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https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_II%3A_Radical_Reactions_of_Carbohydrates/Appendix_I%3A_Hydrogen-Atom_Donors/V._Compounds_with_Boron%E2%80%93Hydrogen_Bonds

Phosphine-boranes ( ) ( ) are a group of compounds that have the abil­ity to react selectively with xanthates in the presence of compounds containing bromine or chlorine (but not iodine). For example, cyclo­hexyl bromide is recovered without change when it is added to the reaction shown in eq 14; in contrast, ­tri- -butyltin hydride and most other hydrogen-atom transfers used in radical reactions readily dehalo­genate bromides. If this lack of reactivity between alkyl bromides and phosphine-boranes extends to halo­gen­ated carbo­hydrates, it will make possible their chemo­selective deoxy­gen­ation without dehalogenation.

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties/Electronegativity/Allred-Rochow_Electronegativity

Allred-Rochow Electronegativity is a measure that determines the values of the electrostatic force exerted by the effective nuclear charge on the valence electrons. The value of the effective nuclear charges is estimated from . The higher charge, the more likely it will attract electrons. Although, Slater's rule are partly empirical. So the Allred-Rochow electronegativity is no more rigid than the . Pauling established Electronegativity as the "power" of an atom in a molecule to attract electron to itself. It is a measure of the atom's ability to attract electron to itself while the electron is still attached to another atom. The higher the values, the more likely that atom can pull electron from another atom and into itself. Electronegativity correlates with bond polarity ionization energy, electron affinity, effective nuclear charge, and atomic size. The periodic trend for electronegativity generally increases from left to right and decreases as it go down the group. The exception are Hydrogen and the noble gases because the noble gases are content with their filled outermost shells, and hydrogen cannot bear to lose a valence electron unlike the rest of the group 1 metals. The elements in the halogen group usually have the highest electronegativity values because they only need to attract one valence electron to complete the octet in their outer shell. Whereas the except for Hydrogen, are willing to give up their only valence electron so they can fulfill having a complete, filled outer shell. are rules that provides the values for the effective nuclear charge concept, or \(Z_{eff}\). These rules are based on experimental data for electron promotion and ionization energies, and \(Z_{eff}\) is determined from this equation: \[Z_{eff} = Z - S \label{A}\] where Through this equation, this tells us that electron may get reduced nuclear charge due to high shielding. Allred and Rochow used \(Z_{eff}\) because it is accurate due to the involvement of shielding that prevents electron to reach its true nuclear charge: \(Z\). When an atom with filled s-shell attracts electrons, those electrons will go to the unfilled p-orbital. Since the electrons have the same negative charge, they will not only repel each other, but also repel the electrons from the filled s-shell. This creates a shielding effect where the inner core electrons will shield the outer core electrons from the nucleus. Not only would the outer core electrons experience effective nuclear charge, but it will make them easily removed from the outer shell. Thus, It is easier for outer electrons to penetrate the p shell, which has little likelihood of being near the nuclear, rather than the s shell. Consider this, each of the outer electron in the (ns, np) group contributes S = 0.35, S = 0.85 in the (n - 1) shell, and S = 1.00 in the (n - 2) or lower shells. What is the \(Z_{eff}\) for the 4s electrons in Ca. Since \(\ce{Ca}\) has atomic number of 20, \(Z = 20\). : \[\begin{align*} Z_{eff} &= Z - S \\[4pt] &= 20 - ((8\times 0.85) + (10 \times 1.00)) \\[4pt] &= 3.2 \end{align*}\] So, Ca has a \(Z_{eff}\) of 3.2. Allred and Rochow were two chemists who came up with the Allred-Rochow Electronegativity values by taking the electrostatic force exerted by effective nuclear charge, Z , on the valence electron. To do so, they came up with an equation: \[\chi^{AR} = \left(\dfrac{3590 \times Z_{eff}}{r^2_{cov}}\right) + 0.744 \label{1}\] At the time, the values for the covalent radius, \(r_{cov}\), were inaccurate. Allred and Rochow added certain perimeters so that it would more closely correspond to Pauling's electronegativity scale. In this table, the electronegativities increases from left to right just like Pauling's scale because the \(Z\) is increasing. As we go down the group, it decreases because of the larger atomic size that increases the distance between the electrons and nucleus.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Part_V%3A__Reactivity_in_Organic%2C_Biological_and_Inorganic_Chemistry_3/RO._Reduction_and_Oxidation/RO9._Electron_Transfer%3A_Inner_Sphere

In some cases, electron transfers occur much more quickly in the presence of certain ligands. For example, compare the rate constants for the following two electron transfer reactions, involving almost exactly the same complexes: \[ Co (NH_3)_6^{3+} + Cr^{2+} \rightarrow Co^{2+} + Cr^{3+} + 6 NH_3 K = 10^{-4} M^{-1} s^{-1} \] \[ Co (NH_3)_5 Cl^{2+} + Cr^{2+} \rightarrow Co^{2+} + CrCl^{2+} + 6 NH_3 K = 6 \times 10^5 M^{-1} s^{-1} \] (Note: aqua ligands are omitted for simplicity. Ions, unless noted otherwise, are aqua complexes.) Notice two things: first, when there is a chloride ligand involved, the reaction is much faster. Second, after the reaction, the chloride ligand has been transferred to the chromium ion. Possibly, those two events are part of the same phenomenon. Similar rate enhancements have been reported for reactions in which other halide ligands are involved in the coordination sphere of one of the metals. In the 1960’s, of Stanford University proposed that halides (and other ligands) may promote electron transfer via bridging effects. What he meant was that the chloride ion could use one of its additional lone pairs to bind to the chromium ion. It would then be bound to both metals at the same time, forming a bridge between them. Perhaps the chloride could act as a conduit for electron transfer. The chloride might then remain attached to the chromium, to which it had already formed a bond, leaving the cobalt behind. Electron transfers that occur via ligands shared by the two metals undergoing oxidation and reduction are termed "inner sphere" electron transfers. Taube was awarded the Nobel Prize in chemistry in 1983; the award was based on his work on the mechanism of electron transfer reactions. Take another look at the two electron transfer reactions involving the cobalt and chromium ion, above. Other ligands can be involved in inner sphere electron transfers. These ligands include carboxylates, oxalate, azide, thiocyanate, and pyrazine ligands. All of these ligands have additional lone pairs with which to bind a second metal ion. Draw an example of each of the ligands listed above bridging between a cobalt(III) and chromium(II) aqua complex. Once the bridge is in place, the electron transfer may take place via either of two mechanisms. Suppose the bridging ligand is a chloride. The first step might actually involve an electron transfer from chlorine to the metal; that is, the chloride could donate one electron from one of its idle lone pairs. This electron could subsequently be replaced by an electron transfer from metal to chlorine. Alternatively, an electron might first be transferred from metal to chlorine, which subsequently passes an electron along to the other metal. In the case of chlorine, this idea may be unsatisfactory, becuase chlorine already has a full octet. Nevertheless, some of the other bridging ligands may have low-lying unoccupied molecular orbitals that could be populated by this extra electron, temporarily. ,

https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/01%3A_Introduction_-_The_Ambit_of_Chemistry/1.06%3A_Measurements_Quantities_and_Unity_Factors/1.6.01%3A_Measurements_Quantities_and_Unity_Factors_Lecture_Demonstrations

Convert 12 lb weight of bowling ball to g, showing unity factors 12.00 lb x 453.59237 g/lb = 5443 g V=4/3 π r V = 4/3 * 3.1416 * (1/2 * 8.59 in) = 4/3 * 3.1416 * 79.23 in =331.9 in Unity Factor: 2.54 cm = 1 in 331.9 in x (2.54 cm / 1 in) Note!!! 331.9 in x 16.39cm /in = 5439 cm Will the bowling ball float in water? Demo D = 5443 g / 5439 cm Too close to call. See What is the mass of hydrogen? Density of hydrogen at room temperature and 1 Atm = 0.082 g/L What is the volume in L, assuming same size as bowling ball? Unity Factors? 1 cm = 1 mL = 10 L (Note: 1 mL = 1 cm = “1 cc”) V (L) = 5500 cm x (1 L / 1000 cm ) m (g) = V (L) * D (g/L) = 5.500 L x 0.082 g/cm = 0.451 g Why does the Hydrogen balloon float? F = W = m g F = W = (0.451 g x 1 kg /1000g) * 9.8 m*s = = 0. 0044 N Force Upward: (Archimedes) D of air = 1.2 g /L; Archimedes Principle: buoyancy = mass of air displaced (6.6 g) F = m g F = ( 6.6 g x 1 kg /1000g) * 9.8 m*s = = 0. 065 N Net force = 0.065 N - 0.044 N upwards.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/16%3A_Solubility_and_Precipitation_Equilibria/16.6%3A_A_Deeper_Look%3A_Selective_Precipitation_of_Ions

The composition of relatively complex mixtures of metal ions can be determined using , a procedure for discovering the identity of metal ions present in the mixture (rather than quantitative information about their amounts). The procedure used to separate and identify more than 20 common metal cations from a single solution consists of selectively precipitating only a few kinds of metal ions at a time under given sets of conditions. Consecutive precipitation steps become progressively less selective until almost all of the metal ions are precipitated, as illustrated in \(\Page {1}\). Most metal chloride salts are soluble in water; only \(\ce{Ag^{+}}\), \(\ce{Pb^{2+}}\), and \(\ce{Hg2^{2+}}\) form chlorides that precipitate from water. Thus the first step in a qualitative analysis is to add about 6 M \(\ce{HCl}\), thereby causing \(\ce{AgCl}\), \(\ce{PbCl2}\), and/or \(\ce{Hg2Cl2}\) to precipitate. If no precipitate forms, then these cations are not present in significant amounts. The precipitate can be collected by filtration or centrifugation. Next, the acidic solution is saturated with \(\ce{H2S}\) gas. Only those metal ions that form very insoluble sulfides, such as \(\ce{As^{3+}}\), \(\ce{Bi^{3+}}\), \(\ce{Cd^{2+}}\), \(\ce{Cu^{2+}}\), \(\ce{Hg^{2+}}\), \(\ce{Sb^{3+}}\), and \(\ce{Sn^{2+}}\), precipitate as their sulfide salts under these acidic conditions. All others, such as \(\ce{Fe^{2+}}\) and \(\ce{Zn^{2+}}\), remain in solution. Once again, the precipitates are collected by filtration or centrifugation. Ammonia or \(\ce{NaOH}\) is now added to the solution until it is basic, and then \(\ce{(NH4)2S}\) is added. This treatment removes any remaining cations that form insoluble hydroxides or sulfides. The divalent metal ions \(\ce{Co^{2+}}\), \(\ce{Fe^{2+}}\), \(\ce{Mn^{2+}}\), \(\ce{Ni^{2+}}\), and \(\ce{Zn^{2+}}\) precipitate as their sulfides, and the trivalent metal ions \(\ce{Al^{3+}}\) and \(\ce{Cr^{3+}}\) precipitate as their hydroxides: \(\ce{Al(OH)3}\) and \(\ce{Cr(OH)3}\). If the mixture contains \(\ce{Fe^{3+}}\), sulfide reduces the cation to \(\ce{Fe^{2+}}\), which precipitates as \(\ce{FeS}\). The next metal ions to be removed from solution are those that form insoluble carbonates and phosphates. When \(\ce{Na2CO3}\) is added to the basic solution that remains after the precipitated metal ions are removed, insoluble carbonates precipitate and are collected. Alternatively, adding \(\ce{(NH4)2HPO4}\) causes the same metal ions to precipitate as insoluble phosphates. At this point, we have removed all the metal ions that form water-insoluble chlorides, sulfides, carbonates, or phosphates. The only common ions that might remain are any alkali metals (\(\ce{Li^{+}}\), \(\ce{Na^{+}}\), \(\ce{K^{+}}\), \(\ce{Rb^{+}}\), and \(\ce{Cs^{+}}\)) and ammonium (\(\ce{NH4^{+}}\)). We now take a second sample from the original solution and add a small amount of \(\ce{NaOH}\) to neutralize the ammonium ion and produce \(\ce{NH3}\). (We cannot use the same sample we used for the first four groups because we added ammonium to that sample in earlier steps.) Any ammonia produced can be detected by either its odor or a litmus paper test. A flame test on another original sample is used to detect sodium, which produces a characteristic bright yellow color. The other alkali metal ions also give characteristic colors in flame tests, which allows them to be identified if only one is present. Metal ions that precipitate together are separated by various additional techniques, such as forming complex ions, changing the pH of the solution, or increasing the temperature to redissolve some of the solids. For example, the precipitated metal chlorides of group 1 cations, containing \(\ce{Ag^{+}}\), \(\ce{Pb^{2+}}\), and \(\ce{Hg2^{2+}}\), are all quite insoluble in water. Because \(\ce{PbCl2}\) is much more soluble in hot water than are the other two chloride salts, however, adding water to the precipitate and heating the resulting slurry will dissolve any \(\ce{PbCl2}\) present. Isolating the solution and adding a small amount of \(\ce{Na2CrO4}\) solution to it will produce a bright yellow precipitate of \(\ce{PbCrO4}\) if \(\ce{Pb^{2+}}\) were in the original sample ( \(\Page {2}\)). As another example, treating the precipitates from group 1 cations with aqueous ammonia will dissolve any \(\ce{AgCl}\) because \(\ce{Ag^{+}}\) forms a stable complex with ammonia: \(\ce{[Ag(NH3)2]^{+}}\). In addition, \(\ce{Hg2Cl2}\) in ammonia \[\ce{2Hg2^{2+} \rightarrow Hg + Hg^{2+}} \nonumber \] to form a black solid that is a mixture of finely divided metallic mercury and an insoluble mercury(II) compound, which is separated from solution: \[\ce{Hg2Cl2(s) + 2NH3(aq) \rightarrow Hg(l) + Hg(NH_2)Cl(s) + NH^{+}4(aq) + Cl^{−}(aq)} \nonumber \] Any silver ion in the solution is then detected by adding \(\ce{HCl}\), which reverses the reaction and gives a precipitate of white \(\ce{AgCl}\) that slowly darkens when exposed to light: \[\ce{[Ag(NH3)2]^{+} (aq) + 2H^{+}(aq) + Cl^{−}(aq) \rightarrow AgCl(s) + 2NH^{+}4(aq)} \nonumber \] Similar but slightly more complex reactions are also used to separate and identify the individual components of the other groups. In qualitative analysis, the identity, not the amount, of metal ions present in a mixture is determined. The technique consists of selectively precipitating only a few kinds of metal ions at a time under given sets of conditions. Consecutive precipitation steps become progressively less selective until almost all the metal ions are precipitated. Other additional steps are needed to separate metal ions that precipitate together.

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/02%3A_Reaction_Rates/2.01%3A_Experimental_Determination_of_Kinetics/2.1.02%3A_Measuring_Reaction_Rates

The method for determining a reaction rate is relatively straightforward. Since a reaction rate is based on change over time, it must be determined from tabulated values or found experimentally. With the obtained data, it is possible to calculate the reaction rate either algebraically or graphically. What follows is general guidance and examples of measuring the rates of a reaction. Measuring time change is easy; a stopwatch or any other time device is sufficient. However, determining the change in concentration of the reactants or products involves more complicated processes. The change of concentration in a system can generally be acquired in two ways: For supplemental information relating to measuring reaction rates, such as the concentration of reactants, the role of catalysts, the characteristics of the . It does not matter whether an experimenter monitors the reagents or products because there is no effect on the overall reaction. However, since reagents decrease during reaction, and products increase, there is a sign difference between the two rates. Reagent concentration decreases as the reaction proceeds, giving a negative number for the change in concentration. The products, on the other hand, increase concentration with time, giving a positive number. Since the convention is to express the rate of reaction as a positive number, to solve a problem, set the overall rate of the reaction equal to the negative of a reagent's disappearing rate. The overall rate also depends on stoichiometric coefficients. It is worth noting that the process of measuring the concentration can be greatly simplified by taking advantage of the different physical or chemical properties (ie: phase difference, reduction potential, etc.) of the reagents or products involved in the reaction by using the above methods. We have emphasized the importance of taking the sign of the reaction into account to get a positive reaction rate. Now, we will turn our attention to the importance of stoichiometric coefficients. rate of reaction = \( - \dfrac{1}{a}\dfrac{ \Delta [A]}{ \Delta t} = - \dfrac{1}{b} \dfrac{\Delta [B]}{\Delta t} = \dfrac{1}{c}\dfrac{ \Delta [C]}{\Delta t} = \dfrac{1}{d}\dfrac{ \Delta [D]}{\Delta t} \) This formula can also be written as: rate of reaction = \( - \dfrac{1}{a} \) (rate of disappearance of A) = \( - \dfrac{1}{b} \) (rate of disappearance of B) = \( \dfrac{1}{c} \) (rate of formation of C) = \( \dfrac{1}{d} \) (rate of formation of D) Even though the concentrations of A, B, C and D may all change at different rates, there is only one average rate of reaction. To get this unique rate, choose any one rate and divide it by the stoichiometric coefficient. When the reaction has the formula: The general case of the unique average rate of reaction has the form: rate of reaction = \( - \dfrac{1}{C_{R1}}\dfrac{\Delta [R_1]}{\Delta t} = \dots = - \dfrac{1}{C_{Rn}}\dfrac{\Delta [R_n]}{\Delta t} = \dfrac{1}{C_{P1}}\dfrac{\Delta [P_1]}{\Delta t} = \dots = \dfrac{1}{C_{Pn}}\dfrac{\Delta [P_n]}{\Delta t} \)

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Chemical_Bonding/Fundamentals_of_Chemical_Bonding/Chemical_Bonds/Valence_Bond_Theory_and_Hybrid_Atomic_Orbitals

By applying the VSEPR theory, one deduces the following results: Hint: A chemical bond is due to the overlap of atomic orbitals. Molecular orbital theory considers the energy states of the molecule. Hint: Using three atomic orbitals generates three hybrid orbitals. Number of orbitals does not change in hybridization of atomic orbitals. Hint: The bond angles are expected to be less than 120 degrees. Since the lone electron pair in \(\ce{:SO2}\) and lone electron in \(\ce{.NO2}\) take up more space, we expect the structure to distort leaving a smaller angle than 120 between the bonds. Hint: Methane molecules are tetrahedral. The 4 \(\ce{H}\) atoms form a tetrahedron, and methane has a tetrahedral shape. Hint: All bond angles are 109.5 degrees, the ideal value for a symmetric tetrahedral structure. The structure of this ion is very similar to that of \(\ce{CH4}\). Hint: The sp hybrid orbitals are used by the \(\ce{C}\) atom. Sigma ( ) bonds are due to hybrid orbitals, and 2 orbitals are used for ( ) bonds. The two sigma bonds for each \(\ce{C}\) are due to overlap of hybrid orbitals of each \(\ce{C}\) atom. The \(\ce{C}\) atom has 3 ( ) bonds by using three hybrid orbitals and a ( ) bond, due to one 2 orbital. Hint: Its shape is octahedral. Since the \(\ce{S}\) atom uses hybrid orbitals, you expect the shape to be octahedral. The \(\ce{F}\) atoms form an octahedron around the sulfur. Hint: The \(\ce{P}\) atom uses hybrid orbitals. A total of 5 atomic orbitals are used in the hybridization: one 3 , one 3 and three 3 orbitals. The hybrid orbitals of \(\ce{P}\) give rise to a trigonal bipyramidal coordination around the \(\ce{P}\) atom. The energy of d orbitals in \(\ce{N}\) is not compatible with 2 and 2 orbitals for hybridization. Thus, you seldom encounter a compound with formula \(\ce{NX5}\) with \(\ce{N}\) as the central atom.

