemical formula of a molecular compound that shows the kinds and numbers of atoms present in a molecule of the compound. Ammonia is a compound of nitrogen and hydrogen as shown below: Note from the example that there are some standard rules to follow in writing molecular formulas. The number of atoms of each kind is indicated by a subscript following the atom. If there is only one atom, no number is written. If there is more than one atom of a specific kind, the number is written as a subscript following the atom. We would not write $\ce{N_3H}$ for ammonia, because that would mean that there are three nitrogen atoms and one hydrogen atom in the molecule, which is incorrect. Although it is useful for describing a molecule, the molecular formula does not tell us anything about the shape of the molecule, where the different atoms are, or what kinds of bonds are formed. Structural formulas are much more useful to communicate more detailed information about a molecule because they show which atoms are bonded to one another and, in some cases, the approximate arrangement of the atoms in space. Knowing the structural formula of a compound enables chemists to create a three-dimensional model, which provides information about how that compound will behave physically and chemically. Figure $\PageIndex{3}$ shows some of the different ways to portray the structure of a slightly more complex molecule: methanol. These representations differ greatly in their information content. For example, the molecular formula for methanol ( Figure $\PageIndex{3a}$ ) gives only the number of each kind of atom; writing methanol as CH 4 O tells nothing about its structure. In contrast, the structural formula ( Figure $\PageIndex{3b}$ ) indicates how the atoms are connected, but it makes methanol look as if it is planar (which it is not). Both the ball-and-stick model (part (c) in Figure $\PageIndex{3}$ ) and the perspective drawing ( Figure $\PageIndex{3d}$ ) show the three-dimensional structure of the molecule. The latter (also called a wedge-and-dash representation) is the easiest way to sketch the structure of a molecule in three dimensions. It shows which atoms are above and below the plane of the paper by using wedges and dashes, respectively; the central atom is always assumed to be in the plane of the paper. The space-filling model (part (e) in Figure $\PageIndex{3}$) illustrates the approximate relative sizes of the atoms in the molecule, but it does not show the bonds between the atoms. In addition, in a space-filling model, atoms at the “front” of the molecule may obscure atoms at the “back.” Although a structural formula, a ball-and-stick model, a perspective drawing, and a space-filling model provide a significant amount of information about the structure of a molecule, each requires time and effort. Consequently, chemists often use a condensed structural formula (part (f) in Figure $\PageIndex{3}$ ), which omits the lines representing bonds between atoms and simply lists the atoms bonded to a given atom next to it. Multiple groups attached to the same atom are shown in parentheses, followed by a subscript that indicates the number of such groups. For example, the condensed structural formula for methanol is CH 3 OH, which indicates that the molecule contains a CH 3 unit that looks like a fragment of methane (CH 4 ). Methanol can therefore be viewed either as a methane molecule in which one hydrogen atom has been replaced by an –OH group or as a water molecule in which one hydrogen atom has been replaced by a –CH 3 fragment. Because of their ease of use and information content, we use condensed structural formulas for molecules throughout this text. Ball-and-stick models are used when needed to illustrate the three-dimensional structure of molecules, and space-filling models are used only when it is necessary to visualize the relative sizes of atoms or molecules to understand an important point. Example $\PageIndex{1}$: Molecular Formulas Write the molecular formula for each compound. The condensed structural formula is given. Sulfur monochloride (also called disulfur dichloride) is a vile-smelling, corrosive yellow liquid used in the production of synthetic rubber. Its condensed structural formula is ClSSCl. Ethylene glycol is the major ingredient in antifreeze. Its condensed structural formula is HOCH 2 CH 2 OH. Trimethylamine is one of the substances responsible for the smell of spoiled fish. Its condensed structural formula is (CH 3 ) 3 N. Given : condensed structural formula Asked for : molecular formula Strategy : Identify every element in the condensed structural formula and then determine whether the compound is organic or inorganic. As appropriate, use either organic or inorganic convention to list the elements. Then add appropriate subscripts to indicate the number of atoms of each element present in the molecular formula. Solution : The molecular formula lists the elements in the molecule and the number of atoms of each. A Each molecule of sulfur monochloride has two sulfur atoms and two chlorine atoms. Because it does not contain mostly carbon and hydrogen, it is an inorganic compound. B Sulfur lies to the left of chlorine in the periodic table, so it is written first in the formula. Adding subscripts gives the molecular formula S 2 Cl 2 . A Counting the atoms in ethylene glycol, we get six hydrogen atoms, two carbon atoms, and two oxygen atoms per molecule. The compound consists mostly of carbon and hydrogen atoms, so it is organic. B As with all organic compounds, C and H are written first in the molecular formula. Adding appropriate subscripts gives the molecular formula C 2 H 6 O 2 . A The condensed structural formula shows that trimethylamine contains three CH 3 units, so we have one nitrogen atom, three carbon atoms, and nine hydrogen atoms per molecule. Because trimethylamine contains mostly carbon and hydrogen, it is an organic compound. B According to the convention for organic compounds, C and H are written first, giving the molecular formula C 3 H 9 N. Exercise $\PageIndex{1}$: Molecular Formulas Write the molecular formula for each molecule. Chloroform, which was one of the first anesthetics and was used in many cough syrups until recently, contains one carbon atom, one hydrogen atom, and three chlorine atoms. Its condensed structural formula is $\ce{CHCl3}$. Hydrazine is used as a propellant in the attitude jets of the space shuttle. Its condensed structural formula is $\ce{H2NNH2}$. Putrescine is a pungent-smelling compound first isolated from extracts of rotting meat. Its condensed structural formula is H2NCH2CH2CH2CH2NH2. This is often written as $\ce{H2N(CH2)4NH2}$ to indicate that there are four CH2 fragments linked together. Answer a CHCl 3 Answer b N 2 H 4 Answer c
Courses/University_of_Kansas/General_Organic_and_Biological_Chemistry/07%3A_Energy_and_Chemical_Processes/7.00%3A_Prelude_to_Energy_and_Chemical_Processes	Metabolism is the collective term for the chemical reactions that occur in cells and provide energy to keep cells alive. Some of the energy from metabolism is in the form of heat, and some animals use this heat to regulate their body temperatures. Such warm-blooded animals are called endotherms . In endotherms, problems with metabolism can lead to fluctuations in body temperature. When humans get sick, for instance, our body temperatures can rise higher than normal; we develop a fever. When food is scarce (especially in winter), some endotherms go into a state of controlled decreased metabolism called hibernation . During hibernation, the body temperatures of these endotherms actually decrease. In hot weather or when feverish, endotherms will pant or sweat to rid their bodies of excess heat. Endotherm Body Temperature (°F) Body Temperature (°C) bird up to 110 up to 43.5 cat 101.5 38.6 dog 102 38.9 horse 100.0 37.8 human 98.6 37.0 pig 102.5 39.2 Ectotherms , sometimes called cold-blooded animals, do not use the energy of metabolism to regulate body temperature. Instead, they depend on external energy sources, such as sunlight. Fish, for example, will seek out water of different temperatures to regulate body temperature. The amount of energy available is directly related to the metabolic rate of the animal. When energy is scarce, ectotherms may also hibernate. The connection between metabolism and body temperature is a reminder that energy and chemical reactions are intimately related. A basic understanding of this relationship is especially important when those chemical reactions occur within our own bodies.
Courses/Sacramento_City_College/SCC%3A_Chem_420_-_Organic_Chemistry_I/05%3A_An_Introduction_to_Organic_Reactions_using_Free_Radical_Halogenation_of_Alkanes/5.02%3A_5.2_Reaction_Mechanism_Notation_and_Symbols	Learning Objective accurately and precisely use reaction mechanism notation and symbols including curved arrows to show the flow of electrons The Arrow Notation in Mechanisms Since chemical reactions involve the breaking and making of bonds, a consideration of the movement of bonding (and non-bonding) valence shell electrons is essential to this understanding. It is now common practice to show the movement of electrons with curved arrows, and a sequence of equations depicting the consequences of such electron shifts is termed a mechanism . In general, two kinds of curved arrows are used in drawing mechanisms: 0 1 2 A full head on the arrow indicates the movement or shift of an electron pair: NaN NaN A partial head (fishhook) on the arrow indicates the shift of a single electron: NaN NaN The use of these symbols in bond-breaking and bond-making reactions is illustrated below. If a covalent single bond is broken so that one electron of the shared pair remains with each fragment, as in the first example, this bond-breaking is called homolysis . If the bond breaks with both electrons of the shared pair remaining with one fragment, as in the second and third examples, this is called heterolysis . 0 1 2 Bond-Breaking NaN Bond-Making NaN NaN NaN Other Arrow Symbols Chemists also use arrow symbols for other purposes, and it is essential to use them correctly. 0 1 2 The Reaction Arrow The Equilibrium Arrow The Resonance Arrow NaN NaN NaN The following equations illustrate the proper use of these symbols: Reactive Intermediates The products of bond breaking, shown above, are not stable in the usual sense, and cannot be isolated for prolonged study. Such species are referred to as reactive intermediates , and are believed to be transient intermediates in many reactions. The general structures and names of four such intermediates are given below. A pair of widely used terms, related to the Lewis acid-base notation, should also be introduced here. Electrophile: An electron deficient atom, ion or molecule that has an affinity for an electron pair, and will bond to a base or nucleophile. Nucleophile: An atom, ion or molecule that has an electron pair that may be donated in bonding to an electrophile (or Lewis acid). Using these definitions, it is clear that carbocations ( called carbonium ions in the older literature ) are electrophiles and carbanions are nucleophiles. Carbenes have only a valence shell sextet of electrons and are therefore electron deficient. In this sense they are electrophiles, but the non-bonding electron pair also gives carbenes nucleophilic character. As a rule, the electrophilic character dominates carbene reactivity. Carbon radicals have only seven valence electrons, and may be considered electron deficient; however, they do not in general bond to nucleophilic electron pairs, so their chemistry exhibits unique differences from that of conventional electrophiles. Radical intermediates are often called free radicals . The importance of electrophile / nucleophile terminology comes from the fact that many organic reactions involve at some stage the bonding of a nucleophile to an electrophile, a process that generally leads to a stable intermediate or product. Reactions of this kind are sometimes called ionic reactions , since ionic reactants or products are often involved. Some common examples of ionic reactions and their mechanisms may be examined below. The shapes ideally assumed by these intermediates becomes important when considering the stereochemistry of reactions in which they play a role. A simple tetravalent compound like methane, CH 4 , has a tetrahedral configuration. Carbocations have only three bonds to the charge bearing carbon, so it adopts a planar trigonal configuration. Carbanions are pyramidal in shape ( tetrahedral if the electron pair is viewed as a substituent), but these species invert rapidly at room temperature, passing through a higher energy planar form in which the electron pair occupies a p-orbital. Radicals are intermediate in configuration, the energy difference between pyramidal and planar forms being very small. Since three points determine a plane, the shape of carbenes must be planar; however, the valence electron distribution varies. Ionic Reactions The principles and terms introduced in the previous sections can now be summarized and illustrated by the following three examples. Reactions such as these are called ionic or polar reactions, because they often involve charged species and the bonding together of electrophiles and nucleophiles . Ionic reactions normally take place in liquid solutions, where solvent molecules assist the formation of charged intermediates. 0 1 NaN The substitution reaction shown on the left can be viewed as taking place in three steps. The first is an acid-base equilibrium, in which HCl protonates the oxygen atom of the alcohol. The resulting conjugate acid then loses water in a second step to give a carbocation intermediate. Finally, this electrophile combines with the chloride anion nucleophile to give the final product. NaN The addition reaction shown on the left can be viewed as taking place in two steps. The first step can again be considered an acid-base equilibrium, with the pi-electrons of the carbon-carbon double bond functioning as a base. The resulting conjugate acid is a carbocation, and this electrophile combines with the nucleophilic bromide anion. NaN The elimination reaction shown on the left takes place in one step. The bond breaking and making operations that take place in this step are described by the curved arrows. The initial stage may also be viewed as an acid-base interaction, with hydroxide ion serving as the base and a hydrogen atom component of the alkyl chloride as an acid. NaN There are many kinds of molecular rearrangements called isomerizations. The examples shown on the left are from an important class called tautomerization or, more specifically, keto-enol tautomerization. Tautomers are rapidly interconverted constitutional isomers, usually distinguished by a different bonding location for a labile hydrogen atom (colored red here) and a differently located double bond. The equilibrium between tautomers is not only rapid under normal conditions, but it often strongly favors one of the isomers (acetone, for example, is 99.999% keto tautomer). Even in such one-sided equilibria, evidence for the presence of the minor tautomer comes from the chemical behavior of the compound. Tautomeric equilibria are catalyzed by traces of acids or bases that are generally present in most chemical samples. Exercise 1. Add curved arrows to explain the indicated reactivity and classify the reaction as "homolytic cleavage" or "heterolytic cleavage". 2. Add the correct arrow to each expression below using your knowledge of chemistry. 3. Classify the following reactions as substituion, addition, elimination, or tautomerization (an example of isomerization). Answer 1. 2. 3.
Bookshelves/Organic_Chemistry/Organic_Chemistry_(OpenStax)/29%3A_The_Organic_Chemistry_of_Metabolic_Pathways/29.09%3A_Catabolism_of_Proteins-_Deamination	The catabolism of proteins is much more complex than that of fats and carbohydrates because each of the 20 α -amino acids is degraded through its own unique pathway. The general idea, however, is that ( 1 ) the α amino group is first removed as ammonia by a deamination process, ( 2 ) the ammonia is converted into urea, and ( 3 ) the remaining amino acid carbon skeleton (usually an α -keto acid) is converted into a compound that enters the citric acid cycle. Transamination Deamination is usually accomplished by a transamination reaction in which the –NH 2 group of the amino acid is exchanged with the keto group of α -ketoglutarate, forming a new α -keto acid plus glutamate. The overall process occurs in two parts, is catalyzed by aminotransferases, and involves participation of the coenzyme pyridoxal phosphate, abbreviated PLP, a derivative of pyridoxine (vitamin B 6 ). The aminotransferases differ in their specificity for amino acids, but the mechanism remains the same. The mechanism of the first part of transamination is shown in Figure $\PageIndex{1}$. The process begins with reaction between the α -amino acid and pyridoxal phosphate, which is covalently bonded to the aminotransferase by an imine linkage between the side-chain –NH 2 group of a lysine residue in the enzyme and the PLP aldehyde group. Deprotonation/reprotonation of the PLP–amino acid imine effects tautomerization of the imine C ═ N C ═ N bond, and hydrolysis of the tautomerized imine gives an α -keto acid plus pyridoxamine phosphate (PMP). Step 1 : Transimination The first step in trans amination is trans imination —the reaction of the PLP–enzyme imine with an α-amino acid to give a PLP–amino acid imine plus expelled enzyme as the leaving group. The reaction occurs by nucleophilic addition of the amino acid –NH2 group to the $\ce{C═N}$ bond of the PLP imine, much as an amine adds to the $\ce{C═O}$ bond of a ketone or aldehyde in a nucleophilic addition reaction (Section 19.8). The protonated diamine intermediate undergoes a proton transfer and expels the lysine amino group in the enzyme to complete the step. Steps 2 – 4 : Tautomerization and Hydrolysis Following formation of the PLP–amino acid imine in step 1 , a tautomerization of the C ═ N C ═ N bond occurs in step 2 . The basic lysine residue in the enzyme that was expelled as a leaving group during transimination deprotonates the acidic α position of the amino acid, with the protonated pyridine ring of PLP acting as the electron acceptor. Reprotonation occurs on the carbon atom next to the ring, generating a tautomeric product that is the imine of an α -keto acid with pyridoxamine phosphate, abbreviated PMP. Hydrolysis of this PMP– α -keto acid imine then completes the first part of the transamination reaction. This is the mechanistic reverse of imine formation and occurs by nucleophilic addition of water to the imine, followed by proton transfer and expulsion of PMP as leaving group. Regeneration of PLP from PMP With PLP plus the α -amino acid now converted into PMP plus an α -keto acid, PMP must be transformed back into PLP to complete the catalytic cycle. The conversion occurs by another transamination reaction, this one between PMP and an α -keto acid, usually α -ketoglutarate. The products are PLP plus glutamate, and the mechanism is the exact reverse of that shown in Figure $\PageIndex{1}$. That is, PMP and α -ketoglutarate give an imine; the PMP– α -ketoglutarate imine undergoes tautomerization of the C ═ N C ═ N bond to give a PLP–glutamate imine; and the PLP–glutamate imine reacts with a lysine residue on the enzyme in a transimination process to yield PLP–enzyme imine plus glutamate. Exercise $\PageIndex{1}$ Write all the steps in the transamination reaction of PMP with α -ketoglutarate plus a lysine residue in the enzyme to give the PLP–enzyme imine plus glutamate. Answer Initial imine formation between PMP and α -ketoglutarate is followed by double-bond rearrangement to an isomeric imine and hydrolysis. Exercise $\PageIndex{2}$ What α -keto acid is formed on transamination of leucine? Answer $\ce{(CH3)2CHCH2COCO2^{-}}$ Exercise $\PageIndex{3}$ From what amino acid is the following α -keto acid derived? Answer Asparagine
Courses/Los_Medanos_College/General_College_Chemistry_Lab_Manual_(Semester_1)/01%3A_Part_1_Experiments/1.01%3A_2500_Safety_in_Chemistry_Laboratory	SAFETY AND THE CHEMICAL LABORATORY 1.0 INTRODUCTION Teaching and research laboratories are locations for learning and exploring chemical reactions and properties. Like all activities in daily life, a risk of injury exists. Safety precautions and work practices reduce this risk. There are some hazards that are specific for chemicals. This document cannot cover all laboratory safety guidelines for all future chemistry operations students will encounter. A whole field of industrial hygiene exists to create safer work practices. The objectives for this experiment are to · Establish a culture of safety awareness · Introduce the RAMP concept · State the general rules for working safely in a chemical laboratory · Describe differences between a hazard and a risk and acute and chronic toxicity · Introduce a safety data sheet (SDS) and identify information it contains · Present three common labeling and marking systems · Build safe work practices into good habits · Introduce regulatory agencies · Prepare for emergency responses for when things happen Chemical experiments in the teaching laboratory should contain a section on safety precautions and waste disposal. It is important to review these before entering the laboratory. Safety precautions have been developed over many decades and are based on much experience and many events. The American Chemical Society (ACS) guidelines for undergraduate laboratory safety recommend the R. A. M. P. method. · R ecognize the hazards · A ssess the risks of hazards · M inimize the risks of the hazards · P repare for emergencies from uncontrolled hazards References and further reading Guidelines of Chemical Laboratory Safety in Academic Institutions, ACS Committee on Chemical Safety (CCS) Task Force for Safety Education Guidelines, 2016. National Fire Protection Association (NFPA), 2017 https://www.nfpa.org/Assets/files/Ab...4/704_FAQs.pdf (accessed 06/23/2020) Emergency Response Guidebook (2016), U. S. Department of Transportation. https://www.phmsa.dot.gov/sites/phms...cs/ERG2016.pdf (accessed 06/23/2020) Prudent Practices in the Laboratory, Handling and Management of Chemical Hazards, National Research Council Committee on Prudent Practices in the Laboratory, 2011 Chemical Hygiene Plan, Los Medanos College, (the current version in effect). 2.0 RECOGNIZE THE HAZARDS 2.1 CHEMICAL LABELING To recognize the hazards of chemical, one of the most important things is to read the label, documents that are shipped with the chemical, and additional references. Several types of labeling systems are in use. Each has been developed over time and for specific reasons. Three labeling systems are described below, but there are a few more systems in specific industries: A: National Fire Protection Association (NFPA) The NFPA is an international non-profit organization that develops standards to reduce injury due to fire, electrical and related hazards. The NFPA developed a labeling system that is in common use from a fire service perspective. Federal, state or local regulations may require the use of the NFPA labeling system. It is common to see it on buildings such as water treatment facilities and medical facilities where elevated levels of oxygen are present. The NFPA labeling consists of a diamond made of red, blue, yellow and white squares: Figure 1: NFPA Diamond Each color represents a specific and different hazard. Red is for flammability, blue is for health, yellow is for chemical reactivity and white is for special hazards. A number is shown from 0 (low) to 4 (high) in each of the squares that indicates a general assessment of the specific hazard. The NFPA diamond for acetone (formula C 3 H 6 O), a common organic solvent, is shown below: Figure 2: NFPA Diamond for Acetone Acetone is moderately to highly flammable, and so is rated a '3' in the flammability diamond. Direct exposure can cause minor skin irritation, so it is rated a '1' in the health diamond. Acetone is not particularly chemically reactive, and so is rated as a '0' in the reactivity diamond. It has no special hazards, so the white diamond is left blank. The NFPA labeling is designed for a quick and general overview of the safety concerns, especially for first responders. It is often displayed on buildings and storage facilities. Sometimes it can be seen on smaller containers. B: U. S. Department of Transportation (DOT) Pipeline and Hazardous Materials Safety Administration (PHMSA) When chemicals are transported along public highways, it is important for external packaging to display the hazards posed by the contents in the event of an accident and potential release. The PHMSA, a part of the DOT, regulates many aspects of this transportation, including but not limited to packaging, quantities and placarding. Some examples of the placards include the following: Placards are often placed on trucks and railcars. These are often also displayed on shipping boxes for smaller quantities (in general for air transport) and for compressed gas cylinders. Specific information for the transportation of chemicals can be found in the Emergency Response Guidebook (ERG). The ERG is updated every 4 years by the PHMSA. Most public safety agencies are required to have a hardcopy of the ERG or immediate access to the ERG in the event of an emergency. Figure 3 ERG Covers A link to the current ERG is included in the reference section. A mobile app is available. If the placard contains a number, it is cross-referenced in the ERG with guides to additional information about the specific hazards for the classes of chemicals present. An example would be a placard showing corrosive with the number "1830". The "1830" is an identifier for concentrated sulfuric acid (in the yellow bordered pages of the ERG). It corresponds to Guide No. 137 in the orange pages which has information about "Substances - Water-Reactive Corrosive". The ERG is a useful resource to cross-reference trade and industry names. It provides more specific information about classes of compounds than the NFPA diamond. C: U. S. Department of Labor (DOL) Occupational Safety and Health Administration (OSHA) Globally Harmonized System (GHS) The GHS is an international system of hazard identification. The GHS was required by OSHA to be fully implemented by 2016. GHS labels are not mutually exclusive with the NFPA and DOT systems. It is possible to see a GHS symbol, a DOT symbol and an NFPA diamond side by side. Because GHS is used internationally, the same set of symbols should be used on the same material anywhere on the globe. There are 3 components to the GHS: · Hazard Classification Statements o Physical o Health o Environmental Hazard · Labels and Standard Symbols · Safety Data Sheets (SDS) The chemical manufacturer or importer is responsible for determining the hazard classification for the material. There are physical , health , and environmental hazard statement general categories. Some materials may fall into multiple categories. A listing of the 9 specific hazard classes and GHS pictograms are below: The GHS standardized the format and contents of labels on chemicals from the manufacturer. Within the laboratory environment, the GHS hazard labels will be the system most commonly encountered. The GHS pictograms provide a way to rapidly assess the general chemical hazard of material. Additional information is contained within documents called the Safety Data Sheets (SDSs). The GHS specifies a standard set of 16 sections and content for SDSs. The chemical manufacturer is responsible for producing the SDS that accurately describes the known hazards of the material within the GHS format. Employees working with hazardous material must have access to the SDS as part of several labor laws (both Federal and State). Several online sources of SDS are available. Labor laws require that the documentation is available and accurately describes the hazard of the material. Although it is best to have the SDS from the manufacturer, if a SDS is missing or lost, a replacement SDS can be accessed from an alternative source. Most manufacturers provide SDS free of charge from their catalog website. Some possible sources include: https://www.sigmaaldrich.com/united-states.html https://www.fishersci.com/us/en/home.html https://www.flinnsci.com/ A full SDS for sulfuric acid is included in Appendix A. The SDS has 16 parts: Table 1 Safety Data Sheet Sections 0 1 1. Name 9. Physical and Chemical Properties 2. Hazard(s) 10. Stability and Reactivity 3. Compositions 11. Toxicological information 4. First Aid 12. Ecological Information 5. Firefighting Measures 13. Disposal Considerations 6. Accidental Release Measures 14. Transport information 7. Handling and Storage 15. Regulatory Information 8. Exposure Controls/Personal Protection 16. Other information One of the most important pieces of information provided in the SDS is the proper name and Chemical Abstracts Service (CAS) number for each pure component of the material. The CAS is a unique identifier for a compound and avoids confusion. Many industries use legacy names for chemicals, and these names are often used interchangeably. For example, isopropyl alcohol (also called 'rubbing alcohol') is the common name for 2-propanol. Even though these names are different, they all have the same CAS number of 67-63-0. The Los Medanos College Chemistry Stockroom maintains a SDS for all pure compounds within the stockroom. A SDS will be provided upon request. Reagents produced and distributed for use within the chemistry courses will have a GHS warning label, or the words 'No GHS Label Required', as appropriate. Critical thinking skills are essential when reviewing a SDS. Unusual, rare or new compounds may have unknown hazards that are not included on the SDS. Manufacturers are conservative when describing hazards. Sometimes different manufacturers will assign different hazard classes to the same material. Consider the difference between sulfuric acid at different concentrations. Appendix B shows the SDS pages for different dilutions of sulfuric acid. While it would be expected that concentrated sulfuric acid will have the same GHS hazard class from all manufacturers, with diluted sulfuric acid there may be a judgment call made by the manufacturer. 2.2 ADDITIONAL HAZARDS Like many work and school environments, laboratory hazards are not limited to chemical hazards. Wet floors are potential slip areas. Glassware can break, leaving sharp edges which can cut. Hot items often look like cold items and could result in accidental burns. Electrical hazards from equipment are commonly present. Less common hazards include, but are not limited to, radioactive materials, cryogens (cold materials), high pressure systems, vacuum systems, energetic compounds (explosives), and air or water reactive materials. It is beyond the scope of this document to comprehensively describe safety procedures for less common hazards. Specific knowledge, safety procedures, equipment and careful consultation with experienced personnel is recommended before working with any of these hazards. 3.0 ASSESS THE RISKS OF HAZARDS The amount of information included in safety documentation can be overwhelming. By establishing 9 hazard classes, the GHS system quickly allows a general assessment of the hazard of a material. Whenever an unfamiliar chemical is to be used in a new procedure, a thorough review of the SDS is recommended to accurately evaluate the risk of hazard. Laboratory experiments within introductory chemistry courses have been tested and evaluated to minimize the risk to students, faculty and staff. Maintaining proper labeling and documentation is part of the training for all students and staff for assessing risks of materials. Experiments sometimes have special warnings that are set apart from the main text. Personnel should be certain to read and understand these warnings. An example is shown below: 0 !!Special Warnings Will be Stated Here!! Experiments may include descriptions of specific hazards relevant to the experiment being planned. Be certain to read and understand these instructions. !!Disposal Instructions!! A description of how to dispose of the waste generated in the experiment should be included. Some waste may be permitted for drain disposal. Other waste must be captured for off-site disposal. Waste types are often segregated to prevent side reactions. Some waste is acutely hazardous and is easier and cheaper to manage in smaller quantities. Refer to this section in the specific experiment. Consider the generation, management and disposal of waste generated by laboratory processes. Sometimes, the disposal costs for chemical reagents can be several times the purchase price of the original quantity. 4.0 MINIMIZE THE RISKS OF THE HAZARDS Work practices can be designed to minimize the risks of many types of hazards within the chemistry laboratory. These measures include but are not limited to: · Creation of experimental procedures and standard operating procedures · Evaluation of experimental design · Substitution of less hazardous materials when feasible · Engineering controls such as fume hoods, separate laboratory classrooms · Personal Protective Equipment (PPE) appropriate to the operation By building good work habits within the chemistry laboratory course, risks are minimized. Some good habits include: · Read and learn about the experiment before the course session · Wear protective eyewear · Wear protective clothing o Closed-toe shoes o Long pants o Lab coat · Tie loose clothing and long hair · Use specialized waste containers as appropriate o Broken glass waste containers o Biohazard disposal bags o Sharps disposal containers (for syringe needles) · Segregate waste for disposal · Keep areas clear from clutter · Never leave open flames unattended · Use fume hoods if appropriate The creation, review, and incorporation of this Safety in the Chemistry Laboratory procedure and familiarization by personnel is part of the minimization of risk. 5.0 PREPARE FOR EMERGENCIES FROM UNCONTROLLED HAZARDS While the previously described documentation, labels, procedures and good work practices are very effective at reducing emergencies, failures and mishaps will occur. Glass equipment may experience hairline fractures resulting in breaks. Spills may occur at any time. Electrical equipment may provide sparks and ignition sources for flammable materials. All personnel should be prepared for emergencies and should be familiar with safety equipment including: · Fire extinguishers · Eyewash stations · Safety showers · Spill kits · First aid kits · Fire alarms · Evacuation routes · Emergency telephone numbers Equipment should be inspected on a regular basis. Inspection tags and other documentation should be retained for a time to indicate that equipment has been inspected and is operational. At some facilities, emergency drills are used to test and exercise the emergency response system. 6.0 ACCIDENT INVESTIGATION When an accident does occur, a report should be made to the instructor or supervisor. A review may be initiated. The review may be simple or complex and depends upon the severity of the accident. For example, a dropped beaker that breaks and does not injure anyone may need a simple cleanup, disposal and replacement. However, a chemical splash into a person's eye will require at least 15 minutes of flushing with water followed by a medical evaluation and a more comprehensive investigation. The accident investigation process should be viewed to determine a root cause of the underlying accident. In some cases, the root cause may not be identified, or it cannot be addressed (such as with types of equipment failures). It is important for all people involved in an accident investigation to understand that the purpose is not to establish blame or responsibility, but the root cause (if any) so conditions that led to the accident can be changed to avoid future recurrences. 7.0 Activities 1. Choose one experiment in this laboratory manual that uses chemical reagents. Find a Safety Data Sheet for each reagent in the experiment. List the reagent, CAS number and GHS Warning Symbol for all reagents below: 0 1 2 3 Experiment Number Experiment Number Experiment Number Experiment Number Chemical CAS GHS Warning Symbol Notes NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN 2. Select one of the above reagents. Find a second SDS from an alternative supplier. Are they any significant differences between the two SDS documents? Would you expect any differences? Why or why not? 3. On the SDS provided in Appendix A, select an abbreviation or acronym. Find the full definition and share with your class what it is, what it means, and how should it be interpreted (e.g, LD50, IDLH, TWA, ACGIH, NIOSH, CFR, OSHA, etc.). 4. Draw a map of the laboratory classroom. Include the following: · Where your lab station is located · Exit/entrance doors · Sinks · Fume Hoods · Eyewash station · Safety Showers · Fire Extinguishers · Fire Alarm locations · Emergency notification equipment Appendix A: Safety Data Sheet for Concentrated Sulfuric Acid Appendix B: First Pages of Additional SDS for Sulfuric Acid (various dilutions)
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Topics_in_Thermodynamics_of_Solutions_and_Liquid_Mixtures/01%3A_Modules/1.22%3A_Volume/1.22.10%3A_Volume%3A_Aqueous_Binary_Liquid_Mixtures	For binary aqueous mixtures (at ambient pressure and fixed temperature) there are two interesting reference points. The molar volume of the pure liquid component 2, $\mathrm{V}_{2}^{}(\lambda)$. In the latter case we imagine that each molecule of liquid 2 is surrounded by an infinite expanse of water. With gradual increase in $\mathrm{x}_{2}$, so (on average) the molecules of liquid 2 move closer together. Typically Aqueous Mixtures For these systems $\left[\mathrm{V}_{2}^{\infty}(\operatorname{mix})-\mathrm{V}_{2}^{}(\lambda)\right]$ is negative. But this pattern is not unique to aqueous systems. The unique feature is the decrease in $\left[\mathrm{V}_{2}(\operatorname{mix})-\mathrm{V}_{2}^{}(\lambda)\right]$ with increase in $\mathrm{x}_{2}$ at low $\mathrm{x}_{2}$ [1]. In fact with increase in hydrophobicity of chemical substance 2, the decrease is more striking and the minimum in $\left[\mathrm{V}_{2}(\operatorname{mix})-\mathrm{V}_{2}^{}(\lambda)\right]$ occurs at lower $\mathrm{x}_{2}$. At mole fractions beyond $\mathrm{x}_{2}\left[\mathrm{~V}_{2}(\mathrm{mix})-\mathrm{V}_{2}^{}(\lambda)\right]$ increases with increase in $\mathrm{x}_{2}$. Many explanations have been offered for this complicated pattern. The following is one explanation. The negative $\left[\mathrm{V}_{2}^{\infty}(\operatorname{mix})-\mathrm{V}_{2}^{}(\lambda)\right]$ is accounted for in terms of a liquid clathrate in which part of the hydrophobic group ‘occupies’ a guest site in the liquid water ‘lattice’. The decrease in $\left[\mathrm{V}_{2}(\operatorname{mix})-\mathrm{V}_{2}^{}(\lambda)\right]$ is accounted for in terms of an increasing tendency towards a liquid clathrate hydrate structure. With increase in $\mathrm{x}_{2}$ there comes a point where there is insufficient water to construct the liquid clathrate host. Hence $\left[\mathrm{V}_{2}(\operatorname{mix})-\mathrm{V}_{2}^{}(\lambda)\right]$ increases [2,3]. Typically Non-Aqueous Systems Although $\left[\mathrm{V}_{2}^{\infty}(\operatorname{mix})-\mathrm{V}_{2}^{}(\lambda)\right]$ is negative no minimum is observed in $\left[\mathrm{V}_{2}(\operatorname{mix})-\mathrm{V}_{2}^{}(\lambda)\right]$. Footnotes [1] See for example; fluoroalcohols(aq); C. H. Rochester and J. R. Symonds, J. Fluorine Chem.,1974, 4 ,141. [2] F. Franks, Ann. N. Y. Acad. Sci.,1955, 125 ,277. [3] For many binary aqueous mixtures the patterns in volume related properties often identify transition points at ‘structurally interesting compositions’; G. Roux, D. Roberts, G. Perron and J. E. Desnoyers, J. Solution Chem.,1980, 9 ,629.
Courses/Alma_College/Organic_Chemistry_I_(Alma_College)/01%3A_Structure_and_Bonding/1.11%3A_Hybridization_of_Nitrogen_Oxygen_Phosphorus_and_Sulfur	Objective After completing this section, you should be able to apply the concept of hybridization of atoms such as N, O, P and S to explain the structures of simple species containing these atoms. Key Terms Make certain that you can define, and use in context, the key term below. lone pair electrons Study Notes Nitrogen is frequently found in organic compounds. As with carbon atoms, nitrogen atoms can be sp 3 -, sp 2 - or sp ‑hybridized. Note that, in this course, the term “lone pair” is used to describe an unshared pair of electrons. The valence-bond concept of orbital hybridization can be extrapolated to other atoms including nitrogen, oxygen, phosphorus, and sulfur. In other compounds, covalent bonds that are formed can be described using hybrid orbitals. Nitrogen Bonding in NH 3 The nitrogen in NH 3 has five valence electrons. After hybridization these five electrons are placed in the four equivalent sp 3 hybrid orbitals. The electron configuration of nitrogen now has one sp 3 hybrid orbital completely filled with two electrons and three sp 3 hybrid orbitals with one unpaired electron each. The two electrons in the filled sp 3 hybrid orbital are considered non-bonding because they are already paired. These electrons will be represented as a lone pair on the structure of NH 3 . The three unpaired electrons in the hybrid orbitals are considered bonding and will overlap with the s orbitals in hydrogen to form N-H sigma bonds. Note! This bonding configuration was predicted by the Lewis structure of NH 3 . The four sp 3 hybrid orbitals of nitrogen orientate themselves to form a tetrahedral geometry. The three N-H sigma bonds of NH 3 are formed by sp 3 (N)-1s(H) orbital overlap. The fourth sp 3 hybrid orbital contains the two electrons of the lone pair and is not directly involved in bonding. Methyl amine The nitrogen is sp 3 hybridized which means that it has four sp 3 hybrid orbitals. Two of the sp 3 hybridized orbitals overlap with s orbitals from hydrogens to form the two N-H sigma bonds. One of the sp 3 hybridized orbitals overlap with an sp 3 hybridized orbital from carbon to form the C-N sigma bond. The lone pair electrons on the nitrogen are contained in the last sp 3 hybridized orbital. Due to the sp 3 hybridization the nitrogen has a tetrahedral geometry. However, the H-N-H and H-N-C bonds angles are less than the typical 109.5 o due to compression by the lone pair electrons. Oxygen Bonding in H 2 O The oxygen in H 2 O has six valence electrons. After hybridization these six electrons are placed in the four equivalent sp 3 hybrid orbitals. The electron configuration of oxygen now has two sp 3 hybrid orbitals completely filled with two electrons and two sp 3 hybrid orbitals with one unpaired electron each. The filled sp 3 hybrid orbitals are considered non-bonding because they are already paired. These electrons will be represented as a two sets of lone pair on the structure of H 2 O . The two unpaired electrons in the hybrid orbitals are considered bonding and will overlap with the s orbitals in hydrogen to form O-H sigma bonds. Note! This bonding configuration was predicted by the Lewis structure of H 2 O. The four sp 3 hybrid orbitals of oxygen orientate themselves to form a tetrahedral geometry. The two O-H sigma bonds of H 2 O are formed by sp 3 (O)-1s(H) orbital overlap. The two remaining sp 3 hybrid orbitals each contain two electrons in the form of a lone pair. Methanol The oxygen is sp 3 hybridized which means that it has four sp 3 hybrid orbitals. One of the sp 3 hybridized orbitals overlap with s orbitals from a hydrogen to form the O-H sigma bonds. One of the sp 3 hybridized orbitals overlap with an sp 3 hybridized orbital from carbon to form the C-O sigma bond. Both the sets of lone pair electrons on the oxygen are contained in the remaining sp 3 hybridized orbital. Due to the sp 3 hybridization the oxygen has a tetrahedral geometry. However, the H-O-C bond angles are less than the typical 109.5 o due to compression by the lone pair electrons. Phosphorus Methyl phosphate The bond pattern of phosphorus is analogous to nitrogen because they are both in period 15. However, phosphorus can have have expanded octets because it is in the n = 3 row. Typically, phosphorus forms five covalent bonds. In biological molecules, phosphorus is usually found in organophosphates. Organophosphates are made up of a phosphorus atom bonded to four oxygens, with one of the oxygens also bonded to a carbon. In methyl phosphate, the phosphorus is sp 3 hybridized and the O-P-O bond angle varies from 110° to 112 o . Sulfur Methanethiol & Dimethyl Sulfide Sulfur has a bonding pattern similar to oxygen because they are both in period 16 of the periodic table. Because sulfur is positioned in the third row of the periodic table it has the ability to form an expanded octet and the ability to form more than the typical number of covalent bonds. In biological system, sulfur is typically found in molecules called thiols or sulfides. In a thiol, the sulfur atom is bonded to one hydrogen and one carbon and is analogous to an alcohol O-H bond. In a sulfide, the sulfur is bonded to two carbons. The simplest example of a thiol is methane thiol (CH 3 SH) and the simplest example of a sulfide is dimethyl sulfide [(CH 3 ) 3 S]. In both cases the sulfur is sp 3 hybridized, however the sulfur bond angles are much less than the typical tetrahedral 109.5 o being 96.6 o and 99.1 o respectively. methanethiol dimethyl sulfide Exercises 1) Insert the missing lone pairs of electrons in the following molecules, and tell what hybridization you expect for each of the indicated atoms. a) The oxygen is dimethyl ether: b) The nitrogen in dimethyl amine: c) The phosphorus in phosphine: d) The sulfur in hydrogen sulfide: Solutions 1) a) sp 3 hybridization b) sp 3 hybridization c) sp 3 hybridization d) sp 3 hybridization
Courses/Portland_Community_College/CH104%3A_Allied_Health_Chemisty_I_(2nd_Edition)/08%3A_Counting_Atoms_Ions_and_Molecules	8.1: Counting Atoms by the Gram In chemistry, it is impossible to deal with a single atom or molecule because we can't see them or count them or weigh them. Chemists have selected a number of particles with which to work that is convenient. Since molecules are extremely small, you may suspect this number is going to be very large and you are right. The number of particles in this group is Avagadro's number and the name of this group is the mole. 8.2: Counting Molecules by the Gram The molecular mass of a substance is the sum of the average masses of the atoms in one molecule of a substance. Calculations for formula mass and molecular mass are described. Calculations involving conversions between moles of a material and the mass of that material are described. Calculations are illustrated for conversions between mass and number of particles. 8.3: Chemical Formulas as Conversion Factors Using formulas to indicate how many atoms of each element we have in a substance, we can relate the number of moles of molecules to the number of moles of atoms. In any given formula the ratio of the number of moles of molecules (or formula units) to the number of moles of atoms can be used as a conversion factor. 8.4: Mass Percent Composition of Compounds Chemists often need to know what elements are present in a compound and in what percentage. The percent composition is the percent by mass of each element in a compound. 8.5: Mass Percent Composition from a Chemical Formula The percent composition of a compound can also be determined from the formula of the compound. The subscripts in the formula are first used to calculate the mass of each element in one mole of the compound. This is divided by the molar mass of the compound and multiplied by 100%.
Courses/Lumen_Learning/Book%3A_English_Composition_I-3_(Lumen)/05%3A_Reading%3A_Types_of_Reading_Material/05.4%3A_Characteristics_of_Texts%2C_Part_2	Read the following two passages about starting a business, noting what they have in common and where they differ. Passage #1: “Making a Profit” For-Profit Businesses An organization that aims to earn profit through its operations and is concerned with its own interests and not those of the public (non-profit) is known as a for-profit corporation. Structure A for-profit cooperation is usually an organization operating in the private sector that sets goals which eventually help the organization itself. This kind of a company often makes shares of ownership available to the general public. The purchasers of those shares then become the company’s shareholders; shareholders have bought a portion of ownership of the corporation by giving away a certain amount of money (differentiating from company to company) or assets of a particular value. Such organizations are usually not aided by the government, as they are working for private financial gains, unlike a non-profit organization, which exists to serve a mission. The nature of a for-profit corporation is such that it is required to pay applicable taxes and register with the state. Any donation they receive will also be subject to the tax policies of the concerned country. As these organizations are all corporations and have a separate identity from their owners the owners are not in their personal capacity required to satisfy any debts which the company might owe to anyone. Aims Unlike non-profit organizations, the policies of these organizations are usually profit oriented. Managers (corporate employees) here have a profit-oriented mindset and aim at maximizing the revenue of the firm, which in turn contributes to the profits of the shareholders/owners. Their aim can be accompanied by a goal of serving the society; however, that usually happens only in cases of specific corporations (B-corporations, which we’ll learn about later). Non-Profit Businesses Some organizations are not established solely for the purpose of making and retaining profit; however, they function in much the same way as a business. They establish goals and work to meet them in an effective, efficient manner. Thus, most of the business principles introduced in this text also apply to non-profits. Let’s take a look at some of the characteristics of the non-profit organization. A non-profit business, often referred to as an NPO (non-profit organization), is an organization that uses its surplus revenues to further achieve its purpose or mission, rather than distributing its surplus income to the organization’s directors (or equivalents) as profit or dividends. This is known as the distribution constraint. The decision to adopt a non-profit legal structure is one that will often have taxation implications, particularly where the non-profit seeks income tax exemption, charitable status and so on. Passage #2: “The Ant King: A California Fairy Tale” Sheila split open and the air was filled with gumballs. Yellow gumballs. This was awful for Stan, just awful. He had loved Sheila for a long time, fought for her heart, believed in their love until finally she had come around. They were about to kiss for the first time and then this: yellow gumballs. Stan went to a group to try to accept that Sheila was gone. It was a group for people whose unrequited love had ended in some kind of surrealist moment. There is a group for everything in California. After several months of hard work on himself with the group, Stan was ready to open a shop and sell the thousands of yellow gumballs. He did this because he believed in capitalism, he loved capitalism. He loved the dynamic surge and crash of Amazon’s stock price, he loved the great concrete malls spreading across America like blood staining through a handkerchief, he loved how everything could be tracked and mirrored in numbers. When he closed the store each night he would count the gumballs sold, and he would determine his gross revenue, his operating expenses, his operating margin; he would adjust his balance sheet and learn his debt-to-equity ratio; and after this exercise each night, Stan felt he understood himself and was at peace, and he could go home to his apartment and drink tea and sleep, without shooting himself or thinking about Sheila. On the night before the IPO of gumballs.com, Sheila came to Stan in a dream. She was standing in a kiddie pool; Stan and his brothers and sisters were running around splashing and screaming; she had managed to insert herself into a Super 8 home movie of Stan’s family, shot in the late seventies. She looked terribly sad. “Sheila, where are you?” Stan said. “Why did you leave me, why did you become gumballs?” “The Ant King has me,” Sheila said. “You must rescue me.” Stan woke up, he shaved, he put on his Armani suit, and drove his Lexus to his appointment with his venture capitalists and investment bankers. But the dream would not leave him. . . . “Gumballs are more than candy, isn’t that right, Stan?” said Monique, smiling broadly. Stan nodded. His feet were still wet, inside his argyle socks. “Yes, gumballs have a lot of, ah, a lot greater significance than just candy.” Monique paused and looked at Stan brightly, waiting for him to go on. Across the table, the three Credit Suisse First Boston underwriters—Emilio Toad, Harry Hornpecker, and Moby Pfister—sat stone-faced and unreacting in their gray double-breasted suits. Stan tried to remember the gumballs.com business plan. “They have hard shells,” he said. “People, ah, they want challenge . . . the hardness, the gumminess . . .” Monique broke in smoothly. Monique, all seven post-gender-reassignment-surgery feet of her; Monique, always dressed to the nines and tens; Monique was a Valley legend for her instincts, her suavity, her rapacious, exemplary greed. Stan had sold Monique on the idea of gumballs.com, and she had invested—found him the right contacts, the right team—and here they were at the Big Day, the Exit Strategy. “Stan!” she cried joyously, fixing him with a penetrating stare. “Don’t be shy! Tell them about how gumballs are sex! Tell them about our top-gun semiotics professors, tell them about gumballs as a cultural trope! You see,” she said, swooping onto Hornpecker, Pfister & Toad, “you can’t think of this as a candy thing, a food & bev thing, a consumer cyclic thing; no way, José! Think Pokémon. Think World Wide Wrestling. Think Star Wars !” “Could we get back to the numbers,” said Emilio Toad in a voice that sounded like a cat being liquefied in an industrial-strength mixer. The gray faces of Harry Hornpecker and Moby Pfister twitched in relief. Discussion Question 1 Which of these passages did you enjoy reading more? Why? [practice-area rows=”4″][/practice-area] [reveal-answer q=”498177″]Show Answer[/reveal-answer] [hidden-answer a=”498177″]The answer will vary depending on your personal preferences. Most readers are likely to prefer Passage 2, because of its use of narrative, and the style in which it’s written. If you’re interested in starting your own business, however, the practical knowledge of the first passage will be more appealing to you.[/hidden-answer] Question 2 What differences between the two passages stand out to you? Type these differences here. [practice-area rows=”6″][/practice-area] [reveal-answer q=”196546″]Show Answer[/reveal-answer] [hidden-answer a=”196546″] Some possible answers include Passage 1 Passage 2 Longer paragraphs shorter paragraphs fact-based content story-based content formal tone and vocabulary informal tone and vocabulary informative entertaining Did you note anything else? [/hidden-answer] Question 3 What similarities in the passages are apparent to you? Type these similarities here. [practice-area rows=”6″][/practice-area] [reveal-answer q=”310683″]Show Answer[/reveal-answer] [hidden-answer a=”310683″] Some possible answers include Both passages… share the topic of starting a new business (loosely, admittedly) include facts Did you note anything else? [/hidden-answer] Question 4 Both passages here are excerpts from longer works. Which of these would you want to keep reading to gain a deeper understanding of business practices? learn about one of the more unique business start-ups in a fictional setting? be entertained? be well-informed? [reveal-answer q=”369900″]Show Answer[/reveal-answer] [hidden-answer a=”369900″] Passage #1 Passage #2 Passage #2 Passage #1 [/hidden-answer] CC licensed content, Original Text: Characteristics of Texts, Part 2. Provided by : Lumen Learning. License : CC BY: Attribution CC licensed content, Shared previously Revision and adaptation. Authored by : Linda Williams and Lumen Learning. Provided by : Tidewater Community College. Located at : https://courses.candelalearning.com/masterybusiness2xngcxmasterfall2015/chapter/reading-making-a-profit/ . Project : Introduction to Business. License : CC BY-NC-SA: Attribution-NonCommercial-ShareAlike For-profit corporation. Provided by : Wikipedia. Located at : http://en.wikipedia.org/wiki/For-profit_corporation . License : CC BY-SA: Attribution-ShareAlike Nonprofit Organization. Provided by : Wikipedia. Located at : http://en.wikipedia.org/wiki/Nonprofit_organization . License : CC BY-NC-SA: Attribution-NonCommercial-ShareAlike The Ant King: A California Fairy Tale. Authored by : Benjamin Rosenbaum. Provided by : Small Beer Press. Located at : http://smallbeerpress.com/creative-commons/ . License : CC BY-NC-SA: Attribution-NonCommercial-ShareAlike
Courses/Fullerton_College/Introductory_Chemistry_for_Allied_Health_(Chan)/06%3A_Introduction_to_Chemical_Reactions/6.05%3A_Types_of_Chemical_Reactions_-_Single_and_Double_Replacement_Reactions	Learning Objectives Recognize chemical reactions as single-replacement reactions and double-replacement reactions. Use the periodic table, an activity series, or solubility rules to predict whether single-replacement reactions or double-replacement reactions will occur. Up until now, we have presented chemical reactions as a topic, but we have not discussed how the products of a chemical reaction can be predicted. Here we will begin our study of certain types of chemical reactions that allow us to predict what the products of the reaction will be. A single-replacement reaction is a chemical reaction in which one element is substituted for another element in a compound, generating a new element and a new compound as products. Presented below: \[\ce{2HCl(aq) + Zn(s) → ZnCl2(aq) + H2(g)}\nonumber \] is an example of a single-replacement reaction. The hydrogen atoms in $\ce{HCl}$ are replaced by $\ce{Zn}$ atoms, and in the process a new element—hydrogen—is formed. Another example of a single-replacement reaction is \[\ce{2NaCl(aq) + F2(g) → 2NaF(s) + Cl2(g)}\nonumber \] Here the negatively charged ion changes from chloride to fluoride. A typical characteristic of a single-replacement reaction is that there is one element as a reactant and another element as a product. A double-replacement reaction occurs when parts of two ionic compounds are exchanged, making two new compounds. A characteristic of a double-replacement equation is that there are two compounds as reactants and two different compounds as products. An example is \[\ce{CuCl2(aq) + 2AgNO3(aq) → Cu(NO3)2(aq) + 2AgCl(s)}\nonumber \] There are two equivalent ways of considering a double-replacement equation: either the cations are swapped, or the anions are swapped. (You cannot swap both; you would end up with the same substances you started with.) Either perspective should allow you to predict the proper products, as long as you pair a cation with an anion, and not a cation with a cation or an anion with an anion. Example $\PageIndex{3}$ Predict the products of this double-replacement equation: \[\ce{BaCl2 + Na2SO4 → }\nonumber \] Solution Thinking about the reaction as either switching the cations or switching the anions, we would expect the products to be BaSO 4 and NaCl. Exercise $\PageIndex{3}$ Predict the products of this double-replacement equation: \[\ce{KBr + AgNO3 → }\nonumber \] Answer KNO 3 and AgBr Predicting whether a double-replacement reaction occurs is somewhat more difficult than predicting a single-replacement reaction. However, there is one type of double-replacement reaction that we can predict: the precipitation reaction. A precipitation reaction occurs when two ionic compounds are dissolved in water and form a new ionic compound that does not dissolve; this new compound falls out of solution as a solid precipitate. The formation of a solid precipitate is the driving force that makes the reaction proceed. Key Takeaways A single-replacement reaction replaces one element for another in a compound. A double-replacement reaction exchanges the cations (or the anions) of two ionic compounds. A precipitation reaction is a double-replacement reaction in which one product is a solid precipitate.
Courses/Lumen_Learning/Book%3A_Western_Civilization_II_(Lumen)/02%3A_Week_1%3A_Welcome/02.3%3A_Assignment	Week 1 Assignment Thoroughly read the read me first section to become familiar with the course expectations. Introduction Discussion Points: 20 Check the due dates carefully. All work is due at 11:55 P.M. on the due date. Due dates are subject to change at the discretion of the instructor who will inform the students via the News on the course Homepage. CC licensed content, Original Week 1 Assignment. Provided by : Julianna Wilson, Pima Community College. Located at : http://cc.pima.edu/~lumen/his102/Weeks/week-1-2.html . License : CC BY: Attribution Introduction Discussion. Provided by : Pima Community College. License : CC BY: Attribution
Courses/BridgeValley_Community_and_Technical_College/Fundamentals_of_Chemistry/05%3A_Atoms_and_Molecules/5.03%3A_The_Nuclear_Atom	Learning Objectives Explain the observations that led to Thomson's discovery of the electron. Describe Thomson's "plum pudding" model of the atom and the evidence for it. Draw a diagram of Thomson's "plum pudding" model of the atom and explain why it has this name. Describe Rutherford's gold foil experiment and explain how this experiment altered the "plum pudding" model. Draw a diagram of the Rutherford model of the atom and label the nucleus and the electron cloud. Dalton's Atomic Theory held up well to a lot of the different chemical experiments that scientists performed to test it. In fact, for almost 100 years, it seemed as if Dalton's Atomic Theory was the whole truth. However, in 1897, a scientist named J. J. Thomson conducted some research that suggested that Dalton's Atomic Theory was not the entire story. He suggested that the small, negatively charged particles making up the cathode ray were actually pieces of atoms. He called these pieces "corpuscles," although today we know them as electrons . Thanks to his clever experiments and careful reasoning, J. J. Thomson is credited with the discovery of the electron. Electrons and Plums The electron was discovered by J. J. Thomson in 1897. The existence of protons was also known, as was the fact that atoms were neutral in charge. Since the intact atom had no net charge and the electron and proton had opposite charges, the next step after the discovery of subatomic particles was to figure out how these particles were arranged in the atom. This was a difficult task because of the incredibly small size of the atom. Therefore, scientists set out to design a model of what they believed the atom could look like. The goal of each atomic model was to accurately represent all of the experimental evidence about atoms in the simplest way possible. Following the discovery of the electron, J.J. Thomson developed what became known as the " plum pudding " model in 1904. Plum pudding is an English dessert similar to a blueberry muffin. In Thomson's plum pudding model of the atom, the electrons were embedded in a uniform sphere of positive charge like blueberries stuck into a muffin. The positive matter was thought to be jelly-like or similar to a thick soup. The electrons were somewhat mobile. As they got closer to the outer portion of the atom, the positive charge in the region was greater than the neighboring negative charges, and the electron would be pulled back more toward the center region of the atom. However, this model of the atom soon gave way to a new model developed by New Zealander Ernest Rutherford (1871-1937) about five years later. Thomson did still receive many honors during his lifetime, including being awarded the Nobel Prize in Physics in 1906 and a knighthood in 1908. Atoms and Gold In 1911, Rutherford and coworkers Hans Geiger and Ernest Marsden initiated a series of groundbreaking experiments that would completely change the accepted model of the atom. They bombarded very thin sheets of gold foil with fast moving alpha particles . Alpha particles, a type of natural radioactive particle, are positively charged particles with a mass about four times that of a hydrogen atom. According to the accepted atomic model, in which an atom's mass and charge are uniformly distributed throughout the atom, the scientists expected that all of the alpha particles would pass through the gold foil with only a slight deflection or none at all. Surprisingly, while most of the alpha particles were indeed not deflected, a very small percentage (about 1 in 8000 particles) bounced off the gold foil at very large angles. Some were even redirected back toward the source. No prior knowledge had prepared them for this discovery. In a famous quote, Rutherford exclaimed that it was "as if you had fired a 15-inch [artillery] shell at a piece of tissue and it came back and hit you." Rutherford needed to come up with an entirely new model of the atom in order to explain his results. Because the vast majority of the alpha particles had passed through the gold, he reasoned that most of the atom was empty space. In contrast, the particles that were highly deflected must have experienced a tremendously powerful force within the atom. He concluded that all of the positive charge and the majority of the mass of the atom must be concentrated in a very small space in the atom's interior, which he called the nucleus. The nucleus is the tiny, dense, central core of the atom and is composed of protons and neutrons. Rutherford's atomic model became known as the nuclear model . In the nuclear atom, the protons and neutrons, which comprise nearly all of the mass of the atom, are located in the nucleus at the center of the atom. The electrons are distributed around the nucleus and occupy most of the volume of the atom. It is worth emphasizing just how small the nucleus is compared to the rest of the atom. If we could blow up an atom to be the size of a large professional football stadium, the nucleus would be about the size of a marble. Rutherford's model proved to be an important step towards a full understanding of the atom. However, it did not completely address the nature of the electrons and the way in which they occupy the vast space around the nucleus. It was not until some years later that a full understanding of the electron was achieved. This proved to be the key to understanding the chemical properties of elements. Atomic Nucleus The nucleus (plural, nuclei) is a positively charged region at the center of the atom. It consists of two types of subatomic particles packed tightly together. The particles are protons, which have a positive electric charge, and neutrons, which are neutral in electric charge. Outside of the nucleus, an atom is mostly empty space, with orbiting negative particles called electrons whizzing through it. The figure below shows these parts of the atom. The nucleus of the atom is extremely small. Its radius is only about 1/100,000 of the total radius of the atom. Electrons have virtually no mass, but protons and neutrons have a lot of mass for their size. As a result, the nucleus has virtually all the mass of an atom. Given its great mass and tiny size, the nucleus is very dense. If an object the size of a penny had the same density as the nucleus of an atom, its mass would be greater than 30 million tons! Holding it all Together Particles with opposite electric charges attract each other. This explains why negative electrons orbit the positive nucleus. Particles with the same electric charge repel each other. This means that the positive protons in the nucleus push apart from one another. So why doesn't the nucleus fly apart? An even stronger force—called the strong nuclear force —holds protons and neutrons together in the nucleus. Summary Atoms are the ultimate building blocks of all matter. The modern atomic theory establishes the concepts of atoms and how they compose matter. Bombardment of gold foil with alpha particles showed that some particles were deflected. The nuclear model of the atom consists of a small and dense positively charged interior surrounded by a cloud of electrons.