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Supercritical_Fluids/Case_Study%3A_Removing_caffeine_from_Coffee

molecules are naturally found in coffee beans, tea leaves, cocoa and a variety of exotic berries. When ingested, caffeine can act as a stimulant in humans or a toxin in small animals and insects. A certain portion of the human population can’t tolerate increased levels of caffeine in their body. They can experience extreme side effects including, but not limited to irritability, muscle twitching, dehydration, headaches, increased heart rate, and frequent urination. These side effects can be quite unpleasant, which is why many coffee manufacturers decaffeinate coffee. Decaffeination is a fairly easy process since caffeine is polar and water-soluble. The most popular methods of decaffeinating coffee today are, Swiss Water Processing, Ethyl Acetate Processing, Methylene Chloride Processing (Direct and Indirect), and Supercritical Carbon Dioxide Processing. The Swiss Water Processing method removes caffeine without using any chemicals, but instead applies the law of simple diffusion. First, unroasted (green) coffee beans are soaked in water until caffeine is dissolved in water. The beans are then discarded, and the solution of water, caffeine, and coffee solids is passed through a carbon filter. The carbon filter is made out of activated carbon, carbon that has been made porous through the process of carbonization (reacting carbon in anaerobic conditions until the gaps between carbon atoms are large enough to allow molecules to pass through). The activated carbon filter has holes large enough to allow water and coffee solids (smaller molecules) to pass through, but not caffeine (relatively larger molecule). After filtration, the mixture that is left is water saturated with coffee flavor molecules – referred as “coffee solids” by the manufacturers. The mixture creates a concentration gradient when added to a fresh batch of coffee beans. Concentration gradients take advantage of the law of simple diffusion- the movement of molecules from an area of high solute concentration to an area of low solute concentration in order to 'even out' the uneven distribution of molucules. Since the only difference between the mixture and the fresh coffee beans is the caffeine concentration, caffeine molecules will diffuse out of the beans into the mixture of coffee solids, leaving the coffee beans caffeine free. This method is repeated until the coffee beans are 99.9% decaffeinated, and the flavor is left intact. Ethyl Acetate occurs naturally in many fruits, which is why this method is often referred to as natural decaffeination. It is however much cheaper commercially to use synthetic ethyl acetate. This method requires a thorough steaming of the beans until swell. An ethyl acetate aqueous solution is used to wash the swollen beans repeatedly. Ethyl acetate is a polar molecule, which makes it a good solvent for capturing the polar caffeine molecules from the coffee beans (since 'like dissolves like'). The caffeine molecules bind to the ethyl acetate molecules, and migrate through the cell membranes of cells of the beans. The beans are once again steamed in order to eliminate any ethyl acetate that remains. This method decaffeinates the coffee beans by approximately 97%. Carbon dioxide supercritical fluid (temperature above 31.1 °C and pressure above 73 atm) exhibits both liquid and gas-like behavior. It behaves like gas, and permeates a porous substance, while also exhibiting liquid properties to dissolve substances. Although supercritical carbon dioxide is non-polar, and should only be able to dissolve non-polar substances, certain co-solvents, like water, can be added so that supercritical carbon dioxide can actually dissolve polar molecules like caffeine. Water is more polar than caffeine is, so supercritical carbon dioxide, in the presence of a co-solvent like water, will dissolve the more non-polar substance, in this case, caffeine. In order to use supercritical carbon dioxide to decaffeinate coffee beans, the beans are first steamed until they swell (this is where the co-solvent, water, comes into play). After this, they are immersed in supercritical carbon dioxide which binds to the caffeine molecules and draws them out of the beans, leaving the coffee solids (flavor) embedded in the bean. The resulting coffee beans are about 97% caffeine free. The carbon dioxide is then passed through a charcoal membrane that is selective toward carbon dioxide molecules. Caffeine is stopped by the membrane, because of its larger size relative to carbon dioxide, and collected. Once coffee beans have been decaffeinated, all of the extracted caffeine is made into a white powder and sold to the pharmaceutical or food industries. The pharmaceutical industry adds caffeine into certain drugs, including many pain killers. Food industries add caffeine to certain foods like soda, because of its stimulating effect.

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/06%3A_Structures_and_Energetics_of_Metallic_and_Ionic_solids/6.03%3A_The_Packing_of_Spheres_Model_Applied_to_the_Structures_of_Elements/6.3B%3A_H_and_F_Solids

Hydrogen and Fluorine, as well as Chlorine, Bromine and Iodine, exist as diatomic molecules. Both diatomic Hydrogen and diatomic Florine exist as gasses at room temperature and exhibit extremely low melting and boiling points. If the temperature of the system is lowered below the respective melting points of Hydrogen or Fluorine, molecular motion is reduced and a crystalline lattice is formed, creating a solid. Elemental Hydrogen and Elemental Fluorine, H and F respectively, exist as gases at room temperature. Though the melting point of H is 20.4 K (-252.75 °C), a temperature of 14.0 K (-259.15 °C) is required for the solidification of H . As the solid structure is formed, the diatomic Hydrogen molecules adopt a hexagonal close packing (hcp) structure. Fluorine exhibits similar behaviour. The melting point of Fluorine is 53.54K (-219.61 °C), at which point the diatomic Fluorine adopts a cubic close packed crystalline arrangement (ccp), instead of a hexagonal close packed arrangement. When solid, both dihydrogen and difluorine are small enough to allow for rotation within the solid structure (the radius of Fluorine is about equivalent to that of Hydrogen). This rotation occurs about a central axis midway between the two distal atoms of the molecule, creating two equal radii from the middle of the H-H or F-F bond to the outer boundary of each H or F atom. This rotation occurs 360° in the x, y and z directions of the central axis, as well as all manner of combinations of these three variables. The rotation of the molecule in all directions creates what can be though of as a spherical shell, the boundary of which is created by the distal atoms. Some common uses for solid hydrogen include gamma-ray ablation cages for elemental analysis, and a solid state cage in which to look at hydrogen bonding within molecules. 2. What is the different between a cubic close packing structure and hexagonal close packing? 3. Why do difluorine and dihydrogen exhibit different packing structures.

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Halogenation of saturated aldehydes and ketones usually occurs exclusively by replacement of hydrogens to the carbonyl group: The reagents that commonly are used to halogenate carbonyl compounds are those that are used to halogenate alkanes (e.g. \(\ce{Cl_2}\), \(\ce{Br_2}\), \(\ce{SO_2Cl_2}\), and -bromoamides; see and ). However, the characteristics of the two types of halogenation normally are very different. 2-Propanone has been particularly well studied, and the important features of the halogenation of this compound are summarized as follows: \[v = k \left[ \ce{CH_3COCH_3} \right] \left[ \ce{OH}^\ominus \right]\] \[v = k' \left[ \ce{CH_3COCH_3} \right] \left[ \ce{H}^\oplus \right]\] The ratio of \(k\) to \(k'\) is 12,000, which means that hydroxide ion is a much more effective catalyst than is hydrogen ion. The hydroxide ion is a much more effective catalyst than is hydrogen ion To account for the role of the catalysts and the independence of the rate from the halogen concentration, the ketone necessarily must be slowly converted by the catalysts to something that can react with halogen to give the products. This something is either the enol or the enolate anion of 2-propanone: As long as the first step is slow compared with the steps of Equations 17-2 and 17-3, the overall rate of reaction will be independent of both the concentration of halogen and whether it is chlorine, bromine, or iodine (cf. ). The reaction of either the enol or the enolate anion (Equations 17-2 or 17-3) with \(\ce{Br_2}\) resembles the first step in the electrophilic addition of halogens to carbon-carbon multiple bonds ( ). However, the second step, addition of the nucleophilic halide, if it occurs at all, does not produce any stable product: Unsymmetrical ketones, such as 2-butanone, can form two different enols that will react with halogens to give isomeric halo ketones: The composition of the product mixture will depend on the relative rates of formation of the isomeric enols, provided that the halogenation step is not a reversible reaction. Barring any serious steric effects that influence the rate of reaction, the more rapidly formed enol generally is the more thermodynamically stable enol. The previous discussion of the halogenation of ketones is incomplete in one important respect concerning halogenation. That is, once an \(\alpha\)-halo ketone is formed, the other hydrogens on the same carbon are rendered more acidic by the electron-attracting effect of the halogen and are replaced much more rapidly than the first hydrogen: The result is that, if the monobromoketone is desired, the reaction is carried out best with an catalyst rather than a basic catalyst. A further complication in the base-catalyzed halogenation of a methyl ketone is that the trihaloketone formed is attacked readily by base, thereby resulting in cleavage of a carbon-carbon bond: This sequence is called the because it results in the production of chloroform, bromoform, or iodoform, depending upon the halogen used. The haloform reaction is a useful method for identification of methyl ketones, particularly when iodine is used, because iodoform is a highly insoluble, bright-yellow solid. The reaction also is very effective for the synthesis of carboxylic acids when the methyl ketone is more available than the corresponding acid: Because the haloform reaction is fast, in some cases it can be used to prepare unsaturated acids from unsaturated ketones without serious complications caused by addition of halogen to the double bond: A reaction somewhat similar to the cleavage of haloforms with hydroxide occurs with ketones that do not have \(\alpha\)-hydrogens through the action of sodium amide: This reaction, called the , has utility for the preparation of amides of the types \(\ce{ArCONH_2}\) and -\(\ce{RCONH_2}\), and, through hydrolysis, the corresponding carboxylic acids. The halogen of an \(\alpha\)-halo aldehyde or an \(\alpha\)-halo ketone is exceptionally in \(S_\text{N}1\)-displacement reactions, but is exceptionally in \(S_\text{N}2\) displacements, compared with the halogen of alkyl halides having comparable potential steric effects. Similar behavior is observed with \(\alpha\)-halo carboxylic acids and is discussed in . In some circumstances, the production of a 2-halo alcohol by reduction of the carbonyl group of an \(\alpha\)-halo ketone with metal hydrides is a useful synthetic reaction: When one attempts \(E2\) reactions with \(\alpha\)-halo ketones using strong bases such as alkoxides, an interesting rearrangement pathway may occur called the . In this reaction, the \(\alpha\)-halo ketone is converted to an ester. For example, 2-chlorocyclohexanone is converted to the methyl ester of cyclopentanecarboxylic acid by treatment with sodium methoxide in ether: The mechanism of this reaction has been the subject of many investigations. and (1977)

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Raoult's law states that the vapor pressure of a solvent above a solution is to the vapor pressure of the pure solvent at the same temperature by the mole fraction of the solvent present: In the 1880s, French chemist François-Marie Raoult discovered that when a substance is dissolved in a solution, the vapor pressure of the solution will generally decrease. This observation depends on two variables: At any given temperature for a particular solid or liquid, there is a pressure at which the vapor formed above the substance is in dynamic equilibrium with its liquid or solid form. This is the vapor pressure of the substance at that temperature. At equilibrium, the rate at which the solid or liquid evaporates is equal to the rate that the gas is condensing back to its original form. All solids and liquids have a vapor pressure, and this pressure is constant regardless of how much of the substance is present. Raoult's Law only works for ideal solutions. "An ideal solution shows thermodynamic mixing characteristics identical to those of ideal gas mixtures [except] ideal solutions have intermolecular interactions equal to those of the pure components." Like many other concepts explored in Chemistry, Raoult's Law only applies under ideal conditions in an ideal solution. However, it still works fairly well for the solvent in dilute solutions. In reality though, the decrease in vapor pressure will be greater than that calculated by Raoult's Law for extremely dilute solutions. If you look review the concepts of colligative properties, you will find that adding a solute because the additional solute particles will fill the gaps between the solvent particles and take up space. This means less of the solvent will be on the surface and less will be able to break free to enter the gas phase, resulting in a lower vapor pressure. There are two ways of explaining why Raoult's Law works - a simple visual way, and a more sophisticated way based on entropy. Below is the simple approach. Remember that saturated vapor pressure is what you get when a liquid is in a sealed container. An equilibrium is set up where the number of particles breaking away from the surface is exactly the same as the number sticking on to the surface again. Now suppose solute molecules were added so that the solvent molecules occupied only 50% of the surface of the solution. A certain fraction of the solvent molecules will have sufficient energy to escape from the surface (e.g., 1 in 1000 or 1 in a million). If you reduce the number of solvent molecules on the surface, you are going to reduce the number which can escape in any given time. But it will not make any difference to the ability of molecules in the vapor to stick to the surface again. If a solvent molecule in the vapor hits a bit of surface occupied by the solute particles, it may well stick. There are obviously attractions between solvent and solute otherwise you would not have a solution in the first place. The net effect of this is that when equilibrium is established, there will be solvent molecules in the vapor phase - it is less likely that they are going to break away, but there is not any problem about them returning. However, if there are fewer particles in the vapor at equilibrium, the saturated vapor pressure is lower. In practice, there's no such thing as an ideal solution! However, features of one include: Suppose that in the pure solvent, 1 in 1000 molecules had enough energy to overcome the intermolecular forces and break away from the surface in any given time. In an ideal solution, that would still be exactly the same proportion. Fewer would, of course, break away because there are now fewer solvent molecules on the surface - but of those that are on the surface, the same proportion still break away. If there were strong solvent-solute attractions, this proportion may be reduced to 1 in 2000, or 1 in 5000 or whatever. In any real solution of, say, a salt in water, there are strong attractions between the water molecules and the ions. That would tend to slow down the loss of water molecules from the surface. However, if the solution is sufficiently dilute, there will be good-sized regions on the surface where you still have water molecules on their own. The solution will then approach ideal behavior. There is another thing that you have to be careful of if you are going to do any calculations on Raoult's Law. You may have noticed in the little calculation about mole fraction further up the page, that sugar was as a solute rather than salt. What matters is not actually the number of moles of substance that you put into the solution, but the of moles of particles formed. For each mole of sodium chloride dissolved, you get 1 mole of sodium ions and 1 mole of chloride ions - in other words, you get twice the number of moles of particles as of original salt. So, if you added 0.1 moles of sodium chloride, there would actually be 0.2 moles of particles in the solution - and that's the figure you would have to use in the mole fraction calculation. Unless you think carefully about it, Raoult's Law only works for solutes which change their nature when they dissolve. For example, they must not ionize or associate (e.g., if you put in substance A, it must not form A in solution). If it does either of these things, you have to treat Raoult's law with great care. What matters is not actually the number of moles of substance that you put into the solution, but the of moles of particles formed. The effect of Raoult's Law is that the saturated vapor pressure of a solution is going to be lower than that of the pure solvent at any particular temperature. That has important effects on the phase diagram of the solvent. The next diagram shows the phase diagram for pure water in the region around its normal melting and boiling points. The 1 atmosphere line shows the conditions for measuring the normal melting and boiling points. The line separating the liquid and vapor regions is the set of conditions where liquid and vapor are in equilibrium. It can be thought of as the effect of pressure on the boiling point of the water, but it is also the curve showing the effect of temperature on the saturated vapor pressure of the water. These two ways of looking at the same line are discussed briefly in a note about half-way down the page about phase diagrams (follow the last link above). If you draw the saturated vapor pressure curve for a solution of a non-volatile solute in water, it will always be lower than the curve for the pure water. If you look closely at the last diagram, you will see that the point at which the liquid-vapor equilibrium curve meets the solid-vapor curve has moved. That point is the triple point of the system - a unique set of temperature and pressure conditions at which it is possible to get solid, liquid and vapor all in equilibrium with each other at the same time. Since the triple point has solid-liquid equilibrium present (amongst other equilibria), it is also a melting point of the system - although not the normal melting point because the pressure is not one atmosphere. The curves for the pure water and for the solution are often drawn parallel to each other. That has got to be wrong! Suppose you have a solution where the mole fraction of the water is 0.99 and the vapor pressure of the pure water at that temperature is 100 kPa. The vapor pressure of the solution will be 99 kPa - a fall of 1 kPa. At a lower temperature, where the vapor pressure of the pure water is 10 kPa, the fall will only be 0.1 kPa. For the curves to be parallel the falls would have to be the same over the whole temperature range. They aren't! That must mean that the phase diagram needs a new melting point line (a solid-liquid equilibrium line) passing through the new triple point. That is shown in the next diagram. Now we are finally in a position to see what effect a non-volatile solute has on the melting and freezing points of the solution. Look at what happens when you draw in the 1 atmosphere pressure line which lets you measure the melting and boiling points. The diagram also includes the melting and boiling points of the pure water from the original phase diagram for pure water (black lines). Because of the changes to the phase diagram, you can see that: We have looked at this with water as the solvent, but using a different solvent would make no difference to the argument or the conclusions. The only difference is in the slope of the solid-liquid equilibrium lines. For most solvents, these slope forwards whereas the water line slopes backwards. You could prove to yourself that that does not affect what we have been looking at by re-drawing all these diagrams with the slope of that particular line changed. You will find it makes no difference whatsoever. We can calculate the vapor pressure of the solution in two ways, depending on the volatility of the solute. If the solute is volatile, it will exert its own vapor pressure and this amount is a significant contribution to the overall vapor pressure of the solution, and thus needs to be included in the calculations. On the other hand, if it is nonvolatile, the solute will not produce vapor pressure in solution at that temperature. These calculations are fairly straightforward if you are comfortable with stoichiometric conversions. Because the solute is nonvolatile, you need only determine the change in vapor pressure for the solvent. Using the equation for Raoult's Law, you will need to find the mole fraction of the solvent and the vapor pressure of the pure solvent is typically given. 1.5 moles of cherry Kool-Aid are added to a pitcher containing 2 liters of water on a nice day at 25 C. The vapor pressure of water alone is 23.8 mm Hg at 25 C. What is the new vapor pressure of Kool-Aid? \(P_{H_2O}\) = 23.8 mm Hg To solve for the mole fraction, you must first convert the 2 L of water into moles: 1 L = 1000 mL = 1000 g Knowing this, you can convert the mass of water (2000 g) into moles: 2000 g / 18.02 g (molar mass of water) = 110.9 moles H O Solve for the mole fraction, \(\chi_{H_2O}\): \(\chi_{H_2O}\) = moles H O / total moles = 110.9 moles / 110.9 + 1.5 moles = 0.979 Finally, apply Raoult's Law \(P_{Kool-Aid} = \chi_{H_2O} \, P_{H_2O}\) (0.979)(23.8 mm Hg) = 23.3 mm Hg Calculate the vapor pressure of a solution made by dissolving 50.0 g glucose, \(C_6H_{12}O_6\), in 500 g of water. The vapor pressure of pure water is 47.1 torr at 37°C To use Raoult's Law (Equation \(\ref{RLaw}\)), we need to calculate the mole fraction of water (the solvent) in this sugar-water solution. \[ \chi_{solvent} = \dfrac{ \text{moles of water}}{\text{moles of solute} + \text{moles of solvent}} \nonumber \] \[ \chi_{solvent} = \dfrac{ n_{water}}{ n_{glucose} + n_{water} } \nonumber\] The molar mass of glucose if 180.2 g/mol and of water is 18 g/mol. So \[n_{water} = \dfrac{500\,g}{18\,g /mol} = 27.7 \,mol \nonumber\] and \[n_{glucose} = \dfrac{50\,g}{180.2\,g /mol} = 0.277 \,mol \nonumber\] and \[ \chi_{solvent} = \dfrac{ 27.7 \,mol}{ 0.277 \,mol + 27.7 \,mol } = 0.99 \nonumber\] Note that this still relatively dilute. \[ P_{solution} = 0.99 \times 47.1 = 46.63 \, torr \nonumber\] Calculate the vapor pressure of a solution made by dissolving 50.0 g CaCl , \(C_6H_{12}O_6\), in 500 g of water. The vapor pressure of pure water is 47.1 torr at 37°C To use Raoult's Law (Equation \(\ref{RLaw}\)), we need to calculate the mole fraction of water (the solvent) in this salt-water solution. \[ \chi_{solvent} = \dfrac{ \text{moles of water}}{\text{moles of solute} + \text{moles of solvent}} \nonumber\] \[ \chi_{solvent} = \dfrac{ n_{water}}{ n_{solutes} + n_{water} } \nonumber\] The molar mass of \(\ce{CaCl_2}\) if 111 g/mol and of water is 18 g/mol. So \[n_{water} = \dfrac{500\,g}{18\,g /mol} = 27.7 \,mol \nonumber\] and \[n_{solutes} = \dfrac{50\,g}{111 \,g /mol} = 0.45 \,mol \nonumber\] but this is really: and \[ \chi_{solvent} = \dfrac{ 27.7 \,mol}{ 0.45 \,mol + 0.9 \,mol + 27.7 \,mol } = 0.953 \nonumber\] Note that this still relatively dilute. \[ P_{solution} = 0.953 \times 47.1 = 44.88\, torr \nonumber\] The only difference between volatile and nonvolatile solutes, is that the partial pressure exerted by the vapor pressure of the volatile solute and the vapor pressure of the solvent must be accounted for. The sum of the two will give you the total vapor pressure of the solution. What are the partial pressures of benzene and toluene in a solution in which the mole fraction of benzene is 0.6? What is the total vapor pressure? The vapor pressure of pure benzene is 95.1 mm Hg and the vapor pressure of pure toluene 28.4 mm Hg at 25 C. If \(\chi_{benzene} = 0.6\), than \(\chi_{toluene} = 0.4\) because \(1 - 0.6 = 0.4\). P = x P = (0.6)(95.1 mm Hg) = 57.1 mm Hg P = x P = (0.4)(28.4 mm Hg) = 11.4 mm Hg The total vapor pressure is simply the sum of the partial pressures: P = P + P = 57.1 mm Hg + 11.4 mm Hg = 68.5 mm Hg *MM = molar mass Solve for x . Moles H O = 500 g / 18.02 g/mol = 27.7 moles H O Moles glucose = 78 g / 180.16 g/mol = 0.433 moles glucose x = 27.7 moles / (27.7 + 0.433) moles = 0.985 Now we can apply Raoult's Law. \[P = XP^o = (0.985)(23.8\; mmHg) = 23.4\; mmHg\] Calculate the moles of each component. Determine the mole fraction of ethanol and apply Raoult's Law. = 0.326 moles / (0.326 + 0.297) moles = 0.523 P = xP = (0.523)(52.3 torr) = 27.4 torr If x = 0.577 then x = 0.423 because 1 - 0.577 = 0.423 If you rearrange the Raoult's Law equation, you can solve for P . \[P = XP^o \rightarrow \] \[P^o = \dfrac{P}{X} = \dfrac{698\; mmHg}{0.423} = 1650\; mmHg\]

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a) Ag b) Ni c) Mn d) Cr e) Cu f) Fe g)Os h) Re a) Pb & S b) Sn & 2 O c) Hg & S d) Fe & 2 S e) 2 Fe & 3 O f) 2 Fe & 1 Fe & 4 O Probably Fe , to replace Zn ions. a) C b) C c) d) C e) C a) Mg b) Cu c) Mn d) Ca e) Mn f) Mn a) Cu(II), Fe(II) b) Zn(II), Fe(III) c) Be(II), Al(III) d) Cu(I), Fe(III) e) Cu(II), Al(III) MnO : 4 x O (= 8 ) + Mn = 1 overall MnO : 2 x O (= 4 ) + Mn = neutral overall difference = 3 e a) SO : 4 x O (= 8 ) + S = 2 overall S O : 8 x O (= 16 ) + 2 x S (= 14 ) = 2 overall difference = 1 e per S, or 2 e overall S O + 2 e → 2 SO b) HPO : 3 x O (= 6 ) + H + P = 2 overall P : P(0) difference = 3 e HPO + 3 e + 5 H → P + 3 H O c) Ti O : 3 x O (= 6 ) + 2 x Ti (= 6 ) = neutral overall TiO : O + Ti = neutral overall difference = 1 e per Ti, or 2 e overall Ti O + 2 e + 2 H → 2 TiO + H O d) N : N(0) NH OH : O + 3 x H (= 3 ) + N = neutral overall difference = 1 e per N, or 2 e overall N + 2 e + 2 H + 2 H O → 2 NH OH Lithium is an alkali metal, in the first column of the periodic table. It has a relatively low ionization energy because it has a noble gas configuration as a cation. That noble gas configuration is stable because of the relatively large number of nuclear protons and a relatively short distance between the nucleus and the outermost shell of electrons. In lithium metal, the outermost electron is relatively far from the nucleus and so it is at a relatively high energy, and easily lost. Fluorine is a halogen, with a relatively high electron affinity. It easily gains an electron to get to a noble gas configuration as a fluoride anion. That noble gas configuration is stable because of the relatively large number of nuclear protons and a relatively short distance between the nucleus and the outermost shell of electrons. From most easily oxidized to least easily oxidized: Li > Al > Fe > Cu > Au When the table of standard reduction potentials is displayed with the most negative value at the top and the most positive value at the bottom, any given half-reaction will go forward if it is coupled with the reverse of a half-reaction that lies above it in the table. The opposite is not the case; no half reaction will go forward if it is coupled with the reverse of a half-reaction below it in the table. a) a b) c) a) 2 Li + F → 2 Li + 2 F b) E = +5.91 V c) Things look pretty grim. d) e) This scheme would result in the release of a small amount of energy at each stage. Each step could be harnessed to perform a task more efficiently, with less heat loss. There are really two significant departures from expectation here. Lithium is much more active than expected based on electronegativity. The larger alkali metals, cesium, rubidium and francium, are all less active than expected on that basis. We will see that another factor the influences activity in redox is the stability of ions in aqueous solution. Lithium cation is a small ion; water molecules bind very strongly to the ion because the electrons get relatively close to lithium's nucleus. That strong binding stabilizes this ion especially, tipping the malance of the reaction more strongly towards oxidation of lithium. The larger alkali metal ions are not nearly as stabilized by water ligands in aqueous solution, so the balance of their reactions does not tilt as strongly towards aqueous ions. Li /Li: E = - 3.04 V; ΔH = 147 kJ/mol; IE = 520 kJ/mol; ΔH = -520 kJ/mol Na /Na: E = - 2.71 V; ΔH = 97 kJ/mol; IE = 495 kJ/mol; ΔH = -406 kJ/mol K /K: E = - 2.931 V; ΔH = 77 kJ/mol; IE = 419 kJ/mol; ΔH = -320 kJ/mol Potassium should be the easiest of the three to oxidize. It is easier to oxidize than sodium. However, lithium's high heat of hydration reverses the trend and tips the balance of reaction in favour of ion formation. Cu /Cu: E = + 0.340 V; ΔH = 300 kJ/mol; IE = 745 kJ/mol & 1958 kJ/mol; ΔH = - 2099 kJ/mol Ni /Ni: E = - 0.25 V; ΔH = 377 kJ/mol; IE = 737 kJ/mol & 1753 kJ/mol; ΔH = - 2096 kJ/mol Zn /Zn: E = - 0.7618 V; ΔH = 123 kJ/mol; IE = 906 kJ/mol & 1733 kJ/mol; ΔH = - 2047 kJ/mol In this case, zinc may be considered the outlier. Copper should be easier to reduce than nickel based solely on electronegativity. However, zinc's very low heat of vaporization suggests that formation of the solid metal is less favoured in that case, helping to tilt the balance toward zinc ion instead. The bonds to iron would contract because the increased charge on the iron would attract the ligand donor electrons more strongly. The bonds to copper would lengthen because of the lower charge on the copper. ,

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The oxidation state of an element is related to the number of electrons that an atom loses, gains, or appears to use when joining with another atom in compounds. It also determines the ability of an atom to oxidize (to lose electrons) or to reduce (to gain electrons) other atoms or species. Almost all of the have multiple oxidation states experimentally observed. Filling atomic orbitals requires a set number of electrons. The s-block is composed of elements of Groups I and II, the alkali and alkaline earth metals (sodium and calcium belong to this block). Groups XIII through XVIII comprise of the p-block, which contains the nonmetals, halogens, and noble gases (carbon, nitrogen, oxygen, fluorine, and chlorine are common members). Transition metals reside in the d-block, between Groups III and XII. If the following table appears strange, or if the orientations are unclear, please review the section on . The key thing to remember about electronic configuration is that the most stable noble gas configuration is ideal for any atom. Forming bonds are a way to approach that configuration. In particular, the transition metals form more lenient bonds with anions, cations, and neutral complexes in comparison to other elements. This is because the d orbital is rather diffused (the f orbital of the lanthanide and actinide series more so). Counting through the periodic table is an easy way to determine which electrons exist in which orbitals. As mentioned before, by counting protons (atomic number), you can tell the number of electrons in a neutral atom. Organizing by block quickens this process.  1s (H, He), 2s (Li, Be), 2p (B, C, N, O, F, Ne), 3s (Na, Mg), 3p (Al, Si, P, S, Cl, Ar), 4s (K, Ca), 3d (Sc, Ti, V). If you do not feel confident about this counting system and how electron orbitals are filled, please see the section on . Referring to the periodic table below confirms this organization. We have three elements in the 3d orbital. Therefore, we write in the order the orbitals were filled. 1s 2s 2p 3s 3p 4s 3d or [Ar] 4s 3d . The neutral atom configurations of the fourth period transition metals are in Table \(\Page {2}\). Chromium and copper appear anomalous. Take a brief look at where the element (atomic number 24) lies on the Periodic Table (Figure \(\Page {1}\)). The electronic configuration for chromium is not [Ar] 4s 3d but instead it is [Ar] 4s 3d . This is because the half-filled 3d manifold (with one 4s electron) is more stable than a partially filled d-manifold (and a filled 4s manifold). You will notice from Table \(\Page {2}\) that the copper exhibits a similar phenomenon, although with a fully filled d-manifold. When considering ions, we add or subtract negative charges from an atom. Keeping the atomic orbitals when assigning oxidation numbers in mind helps in recognizing that transition metals pose a special case, but not an exception to this convenient method. An atom that accepts an electron to achieve a more stable configuration is assigned an oxidation number of -1. The donation of an electron is then +1. When a transition metal loses electrons, it tends to lose it's s orbital electrons before any of its d orbital electrons. For more discussion of these compounds form, see . Write the electronic configurations of: The atomic number of iron is 26 so there are 26 protons in the species. Determine the more stable configuration between the following pair: Most transition metals have multiple oxidation states, since it is relatively easy to lose electron(s) for transition metals compared to the alkali metals and alkaline earth metals. Alkali metals have one electron in their valence s-orbital and their ions almost always have oxidation states of +1 (from losing a single electron). Similarly, alkaline earth metals have two electrons in their valences s-orbitals, resulting in ions with a +2 oxidation state (from losing both). However, transitions metals are more complex and exhibit a range of observable oxidation states due primarily to the removal of d-orbital electrons. The following chart describes the most common oxidation states of the period 3 elements. Scandium is one of the two elements in the first transition metal period which has only one oxidation state (zinc is the other, with an oxidation state of +2). All the other elements have at least two different oxidation states. Manganese, which is in the middle of the period, has the highest number of oxidation states, and indeed the highest oxidation state in the whole period since it has five unpaired electrons (see table below). To help remember the stability of higher oxidation states for transition metals it is important to know the trend: the stability of the higher oxidation states progressively increases down a group. For example, in group 6, (chromium) Cr is most stable at a +3 oxidation state, meaning that you will not find many stable forms of Cr in the +4 and +5 oxidation states. By contrast, there are many stable forms of molybdenum (Mo) and tungsten (W) at +4 and +5 oxidation states. What makes zinc stable as Zn ? What makes scandium stable as Sc ? Zinc has the neutral configuration [Ar]4s 3d . Losing 2 electrons does not alter the complete d orbital. Neutral scandium is written as [Ar]4s 3d . Losing 3 electrons brings the configuration to the noble state with valence 3p . Why is iron almost always Fe or Fe ? Iron is written as [Ar]4s 3d . Losing 2 electrons from the s-orbital (3d ) or 2 s- and 1 d-orbital (3d ) electron are fairly stable oxidation states. Write manganese oxides in a few different oxidation states. Which ones are possible and/or reasonable? Although Mn is the most stable ion for manganese, the d-orbital can be made to remove 0 to 7 electrons. Compounds of manganese therefore range from Mn(0) as Mn , Mn(II) as MnO, Mn(II,III) as Mn O , Mn(IV) as MnO , or manganese dioxide, Mn(VII) in the permanganate ion MnO , and so on. When given an ionic compound such as \(\ce{AgCl}\), you can easily determine the oxidation state of the transition metal. In this case, you would be asked to determine the oxidation state of silver (Ag). Since we know that chlorine (Cl) is in the group of the periodic table, we then know that it has a charge of -1, or simply Cl . In addition, by seeing that there is no overall charge for \(\ce{AgCl}\), (which is determined by looking at the top right of the compound, i.e., AgCl # represents Determine the oxidation state of cobalt in \(\ce{CoBr2}\). Similar to chlorine, bromine (\(\ce{Br}\)) is also a halogen with an oxidation charge of -1 (\(\ce{Br^{-}}\)). Since there are two bromines each with a charge of -1. In addition, we know that \(\ce{CoBr2}\) has an overall neutral charge, therefore we can conclude that the cation (cobalt), \(\ce{Co}\) must have an oxidation state of +2 to neutralize the -2 charge from the two bromine anions. What is the oxidation state of zinc in \(\ce{ZnCO3}\). (Note: the \(\ce{CO3}\) anion has a charge state of -2) \(\ce{CO3}\) s us \(\ce{Zn^{2+}}\) an \(\ce{CO3^{-2}}\) Consider the manganese (\(\ce{Mn}\)) atom in the permanganate (\(\ce{MnO4^{-}}\)) ion. Since has an oxidation state of -2 and we know there are four oxygen atoms. In addition, this compound has an overall charge of -1; therefore the overall charge is not neutral in this example. Thus, since the oxygen atoms in the ion contribute a total oxidation state of -8, and since the overall charge of the ion is -1, the sole manganese atom must have an oxidation state of +7. This gives us \(\ce{Mn^{7+}}\) and \(\ce{4 O^{2-}}\), which will result as \(\ce{MnO4^{-}}\). This example also shows that manganese atoms can have an oxidation state of +7, which is the highest possible oxidation state for the fourth period transition metals. is widely studied because it is an important reducing agent in chemical analysis and is also studied in biochemistry for catalysis and in metallurgy in fortifying alloys. In plants, manganese is required in trace amounts; stronger doses begin to react with enzymes and inhibit some cellular function. Due to manganese's flexibility in accepting many oxidation states, it becomes a good example to describe general trends and concepts behind electron configurations. Electron configurations of unpaired electrons are said to be and respond to the proximity of magnets. Fully paired electrons are and do not feel this influence. Manganese, in particular, has paramagnetic and diamagnetic orientations depending on what its oxidation state is. \(\ce{Mn2O3}\) is manganese(III) oxide with manganese in the +3 state. 4 unpaired electrons means this complex is paramagnetic. \[\ce{[Ar]} 4s^{0} 3d^{4}\nonumber\] \(\ce{MnO2}\) is manganese(IV) oxide, where manganese is in the +4 state. 3 unpaired electrons means this complex is less paramagnetic than Mn . \[\ce{[Ar]} 4s^{0} 3d^{3}\nonumber\] \(\ce{KMnO4}\) is potassium permanganate, where manganese is in the +7 state with no electrons in the 4s and 3d orbitals. \[\ce{[Ar]} 4s^{0} 3d^{0}\nonumber\] Since the 3p orbitals are all paired, this complex is diamagnetic. Oxidation states of transition metals follow the general rules for most other ions, except for the fact that the d orbital is degenerated with the s orbital of the higher quantum number. Transition metals achieve stability by arranging their electrons accordingly and are oxidized, or they lose electrons to other atoms and ions. These resulting cations participate in the or synthesis of other compounds. Determine the oxidation states of the transition metals found in these neutral compounds. Note: The transition metal is underlined in the following compounds.