Courses/Lebanon_Valley_College/CHM_312%3A_Physical_Chemistry_II_(Lebanon_Valley_College)/05%3A_Single_Component_Phase_Equilibrium/5.06%3A_Gibbs_Phase_Rule	In chapter 1 , we have already seen that the number of independent variables required to describe an ideal gas is two. This number was derived by counting the total number of variables $(3: P,\overline{V},T)$, and reduce it by one because the ideal gas law constrains the value of one of them, once the other two are fixed. For a generic system potentially containing more than one chemical substance in several different phases, however, the number of independent variables can be different than two. For a system composed of $c$ components (chemical substances) and $p$ phases, the number of independent variables, $f$, is given by the Gibbs phase rule: \[ f=c-p+2. \label{12.2.1} \] The Gibbs phase rule derives from the fact that different phases are in equilibrium with each other at some conditions, resulting in the reduction of the number of independent variables at those conditions. More rigorously, when two phases are in thermodynamic equilibrium, their chemical potentials are equal (see Equation 12.1.4 ). For each equality, the number of independent variables—also called the number of degrees of freedom —is reduced by one. For example, the chemical potentials of the liquid and its vapor depend on both $T$ and $P$. But when these phases are in equilibrium with each other, their chemical potentials must be equal. If either the pressure or the temperature is fixed, the other variable will be uniquely determined by the equality relation. In other terms, when a liquid is in equilibrium with its vapor at a given pressure, the temperature is determined by the fact that the chemical potentials of the two phases is the same, and is denoted as the boiling temperature $T_{\text{b}}$. Similarly, at a given temperature, the pressure of the vapor is uniquely determined by the same equality relation and is denoted as the vapor pressure, $P^$. The Gibbs phase rule is obtained considering that the number of independent variables is given by the total number of variables minus the constraints. The total number of variables is given by temperature, pressure, plus all the variables required to describe each of the phases. The composition of each phase is determined by $(c-1)$ variables.$^1$ The number of constraints is determined by the number of possible equilibrium relations, which is $c(p-1)$ since the chemical potential of each component must be equal in all phases. The number of degrees of freedom $f$ is then given by \[\begin{align} f &=(c-1)p+2-c(p-1) \\[4pt] &=c-p+2 \end{align*} \] which is the Gibbs phase rule, as in Equation \ref{12.2.1}. For a 1-component system $c-1=1-1=0$, and no additional variable is required to determine the composition of each phase. For a 2-component system, however, each phase will contain both components, hence $c-1=2-1=1$ additional variable will be required to describe it–the mole fraction.
Courses/Northern_Michigan_University/CH_215%3A_Chemistry_of_the_Elements_Fall_2023/06%3A_Transition_Metals_and_Coordination_Chemistry/6.17%3A_Complex_Ion_Chemistry/6.17.04%3A_Reactions_of_the_Hexaaqua_Ions_with_Ammonia	This page describes and explains the reactions between complex ions of the type [M(H 2 O) 6 ] n+ and ammonia solution.Reactions of the hexaaqua ions with ammonia solution are complicated by the fact that the ammonia can have two quite different functions. It can act as a base (in the Brønsted-Lowry sense), but it is also a possible ligand which can replace water molecules around the central metal ion. When it acts as a ligand, it is acting as a Lewis base. We need to look at these two functions separately. Ammonia acting as a (Brønsted-Lowry) base This is what happens when you only add small amounts of dilute ammonia solution to any of the hexaaqua ions. The ligand effect only happens with an excess of ammonia or with concentrated ammonia - and with some metals you don't even see it then. We'll talk through what happens if you add a small amount of dilute ammonia solution to a solution containing a 2+ hexaaqua ion. These have the formula [M(H 2 O) 6 ] 2+ , and they are acidic. Their acidity is shown in the reaction of the hexaaqua ions with water molecules from the solution: \[ \ce{ [M(H2O)6]^{2+} (aq) + H2O <=> [M(H2O)5(OH)]^{+} (aq) + H3O^{+} (aq) } \nonumber \] They are acting as acids by donating hydrogen ions to water molecules in the solution. Because of the confusing presence of water from two different sources (the ligands and the solution), it is easier to simplify this: \[ \ce{ [M(H2O)6]^{2+} (aq) <=> [M(H2O)5(OH)]^{+} (aq) + H^{+} (aq) } \nonumber \] Adding ammonia solution to this equilibrium - stage 1 There are two possible reactions. Reaction of ammonia with the hydroxonium ions (hydrogen ions) Ammonia will react with these to produce ammonium ions. According to Le Chatelier's Principle, the position of equilibrium will move to the right, producing more of the new complex ion. Reaction of ammonia with the hexaaqua ion Statistically, there is far more chance of an ammonia molecule hitting a hexaaqua metal ion than of hitting a hydrogen ion. There are far more hexaaqua ions present. If that happens, you get exactly the same new complex ion formed as above. \[ \ce{ [M(H2O)6]^{2+} (aq) + NH3 <=> [M(H2O)5(OH)]^{+} (aq) + NH4^{+} (aq) } \nonumber \] Notice that this is still a reversible change (unlike the corresponding change when you add hydroxide ions). Ammonia is only a weak base. The second stage of the reaction Whichever of the above reactions happens, you end up with [M(H 2 O) 5 (OH)] + ions in solution. These are also acidic, and can lose hydrogen ions from another of the water ligands. Taking the easier version of the equilibrium: \[ \ce{ [M(H2O)5(OH)]^{+} (aq) <=> [M(H2O)4(OH)2] (s) + H^{+} (aq) } \nonumber \] Adding ammonia again tips the equilibrium to the right - either by reacting with the hydrogen ions, or by reacting directly with the complex on the left-hand side. When this happens, the new complex formed no longer has a charge - this is a "neutral complex". It is insoluble in water - and so a precipitate is formed. This precipitate is often written without including the remaining water ligands. In other words we write it as M(OH) 2 . A precipitate of the metal hydroxide has been formed. Summarizing what has happened You can also usefully write the complete change as an overall equilibrium reaction. This will be important for later on. \[ \ce{ [M(H2O)6]^{2+} (aq) + 2NH3 <=> [M(H2O)4(OH)2] (s) + 2NH4^{+} (aq) } \nonumber \] If you did the same reaction with a 3+ ion, the only difference is that you would have to remove a total of 3 hydrogen ions in order to get to the neutral complex. That would give the overall equation: \[ \ce{ [M(H2O)6]^{3+} (aq) + 3NH3 <=> [M(H2O)3(OH)3] (s) + 4NH4^{+} (aq) } \nonumber \] Ions of specific metals Remember that we are concentrating for the moment on the ammonia acting as a base - in other words, on the formation of hydroxide precipitates when you add small amounts of ammonia solution to solutions containing hexaaqua metal The diagrams, however, will show the complete change so I don't have to repeat them later on. Ignore the cases where the precipitate dissolves in excess ammonia for the moment. 2+ ions hexaaquacobalt(II) hexaaquacopper(II) hexaaquairon(II) Iron is very easily oxidized under alkaline conditions. Oxygen in the air oxidizes the iron(II) hydroxide precipitate to iron(III) hydroxide especially around the top of the tube. The darkening of the precipitate comes from the same effect. This is NOT a ligand exchange reaction. hexaaquamanganese(II) I have shown the original solution as very pale pink (the palest I can produce!), but in fact it is virtually colorless. The pale brown precipitate is oxidized to darker brown manganese(III) oxide in contact with oxygen from the air. Again, this isn't a ligand exchange reaction. hexaaquanickel(II) hexaaquazinc You start and finish with colourless solutions, producing a white precipitate on the way. 3+ ions hexaaquaaluminium Starting from a colorless solution, you get a white precipitate. hexaaquachromium(III) hexaaquairon(III) Summary of the effect of adding small amounts of ammonia solution In each case you get a precipitate of the neutral complex - the metal hydroxide. Apart from minor differences in the exact shade of color you get, these are almost all exactly the same as the precipitates you get when you add a little sodium hydroxide solution to the solutions of the hexaaqua ions. The only real difference lies in the color of the cobalt precipitate. Ammonia acting as a ligand The ligand exchange reaction In some cases, ammonia replaces water around the central metal ion to give another soluble complex. This is known as a ligand exchange reaction, and involves an equilibrium such as this one: \[ [Cu(H_2O)_6]^{2+} +4NH_3 \rightleftharpoons [Cu(NH_3)_4(H_2O)_2]^{2+} + 4H_2O \nonumber \] The formation of this new soluble complex causes the precipitate to dissolve. The ammonia attaches to the central metal ion using the lone pair of electrons on the nitrogen atom. Because it is a lone pair donor, it is acting as a Lewis base. Explaining why the precipitate dissolves Almost all text books leave the argument at this point, assuming that it is obvious why the formation of the complex causes some precipitates to dissolve - it is not! Consider the copper case as typical of any of them. There are two equilibria involved in this. The first is the one in which ammonia is acting as a base and producing the precipitate: \[ [Cu(H_2O)_6]^{2+} + 2NH_3 \rightleftharpoons [Cu(H_2O)_4(OH)_2] + 2NH_4^+ \nonumber \] The other one is the ligand exchange reaction: \[ [Cu(H_2O)_6]^{2+} + 4NH_3 \rightleftharpoons [Cu(NH_3)_4(H_2O)_2]^{2+} + 4H_2O \nonumber \] Notice that the hexaaqua ion appears in both of these. There is now an interaction between the two equilibria: Looking at it like this is helpful in explaining why some precipitates dissolve in excess ammonia while others don't. It depends on the positions of the equilibria. To get the precipitate to dissolve, you obviously need the ligand exchange equilibrium to lie well to the right, but you need the acid-base equilibrium to be easy to pull back to the left.
Courses/Pacific_Union_College/Quantum_Chemistry/05%3A_The_Harmonic_Oscillator_and_the_Rigid_Rotor/5.01%3A_A_Harmonic_Oscillator_Obeys_Hooke's_Law	The motion of two atoms in a diatomic molecule can be separated into translational, vibrational, and rotational motions. Both rotation and vibrational motions are internal motions that do not change the center of mass for the molecule (Figure $\PageIndex{1}$), which is described by translational motion. Quantum translational motions can be modeled with the particle in a box model discussed previously and rotation and vibration can be modeled via the rigid rotor and harmonic oscillator models, respectively. Before delving into the quantum mechanical harmonic oscillator, we will introduce the classical harmonic oscillator (i.e., involving classical mechanics) to build an intuition that we will extend to the quantum world. A classical description of the vibration of a diatomic molecule is needed because the quantum mechanical description begins with replacing the classical energy with the Hamiltonian operator in the Schrödinger equation. It also is interesting to compare and contrast the classical description with the quantum mechanical picture. The Classical Harmonic Oscillator Simple harmonic oscillators about a potential energy minimum can be thought of as a ball rolling frictionlessly in a curved dish or a pendulum swinging frictionlessly back and forth (Figure $\PageIndex{2}$). The restoring forces are precisely the same in either horizontal direction. If we consider the bond to behave like a mass on a spring (Figure $\PageIndex{2}$), then this restoring force ($F$) is proportional to the displacement ($x$) from the equilibrium length ($x_o$) - this is Hooke's Law: \[ F = - kx \label {5.1.2}\] where $k$ is the force constant. Hooke's Law says that the force is proportional to, but in opposite direction to, the displacement ($x$). The force constant reflects the stiffness of the spring. The idea incorporated into the application of Hooke's Law to a diatomic molecule is that when the atoms move away from their equilibrium positions, a restoring force is produced that increases proportionally with the displacement from equilibrium. The potential energy for such a system increases quadratically with the displacement. \[ V (x) = \dfrac {1}{2} k x^2 \label {5.1.3}\] Hooke's Law or the harmonic (i.e. quadratic) potential given by Equation $\ref{5.1.3}$ is an excellent approximation for the vibrational oscillations of molecules. The magnitude of the force constant $k$ depends upon the nature of the chemical bond in molecular systems just as it depends on the nature of the spring in mechanical systems. The larger the force constant, the stiffer the spring or the stiffer the bond. Since it is the electron distribution between the two positively charged nuclei that holds them together, a double bond with more electrons has a larger force constant than a single bond, and the nuclei are held together more tightly. A stiff bond with a large force constant is not necessarily a strong bond with a large dissociation energy. A harmonic oscillator has no dissociation energy since it CANNOT be broken - there is always a restoring force to keep the molecule together. This is one flaw in the model in describing real vibrating diatomics. Two atoms or one? You may have questioned the applicability of the harmonic oscillator model involving one moving mass bound to a fix wall via a spring like in Figure $\PageIndex{2}$ for the vibration of a diatomic molecule with two moving masses like in Figure $\PageIndex{1}$. It turned out the two are mathematically the same with internal vibration motion described by a single reduced particle with a reduced mass $μ$. For a diatomic molecule, Figure $\PageIndex{3}$, the vector $\vec{r}$ corresponds to the internuclear axis. The magnitude or length of $r$ is the bond length, and the orientation of $r$ in space gives the orientation of the internuclear axis in space. Changes in the orientation correspond to rotation of the molecule, and changes in the length correspond to vibration. The change in the bond length from the equilibrium bond length is the vibrational coordinate for a diatomic molecule. Example $\PageIndex{1}$ Show that minus the first derivative of the harmonic potential energy function in Equation $\ref{5.1.3}$ with respect to $x$ is the Hooke's Law force. Show that the second derivative is the force constant, $k$. At what value of $x$ is the potential energy a minimum; at what value of $x$ is the force zero? Sketch graphs to compare the potential energy and the force for a system with a large force constant to one with a small force constant. Solution a. Hooke's Law for a spring entails that the force applied on a spring $F$ is equal to the force constant, $-k$ times the distance compressed or stretched, $x$ (Equation \ref{5.1.2}). The derivative of $V(x) = 0.5 k x^2$ is \[V'(x) = (2)\left(\dfrac{1}{2}\right)kx = kx. \nonumber\] The negative of this is $-V'(x) = -kx$ which is exactly equal to Hooke's Law. b. The second derivative \[V"(x) = \dfrac{d}{dx} kx = k \nonumber\] Thus, the second derivative of this equation for potential energy is equal to the force constant, $k$. c. To find the minimum potential energy, it is easiest to set the first derivative equal to zero and solve for x. When $V'(x) = kx = 0$ then x must be equal to zero. Thus, the minimum potential energy is when x=0. Plugging this into Hooke's Law, $F(0) = -k(0) = 0$ so this is also the value for x when the force is zero. d. The force constant has a drastic effect on both the potential energy and the force. A system with a large force constant requires minimal change in $x$ to have a drastic change in potential energy or force, whereas a system with a small force constant is the exact opposite phenomenon. Solving the Harmonic Oscillator Model The classical equation of motion for a one-dimensional simple harmonic oscillator with a particle of mass $m$ attached to a spring having spring constant $k$ is \[ m \dfrac{d^2x(t)}{dt^2} = -kx(t) \label{5.1.4a}\] which can be written in the standard wave equation form: \[ \dfrac{d^2x(t)}{dt^2} + \dfrac{k}{m}x(t) = 0 \label{5.1.4b}\] Equation $\ref{5.1.4a}$ is a linear second-order differential equation that can be solved by the standard method of factoring and integrating. The resulting solution to Equation $\ref{5.1.4a}$ is \[ x(t) = x_o \sin (\omega t + \phi) \label{5.1.5}\] with \[\omega = \sqrt{\dfrac{k}{m}} \label{5.1.6}\] and the momentum has time dependence \[\begin{align} p &= mv \\[4pt] &=mx_o \omega \cos (\omega t + \phi) \label{5.1.7} \end{align}\] Figure $\PageIndex{4}$ show the displacement of the bond from its equilibrium length as a function of time. Such motion is called harmonic . Example $\PageIndex{2}$ Substitute the following functions into Equation $\ref{5.1.4b}$ to demonstrate that they are both possible solutions to the classical equation of motion. $x(t) = x_0 e^{i \omega t} $ $x(t) = x_0 e^{-i \omega t}$ where \[ \omega = \sqrt {\dfrac {k}{m}} \nonumber\] Note that the Greek symbol $\omega$ for frequency represents the angular frequency $2π\nu$. Solution a This requires simply placing the given function $x(t) = x_0 e^{i \omega t} $ into Equation $\ref{5.1.4b}$. \[ \begin{align} \frac{d^2 x(t) }{dt^2} + \frac{k}{m} x(t) &= 0 \\[4pt] \frac{d^2 }{dt^2} \left( x_o e^{i \omega t} \right)+ \frac{k}{m}x_o e^{i \omega t} &= 0\\[4pt] x_o \frac{d^2 }{dt^2} \left( e^{i \omega t} \right)+ \frac{k}{m}x_o e^{i \omega t} &= 0 \\[4pt] x_o i^2 \omega^2 \left( e^{i \omega t} \right)+ \frac{k}{m}x_o e^{i \omega t} &= 0 \\[4pt] x_o i^2 \omega^2 e^{i \omega t} + \frac{k}{m}x_o e^{i \omega t} &= 0 \\[4pt] x_o i^2 \omega^2 + \frac{k}{m}x_o &= 0 \\[4pt] -x_o \frac{k}{m}+ \frac{k}{m}x_o = 0 \; \textrm{ with } \; \omega &= \sqrt{\frac{k}{m}} \end{align}\] Solution b This requires simply placing the given function $x(t) = x_0 e^{-i \omega t}$ into Equation $\ref{5.1.4b}$. \[ \begin{align} \frac{d^2 x(t) }{dt^2} + \frac{k}{m} x(t) &= 0 \\[4pt] \frac{d^2 }{dt^2} \left( x_o e^{-i \omega t} \right)+ \frac{k}{m}x_o e^{-i \omega t}& = 0 \\[4pt] x_o \frac{d^2 }{dt^2} \left( e^{-i \omega t} \right)+ \frac{k}{m}x_o e^{-i \omega t} &= 0 \\[4pt] x_o i^2 \omega^2 \left( e^{-i \omega t} \right)+ \frac{k}{m}x_o e^{-i \omega t} &= 0 \\[4pt] x_o i^2 \omega^2 e^{-i \omega t} + \frac{k}{m}x_o e^{-i \omega t}& = 0\\[4pt] x_o i^2 \omega^2 + \frac{k}{m}x_o &= 0 \\[4pt] -x_o \frac{k}{m}+ \frac{k}{m}x_o = 0 \; \textrm{ with } \; \omega &= \sqrt{\frac{k}{m}} \end{align}\] Example $\PageIndex{3}$ Show that sine and cosine functions also are solutions to Equation $\ref{5.1.4b}$. Answer Using Equation $\ref{5.1.4b}$ \begin{align} \frac{d^2x(t)}{dt^2} +\frac{k}{m}x(t)=0\\ w=(\frac{k}{m})^{1/2}\\ \end{align} For \begin{align} x(t)=x_{o}sin(wt+\phi) \end{align} Take the second derivative of $x(t)$ \begin{align} \frac{d^2x(t)}{dt}=-w^2x_{o}sin(wt+phi)\\ -w^2x_{o}sin(wt+phi)=-\frac{k}{m}x_{o}sin((\frac{k}{m})^{1/2}t+phi)\\ \end{align} Plug in $x(t)$ and the second derivative of $x(t)$ into Equation $\ref{5.1.4b}$ \begin{align} -\frac{k}{m}x_{o}sin((\frac{k}{m})^{1/2}t+phi)+\frac{k}{m}x_{o}sin((\frac{k}{m})^{1/2}t+phi)=0\\ \end{align} The sine equation is a solution to eEquation $\ref{5.1.4b}$ For $x(t)=x_{o}\cos(wt+\phi) $ Take the second derivative of $x(t)$ \begin{align} \frac{d^2x(t)}{dt}=-w^2x_{o}cos(wt+phi)\\ -w^2x_{o}cos(wt+phi)=-\frac{k}{m}x_{o}cos((\frac{k}{m})^{1/2}t+phi)\\ \end{align} Plug in x(t) and the second derivative of x(t) into Equation $\ref{5.1.4b}$ \begin{align} -\frac{k}{m}x_{o}cos((\frac{k}{m})^{1/2}t+phi)+\frac{k}{m}x_{o}cos((\frac{k}{m})^{1/2}t+phi)=0 \end{align} The cosine equation is a solution to Equation $\ref{5.1.4b}$ Example $\PageIndex{4}$ Identify what happens to the frequency of the motion as the force constant increases in one case and as the mass increases in another case. If the force constant is increased 9-fold and the mass is increased by 4-fold, by what factor does the frequency change? Answer This is a simple application of Equation \ref{5.1.6}. As the force constant increases, the frequency of the motion increases, while as the mass increases, the frequency of the motion decreases. If the force constant increased 9-fold and the mass increased 4-fold, \[ω=\sqrt{\dfrac{9k}{4m}}= \dfrac{3}{2} \left(\dfrac{k}{m}\right) \nonumber\] The entire frequency of motion would increase by a factor of 3/2. Harmonic Oscillator Energies The energy of the vibration is the sum of the kinetic energy and the potential energy. The momentum associated with the harmonic oscillator is \[p = m \dfrac {dx}{dt} \label {5.1.8}\] so combining Equations \ref{5.1.8} and \ref{5.1.3}, the total energy can be written as \[ \begin{align} E &= T + V \\[4pt] &= \dfrac {p^2}{2 m} + \dfrac {k}{2} x^2 \label {5.1.9} \end{align}\] The total energy of the harmonic oscillator is equal to the maximum potential energy stored in the spring when $x = \pm A$, called the turning points ( Figure $\PageIndex{5}$). The total energy (Equation $\ref{5.1.9}$) is continuously being shifted between potential energy stored in the spring and kinetic energy of the mass. The motion of a classical oscillator is confined to the region where its kinetic energy is nonnegative, which is what the energy relation Equation \ref{5.1.9} says. Physically, it means that a classical oscillator can never be found beyond its turning points, and its energy depends only on how far the turning points are from its equilibrium position. The energy of a classical oscillator changes in a continuous way. The lowest energy that a classical oscillator may have is zero, which corresponds to a situation where an object is at rest at its equilibrium position. The zero-energy state of a classical oscillator simply means no oscillations and no motion at all (a classical particle sitting at the bottom of the potential well in Figure $\PageIndex{5}$). When an object oscillates, no matter how big or small its energy may be, it spends the longest time near the turning points, because this is where it slows down and reverses its direction of motion. Therefore, the probability of finding a classical oscillator between the turning points is highest near the turning points and lowest at the equilibrium position. (Note that this is not a statement of preference of the object to go to lower energy. It is a statement about how quickly the object moves through various regions.) Example $\PageIndex{5}$ What happens to the frequency of the oscillation as the vibration is excited with more and more energy? What happens to the maximum amplitude of the vibration as it is excited with more and more energy? Solution Dec 25, 2019, 7:48 PM a. Frequency The energy of the harmonic oscillator can be written as \[E_{v}=h v\left(v+\dfrac{1}{2}\right)\] We can see from the formula of energy that frequency is proportional to energy, so an increase in energy would mean higher frequency. b. Amplitude The kinetic and potential terms for energy of the harmonic oscillator can be written as \[\begin{aligned}E=K+V=\frac{1}{2} m \omega^{2} A^{2} \sin ^{2} \omega t+\frac{1}{2} k A^{2} \cos ^{2} \omega t \\\omega=\sqrt{\frac{k}{m}} \\E=\frac{1}{2} k A^{2}\left(\sin ^{2} \omega t+\cos ^{2} \omega t\right)=\frac{1}{2} k A^{2}\end{aligned}\] The maximum amplitude of the vibration will increase as the energy increases. Here, amplitude increases both with energy and with frequency, giving both frequency and amplitude a direct relationship with energy of vibration. Example $\PageIndex{6}$ If a molecular vibration is excited by collision with another molecule and is given a total energy $E_{hit}$ as a result, what is the maximum amplitude of the oscillation? Is there any constraint on the magnitude of energy that can be introduced? Answer The equation that defines the energy of a molecular vibration can be approximated is: \[E_{h i t}=T+V=\frac{p^{2}}{2 m}+\frac{k}{2} x\] The maximum amplitude of a harmonic oscillator is equal to x when the kinetic energy term of total energy equals zero \[E_{hit}=\frac{k}{2}x\] Solving for x gives the maximum amplitude: \[x=\sqrt{\frac{2}{k} E_{h i t}}\] The constraint for the energy that can be introduced cannot be greater than the energy required to break the bond between atoms. Contributors Michael Fowler (Beams Professor, Department of Physics , University of Virginia) David M. Hanson, Erica Harvey, Robert Sweeney, Theresa Julia Zielinski (" Quantum States of Atoms and Molecules ")
Courses/Grand_Rapids_Community_College/CHM_120_-_Survey_of_General_Chemistry(Neils)/9%3A_Electrochemistry/9.5_Driving_ReactantFavored_Reactions_Using_Electricity	Skills to Develop To understand electrolysis and describe it quantitatively. In this chapter, we have described various galvanic cells in which a spontaneous chemical reaction is used to generate electrical energy. In an electrolytic cell, however, the opposite process, called electrolysis , occurs: an external voltage is applied to drive a nonspontaneous reaction. In this section, we look at how electrolytic cells are constructed and explore some of their many commercial applications. Electrolytic Cells If we construct an electrochemical cell in which one electrode is copper metal immersed in a 1 M Cu 2 + solution and the other electrode is cadmium metal immersed in a $\,1\; M\, Cd^{2+}$ solution and then close the circuit, the potential difference between the two compartments will be 0.74 V. The cadmium electrode will begin to dissolve (Cd is oxidized to Cd 2 + ) and is the anode, while metallic copper will be deposited on the copper electrode (Cu 2 + is reduced to Cu), which is the cathode (Figure $\PageIndex{1a}$). Figure $\PageIndex{1}$: An Applied Voltage Can Reverse the Flow of Electrons in a Galvanic Cd/Cu Cell. (a) When compartments that contain a Cd electrode immersed in 1 M Cd 2 + (aq) and a Cu electrode immersed in 1 M Cu 2 + (aq) are connected to create a galvanic cell, Cd(s) is spontaneously oxidized to Cd 2 + (aq) at the anode, and Cu 2 + (aq) is spontaneously reduced to Cu(s) at the cathode. The potential of the galvanic cell is 0.74 V. (b) Applying an external potential greater than 0.74 V in the reverse direction forces electrons to flow from the Cu electrode [which is now the anode, at which metallic Cu(s) is oxidized to Cu 2 + (aq)] and into the Cd electrode [which is now the cathode, at which Cd 2 + (aq) is reduced to Cd(s)]. The anode in an electrolytic cell is positive because electrons are flowing from it, whereas the cathode is negative because electrons are flowing into it. The overall reaction is as follows: \[ \ce{Cd (s) + Cu^{2+} (aq) \rightarrow Cd^{2+} (aq) + Cu (s)} \label{20.9.1}\] with $E°_{cell} = 0.74\; V$ This reaction is thermodynamically spontaneous as written (ΔG° < 0): \[ \begin{align} \Delta G^\circ &=-nFE^\circ_\textrm{cell} \label{20.9.2a} \\[5pt] &=-(\textrm{2 mol e}^-)[\mathrm{96,486\;J/(V\cdot mol)}](\mathrm{0.74\;V}) \\[5pt] &=-\textrm{140 kJ (per mole Cd)} \label{20.9.2b} \end{align}\] In this direction, the system is acting as a galvanic cell. In an electrolytic cell, an external voltage is applied to drive a nonspontaneous reaction. The reverse reaction, the reduction of Cd 2 + by Cu, is thermodynamically nonspontaneous and will occur only with an input of 140 kJ. We can force the reaction to proceed in the reverse direction by applying an electrical potential greater than 0.74 V from an external power supply. The applied voltage forces electrons through the circuit in the reverse direction, converting a galvanic cell to an electrolytic cell. Thus the copper electrode is now the anode (Cu is oxidized), and the cadmium electrode is now the cathode (Cd 2 + is reduced) (Figure $\PageIndex{1b}$). The signs of the cathode and the anode have switched to reflect the flow of electrons in the circuit. The half-reactions that occur at the cathode and the anode are as follows: cathode: \[\ce{Cd^{2+}(aq) + 2e^{−} \rightarrow Cd(s)}\label{20.9.3}\] with $E^°_{cathode} = −0.40 \, V$ anode: \[\ce{Cu(s) \rightarrow Cu^{2+}(aq) + 2e^{−}} \label{20.9.4}\] with $E^°_{anode} = 0.34 \, V $ Overall: \[\ce{Cd^{2+}(aq) + Cu(s) \rightarrow Cd(s) + Cu^{2+}(aq) } \label{20.9.5}\] with $E^°_{cell} = −0.74 \: V$ Because $ E^°_{cell} < 0$, the overall reaction—the reduction of $Cd^{2+}$ by $Cu$—clearly cannot occur spontaneously and proceeds only when sufficient electrical energy is applied. The differences between galvanic and electrolytic cells are summarized in Table $\PageIndex{1}$. Property Galvanic Cell Electrolytic Cell ΔG < 0 > 0 Ecell > 0 < 0 Electrode Process NaN NaN anode oxidation oxidation cathode reduction reduction Sign of Electrode NaN NaN anode − + cathode + − Electrolytic Reactions At sufficiently high temperatures, ionic solids melt to form liquids that conduct electricity extremely well due to the high concentrations of ions. If two inert electrodes are inserted into molten $NaCl$, for example, and an electrical potential is applied, $Cl^−$ is oxidized at the anode, and $Na^+$ is reduced at the cathode. The overall reaction is as follows: \[\ce{ 2NaCl (l) \rightarrow 2Na(l) + Cl2(g)} \label{20.9.6}\] This is the reverse of the formation of NaCl from its elements. The product of the reduction reaction is liquid sodium because the melting point of sodium metal is 97.8°C, well below that of $NaCl$ (801°C). Approximately 20,000 tons of sodium metal are produced commercially in the United States each year by the electrolysis of molten $NaCl$ in a Downs cell (Figure $\PageIndex{2}$). In this specialized cell, $CaCl_2$ (melting point = 772°C) is first added to the $NaCl$ to lower the melting point of the mixture to about 600°C, thereby lowering operating costs. Figure $\PageIndex{2}$: A Downs Cell for the Electrolysis of Molten NaCl. The electrolysis of a molten mixture of NaCl and CaCl 2 results in the formation of elemental sodium and chlorine gas. Because sodium is a liquid under these conditions and liquid sodium is less dense than molten sodium chloride, the sodium floats to the top of the melt and is collected in concentric capped iron cylinders surrounding the cathode. Gaseous chlorine collects in the inverted cone over the anode. An iron screen separating the cathode and anode compartments ensures that the molten sodium and gaseous chlorine do not come into contact. Similarly, in the Hall–Heroult process used to produce aluminum commercially, a molten mixture of about 5% aluminum oxide (Al 2 O 3 ; melting point = 2054°C) and 95% cryolite (Na 3 AlF 6 ; melting point = 1012°C) is electrolyzed at about 1000°C, producing molten aluminum at the cathode and CO 2 gas at the carbon anode. The overall reaction is as follows: \[2Al_2O_{3(l)} + 3C_{(s)} \rightarrow 4Al_{(l)} + 3CO_{2(g)} \label{20.9.7}\] Oxide ions react with oxidized carbon at the anode, producing CO 2 (g). There are two important points to make about these two commercial processes and about the electrolysis of molten salts in general. The electrode potentials for molten salts are likely to be very different from the standard cell potentials listed in Table P2 , which are compiled for the reduction of the hydrated ions in aqueous solutions under standard conditions. Using a mixed salt system means there is a possibility of competition between different electrolytic reactions. When a mixture of NaCl and CaCl 2 is electrolyzed, Cl − is oxidized because it is the only anion present, but either Na + or Ca 2 + can be reduced. Conversely, in the Hall–Heroult process, only one cation is present that can be reduced (Al 3 + ), but there are three species that can be oxidized: C, O 2− , and F − . In the Hall–Heroult process, C is oxidized instead of O 2− or F − because oxygen and fluorine are more electronegative than carbon, which means that C is a weaker oxidant than either O 2 or F 2 . Similarly, in the Downs cell, we might expect electrolysis of a NaCl/CaCl 2 mixture to produce calcium rather than sodium because Na is slightly less electronegative than Ca (χ = 0.93 versus 1.00, respectively), making Na easier to oxidize and, conversely, Na + more difficult to reduce. In fact, the reduction of Na + to Na is the observed reaction. In cases where the electronegativities of two species are similar, other factors, such as the formation of complex ions, become important and may determine the outcome. Electrolysis can also be used to drive the thermodynamically nonspontaneous decomposition of water into its constituent elements: H 2 and O 2 . However, because pure water is a very poor electrical conductor, a small amount of an ionic solute (such as H 2 SO 4 or Na 2 SO 4 ) must first be added to increase its electrical conductivity. Inserting inert electrodes into the solution and applying a voltage between them will result in the rapid evolution of bubbles of H 2 and O 2 (Figure $\PageIndex{3}$). Figure $\PageIndex{3}$: The Electrolysis of Water. Applying an external potential of about 1.7–1.9 V to two inert electrodes immersed in an aqueous solution of an electrolyte such as H 2 SO 4 or Na 2 SO 4 drives the thermodynamically nonspontaneous decomposition of water into H 2 at the cathode and O 2 at the anode. The reactions that occur are as follows: cathode: \[2H^+_{(aq)} + 2e^− \rightarrow H_{2(g)} \;\;\; E^°_{cathode} = 0 V \label{20.9.8}\] anode: \[2H_2O_{(l)} → O_{2(g)} + 4H+_{(aq)} + 4e^−\;\;\; E^°_{anode} = 1.23\; V \label{20.9.9}\] overall: \[2H_2O_{(l)} → O_{2(g)} + 2H_{2(g)}\;\;\; E^°_{cell} = −1.23 \;V \label{20.9.10}\] For a system that contains an electrolyte such as Na 2 SO 4 , which has a negligible effect on the ionization equilibrium of liquid water, the pH of the solution will be 7.00 and [H + ] = [OH − ] = 1.0 × 10 −7 . Assuming that $P_\mathrm{O_2}$ = $P_\mathrm{H_2}$ = 1 atm, we can use the standard potentials to calculate E for the overall reaction: \[\begin{align}E_\textrm{cell} &=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log(P_\mathrm{O_2}P^2_\mathrm{H_2}) \\ E_\textrm{cell} &=-\textrm{1.23 V}-\left(\dfrac{\textrm{0.0591 V}}{4}\right)\log(1)=-\textrm{1.23 V}\end{align} \label{20.9.11}\] Thus E cell is −1.23 V, which is the value of E° cell if the reaction is carried out in the presence of 1 M H + rather than at pH 7.0. In practice, a voltage about 0.4–0.6 V greater than the calculated value is needed to electrolyze water. This added voltage, called an overvoltage , represents the additional driving force required to overcome barriers such as the large activation energy for the formation of a gas at a metal surface. Overvoltages are needed in all electrolytic processes, which explain why, for example, approximately 14 V must be applied to recharge the 12 V battery in your car. In general, any metal that does not react readily with water to produce hydrogen can be produced by the electrolytic reduction of an aqueous solution that contains the metal cation. The p-block metals and most of the transition metals are in this category, but metals in high oxidation states, which form oxoanions, cannot be reduced to the metal by simple electrolysis. Active metals, such as aluminum and those of groups 1 and 2, react so readily with water that they can be prepared only by the electrolysis of molten salts. Similarly, any nonmetallic element that does not readily oxidize water to O 2 can be prepared by the electrolytic oxidation of an aqueous solution that contains an appropriate anion. In practice, among the nonmetals, only F 2 cannot be prepared using this method. Oxoanions of nonmetals in their highest oxidation states, such as NO 3 − , SO 4 2− , PO 4 3− , are usually difficult to reduce electrochemically and usually behave like spectator ions that remain in solution during electrolysis. In general, any metal that does not react readily with water to produce hydrogen can be produced by the electrolytic reduction of an aqueous solution that contains the metal cation. Electroplating In a process called electroplating , a layer of a second metal is deposited on the metal electrode that acts as the cathode during electrolysis. Electroplating is used to enhance the appearance of metal objects and protect them from corrosion. Examples of electroplating include the chromium layer found on many bathroom fixtures or (in earlier days) on the bumpers and hubcaps of cars, as well as the thin layer of precious metal that coats silver-plated dinnerware or jewelry. In all cases, the basic concept is the same. A schematic view of an apparatus for electroplating silverware and a photograph of a commercial electroplating cell are shown in Figure $\PageIndex{4}$. Figure $\PageIndex{3}$: Electroplating. (a) Electroplating uses an electrolytic cell in which the object to be plated, such as a fork, is immersed in a solution of the metal to be deposited. The object being plated acts as the cathode, on which the desired metal is deposited in a thin layer, while the anode usually consists of the metal that is being deposited (in this case, silver) that maintains the solution concentration as it dissolves. (b) In this commercial electroplating apparatus, a large number of objects can be plated simultaneously by lowering the rack into the Ag+ solution and applying the correct potential. The half-reactions in electroplating a fork, for example, with silver are as follows: cathode (fork): \[Ag^+_{(aq)} + e− \rightarrow Ag_{(s)}\;\;\; E°_{cathode} = 0.80 V\label{20.9.12}\] anode (silver bar): \[Ag_{(s)} \rightarrow Ag^+_{(aq)}\;\;\; E°_{anode} = 0.80 V \label{20.9.13}\] The overall reaction is the transfer of silver metal from one electrode (a silver bar acting as the anode) to another (a fork acting as the cathode). Because E° cell = 0 V, it takes only a small applied voltage to drive the electroplating process. In practice, various other substances may be added to the plating solution to control its electrical conductivity and regulate the concentration of free metal ions, thus ensuring a smooth, even coating. Summary In electrolysis, an external voltage is applied to drive a nonspontaneous reaction. The quantity of material oxidized or reduced can be calculated from the stoichiometry of the reaction and the amount of charge transferred. Relationship of charge, current and time: \[ C = A \times t \nonumber\] In electrolysis, an external voltage is applied to drive a nonspontaneous reaction. Electrolysis can also be used to produce H 2 and O 2 from water. In practice, an additional voltage, called an overvoltage, must be applied to overcome factors such as a large activation energy and a junction potential. Electroplating is the process by which a second metal is deposited on a metal surface, thereby enhancing an object’s appearance or providing protection from corrosion. The amount of material consumed or produced in a reaction can be calculated from the stoichiometry of an electrolysis reaction, the amount of current passed, and the duration of the electrolytic reaction.
Courses/Southeast_Missouri_State_University/CH185%3A_General_Chemistry_(Ragain)/08%3A_Chemical_Bonding_I-_Lewis_Structures_and_Determining_Molecular_Shapes/8.04%3A_Ionic_Bonding	Learning Objectives To describe the characteristics of ionic bonding. To quantitatively describe the energetic factors involved in the formation of an ionic bond. Ions are atoms or molecules which are electrically charged. Cations are positively charged and anions carry a negative charge. Ions form when atoms gain or lose electrons. Since electrons are negatively charged, an atom that loses one or more electrons will become positively charged; an atom that gains one or more electrons becomes negatively charged. Ionic bonding is the attraction between positively- and negatively-charged ions. These oppositely charged ions attract each other to form ionic networks (or lattices ). Electrostatics explains why this happens: opposite charges attract and like charges repel. When many ions attract each other, they form large, ordered, crystal lattices in which each ion is surrounded by ions of the opposite charge. Generally, when metals react with non-metals, electrons are transferred from the metals to the non-metals. The metals form positively-charged ions and the non-metals form negatively-charged ions. Generating Ionic Bonds Ionic bonds form when metals and non-metals chemically react. By definition, a metal is relatively stable if it loses electrons to form a complete valence shell and becomes positively charged. Likewise, a non-metal becomes stable by gaining electrons to complete its valence shell and become negatively charged. When metals and non-metals react, the metals lose electrons by transferring them to the non-metals, which gain them. Consequently, ions are formed, which instantly attract each other—ionic bonding. In the overall ionic compound, positive and negative charges must be balanced, because electrons cannot be created or destroyed, only transferred. Thus, the total number of electrons lost by the cationic species must equal the total number of electrons gained by the anionic species. Example $\PageIndex{1}$: Sodium Chloride For example, in the reaction of Na (sodium) and Cl (chlorine), each Cl atom takes one electron from a Na atom. Therefore each Na becomes a Na + cation and each Cl atom becomes a Cl - anion. Due to their opposite charges, they attract each other to form an ionic lattice. The formula (ratio of positive to negative ions) in the lattice is $\ce{NaCl}$. \[\ce{2Na (s) + Cl 2(g) \rightarrow 2NaCl (s)} \nonumber \] These ions are arranged in solid NaCl in a regular three-dimensional arrangement (or lattice): NaCl lattice. (left) 3-D structure and (right) simple 2D slice through lattes. Images used with permission from Wikipedia and Mike Blaber. The chlorine has a high affinity for electrons, and the sodium has a low ionization energy. Thus the chlorine gains an electron from the sodium atom. This can be represented using ewis dot symbols (here we will consider one chlorine atom, rather than Cl 2 ): The arrow indicates the transfer of the electron from sodium to chlorine to form the Na + metal ion and the Cl - chloride ion. Each ion now has an octet of electrons in its valence shell: Na +: 2s 2 2p 6 Cl - : 3s 2 3p 6 Energetics of Ionic Bond Formation Ionic bonds are formed when positively and negatively charged ions are held together by electrostatic forces. Consider a single pair of ions, one cation and one anion. How strong will the force of their attraction be? According to Coulomb's Law , the energy of the electrostatic attraction ($E$) between two charged particles is proportional to the magnitude of the charges and inversely proportional to the internuclear distance between the particles ($r$): \[E \propto \dfrac{Q_{1}Q_{2}}{r} \label{Eq1a} \] \[ E = k\dfrac{Q_{1}Q_{2}}{r} \label{Eq1b} \] where each ion’s charge is represented by the symbol Q . The proportionality constant k is equal to 2.31 × 10 −28 J·m. This value of k includes the charge of a single electron (1.6022 × 10 −19 C) for each ion. The equation can also be written using the charge of each ion, expressed in coulombs (C), incorporated in the constant. In this case, the proportionality constant, k , equals 8.999 × 109 J·m/C 2 . In the example given, Q 1 = +1(1.6022 × 10 −19 C) and Q 2 = −1(1.6022 × 10 −19 C). If Q 1 and Q 2 have opposite signs (as in NaCl, for example, where Q 1 is +1 for Na + and Q 2 is −1 for Cl − ), then E is negative, which means that energy is released when oppositely charged ions are brought together from an infinite distance to form an isolated ion pair. Energy is always released when a bond is formed and correspondingly, it always requires energy to break a bond. As shown by the green curve in the lower half of Figure $\PageIndex{1}$, the maximum energy would be released when the ions are infinitely close to each other, at r = 0. Because ions occupy space and have a structure with the positive nucleus being surrounded by electrons, however, they cannot be infinitely close together. At very short distances, repulsive electron–electron interactions between electrons on adjacent ions become stronger than the attractive interactions between ions with opposite charges, as shown by the red curve in the upper half of Figure $\PageIndex{1}$. The total energy of the system is a balance between the attractive and repulsive interactions. The purple curve in Figure $\PageIndex{1}$ shows that the total energy of the system reaches a minimum at r 0 , the point where the electrostatic repulsions and attractions are exactly balanced. This distance is the same as the experimentally measured bond distance . Consider the energy released when a gaseous $Na^+$ ion and a gaseous $Cl^-$ ion are brought together from r = ∞ to r = r 0 . Given that the observed gas-phase internuclear distance is 236 pm, the energy change associated with the formation of an ion pair from an $Na^+_{(g)}$ ion and a $Cl^-_{(g)}$ ion is as follows: \[ \begin{align} E &= k\dfrac{Q_{1}Q_{2}}{r_{0}} \\[4pt] &= (2.31 \times {10^{ - 28}}\rm{J}\cdot \cancel{m} ) \left( \dfrac{( + 1)( - 1)}{236\; \cancel{pm} \times 10^{ - 12} \cancel{m/pm}} \right) \\[4pt] &= - 9.79 \times 10^{ - 19}\; J/ion\; pair \label{Eq2} \end{align} \] The negative value indicates that energy is released. Our convention is that if a chemical process provides energy to the outside world, the energy change is negative. If it requires energy, the energy change is positive. To calculate the energy change in the formation of a mole of NaCl pairs, we need to multiply the energy per ion pair by Avogadro’s number: \[ E=\left ( -9.79 \times 10^{ - 19}\; J/ \cancel{ion pair} \right )\left ( 6.022 \times 10^{ 23}\; \cancel{ion\; pair}/mol\right )=-589\; kJ/mol \label{Eq3} \] This is the energy released when 1 mol of gaseous ion pairs is formed, not when 1 mol of positive and negative ions condenses to form a crystalline lattice. Because of long-range interactions in the lattice structure, this energy does not correspond directly to the lattice energy of the crystalline solid. However, the large negative value indicates that bringing positive and negative ions together is energetically very favorable, whether an ion pair or a crystalline lattice is formed. We summarize the important points about ionic bonding: At r 0 , the ions are more stable (have a lower potential energy) than they are at an infinite internuclear distance. When oppositely charged ions are brought together from r = ∞ to r = r 0 , the energy of the system is lowered (energy is released). Because of the low potential energy at r 0 , energy must be added to the system to separate the ions. The amount of energy needed is the bond energy. The energy of the system reaches a minimum at a particular internuclear distance (the bond distance). Example $\PageIndex{2}$: LiF Calculate the amount of energy released when 1 mol of gaseous Li + F − ion pairs is formed from the separated ions. The observed internuclear distance in the gas phase is 156 pm. Given: cation and anion, amount, and internuclear distance Asked for: energy released from formation of gaseous ion pairs Strategy: Substitute the appropriate values into Equation $\ref{Eq1b}$ to obtain the energy released in the formation of a single ion pair and then multiply this value by Avogadro’s number to obtain the energy released per mole. Solution: Inserting the values for Li + F − into Equation $\ref{Eq1b}$ (where Q 1 = +1, Q 2 = −1, and r = 156 pm), we find that the energy associated with the formation of a single pair of Li + F − ions is \[ \begin{align} E &=k \dfrac{Q_1Q_2}{r_0} \\[4pt] &=\left(2.31 \times 10^{−28} J⋅\cancel{m} \right) \left(\dfrac{\text{(+1)(−1)}}{156\; pm \times 10^{−12} \cancel{m/pm}} \right)\\[4pt] &=−1.48 \times 10^{−18} \end{align} \nonumber \] Then the energy released per mole of Li + F − ion pairs is \[ \begin{align} E&= \left(−1.48 \times 10^{−18} J/ \cancel{\text{ion pair}}\right) \left(6.022 \times 10^{23} \cancel{\text{ion pair}}/mol\right)\\[4pt] &−891 \;kJ/mol \end{align} \nonumber \] Because Li + and F − are smaller than Na + and Cl − (see Section 7.3 ), the internuclear distance in LiF is shorter than in NaCl. Consequently, in accordance with Equation $\ref{Eq1b}$, much more energy is released when 1 mol of gaseous Li + F − ion pairs is formed (−891 kJ/mol) than when 1 mol of gaseous Na + Cl − ion pairs is formed (−589 kJ/mol). Exercise $\PageIndex{2}$: Magnesium oxide Calculate the amount of energy released when 1 mol of gaseous $\ce{MgO}$ ion pairs is formed from the separated ions. The internuclear distance in the gas phase is 175 pm. Answer −3180 kJ/mol = −3.18 × 10 3 kJ/mol Electron Configuration of Ions How does the energy released in lattice formation compare to the energy required to strip away a second electron from the Na + ion? Since the Na + ion has a noble gas electron configuration, stripping away the next electron from this stable arrangement would require more energy than what is released during lattice formation (Sodium I 2 = 4,560 kJ/mol). Thus, sodium is present in ionic compounds as Na + and not Na 2 + . Likewise, adding an electron to fill a valence shell (and achieve noble gas electron configuration) is exothermic or only slightly endothermic. To add an additional electron into a new subshell requires tremendous energy - more than the lattice energy. Thus, we find Cl - in ionic compounds, but not Cl 2 - . Compound Lattice Energy (kJ/mol) LiF 1024 LiI 744 NaF 911 NaCl 788 NaI 693 KF 815 KBr 682 KI 641 MgF2 2910 SrCl2 2130 MgO 3938 This amount of energy can compensate for values as large as I 3 for valence electrons (i.e. can strip away up to 3 valence electrons). Because most transition metals would require the removal of more than 3 electrons to attain a noble gas core, they are not found in ionic compounds with a noble gas core. A transition metal always loses electrons first from the higher 's' subshell, before losing from the underlying 'd' subshell. (The remaining electrons in the unfilled d subshell are the reason for the bright colors observed in many transition metal compounds!) For example, iron ions will not form a noble gas core: Fe: [Ar]4s 2 3d 6 Fe 2 + : [Ar] 3d 6 Fe 3 + : [Ar] 3d 5 Some metal ions can form a pseudo noble gas core (and be colorless), for example: Ag: [Kr]5s 1 4d 10 Ag + [Kr]4d 10 Compound: AgCl Cd: [Kr]5s 2 4d 10 Cd 2 + [Kr]4d 10 Compound: CdS The valence electrons do not adhere to the "octet rule" in this case (a limitation of the usefulness of this rule). Note: The silver and cadmium atoms lost the 5s electrons in achieving the ionic state. When a positive ion is formed from an atom, electrons are always lost first from the subshell with the largest principle quantum number Polyatomic Ions Not all ionic compounds are formed from only two elements. Many polyatomic ions exist, in which two or more atoms are bound together by covalent bonds. They form a stable grouping which carries a charge (positive or negative). The group of atoms as a whole acts as a charged species in forming an ionic compound with an oppositely charged ion. Polyatomic ions may be either positive or negative, for example: NH 4 + (ammonium) = cation SO 4 2 - (sulfate) = anion The principles of ionic bonding with polyatomic ions are the same as those with monatomic ions. Oppositely charged ions come together to form a crystalline lattice, releasing a lattice energy. Based on the shapes and charges of the polyatomic ions, these compounds may form crystalline lattices with interesting and complex structures. Summary The amount of energy needed to separate a gaseous ion pair is its bond energy. The formation of ionic compounds are usually extremely exothermic . The strength of the electrostatic attraction between ions with opposite charges is directly proportional to the magnitude of the charges on the ions and inversely proportional to the internuclear distance. The total energy of the system is a balance between the repulsive interactions between electrons on adjacent ions and the attractive interactions between ions with opposite charges.
Bookshelves/Environmental_Chemistry/Toxicology_MSDT/6%3A_Principles_of_Toxicology	Introduction ToxTutor is a self-paced tutorial covering key principles of toxicology and was adopted from the National Library of Medicine (NLM) chemical and toxicology databases. While a knowledge of anatomy and physiology is not required for viewing ToxTutor, the Introduction to the Human Body from the National Cancer Institute provides a good introduction to the topic. Topics Covered in this Course ToxTutor is divided into the following sections: Section 1: Introduction to Toxicology Section 2: Dose and Dose Response Section 3: Toxic Effects Section 4: Interactions Section 5: Toxicity Testing Methods Section 6: Risk Assessment Section 7: Exposure Standards and Guidelines Section 8: Basic Physiology Section 9: Introduction to Toxicokinetics Section 10: Absorption Section 11: Distribution Section 12: Biotransformation Section 13: Excretion Section 14: Cellular Toxicology Section 15: Intuitive Toxicology and Risk Communication Section 16: Environmental Toxicology, Environmental Health, and One Health Section 17: Conclusion Each section of ToxTutor contains one or more related content pages. Next and Back buttons are provided to allow you to navigate through these pages. For more information, see the "Getting Around" section below. The basic principles of toxicology described in ToxTutor are similar to those taught in university programs and are well described in toxicology literature. A list of the textbooks used as the primary resources for the tutorials is found in the Bibliography . Using ToxTutor Getting Around It will take approximately three hours to complete this self-paced tutorial. There are a variety of ways you can navigate ToxTutor. You can: Use the Glossary button in the upper right corner of each page to access the glossary . Click the links in the list above (or any underlined words in any module which are linked) to access sections directly. Click the Previous and Next links at the bottom of each page to move through the material. Use a combination of the above methods to explore the course contents. Progress At the top of each page in ToxTutor are links indicating where the current page falls within the overall ToxTutor program. You can click these links to return to this homepage or to the section that contains the page. Links Throughout the course, you will encounter links. Any link to a resource outside of ToxTutor, will typically open in a new tab or window. All other links are to other areas within ToxTutor. Links are bold and underlined as seen here and here . In addition, clicking on some images may open another external website for more information. Credits ToxTutor was adopted from the U.S National Library of Medicine in 2021. More Information can be seen in the ToxTutor Bibliography .
Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/07%3A_Alkyl_Halides-_Nucleophilic_Substitution_and_Elimination/7.03%3A_Preparation_of_Alkyl_Halides	Learning Objective specify the reagents for the most efficient synthesis of alkyl halides using free-radical halogenation of alkanes (Chapter 5) or allylic halogenation of alkenes with NBS Free radical halogenation of alkanes Free radical halogenation of alkanes is the substitution of a single hydrogen on the alkane for a single halogen to form a haloalkane. Light is required to initiate the radical formation and is a good example of a photochemical reaction. The simplest example is shown below for methane reacting with chlorine in the presence of light to form chloromethane and hydrogen chloride gas. \[\ce{CH4 + Cl2 + energy → CH3Cl + HCl}\] Free radical halogenation of alkanes has been thoroughly explained in chapter 5. The structure of the alkane is evaluated to choose between the high reactivity of chlorine (Cl 2 ) and the high selectivity of bromine (Br 2 ). Allylic Bromination When halogens are in the presence of unsaturated molecules such as alkenes, the expected reaction is addition to the double bond carbons resulting in a vicinal dihalide (halogens on adjacent carbons). The reaction is studied in a later chapter. To avoid halogen reactions at the alkene the halogen concentration is kept low enough that a substitution reaction occurs at the allylic position rather than addition at the double bond. The product is an allylic halide (halogen on carbon next to double bond carbons), which is acquired through a radical chain mechanism. Why Substitution of Allylic Hydrogens? As the table below shows, the dissociation energy for the allylic C-H bond is lower than the dissociation energies for the C-H bonds at the vinylic and alkylic positions. This is because the radical formed when the allylic hydrogen is removed is resonance-stabilized. Hence, given that the halogen concentration is low, substitution at the allylic position is favored over competing reactions. However, when the halogen concentration is high, addition at the double bond is favored because a polar reaction out competes the radical chain reaction. Radical Allylic Bromination using NBS and light Preparation of Bromine (low concentration) NBS (N-bromosuccinimide) is the most commonly used reagent to produce low concentrations of bromine. When suspended in tetrachloride (CCl 4 ), NBS reacts with trace amounts of HBr to produce a low enough concentration of bromine to facilitate the allylic bromination reaction. Allylic Bromination Mechanism Step 1: Initiation Once the pre-initiation step involving NBS produces small quantities of Br 2 , the bromine molecules are homolytically cleaved by light to produce bromine radicals. Step 2: Propagation One bromine radical produced by homolytic cleavage in the initiation step removes an allylic hydrogen of the alkene molecule. A radical intermediate is generated, which is stabilized by resonance. The stability provided by delocalization of the radical in the alkene intermediate is the reason that substitution at the allylic position is favored over competing reactions such as addition at the double bond. The intermediate radical then reacts with a Br 2 molecule to generate the allylic bromide product and regenerate the bromine radical, which continues the radical chain mechanism. If the alkene reactant is asymmetric, two distinct product isomers are formed. Step 3: Termination The radical chain mechanism of allylic bromination can be terminated by any of the possible steps shown below. Radical Allylic Chlorination Like bromination, chlorination at the allylic position of an alkene is achieved when low concentrations of Cl 2 are present. The reaction is run at high temperatures to achieve the desired results. Industrial Uses Allylic chlorination has important practical applications in industry. Since chlorine is inexpensive, allylic chlorinations of alkenes have been used in the industrial production of valuable products. For example, 3-chloropropene, which is necessary for the synthesis of products such as epoxy resin, is acquired through radical allylic chlorination (shown below). Exercises Exercises Predict the two products of the allylic chlorination reaction of 1-heptene. What conditions are required for allylic halogenation to occur? Why does this reaction outcompete other possible reactions such as addition when these conditions are met? Predict the product of the allylic bromination reaction of 2-phenylheptane. (Hint: How are benzylic hydrogens similar to allylic hydrogens?) The reactant 5-methyl-1-hexene generates the products 3-bromo-5-methyl-1-hexene and 1-bromo-5-methyl-2-hexene. What reagents were used in this reaction? Predict the products of the following reactions: Solutions 1. 3-chloro-1-heptene and 1-chloro-2-heptene 2. A low concentration of halide radical is sufficient for reaction at the allylic carbon without creating a reactivty environment for the pi bond of the alkene. 3. 2-bromo-2-phenylheptane 4. NBS with light 5. The product (A) is a 1° halogen which is more stable product even though the (B) had a better transition state with a 2° radical. 6. References Djerassi, Carl. "Brominations with N-Bromosuccinimide and Related Compounds - The Wohl-Ziegler Reaction." Organic and Biomolecular Chemistry (2003). RSC Publishing . 9 June 2003. Royal Society of Chemistry. 25 Feb. 2009. Kent, Doug. Allylic Bromination. Chem 118B Workshop. Learning Skills Center. 3 Feb. 2009. Li, Chao-Jun, and Tak-Hang Chan. Organic Chemistry: Structure and Function . 5th ed. New York: W.H. Freeman and Company, 2007.