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When we sample a particular distribution, the value that we obtain depends on chance and on the nature of the distribution described by the function \(f\left(u\right)\). The probability that any given trial will produce \(u\) in the interval \(a<u<b\) is equal to \(f\left(b\right)-f\left(a\right)\). We often find situations in which a second function of \(u\), call it \(g\left(u\right)\), is also of interest. If we sample the distribution and obtain a value of the random variable, \(u_k\), then the value of \(g\) associated with that trial is \(g\left(u_k\right)\). The question arises: Given \(g(u)\) and the distribution function \(f(u)\), what should we expect the value of \(g\left(u_k\right)\) to be? That is, if we get a value of \(u\) from the distribution and then find \(g\left(u\right)\), what value should we expect to find for \(g\left(u\right)\)? While this seems like a reasonable question, it is obvious that we can give a meaningful answer only when we can define more precisely just what we mean by “expect.” To understand our definition of the (sometimes called the ) of \(g\left(u\right)\), let us consider a game of chance. Suppose that we have a needle that rotates freely on a central axis. When spun, the needle describes a circular path, and its point eventually comes to rest at some point on this path. The location at which the needle stops is completely random. Imagine that we divide the circular path into six equal segments, which we number from one to six. When we spin the needle, it is equally likely to stop over any of these segments. Now, let us suppose that we conduct a lottery by selling six tickets, also numbered from one to six. We decide the winner of the lottery by spinning the needle. The holder of the ticket whose number matches the number on which the needle stops receives a payoff of $6000. After the spin, one ticket is worth $6000, and the other five are valueless. We ask: Before the spin, what is any one of the lottery tickets worth? In this context, it is reasonable to define the expected value of a ticket as the amount that we should be willing to pay to buy a ticket. If we buy them all, we receive $6000 when the winning ticket is selected. If we pay $1000 per ticket to buy them all, we get our money back. If we buy all the tickets, the expected value of each ticket is $1000. What if we buy only one ticket? Is it reasonable to continue to say that its expected value is $1000? We argue that it is. One argument is that the expected value of a ticket should not depend on who owns the ticket; so, it should not depend on whether we buy one, two, or all of them. A more general argument supposes that repeated lotteries are held under the same rules. If we spend $1000 to buy one ticket in each of a very large number of such lotteries, we expect that we will eventually “break even.” Since the needle comes to rest at each number with equal probability, we reason that \[\begin{array}{l} \text{Expected value of a ticket} \\ =\$6000\left(fraction\ of\ times\ our\ ticket\ would\ be\ selected\right) \\ =\$6000\left({1}/{6}\right) \\ =\$1000 \end{array}\] Since we assume that the fraction of times our ticket would be selected in a long series of identical lotteries is the same thing as the probability that our ticket will be selected in any given drawing, we can also express the expected value as \[ \begin{array}{l} \text{Expected value of a ticket} \\ =\$6000\left(probability\ that\ our\ ticket\ will\ be\ be\ selected\right) \\ =\$6000\left({1}/{6}\right) \\ =\$1000 \end{array}\] Clearly, the ticket is superfluous. The game depends on obtaining a value of a random variable from a distribution. The distribution is a spin of the needle. The random variable is the location at which the needle comes to rest. We can conduct essentially the same game by allowing any number of participants to bet that the needle will come to rest on any of the six equally probable segments of the circle. If an individual repeatedly bets on the same segment in many repetitions of this game, the total of his winnings eventually matches the total amount that he has wagered. (More precisely, the total of his winnings divided by the total amount he has wagered becomes arbitrarily close to one.) Suppose now that we change the rules. Under the new rules, we designate segment \(1\) of the circle as the payoff segment. Participants pay a fixed sum to be eligible for the payoff for a particular game. Each game is decided by a spin of the needle. If the needle lands in segment \(1\), everyone who paid to participate in that game receives $6000. Evidently, the new rules have no effect on the value of participation. Over the long haul, a participant in a large number of games wins $6000 in one-sixth of these games. We take this to be equivalent to saying that he has a probability of one-sixth of winning $6000 in a given game in which he participates. His expected payoff is \[ \begin{array}{l} \text{Expected value of game} \\ =\$6000\left(probability\ of\ winning\ \$6000\right) \\ =\$6000\left({1}/{6}\right) \\ =\$1000 \end{array}\] Let us change the game again. We sub-divide segment \(2\) into equal-size segments \(2A\) and \(2B\). The probability that the needle lands in \(2A\) or \(2B\) is \({1}/{12}\). In this new game, the payoff is $6000 when the needle lands in either segment \(1\) or segment \(2A\). We can use any of the arguments that we have made previously to see that the expected payoff game is now \(\$6000\left({1}/{4}\right)=\$1500\). However, the analysis that is most readily generalized recognizes that the payoff from this game is just the sum of the payout from the previous game plus the payout from a game in which the sole payout is $6000 whenever the needle lands in segment \(2A\). For the new game, we have \[ \begin{array}{l} \text{Expected value of a game} \\ =\$6000\times P\left(segment\ 1\right)+\$6000\times P\left(segment\ 2A\right) \\ =\$6000\left({1}/{6}\right)+\$6000\left({1}/{12}\right) \\ =\$1500 \end{array}\] We can devise any number of new games by dividing the needle’s circular path into \(\mathrm{\textrm{Ω}}\) non-overlapping segments. Each segment is a possible outcome. We number the possible outcomes \(1\), \(2\), , \(i\), , Ω, label these outcomes \(u_1\), \(u_2\), , \(u_i\), , \(u_{\textrm{Ω}}\), and denote their probabilities as \(P\left(u_1\right)\), \(P\left(u_2\right)\),...,\(P\left(u_i\right)\), , \(P\left(u_{\textrm{Ω}}\right)\). We say that the probability of outcome \(u_i\), \(P\left(u_i\right)\), is the of outcome \(u_i\). We denote the respective payoffs as \(g\left(u_1\right)\), \(g\left(u_2\right)\),...,\(g\left(u_i\right)\), , \(g\left(u_{\textrm{Ω}}\right)\). Straightforward generalization of our last analysis shows that the expected value for participation in any game of this type is \[\sum^{\textrm{Ω}}_{i=1}{g\left(u_i\right)\times P\left(u_i\right)}\] Moreover, the spinner is representative of any distribution, so it is reasonable to generalize further. We can say that the expected value of the outcome of a single trial is always the probability-weighted sum, over all possible outcomes, of the value of each outcome. A common notation uses angular brackets to denote the expected value for a function of the random variable; the expected value of \(g\left(u\right)\) is \(\left\langle g\left(u\right)\right\rangle\). For a discrete distribution with \(\textrm{Ω}\) exhaustive mutually-exclusive outcomes \(u_i\), probabilities \(P\left(u_i\right)\), and outcome values (payoffs) \(g\left(u_i\right)\), we define the expected value of \(g\left(u\right)\) to be \[\left\langle g\left(u\right)\right\rangle \ =\sum^{\textrm{Ω}}_{i=1}{g\left(u_i\right)}\times P\left(u_i\right)\] Now, let us examine the expected value of \(g\left(u\right)\) from a slightly different perspective. Let the number of times that each of the various outcomes is observed in a particular sample of \(N\) observations be \(N_1,\ N_2,\dots ,N_3,\dots ,N_{\textrm{Ω}}\). We have \(N=N_1+\ N_2+\dots +N_i+\dots +N_{\textrm{Ω}}\). The set \(\{N_1,\ N_2,\dots ,N_i,\dots ,N_{\textrm{Ω}}\}\) specifies the way that the possible outcomes are populated in this particular series of \(N\) observations. We call \(\{N_1,\ N_2,\dots ,N_i,\dots ,N_{\textrm{Ω}}\}\) a . If we make a second series of observations, we obtain a second population set. We infer that the best forecast we can make for the number of occurrences of outcome \(u_i\) in any future series of observations is \(N\times P\left(u_i\right)\). We call \(N\times P\left(u_i\right)\) the of observations of outcome \(u_i\) in a sample of size \(N\). In a particular series of \(N\) trials, the number of occurrences of outcome \(u_i\), and hence of \(g\left(u_i\right)\), is \(N_i\). For the set of outcomes \(\{N_1,\ N_2,\dots ,N_3,\dots ,N_{\textrm{Ω}}\}\), the average value of \(g\left(u\right)\) is \[\overline{g\left(u\right)}=\frac{1}{N}\sum^{\textrm{Ω}}_{i=1}{g\left(u_i\right)\times N_i}\] Collecting a second sample of \(N\) observations produces a second estimate of \(\overline{g\left(u\right)}\). If \(N\) is small, successive estimates of \(\overline{g\left(u\right)}\) may differ significantly from one another. If we make a series of \(N\) observations multiple times, we obtain multiple population sets. In general, the population set from one series of \(N\) observations is different from the population set for a second series of \(N\) observations. If \(N\gg \mathit{\Omega}\), collecting such samples of \(N\) a sufficiently large number of times must produce some population sets more than once, and among those that are observed more than once, one must occur more often than any other. We call it the population set. Let the elements of the most probable population set be \(\{N_1,N_2,\dots ,N_i,\dots ,N_{\textrm{Ω}}\}\). We infer that the most probable population set is the best forecast we can make about the outcomes of any future sample of \(N\) from this distribution. Moreover, we infer that the best estimate we can make of \(N_i\) is that it equals the expected number of observations of outcome \(u_i\); that is, \[N_i\approx N\times P\left(u_i\right)\] Now, \(N_i\) and \(N_i\) must be natural numbers, while \(N\times P\left(u_i\right)\) need only be real. In particular, we can have \(0, but \(N_i\) must be \(0\) or \(1\) (or some higher integer). This is a situation of practical importance, because circumstances may limit the sample size to a number, \(N\), that is much less than the number of possible outcomes, \(\mathit{\Omega}\). (We encounter this situation in our discussion of statistical thermodynamics in Chapter 21. We find that the number of molecules in a system can be much smaller than the number of outcomes—observable energy levels—available to any given molecule.) If many more than \(N\) outcomes have about the same probability, repeated collection of samples of \(N\) observations can produce a series of population sets (each population set different from all of the others) in each of which every element is either zero or one. When this occurs, it may be that no single population set is significantly more probable than any of many others. Nevertheless, every outcome occurs with a well-defined probability. We infer that the set \(\left\{N\times P\left(u_1\right),N\times P\left(u_2\right),\dots ,N\times P\left(u_i\right),\dots ,N\times P\left(u_{\textrm{Ω}}\right)\right\}\) is always an adequate proxy for calculating the expected value for the most probable population set. To illustrate this kind of distribution, suppose that there are \(3000\) possible outcomes, of which the first and last thousand have probabilities that are so low that they can be taken as zero, while the middle \(1000\) outcomes have approximately equal probabilities. Then \(P\left(u_i\right)\approx 0\) for \(1 1000 and 2001\( 3000, while \(P\left(u_i\right)\approx {10}^{-3}\) for \(1001<2000\)>. We are illustrating the situation in which the number of outcomes we can observe, \(N\), is much less than the number of outcomes that have appreciable probability, which is \(1000\). So let us take the number of trials to be \(N=4\). If the value of \(g\left(u\right)\) for each of the \(1000\) middle outcomes is the same, say \(g\left(u_i\right)=100\) for \(1001 , then our calculation of the expected value of \(g\left(u\right)\) will be \[\left\langle g\left(u\right)\right\rangle \ =\frac{1}{4}\sum^{3000}_{i=1}{g\left(u_i\right)\times N}\times P\left(u_i\right)=\frac{1}{4}\sum^{2000}_{i=1001}{100\times N_i}=\frac{400}{4}=100\] regardless of which population set results from the four trials. That is, because all of the populations sets that have a significant chance to be observed have \(N_i=1\) and \(g\left(u_i\right)=100\) for exactly four values of \(i\) in the range \(1001 , all of the population sets that have a significant chance to be observed give rise to the same expected value. Let us compute the arithmetic average, \(\overline{g\left(u\right)}\), using the most probable population set for a sample of trials. In this case, the number of observations of the outcome \(u_i\) is \(N_i=N\times P\left(u_i\right).\) \(\overline{g\left(u\right)}\mathrm{\ }\ =\frac{1}{N}\sum^{\textrm{Ω}}_{i=1}{g\left(u_i\right)}\times N_i\)\(=\frac{1}{N}\sum^{\textrm{Ω}}_{i=1}{g\left(u_i\right)}\times N\times P\left(u_i\right)\)\(=\sum^{\textrm{Ω}}_{i=1}{g\left(u_i\right)\times P\left(u_i\right)}\)\(=\ \ \left\langle g\left(u\right)\right\rangle\) For a discrete distribution, \(\left\langle g\left(u\right)\right\rangle\) is the value of \(\overline{g\left(u\right)}\) that we calculate from the most probable population set, \(\left\{N_1,N_2,\dots ,N_i,\dots ,N_{\textrm{Ω}}\right\}\), or its proxy \(\left\{N\times P\left(u_1\right),N\times P\left(u_2\right),\dots ,N\times P\left(u_i\right),\dots ,N\times P\left(u_{\textrm{Ω}}\right)\right\}\). We can extend the definition of the expected value, \(\left\langle g\left(u\right)\right\rangle\), to cases in which the cumulative probability distribution function, \(f\left(u\right)\), and the outcome-value function, \(g\left(u\right)\), are continuous in the domain of the random variable, \(u_{min} . To do so, we divide this domain into a finite number, \(\mathit{\Omega}\), of intervals, \(\Delta u_i\). We let \(u_i\) be the lower limit of \(u\) in the interval \(\Delta u_i\). Then the probability that a given trial yields a value of the random variable in the interval \(\Delta u_i\) is \(P\left({\Delta u}_i\right)=f\left(u_i+\Delta u_i\right)-f\left(u_i\right)\), and we can approximate the expected value of \(g\left(u\right)\) for the continuous distribution by the finite sum \[\left\langle g\left(u\right)\right\rangle \ =\sum^{\textrm{Ω}}_{i=1}{g\left(u_i\right)\times P\left(\Delta u_i\right)}=\sum^{\textrm{Ω}}_{i=1}{g\left(u_i\right)}\times \left[f\left(u_i+\Delta u_i\right)-f\left(u_i\right)\right]=\sum^{\textrm{Ω}}_{i=1}{g\left(u_i\right)\times \left[\frac{f\left(u_i+\Delta u_i\right)-f\left(u_i\right)}{\Delta u_i}\right]}\times \Delta u_i\] In the limit as \(\mathit{\Omega}\) becomes arbitrarily large and all of the intervals \(\Delta u_i\) become arbitrarily small, the expected value of \(g\left(u\right)\) for a continuous distribution becomes \[\left\langle g\left(u\right)\right\rangle \ =\int^{\infty }_{-\infty }{g\left(u\right)\left[\frac{df\left(u\right)}{du}\right]du}\] This integral is the value of \(\left\langle g\left(u\right)\right\rangle\), where \({df\left(u\right)}/{du}\) is the probability density function for the distribution. If is a constant, we have \[\left\langle g\left(cu\right)\right\rangle =c\left\langle g\left(u\right)\right\rangle\] If \(h\left(u\right)\) is a second function of the random variable, we have \[\left\langle g\left(u\right)+h\left(u\right)\right\rangle =\left\langle g\left(u\right)\right\rangle +\left\langle h\left(u\right)\right\rangle\]

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Lead plumbate, also called red lead, minium or Mennige (in German), is a mineral showing colors from light red to brown/yellow tints. As a pure chemical it shows a vivid red. Minium is rare and occurs in lead mineral deposits that have been subjected to severe oxidizing conditions. It also occurs as a result of mine fires. It is most often associated with galena, cerussite, massicot, litharge, native lead, wulfenite and mimetite. Lead plumbate is obtained by heating lead monoxide (\(PbO\)) to 450-480°C in air: \[3 PbO + 1/2 O_2 \rightarrow Pb_3O_4\] or by oxidative annealing of lead white: \[3 Pb_2CO_3(OH)_2 + O_2 \rightarrow 2 Pb_3O_4 + 3 CO_2 + 3 H_2O\] Lead plumbate decomposes into lead monoxide and oxygen above 550°C. \(Pb_3O_4\) can be seen formally as a lead(II)plumbate(IV), \(Pb_2[PbO_4]\), or \(2PbO\cdot PbO_2\). In nitric acid the lead(II) oxide reacts forming lead nitrate, while the insoluble lead(IV) oxide is left unchanged: \[Pb_3O_4 + 4 HNO_3 \rightarrow 2 Pb(NO_3)_2 + PbO_2 + 2 H_2O\] Lead plumbate is virtually insoluble in water. However, it dissolves in hydrochloric acid (which is present in the stomach), and is therefore toxic when ingested. Lead plumbate (in a mixture with linseed oil or other organic adhesives) has been used as an anti-corrosion paint for iron. It forms insoluble iron(II) and iron(III) plumbates when brought into contact with iron oxides and with elementary iron. However, its use as a protective undercoat paint is limited due to its toxicity. Lead plumbate was used as a red pigment in ancient and medieval periods for paintings and the production of illuminated manuscripts (the term miniature is connected to the name of the substance).