Courses/Prince_Georges_Community_College/CHEM_1020%3A_General_Chemistry_II_(S.N._Yasapala)/05%3A_Thermochemistry/5.03%3A_The_Molecular_Interpretation_of_Entropy_(3rd_Law_of_Thermodynamics)	Learning Objectives To understand entropy from a molecular interpretation As discussed previously, the second law of thermodynamics argues that all processes must increase the total entropy of the universe. However, the universe if often separated into the System and the Surroundings differing changes of entropy can be observed from a system level perspective. Many processes results in an increase in a system's entropy $\Delta S > 0$: Increasing the volume that a gas can occupy will increase the disorder of a gas Dissolving a solute into a solution will increase the entropy of the solute - typically resulting in an increase in the entropy of the system. (Note: the solvation of a solute can sometimes result in a significant decrease in the solvent entropy - leading to a net decrease in entropy of the system ) Phase changes from solid to liquid, or liquid to gas, lead to an increase in the entropy of the system Some processes result in a decrease in the entropy of a system $\Delta S < 0$: A gas molecule dissolved in a liquid is much more confined by neighboring molecules than when its in the gaseous state. Thus, the entropy of the gas molecule will decrease when it is dissolved in a liquid A phase change from a liquid to a solid (i.e. freezing), or from a gas to a liquid (i.e. condensation) results in an decrease in the disorder of the substance, and a decrease in the entropy A chemical reaction between gas molecules that results in a net decrease in the overall number of gas molecules will de crease the disorder of the system, and result in a decrease in the entropy \[2NO_{(g)} + O_{2(g)} \rightarrow 2NO_{2(g)} \;\;\; \Delta S < 0 \label{19.3.1} \] What is the molecular basis for the above observations for the change in entropy? Let's first consider the last example, the decrease in entropy associated with a decrease in the number of gas molecules for a chemical reaction The product of this reaction ($NO_2$) involves the formation of a new N-O bond and the O atoms, originally in a separate $O_2$ molecule, are now connected to the $NO$ molecule via a new $N-O$ bond. Since they are now physically bonded to the other molecule (forming a new, larger, single molecule) the O atoms have less freedom to move around The reaction has resulted in a loss of freedom of the atoms (O atoms) There is a reduction in the disorder of the system (i.e. due to the reduction in the degrees of freedom, the system is more ordered after the reaction). $\Delta S < 0$. Molecular Degrees of Freedom The atoms, molecules, or ions that compose a chemical system can undergo several types of molecular motion, including translation, rotation, and vibration (Figure $\PageIndex{2}$). Translational motion . The entire molecule can move in some direction in three dimensions Rotational motion . The entire molecule can rotate around any axis, (even though it may not actually change its position translationally ) Vibrational motion . The atoms within a molecule have certain freedom of movement relative to each other; this displacement can be periodic motion like the vibration of a tuning fork These forms of molecular motion are ways in which molecules can store energy. The greater the molecular motion of a system, the greater the number of possible microstates and the higher the entropy. The Third Law of Thermodynamics These forms of motion are ways in which the molecule can store energy. The greater the molecular motion of a system, the greater the number of possible microstates and the higher the entropy. A perfectly ordered system with only a single microstate available to it would have an entropy of zero. The only system that meets this criterion is a perfect crystal at a temperature of absolute zero (0 K), in which each component atom, molecule, or ion is fixed in place within a crystal lattice and exhibits no motion (ignoring quantum effects). Such a state of perfect order (or, conversely, zero disorder) corresponds to zero entropy. In practice, absolute zero is an ideal temperature that is unobtainable, and a perfect single crystal is also an ideal that cannot be achieved. Nonetheless, the combination of these two ideals constitutes the basis for the third law of thermodynamics : the entropy of any perfectly ordered, crystalline substance at absolute zero is zero. The Third Law of Thermodynamics The entropy of a pure crystalline substance at absolute zero (i.e. 0 Kelvin) is 0. Since S = 0 corresponds to perfect order . The position of the atoms or molecules in the crystal would be perfectly defined As the temperature increases, the entropy of the atoms in the lattice increase Vibrational motions cause the atoms and molecules in the lattice to be less well ordered The third law of thermodynamics has two important consequences: it defines the sign of the entropy of any substance at temperatures above absolute zero as positive, and it provides a fixed reference point that allows us to measure the absolute entropy of any substance at any temperature.In practice, chemists determine the absolute entropy of a substance by measuring the molar heat capacity (C p ) as a function of temperature and then plotting the quantity C p /T versus T. The area under the curve between 0 K and any temperature T is the absolute entropy of the substance at T. In contrast, other thermodynamic properties, such as internal energy and enthalpy, can be evaluated in only relative terms, not absolute terms. In this section, we examine two different ways to calculate ΔS for a reaction or a physical change. The first, based on the definition of absolute entropy provided by the third law of thermodynamics, uses tabulated values of absolute entropies of substances. The second, based on the fact that entropy is a state function, uses a thermodynamic cycle similar to those discussed previously. Motion never stops at absolute zero! From classical kinetic theory, all motions cease at absolute zero. However things are more complicated from the more advanced (and more accurate) quantum mechanics. The Heisenberg Uncertainly Principle of quantum mechanics argues that molecules, even at absolute zero, always always have motion. Nonetheless, this motion is often ignored in the introduction of the third law of thermodynamics (which is incorrect of course). Continued heating of a Solid Lattice One way of calculating ΔS for a reaction is to use tabulated values of the standard molar entropy (S°), which is the entropy of 1 mol of a substance at a standard temperature of 298 K; the units of S° are J/(mol·K). Unlike enthalpy or internal energy, it is possible to obtain absolute entropy values by measuring the entropy change that occurs between the reference point of 0 K [corresponding to S = 0 J/(mol·K)] and 298 K. As shown in Table $\PageIndex{1}$, for substances with approximately the same molar mass and number of atoms, S° values fall in the order S°(gas) > S°(liquid) > S°(solid). For instance, S° for liquid water is 70.0 J/(mol·K), whereas S° for water vapor is 188.8 J/(mol·K). Likewise, S° is 260.7 J/(mol·K) for gaseous I 2 and 116.1 J/(mol·K) for solid I2. This order makes qualitative sense based on the kinds and extents of motion available to atoms and molecules in the three phases. The correlation between physical state and absolute entropy is illustrated in Figure $\PageIndex{2}$, which is a generalized plot of the entropy of a substance versus temperature. Gases Gases.1 Liquids Liquids.1 Solids Solids.1 Substance S° [J/(mol·K)] Substance S° [J/(mol·K)] Substance S° [J/(mol·K)] He 126.2 H2O 70.0 C (diamond) 2.4 H2 130.7 CH3OH 126.8 C (graphite) 5.7 Ne 146.3 Br2 152.2 LiF 35.7 Ar 154.8 CH3CH2OH 160.7 SiO2 (quartz) 41.5 Kr 164.1 C6H6 173.4 Ca 41.6 Xe 169.7 CH3COCl 200.8 Na 51.3 H2O 188.8 C6H12 (cyclohexane) 204.4 MgF2 57.2 N2 191.6 C8H18 (isooctane) 329.3 K 64.7 O2 205.2 NaN NaN NaCl 72.1 CO2 213.8 NaN NaN KCl 82.6 I2 260.7 NaN NaN I2 116.1 Entropy increases with softer, less rigid solids, solids that contain larger atoms, and solids with complex molecular structures. A closer examination of Table $\PageIndex{1}$ also reveals that substances with similar molecular structures tend to have similar S° values. Among crystalline materials, those with the lowest entropies tend to be rigid crystals composed of small atoms linked by strong, highly directional bonds, such as diamond [S° = 2.4 J/(mol·K)]. In contrast, graphite, the softer, less rigid allotrope of carbon, has a higher S° [5.7 J/(mol·K)] due to more disorder in the crystal. Soft crystalline substances and those with larger atoms tend to have higher entropies because of increased molecular motion and disorder. Similarly, the absolute entropy of a substance tends to increase with increasing molecular complexity because the number of available microstates increases with molecular complexity. For example, compare the S° values for CH 3 OH(l) and CH 3 CH 2 OH(l). Finally, substances with strong hydrogen bonds have lower values of S°, which reflects a more ordered structure. Entropy: Entropy(opens in new window) [youtu.be] Summary In general, the entropy is expected to increase for the following types of processes: The melting of a solid to form a liquid The vaporization of a liquid (or solid) to produce a gas Chemical reactions that involve phase changes of $\text{solid} \rightarrow \text{liquid/gas}$, or $\text{liquid} \rightarrow \text{gas}$ Chemical reactions that result in an increase in the number of gaseous molecules Any time the temperature of a substance is increased
Courses/SUNY_Oneonta/Chem_322_Lecture_Content/07%3A_Reactions_at_the_-Carbon_Part_I	7.1: Prelude to Reactions at the α-carbon, part I There is a connection between the killer platypus in Australia and the 'hunting magic' in the Amazon, and it has to do with the structure and reactivity of what organic chemists refer to as the α -carbon: the carbon atom positioned adjacent to a carbonyl or imine group in an organic molecule. 7.2: Review of Acidity at the α-Carbon Let's review what we learned in section 7.6 about the acidity of a proton on an a-carbon and the structure of the relevant conjugate base, the enolate ion. Remember that this acidity can be explained by the fact that the negative charge on the enolate conjugate base is delocalized by resonance to both the α -carbon and the carbonyl oxygen. 7.3: Isomerization at the α-Carbon Enolate ions are the key reactive intermediates in many biochemical isomerization reactions. Isomerizations can involve either the interconversion of constitutional isomers, in which bond connectivity is altered, or of stereoisomers, where the stereochemical configuration is changed. Enzymes that interconvert constitutional isomers are usually called isomerases, while those that interconvert the configuration of a chiral carbon are usually referred to as racemases or epimerases. 7.4: Aldol Addition Along with Claisen condensation reactions, which we will study in the next chapter, aldol additions are responsible for most of the carbon-carbon bond forming events that occur in a living cell. Because biomolecules are built upon a framework of carbon-carbon bonds, it is difficult to overstate the importance of aldol addition and Claisen condensation reactions in the chemistry of living things! 7.5: α-Carbon Reactions in the Synthesis Lab - Kinetic vs. Thermodynamic Alkylation Products While aldol addition reactions are widespread in biochemical pathways as a way of forming carbon-carbon bonds, synthetic organic chemists working the lab also make use of aldol-like reactions for the same purpose. 7.P: Reactions at the α-Carbon, Part I (Problems) 7.S: Reactions at the α-Carbon, Part I (Summary)
Courses/Modesto_Junior_College/Chemistry_143_-_Bunag/Chemistry_143_-_Introductory_Chemistry_(Bunag)/14%3A_Thermochemistry/14.11%3A_Heats_of_Vaporization_and_Condensation	Natural resources for electric power generation have traditionally been waterfalls, oil, coal, or nuclear power. Research is being carried out to look for other renewable sources to run the generators. Geothermal sites (such as geysers) are being considered because of the steam they produce. Capabilities can be estimated by knowing how much steam is released in a given time at a particular site. Heat of Vaporization and Condensation Energy is absorbed in the process of converting a liquid at its boiling point into a gas. As with the melting point of a solid, the temperature of a boiling liquid remains constant and the input of energy goes into changing the state. The molar heat of vaporization $\left( \Delta H_\text{vap} \right)$ of a substance is the heat absorbed by one mole of that substance as it is converted from a liquid to a gas. As a gas condenses to a liquid, heat is released. The molar heat of condensation $\left( \Delta H_\text{cond} \right)$ of a substance is the heat released by one mole of that substance as it is converted from a gas to a liquid. Since vaporization and condensation of a given substance are the exact opposite processes, the numerical value of the molar heat of vaporization is the same as the numerical value of the molar heat of condensation, but opposite in sign. In other words, $\Delta H_\text{vap} = -\Delta H_\text{cond}$. When $1 \: \text{mol}$ of water at $100^\text{o} \text{C}$ and $1 \: \text{atm}$ pressure is converted to $1 \: \text{mol}$ of water vapor at $100^\text{o} \text{C}$, $40.7 \: \text{kJ}$ of heat is absorbed from the surroundings. When $1 \: \text{mol}$ of water vapor at $100^\text{o} \text{C}$ condenses to liquid water at $100^\text{o} \text{C}$, $40.7 \: \text{kJ}$ of heat is released into the surroundings. \[\begin{array}{ll} \ce{H_2O} \left( l \right) \rightarrow \ce{H_2O} \left( g \right) & \Delta H_\text{vap} = 40.7 \: \text{kJ/mol} \\ \ce{H_2O} \left( g \right) \rightarrow \ce{H_2O} \left( l \right) & \Delta H_\text{cond} =-40.7 \: \text{kJ/mol} \end{array}\nonumber \] Other substances have different values for their molar heats of fusion and vaporization; these substances are summarized in the table below. Substance $\Delta H_\text{fus}$ $\left( \text{kJ/mol} \right)$ $\Delta H_\text{vap}$ $\left( \text{kJ/mol} \right)$ Ammonia $\left( \ce{NH_3} \right)$ 5.65 23.40 Ethanol $\left( \ce{C_2H_5OH} \right)$ 4.60 43.50 Methanol $\left( \ce{CH_3OH} \right)$ 3.16 35.30 Oxygen $\left( \ce{O_2} \right)$ 0.44 6.82 Water $\left( \ce{H_2O} \right)$ 6.01 40.70 Notice that for all substances, the heat of vaporization is substantially higher than the heat of fusion. Much more energy is required to change the state from a liquid to a gas than from a solid to a liquid. This is because of the large separation of the particles in the gas state. The values of the heats of fusion and vaporization are related to the strength of the intermolecular forces. All of the substances in the table above, with the exception of oxygen, are capable of hydrogen bonding. Consequently, the heats of fusion and vaporization of oxygen are far lower than the others. Example $\PageIndex{1}$ What mass of methanol vapor condenses to a liquid as $20.0 \: \text{kJ}$ of heat is released? Solution Step 1: List the known quantities and plan the problem. Known $\Delta H = 20.0 \: \text{kJ}$ $\Delta H_\text{cond} = -35.3 \: \text{kJ/mol}$ Molar mass $\ce{CH_3OH} = 32.05 \: \text{g/mol}$ Unknown First the $\text{kJ}$ of heat released in the condensation is multiplied by the conversion factor $\left( \frac{1 \: \text{mol}}{-35.3 \: \text{kJ}} \right)$ to find the moles of methanol that condensed. Then, moles are converted to grams. Step 2: Solve. \[-20.0 \: \text{kJ} \times \frac{1 \: \text{mol} \: \ce{CH_3OH}}{-35.3 \: \text{kJ}} \times \frac{32.05 \: \text{g} \: \ce{CH_3OH}}{1 \: \text{mol} \: \ce{CH_3OH}} = 18.2 \: \text{g} \: \ce{CH_3OH}\nonumber \] Step 3: Think about your result. Condensation is an exothermic process, so the enthalpy change is negative. Slightly more than one-half mole of methanol is condensed. Summary The molar heat of vaporization $\left( \Delta H_\text{vap} \right)$ is the heat absorbed by one mole of a substance as it is converted from a liquid to a gas. The molar heat of condensation $\left( \Delta H_\text{cond} \right)$ is the heat released by one mole of a substance as it is converted from a gas to a liquid. Examples of calculations involving the molar heat of vaporization and condensation are illustrated.
Courses/University_of_California_Davis/Chem_124A%3A_Fundamentals_of_Inorganic_Chemistry/05%3A_Molecular_Orbitals/5.03%3A_Heteronuclear_Diatomic_Molecules/5.3.01%3A_Polar_bonds	Molecular orbital diagrams for heteronuclear diatomic molecules The molecular orbital diagram of a heteronuclear diatomic molecule is approached in a way similar to that of a homonuclear diatomic molecule. The orbital diagrams may also look similar. A major difference is that the more electronegative atom will have orbitals at a lower energy level. Two examples of heteronuclear diatomic molecules will be explored below as illustrative examples. Carbon monoxide MO diagram Carbon monoxide is an example of a heteronuclear diatomic molecule where both atoms are second-row elements. The valence molecular orbitals in both atoms are the $2s$ and $2p$ orbitals. The molecular orbital diagram for carbon monoxide (Figure $\PageIndex{1}$) is constructed in a way similar to how you would construct dicarbon or dioxygen, except that the oxygen orbitals have a lower potential energy than analogous carbon orbitals. The labeling of molecular orbitals in this diagram follows a convention by which orbitals are given serial labels according to type of orbital ($\sigma$, $\pi$, etc.). The lowest energy orbitals of any type are assigned a value of 1 and higher energy orbitals of the same type are assigned by increasing intervals (..2, 3, 4...). The orbital labeling system described previously is inappropriate for heteronuclear diatomic molecules that cannot be assigned $g$ and $u$ subscripts. A consequence of unequal atomic orbital energy levels is that orbital mixing is significant. Notice the order of the molecular orbitals labeled $1\pi$ and $3\sigma$ in Figure $\PageIndex{1}$. This is a similar order of $\pi$ and $\sigma$ orbitals to the one we saw in the case of the $\sigma_g$ and $\pi_u$ orbitals of $N_2$ and lighter diatomics of the second period. Because the oxygen $2p_z$ orbital is close in energy to both the carbon $2p_z$ and carbon $2s$, these three orbitals will have significant interaction (mixing). The result is an increase in the energy of the $3\sigma$ orbital and a decrease in energy of the $2\sigma^$ orbital, resulting in the diagram shown in Figure $\PageIndex{1}$. In the case of carbon monoxide (Figure $\PageIndex{1}$), atomic orbitals contribute unequally to each molecular orbital. For example, because the $2s$ orbital of oxygen is very close in energy to the $2\sigma$ moelcular orbital, it contributes to that molecular orbital more than the $2s$ orbital from carbon. Notice the shape of this $2\sigma$ orbital and how it is unevenly distributed over the two atoms; it is more heavily distributed on the oxygen because it is most like the oxygen $2s$. This is in line with the assumption that electron density is distributed more on oxygen because it is more electronegative than carbon. Likewise, the $1\pi$ orbitals are unevenly distributed, with more distribution close to the oxygen. Exercise $\PageIndex{1}$ Examine the shape of the $3\sigma$ orbital of carbon monoxide in Figure $\PageIndex{1}$. Describe what ways this shape is different from the shape of the $\sigma_g$ orbitals from second period homonuclear diatomic molecules ( see Fig. 5.2.1.1 ). Rationalize these differences. Both orbitals are re-created below for convenience. Answer The 3$\sigma$ orbital is like the $\sigma_g$ in that it has three lobes and two nodes distributed along the internuclear bond. They are different in their distribution. The two external lobes of $\sigma_g$ are evenly distributed because they are an equal combination of two $p_z$ orbitals (one from each atom). The $3\sigma$ orbital is more heavily distributed toward the carbon atom, the less electronegative atom, than toward the oxygen. The unequal distribution of $3\sigma$ is apparent in the unequal sizes of its exterior lobes and the uneven shape of the interior lobe. The heavier distribution within the exterior lobe on carbon is caused by the mixing of the carbon $2s$ orbital with carbon and oxygen $2p_z$ orbitals. The uneven shape of the interior lobe, where it leans toward oxygen, is best explained by the fact that the $3\sigma$ orbital is closer in energy to the oxygen $2p_z$ than the carbon $2p_z$. Hydrogen fluoride MO diagram Hydrogen fluoride is an example of a heteronuclear diatomic molecule in which the two atoms are from different periods. In this case, the valence orbital of H is $1s$ while those of F are $2s$ and $2p$. The molecular orbital diagram for HF is shown in Figure $\PageIndex{2}$. Three of these orbitals have compatible symmetry for mixing; these are the hydrogen $1s$, fluorine $2s$, and fluorine $2p$. However, the extent to which they will interact depends on their relative energies. Fluorine is more electronegative than H, and the fluorine atom has a higher first ionization energy than does hydrogen. From these trends, we can expect that the fluorine valence orbitals are lower in energy than that of hydrogen. From Table 5.3.1 , we find that the 1s orbital of H (-13.6 eV) is higher in energy than both fluorine orbitals (-18.7 and -40.2 eV, respectively for $2p$ and $2s$). The energies of hydrogen $1s$ and fluorine $2p$ are a good match for combination; however, the fluorine $2s$ orbital is much too different to create a productive interaction. Therefore, we expect that the fluorine $2s$ will create a non-bonding molecular orbital, while the $1s$ and $2p_z$ orbitals combine to make $\sigma$ bonding and $\sigma^$ antibonding molecular orbitals. The remaining $2p_x$ and $2p_y$ orbitals do not have compatible symmetry for bonding with hydrogen, and they will form non-bonding $\pi$ molecular orbitals. The non-bonding orbitals will have similar energy and character as their component atomic orbitals. Chemical reactions take place at the HOMO and LUMO orbitals Knowledge of molecular orbital diagrams, and the shapes of molecular orbitals, can be used to accurately explain and predict chemical reactivity. Chemical reactions take place using the highest occupied molecular orbitals (HOMO) of a nucleophile or Lewis base, and the lowest unoccupied molecular orbital (LUMO) of an electrophile or Lewis acid. Lewis bases react using electrons in the HOMO, while Lewis acids react using the empty LUMO. Example: Reactivity of CO with metal ions CO is an excellent ligand for many metal ions. In fact, the strong affinity between CO and the heme iron (Fe) ions in hemoglobin can explain the mechanism of carbon monoxide poisoning. When CO binds in place of $O_2$ to hemoglobin, that hemoglobin can no longer carry $O_2$ to tissue cells. CO binding to hemoglobin is strong and practically irreversible. When CO binds to metal ions, it does so through the carbon atom. This is contrary to expectations based on the Lewis structure and the known bond polarity, where electron density is polarized toward oxygen. The distribution of the electron density of the HOMO of CO can explain this observation! Exercise $\PageIndex{2}$ Refer to the MO diagram for CO. Identify the HOMO and explain why CO bonds to metal ions through the carbon atom rather than through the oxygen atom. Answer In the interaction between CO and a metal ion, CO would act as a Lewis base; thus it will react using electrons in its HOMO. The MO diagram for CO is shown in Figure $\PageIndex{1}$: the HOMO is the $3\sigma$ orbital (also discussed in Exercise $\PageIndex{1}$) . The electron density of that MO is centered around the carbon atom, thus the carbon atom will be a better Lewis base than the O atom.
Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/21%3A_The_p-Block_Elements/21.06%3A_The_Elements_of_Group_18_(The_Noble_Gases)	Learning Objectives To understand the trends in properties and reactivity of the group 18 elements: the noble gases. The noble gases were all isolated for the first time within a period of only five years at the end of the 19th century. Their very existence was not suspected until the 18th century, when early work on the composition of air suggested that it contained small amounts of gases in addition to oxygen, nitrogen, carbon dioxide, and water vapor. Helium was the first of the noble gases to be identified, when the existence of this previously unknown element on the sun was demonstrated by new spectral lines seen during a solar eclipse in 1868. Actual samples of helium were not obtained until almost 30 years later, however. In the 1890s, the English physicist J. W. Strutt (Lord Rayleigh) carefully measured the density of the gas that remained after he had removed all O 2 , CO 2 , and water vapor from air and showed that this residual gas was slightly denser than pure N 2 obtained by the thermal decomposition of ammonium nitrite. In 1894, he and the Scottish chemist William Ramsay announced the isolation of a new “substance” (not necessarily a new element) from the residual nitrogen gas. Because they could not force this substance to decompose or react with anything, they named it argon (Ar), from the Greek argos, meaning “lazy.” Because the measured molar mass of argon was 39.9 g/mol, Ramsay speculated that it was a member of a new group of elements located on the right side of the periodic table between the halogens and the alkali metals. He also suggested that these elements should have a preferred valence of 0, intermediate between the +1 of the alkali metals and the −1 of the halogens. J. W. Strutt (Lord Rayleigh) (1842–1919) Lord Rayleigh was one of the few members of British higher nobility to be recognized as an outstanding scientist. Throughout his youth, his education was repeatedly interrupted by his frail health, and he was not expected to reach maturity. In 1861 he entered Trinity College, Cambridge, where he excelled at mathematics. A severe attack of rheumatic fever took him abroad, but in 1873 he succeeded to the barony and was compelled to devote his time to the management of his estates. After leaving the entire management to his younger brother, Lord Rayleigh was able to devote his time to science. He was a recipient of honorary science and law degrees from Cambridge University. Sir William Ramsay (1852–1916) Born and educated in Glasgow, Scotland, Ramsay was expected to study for the Calvanist ministry. Instead, he became interested in chemistry while reading about the manufacture of gunpowder. Ramsay earned his PhD in organic chemistry at the University of Tübingen in Germany in 1872. When he returned to England, his interests turned first to physical chemistry and then to inorganic chemistry. He is best known for his work on the oxides of nitrogen and for the discovery of the noble gases with Lord Rayleigh. In 1895, Ramsey was able to obtain a terrestrial sample of helium for the first time. Then, in a single year (1898), he discovered the next three noble gases: krypton (Kr), from the Greek kryptos, meaning “hidden,” was identified by its orange and green emission lines; neon (Ne), from the Greek neos, meaning “new,” had bright red emission lines; and xenon (Xe), from the Greek xenos, meaning “strange,” had deep blue emission lines. The last noble gas was discovered in 1900 by the German chemist Friedrich Dorn, who was investigating radioactivity in the air around the newly discovered radioactive elements radium and polonium. The element was named radon (Rn), and Ramsay succeeded in obtaining enough radon in 1908 to measure its density (and thus its atomic mass). For their discovery of the noble gases, Rayleigh was awarded the Nobel Prize in Physics and Ramsay the Nobel Prize in Chemistry in 1904. Because helium has the lowest boiling point of any substance known (4.2 K), it is used primarily as a cryogenic liquid. Helium and argon are both much less soluble in water (and therefore in blood) than N 2 , so scuba divers often use gas mixtures that contain these gases, rather than N 2 , to minimize the likelihood of the “bends,” the painful and potentially fatal formation of bubbles of N 2 (g) that can occur when a diver returns to the surface too rapidly. Preparation and General Properties of the Group 18 Elements Fractional distillation of liquid air is the only source of all the noble gases except helium. Although helium is the second most abundant element in the universe (after hydrogen), the helium originally present in Earth’s atmosphere was lost into space long ago because of its low molecular mass and resulting high mean velocity. Natural gas often contains relatively high concentrations of helium (up to 7%), however, and it is the only practical terrestrial source. The elements of group 18 all have closed-shell valence electron configurations, either ns 2 np 6 or 1s 2 for He. Consistent with periodic trends in atomic properties, these elements have high ionization energies that decrease smoothly down the group. From their electron affinities, the data in Table $\PageIndex{1}$ indicate that the noble gases are unlikely to form compounds in negative oxidation states. A potent oxidant is needed to oxidize noble gases and form compounds in positive oxidation states. Like the heavier halogens, xenon and perhaps krypton should form covalent compounds with F, O, and possibly Cl, in which they have even formal oxidation states (+2, +4, +6, and possibly +8). These predictions actually summarize the chemistry observed for these elements. Property Helium Neon Argon Krypton Xenon Radon atomic symbol He Ne Ar Kr Xe Rn atomic number 2 10 18 36 54 86 atomic mass (amu) 4.00 20.18 39.95 83.80 131.29 222 valence electron configuration* 1s2 2s22p6 3s23p6 4s24p6 5s25p6 6s26p6 triple point/boiling point (°C) —/−269† −249 (at 43 kPa)/−246 −189 (at 69 kPa)/−189 −157/−153 −112 (at 81.6 kPa)/−108 −71/−62 density (g/L) at 25°C 0.16 0.83 1.63 3.43 5.37 9.07 atomic radius (pm) 31 38 71 88 108 120 first ionization energy (kJ/mol) 2372 2081 1521 1351 1170 1037 normal oxidation state(s) 0 0 0 0 (+2) 0 (+2, +4, +6, +8) 0 (+2) electron affinity (kJ/mol) > 0 > 0 > 0 > 0 > 0 > 0 electronegativity — — — — 2.6 — product of reaction with O2 none none none none not directly with oxygen, but $\ce{XeO3}$ can be formed by Equation \ref{Eq5}. none type of oxide — — — — acidic — product of reaction with N2 none none none none none none product of reaction with X2 none none none KrF2 XeF2, XeF4, XeF6 RnF2 product of reaction with H2 none none none none none none The configuration shown does not include filled d and f subshells. †This is the normal boiling point of He. Solid He does not exist at 1 atm pressure, so no melting point can be given. The configuration shown does not include filled d and f subshells. †This is the normal boiling point of He. Solid He does not exist at 1 atm pressure, so no melting point can be given. The configuration shown does not include filled d and f subshells. †This is the normal boiling point of He. Solid He does not exist at 1 atm pressure, so no melting point can be given. The configuration shown does not include filled d and f subshells. †This is the normal boiling point of He. Solid He does not exist at 1 atm pressure, so no melting point can be given. The configuration shown does not include filled d and f subshells. †This is the normal boiling point of He. Solid He does not exist at 1 atm pressure, so no melting point can be given. The configuration shown does not include filled d and f subshells. †This is the normal boiling point of He. Solid He does not exist at 1 atm pressure, so no melting point can be given. *The configuration shown does not include filled d and f subshells. †This is the normal boiling point of He. Solid He does not exist at 1 atm pressure, so no melting point can be given. Reactions and Compounds of the Noble Gases For many years, it was thought that the only compounds the noble gases could form were clathrates. Clathrates are solid compounds in which a gas, the guest, occupies holes in a lattice formed by a less volatile, chemically dissimilar substance, the host (Figure $\PageIndex{1}$). Because clathrate formation does not involve the formation of chemical bonds between the guest (Xe) and the host molecules (H 2 O, in the case of xenon hydrate), the guest molecules are immediately released when the clathrate is melted or dissolved. Methane Clathrates In addition to the noble gases, many other species form stable clathrates. One of the most interesting is methane hydrate, large deposits of which occur naturally at the bottom of the oceans. It is estimated that the amount of methane in such deposits could have a major impact on the world’s energy needs later in this century. The widely held belief in the intrinsic lack of reactivity of the noble gases was challenged when Neil Bartlett, a British professor of chemistry at the University of British Columbia, showed that PtF 6 , a compound used in the Manhattan Project, could oxidize O 2 . Because the ionization energy of xenon (1170 kJ/mol) is actually lower than that of O 2 , Bartlett recognized that PtF 6 should also be able to oxidize xenon. When he mixed colorless xenon gas with deep red PtF 6 vapor, yellow-orange crystals immediately formed (Figure $\PageIndex{3}$). Although Bartlett initially postulated that they were $\ce{Xe^{+}PtF6^{−}}$, it is now generally agreed that the reaction also involves the transfer of a fluorine atom to xenon to give the $\ce{XeF^{+}}$ ion: \[\ce{Xe(g) + PtF6(g) -> [XeF^{+}][PtF5^{−}](s)} \label{Eq1}\] Subsequent work showed that xenon reacts directly with fluorine under relatively mild conditions to give XeF 2 , XeF 4 , or XeF 6 , depending on conditions; one such reaction is as follows: \[\ce{Xe(g) + 2F2(g) -> XeF4(s)} \label{Eq2}\] The ionization energies of helium, neon, and argon are so high (Table $\PageIndex{1}$) that no stable compounds of these elements are known. The ionization energies of krypton and xenon are lower but still very high; consequently only highly electronegative elements (F, O, and Cl) can form stable compounds with xenon and krypton without being oxidized themselves. Xenon reacts directly with only two elements: F 2 and Cl 2 . Although $\ce{XeCl2}$ and $\ce{KrF2}$ can be prepared directly from the elements, they are substantially less stable than the xenon fluorides. The ionization energies of helium, neon, and argon are so high that no stable compounds of these elements are known. Because halides of the noble gases are powerful oxidants and fluorinating agents, they decompose rapidly after contact with trace amounts of water, and they react violently with organic compounds or other reductants. The xenon fluorides are also Lewis acids; they react with the fluoride ion, the only Lewis base that is not oxidized immediately on contact, to form anionic complexes. For example, reacting cesium fluoride with XeF 6 produces CsXeF 7 , which gives Cs 2 XeF 8 when heated: \[\ce{XeF6(s) + CsF(s) -> CsXeF7(s)} \label{Eq3}\] \[\ce{2CsXeF7(s) ->[\Delta] Cs2XeF8(s) + XeF6(g)} \label{Eq4}\] The $\ce{XeF8^{2-}}$ ion contains eight-coordinate xenon and has the square antiprismatic structure, which is essentially identical to that of the IF 8 − ion. Cs 2 XeF 8 is surprisingly stable for a polyatomic ion that contains xenon in the +6 oxidation state, decomposing only at temperatures greater than 300°C. Major factors in the stability of Cs 2 XeF 8 are almost certainly the formation of a stable ionic lattice and the high coordination number of xenon, which protects the central atom from attack by other species. (Recall from that this latter effect is responsible for the extreme stability of SF 6 .) For a previously “inert” gas, xenon has a surprisingly high affinity for oxygen, presumably because of π bonding between $O$ and $Xe$. Consequently, xenon forms an extensive series of oxides and oxoanion salts. For example, hydrolysis of either $XeF_4$ or $XeF_6$ produces $XeO_3$, an explosive white solid: \[\ce{XeF6(aq) + 3H2O(l) -> XeO3(aq) + 6HF(aq)} \label{Eq5}\] Treating a solution of XeO 3 with ozone, a strong oxidant, results in further oxidation of xenon to give either XeO 4 , a colorless, explosive gas, or the surprisingly stable perxenate ion (XeO 6 4− ), both of which contain xenon in its highest possible oxidation state (+8). The chemistry of the xenon halides and oxides is best understood by analogy to the corresponding compounds of iodine. For example, XeO 3 is isoelectronic with the iodate ion (IO 3 − ), and XeF 8 2− is isoelectronic with the IF 8 − ion. Xenon has a high affinity for both fluorine and oxygen. Because the ionization energy of radon is less than that of xenon, in principle radon should be able to form an even greater variety of chemical compounds than xenon. Unfortunately, however, radon is so radioactive that its chemistry has not been extensively explored. Example $\PageIndex{1}$ On a virtual planet similar to Earth, at least one isotope of radon is not radioactive. A scientist explored its chemistry and presented her major conclusions in a trailblazing paper on radon compounds, focusing on the kinds of compounds formed and their stoichiometries. Based on periodic trends, how did she summarize the chemistry of radon? Given: nonradioactive isotope of radon Asked for: summary of its chemistry Strategy: Based on the position of radon in the periodic table and periodic trends in atomic properties, thermodynamics, and kinetics, predict the most likely reactions and compounds of radon. Solution We expect radon to be significantly easier to oxidize than xenon. Based on its position in the periodic table, however, we also expect its bonds to other atoms to be weaker than those formed by xenon. Radon should be more difficult to oxidize to its highest possible oxidation state (+8) than xenon because of the inert-pair effect. Consequently, radon should form an extensive series of fluorides, including RnF 2 , RnF 4 , RnF 6 , and possibly RnF 8 (due to its large radius). The ion RnF 8 2− should also exist. We expect radon to form a series of oxides similar to those of xenon, including RnO 3 and possibly RnO 4 . The biggest surprise in radon chemistry is likely to be the existence of stable chlorides, such as RnCl 2 and possibly even RnCl 4 . Exercise $\PageIndex{1}$ Predict the stoichiometry of the product formed by reacting XeF 6 with a 1:1 stoichiometric amount of KF and propose a reasonable structure for the anion. Answer $\ce{KXeF7}$; the xenon atom in XeF 7 − has 16 valence electrons, which according to the valence-shell electron-pair repulsion model could give either a square antiprismatic structure with one fluorine atom missing or a pentagonal bipyramid if the 5s 2 electrons behave like an inert pair that does not participate in bonding. Summary The noble gases are characterized by their high ionization energies and low electron affinities. Potent oxidants are needed to oxidize the noble gases to form compounds in positive oxidation states. The noble gases have a closed-shell valence electron configuration. The ionization energies of the noble gases decrease with increasing atomic number. Only highly electronegative elements can form stable compounds with the noble gases in positive oxidation states without being oxidized themselves. Xenon has a high affinity for both fluorine and oxygen, which form stable compounds that contain xenon in even oxidation states up to +8.
Courses/Pasadena_City_College/CHEM_001A%3A_General_Chemistry_and_Chemical_Analysis/19%3A_Gibbs_Energy_and_Thermodynamics	Template:HideTOC 19.1: Nature’s Heat Tax- You Can’t Win and You Can’t Break Even 19.2: Spontaneous and Nonspontaneous Processes Chemical and physical processes have a natural tendency to occur in one direction under certain conditions. A spontaneous process occurs without the need for a continual input of energy from some external source, while a nonspontaneous process requires such. Systems undergoing a spontaneous process may or may not experience a gain or loss of energy, but they will experience a change in the way matter and/or energy is distributed within the system. 19.3: Entropy and the Second Law of Thermodynamics Entropy (S) is a state function whose value increases with an increase in the number of available microstates.For a given system, the greater the number of microstates, the higher the entropy. During a spontaneous process, the entropy of the universe increases. 19.4: Entropy Changes Associated with State Changes under construction 19.5: Heat Transfer and Changes in the Entropy of the Surroundings 19.6: Gibbs Energy We can predict whether a reaction will occur spontaneously by combining the entropy, enthalpy, and temperature of a system in a new state function called Gibbs free energy (G). The change in free energy (ΔG) is the difference between the heat released during a process and the heat released for the same process occurring in a reversible manner. If a system is at equilibrium, ΔG = 0. If the process is spontaneous, ΔG < 0. If the process is not spontaneous as written. 19.7: Entropy Changes in Chemical Reactions Changes in internal energy, that are not accompanied by a temperature change, might reflect changes in the entropy of the system. 19.8: Gibbs Energy Changes in Chemical Reactions We can predict whether a reaction will occur spontaneously by combining the entropy, enthalpy, and temperature of a system in a new state function called Gibbs free energy (G). The change in free energy (ΔG) is the difference between the heat released during a process and the heat released for the same process occurring in a reversible manner. If a system is at equilibrium, ΔG = 0. If the process is spontaneous, ΔG < 0. If the process is not spontaneous as written. 19.9: Gibbs Energy Changers for Non-Standard States For a reversible process (with no external work), the change in free energy can be expressed in terms of volume, pressure, entropy, and temperature. If ΔG° < 0, then K > 1, and products are favored over reactants. If ΔG° > 0, then K < 1, and reactants are favored over products. If ΔG° = 0, then K = 1, and the system is at equilibrium. We can use the measured equilibrium constant K at one temperature and ΔH° to estimate the equilibrium constant for a reaction at any other temperature. 19.10: Gibbs Energy and Equilibrium For a reversible process (with no external work), the change in free energy can be expressed in terms of volume, pressure, entropy, and temperature. If ΔG° < 0, then K > 1, and products are favored over reactants. If ΔG° > 0, then K < 1, and reactants are favored over products. If ΔG° = 0, then K = 1, and the system is at equilibrium. We can use the measured equilibrium constant K at one temperature and ΔH° to estimate the equilibrium constant for a reaction at any other temperature.
Courses/Saint_Francis_University/Chem_114%3A_Human_Chemistry_II_(Hargittai)/15%3A_Amines/15.03%3A_Heterocyclic_Nitrogen_Compounds	Learning Objectives Objective 1 Objective 2 Heterocyclic Amines Looking back at the various cyclic hydrocarbons discussed previously, we see that all the atoms in the rings of these compounds are carbon atoms. In other cyclic compounds, called heterocyclic compounds (Greek heteros , meaning “other”), nitrogen, oxygen, sulfur, or some other atom is incorporated in the ring. Many heterocyclic compounds are important in medicine and biochemistry. Some compose part of the structure of the nucleic acids, which in turn compose the genetic material of cells and direct protein synthesis. Many heterocyclic amines occur naturally in plants. Like other amines, these compounds are basic. Such a compound is an alkaloid , a name that means “like alkalis.” Many alkaloids are physiologically active, including the familiar drugs caffeine, nicotine, and cocaine. To Your Health: Three Well-Known Alkaloids Caffeine is a stimulant found in coffee, tea, and some soft drinks. Its mechanism of action is not well understood, but it is thought to block the activity of adenosine, a heterocyclic base that acts as a neurotransmitter, a substance that carries messages across a tiny gap (synapse) from one nerve cell (neuron) to another cell. The effective dose of caffeine is about 200 mg, corresponding to about two cups of strong coffee or tea. Nicotine acts as a stimulant by a different mechanism; it probably mimics the action of the neurotransmitter acetylcholine. People ingest this drug by smoking or chewing tobacco. Its stimulant effect seems transient, as this initial response is followed by depression. Nicotine is highly toxic to animals. It is especially deadly when injected; the lethal dose for a human is estimated to be about 50 mg. Nicotine has also been used in agriculture as a contact insecticide. Cocaine acts as a stimulant by preventing nerve cells from taking up dopamine, another neurotransmitter, from the synapse. High levels of dopamine are therefore available to stimulate the pleasure centers of the brain. The enhancement of dopamine action is thought to be responsible for cocaine’s “high” and its addictive properties. After the binge, dopamine is depleted in less than an hour. This leaves the user in a pleasureless state and (often) craving more cocaine. Cocaine is used as the salt cocaine hydrochloride and in the form of broken lumps of the free (unneutralized) base, which is called crack cocaine . Because it is soluble in water, cocaine hydrochloride is readily absorbed through the watery mucous membranes of the nose when it is snorted. Crack cocaine is more volatile than cocaine hydrochloride. It vaporizes at the temperature of a burning cigarette. When smoked, cocaine reaches the brain in 15 s. Summary Amines are bases; they react with acids to form salts. Salts of aniline are properly named as anilinium compounds, but an older system is used to name drugs: the salts of amine drugs and hydrochloric acid are called “hydrochlorides.” Heterocyclic amines are cyclic compounds with one or more nitrogen atoms in the ring.
Courses/UW-Whitewater/UWX_CH114%3A_Chemistry_in_the_Kitchen/06%3A_Macronutrients_-_Fats/6.04%3A_Steroids	Learning Objectives Use the structure of the steroid skeleton to recognize the presence of a steroid Identify the functions of steroids produced in mammals. All the lipids discussed so far are saponifiable , reacting with aqueous base solution to yield simpler components, such as glycerol, fatty acids, amino alcohols, and sugars. Lipid samples extracted from cellular material, however, also contain a small but important fraction that does not react with base. The most important nonsaponifiable lipids are the steroids. These compounds include the bile salts , cholesterol and related compounds, and certain hormones (such as cortisone and the sex hormones). Steroids occur in plants, animals, yeasts, and molds but not in bacteria. They may exist in free form or combined with fatty acids or carbohydrates. All steroids have a characteristic structural component consisting of four fused rings. Chemists identify the rings by capital letters and number the carbon atoms as shown in Figure $\PageIndex{1a}$. Slight variations in this structure or in the atoms or groups attached to it produce profound differences in biological activity. Cholesterol Cholesterol (Figure $\PageIndex{1b}$) does not occur in plants, but it is the most abundant steroid in the human body (240 g is a typical amount). Excess cholesterol is believed to be a primary factor in the development of atherosclerosis and heart disease, which are major health problems in the United States today. About half of the body’s cholesterol is interspersed in the lipid bilayer of cell membranes. Much of the rest is converted to cholic acid, which is used in the formation of bile salts. Cholesterol is also a precursor in the synthesis of sex hormones, adrenal hormones, and vitamin D. Excess cholesterol not metabolized by the body is released from the liver and transported by the blood to the gallbladder. Normally, it stays in solution there until being secreted into the intestine (as a component of bile) to be eliminated. Sometimes, however, cholesterol in the gallbladder precipitates in the form of gallstones (Figure $\PageIndex{2}$). Indeed, the name cholesterol is derived from the Greek chole , meaning “bile,” and stereos , meaning “solid.” To Your Health: Cholesterol and Heart Disease Heart disease is the leading cause of death in the United States for both men and women. The Centers for Disease Control and Prevention reported that heart disease claimed 631,636 lives in the United States (26% of all reported deaths) in 2006. Scientists agree that elevated cholesterol levels in the blood, as well as high blood pressure, obesity, diabetes, and cigarette smoking, are associated with an increased risk of heart disease. A long-term investigation by the National Institutes of Health showed that among men ages 30 to 49, the incidence of heart disease was five times greater for those whose cholesterol levels were above 260 mg/100 mL of serum than for those with cholesterol levels of 200 mg/100 mL or less. The cholesterol content of blood varies considerably with age, diet, and sex. Young adults average about 170 mg of cholesterol per 100 mL of blood, whereas males at age 55 may have cholesterol levels at 250 mg/100 mL or higher because the rate of cholesterol breakdown decreases with age. Females tend to have lower blood cholesterol levels than males. To understand the link between heart disease and cholesterol levels, it is important to understand how cholesterol and other lipids are transported in the body. Lipids, such as cholesterol, are not soluble in water and therefore cannot be transported in the blood (an aqueous medium) unless they are complexed with proteins that are soluble in water, forming assemblages called lipoproteins . Lipoproteins are classified according to their density, which is dependent on the relative amounts of protein and lipid they contain. Lipids are less dense than proteins, so lipoproteins containing a greater proportion of lipid are less dense than those containing a greater proportion of protein. Research on cholesterol and its role in heart disease has focused on serum levels of low-density lipoproteins (LDLs) and high-density lipoproteins (HDLs). One of the most fascinating discoveries is that high levels of HDLs reduce a person’s risk of developing heart disease, whereas high levels of LDLs increase that risk. Thus the serum LDL:HDL ratio is a better predictor of heart disease risk than the overall level of serum cholesterol. Persons who, because of hereditary or dietary factors, have high LDL:HDL ratios in their blood have a higher incidence of heart disease. How do HDLs reduce the risk of developing heart disease? No one knows for sure, but one role of HDLs appears to be the transport of excess cholesterol to the liver, where it can be metabolized. Therefore, HDLs aid in removing cholesterol from blood and from the smooth muscle cells of the arterial wall. Dietary modifications and increased physical activity can help lower total cholesterol and improve the LDL:HDL ratio. The average American consumes about 600 mg of cholesterol from animal products each day and also synthesizes approximately 1 g of cholesterol each day, mostly in the liver. The amount of cholesterol synthesized is controlled by the cholesterol level in the blood; when the blood cholesterol level exceeds 150 mg/100 mL, the rate of cholesterol biosynthesis is halved. Hence, if cholesterol is present in the diet, a feedback mechanism suppresses its synthesis in the liver. However, the ratio of suppression is not a 1:1 ratio; the reduction in biosynthesis does not equal the amount of cholesterol ingested. Thus, dietary substitutions of unsaturated fat for saturated fat, as well as a reduction in consumption of trans fatty acids, is recommended to help lower serum cholesterol and the risk of heart disease. Steroid Hormones Hormones are chemical messengers that are released in one tissue and transported through the circulatory system to one or more other tissues. One group of hormones is known as steroid hormones because these hormones are synthesized from cholesterol, which is also a steroid. There are two main groups of steroid hormones: adrenocortical hormones and sex hormones. The adrenocortical hormones, such as aldosterone and cortisol (Table $\PageIndex{1}$), are produced by the adrenal gland, which is located adjacent to each kidney. Aldosterone acts on most cells in the body, but it is particularly effective at enhancing the rate of reabsorption of sodium ions in the kidney tubules and increasing the secretion of potassium ions and/or hydrogen ions by the tubules. Because the concentration of sodium ions is the major factor influencing water retention in tissues, aldosterone promotes water retention and reduces urine output. Cortisol regulates several key metabolic reactions (for example, increasing glucose production and mobilizing fatty acids and amino acids). It also inhibits the inflammatory response of tissue to injury or stress. Cortisol and its analogs are therefore used pharmacologically as immunosuppressants after transplant operations and in the treatment of severe skin allergies and autoimmune diseases, such as rheumatoid arthritis. Hormone Effect NaN regulates salt metabolism; stimulates kidneys to retain sodium and excrete potassium NaN stimulates the conversion of proteins to carbohydrates NaN regulates the menstrual cycle; maintains pregnancy NaN stimulates female sex characteristics; regulates changes during the menstrual cycle NaN stimulates and maintains male sex characteristics The sex hormones are a class of steroid hormones secreted by the gonads (ovaries or testes), the placenta, and the adrenal glands. Testosterone and androstenedione are the primary male sex hormones, or androgens , controlling the primary sexual characteristics of males, or the development of the male genital organs and the continuous production of sperm. Androgens are also responsible for the development of secondary male characteristics, such as facial hair, deep voice, and muscle strength. Two kinds of sex hormones are of particular importance in females: progesterone, which prepares the uterus for pregnancy and prevents the further release of eggs from the ovaries during pregnancy, and the estrogens, which are mainly responsible for the development of female secondary sexual characteristics, such as breast development and increased deposition of fat tissue in the breasts, the buttocks, and the thighs. Both males and females produce androgens and estrogens, differing in the amounts of secreted hormones rather than in the presence or absence of one or the other. Sex hormones, both natural and synthetic, are sometimes used therapeutically. For example, a woman who has had her ovaries removed may be given female hormones to compensate. Some of the earliest chemical compounds employed in cancer chemotherapy were sex hormones. For example, estrogens are one treatment option for prostate cancer because they block the release and activity of testosterone. Testosterone enhances prostate cancer growth. Sex hormones are also administered in preparation for sex-change operations, to promote the development of the proper secondary sexual characteristics. Oral contraceptives are synthetic derivatives of the female sex hormones; they work by preventing ovulation. Bile Salts Bile is a yellowish green liquid (pH 7.8–8.6) produced in the liver. The most important constituents of bile are bile salts, which are sodium salts of amidelike combinations of bile acids, such as cholic acid (part (a) of Figure $\PageIndex{3}$) and an amine such as the amino acid glycine (part (b) of Figure $\PageIndex{3}$). They are synthesized from cholesterol in the liver, stored in the gallbladder, and then secreted in bile into the small intestine. In the gallbladder, the composition of bile gradually changes as water is absorbed and the other components become more concentrated. Because they contain both hydrophobic and hydrophilic groups, bile salts are highly effective detergents and emulsifying agents; they break down large fat globules into smaller ones and keep those smaller globules suspended in the aqueous digestive environment. Enzymes can then hydrolyze fat molecules more efficiently. Thus, the major function of bile salts is to aid in the digestion of dietary lipids. Surgical removal is often advised for a gallbladder that becomes infected, inflamed, or perforated. This surgery does not seriously affect digestion because bile is still produced by the liver, but the liver’s bile is more dilute and its secretion into the small intestine is not as closely tied to the arrival of food. Can you identify steroids at a glance? A helpful way to remember the basic structure of a steroid is to focus on the four rings in the structure.The presence of the rings indicate that you are looking at a compound that belongs to the steroid family. Some chemists nickname the ring structure "three rooms and a garage". You should be able to identify steroids at a glance, and recall they are a member of the lipid family of biomolecules. You are not required to name any of the compounds or classify the biomolecules formed from the basic steroid ring structure. Summary Steroids have a four-fused-ring structure and have a variety of functions. Cholesterol is a steroid found in mammals that is needed for the formation of cell membranes, bile acids, and several hormones. Bile salts are secreted into the small intestine to aid in the digestion of fats.
Ancillary_Materials/Worksheets/Worksheets%3A_Inorganic_Chemistry/Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry/12%3A_Molecular_Orbitals/12.17%3A_Heteroaromatics	Many important biomolecules behave as aromatics, even though they don't contain benzene. Histidine is a very common amino acid with a cyclic side chain. The bases in nucleic acids, DNA and RNA, are also aromatic. Closer examination shows that these compounds follow some variation of Huckel's rules for aromaticity. The variations involve the inclusion of lone pairs in the pi bonding system. Pyrrole is a five-membered ring with a nitrogen in it. It contains two C=C bonds. Pyrrole is aromatic. It usually reacts in a way similar to benzene, rather than like alkenes. But how does it fit the rules for aromaticity? Pyrrole is cyclic. It is flat. Is it conjugated? It does have alternating double bond-single bond-double-single- lone pair . We have seen before that lone pairs can be conjugated with pi bonds. So, pyrrole is conjugated all the way around the ring. Counting the lone pair as a pair of pi electrons gives it an odd number. a lone pair can act like a pair of pi electrons if the molecule will become aromatic as a result. Does a lone pair always have to act as a pi pair? Pyridine is a six membered ring, exactly like benzene, except that one CH unit is replaced by a nitrogen. The nitrogen has a lone pair. Pyridine already has three pi bonds. It is flat and fully conjugated. Does the lone pair add in to the system and make it anti-aromatic? It turns out it can't. There is already a pi bond on that nitrogen, so it would be difficult for the lone pair to occupy the same space, above and below the ring. Instead, the lone pair remains orthogonal to the pi system. It is in the plane of the ring, rather than above and below it. It completely avoids interacting with the pi system. a lone pair can't act like a pair of pi electrons if there is already another pair of pi electrons on the same atom. Problem MO17.1. Create a Hückel MO diagram for each of the following 5-membered rings. Add electrons. Label the HOMO and the LUMO. Draw pictures of the molecular orbitals. Indicate the number of nodes for each orbital energy level. Label the energy levels (π, σ, n, π, σ) Problem MO17.2. Create a Hückel MO diagram for each of the following 6-membered rings. Add electrons. Label the HOMO and the LUMO. Draw pictures of the molecular orbitals. Indicate the number of nodes for each orbital energy level. Label the energy levels (π, σ, n, π, σ)
Courses/San_Francisco_State_University/General_Physical_Chemistry_I_(Gerber)/04%3A_The_First_Law_of_Thermodynamics/4.05%3A_An_Adiabatic_Process_is_a_Process_in_which_No_Energy_as_Heat_is_Transferred	Isothermal expansion of an ideal gas For a monatomic ideal gas we have seen that energy $\langle E \rangle$ observed as $U= 3/2 nRT$. This means that energy is only dependent on temperature and if a gas is compressed isothermally , then the internal energy does not change: \[ΔU_{isothermal-ideal gas} = 0 \nonumber \] This means that the reversible work must cancel the reversible heat: \[ΔU_{rev} = w_{rev} + q_{rev} = 0 \nonumber \] Therefore \[w_{rev} = - q_{rev} \nonumber \] so from the expression of the reversible work for expansion in the last section \[ q_{rev}= nRT\ln \dfrac{V_2}{V_1} \nonumber \] If $V_2>V_1$ (expansion), then you (or the environment) must put heat into the system because this is a positive number. Adiabatic expansion of an ideal gas Now suppose you make sure that no heat can enter the cylinder. (Put it in styrofoam or so). Then the path can still be reversible (slow pulling) but the process is then adiabatic . This bat- part comes from a Greek verb βαινω (baino) that means walking, compare acro bat , someone who goes high places (acro-). The δια (dia) part means 'through' (cf. diagram, diorama, diagonal etc.) and the prefix α- (a-) denies it all (compare atypical versus typical ). So the styrofoam prevents the heat from walking through the wall. When expanding the gas from V 1 to V 2 it still does reversible work but where does that come from? It can only come from the internal energy itself. So in this case any energy change should consist of work (adiabatic means: $δq=0$). \[dU = δw_{rev} \nonumber \] This implies that the temperature must drop, because if $U$ changes, then $T$ must change. The change of energy with temperature at constant volume is known as the heat capacity (at constant volume) $C_v$ \[ C_v =\left( \dfrac{\partial U}{\partial T} \right)_V \nonumber \] For an ideal gas $U$ only changes with temperature, so that or: We can now compare two paths to go from state $P_1,V_1,T_1$ to state $P_2,V_2,T_1$: Reversible isothermal expansion A Reversible adiabatic expansion B followed by reversible isochoric heating C Notice that the temperature remains $T_1$ for path A (isotherm!), but that it drops to $T_2$ on the adiabat B, so that the cylinder has to be isochorically warmed up, C, to regain the same temperature. $\Delta U_{tot}$ should be the same for both path A and the combined path B+C, because the end points are the same ($U$ is a state function!). As the and points are at the same temperature and $U$ only depends on $T$: \[\Delta U_{tot}=0 \nonumber \] Along adiabat B: \[q_{rev}=0 \nonumber \] Along isochoric heating C, there is no volume work because the volume is kept constant, so that: This is the only reversible heat involved in path B+C. However, we know that $\Delta U_{tot}$ for path A is zero (isothermal!). This means that the volume work along B must cancel the heat along C: The book keeping looks as follows, all paths are reversible: \[\Delta U_{B+C} = \Delta U_A = 0 = q_B + w_B + q_C + w_C \nonumber \] We know that $q_B=0$ since it is an adiabat and $w_C=0$ since it is an isochore: \[\Delta U_{B+C} = \Delta U_A = 0 = 0 + w_B + q_C + 0 \nonumber \] Therefore: \[w_B=-q_C \nonumber \] We had already seen before that along the isotherm A: \[w_A = - q_A = - nRT \ln \dfrac{V_1}{V_2} \nonumber \] As you can see \w_A\) and $w_B$ are not the same. Work is a path function, even if reversible. As we are working with an ideal gas we can be more precise about $w_B$ and $q_c$ as well. The term $w_B$ along the adiabat is reversible volume work. Since there is no heat along B we can write a straight $d$ instead of $\delta$ for the work contributions (It is the only contribution and must be identical to the state function $dU$):
Bookshelves/Organic_Chemistry/Organic_Chemistry_III_(Morsch_et_al.)/25%3A_Carbohydrates/25.05%3A_Cyclic_Structures_of_Monosaccharides_-_Anomers	Objectives After completing this section, you should be able to determine whether a given monosaccharide will exist as a pyranose or furanose. draw the cyclic pyranose form of a monosaccharide, given its Fischer projection. draw the Fischer projection of a monosaccharide, given its cyclic pyranose form. draw, from memory, the cyclic pyranose form of D-glucose. determine whether a given cyclic pyranose form represents the D or L form of the monosaccharide concerned. describe the phenomenon known as mutarotation. explain, through the use of chemical equations, exactly what happens at the molecular level during the mutarotation process. Key Terms Make certain that you can define, and use in context, the key terms below. alpha anomer anomer anomeric centre beta anomer furanose mutarotation pyranose Study Notes If necessary, before you attempt to study this section, review the formation of hemiacetals discussed in Section 19.10 . Cyclic Monosaccharides In Section 19-10 it was discussed that the reaction of one equivalent of an alcohol, in the presence of an acid catalyst, adds reversibly to aldehydes and ketones to form a hydroxy ether called a hemiacetal (R 2 COHOR') (h emi , Greek, half). Molecules which have both an alcohol and a carbonyl can undergo an intramolecular version of the same reaction forming a cyclic hemiacetal. Because sugars often contain alcohol and carbonyl functional groups, intramolecular hemiacetal formation is common in carbohydrate chemistry. Five and six-membered rings are favored over other ring sizes because of their low angle and eclipsing strain. Cyclic structures of this kind are termed furanose (five-membered) or pyranose (six-membered), reflecting the ring size relationship to the common heterocyclic compounds furan and pyran shown below. Furan (5-membered ring) and pyran (6-membered ring) structures Unlike most of the biochemical reactions you will see in this text, sugar cyclization reactions are not catalyzed by enzymes: they occur spontaneously and reversibly in aqueous solution. Sugars are often shown in their open-chain form, however, in aqueous solution, glucose, fructose, and other sugars of five or six carbons rapidly interconvert between straight-chain and cyclic forms. For most five- and six-carbon sugars, the cyclic forms predominate in equilibrium since they are more stable. The size of the cyclic hemiacetal ring adopted by a given sugar is not constant, but may vary with substituents and other structural features. Aldohexoses usually form pyranose rings and their pentose homologs tend to prefer the furanose form, but there are many counter examples. At equilibrium less than 1% of glucose is in an open chain form with the rest being almost exclusively in its cyclic pyranose form. The pyranose ring is formed by attack of the hydroxyl on carbon 5 of glucose to the aldehyde carbon (carbon #1, also called the anomeric carbon in carbohydrate terminology). The cyclic form of glucose is called glucopyranose. Notice that for glucose and other aldohexoses the hydroxyl that forms the cyclic hemiacetal is also the one that determines the D/L designation of a sugar. Pyranose rings are often drawn in a chair conformation like cyclohexane rings ( Section 4-6) with substituents being either an axial or equatorial position. Pyranose rings are even capable of undergoing a ring flip to change between chair conformations. By convention the ring oxygen is placed to the right and to the rear of the structure (top right of the drawing). Groups which go to the right in a Fischer projection will be orientend 'down' of the pyranose ring while groups to the left are oriented 'up' in the chair structure. Also, the terminal -CH 2 OH group is oriented up on the pyranose ring for D-sugars and down for L-Sugars. When D-glucose cyclizes it forms a 37/63 mixture of the alpha and beta anomer respectively. The beta anomer is preferred because β -D-glucopyranose is the only aldohexose which can be drawn with all its bulky substituents (-OH and -CH 2 OH) in equatorial positions, making it the most stable of the eight D-aldohexoses, which probably accounts for its widespread prevalence in nature. It is possible to obtain a sample of crystalline glucose in which all the molecules have the α structure or all have the β structure. The α form melts at 146°C and has a specific rotation of +112°, while the β form melts at 150°C and has a specific rotation of +18.7°. When the sample is dissolved in water, however, a mixture is soon produced containing both anomers as well as the straight-chain form, in dynamic equilibrium. You can start with a pure crystalline sample of glucose consisting entirely of either anomer, but as soon as the molecules dissolve in water, they open to form the carbonyl group and then re-close to form either the α or the β anomer. The opening and closing repeats continuously in an ongoing interconversion between anomeric forms and is referred to as mutarotation (Latin mutare , meaning “to change”). At equilibrium, the mixture consists of about 36% α-D-glucose, 64% β-D-glucose, and less than 0.02% of the open-chain aldehyde form. The observed rotation of this solution is +52.7°. Example $\PageIndex{1}$ Fructose in aqueous solution forms a six-membered cyclic hemiketal called fructopyranose when the hydroxyl oxygen on carbon #6 attacks the ketone carbon (carbon #2, the anomeric carbon in fructose). In this case, the β anomer is heavily favored in equilibrium by a ratio of 70:1, because in the minor α anomer the bulkier -CH 2 OH group occupies an axial position. Notice in the above figure that the percentages of α and β anomers present at equilibrium do not add up to 100%. Fructose also exists in solution as a five-membered cyclic hemiketal, referred to in carbohydrate nomenclature as fructofuranose. In the formation of fructofuranose from open-chain fructose, the hydroxyl group on the fifth carbon attacks the ketone. In aqueous solution, then, fructose exists as an equilibrium mixture of 70% β-fructopyranose, 23% β-fructofuranose, and smaller percentages of the open chain and cyclic α-anomers. The β-pyranose form of fructose is one of the sweetest compounds known, and is the main component of high-fructose corn syrup. The β-furanose form is much less sweet. Although we have been looking at specific examples for glucose and fructose, other five- and six-carbon monosaccharides also exist in solution as equilibrium mixtures of open chains and cyclic hemiacetals and hemiketals. Shorter monosaccharides are unlikely to undergo analogous ring-forming reactions, however, due to the inherent instability of three and four-membered rings. Drawing Cyclic Structures of Monosaccharides The cyclic forms of sugars are commonly depicted as Haworth projections . This convention, first suggested by the English chemist Walter N. Haworth, shows molecules drawn as planar rings with darkened edges representing the side facing toward the viewer. The structure is simplified to show only the functional groups attached to the carbon atoms. Any group written to the right in a Fischer projection appears below (bottom face) the plane of the ring in a Haworth projection, and any group written to the left in a Fischer projection appears above (top face) the plane in a Haworth projection. Figure: Conversion of the Fischer projection of D-glucose to the Haworth projection of ß-D-glucose. When converting a Fischer projection (line) to a Haworth projection, you must first identify the type of monosaccharide involved. If the carbohydrate represents an aldohexose, the pyranose ring is typically used. A pyranose is a cyclic structure that contains five carbon atoms and an oxygen. If the carbohydrate represents a ketohexose, the furanose ring is typically used. The furanose ring contains four carbon atoms and an oxygen. Indicate the arrangement of the hydroxyl group attached to the anomeric carbon to identify the sugar as an alpha or beta anomer. The α and β anomers are determined with respect to carbon 6. If the molecule represents a D-sugar, carbon 6 will be above the plane of the ring (top face) and form an L-sugar, carbon 6 will be below the plane of the ring (ring). The α anomer occurs when the OH on the anomeric carbon is trans to carbon 6 and the β anomer occurs when the OH on the anomeric carbon is cis to carbon 6. If the cyclic structure contains a furanose, since carbon 1 is not included within the ring, that carbon group would be arranged in the opposite direction of the OH group. The remaining chiral centers (carbons 2, 3 and 4 of the pyranose or carbons 3 and 4 of the furanose) are arranged based on the directions of the hydroxyl from the Fischer projection structures. Groups to the left of the Fischer projection would point up (top face), while groups to the right would point down (bottom face). Since the Fischer Projection of any given carbohydrate is always the same, the Haworth Projection is essentially always the same. The only differences between the Haworth Projection of the alpha or beta form of a single carbohydrate, is how the OH (and carbon 1 if furanose ring) is arranged around the anomeric carbon to determine whether the molecule is alpha or beta. Stability of Chair Conformation in Pyranose Sugars Previously, we have seen the six ring atoms of cyclic glucose drawn in two dimensions. A more accurate depiction shows that the molecule adopts, as expected, a chair conformation. The conformation in which all substituents are equatorial is lower in energy. The two isomeric forms are referred to by the Greek letters alpha ( α ) and beta ( β ). We have not learned about stereoisomerism quite yet, but you can still recognize that the bonding configuration on one carbon is different. On the alpha isomer, one of the hydroxyl groups is axial – this isomer is not able to adopt a chair conformation in which all non-hydrogen substituents are equatorial. The lower energy conformation is the one in which four of the five substituents are equatorial, but the presence of the one axial hydroxyl group means that the alpha isomer is, overall, less stable than the beta isomer. The most abundant form of fructose in aqueous solution is also a six-membered ring. The lower energy chair conformation is the one with three of the five substituents (including the bulky –CH 2 OH group) in the equatorial position. Exercise $\PageIndex{1}$ Draw the following in their most stable chair conformation: α-D-galactopyranose and α-D-mannopyranose. Which is expected to be the more stable? Answer Because the both have two axial OH's their chair conformations should be roughly the same stability. Exercise $\PageIndex{2}$ Draw the two chair conformations of the sugar called mannose, being sure to clearly show each non-hydrogen substituent as axial or equatorial. Predict which conformation is likely to be more stable, and explain why. Answer Exercise $\PageIndex{3}$ Draw the cyclic structure of α-D-altrose. Answer Exercise $\PageIndex{4}$ Draw the cyclic structure for β-D-galactose. Identify the anomeric carbon. Answer To identify the structure, we should first start with the Fischer projection of D-galactose. Since it is an aldohexose, we will start with the pyranose ring. The beta anomer was requested, so the OH on the anomeric carbon (C1) is cis to C6. Since C6 is top face (pointing up), the OH will be top face. Carbons 2, 3, and 4 are then arranged based on the Fischer projection arrangement at those carbons (C2 right, C3 left, and C4 left). Exercise $\PageIndex{5}$ Given that the aldohexose D-mannose differs from D-glucose only in the configuration at the second carbon atom, draw the cyclic structure for α-D-mannose. Answer Exercise $\PageIndex{6}$ Draw the cyclic structure for β-D-glucose. Identify the anomeric carbon. Answer Exercise $\PageIndex{7}$ a) Identify the anomeric carbon of each of the sugars shown below, and specify whether the structure shown is a hemiacetal or hemiketal. b) Draw mechanisms for cyclization of the open-chain forms to the cyclic forms shown. Answer Exercise $\PageIndex{8}$ Draw a mechanism for the conversion of α-glucopyranose t o open-chain glucose. Answer 6. Exercise $\PageIndex{9}$ Identify the following monosaccharide, write its full name, and draw its open-chain form as a Fischer projection. Answer β-D-idopyranose.
Bookshelves/General_Chemistry/Chemistry_2e_(OpenStax)/19%3A_Transition_Metals_and_Coordination_Chemistry/19.00%3A_Introduction	We have daily contact with many transition metals. Iron occurs everywhere—from the rings in your spiral notebook and the cutlery in your kitchen to automobiles, ships, buildings, and in the hemoglobin in your blood. Titanium is useful in the manufacture of lightweight, durable products such as bicycle frames, artificial hips, and jewelry. Chromium is useful as a protective plating on plumbing fixtures and automotive detailing. In addition to being used in their pure elemental forms, many compounds containing transition metals have numerous other applications. Silver nitrate is used to create mirrors, zirconium silicate provides friction in automotive brakes, and many important cancer-fighting agents, like the drug cisplatin and related species, are platinum compounds. The variety of properties exhibited by transition metals is due to their complex valence shells. Unlike most main group metals where one oxidation state is normally observed, the valence shell structure of transition metals means that they usually occur in several different stable oxidation states. In addition, electron transitions in these elements can correspond with absorption of photons in the visible electromagnetic spectrum, leading to colored compounds. Because of these behaviors, transition metals exhibit a rich and fascinating chemistry.
Courses/Los_Angeles_Trade_Technical_College/DMA_Chem_51/2%3A_Beginning_Chemistry_(Ball)/13%3A_Nuclear_Chemistry	Most chemists pay little attention to the nucleus of an atom except to consider the number of protons it contains because that determines an element’s identity. However, in nuclear chemistry, the composition of the nucleus and the changes that occur there are very important. Applications of nuclear chemistry may be more widespread than you realize. Many people are aware of nuclear power plants and nuclear bombs, but nuclear chemistry also has applications ranging from smoke detectors to medicine, from the sterilization of food to the analysis of ancient artifacts. In this chapter, we will examine some of the basic concepts of nuclear chemistry and some of the nuclear reactions that are important in our everyday lives. 13.1: Prelude to Nuclear Chemistry Many people think of nuclear chemistry in connection with the nuclear power industry and atomic bombs but do not realize that most smoke detectors rely on nuclear chemistry and save countless lives every year. The applications of nuclear chemistry may be more widespread than you think. 13.2: Radioactivity The major types of radioactivity include alpha particles, beta particles, and gamma rays. Fission is a type of radioactivity in which large nuclei spontaneously break apart into smaller nuclei. 13.3: Uses of Radioactive Isotopes Radioactivity has several practical applications, including tracers, medical applications, dating once-living objects, and preservation of food. 13.4: Nuclear Chemistry (Exercises) These are exercises and select solutions to company Chapter 15 of the "Beginning Chemistry" Textmap formulated around the Ball et al. textbook. Thumbnail: Part of carbon–nitrogen–oxygen (CNO) reaction chain diagram, made just to be illustrative for nuclear reactions in general. Image used with permission (CC BY-SA 3.0; Michalsmid ).
Courses/Lumen_Learning/Book%3A_US_History_II_(OS_Collection)_(Lumen)/07%3A_Politics_in_the_Gilded_Age%2C_1870-1900/07.1%3A_Introduction	L. Frank Baum’s story of a Kansas girl and the magical land of Oz has become a classic of both film and screen, but it may have originated in part as an allegory of late nineteenth-century politics and the rise of the Populist movement. L. Frank Baum was a journalist who rose to prominence at the end of the nineteenth century. Baum’s most famous story, The Wizard of Oz , was published in 1900, but “Oz” first came into being years earlier, when he told a story to a group of schoolchildren visiting his newspaper office in South Dakota. He made up a tale of a wonderful land, and, searching for a name, he allegedly glanced down at his file cabinet, where the bottom drawer was labeled “O-Z.” Thus was born the world of Oz, where a girl from struggling Kansas hoped to get help from a “wonderful wizard” who proved to be a fraud. Since then, many have speculated that the story reflected Baum’s political sympathies for the Populist Party, which galvanized midwestern and southern farmers’ demands for federal reform. Whether he intended the story to act as an allegory for the plight of farmers and workers in late nineteenth-century America, or whether he simply wanted to write an “American fairy tale” set in the heartland, Populists looked for answers much like Dorothy did. And the government in Washington proved to be meek rather than magical. 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/
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Solubilty/Solubility	One of the general properties of ionic compounds is water solubility. The oceans are solutions of salt in water. In a mixture, two or more materials are mixed together but they remain essentially separate, like sand and water. The sand can be easily distinguished from the water, because even if a mixture of the two is shaken it will spontaneously separate over time. In a suspension, one or more materials is mixed into a liquid, and the mixture becomes somewhat homogeneous. Instead of having easily identifiable layers, the liquid has a uniform appearance throughout. However, suspensions are generally cloudy liquids. Milk is a suspension containing water, fats and proteins. They may settle out into separate layers eventually, but it takes time. In a solution, one or more materials is mixed into a liquid, and the mixture becomes a completely homogeneous liquid. Solutions are transparent, not cloudy, and can be colored or colorless. Saltwater, an example of a solution, is diagrammed in the figure below. Pieces of salt are not visible in the solution; the salt particles are too small. The salt is separated into individual ions, surrounded by water molecules. This change from Figure IC4.3 to Figure IC4.2 is not instantaneous upon adding salt to water; stirring is required to produce the solution. Eventually more of the salt dissolves in the water, as shown: If enough salt is added, the system might come to "equilibrium": the water has dissolved all of the salt that it can, so the rest of the salt remains solid. This equilibrium is "dynamic": ions are dissolved in the water at the same time that ions are deposited from solution into the solid state. However, the overall ratio of dissolved ions to water stays the same. Problem IC4.1. Consider further the idea that a given amount of water is only able to dissolve a specific amount of salt. In the diagram above, how many water molecules are there? How many units of salt (an anion and a cation) are dissolved? If there were only a dozen water molecules present, how many units of salt would dissolve? If a hundred water molecules were present, how many units of salt would dissolve? Why do salts dissolve in water? Water is a molecular compound; the atoms are directly attached to each other, rather than being ions that are attracted to each other. Because of electronegativity differences, the oxygen atom in water has a partial negative charge and the hydrogen atoms have partial positive charges. Ionic compounds can dissolve in polar liquids like water because the ions are attracted to either the positive or negative part of the molecule. There is a sort of tug-of-war involved with species dissolved in water. The water pulls individual ions away from the solid. The solid is pulling individual ions back out of the water. There exists an equilibrium based on how strongly the water attracts the ions, versus how strong the ionic solid attracts the ions. It is possible to predict varying degrees of solubility in water for different ionic compounds using the principles of Coulomb's law. The smaller the ions, the closer together they are, and the harder it is for the water molecules to pull the ions away from each other. Problem IC4.2. Predict which of the following pairs should be more soluble in water, based on the Coulombic attraction between ions. LiF or NaF NaK or KF BeO or LiF Problem IC4.3. Although lithium fluoride and magnesium oxide contain cations and anions of roughly the same size, lithium fluoride is much more soluble in water (2.7 g/L) than magnesium oxide (0.087 g/L) at room temperature. Propose a reason why. The trends in melting points in ionic compounds are more complicated with regard to solubility. The water solubility of alkali chlorides does not follow a simple trend (as shown in Table IC4.1). Table IC4.1 Water solubility among alkali chlorides. 0 1 Compound Water Solubility in g/L at 20oC LiCl 83 NaCl 359 KCl 344 Lithium chloride is the least water-soluble of the three compounds. This is feasible, as the lithium ions are small and the attraction for the chloride would be stronger over that shorter distance. However, potassium chloride would be expected to be the most soluble of the three compounds, and it is slightly less soluble than sodium chloride. Problem IC4.4. Propose an explanation for why the water solubility of the alkali chlorides does not simply increase as the cation gets larger. If the halide is varied is used, similar trends are observed:. Table IC4.2 Water solubility among lithium halides. Compound Water Solubility in g/L at 20oC LiCl 83 LiBr 166 LiI 150 Again, it is unsurprising that the lithium chloride is the least soluble, but the most soluble is the lithium bromide, not the lithium iodide. This type of behavior indicates that there is more than one factor influencing the phenomenon of interest. In the case of solubility, there are several other factors, some of which are more complicated. One simply involves the fact that there are two interactions to consider. The dissolution of these compounds requires more than simply overcoming the attraction of the ionic solid for individual ions, as does melting: the attraction of the water to the ion must also be considered. That attraction is also governed by Coulomb's Law. At some stage, there is a tipping point, when the factors that increase attraction between the ions also increase the attraction between the ion and the water. One or the other of these factors becomes the dominant influence under different circumstances. Several interactions are involved in dissolution. Cation-anion attraction is just one of these interactions. Cation-water and anion-water interactions are important, too. Water-water interactions also play a role.
Courses/Georgian_College/Chemistry_-_Academic_and_Career_Preparation/zz%3A_Back_Matter/20%3A_Glossary	Words (or words that have the same definition) The definition is case sensitive (Optional) Image to display with the definition [Not displayed in Glossary, only in pop-up on pages] (Optional) Caption for Image (Optional) External or Internal Link (Optional) Source for Definition (Eg. "Genetic, Hereditary, DNA ...") (Eg. "Relating to genes or heredity") NaN The infamous double helix https://bio.libretexts.org/ CC-BY-SA; Delmar Larsen Word(s) Definition Image Caption Link Source Sample Word 1 Sample Definition 1 NaN NaN NaN NaN
Courses/Oregon_Tech_PortlandMetro_Campus/OT_-_PDX_-_Metro%3A_General_Chemistry_III/02%3A_Chemical_Equilibrium/2.03%3A_Expressing_the_Equilibrium_Constant_in_Terms_of_Pressure	Learning Objectives To understand how different phases affect equilibria. When the products and reactants of an equilibrium reaction form a single phase, whether gas or liquid, the system is a homogeneous equilibrium. In such situations, the concentrations of the reactants and products can vary over a wide range. In contrast, a system whose reactants, products, or both are in more than one phase is a heterogeneous equilibrium, such as the reaction of a gas with a solid or liquid. As noted in the previous section, the equilibrium constant expression is actually a ratio of activities. To simplify the calculations in general chemistry courses, the activity of each substance in the reaction is often approximated using a ratio of the molarity of a substance compared to the standard state of that substance. For substances that are liquids or solids, the standard state is just the concentration of the substance within the liquid or solid. Because the molar concentrations of pure liquids and solids normally do not vary greatly with temperature, the ratio of the molarity to the standard state for substances that are liquids or solids always has a value of 1. For example, for a compound such as CaF 2 (s), the term going into the equilibrium expression is [CaF 2 ]/[CaF 2 ] which cancels to unity. Thus, when the activities of the solids and liquids (including solvents) are incorporated into the equilibrium expression, they do not change the value. Consider the following reaction, which is used in the final firing of some types of pottery to produce brilliant metallic glazes: \[\ce{CO2(g) + C(s) \rightleftharpoons 2CO(g)} \label{Eq14.4.1} \] The glaze is created when metal oxides are reduced to metals by the product, carbon monoxide. The equilibrium constant expression for this reaction is as follows: \[K=\dfrac{a_{\ce{CO}}^2}{a_{\ce{CO2}}a_{C}}=\dfrac{[\ce{CO}]^2}{[\ce{CO2}][1]}=\dfrac{[\ce{CO}]^2}{[\ce{CO_2}]}\label{Eq14.4.2} \] The equilibrium constant for this reaction can also be written in terms of the partial pressures of the gases: \[K_p=\dfrac{(P_{CO})^2}{P_{CO_2}} \label{Eq14.4.3} \] Incorporating all the constant values into $K′$ or $K_p$ allows us to focus on the substances whose concentrations change during the reaction. Although the activities of pure liquids or solids are not written explicitly in the equilibrium constant expression, these substances must be present in the reaction mixture for chemical equilibrium to occur. Whatever the concentrations of $\ce{CO}$ and $\ce{CO_2}$, the system described in Equation $\ref{Eq14.4.1}$ will reach chemical equilibrium only if a stoichiometric amount of solid carbon or excess solid carbon has been added so that some is still present once the system has reached equilibrium. As shown in Figure $\PageIndex{1}$, it does not matter whether 1 g or 100 g of solid carbon is present; in either case, the composition of the gaseous components of the system will be the same at equilibrium. Example $\PageIndex{1}$ Write each expression for $K$, incorporating all constants, and $K_p$ for the following equilibrium reactions. $\ce{PCl3(l) + Cl2(g) <=> PCl5(s)}$ $\ce{Fe3O4(s) + 4H2(g) <=> 3Fe(s) + 4H2O(g)}$ Given : balanced equilibrium equations. Asked for : expressions for $K$ and $K_p$. Strategy : Find $K$ by writing each equilibrium constant expression as the ratio of the concentrations of the products and reactants, each raised to its coefficient in the chemical equation. Then express $K_p$ as the ratio of the partial pressures of the products and reactants, each also raised to its coefficient in the chemical equation. Solution This reaction contains a pure solid ($PCl_5$) and a pure liquid ($PCl_3$). Their activities are equal to 1, so when incorporated into the equilibrium constant expression, they do not change the value. So \[K=\dfrac{1}{(1)[Cl_2]} \nonumber \] and \[K_p=\dfrac{1}{(1)P_{Cl_2}} \nonumber \] This reaction contains two pure solids ($Fe_3O_4$ and $Fe$), which are each assigned a value of 1 in the equilibrium constant expressions: \[K=\dfrac{(1)[H_2O]^4}{(1)[H_2]^4} \nonumber \] and \[K_p=\dfrac{(1)(P_{H_2O})^4}{(1)(P_{H_2})^4} \nonumber \] Exercise $\PageIndex{1}$ Write the expressions for $K$ and $K_p$ for the following reactions. $\ce{CaCO3(s) <=> CaO(s) + CO2(g)}$ $ \underset{glucose}{\ce{C6H12O6(s)}} + \ce{6O2(g) <=> 6CO2(g) + 6H2O(g)}$ Answer a $K = [\ce{CO_2}]$ and $K_p = P_{\ce{CO_2}}$ Answer b $K=\dfrac{[CO_2]^6[H_2O]^6}{[O_2]^6}$ and $K_p=\dfrac{(P_{CO_2})^6(P_{H_2O})^6}{(P_{O_2})^6}$ For reactions carried out in solution, the solvent is assumed to be pure, and therefore is assigned an activity equal to 1 in the equilibrium constant expression. The activities of the solutes are approximated by their molarities. The result is that the equilibrium constant expressions appear to only depend upon the concentrations of the solutes. The activities of pure solids, pure liquids, and solvents are defined as having a value of '1'. Often, it is said that these activities are "left out" of equilibrium constant expressions. This is an unfortunate use of words. The activities are not "left out" of equilibrium constant expressions. Rather, because they have a value of '1', they do not change the value of the equilibrium constant when they are multiplied together with the other terms. The activities of the solutes are approximated by their molarities. Summary An equilibrated system that contains products and reactants in a single phase is a homogeneous equilibrium; a system whose reactants, products, or both are in more than one phase is a heterogeneous equilibrium.
Courses/Saint_Francis_University/CHEM_113%3A_Human_Chemistry_I_(Muino)/12%3A_Introduction_to_Organic_Chemistry_-_Alkanes/12.09%3A_Cycloalkanes	Learning Objectives Identify the structures of cycloalkanes. A cyclic hydrocarbon is a hydrocarbon in which the carbon chain joins to itself in a ring. A cycloalkane is a cyclic hydrocarbon in which all of the carbon-carbon bonds are single bonds and each carbon is bonded to two hydrogen atoms, they are saturated compounds. Cycloalkanes have the general formula $\ce{C_{n}H_{2n}}$. The simplest of these cyclic hydrocarbons, cyclopropane, has the formula C 3 H 6 , which makes a three-carbon ring. The structural formulas of cyclic hydrocarbons can be represented in multiple ways, two of which are shown above. Each atom can be shown as in the structure on the left from the figure above. A convenient shorthand is to omit the element symbols and only show the shape, as in the triangle on the right. Carbon atoms are understood to be the vertices of the triangle. The carbon atoms in cycloalkanes have a bond angle of $109.5^\text{o}$. However, an examination of the cyclopropane structure shows that the triangular structure results in a $\ce{C-C-C}$ bond angle of $60^\text{o}$. This deviation from the ideal angle is called ring strain and makes cyclopropane a fairly unstable and reactive molecule. Ring strain is decreased for cyclobutane, with a bond angle of $90^\text{o}$, but is still significant. Cyclopentane has a bond angle of about $108^\text{o}$. This minimal ring strain for cyclopentane makes it a more stable compound. Cyclohexane is a six-carbon cycloalkane, shown below. All three of the depictions of cyclohexane above are somewhat misleading, because the molecule is not planar. In order to reduce the ring strain and attain a bond angle of approximately $109.5^\text{o}$, the molecule is actually puckered. The ring structure in cycloalkanes also prevents rotation around the carbon–carbon bonds without breaking open the ring, thus they are more rigid and less flexible than acyclic alkanes. This property is called restricted rotation . Note To Your Health: Cyclopropane as an Anesthetic With its boiling point of −33°C, cyclopropane is a gas at room temperature. It is also a potent, quick-acting anesthetic with few undesirable side effects in the body. It is no longer used in surgery, however, because it forms explosive mixtures with air at nearly all concentrations.
Courses/Tusculum_University/CHEM_411%3A_Introductory_Chemical_Thermodynamics_(Pearson)/1%3A_Measurement_energy_and_the_quantites_of_state	This introduction is taken from my first introduction of the content to students in CHEM 411 at Tusculum University in Fall 2019 - the first to study from this material. I still am trying to build this material with these principles in mind. You have had enough courses that ask you to remember a ton of facts and terms. And you will still need to make sure you know old and new terms for this course, and you will need to be certain you are using those terms precisely, but there’s no good means to test you on the terms themselves as a means to make them meaningful. You’ve also had enough experience being given equations that are rules and having to use those equations (frequently without much context) to solve simple one- or two-step problems. Those skills are important to build, and I will check skills on equation sets from time to time here, especially when those equations involve calculus. But they don’t need to be our primary business either. The development of physical chemistry, especially chemical thermodynamics, is a necessarily theoretical discipline. Energy is a constructed physical quantity we use to make our language understandable across disciplines - it doesn’t have the tangibility of force or the broad common meaning of temperature. We can take some of that language and directly apply it to behavior we do see - but even then, much of that is the behavior of gases, which are atoms and molecules spread out over wide distances from one another. We have to make some effort to describe that behavior meaningfully. That effort to describe that behavior demands the use of our imaginations, and demands development of a new language, and demands mathematical descriptions to fit that language - and all must fit together to make a coherent whole. Our primary business in chemical thermodynamics is the development of a good theory of the transfer of energy through physical and chemical change . My writing in this text, and the problems I assign for you to solve, will all be in support of developing this theory in a meaningful way, in a way that allows you both to understand the richness of the fundamental ideas of energy and to describe energy transfer in practical contexts. There will be both English-language concepts and mathematical concepts that we will use precisely . I can’t emphasize enough how important it is to not be careless with how you use the mathematics you’ll be developing in this course - tricks you’ve employed in the past are much less likely to work, and points where you’ve taken mathematical shortcuts in the past are much more likely to lead to dead ends now. But our process in this course is going to be very intentional as well. I will do my best to not assume that you automatically understand some of the core math concepts, particularly calculus concepts. For many - perhaps most - of you, this is the first serious physical science problem solving you will ever do that involves calculus. My development of methods for this kind of problem solving will be patient and total. I joke that I’m interested in making good theoreticians in this course. But you’ve had a host of good experience in lab sciences where your work is very practical. I want you to master the symbolic and conceptual logic necessary to develop physical ideas using the most powerful scientific tools: pencil, paper, and your mind. 1.1: The language of measurement in physical chemistry 1.2: The meaning and measurement of temperature 1.3: The variables of state, pressure units, and the ideal gas law 1.4: The ideal gas law, functions and derivatives 1.5: Ideal vs. real gases and the van der Waals equation 1.6: Another way of dealing with real gases: the virial equation 1.7: Connecting the van der Waals and the viral equations: the Boyle temperature
Courses/Smith_College/CHM_222_Chemistry_II%3A_Organic_Chemistry_(2025)/07%3A_An_Overview_of_Organic_Reactions/7.09%3A_Describing_a_Reaction_-_Bond__Dissociation_Energies	Objectives After completing this section, you should be able to predict the value of ΔH° for a gas-phase reaction, given the necessary bond dissociation energy data. predict the dissociation energy of a particular bond, given ΔH° for a reaction involving the bond and any other necessary bond dissociation energy data. outline the limitations of using bond dissociation energies to predict whether or not a given reaction will occur. Key Terms Make certain that you can define, and use in context, the key terms below. bond dissociation energy solvation Study Notes The idea of calculating the standard enthalpy of a reaction from the appropriate bond dissociation energy data should be familiar to you from your first-year chemistry course. Solvation is the interaction between solvent molecules and the ions or molecules dissolved in that solvent. The homolytic bond dissociation energy is the amount of energy needed to break apart one mole of covalently bonded gases into a pair of radicals. The SI units used to describe bond energy are kiloJoules per mole of bonds (kJ/Mol). It indicates how strongly the atoms are bonded to each other. Introduction Breaking a covalent bond between two partners, A-B, can occur either heterolytically, where the shared pair of electron goes with one partner or another \[A-B \rightarrow A^+ + B:^- \nonumber \] or \[A-B \rightarrow A:^- + B^+ \nonumber \] or homolytically, where one electron stays with each partner. \[A-B \rightarrow A^• + B^• \nonumber \] The products of homolytic cleavage are radicals and the energy that is required to break the bond homolytically is called the Bond Dissociation Energy (BDE) and is a measure of the strength of the bond. Calculation of the BDE The BDE for a molecule A-B is calculated as the difference in the enthalpies of formation of the products and reactants for homolysis \[BDE = \Delta_fH(A^•) + \Delta_fH(B^•) - \Delta_fH(A-B) \nonumber \] Officially, the IUPAC definition of bond dissociation energy refers to the energy change that occurs at 0 K, and the symbol is $D_o$. However, it is commonly referred to as BDE, the bond dissociation energy, and it is generally used, albeit imprecisely, interchangeably with the bond dissociation enthalpy , which generally refers to the enthalpy change at room temperature (298K). Although there are technically differences between BDEs at 0 K and 298 K, those difference are not large and generally do not affect interpretations of chemical processes. Bond Breakage/Formation Bond dissociation energy (or enthalpy) is a state function and consequently does not depend on the path by which it occurs. Therefore, the specific mechanism in how a bond breaks or is formed does not affect the BDE. Bond dissociation energies are useful in assessing the energetics of chemical processes. For chemical reactions, combining bond dissociation energies for bonds formed and bonds broken in a chemical reaction using Hess's Law can be used to estimate reaction enthalpies. Example 6.8.1: Chlorination of Methane Consider the chlorination of methane \[CH_4 + Cl_2 \rightarrow CH_3Cl + HCl \nonumber \] the overall reaction thermochemistry can be calculated exactly by combining the BDEs for the bonds broken and bonds formed CH 4 → CH 3 • + H• BDE(CH 3 -H) Cl 2 → 2Cl• BDE(Cl 2 ) \nonumber \] H• + Cl• → HCl -BDE(HCl) CH 3 • + Cl• → CH 3 Cl -BDE(CH 3 -Cl) --------------------------------------------------- \[CH_4 + Cl_2 \rightarrow CH_3Cl + HCl \nonumber \] \[\Delta H = BDE(R-H) + BDE(Cl_2) - BDE(HCl) - BDE(CH_3-Cl) \nonumber \] Because reaction enthalpy is a state function, it does not matter what reactions are combined to make up the overall process using Hess's Law. However, BDEs are convenient to use because they are readily available. Alternatively, BDEs can be used to assess individual steps of a mechanism. For example, an important step in free radical chlorination of alkanes is the abstraction of hydrogen from the alkane to form a free radical. RH + Cl• → R• + HCl The energy change for this step is equal to the difference in the BDEs in RH and HCl \[\Delta H = BDE(R-H) - BDE(HCl) \nonumber \] This relationship shows that the hydrogen abstraction step is more favorable when BDE(R-H) is smaller. The difference in energies accounts for the selectivity in the halogenation of hydrocarbons with different types of C-H bonds. R-H Do, kJ/mol D298, kJ/mol R-H.1 Do, kJ/mol.1 D298, kJ/mol CH3-H 432.7±0.1 439.3±0.4 H2C=CH-H 456.7±2.7 463.2±2.9 CH3CH2-H NaN 423.0±1.7 C6H5-H 465.8±1.9 472.4±2.5 (CH3)2CH-H NaN 412.5±1.7 HCCH 551.2±0.1 557.8±0.3 (CH3)3C-H NaN 403.8±1.7 NaN NaN NaN NaN NaN NaN H2C=CHCH2-H NaN 371.5±1.7 HC(O)-H NaN 368.6±0.8 C6H5CH2-H NaN 375.3±2.5 CH3C(O)-H NaN 374.0±1.2 NaN NaN NaN Trends in C-H BDEs It is important to remember that C-H BDEs refer to the energy it takes to break the bond, and is the difference in energy between the reactants and the products. Therefore, it is not appropriate to interpret BDEs solely in terms of the "stability of the radical products" as is often done. Analysis of the BDEs shown in the table above shows that there are some systematic trends: BDEs vary with hybridization: Bonds with sp 3 hybridized carbons are weakest and bonds with sp hybridized carbons are much stronger. The vinyl and phenyl C-H bonds are similar, reflecting their sp 2 hybridization. The correlation with hybridization can be viewed as a reflection of the C-H bond lengths. Longer bonds formed with sp 3 orbitals are consequently weaker. Shorter bonds formed with orbitals that have more s-character are similarly stronger. C-H BDEs vary with substitution: Among sp 3 hybridized systems, methane has the strongest C-H bond. C-H bonds on primary carbons are stronger than those on secondary carbons, which are stronger than those on tertiary carbons. Interpretation of C-H BDEs for sp 3 Hybridized Carbons The interpretation of the BDEs in saturated molecules has been subject of recent controversy. As indicated above, the variation in BDEs with substitution has traditionally been interpreted as reflecting the stabilities of the alkyl radicals, with the assessment that more highly substituted radicals are more stable, as with carbocations. Although this is a popular explanation, it fails to account fo the fact the bonds to groups other than H do not show the same types of variation. R BDE(R-CH3) BDE(R-Cl) BDE(R-Br) BDE(R-OH) CH3- 377.0±0.4 350.2±0.4 301.7±1.3 385.3±0.4 CH3CH2- 372.4±1.7 354.8±2.1 302.9±2.5 393.3±1.7 (CH3)2CH- 370.7±1.7 356.5±2.1 309.2±2.9 399.6±1.7 (CH3)3C- 366.1±1.7 355.2±2.9 303.8±2.5 400.8±1.7 Therefore, although C-CH 3 bonds get weaker with more substitution, the effect is not nearly as large as that observed with C-H bonds. The strengths of C-Cl and C-Br bonds are not affected by substitution, despite the fact that the same radicals are formed as when breaking C-H bonds, and the C-OH bonds in alcohols actually increase with more substitution. Gronert has proposed that the variation in BDEs is alternately explained as resulting from destabilization of the reactants due to steric repulsion of the substituents, which is released in the nearly planar radicals. 1 Considering that BDEs reflect the relative energies of reactants and products, either explanation can account for the trend in BDEs. Another factor that needs to be considered is the electronegativity. The Pauling definition of electronegativity says that the bond dissociation energy between unequal partners is going to be dependent on the difference in electrongativities, according to the expression \[D_o(A-B) = \dfrac{D_o(A-A) + D_o(B-B)}{2} + (X_A - X_B)^2 \nonumber \] where $X_A$ and $X_B$ are the electronegativities and the bond energies are in eV. Therefore, the variation in BDEs can be interpreted as reflecting variation in the electronegativities of the different types of alkyl fragments. There is likely some merit in all three interpretations. Since Gronert's original publication of his alternate explanation, there have been many desperate attempts to defend the radical stability explanation. References 1. Gronert, S. J. Org. Chem. 2006 , 13 , 1209 Further Reading MasterOrganicChemistry Bond Strengths And Radical Stability Exercises Given that ΔH° for the reaction CH 4 (g) + 4F 2 (g) → CF 4 (g) + 4HF (g) is −1936 kJ, use the following data to calculate the average bond energy of the C-F bonds in CF 4 . Bond Average Bond Energy $\ce{\sf{C-H}}$ 413 kJ · mol−1 $\ce{\sf{F-F}}$ 155 kJ · mol−1 $\ce{\sf{H-F}}$ 567 kJ · mol−1 Calculate ΔH° for the reactions given below. [Access bond energy tables via left-hand blue column Resources > Reference Tables > Reference Tables > Thermodynamic Tables > T3: Bond Energies] CH 3 CH 2 OCH 3 + HI -> CH 3 CH 2 OH + CH 3 I CH 3 Cl + NH 3 -> CH 3 NH 2 + HCl Answers: Bonds broken: 4 mol C-H bonds × ( 413 kJ ) ( 1 mol ) = 1652 kJ 4 mol F-F bonds × ( 155 kJ ) ( 1 mol ) = 620 kJ Bonds formed: 4 mol CF bonds × ( x kJ ) ( 1 mol ) = 4 x kJ (where x = the average energy of one mole of C-F bonds in CF 4 , expressed in kJ) 4 mol H-F bonds × ( 567 kJ ) ( 1 mol ) = 2268 kJ Δ H ° = Δ H ° ( bonds broken ) − Δ H ° ( bonds formed ) = ( 1652 kJ + 620 kJ ) − ( 4 x + 2268 kJ ) = 1652 kJ + 620 kJ − 4 x − 2268 kJ = − 1936 kJ Thus, 4 x = 1936 kJ − 2268 kJ + 620 kJ + 1652 kJ = 1940 kJ and x = 1940 kJ 4 mol = 385 kJ ⋅ mol − 1 The average energy of a C-F bond in CF 4 is 385 kJ · mol -1 CH 3 CH 2 OCH 3 + HI → CH 3 CH 2 OH + CH 3 I $\dfrac { \begin{array}{cc} \textrm{Reactant bonds broken} & D \\[6pt] \ce{CH3CH2O-CH3} & {339 \,\, \textrm{kJ/mol}} \\ \ce{H-I} & {298 \,\, \textrm{kJ/mol}} \\ \end{array} } { \begin{array}{cc} {\phantom{\textrm{Reactant bonds broken}}} & {637 \,\, \textrm{kJ/mol}} \end{array} }$ $\quad$ $\dfrac { \begin{array}{cc} \textrm{Product bonds formed} & D \\[6pt] \ce{CH3CH2O-H} & {438 \,\, \textrm{kJ/mol}} \\ \ce{CH3-I} & {238 \,\, \textrm{kJ/mol}} \\ \end{array} } { \begin{array}{cc} {\phantom{\textrm{Product bonds formed}}} & {676 \,\, \textrm{kJ/mol}} \end{array} }$ \begin{align} \Delta H^{\circ} & = D_{\textrm{bonds broken}} + D_{\textrm{bonds formed}}\\ & = {637 \,\, \textrm{kJ/mol} - 676 \,\, \textrm{kJ/mol}} \\ & = {-39 \,\, \textrm{kJ/mol}} \end{align} CH 3 Cl + NH 3 → CH 3 NH 2 + HCl $\dfrac { \begin{array}{cc} \textrm{Reactant bonds broken} & D \\[6pt] \ce{CH3-Cl} & {356 \,\, \textrm{kJ/mol}} \\ \ce{NH2-H} & {450 \,\, \textrm{kJ/mol}} \\ \end{array} } { \begin{array}{cc} {\phantom{\textrm{Reactant bonds broken}}} & {806 \,\, \textrm{kJ/mol}} \end{array} }$ $\quad$ $\dfrac { \begin{array}{cc} \textrm{Product bonds formed} & D \\[6pt] \ce{CH3-NH2} & {364 \,\, \textrm{kJ/mol}} \\ \ce{H-Cl} & {432 \,\, \textrm{kJ/mol}} \\ \end{array} } { \begin{array}{cc} {\phantom{\textrm{Product bonds formed}}} & {796 \,\, \textrm{kJ/mol}} \end{array} }$ \begin{align} \Delta H^{\circ} & = D_{\textrm{bonds broken}} + D_{\textrm{bonds formed}}\\ & = {806 \,\, \textrm{kJ/mol} - 796 \,\, \textrm{kJ/mol}} \\ & = {+10 \,\, \textrm{kJ/mol}} \end{align}
Courses/CSU_San_Bernardino/CHEM_2100%3A_General_Chemistry_I_(Mink)/01%3A_Essential_Ideas/1.06%3A_Measurement_Uncertainty_Accuracy_and_Precision	Learning Objectives By the end of this section, you will be able to: Define accuracy and precision Distinguish exact and uncertain numbers Correctly represent uncertainty in quantities using significant figures Apply proper rounding rules to computed quantities Counting is the only type of measurement that is free from uncertainty, provided the number of objects being counted does not change while the counting process is underway. The result of such a counting measurement is an example of an exact number . By counting the eggs in a carton, one can determine exactly how many eggs the carton contains. The numbers of defined quantities are also exact. By definition, 1 foot is exactly 12 inches, 1 inch is exactly 2.54 centimeters, and 1 gram is exactly 0.001 kilogram. Quantities derived from measurements other than counting, however, are uncertain to varying extents due to practical limitations of the measurement process used. Significant Figures in Measurement The numbers of measured quantities, unlike defined or directly counted quantities, are not exact. To measure the volume of liquid in a graduated cylinder, you should make a reading at the bottom of the meniscus, the lowest point on the curved surface of the liquid. Refer to the illustration in Figure $\PageIndex{1}$. The bottom of the meniscus in this case clearly lies between the 21 and 22 markings, meaning the liquid volume is certainly greater than 21 mL but less than 22 mL. The meniscus appears to be a bit closer to the 22-mL mark than to the 21-mL mark, and so a reasonable estimate of the liquid’s volume would be 21.6 mL. In the number 21.6, then, the digits 2 and 1 are certain, but the 6 is an estimate. Some people might estimate the meniscus position to be equally distant from each of the markings and estimate the tenth-place digit as 5, while others may think it to be even closer to the 22-mL mark and estimate this digit to be 7. Note that it would be pointless to attempt to estimate a digit for the hundredths place, given that the tenths-place digit is uncertain. In general, numerical scales such as the one on this graduated cylinder will permit measurements to one-tenth of the smallest scale division. The scale in this case has 1-mL divisions, and so volumes may be measured to the nearest 0.1 mL. This concept holds true for all measurements, even if you do not actively make an estimate. If you place a quarter on a standard electronic balance, you may obtain a reading of 6.72 g. The digits 6 and 7 are certain, and the 2 indicates that the mass of the quarter is likely between 6.71 and 6.73 grams. The quarter weighs about 6.72 grams, with a nominal uncertainty in the measurement of ± 0.01 gram. If the coin is weighed on a more sensitive balance, the mass might be 6.723 g. This means its mass lies between 6.722 and 6.724 grams, an uncertainty of 0.001 gram. Every measurement has some uncertainty , which depends on the device used (and the user’s ability). All of the digits in a measurement, including the uncertain last digit, are called significant figures or significant digits . Note that zero may be a measured value; for example, if you stand on a scale that shows weight to the nearest pound and it shows “120,” then the 1 (hundreds), 2 (tens) and 0 (ones) are all significant (measured) values. A measurement result is properly reported when its significant digits accurately represent the certainty of the measurement process. But what if you were analyzing a reported value and trying to determine what is significant and what is not? Well, for starters, all nonzero digits are significant, and it is only zeros that require some thought. We will use the terms “leading,” “trailing,” and “captive” for the zeros and will consider how to deal with them. Starting with the first nonzero digit on the left, count this digit and all remaining digits to the right. This is the number of significant figures in the measurement unless the last digit is a trailing zero lying to the left of the decimal point. Captive zeros result from measurement and are therefore always significant. Leading zeros, however, are never significant—they merely tell us where the decimal point is located. The leading zeros in this example are not significant. We could use exponential notation (as described in Appendix B) and express the number as $8.32407 \times 10^{−3}$; then the number 8.32407 contains all of the significant figures, and $10^{−3}$ locates the decimal point. The number of significant figures is uncertain in a number that ends with a zero to the left of the decimal point location. The zeros in the measurement 1,300 grams could be significant or they could simply indicate where the decimal point is located. The ambiguity can be resolved with the use of exponential notation: $1.3\times 10^3$ (two significant figures), $1.30 \times 10^3$ (three significant figures, if the tens place was measured), or $1.300 \times 10^3$ (four significant figures, if the ones place was also measured). In cases where only the decimal-formatted number is available, it is prudent to assume that all trailing zeros are not significant. When determining significant figures, be sure to pay attention to reported values and think about the measurement and significant figures in terms of what is reasonable or likely when evaluating whether the value makes sense. For example, the official January 2014 census reported the resident population of the US as 317,297,725. Do you think the US population was correctly determined to the reported nine significant figures, that is, to the exact number of people? People are constantly being born, dying, or moving into or out of the country, and assumptions are made to account for the large number of people who are not actually counted. Because of these uncertainties, it might be more reasonable to expect that we know the population to within perhaps a million or so, in which case the population should be reported as 3.17 10 8 people. Significant Figures in Calculations A second important principle of uncertainty is that results calculated from a measurement are at least as uncertain as the measurement itself. Take the uncertainty in measurements into account to avoid misrepresenting the uncertainty in calculated results. One way to do this is to report the result of a calculation with the correct number of significant figures, which is determined by the following three rules for rounding numbers: When adding or subtracting numbers, round the result to the same number of decimal places as the number with the least number of decimal places (the least certain value in terms of addition and subtraction). When multiplying or dividing numbers, round the result to the same number of digits as the number with the least number of significant figures (the least certain value in terms of multiplication and division). If the digit to be dropped (the one immediately to the right of the digit to be retained) is less than 5, “round down” and leave the retained digit unchanged; if it is more than 5, “round up” and increase the retained digit by 1. If the dropped digit is 5, and it’s either the last digit in the number or it’s followed only by zeros, round up or down, whichever yields an even value for the retained digit. If any nonzero digits follow the dropped 5, round up. (The last part of this rule may strike you as a bit odd, but it’s based on reliable statistics and is aimed at avoiding any bias when dropping the digit “5,” since it is equally close to both possible values of the retained digit.) The following examples illustrate the application of this rule in rounding a few different numbers to three significant figures: 0.028675 rounds “up” to 0.0287 (the dropped digit, 7, is greater than 5) 18.3384 rounds “down” to 18.3 (the dropped digit, 3, is less than 5) 6.8752 rounds “up” to 6.88 (the dropped digit is 5, and a nonzero digit follows it) 92.85 rounds “down” to 92.8 (the dropped digit is 5, and the retained digit is even) Let’s work through these rules with a few examples. Example $\PageIndex{1}$: Rounding Numbers Round the following to the indicated number of significant figures: 31.57 (to two significant figures) 8.1649 (to three significant figures) 0.051065 (to four significant figures) 0.90275 (to four significant figures) Solution 31.57 rounds “up” to 32 (the dropped digit is 5, and the retained digit is even) 8.1649 rounds “down” to 8.16 (the dropped digit, 4, is less than 5) 0.051065 rounds “down” to 0.05106 (the dropped digit is 5, and the retained digit is even) 0.90275 rounds “up” to 0.9028 (the dropped digit is 5, and the retained digit is even) Exercise $\PageIndex{1}$ Round the following to the indicated number of significant figures: 0.424 (to two significant figures) 0.0038661 (to three significant figures) 421.25 (to four significant figures) 28,683.5 (to five significant figures) Answer 0.42 0.00387 421.2 28,684 Example $\PageIndex{2}$: Addition and Subtraction with Significant Figures Rule: When adding or subtracting numbers, round the result to the same number of decimal places as the number with the fewest decimal places (i.e., the least certain value in terms of addition and subtraction). Add 1.0023 g and 4.383 g. Subtract 421.23 g from 486 g. Solution (a) \[\begin{align} 1&.0023 \, \text{g} \\ +\quad 4&.383 \, \text{g}\\ \hline 5&.3853\, \text{g} \end{align} \] Answer is 5.385 g (round to the thousandths place; three decimal places) (b) \[\begin{align} 486& \, \text{g} \\ -\quad 421&.23 \, \text{g}\\ \hline 64&.77\, \text{g} \end{align} \] Answer is 65 g (round to the ones place; no decimal places) Exercise $\PageIndex{2}$ Add 2.334 mL and 0.31 mL. Subtract 55.8752 m from 56.533 m. Answer 2.64 mL 0.658 m Example $\PageIndex{3}$: Multiplication and Division with Significant Figures Rule: When multiplying or dividing numbers, round the result to the same number of digits as the number with the fewest significant figures (the least certain value in terms of multiplication and division). Multiply 0.6238 cm by 6.6 cm. Divide 421.23 g by 486 mL. Solution (a) \[\underbrace{0.6238\,\text{cm}}_{\text{four significant figures}} \times \underbrace{6.6\, \text{cm}}_{\text{two significant figures}} =4.11708\, \text{cm}^2 \rightarrow \underbrace{4.1\, \text{cm}^2}_{\text{round to two significant figures}} \nonumber \] (b) \[\underbrace{421.23\,\text{g}}_{\text{five significant figures}} ÷ \underbrace{486\, \text{mL}}_{\text{three significant figures}} = 0.866728... \, \text{g}/\text{mL} \rightarrow \underbrace{0.867\, \text{g}/\text{mL} }_{\text{round to three significant figures}} \nonumber \] Exercise $\PageIndex{3}$ Multiply 2.334 cm and 0.320 cm. Divide 55.8752 m by 56.53 s. Answer 0.747 cm 2 0.9884 m/s In the midst of all these technicalities, it is important to keep in mind the reason for these rules about significant figures and rounding—to correctly represent the certainty of the values reported and to ensure that a calculated result is not represented as being more certain than the least certain value used in the calculation. Example $\PageIndex{4}$: Calculation with Significant Figures One common bathtub is 13.44 dm long, 5.920 dm wide, and 2.54 dm deep. Assume that the tub is rectangular and calculate its approximate volume in liters. Solution \[\begin{aligned} V & =l \times w \times d \\ & =13.44 \, \text{dm} \times 5.920 \, \text{dm} \times 2.54 \, \text{dm} \\ & =202.09459 \ldots \, \text{dm}^3 \text { (value from calculator) } \\ & =202 \, \text{dm}^3 \text { or } 202\, \text{L} \text { (answer rounded to three significant figures) } \end{aligned} \nonumber \] Exercise $\PageIndex{4}$ What is the density of a liquid with a mass of 31.1415 g and a volume of 30.13 cm 3 ? Answer 1.034 g/mL Example $\PageIndex{5}$: Experimental Determination of Density Using Water Displacement A piece of rebar is weighed and then submerged in a graduated cylinder partially filled with water, with results as shown. Use these values to determine the density of this piece of rebar. Rebar is mostly iron. Does your result in (a) support this statement? How? Solution The volume of the piece of rebar is equal to the volume of the water displaced: \[\text { volume }=22.4\, \text{mL} - 13.5\, \text{mL} =8.9\, \text{mL} = 8.9\, \text{cm}^3 \nonumber \] (rounded to the nearest 0.1 mL, per the rule for addition and subtraction) The density is the mass-to-volume ratio: \[\text { density }=\frac{\text { mass }}{\text { volume }}=\frac{69.658\, \text{g} }{8.9\, \text{cm}^3}=7.8\, \text{g} / \text{cm}^3 \nonumber \] (rounded to two significant figures, per the rule for multiplication and division) The density of iron is 7.9 g/cm 3 , very close to that of rebar, which lends some support to the fact that rebar is mostly iron. Exercise $\PageIndex{5}$ An irregularly shaped piece of a shiny yellowish material is weighed and then submerged in a graduated cylinder, with results as shown. Use these values to determine the density of this material. Do you have any reasonable guesses as to the identity of this material? Explain your reasoning. Answer 19 g/cm 3 It is likely gold; the right appearance for gold and very close to the known density of gold at $19.3 \, \text{g}/\text{cm}^3$ Accuracy and Precision Scientists typically make repeated measurements of a quantity to ensure the quality of their findings and to evaluate both the precision and the accuracy of their results. Measurements are said to be precise if they yield very similar results when repeated in the same manner. A measurement is considered accurate if it yields a result that is very close to the true or accepted value. Precise values agree with each other; accurate values agree with a true value. These characterizations can be extended to other contexts, such as the results of an archery competition (Figure $\PageIndex{2}$). Suppose a quality control chemist at a pharmaceutical company is tasked with checking the accuracy and precision of three different machines that are meant to dispense 10 ounces (296 mL) of cough syrup into storage bottles. She proceeds to use each machine to fill five bottles and then carefully determines the actual volume dispensed, obtaining the results tabulated in Table $\PageIndex{1}$. Dispenser #1 Dispenser #2 Dispenser #3 283.3 298.3 296.1 284.1 294.2 295.9 283.9 296.0 296.1 284.0 297.8 296.0 284.1 293.9 296.1 Considering these results, she will report that dispenser #1 is precise (values all close to one another, within a few tenths of a milliliter) but not accurate (none of the values are close to the target value of 296 mL, each being more than 10 mL too low). Results for dispenser #2 represent improved accuracy (each volume is less than 3 mL away from 296 mL) but worse precision (volumes vary by more than 4 mL). Finally, she can report that dispenser #3 is working well, dispensing cough syrup both accurately (all volumes within 0.1 mL of the target volume) and precisely (volumes differing from each other by no more than 0.2 mL).
Bookshelves/Inorganic_Chemistry/Introduction_to_Inorganic_Chemistry_(Wikibook)/11%3A_Basic_Science_of_Nanomaterials/11.04%3A__Synthesis_of_Semiconductor_Nanocrystals	Early work on the quantum size effect in semiconductor nanoparticles used simple metathesis reactions in the synthesis. For example, CdSe and PbS can be precipitated at ambient temperature by the reactions: $\ce{CdCl2_{(aq)} + H2Se_{(g)} = CdSe_{(s)} + 2HCl_{(aq)}}$ $\ce{Pb(NO3)2 + H2S_{(g)} = PbS_{(s)} + 2HNO3_{(aq)}}$ 0 Complete atomistic model of a 5 nm diameter colloidal lead sulfide nanoparticle with surface passivation by oleic acid, oleyl and hydroxyl groups. The growth of the particles was restricted by carrying out these reactions in different matrices, such as in polymer films or the silicate cages of zeolites, and capping ligands were also sometimes used to limit particle growth. While these reactions did produce nanoparticles, in general a broad distribution of particle sizes was obtained. The particles were also unstable to Ostwald ripening - in which large particles grow at the expense of smaller ones in order to minimize the total surface energy - because of the reversibility of the acid-forming synthetic reactions in aqueous media. The lack of good samples prevented detailed studies and the development of applications for semiconductor quantum dots. A very important development in nanoparticle synthesis came in the early 1990's, when Murray, Norris, and Bawendi introduced the first non-aqueous, controlled growth process for II-VI semiconductor quantum dots. [5] The keys to this synthesis were (1) to use non-aqueous solvents and capping ligands to stabilize the products against ripening, (2) to carry out the reaction at high temperature to ensure good crystallinity, and (3) to separate the steps of particle nucleation and growth , and thereby obtain particles of uniform size. This procedure is illustrated below: The synthesis is carried out in a coordinating, high boiling solvent that is a mixture of trioctylphosphine ( TOP ) and trioctylphosphine oxide ( TOPO ). In early experiments, organometallic cadmium compounds such as diethylcadmium were used as the metal source, but it was later found that these highly toxic and pyrophoric compounds could be replaced by CdO. At the start of the reaction, a selenium source, typically bis(trimethylsilyl)selenium, [(CH 3 ) 3 Si] 2 Se, dissolved in TOP is rapidly injected into the hot (350 °C) reaction mixture. The reaction causes a rapid burst of nanoparticle nucleation, but the temperature also drops as the cold solvent is injected and so the nucleation event ends quickly. The cooled solution now contains nanocrystal seeds. It is supersaturated in TOPO-Cd and TOP-Se, but particle growth proceeds slowly until the solution is heated again to the growth temperature, ca. 250°C. Particle growth and size-focusing occurs because small particles require less added material to grow by an amount ΔR than larger particles. This is because the volume of an added shell around a spherical seed is 4πR 2 ΔR, so for larger R, ΔR is smaller. Very narrow particle size distributions can be obtained under conditions of high supersaturation, where the rate of nanoparticle growth is fast relative to particle dissolution and Ostwald ripening. [6] The size distribution can then be narrowed further by adding a non-solvent such as hexane to the cooled reaction mixture. The largest particles precipitate first, followed by smaller particles. Because the nanoparticles are capped with a ligand shell of TOP, they can be re-suspended in organic solvents once they are size-separated. The high temperature synthesis of semiconductor quantum dots has been applied to a broad variety of materials including II-VI, III-V, and IV-VI semiconductors. Monodisperse nanoparticles of controlled shapes can be made by variants of this method. For example, it is possible to adjust the conditions so that CdSe nucleates in the zincblende polymorph as tetrahedrally shaped seeds, and then grow polar wurtzite "arms" onto each triangular face, resulting in nanocrystal tetrapods. Numerous other nanocrystal shapes such as rods, arrowheads, rice (tapered rods), and polar structures such as Janus rods can be made by variants of this technique. These shape-control strategies often involve the use of ligands that adsorb specifically to certain crystal faces and inhibit their growth. For example, hexylphosphonic acid ligands adsorb selectively to Cd-rich crystal faces and thus lead to the growth of prismatic wurtzite-phase CdSe nanocrytals. A second widely used method of semiconductor nanocrystal synthesis involves growth from molecular precursors and molten metal droplets, as shown in the figure at the right. The vapor-liquid-solid (VLS) and related solution-liquid-solid (SLS) growth processes rely on the fact that semiconductors such as Si, Ge, GaAs, InP, and others are soluble at high temperatures in liquid metals such as Au and Cu. The catalytic reduction of a molecule such as silicon tetrachloride (SiCl 4 ) at the surface of a gold nanocrystal liberates HCl gas and creates a solid solution of Si in Au. The presence of Si lowers the melting point of Au, and as more SiCl 4 reacts, a liquid eutectic droplet of the Si-Au alloy is formed. When this droplet becomes supersaturated in Si, a silicon nanocrystal nucleates and grows. This reaction can be performed on the surface of a macroscopic Si crystal, in which case nanocrystal "whiskers" grow from the surface, typically as single crystals and with an epitaxial orientation that is determined by the Si crystal face of the substrate. The diameter of the whiskers is controlled by the radius of the Au drops, which can be as small as a few nanometers and as large as several microns. Using this technique, "forests" of nanowires or microwires can be grown. Because the composition of the nanowire depends on the precursor being fed to the Au droplets, it is possible to make "totem pole" structures with varied compositions along the nanowire axis. Semiconductor shells can also be grown around the wires by chemical vapor deposition (CVD) . The VLS process can also be adapted to complex compositions for which no molecular precursor is available by using a laser to ablate the semiconductor from a solid target. 0 Epitaxial growth of silicon nanowires catalyzed by gold nanoparticles in the vapor-liquid-solid (VLS) process. Semiconductor nanowires made by this method are the basis of extremely sensitive biosensors , in which a molecular binding event anywhere along the wire strongly affects its electrical conductivity. [7] Nanowire and microwire arrays are also being studied as solar cell and lithium battery materials, as well as nanoscale electronic and optoelectronic devices.