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In this section, consider the differences between two types of changes in a system: Those that occur spontaneously and those that occur only with the continuous input of energy. In doing so, we’ll gain an understanding as to why some systems are naturally inclined to change in one direction under certain conditions. We’ll also gain insight into how the spontaneity of a process affects the distribution of energy and matter within the system. Processes have a natural tendency to occur in one direction under a given set of conditions. Water will naturally flow downhill, but uphill flow requires outside intervention such as the use of a pump. A is one that occurs naturally under certain conditions. A , on the other hand, will not take place unless it is “driven” by the continual input of energy from an external source. A process that is spontaneous in one direction under a particular set of conditions is nonspontaneous in the reverse direction. At room temperature and typical atmospheric pressure, for example, ice will spontaneously melt, but water will not spontaneously freeze. The spontaneity of a process is correlated to the speed of the process. A spontaneous change may be so rapid that it is essentially instantaneous or so slow that it cannot be observed over any practical period of time. To illustrate this concept, consider the decay of radioactive isotopes, a topic more thoroughly treated in the chapter on nuclear chemistry. Radioactive decay is by definition a spontaneous process in which the nuclei of unstable isotopes emit radiation as they are converted to more stable nuclei. All the decay processes occur spontaneously, but the rates at which different isotopes decay vary widely. Technetium-99m is a popular radioisotope for medical imaging studies that undergoes relatively rapid decay and exhibits a half-life of about six hours. Uranium-238 is the most abundant isotope of uranium, and its decay occurs much more slowly, exhibiting a half-life of more than four billion years (Figure \(\Page {1}\)). As another example, consider the conversion of diamond into graphite (Figure \(\Page {2}\)). \[\ce{C(s, diamond)}⟶\ce{C(s, graphite)} \label{Eq1} \] The phase diagram for carbon indicates that graphite is the stable form of this element under ambient atmospheric pressure, while diamond is the stable allotrope at very high pressures, such as those present during its geologic formation. Thermodynamic calculations of the sort described in the last section of this chapter indicate that the conversion of diamond to graphite at ambient pressure occurs spontaneously, yet diamonds are observed to exist, and persist, under these conditions. Though the process is spontaneous under typical ambient conditions, its rate is extremely slow, and so for all practical purposes diamonds are indeed “forever.” Situations such as these emphasize the important distinction between the thermodynamic and the kinetic aspects of a process. In this particular case, diamonds are said to be but under ambient conditions. As we extend our discussion of thermodynamic concepts toward the objective of predicting spontaneity, consider now an isolated system consisting of two flasks connected with a closed valve. Initially there is an ideal gas on the left and a vacuum on the right (Figure \(\Page {3}\)). When the valve is opened, the gas spontaneously expands to fill both flasks. Recalling the definition of pressure-volume work from the chapter on thermochemistry, note that no work has been done because the pressure in a vacuum is zero. \[ \begin{align} w&=−PΔV \\[4pt]&=0 \,\,\, \mathrm{(P=0\: in\: a\: vaccum)} \label{Eq2} \end{align} \] Note as well that since the system is isolated, no heat has been exchanged with the surroundings (q = 0). The first law of thermodynamics confirms that there has been no change in the system’s internal energy as a result of this process. \[ \begin{align} ΔU&=q+w \tag{First Law of Thermodynamics} \\[4pt] &=0+0=0 \label{Eq3}\end{align} \] The spontaneity of this process is therefore not a consequence of any change in energy that accompanies the process. Instead, the movement of the gas appears to be related to the greater, more that results when the gas is allowed to expand. Initially, the system was comprised of one flask containing matter and another flask containing nothing. After the spontaneous process took place, the matter was distributed both more widely (occupying twice its original volume) and more uniformly (present in equal amounts in each flask). Now consider two objects at different temperatures: object X at temperature and object Y at temperature , with > (Figure \(\Page {4}\)). When these objects come into contact, heat spontaneously flows from the hotter object (X) to the colder one (Y). This corresponds to a loss of thermal energy by X and a gain of thermal energy by Y. \[q_\ce{X}<0 \hspace{20px} \ce{and} \hspace{20px} q_\ce{Y}=−q_\ce{X}>0 \label{Eq4} \] From the perspective of this two-object system, there was no net gain or loss of thermal energy, rather the available thermal energy was redistributed among the two objects. This spontaneous process resulted in a . As illustrated by the two processes described, an important factor in determining the spontaneity of a process is the extent to which it changes the dispersal or distribution of matter and/or energy. In each case, a spontaneous process took place that resulted in a more uniform distribution of matter or energy. Describe how matter and energy are redistributed when the following spontaneous processes take place: Describe how matter and energy are redistributed when you empty a canister of compressed air into a room. This process entails both a greater and more uniform dispersal of matter as the compressed air in the canister is permitted to expand into the lower-pressure air of the room. The process also requires an input of energy to disrupt the intermolecular forces between the closely-spaced gas molecules that are originally compressed into the container. If you were to touch the nozzle of the canister, you would notice that it is cold because the exiting molecules are taking energy away from their surroundings, and the canister is part of the surroundings. 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. In this section we have only discussed nuclear decay, physical changes of pure substances, and macroscopic events such as water flowing downhill. In the following sections we will discuss mixtures and chemical reactions, situations in which the description of sponteneity becomes more challenging. 

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In the early 19th century, steam engines came to play an increasingly important role in industry and transportation. However, a systematic set of theories of the conversion of thermal energy to motive power by steam engines had not yet been developed. Nicolas Léonard Sadi Carnot (1796-1832), a French military engineer, published in 1824. The book proposed a generalized theory of heat engines, as well as an idealized model of a thermodynamic system for a heat engine that is now known as the Carnot cycle. Carnot developed the foundation of the second law of thermodynamics, and is often described as the "Father of thermodynamics." The Carnot cycle consists of the following four processes: The P-V diagram of the Carnot cycle is shown in Figure \(\Page {2}\). In isothermal processes I and III, ∆U=0 because ∆T=0. In adiabatic processes II and IV, q=0. Work, heat, ∆U, and ∆H of each process in the Carnot cycle are summarized in Table \(\Page {1}\). The T-S diagram of the Carnot cycle is shown in Figure \(\Page {3}\). In isothermal processes I and III, ∆T=0. In adiabatic processes II and IV, ∆S=0 because dq=0. ∆T and ∆S of each process in the Carnot cycle are shown in Table \(\Page {2}\). The Carnot cycle is the most efficient engine possible based on the assumption of the absence of incidental wasteful processes such as friction, and the assumption of no conduction of heat between different parts of the engine at different temperatures. The efficiency of the carnot engine is defined as the ratio of the energy output to the energy input. \[\begin{align*} \text{efficiency} &=\dfrac{\text{net work done by heat engine}}{\text{heat absorbed by heat engine}} =\dfrac{-w_{sys}}{q_{high}} \\[4pt] &=\dfrac{nRT_{high}\ln\left(\dfrac{V_{2}}{V_{1}}\right)+nRT_{low}\ln \left(\dfrac{V_{4}}{V_{3}}\right)}{nRT_{high}\ln\left(\dfrac{V_{2}}{V_{1}}\right)} \end{align*}\] Since processes II (2-3) and IV (4-1) are adiabatic, \[\left(\dfrac{T_{2}}{T_{3}}\right)^{C_{V}/R}=\dfrac{V_{3}}{V_{2}}\] and \[\left(\dfrac{T_{1}}{T_{4}}\right)^{C_{V}/R}=\dfrac{V_{4}}{V_{1}}\] And since = and = , \[\dfrac{V_{3}}{V_{4}}=\dfrac{V_{2}}{V_{1}}\] Therefore, \[\text{efficiency}=\dfrac{nRT_{high}\ln\left(\dfrac{V_{2}}{V_{1}}\right)-nRT_{low}\ln\left(\dfrac{V_{2}}{V_{1}}\right)}{nRT_{high}\ln\left(\dfrac{V_{2}}{V_{1}}\right)}\] \[\boxed{\text{efficiency}=\dfrac{T_{high}-T_{low}}{T_{high}}}\] The Carnot cycle has the greatest efficiency possible of an engine (although other cycles have the same efficiency) based on the assumption of the absence of incidental wasteful processes such as friction, and the assumption of no conduction of heat between different parts of the engine at different temperatures.

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Suppose that we have a probability density function like that sketched in Figure 8 and that the area under the curve in the interval \(a<u<b\) is 0.25. If we draw a large number of samples from the distribution, our definitions of probability and the probability density function mean that about 25% of the values we draw will lie in the interval \(a<u<b\). We expect the percentage to become closer and closer to 25% as the total number of samples drawn becomes very large. The same would be true of any other interval, \(c<u<d\), where the area under the curve in the interval \(c<u<d\) is 0.25. If we draw exactly four samples from this distribution, the values can be anywhere in the domain of \(u\). However, if we ask what arrangement of four values best approximates the result of drawing a large number of samples, it is clear that this arrangement must have a value in each of the four, mutually-exclusive, 25% probability zones. We can extend this conclusion to any number of representative points. If we ask what arrangement of points would best represent the arrangement of a large number of points drawn from the distribution, the answer is clearly that one of the \(N\) representative points should lie within each of \(N\), mutually-exclusive, equal-area segments that span the domain of \(u\).) We can turn this idea around. In the absence of information to the contrary, the best assumption we can make about a set of \(N\) values of a random variable is that each represents an equally probable outcome. If our entire store of information about a distribution consists of four data points drawn from the distribution, the best description that we can give of the probability density function is that one-fourth of the area under the curve lies above a segment of the domain that is associated with each point. If we have \(N\) points, the best estimate we can make of the distribution from which the \(N\) points are drawn is that \({\left({1}/{N}\right)}^{th}\) of the area lies above each of them. This view tells us to associate a probability of \({1}/{N}\) with an interval around each data point, but it does not tell us where to begin or end the interval. If we could decide where the interval about each data point began and ended, we could estimate the shape of the probability density function. For a small number of points, we could not expect this estimate to be very accurate, but it would be the best possible estimate based on the given data. Now, instead of trying to find the best interval to associate with each data point, let us think about the intervals into which the data points divide the domain. This small change of perspective leads us to a logical way to divide the domain of \(u\) into specific intervals of equal probability. If we put \(N\) points on any line, these points divide the line into \(N+1\) segments. There is a segment to the left of every point; there are \(N\) such segments. There is one final segment to the right of the right-most point, and so there are \(N+1\) segments in all. In the absence of information to the contrary, the best assumption we can make is that \(N\) data points divide their domain into \(N+1\) segments, each of which is associated with equal probability. The fraction of the area above each of these segments is \({1}/{\left(N+1\right)}\); also, the probability associated with each segment is \({1}/{\left(N+1\right)}\). If, as in the example above, there are four data points, the best assumption we can make about the probability density function is that 20% of its area lies between the left boundary and the left-most data point, and 20% lies between the right-most data point and the right boundary. The three intervals between the four data points each represent an additional 20% of the area. Figure 9 indicates the \(N\) data points that best approximate the distribution sketched in Figure 8. The sketches in Figure 10 describe the probability density functions implied by the indicated sets of data points.

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This page looks at some of the problems with the usual way of explaining the electronic structures of the d-block elements based on the order of filling of the d and s orbitals. The way that the order of filling of orbitals is normally taught gives you an easy way of working out the electronic structures of elements. However, it does throw up problems when you come to explain various properties of the transition elements. This page takes a closer look at this, and offers a more accurate explanation which avoids the problems. The explains how electrons fill low energy orbitals (closer to the nucleus) before they fill higher energy ones. Where there is a choice between orbitals of equal energy, they fill the orbitals singly as far as possible . The diagram (not to scale) summarizes the energies of the orbitals up to the 4p level. The oddity is the position of the 3d orbitals, which are shown at a slightly higher level than the 4s. This means that the 4s orbital which will fill first, followed by all the 3d orbitals and then the 4p orbitals. Similar confusion occurs at higher levels, with so much overlap between the energy levels that the 4f orbitals do not fill until after the 6s, for example. Everything is straightforward up to this point, but the 3-level orbitals are not all full - the 3d levels have not been used yet. But if you refer back to the energies of the orbitals, you will see that the next lowest energy orbital is the 4s - so that fills first. d-block elements are thought of as elements in which the last electron to be added to the atom is in a d orbital (actually, that turns out not to be true! We will come back to that in detail later.) The electronic structures of the d-block elements are shown in the table below. Each additional electron usually goes into a 3d orbital. For convenience, [Ar] is used to represent 1s 2s 2p 3s 3p . This is probably the most unsatisfactory thing about this approach to the electronic structures of the . In all the chemistry of the transition elements, the 4s orbital behaves as the outermost, highest energy orbital. The reversed order of the 3d and 4s orbitals only seems to apply to atom up in the first place. In all other respects, the 4s electrons are always the electrons you need to think about first. When d-block (first row) elements form ions, the 4s electrons are lost first. Consider the electronic structure of neutral iron and iron (III). To write the electronic structure for Fe : The 4s electrons are lost first followed by one of the 3d electrons. This last bit about the formation of the ions is clearly unsatisfactory. When discussing for these elements, you talk in terms of the 4s electrons as the outer electrons being shielded from the nucleus by the inner 3d levels. We say that the first ionization energies do not change much across the transition series, because each additional 3d electron more or less screens the 4s electrons from the extra proton in the nucleus. The explanations around ionization energies are based on the 4s electrons having the higher energy, and so being removed first. The usual way of teaching this is an easy way of working out what the electronic structure of any atom is - with a few odd cases to learn like chromium or copper. The problems arise when you try to take it too literally. It is way of working out structures - no more than that. The flaw lies in the diagram we started with (Figure 1) and assuming that it applies to all atoms. In other words, we assume that the energies of the various levels are going to be those we draw in this diagram. If you stop and think about it, that has got to be wrong. As you move from element to element across the Periodic Table, protons are added to the nucleus and electrons surrounding the nucleus. The various attractions and repulsions in the atoms are bound to change as you do this - and it is those attractions and repulsions which govern the energies of the various orbitals. That means that student must rethink this on the basis that what we drew above is likely to look the same for all elements. So rather than working out the electronic structure of scandium by imagining that you just throw another electron into a calcium atom, with the electron going into a 3d orbital because the 4s is already full, you really need to look more carefully at it. Remember that, in reality, for Sc through to Zn the 3d orbitals have the lower energy - not the 4s. So why is not the electronic configuration of [Ar] 3d rather than [Ar] 3d 4s ? Imagine you are building a scandium atom from boxes of protons, neutrons and electrons. You have built the nucleus from 21 protons and 24 neutrons, and are now adding electrons around the outside. So far you have added 18 electrons to fill all the levels up as far as 3p. Essentially you have made the ion Sc . Now you are going to add the next electron to make Sc . Where will the electron go? The 3d orbitals at scandium have a lower energy than the 4s, and so the next electron will go into a 3d orbital. The structure is [Ar] 3d . You might expect the next electron to go into a lower energy 3d orbital as well, to give [Ar] 3d . But it doesn't. You have something else to think about here as well. If you add another electron to any atom, you are bound to increase the amount of repulsion. Repulsion raises the energy of the system, making it less energetically stable. It obviously helps if this effect can be kept to a minimum. The 3d orbitals are quite compactly arranged around the nucleus. Introducing a second electron into a 3d orbital produces more repulsion than if the next electron went into the 4s orbital. There is not a very big gap between the energies of the 3d and 4s orbitals. The reduction in repulsion more than compensates for the energy needed to do this. The energetically most stable structure for Sc is therefore [Ar] 3d 4s . In each of these cases we have looked at, the 3d orbitals have the lowest energy, but as we add electrons, repulsion can push some of them out into the higher energy 4s level. The difficulty with this approach is that you cannot use it to the structures of the rest of the elements in the transition series. In fact, what you have to do is to look at the actual electronic structure of a particular element and its ions, and then work out what must be happening in terms of the energy gap between the 3d and 4s orbitals and the repulsions between the electrons. The common way of teaching this (based on the wrong order of filling of the 3d and 4s orbitals for transition metals) gives a method which lets you predict the electronic structure of an atom correctly most of the time. The better way of looking at it from a theoretical point of view no longer lets you do that. You can get around this, of course. If you want to work out a structure, use the old method. But remember that it is based on a false idea, and do not try to use it for anything else - like working out which electrons will be lost first from a transition element, for example. Thinking about the other elements in the series in the same way as we did with scandium, in each case the 3d orbitals will take the first electron(s). Then at some point repulsion will push the next ones into the 4s orbital. When this happens varies from element to element. has two more electrons than scandium, and two more protons as well, of course. Think about building up a vanadium atom in exactly the same way that we did scandium. We have the nucleus complete and now we are adding electrons. When we have added 18 electrons to give the argon structure, we have then built a V ion. Now look at what happens when you add the next 5 electrons. The energy gap between the 3d and 4s levels has widened. In this case, it is not energetically profitable to promote any electrons to the 4s level until the very end. In the ions, all the electrons have gone into the 3d orbitals. You couldn't predict this just by looking at it. Why is the electronic structure of [Ar]3d 4s instead of [Ar]3d 4s ? Because that is the structure in which the balance of repulsions and the size of the energy gap between the 3d and 4s orbitals happens to produce the lowest energy for the system. Many chemistry textbooks and teachers try to explain this by saying that the half-filled orbitals minimize repulsions, but that is a flawed, incomplete argument. You are not taking into account the size of the energy gap between the lower energy 3d orbitals and the higher energy 4s orbital. Two rows directly underneath chromium in the Periodic Table is tungsten. Tungsten has exactly the same number of outer electrons as chromium, but its outer structure is 5d 6s , 5d 6s . In this case, the most energetically stable structure is not the one where the orbitals are half-full. You cannot make generalizations like this!