Courses/Southeast_Missouri_State_University/CH185%3A_General_Chemistry_(Ragain)/07%3A_Periodic_Properties_of_the_Elements/7.01%3A_Nerve_Signal_Transmission	Electric currents in the vastly complex system of billions of nerves in our body allow us to sense the world, control parts of our body, and think. These are representative of the three major functions of nerves. First, nerves carry messages from our sensory organs and others to the central nervous system, consisting of the brain and spinal cord. Second, nerves carry messages from the central nervous system to muscles and other organs. Third, nerves transmit and process signals within the central nervous system. The sheer number of nerve cells and the incredibly greater number of connections between them makes this system the subtle wonder that it is. Nerve conduction is a general term for electrical signals carried by nerve cells. It is one aspect of bioelectricity, or electrical effects in and created by biological systems. Nerve cells, properly called neurons , look different from other cells—they have tendrils, some of them many centimeters long, connecting them with other cells. (See Figure 1.) Signals arrive at the cell body across synapses or through dendrites , stimulating the neuron to generate its own signal, sent along its long axon to other nerve or muscle cells. Signals may arrive from many other locations and be transmitted to yet others, conditioning the synapses by use, giving the system its complexity and its ability to learn. The method by which these electric currents are generated and transmitted is more complex than the simple movement of free charges in a conductor, but it can be understood with principles already discussed in this text. The most important of these are the Coulomb force and diffusion. Figure 2 illustrates how a voltage (potential difference) is created across the cell membrane of a neuron in its resting state. This thin membrane separates electrically neutral fluids having differing concentrations of ions, the most important varieties being $Na^{+}$, $K^{+}$, and $Cl^{-}$. (these are sodium, potassium, and chlorine ions with single plus or minus charges as indicated). As discussed in Molecular Transport Phenomena: Diffusion, Osmosis, and Related Processes, free ions will diffuse from a region of high concentration to one of low concentration. But the cell membrane is semipermeable , meaning that some ions may cross it while others cannot. In its resting state, the cell membrane is permeable to $K^{+}$ and $Cl^{-}$, and impermeable to $Na^{+}$. Diffusion of $K^{+}$ and $Cl^{-}$ thus creates the layers of positive and negative charge on the outside and inside of the membrane. The Coulomb force prevents the ions from diffusing across in their entirety. Once the charge layer has built up, the repulsion of like charges prevents more from moving across, and the attraction of unlike charges prevents more from leaving either side. The result is two layers of charge right on the membrane, with diffusion being balanced by the Coulomb force. A tiny fraction of the charges move across and the fluids remain neutral (other ions are present), while a separation of charge and a voltage have been created across the membrane. OpenStax
Courses/San_Diego_Miramar_College	Chem 100: Fundamentals of G(OB) Chemistry (Garces) This course is a one-semester survey of general, organic, and biological chemistry for nursing and other health-related fields. Topics include general chemistry, organic chemistry, and biological chemistry as they apply to the human body. This course is intended for students majoring in nursing, nutrition, or allied health fields. Front Matter 1: Chemistry Around Us 2: Measurements 3: Matter and Energy 4: Atoms and Elements 5: Nuclear Chemistry 6: Molecules and Compounds 7: Stoichiometry 8: Gases 9: Solutions 10: Kinetics and Equilibrium 11: Acids and Bases Back Matter Chemistry 201: General Chemistry II (Garces) Emphasis is placed on the fundamental principles of physical and inorganic chemistry. Topics include techniques of data analysis, chemical kinetics, chemical equilibrium, acids, bases, and salts, thermochemistry, electrochemistry, coordination chemistry. Computer skills are introduced and applied to data analysis, laboratory simulations, and computer interfacing with laboratory equipment. Front Matter 1: Liquids, Solids, and Intermolecular Forces 2: Solutions 3: Chemical Kinetics 4: Chemical Equilibrium 5: Acids and Bases 6: Free Energy and Thermodynamics 7: Electrochemistry 8: Transition Metals and Coordination Compounds 9: Radioactivity and Nuclear Chemistry Back Matter
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Fundamentals/Stirlings_Approximation	Stirling's approximation is named after the Scottish mathematician James Stirling (1692-1770). In confronting statistical problems we often encounter factorials of very large numbers. The factorial $N!$ is a product $N(N-1)(N-2)...(2)(1)$. Therefore, $\ln \,N!$ is a sum \[\left.\ln N!\right. = \ln 1 + \ln 2 + \ln 3 + ... + \ln N = \sum_{k=1}^N \ln k. \label{1} \] where we have used the property of logarithms that $\log(abc) =\log(a) + \log(b) +\log(c)$. The sum is shown in figure below. Using Euler-MacLaurin formula one has \[\sum_{k=1}^N \ln k=\int_1^N \ln x\,dx+\sum_{k=1}^p\frac{B_{2k}}{2k(2k-1)}\left(\frac{1}{n^{2k-1}}-1\right)+R , \label{2} \] where B 1 = −1/2, B 2 = 1/6, B 3 = 0, B 4 = −1/30, B 5 = 0, B 6 = 1/42, B 7 = 0, B 8 = −1/30, ... are the Bernoulli numbers , and $ R$ is an error term which is normally small for suitable values of $ p$ . Then, for large $N$, \[\ln N! \sim \int_1^N \ln x\,dx \approx N \ln N -N . \label{3} \] after some further manipulation one arrives at (apparently Stirling's contribution was the prefactor of $\sqrt{2\pi})$ \[N! = \sqrt{2 \pi N} \; N^{N} e^{-N} e^{\lambda_N} \label{4} \] where \[\dfrac{1}{12N+1} < \lambda_N < \frac{1}{12N}. \label{5} \] The sum of the area under the blue rectangles shown below up to $N$ is $\ln N!$. As you can see the rectangles begin to closely approximate the red curve as $m$ gets larger. The area under the curve is given the integral of $\ln x$. \[ \ln N! = \sum_{m=1}^N \ln m \approx \int_1^N \ln x\, dx \label{6} \] To solve the integral use integration by parts \[ \int u\,dv=uv-\int v\,dy \label{7A} \] Here we let $u = \ln x$ and $dv = dx$. Then $v = x$ and $du = \frac{dx}{x}$. \[ \int_0^N \ln x \, dx = x \ln x\|_0^N - \int_0^N x \dfrac{dx}{x} \label{7B} \] Notice that $x/x = 1$ in the last integral and $x \ln x$ is 0 when evaluated at zero, so we have \[ \int_0^N \ln x \, dx = N \ln N - \int_0^N dx \label{8} \] Which gives us Stirling’s approximation: $\ln N! = N \ln N – N$. As is clear from the figure above Stirling’s approximation gets better as the number N gets larger (Table $\PageIndex{1}$). N N! ln N! N ln N – N Error 10 3.63 x 106 15.1 13.02 13.8% 50 3.04 x 1064 148.4 145.60 1.88% 100 9.33 x 10157 363.7 360.50 0.88% 150 5.71 x 10262 605.0 601.60 0.56% Calculators often overheat at 200!, which is all right since clearly result are converging. In thermodynamics, we are often dealing very large N (i.e., of the order of Avagadro’s number) and for these values Stirling’s approximation is excellent. References J. Stirling "Methodus differentialis, sive tractatus de summation et interpolation serierum infinitarium", London (1730). English translation by J. Holliday "The Differential Method: A Treatise of the Summation and Interpolation of Infinite Series" (1749)
Courses/Heartland_Community_College/CHEM_120%3A_Fundamentals_of_Chemistry/01%3A_Measurement_and_Problem_Solving/1.06%3A_Significant_Figures_in_Calculations	Learning Objectives Use significant figures correctly in mathematical operations. It is important to be aware of significant figures when performing mathematical operations. For example, dividing 125 by 307 on a calculator gives 0.4071661238… to an infinite number of digits. However, not all of the digits in this answer have a practical meaning. Remember that calculators do not understand significant figures, and the person who is performing the calculation must apply the rules of significant figures to a result obtained from using a calculator. When performing mathematical operations, there are two rules for limiting the number of significant figures in an answer, based on the operation being performed: One rule is applied for multiplication and division, and a different rule is applied for addition and subtraction. In operations involving significant figures, the answer is reported in such a way that it reflects the reliability of the lowest quality value involved in the operation. In other words, an answer can be no more precise than the least precise number used to get that answer. Multiplication and Division For multiplication and division, the number of significant figures in each number being multiplied or divided must be counted, and the answer must be limited to the lesser count of significant figures. When comparing the number of digits that the answer should have to the number of digits that it actually has, three possible scenarios could arise. Specifically, if the number of digits that the answer actually has is exactly equal to the number of digits that it should have, no adjustments need to be made, and the calculated answer is the final answer. the number of digits that the answer actually has is less than the number of digits that it should have, significant zeroes must be added to the calculated answer, until it contains the correct number of significant figures. the number of digits that the answer actually has is more than the number of digits that it should have, the calculated answer must be rounded. Rounding To round a number, the number of significant figures that the number should have must be known and are selected starting from the left-most digit in the calculated number. If the number immediately to the right of the last significant digit is less than 5, the value of the last significant digit remains the same and all subsequent digits are dropped. This is called rounding down . If the number immediately to the right of the last significant digit is greater than or equal to 5, the last significant digit is increased by 1 and all subsequent digits are again dropped. This is called rounding up . Regardless of whether a number is rounded up or down, the value of the final must remain approximately equal to that of the original, calculated number. For example, consider the number "207.518". Right now, the measurement contains six significant figures. How would this number be rounded to have five significant figures? Four significant figures? Two significant figures? The answers, as well as the reasoning behind them, is outlined in Table $\PageIndex{1}$. Number of Significant Figures Rounded Value Reasoning 5 207.52 The left-most five digits are selected, and the last selected digit (1) is rounded up to a 2, since the next digit (8) is greater than 5. 4 207.50 The left-most four digits are selected, and the last selected digit (5) is rounded down to remain a 5, since the next digit (1) is less than 5. 2 210.00 The left-most two digits are selected, and the last selected digit (0) is rounded up to a 1, since the next digit (7) is greater than 5. However, the resultant value of "21" is not similar to "207.518". A non-significant ("placeholder") zero is added, as "210" is approximately equal to "207.518". Scientific Notation Remember that scientific notation provides a way of communicating significant figures without ambiguity, since all significant figures can simply be included in the coefficient. For example, the number 450 has two significant figures and would be written in scientific notation as 4.5 × 10 2 , whereas 450.0 has four significant figures and would be written as 4.500 × 10 2 . In scientific notation, all significant figures are listed explicitly. A full multiplication example is shown below. The first number involved in the calculation contains 4 significant figures, and the second number contains 5 significant figures. Therefore, final answer must be limited to 4 significant figures, and the digits "4,094" are selected to keep. Since the next digit (1) is less than 5, the number is rounded down and remains "4,094". Addition and Subtraction For addition and subtraction, the right-most significant digit in each number being added or subtracted must be selected, and the answer must be limited to the left-most of those options. When comparing the number of digits that the answer should have to the number of digits that it actually has, the same three scenarios that were considered for multiplication and division could arise. For example, consider adding 1.2 and 4.41. The right-most significant digit in "1.2" is the "2", located in the tenths column, while the right-most significant digit in "4.41" is the "1" in the hundredths column. The answer would be limited to the tenths column, since the tenths column is further to the left, relative to the hundredths column. Since the next digit in the calculated answer (1) is less than 5, the last selected digit (6) is rounded down to remain a 6. Exercise $\PageIndex{1}$ Perform the following mathematical operations, expressing each answer with the appropriate number of significant figures. 217 ÷ 903.0 = 13.77 + 908.226 + 515 = Answer a Calculated answer: 0.240310075... Correct answer: 0.240 Rationale: The operation being performed is division, so the number of significant digits in each given number must be counted. The first number involved in the calculation contains 3 significant figures, and the second number contains 4 significant figures. Therefore, the answer must be limited to 3 significant figures, and the digits "0.240" are selected to keep. (Note that the first zero is not significant, as it is before the first non-zero digit, so does not count toward the total number of selected digits.) Since the next digit (3) is less than 5, the number is rounded down and remains "0.240". Answer b Calculated answer: 1,436.996 Correct answer: 1,437 Rationale: The operation being performed is addition, so the right-most significant digit in each number being added must be selected. The right-most significant digit in the first number involved in the calculation is the final 7, in the hundredths place. The right-most significant digit in the second number involved in the calculation is the 6, in the thousandths place. The right-most significant digit in the third number involved in the calculation is the final 5, in the ones place. Therefore, the answer must be limited to the ones place, since the ones place is the furthest to the left of the relevant options. Since the next digit in the calculated answer (9) is greater than 5, the number is rounded up to become "1,437". Calculations Involving Multiplication/Division and Addition/Subtraction When performing calculations that involve either a multiplication or a division and an addition or a subtraction, the correct order of operations must be applied. Any operation written in parentheses must be performed first, followed by any multiplication and division. Addition and subtraction is performed last. Ideally, the significance of each intermediate answer would be mentally-determined, but ultimately only applied after performing the final calculation, because successive rounding can compound inaccuracies. When working on paper, however, it is often desirable to minimize the number of digits that must be written out. Therefore, in the worked examples in this text, the results of intermediate steps in a calculation will be shown to the correct number of significant figures allowed for that step, effectively treating each step as a separate calculation. This procedure is intended to reinforce the rules for determining the number of significant figures, but in some cases it may give a final answer that differs in the last digit from that obtained using a calculator, where all digits are carried through to the last step.
Courses/Tennessee_State_University/CHEM_4210%3A_Inorganic_Chem_II_(Siddiquee)/02%3A_d-Block_Metal_Chemistry-_General_Considerations/2.05%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 as it experiences the effective nuclear charge of the metal, although a few instances involve stabilization of metal electrons by ligand nuclei (inverse ligand fields). Regardless, metal-ligand bond formation is stabilizing and classified by the way it preferences the addition of ligands to the complex. Steric effects, specifically steric repulsions between ligands. One reason coordination numbers do not increase indefinitely is that only so many ligands can fit around a metal. Exactly how many depends on the size of the metal center. This is one of the more important factors. Many metals tend to exhibit preferred coordination numbers, which depend on their oxidation state and size as shown in Figure $\sf{\PageIndex{1}}$. 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 low valent metals and as size increases on moving from right to left across the transition metal block of the periodic table. Thus, the early transition metal molybdenum forms seven- and eight-coordinate [Mo III (CN) 7 ] 4- and [Mo IV (CN) 8 ] 4- with the sterically undemanding cyano ligand. the size of the 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. how the ligand bonds to the metal. Again, all other things being equal, ligands that are more sterically demanding in the vicinity of the metal center tend to limit the ability of other ligands to bind more than those which bind through a small extended group. For example, a bulky isocyanide like t-BuCN 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 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 on different ligands. 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 coordination sphere. With a given metal, ligands that are more electron donating have a greater tendency to form complexes with lower coordination numbers with a given metal 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 from 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 maximizes the intra-electron group repulsions. 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 $\sf{\PageIndex{2}}$. As may be seen from the geometries listed in Figure $\sf{\PageIndex{2}}$, 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 as ligands become increasingly dispersed across the metal's coordination sphere. In complexes containing five, seven, eight, or higher coordinate metals, there are 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 $\sf{\PageIndex{2}}$. d-electron effects A few coordination geometries are noticeably absent from the Kepert-preferred and Kepert-accessible geometries in Figure $\sf{\PageIndex{1}}$. 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 a 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 depends on 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, differs from case to case and can include distortions of the complex's geometry. For instance, an ideal octahedral coordination geoemtry might be tetragonally distorted by flattening or elongating it. imparting a strong preference for 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 $\sf{\PageIndex{2}}$ may be extended to that shown in Figure $\sf{\PageIndex{3}}$. Ligand constraints imposed by rigid or semirigid ligands 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. Thus 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 $\sf{\PageIndex{4}}$. 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 particularly spectacular case of an entactic state involves the blue copper proteins azurin and plastocyanin, the structure of which is given in Figure $\sf{\PageIndex{5}}$. As may be seen from the structure in Figure $\sf{\PageIndex{5}}$, 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 it easier for the copper center to undergo facile redox reactions, enabling it to better function as an electron carrier. Crystal packing effects, in which the energy-lowering packing of molecules and ions in a crystal drives the distortion of a complex's structure away from what it would adopt in the gas phase or solution This effect is similar to that of ligand constraints except that in this case it arises 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 dodecahedral coordination geometry in the crystals of many of its salts. Relativistic effects on orbital energies The proximity of fast moving electrons to massive nuclei in the heavier transition elements results in relativistic 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 $\sf{\PageIndex{6A}}$. The combination of smaller sizes and altered orbital energies affects coordination preferences. Relativistic 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 $\sf{\PageIndex{6B}}$, 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/(^{\circ}\) 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-triisopropylphenyl)benzene as shown in Figure $\sf{\PageIndex{7}}$. Coordination Number 2. A coordination number of 2 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 slats 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 $\sf{\PageIndex{8}}$. 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. \[\sf{4~Au~~+~~8~CN^-~~+~~O_2~~+~~2~H_2O~~\longrightarrow~~4~[Au(CN)_2]^-~~+~~4~OH^-} \nonumber \] 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. And by means of honorary mention, the mercury(I) ion, 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. The classic examples are given in Figure $\sf{\PageIndex{9}}$. 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. Coordination Number 4 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 d 0 configurations 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- , and [CdI 4 ] 2- . For other electron configurations, 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 + , Pt 2 + , and 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 ]. Coordination Number 5 The two common coordination geometries for five-coordinate complexes are trigonal bipyramidal and square pyramidal. 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 $\sf{\PageIndex{4}}$ and for complexes containing oxo and nitrido ligands, examples of which are shown in Figure $\sf{\PageIndex{12}}$. 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 $\sf{\PageIndex{13}}$. Coordination Number 6 The two common coordination geometries for coordination number 8 are octahedral and trigonal prismatic. Trigonal prismatic coordination is related to octahedral coordination as shown in Figure $\sf{\PageIndex{15}}$. As may be seen in Figure $\sf{\PageIndex{15}}$, 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 trigonal 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 $\sf{\PageIndex{4C}}$ may be used to encourage the adoption of a trigonal prismatic geometry, although once the number of d electrons present exceeds two, 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. 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 $\sf{\PageIndex{18}}$. 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, and 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 ]. As with other high-coordinate structures, the energy difference between these eightfold coordination geometries is small enough that packing effects can significantly influence 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 idiosyncratically determined by the ligands. Simple examples 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- . 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 $\sf{\PageIndex{23}}$, [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 ligands 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 to involve a Co center in the midst of a B 16 - "molecular drum" held together by cluster bonds. References 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. 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. Cremades, E.; Echeverría, J.; Alvarez, S., The Trigonal Prism in Coordination Chemistry. Chemistry – A European Journal 2010, 16 (34), 10380-10396. 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. Concepción Gimeno, M. The Chemistry of Gold in Laguna, Antonio (ed.) Modern Supramolecular Gold Chemistry: Gold-Metal Interactions and Applications . Wiley, 2008. 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. Persson, I., Hydrated metal ions in aqueous solution: How regular are their structures? Pure and Applied Chemistry 2010, 8 2(10) , 1901. 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. 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. 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. 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 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. 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. 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 cyclopentadienyl 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, and so are better thought of as a metal trapped in a spacious sixty-carbon cage. Zalkin, A.; Forrester, J.D.; Templeton, D.H. Crystal structure of cerium magnesium nitrate hydrate Journal of Chemical Physics , 1963 , 39 , 2881-2891. 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. 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).
Courses/Modesto_Junior_College/Chemistry_143%3A_Introductory_College_Chemistry_(Brzezinski)/CHEM_143%3A_Text_(Brzezinski)/05%3A_Nuclear_Chemistry	5.1: Radioactivity Atoms are composed of subatomic particles—protons, neutrons, and electrons. Protons and neutrons are located in the nucleus and provide most of the mass of the atom, while electrons circle the nucleus in shells and subshells and account for an atom’s size. There are three main forms of radioactive emissions and are alpha particles, beta particles, and gamma rays. 5.2: Nuclear Bombardment Reactions 5.3: Half-Life Natural radioactive processes are characterized by a half-life, the time it takes for half of the material to decay radioactively. The amount of material left over after a certain number of half-lives can be easily calculated. 5.3.1: Practice Half-Life 5.4: Units of Radioactivity Radioactivity can be expressed in a variety of units, including rems, rads, and curies. 5.5: Uses of Radioactive Isotopes Radioactivity has several practical applications, including tracers, medical applications, dating once-living objects, and the preservation of food. 5.6: Nuclear Energy Nuclear energy comes from tiny mass changes in nuclei as radioactive processes occur. In fission, large nuclei break apart and release energy; in fusion, small nuclei merge together and release energy. 5.6.1: Practice Nuclear Chemistry
Courses/National_Yang_Ming_Chiao_Tung_University/Chemical_Principles_for_Medical_Students/05%3A_Gases/5.08%3A_Molecular_Effusion_and_Diffusion	Learning Objectives To understand the significance of the kinetic molecular theory of gases We now describe how the kinetic molecular theory of gases explains some of the important relationships we have discussed previously. Diffusion and Effusion As you have learned, the molecules of a gas are not stationary but in constant and random motion. If someone opens a bottle of perfume in the next room, for example, you are likely to be aware of it soon. Your sense of smell relies on molecules of the aromatic substance coming into contact with specialized olfactory cells in your nasal passages, which contain specific receptors (protein molecules) that recognize the substance. How do the molecules responsible for the aroma get from the perfume bottle to your nose? You might think that they are blown by drafts, but, in fact, molecules can move from one place to another even in a draft-free environment. Diffusion is the gradual mixing of gases due to the motion of their component particles even in the absence of mechanical agitation such as stirring. The result is a gas mixture with uniform composition. Diffusion is also a property of the particles in liquids and liquid solutions and, to a lesser extent, of solids and solid solutions. The related process, effusion , is the escape of gaseous molecules through a small (usually microscopic) hole, such as a hole in a balloon, into an evacuated space. The phenomenon of effusion had been known for thousands of years, but it was not until the early 19th century that quantitative experiments related the rate of effusion to molecular properties. The rate of effusion of a gaseous substance is inversely proportional to the square root of its molar mass. This relationship is referred to as Graham’s law , after the Scottish chemist Thomas Graham (1805–1869). The ratio of the effusion rates of two gases is the square root of the inverse ratio of their molar masses : \[\dfrac{\text{rate of effusion A}}{\text{rate of effusion B}}=\sqrt{\dfrac{M_B}{M_A}} \label{10.8.1} \] Helium ( M = 4.00 g/mol) effuses much more rapidly than ethylene oxide ( M = 44.0 g/mol). Because helium is less dense than air, helium-filled balloons “float” at the end of a tethering string. Unfortunately, rubber balloons filled with helium soon lose their buoyancy along with much of their volume. In contrast, rubber balloons filled with air tend to retain their shape and volume for a much longer time. Because helium has a molar mass of 4.00 g/mol, whereas air has an average molar mass of about 29 g/mol, pure helium effuses through the microscopic pores in the rubber balloon $\sqrt{\dfrac{29}{4.00}}=2.7$ times faster than air. For this reason, high-quality helium-filled balloons are usually made of Mylar, a dense, strong, opaque material with a high molecular mass that forms films that have many fewer pores than rubber. Hence, mylar balloons can retain their helium for days. At a given temperature, heavier molecules move more slowly than lighter molecules. Example $\PageIndex{1}$ During World War II , scientists working on the first atomic bomb were faced with the challenge of finding a way to obtain large amounts of $\ce{^{235}U}$. Naturally occurring uranium is only 0.720% $\ce{^{235}U}$, whereas most of the rest (99.275%) is $\ce{^{238}U}$, which is not fissionable (i.e., it will not break apart to release nuclear energy) and also actually poisons the fission process. Because both isotopes of uranium have the same reactivity, they cannot be separated chemically. Instead, a process of gaseous effusion was developed using the volatile compound $UF_6$ (boiling point = 56°C). Calculate the ratio of the rates of effusion of 235 UF 6 and 238 UF 6 for a single step in which UF 6 is allowed to pass through a porous barrier. (The atomic mass of 235 U is 235.04, and the atomic mass of 238 U is 238.05.) If n identical successive separation steps are used, the overall separation is given by the separation in a single step (in this case, the ratio of effusion rates) raised to the n th power. How many effusion steps are needed to obtain 99.0% pure 235 UF 6 ? Given: isotopic content of naturally occurring uranium and atomic masses of 235 U and 238 U Asked for: ratio of rates of effusion and number of effusion steps needed to obtain 99.0% pure 235 UF 6 Strategy: Calculate the molar masses of 235 UF 6 and 238 UF 6 , and then use Graham’s law to determine the ratio of the effusion rates. Use this value to determine the isotopic content of 235 UF 6 after a single effusion step. Divide the final purity by the initial purity to obtain a value for the number of separation steps needed to achieve the desired purity. Use a logarithmic expression to compute the number of separation steps required. Solution: A The first step is to calculate the molar mass of UF 6 containing 235 U and 238 U. Luckily for the success of the separation method, fluorine consists of a single isotope of atomic mass 18.998. The molar mass of 235 UF 6 is 234.04 + (6)(18.998) = 349.03 g/mol The molar mass of 238 UF 6 is 238.05 + (6)(18.998) = 352.04 g/mol The difference is only 3.01 g/mol (less than 1%). The ratio of the effusion rates can be calculated from Graham’s law using Equation $\ref{10.8.1}$: \[\rm\dfrac{\text{rate }^{235}UF_6}{\text{rate }^{238}UF_6}=\sqrt{\dfrac{352.04\;g/mol}{349.03\;g/mol}}=1.0043 \nonumber \] B To obtain 99.0% pure 235 UF 6 requires many steps. We can set up an equation that relates the initial and final purity to the number of times the separation process is repeated: final purity = (initial purity)(separation) n In this case, 0.990 = (0.00720)(1.0043) n , which can be rearranged to give \[1.0043^n=\dfrac{0.990}{0.00720}=137.50 \nonumber \] Taking the logarithm of both sides gives \[\begin{align} n\ln(1.0043)&=\ln(137.50) \\[4pt] n &=\dfrac{\ln(137.50)}{\ln(1.0043)} \\[4pt]&=1148 \end{align} \nonumber \] Thus at least a thousand effusion steps are necessary to obtain highly enriched 235 U. Below is a small part of a system that is used to prepare enriched uranium on a large scale. Exercise $\PageIndex{1}$ Helium consists of two isotopes: 3 He (natural abundance = 0.000134%) and 4 He (natural abundance = 99.999866%). Their atomic masses are 3.01603 and 4.00260, respectively. Helium-3 has unique physical properties and is used in the study of ultralow temperatures. It is separated from the more abundant 4 He by a process of gaseous effusion. Calculate the ratio of the effusion rates of 3 He and 4 He and thus the enrichment possible in a single effusion step. How many effusion steps are necessary to yield 99.0% pure 3 He? Answer a ratio of effusion rates = 1.15200; one step gives 0.000154% 3 He Answer b 96 steps Rates of Diffusion or Effusion Graham’s law is an empirical relationship that states that the ratio of the rates of diffusion or effusion of two gases is the square root of the inverse ratio of their molar masses. The relationship is based on the postulate that all gases at the same temperature have the same average kinetic energy. We can write the expression for the average kinetic energy of two gases with different molar masses: \[KE=\dfrac{1}{2}\dfrac{M_{\rm A}}{N_A}v_{\rm rms,A}^2=\dfrac{1}{2}\dfrac{M_{\rm B}}{N_A}v_{\rm rms,B}^2\label{10.8.2} \] Multiplying both sides by 2 and rearranging give \[\dfrac{v_{\rm rms, B}^2}{v_{\rm rms,A}^2}=\dfrac{M_{\rm A}}{M_{\rm B}}\label{10.8.3} \] Taking the square root of both sides gives \[\dfrac{v_{\rm rms, B}}{v_{\rm rms,A}}=\sqrt{\dfrac{M_{\rm A}}{M_{\rm B}}}\label{10.8.4} \] Thus the rate at which a molecule, or a mole of molecules, diffuses or effuses is directly related to the speed at which it moves. Equation $\ref{10.8.4}$ shows that Graham’s law is a direct consequence of the fact that gaseous molecules at the same temperature have the same average kinetic energy. Typically, gaseous molecules have a speed of hundreds of meters per second (hundreds of miles per hour). The effect of molar mass on these speeds is dramatic, as illustrated in Figure $\PageIndex{3}$ for some common gases. Because all gases have the same average kinetic energy, according to the Boltzmann distribution , molecules with lower masses, such as hydrogen and helium, have a wider distribution of speeds. The lightest gases have a wider distribution of speeds and the highest average speeds. Molecules with lower masses have a wider distribution of speeds and a higher average speed. Gas molecules do not diffuse nearly as rapidly as their very high speeds might suggest. If molecules actually moved through a room at hundreds of miles per hour, we would detect odors faster than we hear sound. Instead, it can take several minutes for us to detect an aroma because molecules are traveling in a medium with other gas molecules. Because gas molecules collide as often as 10 10 times per second, changing direction and speed with each collision, they do not diffuse across a room in a straight line, as illustrated schematically in Figure $\PageIndex{4}$. The average distance traveled by a molecule between collisions is the mean free path . The denser the gas, the shorter the mean free path; conversely, as density decreases, the mean free path becomes longer because collisions occur less frequently. At 1 atm pressure and 25°C, for example, an oxygen or nitrogen molecule in the atmosphere travels only about 6.0 × 10 −8 m (60 nm) between collisions. In the upper atmosphere at about 100 km altitude, where gas density is much lower, the mean free path is about 10 cm; in space between galaxies, it can be as long as 1 × 10 10 m (about 6 million miles). The denser the gas, the shorter the mean free path. Example $\PageIndex{2}$ Calculate the rms speed of a sample -butene (C 4 H 8 ) at 20°C. Given: compound and temperature Asked for: rms speed Strategy: Calculate the molar mass of cis-2-butene. Be certain that all quantities are expressed in the appropriate units and then use Equation 10.8.5 to calculate the rms speed of the gas. Solution: To use Equation 10.8.4, we need to calculate the molar mass of cis -2-butene and make sure that each quantity is expressed in the appropriate units. Butene is C 4 H 8 , so its molar mass is 56.11 g/mol. Thus \[\begin{align} u_{\rm rms} &= \sqrt{\dfrac{3RT}{M}} \\[4pt] &=\rm\sqrt{\dfrac{3\times8.3145\;\dfrac{J}{K\cdot mol}\times(20+273)\;K}{56.11\times10^{-3}\;kg}}\\[4pt] &=361\;m/s \end{align} \nonumber \] or approximately 810 mi/h. Exercise $\PageIndex{1}$ Calculate the rms speed of a sample of radon gas at 23°C. Answer $1.82 \times 10^2\; m/s$ (about 410 mi/h) The kinetic molecular theory of gases demonstrates how a successful theory can explain previously observed empirical relationships (laws) in an intuitively satisfying way. Unfortunately, the actual gases that we encounter are not “ideal,” although their behavior usually approximates that of an ideal gas. Summary The kinetic molecular theory of gases provides a molecular explanation for the observations that led to the development of the ideal gas law. Average kinetic energy: \[\overline{e_K}=\dfrac{1}{2}m{u_{\rm rms}}^2=\dfrac{3}{2}\dfrac{R}{N_A}T, \nonumber \] Root mean square speed: \[u_{\rm rms}=\sqrt{\dfrac{u_1^2+u_2^2+\cdots u_N^2}{N}}, \nonumber \] Kinetic molecular theory of gases : \[u_{\rm rms}=\sqrt{\dfrac{3RT}{M}}. \nonumber \] Graham’s law for effusion: \[\dfrac{v_{\rm rms, B}}{v_{\rm rms,A}}=\sqrt{\dfrac{M_{\rm A}}{M_{\rm B}}} \nonumber \] Diffusion is the gradual mixing of gases to form a sample of uniform composition even in the absence of mechanical agitation. In contrast, effusion is the escape of a gas from a container through a tiny opening into an evacuated space. The rate of effusion of a gas is inversely proportional to the square root of its molar mass ( Graham’s law ), a relationship that closely approximates the rate of diffusion. As a result, light gases tend to diffuse and effuse much more rapidly than heavier gases. The mean free path of a molecule is the average distance it travels between collisions.
Courses/can	Chemistry is the science of matter, one of the oldest and most fundamental of the physical sciences. Before you study solar systems and stars, you start with helium and hydrogen. If you want to know nutrition and respiration, you begin with carbon and oxygen. Skyscrapers stand because of iron and the semiconductor industry is the business of silicon. For most college students, the first collegiate science course is chemistry. Chemistry courses at Cañada College are designed to parallel those in most major colleges and universities, allowing you to complete many of the foundation courses you will need for a four-year degree before transfer. There are three "first" chemistry courses to get started with chemistry at Cañada College CHEM 210 General Chemistry I (Puenzo) CHEM 210: General Chemistry I - An Atoms-Up Approach CHEM 231: Organic Chemistry I Textbook CHEM 220: General Chemistry II - Chemical Dynamics CHEM 232 - Organic Chemistry II (Puenzo)
Bookshelves/Organic_Chemistry/Organic_Chemistry_I_(Morsch_et_al.)/01%3A_Structure_and_Bonding/1.04%3A_Development_of_Chemical_Bonding_Theory	Objectives After completing this section, you should be able to draw Lewis Dot Symbols for main group elements and ions. describe the three-dimensional nature of molecules. sketch a tetrahedral molecule, CX 4 , using the “wedge-and-broken-line” method of representation. make a ball-and-stick model of a simple tetrahedral molecule such as methane, CH 4 . draw Lewis Dot Structures for 2 electron group molecules. draw Lewis Dot Structures for 3 electron group molecules. draw Lewis Dot Structures for 4 electron group molecules. Key Terms Make certain that you can define, and use in context, the key terms below. bond strength covalent bond ionic bond Lewis structure lone-pair electron non-bonding electron Study Notes To draw Lewis structures successfully, you need to know the number of valence electrons present in each of the atoms involved. Memorize the number of valence electrons possessed by each of the elements commonly encountered in organic chemistry: C, H, O, N, S, P and the halogens. When drawing any organic structure, you must remember that a neutral carbon atom will almost always have four bonds. Similarly, hydrogen always has one bond; neutral oxygen atoms have two bonds; and neutral nitrogen atoms have three bonds. By committing these simple rules to memory, you can avoid making unnecessary mistakes later in the course. The “wedge-and-broken-line” type of representation, which helps to convey the three-dimensional nature of organic compounds, will be used throughout the course. Bonding Overview Why are some substances chemically bonded molecules and others are an association of ions? The answer to this question depends upon the electronic structures of the atoms and nature of the chemical forces within the compounds. Although there are no sharply defined boundaries, chemical bonds are typically classified into three main types: ionic bonds, covalent bonds, and metallic bonds. In this chapter, each type of bond and the general properties found in typical substances in which the bond type occurs will be discussed. Ionic bonds results from electrostatic forces that exist between ions of opposite charge . These bonds typically involve a metal with a nonmetal Covalent bonds result from the sharing of electrons between two atoms . The bonds typically involve one nonmetallic element with another Metallic bonds are found in solid metals (copper, iron, aluminum) with each metal atom bonded to several neighboring metal atoms and the bonding electrons are free to move throughout the 3-dimensional structure. Each bond classification is discussed in detail in subsequent sections of the chapter. Let's look at the preferred arrangements of electrons in atoms when they form chemical compounds. Lewis Symbols At the beginning of the 20 th century, the American chemist G. N. Lewis (1875–1946) devised a system of symbols—now called Lewis electron dot symbols, often shortened to Lewis dot symbols —that can be used for predicting the number of bonds formed by most elements in their compounds. Each Lewis dot symbol consists of the chemical symbol for an element surrounded by dots that represent its valence electrons. Lewis Dot symbols: provide a convenient representation of valence electrons allows you to keep track of valence electrons during bond formation consists of the chemical symbol for the element plus a dot for each valence electron To write an element’s Lewis dot symbol, we place dots representing its valence electrons, one at a time, around the element’s chemical symbol. Up to four dots are placed above, below, to the left, and to the right of the symbol (in any order, as long as elements with four or fewer valence electrons have no more than one dot in each position). The next dots, for elements with more than four valence electrons, are again distributed one at a time, each paired with one of the first four. Fluorine, for example, with the electron configuration [He]2 s 2 2 p 5 , has seven valence electrons, so its Lewis dot symbol is constructed as follows: Lewis used the unpaired dots to predict the number of bonds that an element will form in a compound. Consider the symbol for nitrogen in Figure $\PageIndex{2}$. The Lewis dot symbol explains why nitrogen, with three unpaired valence electrons, tends to form compounds in which it shares the unpaired electrons to form three bonds. Boron, which also has three unpaired valence electrons in its Lewis dot symbol, also tends to form compounds with three bonds, whereas carbon, with four unpaired valence electrons in its Lewis dot symbol, tends to share all of its unpaired valence electrons by forming compounds in which it has four bonds. Lewis symbols are a tool to help draw structures. We will see why bonding in molecular compounds follow Lewis' theory in the next section. Elements in the same group have the same number of valence electrons and similar Lewis symbols. For example, the electron configuration for atomic sulfur is [Ne]3s 2 3p 4 , thus there are six valence electrons. Its Lewis symbol would therefore be similar to oxygen and look like: The Octet Rule Lewis’s major contribution to bonding theory was to recognize that atoms tend to lose, gain, or share electrons to reach a total of eight valence electrons, called an octet . This so-called octet rule explains the stoichiometry of most compounds in the s and p blocks of the periodic table. We now know from quantum mechanics that the number eight corresponds to one ns and three np valence orbitals, which together can accommodate a total of eight electrons. Remarkably, though, Lewis’s insight was made nearly a decade before Rutherford proposed the nuclear model of the atom. Common exceptions to the octet rule are helium, whose 1 s 2 electron configuration gives it a full n = 1 shell, and hydrogen, which tends to gain or share its one electron to achieve the electron configuration of helium. Lewis's idea of an octet explains why noble gases rarely form compounds. They have the stable s 2 p 6 configuration (full octet, no charge), so they have no reason to react and change their configuration. All other elements attempt to gain, lose, or share electrons to achieve a noble gas configuration. This explains why atom combine together to form compounds. By forming bond the makes the atoms more stable and lower in energy. Making bonds releases energy and represents a driving force for the formation of compounds. Atoms often gain, lose, or share electrons to achieve the same number of electrons as the noble gas closest to them in the periodic table. Lewis Structures Lewis structures represent how Lewis symbols gain, lose, or share electrons to obtain an octet by forming compounds. Lewis Structures of Ionic Compounds Whenever there is a metal present in the structure of an organic compound, there is a high likelihood that at least one ionic bond is present. Ionic bonds are represented differently in Lewis structures than covalent bonds. Great care should be taken whenever drawing the Lewis structure of an organic compounds which contains and ionic bond. Ionic bonds typically are formed when a metal and a nonmetal are part of a compound. Some atoms achieve an octet by fully gaining or losing electrons to form ions. Ionic bonds form through the eletrostatic attraction of the created ions. The formula for table salt is NaCl. It is the result of Na + ions and Cl - ions bonding together. If sodium metal and chlorine gas mix under the right conditions, they will form salt. The sodium loses an electron, and the chlorine gains that electron. In the process, a great amount of light and heat is released. The resulting salt is mostly unreactive — it is stable. It will not undergo any explosive reactions, unlike the sodium and chlorine that it is made of. Why? Referring to the octet rule, atoms attempt to get a noble gas electron configuration, which is eight valence electrons. Sodium (1s 2 2s 2 2p 6 3s 1 ) has one valence electron, so giving it up would result in the same electron configuration as neon (1s 2 2s 2 2p 6 ). Chlorine (1s 2 2s 2 2p 6 3s 2 3p 7 )has seven valence electrons, so if it takes one it will have eight (an octet). Chlorine has the electron configuration of argon (1s 2 2s 2 2p 6 3s 2 3p 8 ) when it gains an electron. The Lewis structure of an ionic compound show the movement of electrons. For NaCl, sodium is in group 1 and has one valence electron and chlorine is in group 17 and has seven valence electrons. Sodium loses its sole valence electron thereby becomes positively charged. Chlorine gains this electron, gaining a full octet, and a negative charge. After the gain/loss of an electron the new Lewis structures of Na + and Cl - are written next to each other representing the ionic bond in NaCl. Examples of Lewis Structures of Ionic Compounds Covalent Bonds and the Lewis Structures of Molecular Compounds While alkali metals (such as sodium and potassium), alkaline earth metals (such as magnesium and calcium), and halogens (such as fluorine and chlorine) often form ions in order to achieve a full octet, the principle elements of organic chemistry - carbon, hydrogen, nitrogen, and oxygen - instead tend to fill their octet by sharing electrons with other atoms, forming covalent bonds. Consider the simplest case of hydrogen gas. An isolated hydrogen atom has only one electron, located in the 1 s orbital. If two hydrogen atoms come close enough so that their respective 1 s orbitals overlap, the two electrons can be shared between the two nuclei, and a covalently bonded H 2 molecule is formed. In the Lewis structure of H 2 , each pair of electrons that is shared between two atoms is drawn as a single line, designating a single covalent bond. Hydrogen represents a special case, because a hydrogen atom cannot fulfill the octet rule; it needs only two electrons to have a full shell. This is often called the ‘doublet rule’ for hydrogen. One of the simplest organic molecules is methane with the molecular formula CH 4 . Methane is the ‘natural gas’ burned in home furnaces and hot water heaters, as well as in electrical power generating plants. To illustrate the covalent bonding in methane using a Lewis structure, we first must recognize that, although a carbon atom has a total of six electrons it's Lewis symbol has four unpaired electrons. Following Lewis' theory the carbon atom wants to form four covalent bonds to fill its octet. In a methane molecule, the central carbon atom shares its four valence electrons with four hydrogen atoms, thus forming four bonds and fulfilling the octet rule (for the carbon) and the ‘doublet rule’ (for each of the hydrogens). The next relatively simple organic molecule to consider is ethane, which has the molecular formula C 2 H 6 . If we draw each atom's Lewis symbol separately, we can see that the octet/doublet rule can be fulfilled for all of them by forming one carbon-carbon bond and six carbon-hydrogen bonds. The same approach can be used for molecules in which there is no carbon atom. In a water molecule, the Lewis symbol of the oxygen atom has two unpaired electrons. These are paired with the single electron in the Lewis symbols of the hydrogens' two O-H covalent bonds. The remaining four non-bonding electrons on oxygen called ‘lone pairs’. Since the lone pair electrons are often NOT shown in chemical structures, it is important to see mentally add the lone pairs. In the beginning, it can be helpful to physically add the lone pair electrons. 0 1 2 NaN NaN NaN methylamine ethanol chloromethane Exercise For the following structure, please fill in all of the missing lone pair electrons. Answer When two or more electrons are shared between atoms a multiple covalent bond is formed. The molecular formula for ethene (also known as ethylene, a compound found in fruits, such as apples, that signals them to ripen) is C 2 H 4 . Arranging Lewis symbols of the atoms, you can see that the octet/doublet rule can be fulfilled for all atoms only if the two carbons share two pairs of electrons between them. Ethene contains a carbon/carbon double bond. Following this pattern, the triple bond in ethyne molecular formula C 2 H 2 , (also known as acetylene, the fuel used in welding torches), is formed when the two carbon atoms share three pairs of electrons between them. Exercise $\PageIndex{1}$ Draw the Lewis structure for ammonia, NH 3 . Answer Molecular Shape A stick and wedge drawing of methane shows the tetrahedral angles...(The wedge is coming out of the paper and the dashed line is going behind the paper. The solid lines are in the plane of the paper.) The following examples make use of this notation, and also illustrate the importance of including non-bonding valence shell electron pairs when viewing such configurations. 0 1 2 NaN NaN NaN Methane Ammonia Water Bonding configurations are readily predicted by valence-shell electron-pair repulsion theory, commonly referred to as VSEPR in most introductory chemistry texts. This simple model is based on the fact that electrons repel each other, and that it is reasonable to expect that the bonds and non-bonding valence electron pairs associated with a given atom will prefer to be as far apart as possible. The bonding configurations of carbon are easy to remember, since there are only three categories. Configuration Bonding Partners Bond Angles Example Tetrahedral 4 109.5º NaN Trigonal Planar 3 120º NaN Linear 2 180º NaN Figure 1.4.3 In the three examples shown above, the central atom (carbon) does not have any non-bonding valence electrons; consequently the configuration may be estimated from the number of bonding partners alone. However, for molecules of water and ammonia, the non-bonding electrons must be included in the calculation. In each case there are four regions of electron density associated with the valence shell so that a tetrahedral bond angle is expected. The measured bond angles of these compounds (H 2 O 104.5º & NH 3 107.3º) show that they are closer to being tetrahedral than trigonal planar or linear. Of course, it is the configuration of atoms (not electrons) that defines the the shape of a molecule, and in this sense ammonia is said to be pyramidal (not tetrahedral). The compound boron trifluoride, BF 3 , does not have non-bonding valence electrons and the configuration of its atoms is trigonal. Nice treatments of VSEPR theory have been provided by Oxford and Purdue. The best way to study the three-dimensional shapes of molecules is by using molecular models. Many kinds of model kits are available to students and professional chemists. Two Electron Groups Our first example is a molecule with two bonded atoms and no lone pairs of electrons, B e H 2 . AX 2 : BeH 2 1. The central atom, beryllium, contributes two valence electrons, and each hydrogen atom contributes one. The Lewis electron structure is 2. There are two electron groups around the central atom. We see from Figure 1.4.3 that the arrangement that minimizes repulsions places the groups 180° apart. 3. Both groups around the central atom are bonding pairs (BP). Thus BeH 2 is designated as AX 2 . 4. From Figure 1.4.3 we see that with two bonding pairs, the molecular geometry that minimizes repulsions in BeH 2 is linear . AX 2 : CO 2 1. The central atom, carbon, contributes four valence electrons, and each oxygen atom contributes six. The Lewis electron structure is 2. The carbon atom forms two double bonds. Each double bond is a group, so there are two electron groups around the central atom. Like BeH 2 , the arrangement that minimizes repulsions places the groups 180° apart. 3. Once again, both groups around the central atom are bonding pairs (BP), so CO 2 is designated as AX 2 . 4. VSEPR only recognizes groups around the central atom. Thus the lone pairs on the oxygen atoms do not influence the molecular geometry. With two bonding pairs on the central atom and no lone pairs, the molecular geometry of CO 2 is linear (Figure 1.4.3). Three Electron Groups AX 3 : BCl 3 1. The central atom, boron, contributes three valence electrons, and each chlorine atom contributes seven valence electrons. The Lewis electron structure is 2. There are three electron groups around the central atom. To minimize repulsions, the groups are placed 120° apart (Figure 1.4.3). 3. All electron groups are bonding pairs (BP), so the structure is designated as AX 3 . 4. From Figure 1.4.3 we see that with three bonding pairs around the central atom, the molecular geometry of BCl 3 is trigonal planar . AX 3 : CO 3 2 − 1. The central atom, carbon, has four valence electrons, and each oxygen atom has six valence electrons. As you learned previously, the Lewis electron structure of one of three resonance forms is represented as 2. The structure of CO 3 2 − is a resonance hybrid. It has three identical bonds, each with a bond order of 4/3 . We minimize repulsions by placing the three groups 120° apart (Figure 1.4.3). 3. All electron groups are bonding pairs (BP). With three bonding groups around the central atom, the structure is designated as AX 3 . 4. We see from Figure 1.4.3 that the molecular geometry of CO 3 2 − is trigonal planar. In our next example we encounter the effects of lone pairs and multiple bonds on molecular geometry for the first time. AX 2 E: SO 2 1. The central atom, sulfur, has 6 valence electrons, as does each oxygen atom. With 18 valence electrons, the Lewis electron structure is shown below. 2. There are three electron groups around the central atom, two double bonds and one lone pair. We initially place the groups in a trigonal planar arrangement to minimize repulsions (Figure 1.4.3). 3. There are two bonding pairs and one lone pair, so the structure is designated as AX 2 E. This designation has a total of three electron pairs, two X and one E. Because a lone pair is not shared by two nuclei, it occupies more space near the central atom than a bonding pair (Figure 1.4.4). Thus bonding pairs and lone pairs repel each other electrostatically in the order BP–BP < LP–BP < LP–LP. In SO 2 , we have one BP–BP interaction and two LP–BP interactions. 4. The molecular geometry is described only by the positions of the nuclei, not by the positions of the lone pairs. Thus with two nuclei and one lone pair the shape is bent , or V shaped , which can be viewed as a trigonal planar arrangement with a missing vertex (Figures 1.4.2.1 and 1.4.3). As with SO 2 , this composite model of electron distribution and negative electrostatic potential in ammonia shows that a lone pair of electrons occupies a larger region of space around the nitrogen atom than does a bonding pair of electrons that is shared with a hydrogen atom. Like lone pairs of electrons, multiple bonds occupy more space around the central atom than a single bond, which can cause other bond angles to be somewhat smaller than expected. This is because a multiple bond has a higher electron density than a single bond, so its electrons occupy more space than those of a single bond. For example, in a molecule such as CH 2 O (AX 3 ), whose structure is shown below, the double bond repels the single bonds more strongly than the single bonds repel each other. This causes a deviation from ideal geometry (an H–C–H bond angle of 116.5° rather than 120°). Four Electron Groups One of the limitations of Lewis structures is that they depict molecules and ions in only two dimensions. With four electron groups, we must learn to show molecules and ions in three dimensions. AX 4 : CH 4 1. The central atom, carbon, contributes four valence electrons, and each hydrogen atom has one valence electron, so the full Lewis electron structure is 2. There are four electron groups around the central atom. As shown in Figure 1.4.2, repulsions are minimized by placing the groups in the corners of a tetrahedron with bond angles of 109.5°. 3. All electron groups are bonding pairs, so the structure is designated as AX 4 . 4. With four bonding pairs, the molecular geometry of methane is tetrahedral (Figure 1.4.3). AX 3 E: NH 3 1. In ammonia, the central atom, nitrogen, has five valence electrons and each hydrogen donates one valence electron, producing the Lewis electron structure 2. There are four electron groups around nitrogen, three bonding pairs and one lone pair. Repulsions are minimized by directing each hydrogen atom and the lone pair to the corners of a tetrahedron. 3. With three bonding pairs and one lone pair, the structure is designated as AX 3 E. This designation has a total of four electron pairs, three X and one E. We expect the LP–BP interactions to cause the bonding pair angles to deviate significantly from the angles of a perfect tetrahedron. 4. There are three nuclei and one lone pair, so the molecular geometry is trigonal pyramidal . In essence, this is a tetrahedron with a vertex missing (Figure 1.4.3). However, the H–N–H bond angles are less than the ideal angle of 109.5° because of LP–BP repulsions. AX 2 E 2 : H 2 O 1. Oxygen has six valence electrons and each hydrogen has one valence electron, producing the Lewis electron structure 2. There are four groups around the central oxygen atom, two bonding pairs and two lone pairs. Repulsions are minimized by directing the bonding pairs and the lone pairs to the corners of a tetrahedron Figure 1.4.3. 3. With two bonding pairs and two lone pairs, the structure is designated as AX 2 E 2 with a total of four electron pairs. Due to LP–LP, LP–BP, and BP–BP interactions, we expect a significant deviation from idealized tetrahedral angles. 4. With two hydrogen atoms and two lone pairs of electrons, the structure has significant lone pair interactions. There are two nuclei about the central atom, so the molecular shape is bent , or V shaped , with an H–O–H angle that is even less than the H–N–H angles in NH 3 , as we would expect because of the presence of two lone pairs of electrons on the central atom rather than one.. This molecular shape is essentially a tetrahedron with two missing vertices.
Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/01%3A_Introduction_to_Chemistry/1.12%3A_Scientific_Problem_Solving	How can we use problem solving in our everyday routines? One day you wake up and realize your clock radio did not turn on to get you out of bed. You are puzzled, so you decide to find out what happened. You list three possible explanations: There was a power failure and your radio cannot turn on. Your little sister turned it off as a joke. You did not set the alarm last night. Upon investigation, you find that the clock is on, so there is no power failure. Your little sister was spending the night with a friend and could not have turned the alarm off. You notice that the alarm is not set—your forgetfulness made you late. You have used the scientific method to answer a question. Scientific Problem Solving Humans have always wondered about the world around them. One of the questions of interest was (and still is): what is this world made of? Chemistry has been defined in various ways as the study of matter. What matter consists of has been a source of debate over the centuries. One of the key areas for this debate in the Western world was Greek philosophy. The basic approach of the Greek philosophers was to discuss and debate the questions they had about the world. There was no gathering of information to speak of, just talking. As a result, several ideas about matter were put forth, but never resolved. The first philosopher to carry out the gathering of data was Aristotle (384-322 B.C.). He recorded many observations on the weather, on plant and animal life and behavior, on physical motions, and a number of other topics. Aristotle could probably be considered the first "real" scientist, because he made systematic observations of nature and tried to understand what he was seeing. Inductive and Deductive Reasoning Two approaches to logical thinking developed over the centuries. These two methods are inductive reasoning and deductive reasoning . Inductive reasoning involves getting a collection of specific examples and drawing a general conclusion from them. Deductive reasoning takes a general principle and then draws a specific conclusion from the general concept. Both are used in the development of scientific ideas. Inductive reasoning first involves the collection of data: "If I add sodium metal to water, I observe a very violent reaction. Every time I repeat the process, I see the same thing happen." A general conclusion is drawn from these observations: the addition of sodium to water results in a violent reaction. In deductive reasoning, a specific prediction is made based on a general principle. One general principle is that acids turn blue litmus paper red. Using the deductive reasoning process, one might predict: "If I have a bottle of liquid labeled 'acid', I expect the litmus paper to turn red when I immerse it in the liquid." The Idea of the Experiment Inductive reasoning is at the heart of what is now called the " scientific method ." In European culture, this approach was developed mainly by Francis Bacon (1561-1626), a British scholar. He advocated the use of inductive reasoning in every area of life, not just science. The scientific method, as developed by Bacon and others, involves several steps: Ask a question - identify the problem to be considered. Make observations - gather data that pertains to the question. Propose an explanation (a hypothesis) for the observations. Make new observations to test the hypothesis further. Note that this should not be considered a "cookbook" for scientific research. Scientists do not sit down with their daily "to do" list and write down these steps. The steps may not necessarily be followed in order. But this does provide a general idea of how scientific research is usually done. When a hypothesis is confirmed repeatedly, it eventually becomes a theory—a general principle that is offered to explain natural phenomena. Note a key word— explain , or explanation . A theory offers a description of why something happens. A law, on the other hand, is a statement that is always true, but offers no explanation as to why. The law of gravity says a rock will fall when dropped, but does not explain why (gravitational theory is very complex and incomplete at present). The kinetic molecular theory of gases, on the other hand, states what happens when a gas is heated in a closed container (the pressure increases), but also explains why (the motions of the gas molecules are increased due to the change in temperature). Theories do not get "promoted" to laws, because laws do not answer the "why" question. Summary The early Greek philosophers spent their time talking about nature, but did little or no actual exploration or investigation. Inductive reasoning - to develop a general conclusion from a collection of observations. Deductive reasoning - to make a specific statement based on a general principle. Scientific method - a process of observation, developing a hypothesis, and testing that hypothesis. Review What was the basic shortcoming of the Greek philosophers approach to studying the material world? How did Aristotle improve the approach? Define “inductive reasoning” and give an example. Define “deductive reasoning” and give an example. What is the difference between a hypothesis and a theory? What is the difference between a theory and a law?