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Silicates are some of the most abundant minerals on Earth. They are some of the most common raw material that takes over 75% of the Earth's crust. A majority of the igneous rocks and sedimentary rocks are made of silicate minerals. The most common type of silicate is (SiO ) . There are many different types of silicates. Most of them have a general chemical formula of X Y (Z O )W . Some of the subcategories of silicates are the following: Nesosilicates are made up of units of independent tetrahedral. Some of the minerals that contain nesosilicates are olivine, garnet, zircon, kyanite, topaz, and staurolite. Olivine is important in the processes of igneous rock forming. It has a general formula of (Mg, Fe) SiO . As for garnet, it belongs to the isomorphic group, where it often occurs as dodecahedron crystals, such as pyrope, almandine, and grossularite. It is usually found in metamorphic rocks, and is known for being the January birthstone. Zircon, on the other hand, is marketed as gemstone and is oxidized to produce gemstones that are similar to diamonds known as cubic zirconia. Kyanite is a part of a polymorphic group (Al OSiO ). Ex. Sorosilicate is made up of two tetrahedrals shared by an oxygen. Some of the minerals that are classified as sorosilicates are hemimorphite, epidote, and allanite. Hemimorphite is usually found as bladed crystals. Epidote belongs to the isomorphic group, which is important in forming mineral. Lastly, allanite have metamict structure that is usually black with no cleavage. ex. Cyslosilicates are made up of closed ring units of tetrahedral sharing two oxygen atoms. They are known for their hardness and consists a variety of gemstones. They also have poor cleavage. Some minerals that are classified as cyclosilicates are beryl, cordierite, and troumaline. The gemstones that are classified as beryl include emerald (deep green), aquamarine (greenish-blue), and morganite (red). Tourmalines also have a variety of gemstones, which include rubellite (red-pink) and indicolite (dark blue). As for cordierite, it often show dichroism, meaning that it shows different colors different concentrations. Ex. Inosilicates are made up of continuous double chain units of tetrahedral, each sharing 2 and 3 oxygen. They include the pyroxene group, which are single chain minerals without hydroxide, and the amphibole group, which are double chain with hydroxide. The pyroxene group has two directional 90 degree cleavages. Some examples are enstatite-ferrosilite, diopside-hedenbergite, augite, and spodumene. As for amphibole group, it has two directional cleavages at 124-56 degrees. Some examples are tremolite-actinolite and hornblende. Both of these groups are rock-forming minerals. Ex. Phyllosilicates comprise continuous sheet units of tetrahedral, each sharing 3 oxygen atoms. They include the clay and mica minerals, which are rock-forming minerals. The clay group is made of hydrous aluminum layered silicates. Some examples are kaolinite and talc. On the other hand, the mica group consists of thin sheets and a multitude of ionic substitutions of Al and Si . Some examples are muscovite (light color), biotite (black or dark colored), and lepidolite (pink colored and a source of lithium). There is also the serpentine group that belongs to the phyllosilicates. Some examples are serpentine and crysotile. Ex. Tectosilicates consists of continuous framework of tetrahedrals, each sharing all 4 oxygen atoms. Its structure has a great amount of Al-Si substitution. Some of the groups that are classified as tectosilicates are SiO polymorphic group, K-feldspar polymorphic group, feldspathoid group and zeolite group. SiO polymorphic group has a variety of quartz, such as smoky quartz, amethyst, and jasper. Some minerals in the K-felspar polymorphic group include orthoclase and microcline. Microcline has 1 Pb ion replaced for every 2 K ion, showing an omission solid solution and causing a blue green color in the mineral. The felspathoid group minerals are similar to feldspars but only have two-thirds of the amount of silica; they form a silica deficient magma. Some examples of it are leucite and sodalite. Lastly, the zeolite group has hydrous silicates with ionic exchange and absorption properties that can act as water softeners by exchanging Na ion for Ca ion in solution. An example of it is this: \[ Na_2Al_2Si_3O_{10}-2H_2O \rightarrow CaAl_2Si_3O_{10}-2H_2O. \] Ex.

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It generally requires much work and time to evaluate integrals analytically or even numerically on a computer. Our time, computer time, and work can be saved if we can identify by inspection when integrals are zero. The determination of when integrals are zero leads to spectroscopic selection rules and provides a better understanding of them. Here we use graphs to examine properties of the wavefunctions for a particle-in-a-box to determine when the transition dipole moment integral is zero and thereby obtain the spectroscopic selection rules for this system. These considerations are completely general and can be applied to any integrals. We essentially determine whether integrals are zero or not by drawing pictures and thinking. What could be easier? Consider the case for a transition from orbital n = 1 to orbital n = 2 of a molecule described by the particle-in-a-box model. These two wavefunctions are shown in Figure \(\Page {1}\) as f1 and f2, respectively. For the curves shown on the left in the figure, we defined the box to have unit length, L = 1, and infinite potential barriers at x = 0 and x = L as we did previously, so the particle is trapped between 0 and L. For the curves shown on the right, g1 and g2, we put the origin of the coordinate system halfway between the potential barriers, i.e. at the center of the box. The barriers have not moved and the particle has not changed, but our description of the position of the barriers and the particle has changed. We now say the barriers are located at x = -L/2 and x = +L/2, and the particle is trapped between -L/2 and +L/2. Clearly the wavefunctions in Figure \(\Page {1}\) look the same for these two choices of coordinate systems. The appearance of the wavefunctions doesn’t depend on the coordinate system we have chosen or on our labels since the wavefunctions tell us about the probability of finding the particle. This probability does not change when we change the coordinate system or relabel the axis. The names of these functions do change, however. In Figure \(\Page {1a}\), they both are sine functions. In Figure \(\Page {1b}\), one is a cosine function and the other is a sine function multiplied by -1. Sketch \((f_1(x))^2\) and \((g_1(x))^2\). What do you observe? Sketch \((f_2(x))^2\) and \((g_2(x))^2\). What do you observe? What is the significance with respect to the probability given by both f(x) and g(x)? We moved the origin of the coordinate system to the center of the box to take advantage of the symmetry properties of these functions. By symmetry, we mean the correspondence in form on either side of a dividing point, line, or plane. As we shall see, the analysis of the symmetry is straightforward if the origin of the coordinate systems coincides with the dividing point, line, or plane. Since the right and left halves of the box or molecule represented by the box are the same, the square of the wavefunction for x > 0 must be the same as the square of the wavefunction for x < 0. Since the box is symmetrical, the probability density, Ψ2, for the particle distribution also must be symmetrical because there is no reason for the particle to be located preferentially on one side or the other. The transition moment integral for the particle-in-a-box involves three functions (\(ψ_f, ψ_i\), and x) that are multiplied together at each point x to form the integrand. These three functions for i = 1 and f = 2 are plotted on the left in Figure \(\Page {2}\). The integrand is the product of these three functions and is shown on the right in the figure. The integral is the area between the integrand and the zero on the y-axis. Clearly this area and thus also the value of the integral is not zero. The integral is negative because \(ψ_2\) is negative for x > 0 and x is negative for x < 0. Since \(μ_T ≠ 0\), the transition from \(ψ_1\) to \(ψ_2\) is allowed. As we previously mentioned in this chapter, and will see again later, the absorption coefficient is proportional to the absolute square of \(μ_T\) so it is acceptable for the transition moment integral to be negative. It even could involve \(\sqrt {-1}\). Taking the absolute square makes both negative and imaginary quantities positive. Write the expression or function for the integrand that is plotted on the right side of Figure \(\Page {7}\) in terms of x, sine, and cosine functions. Use your function to explain why the integrand is 0 at x = 0 and has minima at x = + 0.25L and - 0.25L. Sketch the corresponding probability function.Where are the peaks in the probability function? Now consider the transition moment integral for quantum state n = 1 to quantum state n = 3. In Figure \(\Page {3}\), the wavefunctions and the x operator are shown on the left side, and the integrand is shown on the right side. For this case we see that the integrand for x < 0 is the negative of the integrand for x > 0. This difference in sign means the net positive area for x > 0 is canceled by the net negative area for x < 0, so the total area and the transition moment integral are zero. We therefore conclude that the transition from n = 1 to n = 3 is forbidden. Write the expression or function for the integrand that is plotted on the right side of Figure \(\Page {8}\) in terms of x and cosine functions. Use your function to explain why the integrand is zero at x = 0, why is it negative just above x = 0, and why as x goes from 0 to –0.5, the integrand first is positive and then negative. In spectroscopy some special terms are used to describe the symmetry properties of wavefunctions. The terms symmetric, gerade, and even describe functions like \(f(x) = x^2\) and \(ψ_1(x)\) for the particle-in-a-box that have the property f(x) = f(-x), i.e. the function has the same values for x > 0 and for x < 0. The terms antisymmetric, ungerade, and odd describe functions like \(f(x) = x\) and \(ψ_2(x)\) for the particle-in-a-box that have the property f(x) = -f(-x), i.e. the function for x > 0 is has values that are opposite in sign compared to the function for x < 0. Gerade and ungerade are German words meaning even and odd and are abbreviated as g and u. Note that antisymmetric doesn’t mean non-symmetric. If an integrand is u, then the integral is zero! It is zero because the contribution from x > 0 is cancelled by the contribution from x < 0, as shown by the example in Figure \(\Page {8}\). An integrand will be u if the product of the functions comprising it is u. The following rules make it possible to quickly identify whether a product of two functions is u. \[g \cdot g = g, u \cdot u = g, g \cdot u = u \label {4-33}\] These rules are the same as those for multiplying +1 for g and -1 for u. The validity of these rules can be seen by examining Figures \(\Page {1}\) and \(\Page {1}\). If an integrand consists of more than two functions, the rules are applied to pairs of functions to obtain the symmetry of their product, and then applied to pairs of the product functions, and so forth, until one obtains the symmetry of the integrand. Use Mathcad or some other software to draw graphs of \(x^2, -x^2, x^3, and -x^3\) as a function of x. Which of these functions are g and which are u? Is the product function \(x^2 \cdot x^3 g\) or u? How about \(x^2 \cdot -x^2 \cdot x^3 \text {and} x^2 \cdot x^3 \cdot -x^3\)? Label each function in Figures 4.7 and 4.8 as g or u. Also label the integrands. Use symmetry arguments to determine which of the following transitions between quantum states are allowed for the particle-in-a-box: n = 2 to 3 or n = 2 to 4. Symmetry properties of functions allow us to identify when the transition moment integral and other integrals are zero. This symmetry-based approach to integration can be generalized and becomes even more powerful when concepts taken from mathematical Group Theory are used. With the tools of Group Theory, one can examine symmetry properties in three-dimensional space for complicated molecular structures. A group-theoretical analysis helps understand features in molecular spectra, predict products of chemical reactions, and simplify theoretical calculations of molecular structures and properties.

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The following are animations of gas-phase trajectories for reactants with different amounts and types of energies. The reactant energy includes a relative translational energy, E , between Cl + CH Br, and a temperature, T , for the CH Br vibrational and rotational energies. Nucleophilic substitution by a ; E = 50 kcal/mol and T = 300K Nucleophilic substitution by an , involving the Cl ---CH complex; E = 1.0 kcal/mol and T = 300K Nucleophilic substitution by an , involving both Cl ---CH Br and ClCH ---Br complex; E = 1.0 kcal/mol and T = 300K A non-reactive collision; E = 1.0 kcal/mol and T = 300K A non-reactive collision forming the Cl ---CH Br complex, which dissociates back to reactants; E = 1.0 kcal/mol and T = 300K. Nucleophilic substitution by an indirect mechanism involving both the Cl ---CH Br and ClCH ---Br complex and recrossing the transition state separating these complexes; E = 1.0 kcal/mol and T = 300K.