Courses/Cameron_University/CHEM1004%3A_Descriptive_Chemistry_(Cameron_University)/01%3A_Chemistry/1.02%3A_Science-_Reproducible_Testable_Tentative_Predictive_and_Explanatory	Learning Objectives Describe the differences between hypothesis and theory as scientific terms. Describe the difference between a theory and scientific law. Identify the components of the scientific method. Although many have taken science classes throughout their course of studies, incorrect or misleading ideas about some of the most important and basic principles in science are still commonplace. Most students have heard of hypotheses , theories , and laws , but what do these terms really mean? Before you read this section, consider what you have learned about these terms previously, and what they mean to you. When reading, notice if any of the text contradicts what you previously thought. What do you read that supports what you thought? What is a Fact? A fact is a basic statement established by experiment or observation. All facts are true under the specific conditions of the observation. What is a Hypothesis? One of the most common terms used in science classes is a " hypothesis ". The word can have many different definitions, dependent on the context in which it is being used: An educated guess: a scientific hypothesis provides a suggested solution based on evidence. Prediction: if you have ever carried out a science experiment, you probably made this type of hypothesis, in which you predicted the outcome of your experiment. Tentative or proposed explanation: hypotheses can be suggestions about why something is observed. In order for a hypothesis to be scientific, a scientist must be able to test the explanation to see if it works, and if it is able to correctly predict what will happen in a situation. For example, "if my hypothesis is correct, I should see _____ result when I perform _____ test." A hypothesis is tentative; it can be easily changed. What is a Theory? The United States National Academy of Sciences describes a theory as: "Some scientific explanations are so well established that no new evidence is likely to alter them. The explanation becomes a scientific theory. In everyday language a theory means a hunch or speculation. Not so in science. In science, the word theory refers to a comprehensive explanation of an important feature of nature supported by facts gathered over time. Theories also allow scientists to make predictions about as yet unobserved phenomena." "A scientific theory is a well-substantiated explanation of some aspect of the natural world, based on a body of facts that have been repeatedly confirmed through observation and experimentation. Such fact-supported theories are not "guesses," but reliable accounts of the real world. The theory of biological evolution is more than "just a theory." It is as factual an explanation of the universe as the atomic theory of matter (stating that everything is made of atoms) or the germ theory of disease (which states that many diseases are caused by germs). Our understanding of gravity is still a work in progress. But the phenomenon of gravity, like evolution, is an accepted fact." Note some key features of theories that are important to understand from this description: Theories are explanations of natural phenomenon. They aren't predictions (although we may use theories to make predictions). They are explanations of why something is observed. Theories aren't likely to change. They have a lot of support and are able to explain many observations satisfactorily. Theories can, indeed, be facts. Theories can change in some instances, but it is a long and difficult process. In order for a theory to change, there must be many observations or evidence that the theory cannot explain. Theories are not guesses. The phrase "just a theory" has no room in science. To be a scientific theory carries a lot of weight—it is not just one person's idea about something Theories aren't likely to change. What is a Law? Scientific laws are similar to scientific theories in that they are principles that can be used to predict the behavior of the natural world. Both scientific laws and scientific theories are typically well-supported by observations and/or experimental evidence. Usually, scientific laws refer to rules for how nature will behave under certain conditions, frequently written as an equation. Scientific theories are overarching explanations of how nature works, and why it exhibits certain characteristics. As a comparison, theories explain why we observe what we do, and laws describe what happens. For example, around the year 1800, Jacques Charles and other scientists were working with gases to, among other reasons, improve the design of the hot air balloon. These scientists found, after numerous tests, that certain patterns existed in their observations of gas behavior. If the temperature of the gas increased, the volume of the gas increased. This is known as a natural law. A law is a relationship that exists between variables in a group of data. Laws describe the patterns we see in large amounts of data, but do not describe why the patterns exist. Laws vs Theories A common misconception is that scientific theories are rudimentary ideas that will eventually graduate into scientific laws when enough data and evidence has been accumulated. A theory does not change into a scientific law with the accumulation of new or better evidence. Remember, theories are explanations; laws are patterns seen in large amounts of data, frequently written as an equation. A theory will always remain a theory, a law will always remain a law. Video $\PageIndex{1}$ What is the difference between scientific law and theory? The Scientific Method Scientists search for answers to questions and solutions to problems by using a procedure called the scientific method . This procedure consists of making observations, formulating hypotheses, and designing experiments, which in turn lead to additional observations, hypotheses, and experiments in repeated cycles (Figure $\PageIndex{1}$). Step 1: Make observations. Observations can be qualitative or quantitative. Qualitative observations describe properties or occurrences in ways that do not rely on numbers. Examples of qualitative observations include the following: "the outside air temperature is cooler during the winter season," "table salt is a crystalline solid," "sulfur crystals are yellow," and "dissolving a penny in dilute nitric acid forms a blue solution and a brown gas." Quantitative observations are measurements, which by definition consist of both a number and a unit. Examples of quantitative observations include the following: "the melting point of crystalline sulfur is 115.21° Celsius," and "35.9 grams of table salt—the chemical name of which is sodium chloride—dissolve in 100 grams of water at 20° Celsius." For the question of the dinosaurs’ extinction, the initial observation was quantitative: iridium concentrations in sediments dating to 66 million years ago were 20–160 times higher than normal. Step 2: Formulate a hypothesis. After deciding to learn more about an observation or a set of observations, scientists generally begin an investigation by forming a hypothesis, a tentative explanation for the observation(s). The hypothesis may not be correct, but it puts the scientist’s understanding of the system being studied into a form that can be tested. For example, the observation that we experience alternating periods of light and darkness which correspond to observed movements of the sun, moon, clouds, and shadows, is consistent with either of two hypotheses: Earth rotates on its axis every 24 hours, alternately exposing one side to the sun. The sun revolves around Earth every 24 hours. Suitable experiments can be designed to choose between these two alternatives. In the case of disappearance of the dinosaurs, the hypothesis was that the impact of a large extraterrestrial object caused their extinction. Unfortunately (or perhaps fortunately), this hypothesis does not lend itself to direct testing by any obvious experiment, but scientists can collect additional data that either supports or refutes it. Step 3: Design and perform experiments. After a hypothesis has been formed, scientists conduct experiments to test its validity. Experiments are systematic observations or measurements, preferably made under controlled conditions—that is, under conditions in which a single variable changes. Step 4: Accept or modify the hypothesis. A properly designed and executed experiment enables a scientist to determine whether the original hypothesis is valid. In the case of validity, the scientist can proceed to step 5. In other cases, experiments may demonstrate that the hypothesis is incorrect or that it must be modified, thus requiring further experimentation. Step 5: Development of a law and/or theory. More experimental data are then collected and analyzed, at which point a scientist may begin to think that the results are sufficiently reproducible (i.e., dependable) to merit being summarized in a law—a verbal or mathematical description of a phenomenon that allows for general predictions. A law simply states what happens; it does not address the question of why. One example of a law, the law of definite proportions (discovered by the French scientist Joseph Proust [1754–1826]), states that a chemical substance always contains the same proportions of elements by mass. Thus, sodium chloride (table salt) always contains the same proportion by mass of sodium to chlorine—in this case, 39.34% sodium and 60.66% chlorine by mass. Sucrose (table sugar) is always 42.11% carbon, 6.48% hydrogen, and 51.41% oxygen by mass. Whereas a law states only what happens, a theory attempts to explain why nature behaves as it does. Laws are unlikely to change greatly over time, unless a major experimental error is discovered. A theory, in contrast, is incomplete and imperfect; it evolves with time to explain new facts as they are discovered. Because scientists can enter the cycle shown in Figure $\PageIndex{1}$ at any point, the actual application of the scientific method to different topics can take many different forms. For example, a scientist may start with a hypothesis formed by reading about work done by others in the field, rather than by making direct observations. Example $\PageIndex{1}$ Classify each statement as a law, theory, experiment, hypothesis, or observation. Ice always floats on liquid water. Birds evolved from dinosaurs. Hot air is less dense than cold air, probably because the components of hot air are moving more rapidly. When 10 g of ice was added to 100 mL of water at 25°C, the temperature of the water decreased to 15.5°C after the ice melted. The ingredients of Ivory soap were analyzed to see whether it really is 99.44% pure, as advertised. Solution This is a general statement of a relationship between the properties of liquid and solid water, so it is a law. This is a possible explanation for the origin of birds, so it is a hypothesis. This is a statement that tries to explain the relationship between the temperature and the density of air based on fundamental principles, so it is a theory. The temperature is measured before and after a change is made in a system, so these are observations. This is an analysis designed to test a hypothesis (in this case, the manufacturer’s claim of purity), so it is an experiment. Exercise $\PageIndex{1}$ Classify each statement as a law, theory, experiment, hypothesis, qualitative observation, or quantitative observation. Measured amounts of acid were added to a Rolaids tablet to see whether it really “consumes 47 times its weight in excess stomach acid.” Heat always flows from hot objects to cooler ones, not in the opposite direction. The universe was formed by a massive explosion that propelled matter into a vacuum. Michael Jordan is the greatest pure shooter ever to play professional basketball. Limestone is relatively insoluble in water, but dissolves readily in dilute acid with the evolution of a gas. Answer 1: experiment Answer 2: law Answer 3: theory Answer 4: hypothesis Answer 5: observation Summary A hypothesis is a tentative explanation that can be tested by further investigation. A theory is a well-supported explanation of observations. A scientific law is a statement that summarizes the relationship between variables. An experiment is a controlled method of testing a hypothesis. The scientific method is a method of investigation involving experimentation and observation to acquire new knowledge, solve problems, and answer questions. The key steps in the scientific method are: Step 1: Make observations. Step 2: Formulate a hypothesis. Step 3: Test the hypothesis through experimentation. Step 4: Accept or modify the hypothesis. Step 5: Development of a law and/or theory.
Courses/SUNY_Oneonta/Chem_221%3A_Organic_Chemistry_I_(Bennett)/1%3ALecture_Textbook/03%3A_Conformations_and_Stereochemistry/3.14%3A_Problems_for_Chapter_3	Solutions to selected problems P3.1: Draw an energy vs dihedral angle graph for rotations about the C 2 -C 3 bond of 2-methylbutane. Start with the highest-energy conformation as the 0 o point. For each energy peak and valley, draw a corresponding Newman projection. P3.2 : a) Which has the highest energy diaxial chair conformation: trans -1,2-dimethylcyclohexane, cis -1,3-dimethylcyclohexane, or trans -1,4-dimethylcyclohexane? Explain. b) Which of the following are trans disubstituted cyclohexanes? c) Draw A-F above in two dimensions (rings in the plane of the page, substituents drawn as solid or dashed wedges). d) Structure D does not have any chiral centers. Explain. e) Draw a diastereomer of structure D (in two dimensions, as in part c). f) Are structure D and its diastereomer chiral? g) Assign R/S designations to the two chiral centers in structure B (hint: making a model will be very helpful!) P 3.3 : The following are structures, drawn in two dimensions, of drugs listed on the products web page of Merck Pharmaceutical. One of the compounds is achiral. a) Circle all chiral centers. (Hint: Don't panic ! Remember - you are looking for sp 3 -hybridized carbons with four different substituents.) b) How many diastereomers are possible for desogestrel? c) Draw two epimeric forms of simvastatin P3.4: Three of the four structures below are chiral. Assign R/S designations to all chiral centers, and identify the achiral molecule. P3.5: Draw the R,R stereoisomers of the structures below. P3.6 : Below are the structures of sucralose, the artificial sweetener with the brand name Splenda ( TM ), and the cancer drug Paclitaxel. Give an R or S designation to chiral centers indicated with an arrow. P3.7 : The four drugs below were featured in a Chemical & Engineering News article (April 16, 2007, p. 42) on new drugs that had been developed in university labs. a) Identify each as chiral or achiral, and identify all stereocenters. Also, state how many possible stereoisomers exist for each structure. b) Two fluorinated Epivar derivatives (structures A and B below) were also mentioned in this article. What is the relationship between structures A and B? (Your choices: not isomers, constitutional isomers, diastereomers but not epimers, epimers, enantiomers, identical) P3.8 : Redraw the following structures in the flat ring, solid/dash wedge convention (the drawings have been started for you). P3.9 : Below is an experimental drug for Alzheimer's disease that was mentioned in the March 13, 2007 issue of Chemical and Engineering News . a) Label all stereocenters as R or S. b) Draw the enantiomer of the molecule shown. P3.10 : The molecules below are potential new drugs for the treatment of Duchenne muscular dystrophy (molecule A) and skin cancer (molecule B) ( Chemical and Engineering News Sept 26, 2005, p. 39). Given the R/S designations, redraw the structure showing the correct stereochemistry. P3.11 : Draw the structure of the following molecules: a) ( R )-3-methyl-3-hexanol b) ( R )-1-chloro-1-phenylethane c) (2 R , 3 R )-2,3-dihydroxybutanedioic acid (tartaric acid) d) ( S )-( E )-4-chloro-3-ethyl-2-pentenoic acid e) (1 S , 3 R )-1-chloro-3-ethylcyclohexane P3.12 : Coelichelin (the structure below to the left) is a natural product from soil bacteria that was identified using a technique known as 'genome mining' ( Chemical and Engineering News Sept. 19, 2005, p. 11). What is the relationship between coelichelin and the compound shown below and to the right? P3.13 : Identify the relationships between the following pairs of structures (Not isomers, constitutional isomers, diastereomers but not epimers, epimers, enantiomers, identical) P3.14 : Identify the relationships between each of the following pairs of pentose sugars (not isomers, constitutional isomers, diastereomers but not epimers, epimers, enantiomers, identical). P3.15 : Identify the relationships between each of the following pairs of hexose sugars (not isomers, constitutional isomers, diastereomers but not epimers, epimers, enantiomers, identical). P3.16 : The compound drawn below (not showing stereochemistry) has been identified as a potential anti-inflammatory agent by scientists at Schering-Plough a pharmaceutical company (see Chemical and Engineering News Nov. 28, 2005 p. 29). How many stereoisomers are possible for the compound? P3.17 : Secramine is a synthetic compound that has been shown to interfere with the transport of newly synthesized proteins in the cell (see Chemical and Engineering News Nov. 28, 2005, p. 27). Also drawn below is a (hypothetical) isomer of secramine. a) Identify the relationship between the two isomers: are they consitutional isomers, confomational isomers, enantiomers, or diastereomers? b) Locate a five-membered ring in the secramine structure. P3.18 : The natural product bistramide A has been shown to bind to actin, an important structural protein in the cell, and supress cell proliferation (see Chemical and Engineering News Nov. 21, 2005, p. 10). a) Label the alkene functional groups as E , Z , or N (no E/Z designation possible) b) Theoretically, how many stereoisomers are possible for bistramide A? P3.19 : a) Draw Newman projections of the gauche and the anti conformations of 1,2-ethanediol. b) Why might the gauche conformation be expected to be the more stable of the two? c) Do you think that gauche is also the most stable conformation of 1,2-dimethoxyethane? Explain. P3.20 : Draw the chair conformation of cis -1,2-dimethylcyclohexane. a) Label the stereochemical configuration at C 1 and C 2 for the structure you drew. b) Build a model of your molecule, and try out different possible boat conformations. Can you find one in which there is a plane of symmetry? c) Is cis -1,2-dimethylcyclohexane a chiral molecule? d) is cis -1,4-dimethylcyclohexane chiral? How about trans -1,4-dimethylcyclohexane? How about trans -1-chloro-4-fluorocyclohexane? P3.21 : In some special cases, a 'chiral center' can be composed of several atoms instead of just one, and molecules which contain such multi-atom chiral centers are indeed chiral. What is the relationship between the two two difluorallene compounds below? It will be very helpful to make models, and review the fundamental definitions in this chapter. Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
Courses/Victor_Valley_College/CHEM100_Victor_Valley_College/05%3A_Compounds_and_Bonding/5.01%3A_Ionic_Compounds/5.1.10%3A_Some_Properties_of_Ionic_Compounds	Learning Objectives Describe the basic physical properties of ionic compounds. The figure below shows just a few examples of the color and brilliance of naturally occurring ionic crystals. The regular and orderly arrangement of ions in the crystal lattice is responsible for the various shapes of these crystals, while transition metal ions give rise to the colors. Melting Points Because of the many simultaneous attractions between cations and anions that occur, ionic crystal lattices are very strong. The process of melting an ionic compound requires the addition of large amounts of energy in order to break all of the ionic bonds in the crystal. For example, sodium chloride has a melting temperature of about 800 o C. As a comparison, the molecular compound water melts at 0 °C. Shattering Ionic compounds are generally hard, but brittle. Why? It takes a large amount of mechanical force, such as striking a crystal with a hammer, to force one layer of ions to shift relative to its neighbor. However, when that happens, it brings ions of the same charge next to each other (see below). The repulsive forces between like-charged ions cause the crystal to shatter. When an ionic crystal breaks, it tends to do so along smooth planes because of the regular arrangement of the ions. Conductivity Another characteristic property of ionic compounds is their electrical conductivity . The figure below shows three experiments in which two electrodes that are connected to a light bulb are placed in beakers containing three different substances. In the first beaker, distilled water does not conduct a current because water is a molecular compound. In the second beaker, solid sodium chloride also does not conduct a current. Despite being ionic and thus composed of charged particles, the solid crystal lattice does not allow the ions to move between the electrodes. Mobile charged particles are required for the circuit to be complete and the light bulb to light up. In the third beaker, the NaCl has been dissolved into the distilled water. Now the crystal lattice has been broken apart and the individual positive and negative ions can move. Example $\PageIndex{1}$ Write the dissociation equation of solid NaCl in water. Solution NaCl( s ) → Na + ( aq ) + Cl – ( aq ) Exercise $\PageIndex{1}$ Write the dissociation equation of solid NH 4 NO 3 in water. Answer NH 4 NO 3 ( s ) → NH 4 + ( aq ) + NO 3 – ( aq ) Key Takeaways Ionic compounds have high melting points. Ionic compounds are hard and brittle. Ionic compounds dissociate into ions when dissolved in water. Solutions of ionic compounds and melted ionic compounds conduct electricity, but solid materials do not. An ionic compound can be identified by its chemical formula: metal + nonmetal or polyatomic ions.
Courses/Indiana_Tech/EWC%3A_CHEM_1000_-_Introductory_Chemistry_(Budhi)/01%3A_The_Chemical_World/1.3%3A_Hypothesis%2C_Theories%2C_and_Laws	Learning Objectives Describe the difference between hypothesis and theory as scientific terms. Describe the difference between a theory and scientific law. Although many have taken science classes throughout the course of their studies, people often have incorrect or misleading ideas about some of the most important and basic principles in science. Most students have heard of hypotheses, theories, and laws, but what do these terms really mean? Prior to reading this section, consider what you have learned about these terms before. What do these terms mean to you? What do you read that contradicts or supports what you thought? What is a Fact? A fact is a basic statement established by experiment or observation. All facts are true under the specific conditions of the observation. What is a Hypothesis? One of the most common terms used in science classes is a "hypothesis". The word can have many different definitions, depending on the context in which it is being used: An educated guess: a scientific hypothesis provides a suggested solution based on evidence. Prediction: if you have ever carried out a science experiment, you probably made this type of hypothesis when you predicted the outcome of your experiment. Tentative or proposed explanation: hypotheses can be suggestions about why something is observed. In order for it to be scientific, however, a scientist must be able to test the explanation to see if it works and if it is able to correctly predict what will happen in a situation. For example, "if my hypothesis is correct, we should see ___ result when we perform ___ test." A hypothesis is very tentative; it can be easily changed. What is a Theory? The United States National Academy of Sciences describes what a theory is as follows: "Some scientific explanations are so well established that no new evidence is likely to alter them. The explanation becomes a scientific theory. In everyday language a theory means a hunch or speculation. Not so in science. In science, the word theory refers to a comprehensive explanation of an important feature of nature supported by facts gathered over time. Theories also allow scientists to make predictions about as yet unobserved phenomena." "A scientific theory is a well-substantiated explanation of some aspect of the natural world, based on a body of facts that have been repeatedly confirmed through observation and experimentation. Such fact-supported theories are not "guesses" but reliable accounts of the real world. The theory of biological evolution is more than "just a theory." It is as factual an explanation of the universe as the atomic theory of matter (stating that everything is made of atoms) or the germ theory of disease (which states that many diseases are caused by germs). Our understanding of gravity is still a work in progress. But the phenomenon of gravity, like evolution, is an accepted fact. Note some key features of theories that are important to understand from this description: Theories are explanations of natural phenomena. They aren't predictions (although we may use theories to make predictions). They are explanations as to why we observe something. Theories aren't likely to change. They have a large amount of support and are able to satisfactorily explain numerous observations. Theories can, indeed, be facts. Theories can change, but it is a long and difficult process. In order for a theory to change, there must be many observations or pieces of evidence that the theory cannot explain. Theories are not guesses. The phrase "just a theory" has no room in science. To be a scientific theory carries a lot of weight; it is not just one person's idea about something Theories aren't likely to change. What is a Law? Scientific laws are similar to scientific theories in that they are principles that can be used to predict the behavior of the natural world. Both scientific laws and scientific theories are typically well-supported by observations and/or experimental evidence. Usually scientific laws refer to rules for how nature will behave under certain conditions, frequently written as an equation. Scientific theories are more overarching explanations of how nature works and why it exhibits certain characteristics. As a comparison, theories explain why we observe what we do and laws describe what happens. For example, around the year 1800, Jacques Charles and other scientists were working with gases to, among other reasons, improve the design of the hot air balloon. These scientists found, after many, many tests, that certain patterns existed in the observations on gas behavior. If the temperature of the gas is increased, the volume of the gas increased. This is known as a natural law. A law is a relationship that exists between variables in a group of data. Laws describe the patterns we see in large amounts of data, but do not describe why the patterns exist. What is a Belief? A belief is a statement that is not scientifically provable. Beliefs may or may not be incorrect; they just are outside the realm of science to explore. Laws vs. Theories A common misconception is that scientific theories are rudimentary ideas that will eventually graduate into scientific laws when enough data and evidence has accumulated. A theory does not change into a scientific law with the accumulation of new or better evidence. Remember, theories are explanations and laws are patterns we see in large amounts of data, frequently written as an equation. A theory will always remain a theory; a law will always remain a law. Video $\PageIndex{1}$: What’s the difference between a scientific law and theory? Summary A hypothesis is a tentative explanation that can be tested by further investigation. A theory is a well-supported explanation of observations. A scientific law is a statement that summarizes the relationship between variables. An experiment is a controlled method of testing a hypothesis. Contributions & Attributions Marisa Alviar-Agnew ( Sacramento City College ) Henry Agnew (UC Davis)
Courses/Arkansas_Northeastern_College/CH14133%3A_Chemistry_for_General_Education/zz%3A_Back_Matter/20%3A_Glossary	Words (or words that have the same definition) The definition is case sensitive (Optional) Image to display with the definition [Not displayed in Glossary, only in pop-up on pages] (Optional) Caption for Image (Optional) External or Internal Link (Optional) Source for Definition (Eg. "Genetic, Hereditary, DNA ...") (Eg. "Relating to genes or heredity") NaN The infamous double helix https://bio.libretexts.org/ CC-BY-SA; Delmar Larsen Word(s) Definition Image Caption Link Source Sample Word 1 Sample Definition 1 NaN NaN NaN NaN
Courses/Duke_University/CHEM_210D%3A_Modern_Applications_of_Chemistry/3%3A_Textbook-_Modern_Applications_of_Chemistry/01%3A_Primer/1.03%3A_Further_Aspects_of_Covalent_Bonding/1.3.03%3A_The_Shapes_of_Molecules/1.3.3.01%3A_Lecture_Demonstrations	Balloon Model for Electrostatic Repulsion Spherical balloons are inflated and tied off, and their inlets are twisted together so that four balloons are attached. They assume tetrahedral geometry by mutual repulsion, modeling VSERPR approach to repulsion of four charge centers. If one balloon is broken, the remaining three assume a planar triangular geometry, and if another balloon is broken, the remaining two assume a linear geometry. "Nitrogen Triiodide" Prepare nitrogen triiodide, See Molecules with Lone Pairs Lecture Demonstrations. The instability of NI 3 (or better, NI 3 *NH 3 ) is due to the same repulsions that are the basis for the VSEPR approach (3 large I atoms and a lone pair bonded to the relatively small N atom) Contributors Ed Vitz (Kutztown University), John W. Moore (UW-Madison), Justin Shorb (Hope College), Xavier Prat-Resina (University of Minnesota Rochester), Tim Wendorff, and Adam Hahn.
Courses/DePaul_University/Thermodynamics_and_Introduction_to_Quantum_Mechanics_(Southern)/02%3A_The_First_Law_of_Thermodynamics/2.05%3A_Exact_and_Inexact_Differentials	Many important thermochemical quantities can be expressed in terms of partial derivatives. Two important examples are the molar heat capacities $C_p$ and $C_V$ which can be expressed as \[ C_p = \left(\dfrac{\partial H}{\partial T}\right)_p \nonumber \] and \[ C_V = \left(\dfrac{\partial U}{\partial T}\right)_V \nonumber \] These are properties that can be measured experimentally and tabulated for many substances. These quantities can be used to calculate changes in quantities since they represent the slope of a surface ($H$ or $U$) in the direction of the specified path (constant $p$ or $V$). This allows us to use the following kinds of relationships: \[ dH = \left(\dfrac{\partial H}{\partial T}\right)_p dT \nonumber \] and \[ \Delta H = \int \left(\dfrac{\partial H}{\partial T}\right)_p dT \nonumber \] Because thermodynamics is kind enough to deal in a number of state variables , the functions that define how those variable change must behave according to some very well determined mathematics. This is the true power of thermodynamics!
Courses/UW-Whitewater/UWX_CH114%3A_Chemistry_in_the_Kitchen/01%3A_Matter_-_Atoms_Elements_and_Compounds/1.01%3A_The_Elements/1.1.02%3A_Measurements	Learning Objectives Express quantities properly using a number and a unit. A coffee maker’s instructions tell you to fill the coffeepot with 4 cups of water and use 3 scoops of coffee. When you follow these instructions, you are measuring. When you visit a doctor’s office, a nurse checks your temperature, height, weight, and perhaps blood pressure (Figure $\PageIndex{1}$); the nurse is also measuring. Chemists measure the properties of matter using a variety of devices or measuring tools, many of which are similar to those used in everyday life. Rulers are used to measure length, balances (scales) are used to measure mass (weight), and graduated cylinders or pipettes are used to measure volume. Measurements made using these devices are expressed as quantities. A quantity is an amount of something and consists of a number and a unit . The number tells us how many (or how much), and the unit tells us what the scale of measurement is. For example, when a distance is reported as “5.2 kilometers,” we know that the quantity has been expressed in units of kilometers and that the number of kilometers is 5.2. \[\color{red} \underbrace{5.2}_{\text{number}} \color{blue} \underbrace{\text{kilometers}}_{\text{unit}} \nonumber \] If you ask a friend how far he or she walks from home to school, and the friend answers “12” without specifying a unit, you do not know whether your friend walks—for example, 12 miles, 12 kilometers, 12 furlongs, or 12 yards. Without units, a number can be meaningless, confusing, or possibly life threatening. Suppose a doctor prescribes phenobarbital to control a patient’s seizures and states a dosage of “100” without specifying units. Not only will this be confusing to the medical professional giving the dose, but the consequences can be dire: 100 mg given three times per day can be effective as an anticonvulsant, but a single dose of 100 g is more than 10 times the lethal amount. Both a number and a unit must be included to express a quantity properly. To understand chemistry, we need a clear understanding of the units chemists work with and the rules they follow for expressing numbers. The next two sections examine the rules for expressing numbers. Exercise $\PageIndex{1}$ Identify the number and the unit in each quantity. one dozen eggs 2.54 centimeters a box of pencils 88 meters per second Answer a The number is one, and the unit is dozen. Answer b The number is 2.54, and the unit is centimeter. Answer c The number 1 is implied because the quantity is only a box. The unit is box of pencils. Answer d The number is 88, and the unit is meters per second. Note that in this case the unit is actually a combination of two units: meters and seconds. Exercise $\PageIndex{2}$ Identify the number and the unit in each quantity. 99 bottles of soda 60 miles per hour 32 fluid ounces 98.6 degrees Fahrenheit Answer a The number is 99, and the unit is bottles of soda. Answer b The number is 60, and the unit is miles per hour. Answer c The number 32, and the unit is fluid ounces Answer d The number is 98.6, and the unit is degrees Fahrenheit Exercise $\PageIndex{2}$ What are the two necessary parts of a quantity? Answer The two necessary parts are the number and the unit. Key Takeaway Identify a quantity properly with a number and a unit.
Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/18%3A_Reactions_of_Aromatic_Compounds/18.14%3A__Solutions_to_Additional_Exercises	18-1 18-2 18-3 This is just one possible way to synthesize it. 18-4 The bromine should be in the meta position. Right now it is in the ortho position, from perhaps having the ethyl group present first and then the having it substituted there. BUT the ethyl group is last to form, and the aldehyde and nitro groups would both encourage a meta substitution. 18-5 Answer: A 18-6 18-7 18-8 18-9 1-chloro-2-methylbenzene and 1-chloro-4-methylbenzene 18-10 18-11 Halogenation, Nitration, and Sulfonation of Benzene 18-12: 18-13: 18-14: Answer: A Activating, Ortho-, Para-Directing Substituents 18-15: 18-16: 18-17: Deactivating, Meta-Directing Substituents 18-18: 18-19: 18-20: Answer: B Halogen Substitutes: Deactivating, but Ortho, Para-Directing 18-21: Answer: B 18-22: 18-23: Effects of Multiple Substituents on Electrophilic Aromatic Substitution 18-24: 18-25: 18-26: Friedel-Crafts Alkylation/Acylation 18-27: 18-28: Friedel-Crafts alkylation using 1-chloropropane is not the best way to synthesize propylbenzene. You will end up with (propan-2-yl)benzene as your main product due to a hydride shift occurring during an intermediate step. A better route of synthesis may be Friedel-Crafts acylation using propanoyl chloride to make 1-phenylpropan-1-one, followed by a Clemmensen reduction to obtain the final product. 18-29: Answer: D Nucleophilic Aromatic Substitution 18-30: 18-31: 18-32: Possible route of synthesis: Aromatic Substitutions Using Organometallic Reagents 18-33: 18-34: Possible route of synthesis: 18-35: Answer: D Side-Chain Reactions of Benzene Derivatives 18-36: 18-37: Answer: A 18-38:
Courses/Lumen_Learning/Book%3A_US_History_II_(OS_Collection)_(Lumen)/08%3A_Leading_the_Way%3A_The_Progressive_Movement%2C_1890-1920/08.8%3A_Video%3A_Progressive_Presidents	This video teaches you about the Progressive Presidents, who are not a super-group of former presidents who create complicated, symphonic, rock soundscapes that transport you into a fantasy fugue state. Although that would be awesome. The presidents most associated with the Progressive Era are Theodore Roosevelt, William Taft, and Woodrow Wilson. During the times these guys held office, trusts were busted, national parks were founded, social programs were enacted, and tariffs were lowered. It wasn’t all positive though, as their collective tenure also saw Latin America invaded A LOT, a split in the Republican party that resulted in a Bull Moose, all kinds of other international intervention, and the end of the Progressive Era saw the United States involved in World War. If all this isn’t enough to entice, two people get shot in this video. Violence sells, they say. A YouTube element has been excluded from this version of the text. You can view it online here: http://pb.libretexts.org/2os/?p=128 All rights reserved content Progressive Presidents: Crash Course US History #29. Provided by : Crash Course. Located at : https://youtu.be/F7flSW1PGsA?t=1s . License : All Rights Reserved . License Terms : Standard YouTube License
Courses/University_of_Arkansas_Little_Rock/Chem_1403%3A_General_Chemistry_2/Homework/16%3A_Acids_and_Bases	Downloadable files Please print and work out the question before looking at the answers Chapter 16: Acids and Bases pdf 16.1: Brønsted-Lowry Concept of Acids and Bases Textbook: Section 16.1 Exercise $\PageIndex{1.a}$ In the reaction, identify the acid and base. $NH_{3}(g)+H_{2}O(l)\rightleftharpoons NH_{4}^{+}(aq)+OH^{-}(aq)$ Answer In this reaction H 2 O is donating a proton so it is the acid $NH_{3}$ is accepting the proton so it is the base Exercise $\PageIndex{1.b}$ What is the conjugate acid of $NH_{3}$? Answer A conjugate acid is a compound formed when an acid donates a proton to a base. So basically it is a base with a hydrogen ion added to it $NH_{3}(g)+H^{+}\rightleftharpoons NH_{4}^{+}(aq)\nonumber $ So the conjugate acid of $NH_{3}$ is $NH_{4}^{+}$ Exercise $\PageIndex{1.c}$ What is the conjugate base of H 2 O? Answer A conjugate base is what is left over after an acid has donated a proton. $ H_{2}O \rightleftharpoons H^{+} + OH^{-} \nonumber $ So the Conjugate base of H 2 O is $OH^{-}$ Exercise $\PageIndex{1.d}$ What is the conjugate acid of NaHSO 3 ? Answer NaHSO 3 is an amphoteric substance, which means it can be an acid or a base $HSO_{3}^{-}+H^{+}\rightarrow H_{2}SO_{3} $ So the conjugate acid of $NaHSO_{3}$ is $H_{2}SO_{3}$ Exercise $\PageIndex{1.e}$ What is the conjugate base of NaHSO 3 ? Answer NaHSO 3 is an amphoteric substance, which means it can be an acid or a base $HSO_{3}^{-}+OH^{-}\rightarrow SO_{3}^{-2}+H_{2}O\nonumber$ So the conjugate base of $NaHSO_{3}$ is $SO_{3}^{-2}$ Exercise $\PageIndex{1.f}$ It is known that the hydride ion H - is a stronger base than OH - , what is(are) the product(s) of the reaction: H - (aq) +H 2 O(l) Answer $H^{-}(aq)+H_{2}O(l)\rightarrow H_{2}(g)+OH^{-}(aq)\nonumber$ base acid C.A. C.B. Since H - is a stronger base than OH - , the reaction is reasonable to take place. Exercise $\PageIndex{1.h}$ Which of the following is a Brønsted-Lowry acid? (CH 3 ) 3 NH + CH 3 COOH HF HNO 2 all of these Answer e. all of these Exercise $\PageIndex{1.i}$ What is the conjugate acid of NH 3 ? Answer NH 4 + Exercise $\PageIndex{1.j}$ What is the conjugate base of OH - ? Answer O 2- Exercise $\PageIndex{1.k}$ What is the conjugate base of HSO 4 - ? Conjugate acid? Answer The conjugate base is SO 4 2 - The conjugate acid is H 2 SO 4 16.2: Water and the pH Scale Textbook: Section 16.2 Exercise $\PageIndex{2.a}$ In a sample of lemon juice, [H+] is 6.2x10 -4 M. What is the pH? Answer \[pH=-log[H^{+}]=-log[6.210^{-4}M]=3.2\nonumber \] Exercise $\PageIndex{2.b}$ A sample of detergent has a pH of 8.20. What is the [H + ]? What is the [OH - ]? Answer a. \[[H^{+}]=10^{-pH}=10^{-8.20}=6.310^-9M\nonumber \] Answer b. \[[OH^{-}]=\frac{K_{w}}{[H^{+}]}=\frac{10^{-14}}{6.310^{-9}}=1.610^{-6}\nonumber \] Exercise $\PageIndex{2.c}$ What is the pH of an aqueous solution at 25°C in which [H + ] is 0.0025 M? Answer \[pH=-log\left [ H^{+} \right ]=-log\left ( 0.0025M \right )=2.60\nonumber\] Exercise $\PageIndex{2.d}$ What is the pH of a solution that contains 3.98 * 10 -9 M hydronium ion at 25°C? Answer \[pH=-log\left [ H^{+} \right ]=-log\left ( 3.9810^{-9}M \right )=8.400\nonumber\] Exercise $\PageIndex{2.e}$ A 0.0035M aqueous solution of a compound has a pH=2.46. The compound is a weak base a weak acid a strong acid a strong base a salt Answer c. a strong acid Exercise $\PageIndex{2.f}$ Calculate the pOH of a solution at 25°C that contains 1.94 10 - 10 M hydronium ions. Answer \[pH=-log\left [ H^{+} \right ]=-log\left ( 1.9410^{-10}M \right )=9.712 \nonumber\] \[pOH=14-pH=14-9.712=4.288\nonumber\] Exercise $\PageIndex{2.g}$ Which solution below has the highest concentration of hydroxide ions? pH = 3.21 pH = 12.59 pH = 7.93 pH = 9.82 pH = 7.00 Answer pH = 12.59 $[OH^{-}]=10^{-12.59}=2.57010^{-13}M \nonumber$ Exercise $\PageIndex{2.h}$ What is the [H + ] of an aqueous solution whose pH is 8.11? Answer $[H^{+}]=10^{-8.11}=7.7610^{-9}M \nonumber$ Exercise $\PageIndex{2.i}$ What is the concentration of H + in a solution at 25°C with a pH of 7.35? Answer $[H^{+}]=10^{-7.35}=4.4710^{-8}M \nonumber$ Exercise $\PageIndex{2.j}$ The magnitude of Kw indicated that water autoionizes very slowly water autoionizes very quickly water autoionizes only to a very small extent the autoionization of water is exothermic the autoionization of water is endothermic Answer c. water autoionizes only to a very small extent Exercise $\PageIndex{2.k}$ What is the concentration of water in pure water? 18 M 100 M 55 M 0.100 M 83 M Answer c. 55 M Exercise $\PageIndex{2.l}$ What is the pH of 10 -9 M HCl (a very dilute strong acid)? Answer pH = 7, because an acid can not have a pH above 7. It is so dilute that the pH is is a consequence of the auto dissociation of the water. Exercise $\PageIndex{2.m}$ What is the pH of 12.0M HCl (super strong acid)? Answer pH = -1.079 Exercise $\PageIndex{2.n}$ What is the pH of 10 -9 M NaOH (very dilute strong base)? Answer pH = 7, because an basevcan not have a pH below 7. It is so dilute that the pH is is a consequence of the auto dissociation of the water. Exercise $\PageIndex{2.o}$ What is the pH of 6.0M NaOH (very concentrated strong base)? Answer pH = 14.78 Exercise $\PageIndex{2.p}$ What is the pH of 10 -6 M HCl (dilute strong acid)? Answer \[pH=-log(10^{-6})=6\nonumber\] Exercise $\PageIndex{2.q}$ What is the pH of 6.0M CH 3 COOH (concentrated acetic acid)? Answer \[pH=-log\sqrt{K_a[HA]_i} = -log(\sqrt{(1.810^{-5})6})=1.98\nonumber\] 16.3: Equilibrium Constants for Acids and Bases Textbook: Section 16.3 Exercise $\PageIndex{3.a}$ The following acids are listed in order of increasing strength. List their conjugate bases in order of increasing strength. HCN, CH 3 COOH, HF, HClO 4 : Answer ClO 4 - < F - < CH 3 COO - < CN - Exercise $\PageIndex{3.b}$ Which one of the following is the weakest acid HF (K a = 6.8 * 10 -4 ) HClO (K a = 3.0 * 10 -8 ) HNO2 (K a = 4.5 10 -4 ) HCN (K a = 4.9 10 -10 ) Acetic Acid (K a = 1.8 10 -5 ) Answer d. HCN (K a = 4.9 10 -10 ) Exercise $\PageIndex{3.c}$ Using the table below, which is the strongest acid? 0 1 Acid Ka HOAc 1.8 * 10-5 HCHO2 1.8 * 10-4 HClO 3.0 * 10-8 HF 6.8 * 10-4 Answer HF with a K a =6.8 * 10 -4 Exercise $\PageIndex{3.d}$ Which species form the following list would be the strongest Bronsted-Lowry base? Cl - Br - NO 3 - F - ClO 4 - Answer d. F - Exercise $\PageIndex{3.e}$ Which of the following ions will act as a weak base in water? OH - Cl - NO 3 - ClO - None of these will act as a weak base in water Answer d. ClO - 16.4: Acid-Base Properties of Salts Textbook: Section 16.4 Exercise $\PageIndex{4.a}$ Given that the Ka for gallic acid, (HC 7 H 5 O 5 ) is 4.5710 -3 , what is the Kb for the gallate ion (C 7 H 5 O 5 - )? T = 25°C Answer \[K_{b}=\frac{K_{w}}{K_{a}}=\frac{10^{-14}}{4.5710^{-3}}=2.1910^{-12} \nonumber \] Exercise $\PageIndex{4.b}$ Kb for C 5 H 5 N is 1.410 -9 . Ka for C 5 H 5 NH + is _____. T = 25°C Answer \[K_{a}=\frac{K_{w}}{K_{b}}=\frac{10^{-14}}{1.410^{-9}}=7.110^{-6} \nonumber \] Exercise $\PageIndex{4.c}$ Ka for HF is 7.010 -4 . Kb for the fluoride ion is _____. Answer \[K_{b}=\frac{K_{w}}{K_{a}}=\frac{10^{-14}}{7.010^{-4}}=1.410^{-11} \nonumber \] 16.5: Acid, Base & Salt Equilibrium Calculations Textbook: Section 16.5 Strong Acids Exercise $\PageIndex{5.1.a}$ What molar concentration of aqueous hydrochloric acid would have a pH = 9.50? Answer It is not possible for a solution of hydrochloric acid to have a pH = 9.50 Exercise $\PageIndex{5.1.b}$ What is the [H + ] and pH of a 0.0037 M HBr solution at 25°C? Answer \[\left [ HBr \right ]=\left [ H^{+} \right ]=0.0037M\nonumber\] \[pH=-log\left [ H^{+} \right ]=-log\left ( 0.0037M \right )=2.43\nonumber\] Exercise $\PageIndex{5.1.c}$ Which of the following acids is not a strong acid? H 2 CO 3 H 2 SO 4 HNO 3 HClO 4 HCl Answer a. H 2 CO 3 Exercise $\PageIndex{5.1.d}$ What is the pH of 0.055M of HCl? Answer \[H^{+}=0.055M\nonumber\] \[pH=-log[H^{+}]=-log[0.055M]=1.26\nonumber \] Exercise $\PageIndex{5.1.e}$ What is the pH of an aqueous solution in which the molar concentration of HCl is 1.3 10 -11 ? Answer You would think \[pH = -log\left [ H^{+} \right ]=-log\left ( 1.310^{-11} \right )=10.89\nonumber\] But it is 7, and the pH is dictated by the water Exercise $\PageIndex{5.1.f}$ The pH of a 0.030 M HCl solution at 25°C is _____. Answer \[pH=-log\left [ H^{+} \right ]=-log\left ( 0.030M \right )=1.52\nonumber\] Strong Bases Exercise $\PageIndex{5.2.a}$ What molar concentration of aqueous barium hydroxide would have pH = 12.25? Answer \[pOH=14-pH=14-12.25=1.75\nonumber\] \[\left [ OH^{-} \right ]=10^{-1.75}=0.01778M\nonumber\] \[\left [ Ba\left ( OH \right )_{2} \right ]=\frac{\left [ OH^{-} \right ]}{2}=\frac{0.01778M}{2}=0.008891M\nonumber\] Exercise $\PageIndex{5.2.b}$ Consider the strong diprotic base Ba(OH) 2 . What is the pH of 0.001M Ba(OH) 2 ? What is the p0H of 0.001M Ba(OH) 2 ? Answer a. \[[OH^{-}]=0.0012=0.002M\nonumber \] \[[H^{+}]=\frac{K_{w}}{[OH^{-}]}=\frac{10^{-14}}{2.010^{-3}}=5.010^{-12}M\nonumber \] \[pH=-log[H^{+}]=-log[5.010^{-12}M]=11.30\nonumber \] Answer b. \[[OH^{-}]=0.0012=0.002M\nonumber \] \[pOH=-log[OH^{-}]=-log[2.010^{-3}M]=2.70\nonumber \] Exercise $\PageIndex{5.2b}$ If 0.56g of CaO is dissolved in the water to make 1.0L solution. What is the pH of the solution? Answer \[CaO(aq)\rightleftharpoons Ca^{+2}(aq) +2O{-2}(aq)\nonumber \] \[O^{-2}(aqs)+H_{2}O(l)\rightleftharpoons Ca^{+2}(aq) +2OH-(aq) \\ \; \\ so \\ \; \\ CaO(aq)+H_{2}O(l)\rightleftharpoons Ca^{+2}(aq) +2OH-(aq)\nonumber ]\] \[\frac{\frac{0.56g CaO}{\frac{56gCaO}{mol}}\frac{2\,mol\,OH^-}{1\,mol\,CaO}}{1.0L}=0.02\,M\,OH^-\nonumber \] \[pOH=-log[OH^{-}]=-log[2.010^{-2}M]=1.70\nonumber \] \[pH=14-pOH=14-1.70=12.3\nonumber \] Exercise $\PageIndex{5.2.c}$ What is the pH of a 0.015 M solution of barium hydroxide? Answer \[Ba(OH)_{2}\rightarrow Ba^{2+} + 2OH^{-}\nonumber\] \[[OH^{-}]=2M=20.015M=0.03M\nonumber\] \[pOH=-log[OH^{-}]=-log(0.03M)=1.52\nonumber\] \[pH=14-pOH=14-1.52=12.48\nonumber\] Exercise $\PageIndex{5.2.d}$ What is the [OH - ] and pH of a 0.035M KOH solution at 25°C? Answer \[\left [KOH\right ]=\left [OH^{-}\right ]=0.035M\nonumber\] \[pH=14-pOH =14-(-log [OH^{-}])\nonumber\] \[14-1.46=12.54\nonumber\] Exercise $\PageIndex{5.2.e}$ The pH of a 0.011 M NaOH solution at 25°C is _____. Answer \[pOH=-log\left (OH^{-}\right )=-log\left (0.011M\right )=1.96\nonumber\] \[pH=14-pOH=14-1.96=12.04\nonumber\] Exercise $\PageIndex{5.2.f}$ What is the pH of a 0.053 M solution of Ca(OH) 2 ? Answer \[\left [ OH^{-} \right ]=2M=20.053M=0.11M\nonumber\] \[pOH=-log\left [ OH^{-} \right ]=-log\left ( 0.11M \right )=0.96\nonumber\] \[pH=14-pOH=14-0.96=13.04\nonumber\] Weak Acids Exercise $\PageIndex{5.3.a}$ What is the pH of 0.20M aqueous HF? K a = 6.8x10 -4 Answer \[HF(aq)\rightleftharpoons H^{+}(aq)+F^{-}(aq) \nonumber \] \[[HF]_{i}>100K_{a} \nonumber \] Therefore, \[[H^{+}]=\sqrt{K_{a}[HF]_{i}}=\sqrt{(6.810^{-4})0.20}=1.210^{-2}M \nonumber \] \[pH=-log[H^{+}]=-log[1.210^{-2}M]=1.93 \nonumber \] Exercise $\PageIndex{5.3.b}$ What is the pH of 0.0050M aqueous HF? K a = 6.8x10 -4 Answer 0 1 2 3 4 R HF(aq) ⇌ H+(aq) + F-(aq) I 0.0050 NaN 0 0 C -x NaN +x +x E 0.0050-x NaN x x $K_{a}=\frac{x^{2}}{0.0050-x}$ Note 100K a is not less than [HA] i so we need to use the quadratic formula. \[K_{a}\left ( 0.0050-x \right )=x^{2}\Rightarrow x^{2}+K_{a}x-K_{a}\left ( 0.0050 \right ) \nonumber \] \[x^{2}+0.00068x-3.410^{-5} \nonumber \] \[x=\frac{-0.00068\pm \sqrt{\left ( 0.00068 \right )^{2}-4(1)(-3.410^{-6})}}{2} \nonumber \] \[x=\frac{-0.00068\pm 0.00375}{2} \nonumber \] \[x=[H^{+}]=0.001535M \nonumber \] \[pH=-log[H^{+}]=-log[0.001535M]=2.8 \nonumber \] Exercise $\PageIndex{5.3.c}$ Which one is more acidic, 0.2MHF or 0.02MHCl? Answer \[0.02HCl:\,pH=-log[0.02]=1.70 \nonumber \] \[0.2HF:\,pH=-log(\sqrt{K_{a}[HA]_{i}})=-log(\sqrt{7.210^{-4}[0.2]})=1.93 \nonumber \] The more dilute strong acid HCl is more acidic than the more concentrated weak acid HF. Exercise $\PageIndex{5.3.d}$ What is the pH of 0.04M of NH 4 Cl? K a =5.6x10 -10 Answer \[NH_{4}^{+}(aq)\rightleftharpoons NH_{3}(aq)+H^{+}(aq)\,\,\,K_{a}=5.610^{-10} \nonumber \] \[\left [NH_{4}^{+}\right ]_{i}>100K_{a}\nonumber\] Therefore, \[[H^{+}]=\sqrt{K_{a}[NH_{4}^{+}]_{i}}=\sqrt{5.610^{-10}(0.04)}=4.7310^{-6}M \nonumber \] \[pH=-log[H^{+}]=-log[4.7310^{-6}M]=5.32 \nonumber \] Exercise $\PageIndex{5.3.e}$ A 0.10M solution of lactic acid (HC 3 H 5 O 3 , one acidic hydrogen) has a pH of 2.45. What is the K a for lactic acid? Answer \[\left [ H^{+} \right ]=10^{-2.45}=0.0035M \nonumber \] 0 1 2 3 4 R HC3H5O3(aq) ⇌ C3H5O3-(aq) + H+(aq) I 0.10M NaN 0 0 C -x NaN +x +x E 0.10-x NaN x x \[K_{a}=\frac{x^{2}}{0.10-x}=\frac{0.0035^{2}}{0.10-0.0035}=1.3010^{-4} \nonumber \] 16.5.3 Exercise $\PageIndex{5.3.f}$ A 0.15M aqueous solution of the weak acid HA at 25°C has a pH of 5.35. What is the value of K a for HA? Answer 0 1 2 3 4 R HA ⇌ H+ + A- I 0.15M NaN 0 0 C -x NaN +x +x E 0.15-x NaN x x \[Ka=\frac{\left [ H^{+} \right ]\left [ A^{-} \right ]}{\left [ HA \right ]}=\frac{(x)(x)}{(0.15-x)} \nonumber \] \[\left [ H^{+} \right ]=x=10^{-pH}=10^{-5.35}=4.4710^{-6}M \nonumber \] \[Ka=\frac{(x)(x)}{(0.15-x)}=\frac{(4.4710^{-6})(4.4710^{-6})}{(0.15-(4.4710^{-6}))}=\frac{2.0010^{-11}}{0.150}=1.3310^{-10} \nonumber \] Exercise $\PageIndex{5.3.g}$ The K a of HClO is 3.010 -8 . What is the pH at 25°C of an aqueous solution that is 0.020M in HClO? Answer \[[HClO]_{i}>100K_{a} \nonumber \] Therefore, \[[H^{+}]=\sqrt{K_{a}[HClO]_{i}}=\sqrt{(3.010^{-8})0.20}=2.510^{-5}M \nonumber \] \[pH=-log[H^{+}]=-log[2.510^{-5}M]=4.61 \nonumber \] Or if you really want to spend the time you can do the RICE diagram..... 0 1 2 3 4 R HClO ⇌ H+ + ClO- I 0.020M NaN 0 0 C -x NaN +x +x E 0.020-x NaN x x \[Ka=\frac{\left [ H^{+} \right ]\left [ ClO^{-} \right ]}{\left [ HClO \right ]}=\frac{(x)(x)}{(0.020-x)} \nonumber \] \[3.010^{-8}=\frac{(x)(x)}{(0.020-x)} \nonumber \] \[\left [ H^{+} \right ]=x=2.510^{-5}M \nonumber \] \[pH=-log\left [ H^{+} \right ]=-log\left ( 2.510^{-5} \right )=4.61 \nonumber \] Exercise $\PageIndex{5.3.h}$ The K a of HF is 6.810 -4 . What is the pH of a 0.35M solution of HF? Answer \[[HF]_{i}>100K_{a} \nonumber \] Therefore, \[[H^{+}]=\sqrt{K_{a}[HF]_{i}}=\sqrt{(6.810^{-4})0.35}=0.0154M \nonumber \] \[pH=-log[H^{+}]=-log[0.1054M]=1.81 \nonumber \] Or if you really want to spend the time you can do the RICE diagram..... 0 1 2 3 4 R HF ⇌ H+ + F- I 0.35M NaN 0 0 C -x NaN +x +x E 0.35-x NaN x x \[Ka=\frac{\left [ H^{+} \right ]\left [ F^{-} \right ]}{\left [ H \right ]}=\frac{(x)(x)}{(0.35-x)} \nonumber \] \[6.810^{-4}=\frac{(x)(x)}{(0.35-x)} \nonumber \] \[\left [ H^{+} \right ]=x=1.510^{-2}M \nonumber \] \[pH=-log\left [ H^{+} \right ]=-log\left ( 1.510^{-2} \right )=1.81 \nonumber \] Exercise $\PageIndex{5.3.i}$ A 0.25M solution of the weak acid HA has a pH of 4.15. What is the value of K a for HA? Answer \[HA\rightleftharpoons H^{+}+A^{-} \nonumber \] \[\left [ H^{+} \right ]=\left [ A^{-} \right ]=10^{-pH}=10^{-4.15}=7.0810^{-5}M \nonumber \] Note how x<<[HA] i and so in the next step we can ignore the extent of reaction (amount dissociated) \[Ka=\frac{\left [ H^{+} \right ]\left [ A^{-} \right ]}{\left [ HA \right ]-x} \approxeq \frac{\left [ H^{+} \right ]\left [ A^{-} \right ]}{\left [ HA \right ]} \nonumber \] \[Ka=\frac{\left [ H^{+} \right ]\left [ A^{-} \right ]}{\left [ HA \right ]}=\frac{\left ( 7.0810^{-5}M \right )\left ( 7.0810^{-5}M \right )}{\left ( 0.25M \right )}=2.010^{-8} \nonumber \] Exercise $\PageIndex{5.3.j}$ In which of the following aqueous solution does the weak acid exhibit the highest percentage ionization? 0.01M HC 2 H 2 C 2 (K a =3.010- 8 ) 0.01M HNO 2 (K a = 4.510 -4 ) 0.01M HF (K a = 6.810 -4 ) 0.01M HClO (K a = 3.010 -8 ) These will all exhibit the same percentage ionization Answer c. 0.01M HF (K a = 6.810 -4 ) Percent Ionization Exercise $\PageIndex{5.4.a}$ What is the percent ionization of 0.25 aqueous HF? Ka = 6.8x10 -4 Answer \[HF(aq)\rightleftharpoons H^{+}(aq)+F^{-}(aq)\nonumber \] \[\left [HF\right ]_{i}>100Ka\nonumber\] Therefore, \[\left [H^{+}\right ]=\sqrt{Ka\left [HF\right ]_{i}}=\sqrt{6.810^{-4}0.25}=1.310^{-2}M\nonumber\] \[\%\text{Ionized}=\frac{\text{amount dissociated}}{\text{Initial concentration}}(100)=\frac{1.310^{-2}}{0.25}=5.2\%\nonumber\] Exercise $\PageIndex{5.4.b}$ What is the percent ionization of 0.0055 aqueous HF? Ka = 6.8x10 -4 Answer 0 1 2 3 4 R HF(aq) ⇌ H+(aq) + F-(aq) I 0.0055M NaN 0 0 C -x NaN +x +x E 0.0055-x NaN x x \[Ka=\frac{x^{2}}{0.0055-x}=6.810^{-4}\nonumber\] \[x=\left [H^{+}\right ]=1.6010^{-3}M\nonumber\] \[\frac{1.6010^{-3}}{0.0055}=29.5\%\nonumber\] Exercise $\PageIndex{5.4.c}$ What is the percent ionization of 0.05M of NH 4 Cl? Ka=5.6x10 -10 Answer $NH_{4}^{+}(aq)\rightleftharpoons NH_{3}(aq)+H^{+}(aq)$, $Ka=5.610^{-10}$ \[\left [NH_{4}^{+}\right ]_{i}>100Ka\nonumber\] \[\left [H^{+}\right ]=\sqrt{Ka\left [NH_{4}^{+}\right ]_{i}}=\sqrt{5.610^{-11}0.005}=5.2910^{-6}M\nonumber\] \[\%\text{Ionized}=\frac{\text{amount dissociated}}{\text{Initial concentration}}(100)=\frac{5.2910^{-6}}{0.05}100\%=0.011\%\nonumber\] Exercise $\PageIndex{5.4.d}$ What is the percent ionization of a 0.002M solution of H 2 CO 3 ? Ka 1 =4.3x10 -7 , Ka 2 = 5.6x10 -11 Answer \[Ka_{1}=4.310^{-7}, Ka_{2}=5.610^{-11}\nonumber\] \[K_{a1}>1000 k_{a2} \;\;\; so \;\; \left [H^{+}\right ]=\left [HCO_{3}^{-}\right ]=\sqrt{Ka_{1}0.002}=2.910^{-5}M\nonumber\] \[\%\text{Ionized}=\frac{\text{amount dissociated}}{\text{Initial concentration}}(100)=\frac{2.910^{-5}}{0.002}100\%=1.45\%\nonumber\] Exercise $\PageIndex{5.4.e}$ What is the percent ionization of 0.04M of hydrazoic acid (HN 3 )? Ka=1.9x10 -5 Answer \[HN_{3}(aq)\rightleftharpoons H^{+}(aq)+N_{3}^{-}(aq)\nonumber\] \[\left [HN_{3}\right ]_{i}>100Ka\nonumber\] Therefore, \[\left [H^{+}\right ]=\sqrt{Ka\left [HN_{3}\right ]_{i}}=\sqrt{1.910^{-5}0.04}=8.710^{-4}M\] \[\%\text{Ionized}=\frac{\text{amount dissociated}}{\text{Initial concentration}}(100)=\frac{8.710^{-4}}{0.04}=2.2\%\nonumber\] Exercise $\PageIndex{5.4.f}$ What is the percent ionization of hypochlorous acid (HClO) in a 0.015 M aqueous solution of HClO at 25C? (K a = 3.0 * 10 -8 ) Answer \[HClO(aq)\rightleftharpoons H^{+}(aq)+ClO^{-}(aq)\nonumber\] \[\left [HClO\right ]_{i}>100Ka\nonumber\] Therefore, \[\left [H^{+}\right ]=\sqrt{Ka\left [HClO\right ]_{i}}=\sqrt{3.010^{-8}0.015}=2.1210^{-5}M\] \[\%\text{Ionized}=\frac{\text{amount dissociated}}{\text{Initial concentration}}(100)=\frac{2.210^{-5}}{0.015}(100)=0.14 \% \nonumber\] Weak Base Exercise $\PageIndex{5.5.a}$ What is the concentration of OH - in 0.10M of ethylamine (C 2 H 5 NH 2 )? Kb=6.4x10 -4 Answer \[\left [ C_{2}H_{5}NH_{2} \right ]>100K_{b} \nonumber \] \[\left [ OH^{-} \right ]=\sqrt{\left (6.410^{-4}\right )0.10}=0.008M \nonumber \] Exercise $\PageIndex{5.5.b}$ What is the concentration of OH - in 0.005M of ethylamine (C 2 H 5 NH 2 )? Kb=6.4x10 -4 Answer 0 1 2 3 4 5 R C2H5NH2(aq) + H2O(l) ⇌ C2H5NH3+(aq) + OH-(aq) I 0.005M NaN NaN 0 0 C -x NaN NaN +x +x E 0.005-x NaN NaN x x \[K_{b}=\frac{x^{2}}{0.005-x}=6.410^{-4} \nonumber \] \[x=1.510^{-3} \nonumber \] Exercise $\PageIndex{5.5.c}$ Which solution is more basic, 0.10M of ethylamine or 0.01M of NaOH? Note, you solved 0.01M ethylamine two problems ago Answer NaOH, 0.01M of NaOH has a pOH of 2.00 Exercise $\PageIndex{5.5.d}$ A solution of NH 3 has a pH of 10.25. What is the concentration of the solution? Kb=1.8x10 -5 Answer \[[H^+]=10^{-10.25}=5.610^{-11} \;\; K_w=[H^+][OH^-] \; \\ \; [OH^-]=\frac{K_w}{[H^+]}=\frac{10^{14}}{5.6x10^{-11}}=1.8x10^{-4} \nonumber \] \[\begin{align}K_{b} & =\frac{x^{2}}{[NH_3]_i-x } \nonumber \\ \; \\ K_b\left( [NH_3]_i -x \right) & = x^2 \nonumber \\ \; \\ K_b [NH_3]_i -K_bx & = x^2 \nonumber \\ \; \\ K_b [NH_3]_i &= x^2 + K_bx \nonumber \\ \; \\ [NH_3]_i &=\frac{x^2 + K_bx}{K_b} \nonumber \\ \; \\ [NH_3]_i &=\frac{\left( 1.8x10^{-4} \right)^2+1.8x10^{-5}}{1.8x10^{-5}} \nonumber \\ \; \\ [NH_3]_i &=2.010^{-3}M \end{align} \] If you want, you can look at the RICE diagram 0 1 2 3 4 5 R NH3(aq) + H2O(l) ⇌ NH4+(aq) OH-(aq) I [NH3]i NaN NaN 0 0 C -x NaN NaN +x +x E [NH3]i-x NaN NaN x x Exercise $\PageIndex{5.5.e}$ If 1.06g of Na 2 CO 3 is dissolved in plenty of water to make 1.0L of the solution, what is the pH of the solution? Kb=1.8x10 -4 Answer \[\frac{\frac{1.06g}{106g/mol}}{1.0L}=0.01M \nonumber \] \[[OH^-]= \sqrt{K_b[B]}=\sqrt{\left( 1.8x10^{-4} \right) 0.01}=0.0013 \;\; so \;\;p[OH]=-log(0.0013)=2.87 \;\;and\;\;pH=14-pOH=11.13 Or if you insist on using the RICD diagram, you can.... 0 1 2 3 4 5 R CO32-(aq) + H2O(l) ⇌ HCO3-(aq) + OH-(aq) I 0.01M NaN NaN 0 0 C -x NaN NaN +x +x E 0.01-x NaN NaN x x \[K_{b}=\frac{x^{2}}{0.01-x}=1.810^{-4} \nonumber \] \[x=1.3146410^{-3}M \nonumber \] \[pOH=2.87 \nonumber \] \[pH=14.00-2.87=11.13 \nonumber \] Exercise $\PageIndex{5.f}$ The pH of a 0.10M solution of a weak base is 9.82. What is the Kb for this base? Answer \[pH+pOH=14 \nonumber \] \[pOH=14-9.82=4.18 \nonumber \] \[ [OH^{-}] = 10^{-pOH} = 10^{-4.18}= 6.6110^{-5} M \nonumber \] d 0 1 2 3 4 5 R B + H2O(l) ⇌ B+ + OH-(aq) I 0.10M NaN NaN 0 0 C -x NaN NaN +x +x E 0.10-x NaN NaN x x \[OH^-]=[B^+]=x ;and\; K_b=\frac{x^2}{[B]_i-x} \approxeq \frac{x^2}{[B]_i} \nonumber\] This is the same eq as $[OH^-]=\sqrt{K_b[B]_i}$ \[K_{b}=\frac{\left ( 6.6110^{-5} \right )^{2}}{0.1}=4.3710^{-8} \nonumber \] pH of Various Salts Exercise $\PageIndex{5.6.a}$ What is the pH of 0.05M of NH 4 Cl? K b (NH 3 ) = 1.8x10 -5 Answer \[NH_{4}^{+}(aq)\rightleftharpoons NH_{3}(aq)+H^{+}(aq) \nonumber \] \[(K_a'(NH_4^+) = \frac{K_w}{K_b}=\frac{10^{-14}}{(1.8x10^{-5}}=5.610^{-10}\nonumber \] \[\left [NH_{4}^{+}\right ]_{i}>100Ka'\nonumber \] Therefore, \[\left [H^{+}\right ]=\sqrt{Ka'\left [NH_{4}^{+}\right ]_{i}}=\sqrt{5.610^{-10}0.05}=5.310^{-6}M\nonumber \] \[pH=-log\left (5.310^{-6}M\right )=5.28\nonumber \] To solve in one equation \[pH=-log\sqrt{\frac{K_w}{K_b}\left[B^+\right]} = 14-(-log\sqrt{\frac{10^{-14}}{1.8x10^{-5}}\left[0.05\right]} ) =5.3 \nonumber \] NOTE: the answer has one sig fig because 0.05 has one sig fig, by doing it in one step you truncate at the end and this gives the best answer. Exercise $\PageIndex{5.6.b}$ What is the pH of 0.05M of NaClO? K a (HClO)=4.0x10 -8 . Answer hypochlorite is the conjugate base of hypochlorous acid and the salt forms hypochlorous acid \[ClO^{-}(aq)+H_{2}O(l)\rightleftharpoons HClO(aq)+OH^{-}(aq) \nonumber\] \[(K_b'(ClO^-) = \frac{K_w}{K_b'}=\frac{10^{-14}}{4.0x10^{-8}}=2.510^{-7}\nonumber \] \[\left [ClO^{-}\right ]_{i}>100K_{b}\nonumber \] \[\left [OH^{-}\right ]=\sqrt{K_{b}\left [ClO^{-}\right ]_{i}}=\sqrt{2.510^{-7}0.05}=1.11810^{-4}M\nonumber \] \[pOH=-log\left (1.310^{-4}M\right )=3.95\nonumber \] \[pH=14.00-3.89=10.0\nonumber \] To solve in one equation \[pH=14-(-log\sqrt{\frac{K_w}{K_a}\left[A^-\right]} )= 14 +log\sqrt{\frac{10^{-14}}{4.0x10^{-8}}\left[0.05\right]} =10.0 \nonumber \] NOTE: the answer has one sig fig because 0.05 has one sig fig, by doing it in one step you truncate at the end and this gives the best answer. Exercise $\PageIndex{5.6.c}$ What is the pH of 0.05M of NaHCO 3 ? K a1 (H 2 CO 3 )=4.5x10 -7 , K a2 (H 2 CO 3 )=4.7x10 -11 , Answer bicarbonate is amphiprotic, and K b ' comes from the first ionization constant \[HCO_{3}^{-}(aq)+H_{2}O(l)\rightleftharpoons H_{2}CO_{3}(aq)+OH^{-}(aq) \nonumber \] \[(K_b'(HCO_3^-) = \frac{K_w}{K_a}=\frac{10^{-14}}{4.5x10^{-7}}=2.2210^{-8}\nonumber \] Since the K b ' (10 -8 ) is greater than K a2 (10 -11 ) it is a stronger base than an acid, so: \[\left [OH^{-}\right ]=\sqrt{K_{b}\left [HCO_{3}^{-}\right ]_{i}}=\sqrt{2.2210^{-8}0.05}=3.3310^{-5}M\nonumber \] \[pOH=-log\left (3.410^{-5}M\right )=4.48\nonumber \] \[pH=14.00-4.48=9.5\nonumber \] To solve in one equation \[pH=14-(-log\sqrt{\frac{K_w}{K_a}\left[A^-\right]} )= 14 + (og\sqrt{\frac{10^{-14}}{4.5x10^{-7}}\left[0.05\right]} =9.5 \nonumber \] NOTE: the answer has one sig fig because 0.05 has one sig fig, by doing it in one step you truncate at the end and this gives the best answer. Exercise $\PageIndex{5.6.d}$ What is the pH of 0.05M of KF? K a (HF)=6.3x10 -4 . Answer fluoride is the conjugate base of hydrofluoric acid \[F^{-}(aq)+H_{2}O(l)\rightleftharpoons HF(aq)+OH^{-}(aq)\nonumber \] \[(K_b'(F^-) = \frac{K_w}{K_a}=\frac{10^{-14}}{6.3x10^{-4}}=1.58710^{-11}\nonumber \] \[\left [F^{-}\right ]_{i}>100K_{b}'\nonumber \] Therefore, \[\left [OH^{-}\right ]=\sqrt{K_{b}\left [F^{-}\right ]_{i}}=\sqrt{1.58710^{-11}0.05}=8.90910^{-7}M\nonumber \] \[pOH=-log\left (8.90910^{-7}M\right )=6.05\nonumber \] \[pH=14.00-6.06=7.95\nonumber \] To solve in one equation \[pH=14-(-log\sqrt{\frac{K_w}{K_a}\left[A^-\right]} )= 14+log\sqrt{\frac{10^{-14}}{6.3x10^{-4}}\left[0.05\right]} =7.95\nonumber \] Exercise $\PageIndex{5.6.e}$ How many grams of NaHCO 3 will be used to make a 1.0L solution that has a pH = 9.0? K a1 (H 2 CO 3 )=4.5x10 -7 , K a2 (H 2 CO 3 )=4.7x10 -11 , Answer bicarbonate is amphiprotic, and K b ' comes from the first ionization constant \[HCO_{3}^{-}(aq)+H_{2}O(l)\rightleftharpoons H_{2}CO_{3}(aq)+OH^{-}(aq) \nonumber \] \[(K_b'(HCO_3^-) = \frac{K_w}{K_a}=\frac{10^{-14}}{4.5x10^{-7}}=2.2210^{-8}\nonumber \] 0 1 2 3 4 5 R HCO3-(aq) + H2O(l) ⇌ H2CO3(aq) + OH-(aq) I [HCO3-(aq)]i NaN NaN 0 0 C -x NaN NaN +x +x E [HCO3-(aq)]i-x NaN NaN x x \[pH+pOH=14 \;\;so\;\;\; pOH=14-pH=14-9=5 \;\;\;[OH^-] = 10^{-pOH}=10^{-5}] \nonumber \] \[K_{b}=\frac{x^{2}}{[HCO_3^-]-x} \approxeq \frac{x^2}{[HCO_3^-]} \nonumber \] \[[HCO_3^-]=\frac{x^2}{K_b}=\frac{[10^{-5}]^2}{2.22x10^{-8}} = 0.0045M \nonumber \] \[\frac{0.0045mol \;NaHCO_3}{l}\left(\frac{84.007g \; NaHCO_3}{mpl}\right)=0.378g=0.38g \nonumber \] Exercise $\PageIndex{5.6.f}$ Ka of HA is 7.510 -9 . What is the pH of a 0.15M solution of NaA? Answer A - is the conjugate base of HA \[A^-(aq) + H_2O(l)\rightleftharpoons HA(aq)+OH^{1}(aq) \nonumber \] \[(K_b'(A^-) = \frac{K_w}{K_a}=\frac{10^{-14}}{(7.5x10^{-9}}=1.33310^{-6}\nonumber \] \[\left [A^-\right ]_{i}>100Ka'\nonumber \] \[\left [OH^{-}\right ]=\sqrt{K_{b}\left [A^{-}\right ]_{i}}=\sqrt{1.33x10^{-6}0.15}=4.47210^{-4}M\nonumber \] \[pOH=-log\left (4.47210^{-4}M\right )=3.35\nonumber \] \[pH=14.00-3.76=10.65\nonumber \] To solve in one equation \[pH=14-(-log\sqrt{\frac{K_w}{K_a}\left[A^-\right]} )= 14 +log\sqrt{\frac{10^{-14}}{7.5x10^{-9}}\left[0.15\right]} =10.65 \nonumber \] NOTE: the answer has two sig figs, by doing it in one step you truncate at the end and this gives the best answer. Exercise $\PageIndex{5.6.g}$ Calculate the pOH of a 0.0827M aqueous sodium cyanide solution at 25°C (for CN - , Kb = 4.910 -10 ). Answer \[pOH=-log[OH-]=-log\sqrt{K_b[CN^-]}=-log\sqrt{4.9x10^{-10}(0.827)}=5.20\nonumber \] Exercise $\PageIndex{5.6.h}$ Determine the pH of a 0.15M solution of KF. For hydrofluoric acid, Ka = 7.010 -4 . Answer First we will do this in one equation, and then do it stepwise, noting we use primes (') to indicate K's of conjugates. \[pH=14-pOH=14-\sqrt{K_b'[A^-]}=14-(-log\sqrt{\frac{K_w}{K_a}\left[A^-\right]} )= 14 +log\sqrt{\frac{10^{-14}}{7.0x10^{-5}}\left[0.15\right]} =8.16 \nonumber \] Now we will do it stepwise \[K_{b}=\frac{K_{w}}{K_{a}}=\frac{10^{-14}}{7.010^{-4}}=1.410^{-11} \nonumber \] \[K_{b}=\frac{\left [ HF \right ]\left [ OH^{-} \right ]}{\left [ F^{-} \right ]}=\frac{(x)(x)}{0.15-x} \approxeq\frac{(x)(x)}{0.15}1.410^{-11} \nonumber \] \[x=\left [ OH^{-} \right ]=1.4510^{-6}M \nonumber \] \[pOH=-log\left [ OH^{-} \right ]=-log\left [ 1.4510^{-6}M \right ]=5.84 \nonumber \] \[pH=14-pOH=14-5.84=8.16 \nonumber \] Exercise $\PageIndex{5.6.i}$ Calculate the pH of 0.726M anilinium hydrochloride, (C 6 H 5 NH 3 Cl) solution in water given that Kb for aniline is 3.8310 -4 . Answer Anilinium is the conjugate acid of aniline (C 6 H 5 N 2 ) \[C_6H_5NH_4^+ \rightleftharpoons C_6H_5NH_2 + H^+\] In one equation: \[pH=-log\sqrt{K_a'[C_6H_5NH_4^+]}=-log\sqrt{\frac{K_w}{K_b}[C_6H_5NH_4^+]}=-log\sqrt{\frac{1x10^{-14}}{3.83x10^{-4}}[0.726]}=5.36 \nonumber \] Taking this out stepwise.... \[K_{a}'=\frac{K_{w}}{K_{b}}=\frac{10^{-14}}{3.8310^{-4}}=2.6110^{-11} \nonumber \] \[K_{a}'=\frac{\left [ C_{6}H_{5}NH_{2} \right ]\left [ H^{+} \right ]}{\left [ C_{6}H_{5}NH_{3}^{+} \right ]}=\frac{(x)(x)}{0.726-x}\approxeq \frac{(x)(x)}{0.726}=2.6110^{-11} \nonumber \] solving for x \[x=\left [ H^{+} \right ]=4.3510^{-6} M \\ pH=-log\left [ H^{+} \right ]=-log\left [ 4.3510^{-6}M \right ]=5.36 \nonumber \] Exercise $\PageIndex{5.6.j}$ The Ka for formic acid (HCHO 2 ) is 1.810 -4 . What is the pH of a 0.35M solution of sodium formate (NaCHO 2 )? Answer Firstst we will do this in one equation, and then do it stepwise, noting we use primes (') to indicate K's of conjugates. \[pH=14-pOH=14-\sqrt{K_b'[A^-]}=14-(-log\sqrt{\frac{K_w}{K_a}\left[A^-\right]} )= 14 +log\sqrt{\frac{10^{-14}}{1.8x10^{-4}}\left[0.35\right]} =8.64 \nonumber \] Now we will do it stepwise. First calculate K b of the conjugate base (K b ') \[K_{b}'=\frac{K_{w}}{K_{a}}=\frac{10^{-14}}{1.810^{-4}}=5.5510^{-11} \nonumber \] \[[OH^-]=\sqrt{K_b'[A^-]}=\sqrt{5.55x10^{-11}[0.35]}=4.41x10^{-6} \nonumber \] \[pOH=-log[OH^-]=-log4.41x10^{-6}=5.36 \] pH=14-pOH=14-5.36= 8.64 \nonumber \] Polyprotic Acids and Bases Exercise $\PageIndex{5.7.a}$ What is the pH of a 0.002M solution of H 2 CO 3 ? Ka 1 =4.3x10 -7 , Ka 2 = 5.6x10 -11 Answer \[Ka_{1}=4.310^{-7},\,Ka_{2}=5.610^{-11} \nonumber \] \[\frac{Ka_{1}}{Ka_{2}}>1000 \nonumber \] \[\left [H^{+}\right ]=\left [HCO_{3}^{-}\right ]=\sqrt{Ka_{1}0.002}=2.910^{-5}M \nonumber \] \[pH=4.53 \nonumber \] Exercise $\PageIndex{5.7.b}$ What is the concentration of CO 3 2 - ion in the solution in Q16.3.11? Answer \[\left [H^{+}\right ]=\left [HCO_{3}^{-}\right ]=2.910^{-5}M \nonumber \] 0 1 2 3 4 R HCO3-(aq) ⇌ H+(aq) + CO32-(aq) I 2.910-5 NaN 2.910-5 0 C -x NaN -x +x E 2.910-5-x NaN 2.910-5-x x \[Ka_{2}=\frac{\left (2.910^{-5}-x\right )x}{2.910^{-5}-x}=5.610^{-11} \nonumber \] \[\left [CO_{3}^{2-}\right ]=5.610^{-11}M \nonumber \] Exercise $\PageIndex{5.7.c}$ What is the pH of a 0.05M of sulfurous acid (H 2 SO 3 )? Ka 1 =1.7x10 -2 , Ka 2 =6.4x10 -8 Answer \[\frac{Ka_{1}}{Ka_{2}}>1000 \nonumber \] 0 1 2 3 4 R H2SO3(aq) ⇌ H+(aq) HSO3-(aq) I 0.05 NaN 0 0 C -x NaN +x +x E 0.05-x NaN x x \[Ka_{1}=\frac{x^{2}}{0.05-x}=1.710^{-2} \nonumber \] \[x=2.210^{-2}M \nonumber \] \[pH=1.66 \nonumber \] Exercise $\PageIndex{5.7.d}$ What is the concentration of SO 3 2 - ion in the solution in Q16.3.13? Answer \[\left [H^{+}\right ]=\left [HSO_{3}^{-}\right ]=2.210^{-2}M \nonumber \] \[Ka_{2}=\frac{\left (2.210^{-2}-x\right )x}{2.210^{-2}-x}=6.410^{-8} \nonumber \] \[\left [SO_{3}^{2-}\right ]=6.410^{-8}M \nonumber \] Exercise $\PageIndex{5.7.e}$ What is the concentration of a sample solution of H 2 CO 3 that has a pH = 4.50? Answer \[10^{-4.50}=3.1610^{-5} \nonumber \] \[\left [H^{+}\right ]=\sqrt{Ka_{1}\left [H_{2}CO_{3}\right ]}=3.1610^{-5}M \nonumber \] \[Ka_{1}=4.310^{-7} \nonumber \] \[\left [H_{2}CO_{3}\right ]=0.0023M \nonumber \] Questions $\PageIndex{5.7.f}$ through $\PageIndex{5.7.l}$ deal with the concentrations of species that exist in aqueous solutions of selenous acid. Exercise $\PageIndex{5.7.f}$ Consider 0.50M of H 2 SeO 3 for the following questions. K a1 =2.7x10 -5 K a2 =2.5x10 -9 What is [H 3 O + ] of H 2 SeO 3 ? Answer The reaction is \[H_2SeO_3 \leftrightharpoons H^+ + HSeO_3^- \;\;K_{a1}=2.7x10^{-5} \\ \; \\ HSeO_3^- \leftrightharpoons H^+ + SeO_3^{-2} \;\; K_{a2} = 2.5x10^{-9} \] Since Ka 1 >1000Ka 2 you can consider all the hydronium to come from the first ionization, and that the second is insignificant. This is logical, as in the first ionization you are pulling a proton from something neutral, while in the second you are pulling it from something negative, which just does not want to happen. \[[H_3O]^+=\sqrt{K_{a1}[HA_i}=\sqrt{2.710^{-5}\left ( 0.50M \right )}=0.0037M\] Exercise $\PageIndex{5.7.g}$ Consider 0.50M of H 2 SeO 3 for the following questions. K a1 =2.7x10 -5 K a2 =2.5x10 -9 What is pH of H 2 SeO 3 ? Answer The reaction is \[H_2SeO_3 \leftrightharpoons H^+ + HSeO_3^- \;\;K_{a1}=2.7x10^{-5} \\ \; \\ HSeO_3^- \leftrightharpoons H^+ + SeO_3^{-2} \;\; K_{a2} = 2.5x10^{-9} \] This is actually the same question as $\PageIndex{5.7.f}$, it is just asking you to express the hydronium ion concentration in terms of pH. Since Ka 1 >1000Ka 2 you can consider all the hydronium to come from the first ionization, and that the second is insignificant. This is logical, as in the first ionization you are pulling a proton from something neutral, while in the second you are pulling it from something negative, which just does not want to happen. \[pH=-log\left [ H^{+} \right ]=-log\sqrt{K_{a1}[HA_i]}=-log\sqrt{2.710^{-5}\left ( 0.50M \right )}=2.43 \nonumber \] Exercise $\PageIndex{5.7.h}$ Consider 0.50M of H 2 SeO 3 for the following questions. K a1 =2.7x10 -5 K a2 =2.5x10 -9 What is the concentration of HSeO 3 - ? Answer The reaction is \[H_2SeO_3 \leftrightharpoons H^+ + HSeO_3^- \;\;K_{a1}=2.7x10^{-5} \\ \; \\ HSeO_3^- \leftrightharpoons H^+ + SeO_3^{-2} \;\; K_{a2} = 2.5x10^{-9} \] Since Ka 1 >1000Ka 2 you can consider the second ionization as negligible and so [HSeO_3^-]=[H^+] \[ HSeO_{3}^{-} =\sqrt{2.710^{-5}\left ( 0.50M \right )}=0.0037M \nonumber \] Exercise $\PageIndex{5.7.j}$ Consider 0.50M of H 2 SeO 3 for the following questions. K a1 =2.7x10 -5 K a2 =2.5x10 -9 What is the concentration of H 2 SeO 3 ? Answer The reaction is \[H_2SeO_3 \leftrightharpoons H^+ + HSeO_3^- \;\;K_{a1}=2.7x10^{-5} \\ \; \\ HSeO_3^- \leftrightharpoons H^+ + SeO_3^{-2} \;\; K_{a2} = 2.5x10^{-9} \] Since the amount dissociated it [H + ]=[HSO 3 - ] = 0.0037M, you simply subtract if from the initial concentration. \[H_{2}SeO_{3}=0.50M-\left [ H^{+} \right ]=0.50M-0.0037M=0.4963M=0.50M \nonumber \] Note, this is a very weak acid and so it does not dissociate. Exercise $\PageIndex{5.7.k}$ Consider 0.50M of H 2 SeO 3 for the following questions. K a1 =2.7x10 -5 K a2 =2.5x10 -9 What is the concentration of OH - ? Answer The reaction is \[H_2SeO_3 \leftrightharpoons H^+ + HSeO_3^- \;\;K_{a1}=2.7x10^{-5} \\ \; \\ HSeO_3^- \leftrightharpoons H^+ + SeO_3^{-2} \;\; K_{a2} = 2.5x10^{-9} \] \[\left [ OH^{-} \right ]=\frac{K_{w}}{\left [ H^{+} \right ]}=\frac{10^{-14}}{\left [ H^{+} \right ]}=2.710^{-12}M \nonumber \] Exercise $\PageIndex{5.7.l}$ Consider 0.50M of H 2 SeO 3 for the following questions. K a1 =2.7x10 -5 K a2 =2.5x10 -9 What is the concentration of SeO 3 2 - ? Answer Look at K 2 \[ K_{a2} = \frac{[H^+]}{[SeO_3^{-2}][HeSeO_3^-]}\] Since $[H^+]=[HeSeO_3^-]$ the above equation becomes \[ K_{a2} = \frac{\cancel{[H^+]}[SeO_3^{-2}]}{\cancel{[HeSeO_3^-]}}\] \[K_{a2}=[SeO_3^{-2}]=2.510^{-9}M \nonumber \] Exercise $\PageIndex{5.7.m}$ An aqueous solution of phosphoric acid has a concentration of 2.5M. (Ka 1 = 7.510 -3 , Ka 2 = 6.210 -8 , Ka 3 = 4.210 -13 ) What is the pH? What is the molar concentration of phosphate ion? Answer a. First dissociation of phosphoric acid: \[H_{3}PO_{4}(aq)\rightleftharpoons H^{+}(aq)+H_{2}PO_{4}^{-}(aq)\,\,\,K_{a1}=7.510^{-3} \nonumber \] \[H_{2}PO_{4}^{-}\rightleftharpoons H^{+}(aq)+HPO_{4}^{2-}(aq)\,\,\,K_{a2}=6.210^{-8} \nonumber \] \[HPO_{4}^{2-}(aq)\rightleftharpoons H^{+}(aq)+PO_{4}^{3-}(aq)\,\,\,K_{a3}=4.210^{-13} \nonumber \] 0 1 2 3 4 R H3PO4(aq) ⇌ H+(aq) + H2PO4-(aq) I 2.5M NaN 0 0 C -x NaN +x +x E 2.5+x NaN x x \[K_{a1}=\frac{\left [H^{+}\right ]\left [H_{2}PO_{4}^{-}\right ]}{\left [H_{3}PO_{4}\right ]}=\frac{(x)(x)}{(2.5-x)}=7.510^{-3} \nonumber \] \[x=\left [H^{+}\right ]=0.13323M \nonumber \] Second dissociation of phosphoric acid: 0 1 2 3 4 R H2PO4-(aq) ⇌ H+(aq) + HPO42-(aq) I 0.13323M NaN 0.13323M 0 C -y NaN +y +y E 0.13323-y NaN 0.13323+y y \[K_{a2}=\frac{\left [ H^{+} \right ]\left [ HPO_{4}^{2-} \right ]}{\left [ H_{2}PO_{4}^{-} \right ]}=\frac{(y)(0.13323+y)}{0.13323-y}=6.210^{-8} \nonumber \] (Assume <<0.13323 M) \[\left [ H^{+} \right ]=K_{a2}=6.210^{-8} \nonumber \] The [H+] from the second step is negligible. Third dissociation of phosphoric acid: 0 1 2 3 4 R HPO42-(aq) ⇌ H+(aq) + PO43-(aq) I 6.210-8M NaN 0.13323M 0 C -z NaN +z +z E 6.210-8-z NaN 0.13323+z y \[K_{a3}=\frac{\left [ H^{+} \right ]\left [ PO_{4}^{3-} \right ]}{\left [ HPO_{4}^{2-} \right ]}=\frac{\left ( z \right )\left ( 0.13323+z \right )}{\left ( 6.210^{-8} \right )-z}=4.210^{-13} \nonumber \] \[z=1.410^{-19}M \nonumber \] The [H+] from the third step is negligible. \[pH=-log\left [ H^{+} \right ]=-log\left [ 0.13323 \right ]=0.88 \nonumber \] Answer b. First dissociation of phosphoric acid: 0 1 2 3 4 R H3PO4(aq) ⇌ H+(aq) + H2PO4-(aq) I 2.5M NaN 0 0 C -x NaN +x +x E 2.5+x NaN x x \[K_{a1}=\frac{\left [H^{+}\right ]\left [H_{2}PO_{4}^{-}\right ]}{\left [H_{3}PO_{4}\right ]}=\frac{(x)(x)}{(2.5-x)}=7.510^{-3} \nonumber \] \[x=\left [H_{2}PO_{4}^{-}\right ]=0.13323M \nonumber \] For the second dissociation: \[K_{a2}=\frac{\left [ H^{+} \right ]\left [ HPO_{4}^{2-} \right ]}{\left [ H_{2}PO_{4}^{-} \right ]}=\frac{(x)(x)}{(0.13323-x)}=6.210^{-8} \nonumber \] \[x=\left [ HPO_{4}^{2-} \right ]=1.9210^{-4}M \nonumber \] For the third dissociation: \[K_{a3}=\frac{\left [ H^{+} \right ]\left [ PO_{4}^{3-} \right ]}{\left [ HPO_{4}^{2-} \right ]}=\frac{(x)(x)}{(1.9210^{-4})-x}=4.210^{-13} \nonumber \] \[x=\left [ PO_{4}^{3-} \right ]=1.9210^{-19}M \nonumber \] Molecular Structure Exercise $\PageIndex{6.a}$ Of the following, which is the strongest acid? HIO HIO 4 HIO 2 HIO 3 The acid strength of all these is nearly the same. Answer b. HIO 4 Exercise $\PageIndex{6.b}$ Of the following, the acid strength of _____ is the greatest. CH 3 COOH ClCH 2 COOH Cl 2 CHCOOH Cl 3 CCOOH BrCH 2 COOH Answer d. Cl 3 CCOOH, these are all based on the structure of acetic acid, where you have one or more halogen relacing the non-titratable proton(s). Chlorine is the most electronegative. In the section on oxy acids with homologous structures we showed that the more electronegative chlorine pulled electron density towards it, and thus made the oxygen-proton bond more polar, and thus more acidic. The same thing is going on here. The trichloroacetic acid has three protons pulling electron density towards it, making it more acidic. This is actually a nearest neighbor effect. Exercise $\PageIndex{6.c}$ Of the following, _____ is the strongest acid. Cl 3 C-COOH H 3 C-COOH Br 2 C-COOH F 3 C-COOH Br 2 ClC-COOH Answer d.F 3 C-COOH Exercise $\PageIndex{6.d}$ Which of the following acids will be the strongest? H 2 SO 4 HSO 4 - H 2 SO 3 H 2 SeO 4 HSO 3 - Answer a. H 2 SO 4 Exercise $\PageIndex{6.e}$ The more electronegative X is, the _____ polar will be the H-X bond and the _____ easily the H-X bond is broken, making HX more _____ acidic. more, less, weakly more, more, weakly more, more, strongly more, less, strongly less, less, strongly Answer c. more, more, strongly Exercise $\PageIndex{6.f}$ Which one is more acidic, HNO 2 or HNO 3 ? Answer HNO 3 Exercise $\PageIndex{6.g}$ Which one is more acidic H 3 AsO 3 or H 3 AsO 4 ? Answer H 3 AsO 4 Exercise $\PageIndex{6.h}$ List the acids in order of increasing acid strength: HClO 2 , HBrO 2 , HIO 2 Answer HIO 2 < HBrO 2 < HClO 2 Exercise $\PageIndex{6.i}$ List the compounds in order of increasing acid strength: AsH 3 , HBr, NaH, H 2 O Answer NaH< AsH 3 < H 2 O< HBr Exercise $\PageIndex{6.j}$ List the compounds in order of increasing acid strength: H 2 TeO 3 , H 2 TeO 4 , H 2 O Answer H 2 O< H 2 TeO 3 < H 2 TeO 4 Acid Anhydrides Exercise $\PageIndex{6.1.a}$ What is the pH of the solution if 0.05mol of SO 3 is dissolved in the water to make 1.0L solution? (SO 3 is very soluble.) Answer \[SO_{3}(g)+H_{2}O(l)\rightleftharpoons H_{2}SO_{4}(aq) \nonumber \] \[H_{2}SO_{4}(aq)\rightleftharpoons H^{+}(aq)+HSO_{4}^{-}(aq) \nonumber \] \[\frac{0.05mol}{1.0L}=0.05M pH=-log\left (0.05M\right )=1.30 \nonumber \] Exercise $\PageIndex{6.1.b}$ How many grams of SO 3 is needed to make a 1.0L solution that has a pH=1.0? Answer \[SO_{3}(g)+H_{2}O(l)\rightleftharpoons H_{2}SO_{4}(aq) \nonumber \] \[H_{2}SO_{4}(aq)\rightleftharpoons H^{+}(aq)+HSO_{4}^{-}(aq) \nonumber \] \[\left [H^{+}\right ]=0.1M \nonumber \] \[0.1M1.0L=0.1mol \nonumber \] \[0.1mol80g/mol=8.0g \nonumber \] Exercise $\PageIndex{6.1.c}$ What is the pH of the solution if 0.002mol of CO 2 is dissolved in the water to make 1.0L solution at 25 o C and 0.1atm? (The solubility of CO 2 in pure water at 25 o C and 0.1atm is 0.0037M.) Ka 1 =4.3x10 -7 , Ka 2 = 5.6x10 -11 Answer \[CO_{2}(g)+H_{2}O(l)\rightleftharpoons H_{2}CO_{3}(aq) \nonumber \] \[H_{2}CO_{3}(aq)\rightleftharpoons H^{+}(aq)+HCO_{3}^{-}(aq) \nonumber \] \[\frac{0.002mol}{1.0L}=0.002M \nonumber \] \[\left [H^{+}\right ]=\sqrt{4.310^{-7}0.002}=2.9310^{-5}M\] \[pH=-log\left (2.9310^{-5}M\right )=4.53 \nonumber \] Exercise $\PageIndex{6.1.d}$ What is the concentration of CO 3 2 - ion in the solution in Q 16.5.3? Answer \[\left [H^{+}\right ]=\left [HCO_{3}^{-}\right ]=2.9310^{-5}M \nonumber \] 0 1 2 3 4 R HCO3-(aq) ⇌ H+(aq) + CO32-(aq) I 2.9310-5 NaN 2.9310-5 0 C -x NaN +x +x E 2.9310-5-x NaN 2.9310-5+x x \[Ka_{2}=\frac{\left (2.9310^{-5}-x\right)x}{2.9310^{-5}+x}=5.610^{-11} \nonumber \] \[\left [CO_{3}^{2-}\right]=5.610^{-11} \nonumber \] Exercise $\PageIndex{6.1.e}$ How many grams of CO 2 is needed at 25 o C and 0.1atm to make a 1.0L solution that has a pH=4.50? Answer \[pH=4.50 \nonumber \] \[\left [H^{+}\right]=10^{-4.50}=3.1610^{-5}M \nonumber \] \[CO_{2}(g)+H_{2}O(l)\rightleftharpoons H_{2}CO_{3}(aq) \nonumber \] 0 1 2 3 4 R H2CO3(aq) ⇌ H+(aq) + HCO3-(aq) I M NaN 0 0 C -x NaN +x +x E M-x NaN 3.1610-5 x \[Ka=\frac{\left ( 3.1610^{-5} \right )^{2}}{M-3.1610^{-5}}=4.310^{-7} \nonumber \] \[M=2.3510^{-3}mol/L \nonumber \] \[2.35M1.0L=2.36mol \nonumber \] \[2.35mol44g/mol=0.10g \nonumber \] Basic Anhydrides Exercise $\PageIndex{6.2.a}$ What is the pH of the solution that is prepared by dissolving 0.62g of Na 2 O in enough water to make 1.0L? Answer \[Na_{2}O(s)+H_{2}O(l)\rightleftharpoons 2NaOH(aq) \nonumber \] \[NaOH(aq)\rightleftharpoons Na^{+}(aq)+OH^{-}(aq) \nonumber \] \[\left [NaOH\right ]=\left [OH^{-}\right ]=\frac{2\frac{0.62g}{62g/mol}}{1.0L}=0.02M \nonumber \] \[pH=14-log\left [OH^{-}\right ]=14.0-\left (-log\left (0.02M\right )\right )=12.3 \nonumber \] Exercise $\PageIndex{6.2.b}$ How many grams of Na 2 O is needed to make a 1.0L solution that has a pH=13.0? Answer \[pH=13.0 \nonumber \] \[\left [OH^{-}\right]=10^{-\left (14-13\right )}=0.10M \nonumber \] \[0.10M1.0L=0.10mol \nonumber \] \[Na_{2}O(s)+H_{2}O(l)\rightleftharpoons 2NaOH(aq) \nonumber \] \[\frac{Na_{2}O}{2NaOH}=\frac{x mol}{0.10mol} \nonumber \] \[x=0.05mol \nonumber \] \[0.05mol62g/mol=3.10g \nonumber \] Exercise $\PageIndex{6.2.c}$ What is the pH of the solution that is prepared by dissolving 0.56g of CaO in the water to make 1.0L? Answer \[CaO(s)+H_{2}O(l)\rightleftharpoons Ca\left (OH\right )_{2}(aq) \nonumber \] \[Ca\left (OH\right )_{2}(aq)\rightleftharpoons Ca^{2+}(aq)+2OH^{-}(aq) \nonumber \] \[2\left [Ca\left (OH\right )_{2}\right]=\left [OH^{-}\right ]=\frac{2\frac{0.56g}{56g/mol}}{1.0L}=0.02M \nonumber \] \[pH=14-\left (-log\left [ OH^{-} \right ]\right )=14.0-1.70=12.3 \nonumber \] Exercise $\PageIndex{6.2.d}$ How many grams of CaO is needed to make a 1.0L solution that has a pH = 13.0? Answer \[pH=13.0 \nonumber \] \[\left [OH^{-}\right]=10^{-\left ( 14-13 \right )}=0.10M \nonumber \] \[0.10M1.0L=0.10mol \nonumber \] \[CaO(s)+H_{2}O(l)\rightleftharpoons Ca\left (OH\right )_{2}(aq) \nonumber \] \[Ca\left (OH\right )_{2}(aq)\rightleftharpoons Ca^{2+}(aq)+2OH^{-}(aq) \nonumber \] \[\frac{2OH^{-}}{Ca\left (OH\right )_{2}}=\frac{Ca\left (OH\right )_{2}}{CaO}=\frac{0.10mol}{xmol} \nonumber \] \[x=0.05mol \nonumber \] \[0.05mol56g/mol=2.80g \nonumber \] Exercise $\PageIndex{6.2.e}$ If 100.0ml of the solution in Q 16.5.9 is transferred to a 500.0ml container, and plenty water was added to fill it up, what is the pH of the solution? Answer \[\left [OH^{-}\right]=10^{-\left ( 14-13 \right )}=0.10M \nonumber \] \[Ca\left (OH\right )_{2}(aq)\rightleftharpoons Ca^{2+}(aq)+2OH^{-}(aq) \nonumber \] \[\frac{\left [ Ca\left (OH\right )_{2} \right ]}{\left [ OH^{-} \right ]}=\frac{1}{2} \nonumber \] \[\left [ Ca\left (OH\right )_{2} \right ]=0.05M \nonumber \] \[\frac{0.05M0.10L}{0.50L}=0.01M \left [ OH^{-} \right ]=0.02M \nonumber \] \[pH=14-\left ( -log\left [ OH^{-} \right ] \right )=14.0-1.7=12.3 \nonumber \] 16.7: Lewis Concept of Acids and Bases Textbook: Section 16.7 Exercise $\PageIndex{7.a}$ CO can form complexes with metals, eg. Fe(CO) 5 , Ni(CO) 4 . Is CO a lewis acid or a lewis base? Answer CO has a lone pair of electrons so it is a Lewis base Exercise $\PageIndex{7.b}$ In Q 16.7.a, Is the metal, such as Fe(II), a lewis acid or a lewis base? Answer Fe(II) accepts the electron pair so it is a Lewis acid Exercise $\PageIndex{7.c}$ In the reaction, $Zn\left ( OH \right )_{2}(s)+2OH^{-}(aq)\rightleftharpoons Zn\left ( OH \right )_{4}^{2-}(aq)$, Which one is the lewis acid? Answer $Zn\left ( OH \right )_{2}$ accepts an electron pair so it is a lewis acid Exercise $\PageIndex{7.d}$ In the reaction, $CO_{2}+O^{2-}\rightarrow CO_{3}^{2-}$, Which one is the lewis acid and lewis base? Answer Lewis acid: $CO_{2}$ Lewis base: $O^{2-}$ Exercise $\PageIndex{7.e}$ Can CH 3 NH 2 be a lewis acid or a lewis base? Answer Lewis base Exercise $\PageIndex{7.f}$ Trimethylamine (CH 3 ) 3 N can react with diborane B 2 H 6 after its dissociation to form (CH 3 ) 3 N-BH 3 . Which one is the lewis acid? Which one is the lewis base? Answer Lewis acid: B 2 H 6 Lewis base: (CH 3 ) 3 N Exercise $\PageIndex{7.g}$ In the reaction, $H_{2}NOH(aq)+HCl(aq)\rightarrow \left [ H_{3}NOH \right ]Cl(aq)$, which one is the lewis acid? Answer $H_{2}NOH$ Exercise $\PageIndex{7.h}$ In the reaction, $SO_{2}(g)+BF_{3}(g)\rightarrow O_{2}S-BF_{3}(s)$, which one is the lewis acid? and the lewis base? Answer Lewis acid: BF 3 Lewis base: SO 2 Exercise $\PageIndex{7.i}$ Which one of the following cannot act as a Lewis base? Cl - NH 3 BF 3 CN - H 2 O Answer C. BF 3 General Questions Textbook: Section 16 Exercise $\PageIndex{8.a}$ A 0.1M aqueous solution of _____ will have the highest pH. KCN, Ka of HCN = 4.010 -10 NH 2 NO 3 , Kb of NH 3 = 1.810 -5 NaOAc, Ka of HOAc = 1.810 -5 NaClO, Ka of HClO = 3.210 -8 NaHS, Kb of HS - = 1.810 -7 Answer a. KCN, Ka of HCN = 4.010 -10 Exercise $\PageIndex{8.b}$ A 0.1M solution of _____ has a pH of 7.0. Na 2 S KF NaNO 3 NH 3 Cl NaF Answer c. NaNO 3 Exercise $\PageIndex{8.c}$ Which of the following possesses the greatest concentration of hydroxide ion? a solution with a pH of 3.0 a 1 10 -4 M solution of HNO 3 a solution with a pOH of 12.0 pure water a 1 * 10 -3 M solution of NH 4 Cl Answer c. solution with pH of 12.0 Textbook: 16: Acids and Bases Robert E. Belford (University of Arkansas Little Rock; Department of Chemistry). The breadth, depth and veracity of this work is the responsibility of Robert E. Belford, [email protected] . You should contact him if you have any concerns. This material has both original contributions, and content built upon prior contributions of the LibreTexts Community and other resources, including but not limited to: Liliane Poirot
Courses/Nassau_Community_College/Principles_of_Chemistry/09%3A_Gases/9.04%3A_Gas_Laws	Learning Objectives Learn what is meant by the term gas laws . Learn and apply Boyle's law. Learn and apply Charles's law. When seventeenth-century scientists began studying the physical properties of gases, they noticed some simple relationships between some of the measurable properties of the gas. Take pressure ( P ) and volume ( V ), for example. Scientists noted that for a given amount of a gas (usually expressed in units of moles [ n ]), if the temperature ( T ) of the gas was kept constant, pressure and volume were related: As one increases, the other decreases. As one decreases, the other increases. We say that pressure and volume are inversely related . There is more to it, however: pressure and volume of a given amount of gas at constant temperature are numerically related. If you take the pressure value and multiply it by the volume value, the product is a constant for a given amount of gas at a constant temperature: P × V = constant at constant n and T If either volume or pressure changes while amount and temperature stay the same, then the other property must change so that the product of the two properties still equals that same constant. That is, if the original conditions are labeled P 1 and V 1 and the new conditions are labeled P 2 and V 2 , we have P 1 V 1 = constant = P 2 V 2 where the properties are assumed to be multiplied together. Leaving out the middle part, we have simply P 1 V 1 = P 2 V 2 at constant n and T This equation is an example of a gas law. A gas law is a simple mathematical formula that allows you to model, or predict, the behavior of a gas. This particular gas law is called Boyle's law , after the English scientist Robert Boyle, who first announced it in 1662. Figure $\PageIndex{1}$ shows two representations of how Boyle's law works. Boyle's law is an example of a second type of mathematical problem we see in chemistry—one based on a mathematical formula. Tactics for working with mathematical formulas are different from tactics for working with conversion factors. First, most of the questions you will have to answer using formulas are word-type questions, so the first step is to identify what quantities are known and assign them to variables. Second, in most formulas, some mathematical rearrangements (i.e., algebra) must be performed to solve for an unknown variable. The rule is that to find the value of the unknown variable, you must mathematically isolate the unknown variable by itself and in the numerator of one side of the equation. Finally, units must be consistent. For example, in Boyle's law there are two pressure variables; they must have the same unit. There are also two volume variables; they also must have the same unit. In most cases, it won't matter what the unit is, but the unit must be the same on both sides of the equation. Example $\PageIndex{1}$ A sample of gas has an initial pressure of 2.44 atm and an initial volume of 4.01 L. Its pressure changes to 1.93 atm. What is the new volume if temperature and amount are kept constant? Solution First, determine what quantities we are given. We are given an initial pressure and an initial volume, so let these values be P 1 and V 1: P 1 = 2.44 atm and V 1 = 4.01 L We are given another quantity, final pressure of 1.93 atm, but not a final volume. This final volume is the variable we will solve for. P 2 = 1.93 atm and V 2 = ? L Substituting these values into Boyle's law, we get (2.44 atm)(4.01 L) = (1.93 atm) V 2 To solve for the unknown variable, we isolate it by dividing both sides of the equation by 1.93 atm—both the number and the unit: \[\frac{(2.44\, atm)(4.01\, L)}{1.93\, atm}=\frac{(1.93\, atm)\, V_{2}}{1.93\, atm}\nonumber \] Note that, on the left side of the equation, the unit atm is in the numerator and the denominator of the fraction. They cancel algebraically, just as a number would. On the right side, the unit atm and the number 1.93 are in the numerator and the denominator, so the entire quantity cancels: \[\frac{(2.44\, \cancel{atm})(4.01\, L)}{1.93\, \cancel{atm}}=\frac{(1.93\, \cancel{atm})\, V_{2}}{1.93\, \cancel{atm}}\nonumber \] What we have left is \[\frac{(2.44)(4.01\, L)}{1.93}=V_{2}\nonumber \] Now we simply multiply and divide the numbers together and combine the answer with the $L$ unit, which is a unit of volume. Doing so, we get $V_2 = 5.07\, L$ Does this answer make sense? We know that pressure and volume are inversely related; as one decreases, the other increases. Pressure is decreasing (from 2.44 atm to 1.93 atm), so volume should be increasing to compensate, and it is (from 4.01 L to 5.07 L). So the answer makes sense based on Boyle's law. Exercise $\PageIndex{1}$ If P 1 = 334 torr, V 1 = 37.8 mL, and P 2 = 102 torr, what is V 2 ? Answer 124 mL As mentioned, you can use any units for pressure or volume, but both pressures must be expressed in the same units, and both volumes must be expressed in the same units. Example $\PageIndex{2}$ A sample of gas has an initial pressure of 722 torr and an initial volume of 88.8 mL. Its volume changes to 0.663 L. What is the new pressure? Solution We can still use Boyle's law to answer this, but now the two volume quantities have different units. It does not matter which unit we change, as long as we perform the conversion correctly. Let us change the 0.663 L to milliliters: \[0.663\, L\times \frac{1000\, ml}{1\, L}=663\, ml\nonumber \] Now that both volume quantities have the same units, we can substitute into Boyle's law: \[(722\, torr)(88.8\, ml)=P_{2}(663\, ml)\nonumber \] \[\frac{(722\, torr)(88.8)\, ml}{(663\, ml)}=P_{2}\nonumber \] The mL units cancel, and we multiply and divide the numbers to get P 2 = 96.7 torr The volume is increasing, and the pressure is decreasing, which is as expected for Boyle's law. Exercise $\PageIndex{2}$ If V 1 = 456 mL, P 1 = 308 torr, and P 2 = 1.55 atm, what is V 2 ? Answer 119 mL There are other measurable characteristics of a gas. One of them is temperature ( T ). Perhaps one can vary the temperature of a gas sample and note what effect it has on the other properties of the gas. Early scientists did just this, discovering that if the amount of a gas and its pressure are kept constant, then changing the temperature changes the volume ( V ). As temperature increases, volume increases; as temperature decreases, volume decreases. We say that these two characteristics are directly related . A mathematical relationship between V and T should be possible except for one thought: what temperature scale should we use? We know from Chapter 2 that science uses several possible temperature scales. Experiments show that the volume of a gas is related to its absolute temperature in Kelvin , not its temperature in degrees Celsius . If the temperature of a gas is expressed in kelvins, then experiments show that the ratio of volume to temperature is a constant: \[\frac{V}{T}=constant\nonumber \] We can modify this equation as we modified Boyle's law: the initial conditions V 1 and T 1 have a certain value, and the value must be the same when the conditions of the gas are changed to some new conditions V 2 and T 2 , as long as pressure and the amount of the gas remain constant. Thus, we have another gas law: \[\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\; at\; constant\; P\; and\; n\nonumber \] This gas law is commonly referred to as Charles's law , after the French scientist Jacques Charles, who performed experiments on gases in the 1780s. The tactics for using this mathematical formula are similar to those for Boyle's law. To determine an unknown quantity, use algebra to isolate the unknown variable by itself and in the numerator; the units of similar variables must be the same. But we add one more tactic: all temperatures must be expressed in the absolute temperature scale (Kelvin). As a reminder, we review the conversion between the absolute temperature scale and the Celsius temperature scale: K = °C + 273 where K represents the temperature in kelvins, and °C represents the temperature in degrees Celsius. Example $\PageIndex{3}$ A sample of gas has an initial volume of 34.8 mL and an initial temperature of 315 K. What is the new volume if the temperature is increased to 559 K? Assume constant pressure and amount for the gas. Solution First, we assign the given values to their variables. The initial volume is V 1 , so V 1 = 34.8 mL, and the initial temperature is T 1 , so T 1 = 315 K. The temperature is increased to 559 K, so the final temperature T 2 = 559 K. We note that the temperatures are already given in kelvins, so we do not need to convert the temperatures. Substituting into the expression for Charles's law yields \[\frac{34.8\, ml}{315\, K}=\frac{V_{2}}{559\, K}\nonumber \] We solve for V 2 by algebraically isolating the V 2 variable on one side of the equation. We do this by multiplying both sides of the equation by 559 K (number and unit). When we do this, the temperature unit cancels on the left side, while the entire 559 K cancels on the right side: \[\frac{(559\cancel{K})(34.8\, ml)}{315\, \cancel{K}}=\frac{V_{2}(\cancel{559\, K})}{\cancel{559\, K}}\nonumber \] The expression simplifies to \[\frac{(559)(34.8\, ml)}{315}=V_{2}\nonumber \] By multiplying and dividing the numbers, we see that the only remaining unit is mL, so our final answer is V 2 = 61.8 mL Does this answer make sense? We know that as temperature increases, volume increases. Here, the temperature is increasing from 315 K to 559 K, so the volume should also increase, which it does. Exercise $\PageIndex{3}$ If V 1 = 3.77 L and T 1 = 255 K, what is V 2 if T 2 = 123 K? Answer 1.82 L It is more mathematically complicated if a final temperature must be calculated because the T variable is in the denominator of Charles's law. There are several mathematical ways to work this, but perhaps the simplest way is to take the reciprocal of Charles's law. That is, rather than write it as \[\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\nonumber \] write the equation as \[\frac{T_{1}}{V_{1}}=\frac{T_{2}}{V_{2}}\nonumber \] It is still an equality and a correct form of Charles's law, but now the temperature variable is in the numerator, and the algebra required to predict a final temperature is simpler. Example $\PageIndex{4}$ A sample of a gas has an initial volume of 34.8 L and an initial temperature of −67°C. What must the temperature of the gas be for its volume to be 25.0 L? Solution Here, we are looking for a final temperature, so we will use the reciprocal form of Charles's law. However, the initial temperature is given in degrees Celsius, not kelvins. We must convert the initial temperature to kelvins: −67°C + 273 = 206 K In using the gas law, we must use T 1 = 206 K as the temperature. Substituting into the reciprocal form of Charles's law, we get \[\frac{206\, K}{34.8\, L}=\frac{T_{2}}{25.0\, L}\nonumber \] Bringing the 25.0 L quantity over to the other side of the equation, we get \[\frac{(25.0\cancel{L})(206\, K)}{34.8\cancel{L}}=T_{2}\nonumber \] The L units cancel, so our final answer is T 2 = 148 K This is also equal to −125°C. As temperature decreases, volume decreases, which it does in this example. Exercise $\PageIndex{4}$ If V 1 = 623 mL, T 1 = 255°C, and V 2 = 277 mL, what is T 2 ? Answer 235 K, or −38°C Summary The behavior of gases can be modeled with gas laws. Boyle's law relates a gas's pressure and volume at constant temperature and amount. Charles's law relates a gas's volume and temperature at constant pressure and amount. In gas laws, temperatures must always be expressed in kelvins.
Courses/Nassau_Community_College/Principles_of_Chemistry/14%3A_Oxidation_and_Reduction/14.02%3A_Oxidation-Reduction_Reactions	Learning Objectives Define oxidation and reduction . Assign oxidation numbers to atoms in simple compounds. Recognize a reaction as an oxidation-reduction reaction. Consider this chemical reaction: \[\ce{Mg(s) + Cl2(g) → MgCl2}\nonumber \] The reactants are two electrically neutral elements; they have the same number of electrons as protons. The product, however, is ionic; it is composed of Mg 2 + and Cl − ions. Somehow, the individual Mg atoms lose two electrons to make the Mg 2 + ion, while the Cl atoms gain an electron to become Cl − ions. This reaction involves the transfer of electrons between atoms. The process of losing and gaining electrons occurs simultaneously. However, mentally we can separate the two processes. Oxidation is defined as the loss of one or more electrons by an atom. Reduction is defined as the gain of one or more electrons by an atom. So oxidation and reduction always occur together; it is only mentally that we can separate them. Chemical reactions that involve the transfer of electrons are called oxidation-reduction (or redox) reactions . Redox reactions require that we keep track of the electrons assigned to each atom in a chemical reaction. How do we do that? We use oxidation numbers to keep track of electrons in atoms. Oxidation numbers are assigned to atoms based on four rules. Oxidation numbers are not necessarily equal to the charge on the atom (although sometimes they can be); we must keep the concepts of charge and oxidation numbers separate. Assigning Oxidation Numbers The rules for assigning oxidation numbers to atoms are as follows: Atoms in their elemental state are assigned an oxidation number of 0. Atoms in monatomic (i.e., one-atom) ions are assigned an oxidation number equal to their charge. Oxidation numbers are usually written with the sign first, then the magnitude, to differentiate them from charges. In compounds, fluorine is assigned a −1 oxidation number; oxygen is usually assigned a −2 oxidation number (except in peroxide compounds [where it is −1] and in binary compounds with fluorine [where it is positive]); and hydrogen is usually assigned a +1 oxidation number (except when it exists as the hydride ion [H − ], in which case rule 2 prevails). In compounds, all other atoms are assigned an oxidation number so that the sum of the oxidation numbers on all the atoms in the species equals the charge on the species (which is zero if the species is neutral). Here are some examples for practice. In H 2 , both H atoms have an oxidation number of 0 by rule 1. In MgCl 2 , magnesium has an oxidation number of +2, while chlorine has an oxidation number of −1 by rule 2. In H 2 O, the H atoms each have an oxidation number of +1, while the O atom has an oxidation number of −2, even though hydrogen and oxygen do not exist as ions in this compound (rule 3). By contrast, by rule 3, each H atom in hydrogen peroxide (H 2 O 2 ) has an oxidation number of +1, while each O atom has an oxidation number of −1. We can use rule 4 to determine oxidation numbers for the atoms in SO 2 . Each O atom has an oxidation number of −2; for the sum of the oxidation numbers to equal the charge on the species (which is zero), the S atom is assigned an oxidation number of +4. Does this mean that the sulfur atom has a 4+ charge on it? No, it means only that the S atom is assigned a +4 oxidation number by our rules of apportioning electrons among the atoms in a compound. Example $\PageIndex{1}$ Assign oxidation numbers to the atoms in each substance. $\ce{Cl2}$ $\ce{GeO2}$ $\ce{Ca(NO3)2}$ Solution $\ce{Cl2}$ is the elemental form of chlorine. Rule 1 states each atom has an oxidation number of 0. By rule 3, oxygen is normally assigned an oxidation number of −2. For the sum of the oxidation numbers to equal the charge on the species (zero), the Ge atom is assigned an oxidation number of +4. $\ce{Ca(NO3)2}$ can be separated into two parts: the $\ce{Ca^{2+}}$ ion and the $\ce{NO_3^{−}}$ ion. Considering these separately, the $\ce{Ca^{2+}}$ ion has an oxidation number of +2 by rule 2. Now consider the $\ce{NO_3^{−}}$ ion. Oxygen is assigned an oxidation number of −2, and there are three of them. According to rule 4, the sum of the oxidation numbers on all atoms must equal the charge on the species, so we have the simple algebraic equation x + 3(−2) = −1 where x is the oxidation number of the N atom and the −1 represents the charge on the species. Evaluating for $x$, \[\begin{align} x + (−6) &= −1\\x &= +5 \end{align}\] Thus the oxidation number on the $\ce{N}$ atom in the $\ce{NO_3^{−}}$ ion is +5. Exercise $\PageIndex{1}$: Phosphoric Acid Assign oxidation numbers to the atoms in H 3 PO 4 . Answer H: +1; O: −2; P: +5 All redox reactions occur with a simultaneous change in the oxidation numbers of some atoms. At least two elements must change their oxidation numbers. When an oxidation number of an atom is increased in the course of a redox reaction, that atom is being oxidized . When an oxidation number of an atom is decreased in the course of a redox reaction, that atom is being reduced . Oxidation and reduction can also be defined in terms of increasing or decreasing oxidation numbers, respectively. Example $\PageIndex{2}$ Identify what is being oxidized and reduced in this redox reaction. \[\ce{2Na + Br2 → 2NaBr} \nonumber \nonumber \] Solution Both reactants are the elemental forms of their atoms, so the Na and Br atoms have oxidation numbers of 0. In the ionic product, the Na + ions have an oxidation number of +1, while the Br − ions have an oxidation number of −1. \[\ce{2\underset{0}{Na} + \underset{0}{Br_{2}} \rightarrow 2\underset{+1 -1}{NaBr} } \nonumber \] Sodium is increasing its oxidation number from 0 to +1, so it is being oxidized; bromine is decreasing its oxidation number from 0 to −1, so it is being reduced: Because oxidation numbers are changing, this is a redox reaction. The total number of electrons being lost by sodium (two, one lost from each Na atom) is gained by bromine (two, one gained for each Br atom). Exercise $\PageIndex{2}$ Identify what is being oxidized and reduced in this redox reaction. \[\ce{C + O2 → CO2}\nonumber \nonumber \] Answer $\ce{C}$ is being oxidized from 0 to +4. $\ce{O}$ is being reduced from 0 to −2. Oxidation reactions can become quite complex, as attested by the following redox reaction: \[\ce{6H^{+}(aq) + 2\underset{+7}{MnO_{4}^{-}}(aq) + 5\underset{-1}{H_{2}O_{2}}(l) -> 2\underset{+2}{Mn^{2+}}(aq) + 5\underset{0}{O_{2}}(g) + 8H_{2}O(l)} \nonumber \] To demonstrate that this is a redox reaction, the oxidation numbers of the species being oxidized and reduced are listed; can you determine what is being oxidized and what is being reduced? This is also an example of a net ionic reaction; spectator ions that do not change oxidation numbers are not displayed in the equation. Eventually, we will need to learn techniques for writing correct (i.e., balanced) redox reactions. Food and Drink Application: Fortifying Food with Iron Iron is an essential mineral in our diet; iron-containing compounds like the heme protein in hemoglobin could not function without it. Most biological iron has the form of the Fe 2 + ion; iron with other oxidation numbers is almost inconsequential in human biology (although the body does contain an enzyme to reduce Fe 3 + to Fe 2 + , so Fe 3 + must have some biological significance, albeit minor). To ensure that we ingest enough iron, many foods are enriched with iron. Although Fe 2 + compounds are the most logical substances to use, some foods use "reduced iron" as an ingredient (bread and breakfast cereals are the most well-known examples). Reduced iron is simply iron metal; iron is added as a fine metallic powder. The metallic iron is oxidized to Fe 2 + in the digestive system and then absorbed by the body, but the question remains: Why are we ingesting metallic iron? Why not just use Fe 2 + salts as an additive? Although it is difficult to establish conclusive reasons, a search of scientific and medical literature suggests a few reasons. One reason is that fine iron filings do not affect the taste of the product. The size of the iron powder (several dozen micrometers) is not noticeable when chewing iron-supplemented foods, and the tongue does not detect any changes in flavor that can be detected when using Fe 2 + salts. Fe 2 + compounds can affect other properties of foodstuffs during preparation and cooking, like dough pliability, yeast growth, and color. Finally, of the common iron substances that might be used, metallic iron is the least expensive. These factors appear to be among the reasons why metallic iron is the supplement of choice in some foods. Key Takeaways Oxidation-reduction (redox) reactions involve the transfer of electrons from one atom to another. Oxidation numbers are used to keep track of electrons in atoms. There are rules for assigning oxidation numbers to atoms. Oxidation is an increase in oxidation number (loss of electrons); reduction is a decrease in oxidation number (gain of electrons).