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Electric vehicles of all kinds are attracting increased interest everywhere, but electric bicycles are very popular in China (where there are over 120 million of them), the Netherlands, and India. They typically have a rechargeable battery pack and electric hub motor. A new electricity source combines a hydrogen fuel cell with a fuel cartridge, winner of a "Green Chemistry Challenge Award" . The sodium silicyde reacts with water to make the hydrogen fuel : The composition of sodium silicide may depend on the method of synthesis. Silicides can be made by the reaction of active metals (like Mg) with sand, or by heating sodium with silicon. Dye et al prepare sodium silicide by the reaction of sodium metal with silica gel, obtaining black powders of (hypothetically) Na Si nanoparticles. Equation (1) not only tells how many molecules of each kind are involved in a reaction, it also indicates the of each substance that is involved. Equation (1) says that 2 NaSi can react with 5 H O to give 1 Na Si O (s) and 5 H . Here we're using the term "formula unit" to indicate that the substance may not be a molecule, but rather an ionic compound or ["network crystal"]. A "formula unit" gives the composition of the substance without specifying the type of bonding. Equation (1) also says that 1 NaSi would react with 5 H O yielding 1 Na Si O (s) and 5 H . The balanced equation does more than this, though. It also tells us that 2 × 2 mol = 4 mol NaSi will react with 2 × 5 mol = 10 mol H O, and that ½ × 2 mol = 1 mol NaSi requires only ½ × 5 = 2.5 mol H O. In other words, the equation indicates that exactly 5 mol H O must react 2 mol NaSi consumed. For the purpose of calculating how much H O is required to react with a certain amount of NaSi therefore, the significant information contained in Eq. (1) is the \(\frac{\text{5 mol H}_{\text{2}}\text{O}}{\text{2 mol NaSi}}\) We shall call such a ratio derived from a balanced chemical equation a and give it the symbol . Thus, for Eq. (1), \(\text{S}\left( \frac{\text{H}_{\text{2}}\text{O}}{\text{NaSi}} \right)=\frac{\text{5 mol H}_{\text{2}}\text{O}}{\text{2 mol NaSi}}~~~~~ \text{(2)}\) The word comes from the Greek words , “element,“ and , “measure.“ Hence the stoichiometric ratio measures one element (or compound) against another. Derive all possible stoichiometric ratios from Eq. (1) Any ratio of amounts of substance given by coefficients in the equation may be used: \(\text{S}\left( \frac{\text{NaSi}}{\text{H}_{\text{2}}\text{O}} \right) =\frac{\text{2 mol NaSi}}{\text{5 mol H}_{\text{2}}\text{O}}\) \(\text{S}\left( \frac{\text{H}_{\text{2}}\text{O}}{\text{Na}_{2}\text{Si}_{2}\text{O}_{5}} \right) =\frac{\text{5 mol H}_{\text{2}}\text{O}}{\text{1 mol Na}_{2}\text{Si}_{2}\text{O}_{2}}\) \(\text{S}\left( \frac{\text{NaSi}}{\text{Na}_{2}\text{Si}_{2}\text{O}_{5}} \right) =\frac{\text{2 mol NaSi}}{\text{1 mol Na}_{2}\text{Si}_{2}\text{O}_{5}}\) \(\text{S}\left( \frac{\text{H}_{\text{2}}\text{O}}{\text{H}_{\text{2}}} \right)=\frac{\text{5 mol H}_{\text{2}}\text{O}}{\text{5 mol H}_{\text{2}}}\) \(\text{S}\left( \frac{\text{NaSi}}{\text{H}_{\text{2}}} \right) =\frac{\text{2 mol NaSi}}{\text{5 mol H}_{\text{2}}}\) \(\text{S}\left( \frac{\text{Na}_{2}\text{Si}_{2}\text{O}_{5}}{\text{H}_{\text{2}}} \right) =\frac{\text{1 mol Na}_{2}\text{Si}_{2}\text{O}_{5}}{\text{5 mol H}_{\text{2}}}\) There are six more stoichiometric ratios, each of which is the reciprocal of one of these. [Eq. (2) gives one of them.] . Using Eq. (1) as an example, this means that the ratio of the amount of H O consumed to the amount of NaSi consumed must be the stoichiometric ratio S(H O/NaSi): \(\frac{n_{\text{H}_{\text{2}}\text{O}\text{ consumed}}}{n_{\text{NaSi}\text{ consumed}}}=\text{S}\left( \frac{\text{H}_{\text{2}}\text{O}}{\text{NaSi}} \right) =\frac{\text{5 mol H}_{\text{2}}\text{O}}{\text{2 mol NaSi}}\) Similarly, the ratio of the amount of H produced to the amount of NaSi consumed must be S(H /NaSi): \(\frac{n_{\text{H}_{\text{2}}\text{ produced}}}{n_{\text{NaSi}\text{ consumed}}} =\text{S}\left( \frac{\text{H}_{\text{2}}}{\text{NaSi}} \right) =\frac{\text{5 mol H}_{\text{2}}}{\text{2 mol NaSi}}\) In general we can say that \(\text{Stoichiometric ratio }\left( \frac{\text{X}}{\text{Y}} \right)=\frac{\text{amount of X consumed or produced}}{\text{amount of Y consumed or produced}}~~~~~\text{(3a)}\) or, in symbols, \(\text{S}\left( \frac{\text{X}}{\text{Y}} \right)=\frac{n_{\text{X consumed or produced}}}{n_{\text{Y consumed or produced}}}~~~~~\text{(3b)}\) Note that in the word Eq. (3a) and the symbolic Eq. (3b), X and Y may represent reactant or product in the balanced chemical equation from which the stoichiometric ratio was derived. No matter how much of each reactant we have, the amounts of reactants and the amounts of products will be in appropriate stoichiometric ratios. Find the amount of hydrogen produced when 3.68 mol NaSi is consumed according to Eq. (1). The amount of hydrogen produced must be in the stoichiometric ratio S(H /NaSi) to the amount of ammonia consumed: \(\text{S}\left( \frac{\text{H}_{\text{2}}}{\text{NaSi}} \right) =\frac{n_{\text{H}_{\text{2}}\text{ produced}}}{n_{\text{NaSi}\text{ consumed}}}\) Multiplying both sides , by we have \(n_{\text{H}_{\text{2}}\text{ produced}} =n_{\text{NaSi}\text{ consumed}}\times \text{S}\left( \frac{\text{H}_{\text{2}}}{\text{NaSi}} \right) =\text{3}\text{.68 mol NaSi}\times \frac{\text{5 mol H}_{\text{2}}}{\text{2 mol NaSi}}=\text{9}\text{.20 mol H}_{\text{2}}\) This is a typical illustration of the use of a stoichiometric ratio as a conversion factor. Example 2 is analogous to Examples 1 and 2 from Conversion Factors and Functions, where density was employed as a conversion factor between mass and volume. Example 2 is also analogous to Examples 2.4 and 2.6, in which the Avogadro constant and molar mass were used as conversion factors. As in these previous cases, there is no need to memorize or do algebraic manipulations with Eq. (3) when using the stoichiometric ratio. Simply remember that the coefficients in a balanced chemical equation give stoichiometric ratios, and that the proper choice results in cancellation of units. In road-map form \(\text{amount of X consumed or produced}\overset{\begin{smallmatrix} \text{stoichiometric} \\ \text{ ratio X/Y} \end{smallmatrix}}{\longleftrightarrow}\text{amount of Y consumed or produced}\) or symbolically. \(n_{\text{X consumed or produced}}\text{ }\overset{S\text{(X/Y)}}{\longleftrightarrow}\text{ }n_{\text{Y consumed or produced}}\) When using stoichiometric ratios, be sure you indicate moles . You can only cancel moles of the same substance. In other words, 1 mol NaSi cancels 1 mol NaSi but does not cancel 1 mol H . The next example shows that stoichiometric ratios are also useful in problems involving the mass of a reactant or product. Suppose it is reasonable to carry about 5 pounds (2268 g) of water on a bicycle for "fuel". Calculate the mass of NaSi that needs to be supplied by a cartridge on the bicycle to completely react with the water. The problem asks that we calculate the mass of NaSi consumed from the mass of H O consumed. As we learned in Example 2 of The Molar Mass, the molar mass can be used to convert from the mass of water to the amount of water. We can then use the appropriate stoichiometric ratio to calculate the amount of NaSi that will react, and finally, use the molar mass to calculate the mass of NaSi. We require the stoichiometric ratio \(\text{S}\left( \frac{\text{NaSi}}{\text{H}_{\text{2}}} \right) =\frac{\text{2 mol NaSi}}{\text{5 mol H}_{\text{2}}}\) The of H present is \(\text{n(mol)}~=~\frac{\text{m(g)}}{\text{M(g/mol)}}\) = 2268 g/18.015 g/mol = 125.9 mol H O The amount of NaSi required is then \(n_{\text{NaSi consumed}} ~=~n_{\text{H}_{2}\text{O consumed}}~~\times~~\text{ conversion factor}\) \(=\text{125.9 mol H}_{2}\text{O}~~\times~~\frac{\text{2 mol NaSi}}{\text{5 mol H}_{\text{2}}\text{O}} =\text{50.36 mol NaSi}\) The of NaSi is \(\text{m}_{\text{NaSi}}=\text{50}\text{.36 mol NaSi}\times \frac{\text{51}\text{.08 g NaSi}}{\text{1 mol NaSI}}=\text{2572 g NaSi}\) This is a reasonably sized cartridge (about 5.67 lb). With practice this kind of problem can be solved in one step by concentrating on the units. The appropriate stoichiometric ratio will convert moles of H O to moles of NaSi and the molar mass will convert moles of NaSi to grams of NaSi. A schematic road map for the one-step calculation can be written as \(n_{\text{H}_{2}\text{O}}~~\xrightarrow{S\text{(NaSi}\text{/H}_{\text{2}}\text{O)}}~~n_{\text{NaSi}}~~\xrightarrow{M_{\text{NaSi}}}~~m_{\text{NaSi}}\) Thus \(\text{m}_{\text{NaSi}}=\text{125.9 mol H}_{\text{2}}\text{O}\times~~\frac{\text{2 mol NaSi}}{\text{5 mol H}_{\text{2}}\text{O}}~~\times~~\frac{\text{51.06 g}}{\text{1 mol NaSi}}=\text{2572 g NaSi}\) These calculations can be organized as a table, with entries below the respective reactants and products in the chemical equation. You may verify the additional calculations that have been done to show the masses of hydrogen product that would be expected. We'll fill in the remaining spots below. Suppose the 2572 g cannister of NaSi in Example 3 is completely depleted, which means completely converted to Na Si O . What mass of the product results? The problem gives the mass of NaSi and asks for the mass of Na Si O that would result from it's complete reaction with water. Thinking the problem through before trying to solve it, we realize that the molar mass of NaSi could be used to calculate the amount of NaSi consumed. Then we need a stoichiometric ratio to get the amount of Na Si O produced. Finally, the molar mass of Na Si O permits calculation of the mass of Na Si O . Symbolically \(m_{\text{NaSi}}~~\) \(\xrightarrow{M_{\text{NaSi}}}~~\) \(~~\xrightarrow{S\text{(Na}_{2}\text{Si}_{2}\text{O}_{5}\text{/NaSi)}}\) \(~~n_{\text{Na}_{\text{2}}\text{Si}_{2}\text{O}_{5}}\) \(~~\xrightarrow{M_{\text{Na}_{2}\text{Si}_{2}\text{O}_{5}}}\) \(~~m_{\text{Na}_{2}\text{Si}_{2}\text{O}_{5}}\) \(m_{\text{Na}_{2}\text{Si}_{2}\text{O}_{5}}\) \(=\text{2572 g }~~\times~~\frac{\text{1 mol NaSi}}{\text{58.08 g}}\) \(~~\times~~\frac{\text{1 mol Na}_{2}\text{Si}_{2}\text{O}_{5}}{\text{2 mol NaSi}}\) \(~~\times~~\frac{\text{182.14 g}}{\text{1 mol Na}_{2}\text{Si}_{2}\text{O}_{5}} =\text{4033 g }\) Now we can complete the table above by adding the amount of Na Si O (25.18 mol, half the amount of NaSi) and its mass, 4033 g (or about 8.9 lb). Will the bike have gained weight, since the cartridge went from 2572 g of NaSi to 4033 g of Na Si O ?

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Polychlorination of methane yields the di-, tri-, and tetrachloromethanes cheaply and efficiently: These substances have excellent solvent properties for nonpolar and slightly polar substances. Chloroform once was used widely as an inhalation anesthetic. However, it has a deleterious effect on the heart and is oxidized slowly by atmospheric oxygen to highly toxic carbonyl dichloride (phosgene, \(\ce{COCl_2}\)). Commercial chloroform contains about \(1\%\) ethanol, which destroys any \(\ce{COCl_2}\) formed by oxidation. Carbon tetrachloride commonly was employed as a cleaning solvent, although its considerable toxicity entails considerable hazard when used indiscriminately. It has been used as a fire-extinguishing fluid for petroleum fires, but its toxicity and tendency to form still more toxic carbonyl dichloride makes it undesirable for confined areas. The common laboratory practice of removing traces of water from solvents with metallic sodium should be applied to halogenated compounds; carbon tetrachloride-sodium mixtures are shock sensitive and can detonate. Trichloroethene ("Tri-Clene", bp \(87^\text{o}\)) is a widely used dry-cleaning solvent. It can be prepared from either ethene or ethyne: Compared with monohaloalkanes, polyhalogen compounds have quite different reactivities and behavior toward nucleophiles and bases. Thus dichloromethane reacts with hydroxide ion by an \(S_\text{N}2\) mechanism much less readily than methyl chloride. The chloromethanol formed then undergoes a rapid \(E2\) elimination to give methanal (formaldehyde), a substance that exists in water largely as dihydroxymethane: Trichloromethane (chloroform) reacts quite differently with base than does chloromethane or dichloromethane - as will be described in the following section. Trihalomethanes, such as trichloromethane (chloroform), are quite reactive toward strong base. The base, such as hydroxide, removes the hydrogen of \(\ce{HCCl_3}\) as a proton much more rapidly than it attacks the carbon in the \(S_\text{N}2\) manner. The carbanion so formed, \(\ce{Cl_3C}^\ominus\), is unstable and loses chloride ion to form a highly reactive neutral intermediate, \(:\ce{CCl_2}\), called : This intermediate has only six valence electrons around carbon and therefore is strongly electrophilic. In aqueous solution it reacts rapidly to form carbon monoxide and methanoate (formate) ion: The formation of \(:\ce{CCl_2}\) from \(\ce{HCCl_3}\) by the reactions of Equation 14-6 results in the elimination of \(\ce{HCl}\) - the leaving groups, \(\ce{H}\) and \(\ce{Cl}\), both originating from the carbon atom. Such reactions are not uncommon and are called \(\alpha\) eliminations or 1,1 eliminations to distinguish them from \(E1\) and \(E2\) reactions, which are \(\beta\) eliminations or 1,2 eliminations. Still other possibilities are reactions such as \(\gamma\) or 1,3 eliminations, but these take on the character of internal \(S_\text{N}2\) reactions and will not be considered in detail here. The product of \(\alpha\) elimination is a neutral species that resembles a carbocation in having only six carbon valence electrons. The simplest carbene is \(:\ce{CH_2}\), methylene. , so much so that they cannot be isolated. Their involvement in reactions usually has to be inferred from the nature of the products or the reaction kinetics. The characteristic carbene reactions involve forming an electron-pair bond to the carbene carbon by reacting with \(\sigma\) bonds, \(\pi\) bonds, or unshared pairs \(\left( n \right)\). Some of these reactions are illustrated here for methylene \(:\ce{CH_2}\).\(^1\) (insertion): ([2 + 1] cycloaddition): (dimerization, addition): Carbenes are much more reactive toward carbon-carbon double bonds than toward single bonds. Without doubt the most useful feature of \(\alpha\) elimination is that it provides a practical route to cyclopropanes and cyclopropenes by [2 + 1] cycloaddition of carbenes to double or triple bonds. These additions are stereospecific additions if they involve singlet carbenes, but can give mixtures with triplet carbenes: Carbene precursors are compounds that have or acquire good leaving groups (e.g., halide ions). Thus, halogen compounds frequently are carbene sources. Trihalomethanes are the oldest known sources of dihalocarbenes; but there are other methods for generating carbenes, and some of these are listed for reference in Table 14-2 (see also ). There is a question as to whether a “free” carbene actually is formed in some of these reactions, particularly those involving metals, but for our purposes we will classify them as routes to carbenes or carbenelike species. Many carbenes, like carbocations, rearrange to more stable structures by the migration of a neighboring group to the electron-deficient carbon. Thus phenylmethylcarbene rearranges to ethenylbenzene (styrene): Replacement of either one or two of the chlorines of carbon tetrachloride by fluorine can be achieved readily with antimony trifluoride containing some antimony pentachloride. The reaction stops after two chlorines have been replaced. The antimony trifluoride can be regenerated continuously from the antimony chloride by addition of anhydrous hydrogen fluoride: Both products are useful as refrigerants, particularly for household refrigerators and air-conditioning units, under the trade name Freon. Difluorodichloromethane (Freon 12) also is employed as a propellant in aerosol bombs, shaving-cream dispensers, and other such containers. It is nontoxic, odorless, nonflammable, and will not react with hot concentrated mineral acids or metallic sodium. This lack of reactivity is generally characteristic of the difluoromethylene group, provided the fluorines are not located on an unsaturated carbon. Attachment of a fluorine atom to a carbon atom bonded to one or more chlorine atoms tends greatly to reduce the reactivity of the chlorines toward almost all types of reagents. Possible environmental problems associated with these substances were discussed in the introduction to this chapter. During World War II, plastics and lubricating compounds of unusual chemical and thermal stability were required for many applications, in particular for pumping apparatus used to separate \(\ce{^{235}U}\) from \(\ce{^{238}U}\) by diffusion of corrosive uranium hexafluoride through porous barriers. It was natural to consider the use of substances made only of carbon and fluorine (fluorocarbons) for such purposes, and considerable effort was spent on methods of preparing compounds such as \(\ce{-(CF_2)}-_n\). Today, many such substances are in common use. These often are called "perfluoro-" compounds, which indicates that all available hydrogens of the parent compound are replaced by fluorine. Thus perfluorocyclohexane is \(\ce{(CF_2)_6}\). A widely used perfluorocarbon is the plastic material \(\ce{-(CF_2)}-_n\), which is produced in quantity by radical polymerization of tetrafluoroethene: The product ("Teflon") is a solid, chemically inert substance that is stable to around \(300^\text{o}\). It makes excellent electrical insulation and gasket materials. It also has self-lubricating properties, which are exploited in the preparation of low-adhesion surfaces (such as "nonstick" fry pans) and light-duty bearings. Tetrafluoroethene can be made on a commercial scale by the following method: The latter reaction involves difluorocarbene \(\left( :\ce{CF_2} \right)\): In the presence of peroxides, tetrafluoroethene polymerizes to the long-chain polymer. If peroxides are excluded, [2 + 2] cycloaddition occurs in high yield to give octafluorocyclobutane (see ): Similar cycloaddition reactions occur with chlorotrifluoroethene and 1,1-dichloro-2,2-difluoroethene. Radical polymerization of chlorotrifluoroethene gives a useful polymer (Kel-F) that is similar to Teflon. An excellent elastomer of high chemical resistance (Viton) can be made by copolymerizing hexafluoropropene with 1,1-difluoroethene. The product is stable to \(300^\text{o}\) and is not attacked by hot concentrated nitric acid. Although expensive, it is unrivaled among elastomers for chemical durability under extreme conditions. The fluorocarbons have extraordinarily low boiling points relative to the hydrocarbons of comparable molecular weight. As seen in Figure 14-3, their boiling points are nearly the same or even lower than those of the alkanes or cycloalkanes with the same number of carbons. Thus octafluorocyclobutane or boils \(17^\text{o}\) lower than cyclobutane, despite an almost fourfold greater molecular weight! Fluorocarbons are very insoluble in most polar solvents and are only slightly soluble in alkanes in the kerosene range. The higher-molecular-weight fluorocarbons are not even miscible in all proportions with their lower-molecular-weight homologs. The physiological properties of organofluorine compounds vary widely. Dichlorodifluoromethane and the saturated fluorocarbons appear to be completely nontoxic. In contrast, perfluoro-2-methylpropene is exceedingly toxic, more so than the war gas, carbonyl dichloride \(\left( \ce{COCl_2} \right)\). Sodium fluoroethanoate \(\left( \ce{CH_2FCO_2Na} \right)\) and 2-fluoroethanol are toxic fluorine derivatives of oxygen-containing organic substances. The fluoroethanoate salt is sold commercially as a rodenticide. Interestingly, sodium trifluoroethanoate is nontoxic. Fluorocarbon derivatives have another interesting and potentially useful property. They dissolve large quantities of oxygen. This fact, combined with their nontoxicity, has led to their use as blood replacements in heart surgery on experimental animals. Mice can live totally immersed in oxygen-saturated liquid fluorocarbons. \(^1\)Life with carbenes is substantially complicated by the fact that there are two different forms (singlet and triplet) of \(:\ce{CH_2}\) and presumably of all other carbenes. The two forms of \(:\ce{CH_2}\) differ considerably in their reactivity. One is the , which has its unshared electrons paired, while the other is the with the same electrons unpaired. For \(:\ce{CH_2}\), the singlet form is the less stable and more reactive, whereas with \(:\ce{CCl_2}\), the triplet is the less stable and more reactive. and (1977)