Courses/University_of_Arkansas_Little_Rock/IOST_Library/Spring_2023%3A_IoST/01%3A_Week_1/01.1%3A_Introduction_to_Python	Why Python? Python is a "high-level" (essentially human readable - not machine code) general purpose open source programming language that is used in a wide variety of applications like scientific computing, data analysis, web development and artificial intelligence. Because it is open source there are vast quantities of freely accessible libraries along containing code that can be integrated into projects. There are also robust programming communities of coders who freely share information over the web and thus there is an ecosystem we can tap into and potentially contribute to, as we develop our projects. Python will not be the only programming language we will need to use in this class, but it is an ideal language to learn coding with. By a high level program language we mean it is not running the hardware, but run in a virtual environment using the python interpreter (note, we capitalize Python when we refer to the language, and do not capitilize python when referring to the interpreter, which is essentially the virtual environment it runs in). Because of this, it can be run on a wide variety of platforms, like Windows, MacOS, Linux, and Unix. How was Python Created? Python is a general-purpose interpreted high-level programming language that was created in the Netherlands by the Dutch programmer Guido Van Rossum in 1991. It started as a hobby project for the programmer to avoid boredom and keep himself busy during the Christmas season. The name "Python" came from Guido's being a big fan of the comedy troupe "Monty Python's Flying Circus" from the 1970s. Python was mainly developed to help programmers express concepts in fewer lines of code and to make their code more readable. Python Versions Today, Python comes in many types among which are: CPython: Which is written in C programming language Jython: Which is written in Java programming language Brython: Which is "Browser Python" and runs in the browser MicroPython: Which runs on a microcontroller Programming in IoT IoT projects mainly consist of physical objects (hardware) interacting with and through commands (Software). The hardware stores the software and executes ("runs") the code. Generally, the hardware consists of electrical circuits that are controlled through instructions called machine code . Definition: Machine code Machine code is a programming language composed of binary numbers or bits (1's and 0's) and is the only language that a computer or a circuit can understand. For example, the letter "A" is represented as 01000001 in machine language, yet it is shown on the screen as "A". By coding with Python, we will not need to learn machine code. We will be writing in code that looks like human language and will leave the translation task to a compiler or an interpreter . Contributors Robert E. Belford (University of Arkansas Little Rock; Department of Chemistry). The breadth, depth and veracity of this work is the responsibility of Robert E. Belford, [email protected] . You should contact him if you have any concerns. This material has both original contributions, and content built upon prior contributions of the LibreTexts Community and other resources, including but not limited to: Emna Bouzid Below will be a series of links to various python resources. As the semester continues this page will be updated. The following are resources that can be of value to learning Python Books Conversational Python : An Introduction to Computer Science Experienced Through Conversational Banter, Jason B. Shepherd Web Resources PEP8 - Python Enhancement Proposals - Style Guide https://docs.python.org/3/tutorial LearnXinYminutes , where X=Python AskPython Google's Python Class https://pythontutor.com/ Python Debugger
Bookshelves/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/13%3A_Carbonyl_Compounds_III%3A_Reactions_at_the_-_Carbon/19.5____Halogenation_of_the_-_Carbon_and_Aldehydes_and_Ketones	A carbonyl containing compound with $\alpha$ hydrogens can undergo a substitution reaction with halogens. This reaction comes about because of the tendency of carbonyl compounds to form enolates in basic condition and enols in acidic condition. In these cases even weak bases, such as the hydroxide anion, is sufficient enough to cause the reaction to occur because it is not necessary for a complete conversion to the enolate. For this reaction Cl 2 , Br 2 or I 2 can be used as the halogens. General reaction Example Acid Catalyzed Mechanism Under acidic conditions the reaction occurs thought the formation of an enol which then reacts with the halogen. Step 1: Protonation of the carbonyl Step 2: Enol formation Step 3: S N 2 attack Step 4: Deprotonation Base Catalyzed Mechanism Under basic conditions the enolate forms and then reacts with the halogen. Note! This is base promoted and not base catalyzed because an entire equivalent of base is required. Step 1: Enolate formation Step 2: S N 2 attack Overreaction during base promoted α halogenation The fact that an electronegative halogen is placed on an α carbon means that the product of a base promoted α halogenation is actually more reactive than the starting material. The electron withdrawing effect of the halogen makes the α carbon even more acidic and therefor promotes further reaction. Because of this multiple halogenations can occur. This effect is exploited in the haloform reaction discussed later. If a monohalo product is required then acidic conditions are usually used. The Haloform Reaction Methyl ketones typically undergo halogenation three times to give a trihalo ketone due to the increased reactivity of the halogenated product as discussed above. This trihalomethyl group is an effective leaving group due to the three electron withdrawing halogens and can be cleaved by a hydroxide anion to effect the haloform reaction. The product of this reaction is a carboxylate and a haloform molecule (CHCl 3 , CHBr 3 , CHI 3 ). Overall the haloform reaction represents an effective method for the conversion of methyl ketones to carboxylic acids. Typically, this reaction is performed using iodine because the subsequent iodoform (CHI 3 ) is a bright yellow precipitate which is easily filtered off. General reaction Example: The Haloform Reaction Mechanism 1) Formation of the trihalo species 2) Nulceophilic attack on the carbonyl carbon 3) Removal of the leaving group 4) Deprotonation Exercise $\PageIndex{1}$ Please draw the products of the following reactions Answer
Courses/Westminster_College/CHE_261_-_Organic_Chemistry_I/08%3A_Substitution_and_Elimination_Reactions/8.02%3A_Two_Mechanistic_Models_for_Nucleophilic_Substitution	As we begin our study of nucleophilic substitution reactions, we will focus at first on simple alkyl halide compounds. While the specific reactions we'll initially consider do not occur in living things, it is nonetheless useful to start with alkyl halides as a model to illustrate some fundamental ideas that we must cover. Later, we will move on to apply what we have earned about alkyl halides to the larger and more complex biomolecules that are undergoing nucleophilic substitution right now in your own cells. The $S_N2$ mechanism You may recall from our brief introduction to the topic in chapter 6 that there are two mechanistic models for how a nucleophilic substitution reaction can proceed. In one mechanism, the reaction is concerted: it takes place in a single step, and bond-forming and bond-breaking occur simultaneously. This is illustrated by the reaction between chloromethane and hydroxide ion: Recall that the hydroxide ion in this reaction is acting as a nucleophile (an electron-rich, nucleus-loving species), the carbon atom of chloromethane is acting as an electrophile (an electron-poor species which is attracted to electrons), and the chloride ion is the leaving group (where the name is self-evident). Organic chemists refer to this mechanism by the term '$S_N2$', where S stands for 'substitution', the subscript N stands for 'nucleophilic', and the number 2 refers to the fact that this is a bimolecular reaction: the overall rate depends on a step in which two separate species collide. A potential energy diagram for this reaction shows the transition state (TS) as the highest point on the pathway from reactants to products. The geometry of an $S_N2$ reaction is specific: the reaction can only occur when the nucleophile collides with the electrophilic carbon from the opposite side relative to the leaving group. This is referred to as backside attack. Approach from the front side simply doesn't work: the leaving group - which, like the nucleophile is an electron-rich group - blocks the way. 2 The result of backside attack is that the bonding geometry at the electrophilic carbon inverts (turns inside-out) as the reaction proceeds. 3 The transition state of the reaction is illustrated by drawing dotted lines to represent the covalent bonds that are in the process of breaking or forming. Because the formal charge on the oxygen nucleophile changes from negative one to zero as the reaction proceeds, and conversely the charge on the chlorine leaving group changes from zero to negative one, at the transition state both atoms are shown bearing a partial negative charge (the symbol $\delta $-). One other drawing convention for transition states is to use brackets, with the double-dagger symbol in subscript. Notice that the transition state for an $S_N2$ reaction has trigonal bipyramidal geometry: the nucleophile, electrophile, and leaving group form a straight line, and the three substituents on carbon (all hydrogen atoms in this case) are arranged in the same plane at $120^{\circ}$ angles. What is the measure in degrees) of the $H-C-O$ angle in the $S_N2$ transition state illustrated above? Consider what would happen if we were to replace one of the hydrogen atoms in chloromethane with deuterium (the $^2H$ isotope), and one with tritium (the radioactive $^3H$ isotope). Now, because it has four different substituents, our carbon electrophile is a chiral center. We'll arbitrarily assume that we start with the $S$ enantiomer. As the hydroxide nucleophile attacks from the backside and the bonding geometry at carbon inverts, we see that the stereochemistry of the product reflects this inversion: we end up with the $R$ enantiomer of the chiral product. $S_N2$ reactions proceed with inversion of stereochemical configuration at the electrophilic carbon. video tutorial/animation: inversion of configuration during SN2 reactions The $S_N1$ mechanism A second model for the nucleophilic substitution reaction is called the $S_N1$ mechanism. The '1' in $S_N1$ indicates that the rate-determining step of the reaction is unimolecular: in other words, the rate-determining step involves a single molecule breaking apart (rather than two molecules colliding as was the case in the $S_N2$ mechanism.) In an $S_N1$ mechanism the carbon-leaving group bond breaks first, before the nucleophile approaches, resulting in formation of a carbocation intermediate (step 1): A carbocation is a powerful electrophile: because the carbon lacks a complete octet of valence electrons, it is 'electron-hungry'. In step 2, a lone pair of electrons on the water nucleophile fills the empty p orbital of the carbocation to form a new bond. Notice that this is actually a three-step mechanism, with a final, rapid acid-base step leading to the alcohol product. A potential energy diagram for this $S_N1$ reaction shows that each of the two positively-charged intermediate stages ($I_1$ and $I_2$ in the diagram) can be visualized as a valley in the path of the reaction, higher in energy than both the reactant and product but lower in energy than the transition states. The first, bond-breaking step is the slowest, rate-determining step - notice it has the highest activation energy and leads to the highest-energy species ($I_1$, the carbocation intermediate). Step 2 is rapid: a new covalent bond forms between a carbocation and and a water nucleophile , and no covalent bonds are broken. Recall from chapter 7 that Bronsted-Lowry proton transfer steps like step 3 are rapid, with low activation energies. The nucleophilic substitution reactions we have seen so far are examples of hydrolysis. This term is one that you will encounter frequently in organic and biological chemistry. Hydrolysis means 'breaking with water': in a hydrolysis reaction, a water molecule (or hydroxide ion) participates in the breaking of a covalent bond. There are many reaction types other than nucleophilic substitution that can accurately be described as hydrolysis, and we will see several examples throughout the remaining chapters of this book. Solvolysis is a more general term, used when a bond in a reagent is broken by a solvent molecule: usually, the solvent in question is water or an alcohol such as methanol or ethanol. Draw a mechanism for the $ S_N1$ solvolysis of tert-butyl chloride in methanol. What new functional group has been formed? We saw that $S_N2$ reactions result in inversion of stereochemical configuration at the carbon center. What about the stereochemical outcome of $S_N1$ reactions? Recall that a carbocation is $sp^2$-hybridized, with an empty p orbital perpendicular to the plane formed by the three sigma bonds: In the second step of an $S_N1$ reaction, the nucleophile can attack from either side of the carbocation (the leaving group is already gone, and thus cannot block attack from one side like in an $S_N2$ reaction). Consider an $S_N1$ reaction with a chiral tertiary alkyl chloride: Because the nucleophile is free to attack from either side of the carbocation electrophile, the reaction leads to a 50:50 mixture of two stereoisomeric products. In other words: In general nonenzymatic SN1 reaction can occur with either retention or inversion of configuration at the electrophilic carbon, leading to racemization if the carbon is chiral. For an example, consider the hydrolysis of (S)-3-chloro-3-methylhexane. The result of this (nonenzymatic) reaction is a racemic mixture of chiral alcohols. It is important to remember, however, that enzymatic reactions are in almost all cases very specific with regard to stereochemical outcome. A biochemical $S_N1$ reaction, as we shall see later, can result in either inversion or retention of configuration at the electrophilic carbon, but generally not a mixture of both: the two reactants are bound with specific geometry in the enzyme's active site, so that the nucleophile can approach from one side only. (The following exercises refer to nonbiological reactions) Draw a complete mechanism for the hydrolysis reaction in the previous figure, showing all bond-breaking and bond-forming steps, and all intermediate species. Draw structures representing TS1 and TS2 in the reaction. Use the solid/dash wedge convention to show three dimensions. What is the expected optical rotation of the product mixture? Could the two organic products be separated on a silica column chromatography? Draw the product(s) of the hydrolysis of (R)-3-chloro-3-methyl heptane. What can you predict, if anything, about the optical rotation of the product(s)? Draw the product(s) of the hydrolysis of (3R,5R)-3-chloro-3,5-dimethyl heptane. What can you predict, if anything, about the optical rotation of the product(s)? Before we go on to look at some actual biochemical nucleophilic substitution reactions, we first need to lay the intellectual groundwork by focusing more closely on the characteristics of the three principal partners in the nucleophilic substitution reaction: the nucleophile, the electrophile, and the leaving group. In addition, we need to consider the carbocation intermediate that plays such a key role in the $S_N1$ mechanism. For the sake of simplicity, we will continue to use simple, non-biological organic molecules and reaction examples as we work through the basic concepts. Video tutorial/animation: SN1 reactions Contributors Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
Courses/Eastern_Mennonite_University/EMU%3A_Chemistry_for_the_Life_Sciences_(Cessna)/12%3A_Organic_Chemistry%3A_Alkanes_and_Halogenated_Hydrocarbons/12.08_Halogenated_Hydrocarbons	Learning Objectives To name halogenated hydrocarbons given formulas and write formulas for these compounds given names. Many organic compounds are closely related to the alkanes. As we noted previously, alkanes react with halogens to produce halogenated hydrocarbons, the simplest of which have a single halogen atom substituted for a hydrogen atom of the alkane. Even more closely related are the cycloalkanes, compounds in which the carbon atoms are joined in a ring, or cyclic fashion. The reactions of alkanes with halogens produce halogenated hydrocarbons , compounds in which one or more hydrogen atoms of a hydrocarbon have been replaced by halogen atoms: The replacement of only one hydrogen atom gives an alkyl halide (or haloalkane) . The common names of alkyl halides consist of two parts: the name of the alkyl group plus the stem of the name of the halogen, with the ending -ide . The IUPAC system uses the name of the parent alkane with a prefix indicating the halogen substituents, preceded by number indicating the substituent’s location. The prefixes are fluoro -, chloro -, bromo -, and iodo -. Thus CH 3 CH 2 Cl has the common name ethyl chloride and the IUPAC name chloroethane. Alkyl halides with simple alkyl groups (one to four carbon atoms) are often called by common names. Those with a larger number of carbon atoms are usually given IUPAC names. Example $\PageIndex{1}$ Give the common and IUPAC names for each compound. CH 3 CH 2 CH 2 Br (CH 3 ) 2 CHCl Solution The alkyl group (CH 3 CH 2 CH 2 –) is a propyl group, and the halogen is bromine (Br). The common name is therefore propyl bromide. For the IUPAC name, the prefix for bromine (bromo) is combined with the name for a three-carbon chain (propane), preceded by a number identifying the carbon atom to which the Br atom is attached, so the IUPAC name is 1-bromopropane. The alkyl group [(CH 3 ) 2 CH–] has three carbon atoms, with a chlorine (Cl) atom attached to the middle carbon atom. The alkyl group is therefore isopropyl, and the common name of the compound is isopropyl chloride. For the IUPAC name, the Cl atom (prefix chloro -) attached to the middle (second) carbon atom of a propane chain results in 2-chloropropane. Exercise $\PageIndex{1}$ Give common and IUPAC names for each compound. CH 3 CH 2 I CH 3 CH 2 CH 2 CH 2 F Example $\PageIndex{2}$ Give the IUPAC name for each compound. Solution The parent alkane has five carbon atoms in the longest continuous chain; it is pentane. A bromo (Br) group is attached to the second carbon atom of the chain. The IUPAC name is 2-bromopentane. The parent alkane is hexane. Methyl (CH 3 ) and bromo (Br) groups are attached to the second and fourth carbon atoms, respectively. Listing the substituents in alphabetical order gives the name 4-bromo-2-methylhexane. Exercise $\PageIndex{2}$ Give the IUPAC name for each compound. A wide variety of interesting and often useful compounds have one or more halogen atoms per molecule. For example, methane (CH 4 ) can react with chlorine (Cl 2 ), replacing one, two, three, or all four hydrogen atoms with Cl atoms. Several halogenated products derived from methane and ethane (CH 3 CH 3 ) are listed in Table $\PageIndex{1}$, along with some of their uses. Formula Common Name IUPAC Name Some Important Uses Derived from CH4 Derived from CH4 Derived from CH4 Derived from CH4 CH3Cl methyl chloride chloromethane refrigerant; the manufacture of silicones, methyl cellulose, and synthetic rubber CH2Cl2 methylene chloride dichloromethane laboratory and industrial solvent CHCl3 chloroform trichloromethane industrial solvent CCl4 carbon tetrachloride tetrachloromethane dry-cleaning solvent and fire extinguishers (but no longer recommended for use) CBrF3 halon-1301 bromotrifluoromethane fire extinguisher systems CCl3F chlorofluorocarbon-11 (CFC-11) trichlorofluoromethane foaming plastics CCl2F2 chlorofluorocarbon-12 (CFC-12) dichlorodifluoromethane refrigerant Derived from CH3CH3 Derived from CH3CH3 Derived from CH3CH3 Derived from CH3CH3 CH3CH2Cl ethyl chloride chloroethane local anesthetic ClCH2CH2Cl ethylene dichloride 1,2-dichloroethane solvent for rubber CCl3CH3 methylchloroform 1,1,1-trichloroethane solvent for cleaning computer chips and molds for shaping plastics To Your Health: Halogenated Hydrocarbons Once widely used in consumer products, many chlorinated hydrocarbons are suspected carcinogens (cancer-causing substances) and also are known to cause severe liver damage. An example is carbon tetrachloride (CCl 4 ), once used as a dry-cleaning solvent and in fire extinguishers but no longer recommended for either use. Even in small amounts, its vapor can cause serious illness if exposure is prolonged. Moreover, it reacts with water at high temperatures to form deadly phosgene (COCl 2 ) gas, which makes the use of CCl 4 in fire extinguishers particularly dangerous. Ethyl chloride, in contrast, is used as an external local anesthetic. When sprayed on the skin, it evaporates quickly, cooling the area enough to make it insensitive to pain. It can also be used as an emergency general anesthetic. Bromine-containing compounds are widely used in fire extinguishers and as fire retardants on clothing and other materials. Because they too are toxic and have adverse effects on the environment, scientists are engaged in designing safer substitutes for them, as for many other halogenated compounds. To Your Health: Chlorofluorocarbons and the Ozone Layer Alkanes substituted with both fluorine (F) and chlorine (Cl) atoms have been used as the dispersing gases in aerosol cans, as foaming agents for plastics, and as refrigerants. Two of the best known of these chlorofluorocarbons (CFCs) are listed in Table $\PageIndex{1}$. Chlorofluorocarbons contribute to the greenhouse effect in the lower atmosphere. They also diffuse into the stratosphere, where they are broken down by ultraviolet (UV) radiation to release Cl atoms. These in turn break down the ozone (O 3 ) molecules that protect Earth from harmful UV radiation. Worldwide action has reduced the use of CFCs and related compounds. The CFCs and other Cl- or bromine (Br)-containing ozone-destroying compounds are being replaced with more benign substances. Hydrofluorocarbons (HFCs), such as CH 2 FCF 3 , which have no Cl or Br to form radicals, are one alternative. Another is hydrochlorofluorocarbons (HCFCs), such as CHCl 2 CF 3 . HCFC molecules break down more readily in the troposphere, and fewer ozone-destroying molecules reach the stratosphere. Figure $\PageIndex{1}$: Ozone in the upper atmosphere shields Earth’s surface from UV radiation from the sun, which can cause skin cancer in humans and is also harmful to other animals and to some plants. Ozone “holes” in the upper atmosphere (the gray, pink, and purple areas at the center) are large areas of substantial ozone depletion. They occur mainly over Antarctica from late August through early October and fill in about mid-November. Ozone depletion has also been noted over the Arctic regions. The largest ozone hole ever observed occurred on 24 September 2006. Source: Image courtesy of NASA, http://ozonewatch.gsfc.nasa.gov/daily.php?date=2006-09-24 . Concept Review Exercises What is the IUPAC name for the HFC that has the formula CH 2 FCF 3 ? (Hint: you must use a number to indicate the location of each substituent F atom.) What is the IUPAC name for the HCFC that has the formula CHCl 2 CF 3 ? Answers 1,1,1,2-tetrafluoroethane 1,1,1-trifluoro-2,2-dichloroethane Key Takeaway The replacement of an hydrogen atom on an alkane by a halogen atom—F, Cl, Br, or I—forms a halogenated compound. Exercises Write the condensed structural formula for each compound. methyl chloride chloroform Write the condensed structural formula for each compound. ethyl bromide carbon tetrachloride Write the condensed structural formulas for the two isomers that have the molecular formula C 3 H 7 Br. Give the common name and the IUPAC name of each. Write the condensed structural formulas for the four isomers that have the molecular formula C 4 H 9 Br. Give the IUPAC name of each. What is a CFC? How are CFCs involved in the destruction of the ozone layer? Explain why each compound is less destructive to the ozone layer than are CFCs. fluorocarbons HCFCs Answers CH 3 Cl CHCl 3 CH 3 CH 2 CH 2 Br, propyl bromide, 1-bromopropane; CH 3 CHBrCH 3 , isopropyl bromide, 2-bromopropane compounds containing Cl, F, and C; by releasing Cl atoms in the stratosphere
Courses/University_of_San_Diego/Fall_2024_Chem_220_Analytical_Chemistry_David_De_Haan/01%3A_Introduction_to_Analytical_Chemsitry	1.1: Introduction to Analytical Chemistry 1.2: Common Analytical Problems Typical problems on which analytical chemists work include qualitative analyses (Is lead present in this sample ?), quantitative analyses (How much lead is present in this sample?), characterization analyses (What are the sample’s chemical and physical properties?), and fundamental analyses (How does this method work and how can it be improved?). 1.3: Selecting an Analytical Method A method is the application of a technique to a specific analyte in a specific matrix. Ultimately, the requirements of the analysis determine the best method. In choosing among the available methods, we give consideration to some or all the following design criteria: accuracy, precision, sensitivity, selectivity, robustness, ruggedness, scale of operation, analysis time, availability of equipment, and cost. 1.4: Developing the Procedure After selecting a method, the next step is to develop a procedure that accomplish our goals for the analysis. In developing a procedure we give attention to compensating for interferences, to selecting and calibrating equipment, to acquiring a representative sample, and to validating the method.
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/09%3A_Chemical_Equilibria	As was discussed in Chapter 6, the natural tendency of chemical systems is to seek a state of minimum Gibbs function. Once the minimum is achieved, movement in any chemical direction will not be spontaneous. It is at this point that the system achieves a state of equilibrium. 9.1: Prelude to Chemical Equilibria The passage discusses the behavior of chemical systems, which naturally seek to achieve a state of minimum Gibbs function. When this minimum is reached, the system is in a state of equilibrium, and spontaneous chemical movement ceases. The direction of spontaneous change is related to the minimization of the partial derivative of Gibbs function with respect to the extent of reaction (??). A negative slope indicates a shift toward products, while a positive slope favors reactants. 9.2: Chemical Potential The page explains the concept of chemical equilibrium, focusing on how the chemical potential of reactants and products balance. Using the reaction A(g) ??? B(g), equilibrium is reached when chemical potentials equate. For ideal gases, this involves using mole fractions and equilibrium constants (K???). It discusses calculating the Gibbs free energy changes and their relation to K???, highlighting that ??G < 0 indicates a spontaneous reaction. 9.3: Activities and Fugacities The text discusses the concept of thermodynamic equilibrium constants, emphasizing that they are unitless and expressed in terms of activities and fugacities rather than concentrations or pressures. Activities of solids and pure liquids are considered unity, assuming ideal behavior, and don't affect the equilibrium constant's magnitude. 9.4: Pressure Dependence of Kp - Le Châtelier's Principle The page discusses the effect of pressure on chemical equilibria, focusing on Le Chatelier's principle which states that a system will adjust to reduce stress when disturbed. While the equilibrium constant K??? remains unchanged with pressure at a fixed temperature, pressure changes impact equilibrium mixtures' composition. 9.5: Degree of Dissociation Reactions such as the one in the previous example involve the dissociation of a molecule. Such reactions can be easily described in terms of the fraction of reactant molecules that actually dissociate to achieve equilibrium in a sample. This fraction is called the degree of dissociation. 9.6: Temperature Dependence of Equilibrium Constants - the van ’t Hoff Equation The page discusses the temperature dependence of the equilibrium constant using the van 't Hoff equation. It explains how the equilibrium constant is temperature-dependent due to changes in standard Gibbs energy, with temperature, unlike its pressure independence. 9.7: The Dumas Bulb Method for Measuring Decomposition Equilibrium The Dumas Bulb experiment is a method to study the temperature-dependent dissociation of N2O4 gas, commonly used in physical chemistry labs. The experiment measures the dissociation degree by equilibrium gas density at varying temperatures in a calibrated glass bulb. A formula calculates dissociation, factoring in measured versus theoretical densities, and an equilibrium constant is derived. The data enables the construction of a van't Hoff plot, providing insights into reaction enthalpy. 9.8: Acid-Base Equilibria This page explains the principles of acid-base reactions and proton transfer processes, emphasizing the importance of pH in biological systems. It covers the dissociation of weak acids using acetic acid as an example, demonstrating the calculation of pH using equilibrium constants. The auto-ionization of water is discussed, illustrating how water dissociates into ions and explaining pH variations with temperature. 9.9: Buffers Buffer solutions are essential for pH control in various processes and involve the equilibrium between a weak acid and its conjugate base. The pH of a buffer can be calculated using an ICE table or the Henderson-Hasselbalch approximation. The accuracy of these calculations relies on the pK??? and concentrations being sufficiently large to maintain equilibrium conditions close to their initial values. 9.10: Solubility of Ionic Compounds This page discusses the solubility of ionic compounds in water using equilibrium concepts. The solubility product, $K_{sp}$, is the equilibrium constant describing the solubility of an electrolyte. Two examples illustrate the calculations for solubility. 9.E: Chemical Equilibria (Exercises) Exercises for Chapter 9 "Chemical Equilibria" in Fleming's Physical Chemistry Textmap. 9.S: Chemical Equilibria (Summary) The page lists various vocabulary terms and concepts related to chemistry, such as common ion effect, conjugate base, degree of dissociation, and others. It also includes references to scientific literature and textbooks where these concepts may be explored in more detail. No further context or summary is provided regarding the terms themselves.
Bookshelves/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_I%3A_Structure_and_Reactivity_of_Carbohydrate_Radicals/07%3A_Radical_Philicity/V._Examples_of_Radical_Philicity_in_Reactions_of_Carbohydrates	A. Hydrogen-atom abstraction The reactions shown in equations 4 and 5 illustrate the importance of radical philicity in hydrogen-abstraction reactions by showing that the nucleophilic radical R· abstracts the electron-deficient hydrogen atom attached to sulfur (eq 5) more rapidly than the electron-rich hydrogen atom bonded to tin (eq 4). 33–35 The differences in rate constants and enthalpies for these two reactions underscore the fact that radical philicity affects the stability of the transition-state structure in a reaction but not the overall energy changes due to bond breaking and bond formation. Comparing the three reactions pictured in Scheme 1 draws attention to the effect of radical philicity on hydrogen-atom abstraction from carbohydrates. In each of these reactions the oxygen-centered radical 22 either abstracts a deuterium atom from Bu 3 SnD to give the deuterated alcohol 24 , or it reacts internally with H-3 to generate the carbon-centered radical 23 . 36 After the radical 23 forms, it then abstracts a deuterium atom for Bu 3 SnD to give the second reaction product ( 25 ). The relative amounts of products 24 and 25 provide a measure of external (deuterium) versus internal (hydrogen) abstraction. As H‑3 becomes less electron rich, internal reaction ( 22 $\rightarrow$ 23 ) becomes less competitive. External abstraction ( 22 $\rightarrow$ 24 ), on the other hand, should not be noticeably affected by changes in substituents at C-3. If, as expected, the transition states for the internal hydrogen-abstraction reactions shown in Scheme 1 are early, these reactions support the idea that polarity matching has a critical role in determining the favored reaction pathway for the radical 22 . B. Radical Addition Pyranos-1-yl radicals add readily to electron-deficient, carbon-carbon double bonds but are much less reactive toward double bonds lacking electron-withdrawing substituents. 37,38 A group of reactions that illustrates this difference in reactivity is found in Scheme 2. 37 The ability of the pyranos-1-yl radical 26 to add to the unsaturated compounds shown in Scheme 2 correlates with the reduction potentials of these compounds; that is, addition to compounds with less negative reduction potentials occurs more rapidly than addition to compounds with more negative reduction potentials. 37 Since the reduction potential in a substituted alkene is a measure of the ease of introducing an electron into a π* orbital, this potential becomes an indicator of energy-level positioning. When comparing two reduction potentials, the less negative one has a lower energy level for the π* orbital. Because this lower energy level causes the π* orbital (LUMO) to interact more effectively with the SOMO of the adding radical, transition-state stabilization due to frontier-orbital interaction increases as the reduction potential for the substituted alkene becomes less negative (Figure 11). Although the addition of a nucleophilic radical to a π bond that is not electron-deficient is too slow to be observed in the reactions shown in Scheme 2 , the situation changes when reactions become intramolecular. For π bonds that are 1,5- or 1,6-related to a radical center, intramolecular addition can take place even if the π bond is not decidedly electron-deficient (Scheme 3). 39 As far as overall reaction rate is concerned, forced, close proximity of the radical center to the π bond can compensate for a small transition-state stabilization caused by a large separation in energy levels of interacting, frontier orbitals. C. Polarity-Reversal Catalysis The philicity of radicals involved in hydrogen-abstraction reactions provides the basis for a phenomenon known as polarity-reversal catalysis. 4,33–35,40 This type of catalysis, which is responsible for the effect that thiols have on the reactions of carbohydrate acetals and ethers, is discussed in Section III of Chapter 5 in Volume II.
Courses/Clackamas_Community_College/CH_112%3A_Chemistry_for_Health_Sciences/17%3A_Amino_Acids_Proteins_and_Enzymes/17.02%3A_Properties_of_Amino_Acids	Learning Objectives To recognize amino acids and classify them based on the characteristics of their side chains. The proteins in all living species, from bacteria to humans, are constructed from the same set of 20 amino acids, so called because each contains an amino group attached to a carboxylic acid. The amino acids in proteins are α-amino acids, which means the amino group is attached to the α-carbon of the carboxylic acid. Humans can synthesize only about half of the needed amino acids; the remainder must be obtained from the diet and are known as essential amino acids. However, two additional amino acids have been found in limited quantities in proteins: Selenocysteine was discovered in 1986, while pyrrolysine was discovered in 2002. The amino acids are colorless, nonvolatile, crystalline solids, melting and decomposing at temperatures above 200°C. These melting temperatures are more like those of inorganic salts than those of amines or organic acids and indicate that the structures of the amino acids in the solid state and in neutral solution are best represented as having both a negatively charged group and a positively charged group. Such a species is known as a zwitterion. Classification In addition to the amino and carboxyl groups, amino acids have a side chain or R group attached to the α-carbon. Each amino acid has unique characteristics arising from the size, shape, solubility, and ionization properties of its R group. As a result, the side chains of amino acids exert a profound effect on the structure and biological activity of proteins. Although amino acids can be classified in various ways, one common approach is to classify them according to whether the functional group on the side chain at neutral pH is nonpolar, polar but uncharged, negatively charged, or positively charged. The structures and names of the 20 amino acids, their one- and three-letter abbreviations, and some of their distinctive features are given in Table $\PageIndex{1}$. Common Name Abbreviation Structural Formula (at pH 6) Molar Mass Distinctive Feature Amino acids with a nonpolar R group Amino acids with a nonpolar R group Amino acids with a nonpolar R group Amino acids with a nonpolar R group Amino acids with a nonpolar R group glycine gly (G) NaN 75 the only amino acid lacking a chiral carbon alanine ala (A) NaN 89 — valine val (V) NaN 117 a branched-chain amino acid leucine leu (L) NaN 131 a branched-chain amino acid isoleucine ile (I) NaN 131 an essential amino acid because most animals cannot synthesize branched-chain amino acids phenylalanine phe (F) NaN 165 also classified as an aromatic amino acid tryptophan trp (W) NaN 204 also classified as an aromatic amino acid methionine met (M) NaN 149 side chain functions as a methyl group donor proline pro (P) NaN 115 contains a secondary amine group; referred to as an α-imino acid Amino acids with a polar but neutral R group Amino acids with a polar but neutral R group Amino acids with a polar but neutral R group Amino acids with a polar but neutral R group Amino acids with a polar but neutral R group serine ser (S) NaN 105 found at the active site of many enzymes threonine thr (T) NaN 119 named for its similarity to the sugar threose cysteine cys (C) NaN 121 oxidation of two cysteine molecules yields cystine tyrosine tyr (Y) NaN 181 also classified as an aromatic amino acid asparagine asn (N) NaN 132 the amide of aspartic acid glutamine gln (Q) NaN 146 the amide of glutamic acid Amino acids with a negatively charged R group Amino acids with a negatively charged R group Amino acids with a negatively charged R group Amino acids with a negatively charged R group Amino acids with a negatively charged R group aspartic acid asp (D) NaN 132 carboxyl groups are ionized at physiological pH; also known as aspartate glutamic acid glu (E) NaN 146 carboxyl groups are ionized at physiological pH; also known as glutamate Amino acids with a positively charged R group Amino acids with a positively charged R group Amino acids with a positively charged R group Amino acids with a positively charged R group Amino acids with a positively charged R group histidine his (H) NaN 155 the only amino acid whose R group has a pKa (6.0) near physiological pH lysine lys (K) NaN 147 — arginine arg (R) NaN 175 almost as strong a base as sodium hydroxide The first amino acid to be isolated was asparagine in 1806. It was obtained from protein found in asparagus juice (hence the name). Glycine, the major amino acid found in gelatin, was named for its sweet taste (Greek glykys , meaning “sweet”). In some cases an amino acid found in a protein is actually a derivative of one of the common 20 amino acids (one such derivative is hydroxyproline). The modification occurs after the amino acid has been assembled into a protein. Configuration Notice in Table $\PageIndex{1}$ that glycine is the only amino acid whose α-carbon is not chiral . Therefore, with the exception of glycine, the amino acids could theoretically exist in either the D- or the L-enantiomeric form and rotate plane-polarized light. As with sugars, chemists used L-glyceraldehyde as the reference compound for the assignment of absolute configuration to amino acids. Its structure closely resembles an amino acid structure except that in the latter, an amino group takes the place of the OH group on the chiral carbon of the L-glyceraldehyde and a carboxylic acid replaces the aldehyde. Modern stereochemistry assignments using the Cahn-Ingold-Prelog priority rules used ubiquitously in chemistry show that all of the naturally occurring chiral amino acids are S except Cys which is R. We learned that all naturally occurring sugars belong to the D series. It is interesting, therefore, that nearly all known plant and animal proteins are composed entirely of L-amino acids. However, certain bacteria contain D-amino acids in their cell walls, and several antibiotics (e.g., actinomycin D and the gramicidins) contain varying amounts of D-leucine, D-phenylalanine, and D-valine. Summary Amino acids can be classified based on the characteristics of their distinctive side chains as nonpolar, polar but uncharged, negatively charged, or positively charged. The amino acids found in proteins are L-amino acids.
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/05%3A_The_Harmonic_Oscillator_and_the_Rigid_Rotor/5.07%3A_Hermite_Polynomials_are_either_Even_or_Odd_Functions	Understand key properties of the Hermite polynomials including orthogonality and symmetry. Be proficient at using symmetries of integrands to quickly solve integrals. Hermite polynomials were defined by Laplace (1810) though in scarcely recognizable form, and studied in detail by Chebyshev (1859). Chebyshev's work was overlooked and they were named later after Charles Hermite who wrote on the polynomials in 1864 describing them as new. They were consequently not new although in later 1865 papers Hermite was the first to define the multidimensional polynomials. The first six Hermite polynomial are plotted in Figure 5.7.1 . Any Hermite polynomial $H_n(x)$ can be generated from a previous one $H_{n-1}(x)$ via the following using the recurrence relation \[ H_{n+1} (x)=2xH_n (x)-2nH_{n-1} (x). \label{5.7.2} \] Hermite Polynomials are Symmetric Let $f(x)$ be a real-valued function of a real variable. Then $f$ is even if the following equation holds for all x and -x in the domain of f \[f(x) = f(-x) \nonumber \] Then $f$ is odd if the following equation holds for all x and -x in the domain of f \[-f(x) = f(-x) \nonumber \] Even and odd are terms used to describe particularly well-behaved functions. An even function is symmetric about the y-axis (Figure 5.7.2 ; left). That is, if we reflect the graph of the function in the $y$-axis, then it does not change. Formally, we say that $f$ is even if, for all $x$ and $−x$ in the domain of $f$, we have \[f(-x)=f(x) \nonumber \] Two examples of even functions are $f(x)=x^2$ and $f(x)=\cos x$. An odd function has rotational symmetry of order two about the origin (Figure 5.7.2 ; middle). That is, if we rotate the graph of the function 180° about the origin, then it does not change. Formally, we say that ff is odd if, for all $x$ and $−x$ in the domain of $f$, we have \[f(-x)=-f(x) \nonumber \] Examples of odd functions are $f(x)=x^3$ and $f(x)=\sin x$. Naturally, not all functions can be classified as even or odd. For example $f=x^3+1$ shown in the right side of Figure 5.7.2 , is neither. You can also think of these properties as symmetry conditions at the origin. More symmetries in 3D space are discussed in Group Theory. Without proof, we can identify several key features involving multiplication properties of even and odd functions: The product of two even functions is an even function. The product of two odd functions is an even function. The product of an even function and an odd function is an odd function. This can be shown graphically as a product table like that in Table 5.7.1 . Product table Odd Function (anti-symmetric) Even Function (symmetric) No symmetry (neither) Odd Function (anti-symmetric) Even Function (symmetric) Odd Function (anti-symmetric) who knows Even Function (symmetric) Odd Function (anti-symmetric) Even Function (symmetric) who knows No symmetry (neither) who knows who knows who knows Notice that the Hermite polynomials in Figure 5.7.1 oscillate from even to odd. We can take advantage of that aspect in our calculation of Harmonic Oscillator wavefunctions. Hermite Polynomial is an even or odd function depends on its degree $n$. Based on \[H_n(-x) = (-1)^n H_n(x) \label{5.7.3} \] $H_n(x)$ is an even function, when $n$ is even. $H_n(x)$ is an odd function, when $n$ is odd. Integration over Symmetric Functions You often consider integrals of the form \[I=\int_{-a}^a f(x)\,\mathrm{d}x \nonumber \] If $f$ is odd or even, then sometimes you can make solving this integral easier. For example, we can rewrite that integral in the following way: \[\begin{align} I=\int_{-a}^a f(x)\,\mathrm{d}x &= \int_{-a}^0 f(x)\,\mathrm{d}x + \int_0^a f(x)\,\mathrm{d}x \\ &= \int_0^a f(-x)\,\mathrm{d}x + \int_0^a f(x)\,\mathrm{d}x \label{intsym} \end{align} \] For an even function, we have $f(-x)=f(x)$ and Equation \ref{intsym} can be simplified \[\begin{align} I &= \int_0^a f(-x)\,\mathrm{d}x + \int_0^a f(x)\,\mathrm{d}x \\[4pt] &=2\int_0^a f(x)\,\mathrm{d}x \end{align} \] For an odd function, we have $f(-x)=-f(x)$ and Equation \ref{intsym} can be simplified \[\begin{align} I &= -\int_0^a f(x)\,\mathrm{d}x + \int_0^a f(x)\,\mathrm{d}x \\[4pt] &=0 \end{align} \] That’s what it means to simplify the integration: the integral of an odd or even function over the interval $[−L,L]$ can be put into a nicer form (and sometimes we can see that it vanishes without ever computing an integral). Technically, evaluating the orthogonality of Hermite polynomials requires integrating over the $\exp(-x^2)$ weight function (Equations $\ref{1}$ and $\ref{2}$). Solution For the Hermite polynomials $H_n(x)$, the relevant inner product (using Dirac Notation) \[ \langle f,g \rangle=\int_{-\infty}^\infty f(x)g(x)\color{red}{\exp(-x^2)}\,\mathrm dx \nonumber \] While the $H_2(x)H_3(x)$ product is indeed an odd function (Table 5.7.1 ), while $exp⁡(−x^2)$ is even. Their product is odd, and thus $\langle f,g \rangle$ certainly ought to be zero. Symmetry is an important aspect of quantum mechanics and mathematics, especially in calculating integrals. Using this symmetry, integrals can be identified to be equal to zero without explicitly solving them. For example, the integral of an odd integrand over all possible values will always be zero irrespective of the exact nature of the function: \[ \int_{-\infty}^{\infty} f(x) \, dx= 0 \nonumber \] This simplifies calculations greatly as demonstrated in the following chapters. Hermite Polynomials are Orthogonal Hermite polynomials $H_n(x)$ are n th-degree polynomials for n = 0, 1, 2, 3 and form an orthogonal set of functions for the weight function $e^{-x^2/2}$. The exact relation is: \[ \int_{-\infty}^{\infty} H_m(x)H_n(x) e^{-x^2/2} dx = 0 \label{1} \] if $m \neq n$ and \[ \int_{-\infty}^{\infty} H_m(x)H_n(x) e^{-x^2/2} dx = 2^n n! \sqrt{\pi} \label{2} \] if $m = n$. This will not be proved, but can the demonstrated using any of the Hermite polynomials listed in the previous section. The orthogonality property becomes important when solving the Harmonic oscillator problems. Note that the integral of Equation \ref{2} is important for normalizing the quantum harmonic oscillator wavefunctions discussed in last Section. Demonstrate that $H_2(x)$ and $H_3(x)$ are orthogonal. Solution We need to confirm \[\int_{-\infty}^{\infty}H_2(x)H_3(x) dx=0 \nonumber \] or when substituted \[\int_{-\infty}^{\infty} (4x^2-2)(8x^3-12x) dx=0 \nonumber \] because it says I need to show it's orthogonal on $ [ -\infty, \infty ] $ or we can just evaluate it on a finite interval $[−L,L]$, where $L$ is a constant. \[ \begin{align} \int_{-L}^{L} (4x^2-2)(8x^3-12x) dx &=\left. 8 \left(\dfrac{2 x^6}{3}-2 x^4+\dfrac{3 x^2}{2}\right)\right\|_{-L}^{L} \\[4pt] &= 8 \left(\dfrac{2 L^6}{3}-2 L^4+\frac{3 L^2}{2}\right)-8 \left(\dfrac{2 (-L)^6}{3}-2 (-L)^4+\dfrac{3 (-L)^2}{2}\right) \\[4pt] &=0. \end{align} \] Concluding Hermite polynomials are a component in the harmonic oscillator wavefunction that dictates the symmetry of the wavefunctions. If your integration interval is symmetric around 0, then the integral over any integrable odd function is zero, no exception. Therefore as soon as you've found that your integrand is odd and your integration interval is symmetric, you're done. Also, for general functions, if you can easily split them into even and odd parts, you only have to consider the integral over the even part for symmetric integration intervals. Another important property is that the product of two even or of two odd functions is even, and the product of an even and an odd function is odd. For example, if ff is even, $x↦f(x)\sin(x)$ is odd, and therefore the integral over it is zero (provided it is well defined).
Courses/Lumen_Learning/Book%3A_Western_Civilization_(Lumen)/12%3A_11%3A_The_Protestant_Reformation/12.2%3A_Luther_and_Protestantism	Martin Luther was a seminal figure in the Protestant Reformation; he strongly disputed the claim that freedom from God’s punishment for sin could be purchased with money, famously argued in his Ninety-five Theses of 1517. LEARNING OBJECTIVE Describe Luther’s criticisms of the Catholic Church KEY POINTS Martin Luther was a German professor of theology, composer, priest, monk and seminal figure in the Protestant Reformation. Luther strongly disputed the claim that freedom from God’s punishment for sin could be purchased with money, called indulgences, which he argued in his Ninety-five Theses of 1517. When confronted by the church for his critiques, he refused to renounce his writings and was excommunicated by the pope and deemed an outlaw by the emperor. Luther’s translation of the Bible into the vernacular made it more accessible to the laity, an event that had a tremendous impact on both the church and German culture. TERMS Ninety-five ThesesA list of propositions for an academic disputation written by Martin Luther in 1517. They advanced Luther’s positions against what he saw as abusive practices by preachers selling plenary indulgences, which were certificates that would reduce the temporal punishment for sins committed by the purchaser or their loved ones in purgatory. excommunicationAn institutional act of religious censure used to deprive, suspend, or limit membership in a religious community or to restrict certain rights within it. indulgencesA way to reduce the amount of punishment one has to undergo for sins, usually through the saying of prayers or good works, which during the middle ages included paying for church buildings or other projects. FULL TEXT Overview Martin Luther (November 10, 1483–February 18, 1546) was a German professor of theology, composer, priest, monk and seminal figure in the Protestant Reformation. Luther came to reject several teachings and practices of the Roman Catholic Church. He strongly disputed the claim that freedom from God’s punishment for sin could be purchased with money, proposing an academic discussion of the practice and efficacy of indulgences in his Ninety-five Theses of 1517. His refusal to renounce all of his writings at the demand of Pope Leo X in 1520 and the Holy Roman Emperor Charles V at the Diet of Worms in 1521 resulted in his excommunication by the pope and condemnation as an outlaw by the emperor. Luther taught that salvation and, subsequently, eternal life are not earned by good deeds but are received only as the free gift of God’s grace through the believer’s faith in Jesus Christ as redeemer from sin. His theology challenged the authority and office of the pope by teaching that the Bible is the only source of divinely revealed knowledge from God, and opposed priestly intervention for the forgiveness of sins by considering all baptized Christians to be a holy priesthood. Those who identify with these, and all of Luther’s wider teachings, are called Lutherans, though Luther insisted on Christian or Evangelical as the only acceptable names for individuals who professed Christ. His translation of the Bible into the vernacular (instead of Latin) made it more accessible to the laity, an event that had a tremendous impact on both the church and German culture. It fostered the development of a standard version of the German language, added several principles to the art of translation, and influenced the writing of an English translation, the Tyndale Bible. His hymns influenced the development of singing in Protestant churches. His marriage to Katharina von Bora, a former nun, set a model for the practice of clerical marriage, allowing Protestant clergy to marry. In two of his later works, Luther expressed antagonistic views toward Jews, writing that Jewish homes and synagogues should be destroyed, their money confiscated, and their liberty curtailed. Condemned by virtually every Lutheran denomination, these statements and their influence on antisemitism have contributed to his controversial status. Portrait of Martin Luther Martin Luther (1528) by Lucas Cranach the Elder. Portrait of Martin Luther’s face. Personal Life Martin Luther was born to Hans Luther and his wife Margarethe on November 10, 1483, in Eisleben, Saxony, then part of the Holy Roman Empire. Hans Luther was ambitious for himself and his family, and he was determined to see Martin, his eldest son, become a lawyer. In 1501, at the age of nineteen, Martin entered the University of Erfurt. In accordance with his father’s wishes, he enrolled in law school at the same university that year, but dropped out almost immediately, believing that law represented uncertainty. Luther sought assurances about life and was drawn to theology and philosophy, expressing particular interest in Aristotle, William of Ockham, and Gabriel Biel. He was deeply influenced by two tutors, Bartholomaeus Arnoldi von Usingen and Jodocus Trutfetter, who taught him to be suspicious of even the greatest thinkers and to test everything himself by experience. Philosophy proved to be unsatisfying, offering assurance about the use of reason but no assurance about loving God, which to Luther was more important. Reason could not lead men to God, he felt, and he thereafter developed a love-hate relationship with Aristotle over the latter’s emphasis on reason. For Luther, reason could be used to question men and institutions, but not God. Human beings could learn about God only through divine revelation, he believed, and scripture therefore became increasingly important to him. Luther dedicated himself to the Augustinian order, devoting himself to fasting, long hours in prayer, pilgrimage, and frequent confession. In 1507, he was ordained to the priesthood, and in 1508, von Staupitz, first dean of the newly founded University of Wittenberg, sent for Luther to teach theology. He was made provincial vicar of Saxony and Thuringia by his religious order in 1515. This meant he was to visit and oversee eleven monasteries in his province. Start of the Reformation In 1516, Johann Tetzel, a Dominican friar and papal commissioner for indulgences, was sent to Germany by the Roman Catholic Church to sell indulgences to raise money to rebuild St. Peter’s Basilica in Rome. Roman Catholic theology stated that faith alone, whether fiduciary or dogmatic, cannot justify man; justification rather depends only on such faith as is active in charity and good works. The benefits of good works could be obtained by donating money to the church. On October 31, 1517, Luther wrote to his bishop, Albert of Mainz, protesting the sale of indulgences. He enclosed in his letter a copy of his “Disputation of Martin Luther on the Power and Efficacy of Indulgences,” which came to be known as the Ninety-five Theses . Historian Hans Hillerbrand writes that Luther had no intention of confronting the church, but saw his disputation as a scholarly objection to church practices, and the tone of the writing is accordingly “searching, rather than doctrinaire.” Hillerbrand writes that there is nevertheless an undercurrent of challenge in several of the theses, particularly in Thesis 86, which asks, “Why does the pope, whose wealth today is greater than the wealth of the richest Crassus, build the basilica of St. Peter with the money of poor believers rather than with his own money?” The first thesis has become famous: “When our Lord and Master Jesus Christ said, ‘Repent,’ he willed the entire life of believers to be one of repentance.” In the first few theses Luther develops the idea of repentance as the Christian’s inner struggle with sin rather than the external system of sacramental confession. In theses 41–47 Luther begins to criticize indulgences on the basis that they discourage works of mercy by those who purchase them. Here he begins to use the phrase, “Christians are to be taught…” to state how he thinks people should be instructed on the value of indulgences. They should be taught that giving to the poor is incomparably more important than buying indulgences, that buying an indulgence rather than giving to the poor invites God’s wrath, and that doing good works makes a person better while buying indulgences does not. Luther objected to a saying attributed to Johann Tetzel that “As soon as the coin in the coffer rings, the soul from purgatory springs.” He insisted that, since forgiveness was God’s alone to grant, those who claimed that indulgences absolved buyers from all punishments and granted them salvation were in error. Luther closes the Theses by exhorting Christians to imitate Christ even if it brings pain and suffering, because enduring punishment and entering heaven is preferable to false security. It was not until January 1518 that friends of Luther translated the Ninety-five Theses from Latin into German and printed and widely copied it, making the controversy one of the first to be aided by the printing press. Within two weeks, copies of the theses had spread throughout Germany; within two months, they had spread throughout Europe. Ninety-five Theses 1517 Nuremberg printing of the Ninety-five Theses as a placard, now in the Berlin State Library. A photo of a 1517 manuscript of the Luther’s Ninety-five Thesis. Excommunication and Later Life On June 15, 1520, the pope warned Luther, with the papal bull Exsurge Domine, that he risked excommunication unless he recanted forty-one sentences drawn from his writings, including the Ninety-five Theses , within sixty days. That autumn, Johann Eck proclaimed the bull in Meissen and other towns. Karl von Miltitz, a papal nuncio , attempted to broker a solution, but Luther, who had sent the pope a copy of On the Freedom of a Christian in October, publicly set fire to the bull and decretals at Wittenberg on December 10, 1520, an act he defended in Why the Pope and his Recent Book are Burned and Assertions Concerning All Articles . As a consequence, Luther was excommunicated by Pope Leo X on January 3, 1521, in the bull Decet Romanum Pontificem . The enforcement of the ban on the Ninety-five Theses fell to the secular authorities. On April 18, 1521, Luther appeared as ordered before the Diet of Worms. This was a general assembly of the estates of the Holy Roman Empire that took place in Worms, a town on the Rhine. It was conducted from January 28 to May 25, 1521, with Emperor Charles V presiding. Prince Frederick III, Elector of Saxony, obtained a safe conduct for Luther to and from the meeting. Johann Eck, speaking on behalf of the empire as assistant of the Archbishop of Trier, presented Luther with copies of his writings laid out on a table and asked him if the books were his, and whether he stood by their contents. Luther confirmed he was their author, but requested time to think about the answer to the second question. He prayed, consulted friends, and gave his response the next day: Unless I am convinced by the testimony of the Scriptures or by clear reason (for I do not trust either in the pope or in councils alone, since it is well known that they have often erred and contradicted themselves), I am bound by the Scriptures I have quoted and my conscience is captive to the Word of God. I cannot and will not recant anything, since it is neither safe nor right to go against conscience. May God help me. Amen. Over the next five days, private conferences were held to determine Luther’s fate. The emperor presented the final draft of the Edict of Worms on May 25, 1521, which declared Luther an outlaw, banned his literature, and required his arrest: “We want him to be apprehended and punished as a notorious heretic.” It also made it a crime for anyone in Germany to give Luther food or shelter, and permitted anyone to kill Luther without legal consequence. By 1526, Luther found himself increasingly occupied in organizing a new church, later called the Lutheran Church, and for the rest of his life would continue building the Protestant movement. An apoplectic stroke on February 18, 1546, deprived him of his speech, and he died shortly afterwards, at 2:45 a.m., aged sixty-two, in Eisleben, the city of his birth. He was buried in the Castle Church in Wittenberg, beneath the pulpit. An interactive or media element has been excluded from this version of the text. You can view it online here: http://pb.libretexts.org/wcw/?p=368 Reluctant Revolutionary PBS Documentary about Martin Luther the “Reluctant Revolutionary.” Luther opposed the Catholic Church’s practices and in 1517 he wrote his Ninety-five Theses, which detailed the church’s failings. His actions led to the start of the Protestant Revolution. CC licensed content, Shared previously Boundless World History. Authored by : Boundless. Located at : https://www.boundless.com/world-history/textbooks/boundless-world-history-textbook/the-protestant-reformation-12/protestantism-56/luther-and-protestantism-1034-17635/ . License : CC BY-SA: Attribution-ShareAlike
Courses/Lumen_Learning/Book%3A_English_Composition_I-3_(Lumen)/20%3A_Research_Process%3A_Finding_Sources/19.3%3A_Style	The stage of proofreading is often focused solely on “correctness”: making sure that all the details are right, and that language is used according to the rules. Proofreading also offers a great opportunity to address more personal concerns, however. It’s a chance to focus on your style , and allows you to craft the final product that best represents your unique perspective. A writer’s style is what sets his or her writing apart. Style is the way writing is dressed up (or down) to fit the specific context, purpose, or audience. Word choice, sentence fluency, and the writer’s voice — all contribute to the style of a piece of writing. How a writer chooses words and structures sentences to achieve a certain effect is also an element of style. When Thomas Paine wrote “These are the times that try men’s souls,” he arranged his words to convey a sense of urgency and desperation. Had he written “These are bad times,” it’s likely he wouldn’t have made such an impact! Style is usually considered to be the province of literary writers. Novelists such as Ernest Hemingway and William Faulkner and poets such as Emily Dickinson and Walt Whitman are well known for their distinctive literary styles. But journalists, scientists, historians, and mathematicians also have distinctive styles, and they need to know how to vary their styles to fit different audiences. For example, the first-person narrative style of a popular magazine like National Geographic is quite different from the objective, third-person expository style of a research journal like Scientific American, even though both are written for informational purposes. Not just right and wrong Style is not a matter of right and wrong but of what is appropriate for a particular setting and audience. Consider the following two passages, which were written by the same author on the same topic with the same main idea, yet have very different styles: “Experiments show that Heliconius butterflies are less likely to ovipost on host plants that possess eggs or egg-like structures. These egg mimics are an unambiguous example of a plant trait evolved in response to a host-restricted group of insect herbivores.” “Heliconius butterflies lay their eggs on Passiflora vines. In defense the vines seem to have evolved fake eggs that make it look to the butterflies as if eggs have already been laid on them.” (Example from Myers, G. (1992). Writing biology: Texts in the social construction of scientific knowledge. Madison: University of Wisconsin Press. p. 150.) What changed was the audience. The first passage was written for a professional journal read by other biologists, so the style is authoritative and impersonal, using technical terminology suited to a professional audience. The second passage, written for a popular science magazine, uses a more dramatic style, setting up a conflict between the butterflies and the vines, and using familiar words to help readers from non-scientific backgrounds visualize the scientific concept being described. Each style is appropriate for the particular audience. Elements of style Many elements of writing contribute to an author’s style, but three of the most important are word choice, sentence fluency, and voice. Word Choice Good writers are concise and precise, weeding out unnecessary words and c hoosing the exact word to convey meaning. Precise words—active verbs, concrete nouns, specific adjectives—help the reader visualize the sentence. Good writers use adjectives sparingly and adverbs rarely, letting their nouns and verbs do the work. Good writers also choose words that contribute to the flow of a sentence. Polysyllabic words, alliteration, and consonance can be used to create sentences that roll off the tongue. Onomatopoeia and short, staccato words can be used to break up the rhythm of a sentence. Sentence Fluency Sentence fluency is the flow and rhythm of phrases and sentences. Good writers use a variety of sentences with different lengths and rhythms to achieve different effects. They use parallel structures within sentences and paragraphs to reflect parallel ideas, but also know how to avoid monotony by varying their sentence structures. Good writers also arrange their ideas within a sentence for greatest effect. They avoid loose sentences, deleting extraneous words and rearranging their ideas for effect. Many students initially write with a looser oral style, adding words on to the end of a sentence in the order they come to mind. This rambling style is often described as a “word dump” where everything in a student’s mind is dumped onto the paper in no particular order. There is nothing wrong with a word dump as a starting point: the advantage of writing over speaking is that writers can return to their words, rethink them, and revise them for effect. Tighter, more readable style results when writers choose their words carefully, delete redundancies, make vague words more specific, and use subordinate clauses and phrases to rearrange their ideas for the greatest effect. Voice Because voice is difficult to measure reliably, it is often left out of scoring formulas for writing tests. Yet voice is an essential element of style that reveals the writer’s personality. A writer’s voice can be impersonal or chatty, authoritative or reflective, objective or passionate, serious or funny. Strategies to Revise for Style Read an essay draft out loud, preferably to another person . Better yet, have another person read your draft to you. Note how that person interprets your words. Does it come across as you had meant it originally? If not, revise. Adopt a persona that’s related to your topic . Write from the perspective of this person you create: what language would a young woman who’d just spent two years in the peace corps use, for instance, if the essay were about the value of volunteer work? How would the words on the page of a project about gun control look coming from the perspective of a very conservative gun owner? Combine (some) short sentences, or break apart (some) long sentences . Sentence length variety is an asset to your readers, as noted above. If you find a stretch of your essay that uses many sentences of approximately the same length close together, focus on combining or breaking apart there. Punch up the word choice . Not every word in an essay can be a “special” word, nor should they be. But if your writing in an area feels a little flat, the injection of a livelier word can have strong rhetorical and emotional impact on your reader. Think of these words as jewels in the right setting. Often swapping out “to be” verbs (is, was, were, etc.) with more action-packed verbs has immediate, positive impact. Adjectives are also good candidates for updating–look for “things” and “stuff,” or “very” and “many,” to replace with more precise terminology. CC licensed content, Original Revision and Adaptation. Provided by : Lumen Learning. License : CC BY-NC-SA: Attribution-NonCommercial-ShareAlike CC licensed content, Shared previously Style. Authored by : Kathleen Cali. Provided by : Learn NC. Located at : http://www.learnnc.org/lp/editions/few/684 . Project : The five features of effective writing. License : CC BY-NC-SA: Attribution-NonCommercial-ShareAlike Image of Style. Authored by : Dr Case. Located at : https://flic.kr/p/b3Qrdt . License : CC BY-NC: Attribution-NonCommercial Image of Inspiration. Authored by : photosteve101. Located at : https://flic.kr/p/8UrMS1 . License : CC BY: Attribution

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