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This page explains what a transition metal is in terms of its electronic structure, and then goes on to look at the general features of transition metal chemistry. These include variable oxidation state (oxidation number), complex ion formation, colored ions, and catalytic activity. The terms transition metal (or element) and d block element are sometimes used as if they mean the same thing. They don't - there's a subtle difference between the two terms. We'll explore d block elements first: You will remember that when you are building the and working out where to put the electrons using the , something odd happens after argon. At argon, the 3s and 3p levels are full, but rather than fill up the 3d levels next, the 4s level fills instead to give potassium and then calcium. Only after that do the 3d levels fill. The elements in the Periodic Table which correspond to the d levels filling are called d block elements. The first row of these is shown in the shortened form of the Periodic Table below. The electronic structures of the d block elements shown are: You will notice that the pattern of filling is not entirely tidy! It is broken at both chromium and copper.Transition metals Not all d block elements count as transition metals! A transition metal is one that forms one or more stable ions which have filled d orbitals. On the basis of this definition, scandium and zinc count as transition metals - even though they are members of the d block. By contrast, copper, [Ar] 3d 4s , forms two ions. In the Cu ion the electronic structure is [Ar] 3d . However, the more common Cu ion has the structure [Ar] 3d . Copper is definitely a transition metal because the Cu ion has an incomplete d level. Here you are faced with one of the most irritating facts in chemistry at this level! When you work out the electronic structures of the first transition series (from scandium to zinc) using the , you do it on the basis that the 3d orbitals have higher energies than the 4s orbitals. That means that you work on the assumption that the 3d electrons are added after the 4s ones. However, in all the chemistry of the transition elements, the 4s orbital behaves as the outermost, highest energy orbital. When these metals form ions, the 4s electrons are always lost first. The 2+ ion is formed by the loss of the two 4s electrons. The 4s electrons are lost first followed by one of the 3d electrons. One of the key features of transition metal chemistry is the wide range of (oxidation numbers) that the metals can show. It would be wrong, though, to give the impression that only transition metals can have variable oxidation states. For example, elements like Sulfur or nitrogen or chlorine have a very wide range of oxidation states in their compounds - and these obviously aren't transition metals. However, this variability is less common in metals apart from the transition elements. Of the familiar metals from the main groups of the Periodic Table, only lead and tin show variable oxidation state to any extent. You will find the above examples and others looked at in detail if you explore the chemistry of individual metals from the transition metal menu. There is a link to this menu at the bottom of the page. We'll look at the formation of simple ions like Fe and Fe . When a metal forms an ionic compound, the formula of the compound produced depends on the energetics of the process. On the whole, the compound formed is the one in which most energy is released. The more energy released, the more stable the compound. There are several energy terms to think about, but the key ones are: The more highly charged the ion, the more electrons you have to remove and the more ionization energy you will have to provide. But off-setting this, the more highly charged the ion, the more energy is released either as lattice enthalpy or the hydration enthalpy of the metal ion. The formula for Calcium chloride is CaCl . Why is that? If you tried to make CaCl, (containing a Ca ion), the overall process is slightly exothermic. By making a Ca ion instead, you have to supply more ionization energy, but you get out lots more lattice energy. There is much more attraction between chloride ions and Ca ions than there is if you only have a 1+ ion. The overall process is very exothermic. Because the formation of CaCl releases much more energy than making CaCl, then CaCl is more stable - and so forms instead. What about CaCl ? This time you have to remove yet another electron from calcium. The first two come from the 4s level. The third one comes from the 3p. That is much closer to the nucleus and therefore much more difficult to remove. There is a large jump in ionization energy between the second and third electron removed. Although there will be a gain in lattice enthalpy, it is not anything like enough to compensate for the extra ionization energy, and the overall process is very endothermic. It definitely is not energetically sensible to make CaCl ! Here are the changes in the electronic structure of iron to make the 2+ or the 3+ ion. The 4s orbital and the 3d orbitals have very similar energies. There is not a huge jump in the amount of energy you need to remove the third electron compared with the first and second. The figures for the first three ionization energies (in kJ mol ) for iron compared with those of calcium are: There is an increase in ionization energy as you take more electrons off an atom because you have the same number of protons attracting fewer electrons. However, there is much less increase when you take the third electron from iron than from calcium. In the iron case, the extra ionization energy is compensated more or less by the extra lattice enthalpy or hydration enthalpy evolved when the 3+ compound is made. The net effect of all this is that the overall enthalpy change is not vastly different whether you make, say, FeCl or FeCl . That means that it is not too difficult to convert between the two compounds. A complex ion has a metal ion at its center with a number of other molecules or ions surrounding it. These can be considered to be attached to the central ion by coordinate ( ) bonds (in some cases, the bonding is actually more complicated). The molecules or ions surrounding the central metal ion are called ligands. Simple ligands include water, ammonia and chloride ions. What all these have got in common is active lone pairs of electrons in the outer energy level. These are used to form co-ordinate bonds with the metal ion. [Fe(H O) ] [Co(NH ) ] [Cr(OH) ] [CuCl ] Other metals also form complex ions - it is not something that only transition metals do. Transition metals do, however, form a very wide range of complex ions. The diagrams show approximate colors for some common transition metal complex ions. You will find these and others discussed if you follow links to individual metals from the transition metal menu (link at the bottom of the page). Alternatively, you could explore the complex ions menu (follow the link in the help box which has just disappeared off the top of the screen). When white light passes through a solution of one of these ions, or is reflected off it, some colors in the light are absorbed. The color you see is how your eye perceives what is left. Attaching ligands to a metal ion has an effect on the energies of the d orbitals. Light is absorbed as electrons move between one d orbital and another. This is explained in detail on another page. Transition metals and their compounds are often good catalysts. A few of the more obvious cases are mentioned below, but you will find catalysis explored in detail elsewhere on the site (follow the link after the examples). Transition metals and their compounds function as catalysts either because of their ability to change oxidation state or, in the case of the metals, to adsorb other substances on to their surface and activate them in the process. All this is explored in the main catalysis section. The combines hydrogen and nitrogen to make ammonia using an iron catalyst. This reaction is at the heart of the manufacture of from vegetable oils. However, the simplest example is the reaction between ethene and hydrogen in the presence of a nickel catalyst. At the heart of the Contact Process is a reaction which converts Sulfur dioxide into Sulfur trioxide. Sulfur dioxide gas is passed together with air (as a source of oxygen) over a solid vanadium(V) oxide catalyst. Persulphate ions (peroxodisulphate ions), S O , are very powerful oxidizing agents. Iodide ions are very easily oxidized to iodine. And yet the reaction between them in solution in water is very slow. The reaction is catalyzed by the presence of either iron(II) or iron(III) ions. \[ S_2O_8^{2-} +2I^- \rightarrow 2SO_4^{2-} + I_2\] Jim Clark ( )

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The Born-Landé equation is a concept originally formulated in 1918 by the scientists Born and Landé and is used to calculate the (measure of the strength of bonds) of a compound. This expression takes into account both the Born interactions as well as the Coulomb attractions. Due to its high simplicity and ease, the Born-Landé equation is commonly used by chemists when solving for lattice energy. This equation proposed by Max Born and Alfred Landé states that lattice energy can be derived from ionic lattice based on electrostatic potential and the potential energy due to repulsion. To solve for the Born-Landé equation, you must have a basic understanding of lattice energy: The Born-Landé equation was derived from these two following equations. the first is the electrostatic potential energy: \[ \Delta U = - \dfrac{N_A M\left | Z^+ \right | \left | Z^- \right |e^2}{4\pi\epsilon_o r} \label{1} \] with The second equation is the repulsive interaction: \[ \Delta U = \dfrac{N_A B}{r^n} \label{2}\] with These equations combine to form: \[ \Delta U (0K) = \dfrac{N_A M\left | Z^+ \right | \left | Z^- \right |e^2}{4\pi\epsilon_or_o} \left ( 1- \dfrac{1}{n} \right) \label{3}\] with Lattice energy, based on the equation from above, is dependent on multiple factors. We see that the charge of ions is proportional to the increase in lattice energy. In addition, as ions come into closer contact, lattice energy also increases. Which compound has the greatest lattice energy? This question requires basic knowledge of lattice energy. Since F gives the compound a +3 positive charge and the Al gives the compound a -1 negative charge, the compound has large electrostatic attraction. The bigger the electrostatic attraction, the greater the lattice energy. What is the lattice energy of NaCl? (Hint: you must look up the values for the constants for this compound) -756 kJ/mol (again, this value is found in a table of constants) Calculate the lattice energy of NaCl.

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Many important molecules have alternating single and double bonds (are conjugated), but have atoms that are more (or less) electron-attracting than carbon. An example is propenal (acrolein), \(18\): With such molecules we need to take into account the fact that the \(\pi\) electrons will be attracted to oxygen from carbon, because oxygen is more electronegative than carbon. With the VB method we can do this by considering ionic electron-pairing schemes, \(18c\) and \(18d\), along with the dienelike structures, \(18a\) and \(18b\). The hybrid, \(18e\), is drawn to reflect the expected relative contributions of the various forms, with \(18a\) being most important. Ionic structures such as \(19a\) and \(19b\) need not be considered for propenal because carbon is much less electron-attracting than oxygen: Analysis of the electronic configuration resulting from the MO calculations accords generally with the VB hybrid \(18e\). An especially important type of carbocation is represented by the 2-propenyl electron-pairing schemes, \(21a\) and \(21b\), which correspond to the hybrid \(21c\). Because \(21a\) and \(21b\) are equivalent and no other single low-energy structure is possible, a sizable delocalization energy is expected. Evidence for this delocalization energy of \(21c\) is available from the comparative ease of reactions involving formation of carbocation intermediates. An example is in \(S_\text{N}1\) ionizations of alkenyl and alkyl halides. The ionization \(\ce{CH_2=CHCH_2Br} \rightarrow \ce{CH_2=CHCH_2^+} + \ce{Br^+}\) proceeds than \(\ce{CH_3CH_2CH_2Br} \rightarrow \ce{CH_3CH_2CH_2^+} + \ce{Br^+}\) (for which no \(\pi\)-electron delocalization is possible). MO treatment of the 2-propenyl cation begins with the atomic-orbital model \(22\): Any \(\pi\) electrons will be delocalized through the orbitals of \(22\), but it is not so easy to be confident that when two electrons are placed into the lowest molecular orbital the resulting electron distribution will be the same as \(21c\) with half of the positive charge on \(\ce{C_1}\) and half on \(\ce{C_3}\). The complete calculation gives the result shown in Figure 21-9. Here the lowest-energy molecular orbital has a higher proportion of the \(p\) orbital of \(\ce{C_2}\) mixed in than the \(p\) orbitals of \(\ce{C_1}\) and \(\ce{C_3}\) - in fact, just the right amount to have \(\ce{C_2}\) neutral and \(\ce{C_1}\) and \(\ce{C_3}\) each with \(\frac{1}{2}^\oplus\) when this MO is filled with two paired electrons. The delocalization energy calculated for the cation is \(\left( 2 \alpha + 2.82 \beta \right) - \left( 2 \alpha + 2 \beta \right) = 0.82 \beta\) or about \(16 \: \text{kcal}\) if \(\beta\) is taken to be \(19 \: \text{kcal}\). Thus in every respect the simple VB and MO methods give the same representation of the 2-propenyl carbocation. You will notice that the 2-propenyl radical and the 2-propenyl carbanion can be formulated by the same set of \(\pi\) molecular orbitals (Figure 21-9) used for the carbocation by putting one or two electrons into the nonbonding MO. The delocalization energies calculated for the radical and anion are the same as for the cation. Thus \(\left( 3 \alpha + 2.82 \beta \right) - \left( 3 \alpha + 2 \beta \right) = 0.82 \beta\) for the radical and \(\left( 4 \alpha + 2.82 \beta \right) - \left( 4 \alpha + 2 \beta \right) = 0.82 \beta\) for the anion. covers qualitative explanations of how the VB method is used to account for the lower-energy (longer-wavelength) radiation required for electron excitation of conjugated polyenes compared to nonconjugated polyenes. Thus 1,3-butadiene has a \(\lambda_\text{max}\) for ultraviolet light at \(217 \: \text{nm}\), whereas 1,5-hexadiene has a corresponding \(\lambda_\text{max}\) at \(185 \: \text{nm}\). We will now consider how the MO approach can be used to understand these differences in excitation energy. The \(\pi\)-energy levels and electronic configurations for delocalized and localized 1,3-butadiene are shown in Figure 21-10 (also see ). Because the double bonds are so far apart, the \(\pi\)-electron system of 1,5-hexadiene by the simple MO approach is identical with that of localized 1,3-butadiene. The calculated energy change for the lowest-energy \(\pi \rightarrow \pi^*\) transition is \(\left( \alpha - 0.62 \beta \right) - \left( \alpha + 0.6 \beta \right) = -1.24 \beta\) for 1,3-butadiene and \(\left( \alpha - \beta \right) - \left( \alpha + \beta \right) = -2 \beta\) for 1,5-hexadiene. In each case the energy of the electron in the orbital (the ) is subtracted from the energy that an electron would have in the orbital (the ). Other transitions are possible, as of an electron from the lowest occupied orbital of energy \(\alpha + 1.62 \beta\) to the highest unoccupied orbital of energy \(\alpha - 1.62 \beta\), but these would have far greater energies. Qualitatively, the \(\pi \rightarrow \pi^*\) transition energy is predicted to be substantially less for 1,3-butadiene than for 1,5-hexadiene. However, any attempt at a quantitative correlation is suspect, because the lowest energy \(\pi \rightarrow \pi^*\) transition calculated for 1,3-butadiene is \(-1.24 \beta\) and, if \(\beta\) is \(19 \: \text{kcal}\) (see ), \(\lambda_\text{max}\) from Equation 9-2 should be \(1214 \: \text{nm}\) instead of the observed \(217 \: \text{nm}\). and (1977)

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There are two fundamental kinds of equilibrium problems: The equilibrium constant for the decomposition of \(CaCO_{3(s)}\) to \(CaO_{(s)}\) and \(CO_{2(g)}\) is \(K = [CO_2]\). At 800°C, the concentration of \(CO_2\) in equilibrium with solid \(CaCO_3\) and \(CaO\) is \(2.5 \times 10^{-3}\; M\). Thus \(K\) at 800°C is \(2.5 \times 10^{-3}\) (remember that equilibrium constants are unitless). A more complex example of this type of problem is the conversion of n-butane, an additive used to increase the volatility of gasoline, into isobutane (2-methylpropane). This reaction can be written as follows: and the equilibrium constant \(K = [\text{isobutane}]/[\text{n-butane}]\). At equilibrium, a mixture of n-butane and isobutane at room temperature was found to contain 0.041 M isobutane and 0.016 M n-butane. Substituting these concentrations into the equilibrium constant expression, Thus the equilibrium constant for the reaction as written is 2.6. The reaction between gaseous sulfur dioxide and oxygen is a key step in the industrial synthesis of sulfuric acid: A mixture of \(SO_2\) and \(O_2\) was maintained at 800 K until the system reached equilibrium. The equilibrium mixture contained Calculate \(K\) and \(K_p\) at this temperature. : balanced equilibrium equation and composition of equilibrium mixture : equilibrium constant Write the equilibrium constant expression for the reaction. Then substitute the appropriate equilibrium concentrations into this equation to obtain \(K\). Substituting the appropriate equilibrium concentrations into the equilibrium constant expression, To solve for \(K_p\), we need to identify \(\Delta n\) where \(Δn = 2 − 3 = −1\) and then \[\begin{align*}K_p &=K(RT)^{Δn} \\[4pt] K_p &=7.9 \times 10^4 [(0.08206\; L⋅atm/mol⋅K)(800 K)]^{−1} \\[4pt] K_p &=1.2 \times 10^3\end{align*}\] Hydrogen gas and iodine react to form hydrogen iodide via the reaction A mixture of \(H_2\) and \(I_2\) was maintained at 740 K until the system reached equilibrium. The equilibrium mixture contained Calculate \(K\) and \(K_p\) for this reaction. \(K = 48.8\) \(K_p = 48.8\) Chemists are not often given the concentrations of all the substances, and they are not likely to measure the equilibrium concentrations of all the relevant substances for a particular system. In such cases, we can obtain the equilibrium concentrations from the initial concentrations of the reactants and the balanced chemical equation for the reaction, as long as the equilibrium concentration of one of the substances is known. Example \(\Page {2}\) shows one way to do this. A 1.00 mol sample of \(NOCl\) was placed in a 2.00 L reactor and heated to 227°C until the system reached equilibrium. The contents of the reactor were then analyzed and found to contain 0.056 mol of \(Cl_2\). Calculate \(K\) at this temperature. The equation for the decomposition of \(NOCl\) to \(NO\) and \(Cl_2\) is as follows: \[\ce{2 NOCl (g) \rightleftharpoons 2NO(g) + Cl2(g)} \nonumber\] : balanced equilibrium equation, amount of reactant, volume, and amount of one product at equilibrium : \(K\) The first step in any such problem is to balance the chemical equation for the reaction (if it is not already balanced) and use it to derive the equilibrium constant expression. In this case, the equation is already balanced, and the equilibrium constant expression is as follows: \[K=\dfrac{[NO]^2[Cl_2]}{[NOCl]^2} \nonumber\] To obtain the concentrations of \(NOCl\), \(NO\), and \(Cl_2\) at equilibrium, we construct a table showing what is known and what needs to be calculated. We begin by writing the balanced chemical equation at the top of the table, followed by three lines corresponding to the initial concentrations, the changes in concentrations required to get from the initial to the final state, and the final concentrations. \[2 NOCl_{(g)} \rightleftharpoons 2NO_{(g)}+Cl_{2(g)} \nonumber\] Initially, the system contains 1.00 mol of \(NOCl\) in a 2.00 L container. Thus \([NOCl]_i = 1.00\; mol/2.00\; L = 0.500\; M\). The initial concentrations of \(NO\) and \(Cl_2\) are \(0\; M\) because initially no products are present. Moreover, we are told that at equilibrium the system contains 0.056 mol of \(Cl_2\) in a 2.00 L container, so \([Cl_2]_f = 0.056 \;mol/2.00 \;L = 0.028\; M\). We insert these values into the following table: \[2 NOCl_{(g)} \rightleftharpoons 2NO_{(g)}+Cl_{2(g)} \nonumber\] We use the stoichiometric relationships given in the balanced chemical equation to find the change in the concentration of \(Cl_2\), the substance for which initial and final concentrations are known: \[Δ[Cl_2] = 0.028 \;M_{(final)} − 0.00\; M_{(initial)}] = +0.028\; M \nonumber\] According to the coefficients in the balanced chemical equation, 2 mol of \(NO\) are produced for every 1 mol of \(Cl_2\), so the change in the \(NO\) concentration is as follows: \[Δ[NO]=\left(\dfrac{0.028\; \cancel{mol \;Cl_2}}{ L}\right)\left(\dfrac{2\; mol\; NO}{1 \cancel{\;mol \;Cl_2}}\right)=0.056\; M \nonumber\] Similarly, 2 mol of \(NOCl\) are consumed for every 1 mol of \(Cl_2\) produced, so the change in the \(NOCl\) concentration is as follows: \[Δ[NOCl]= \left(\dfrac{0.028\; \cancel{mol\; Cl_2}}{L}\right) \left(\dfrac{−2\; mol \;NOCl}{1\; \cancel{mol\; Cl_2}} \right) = -0.056 \;M \nonumber\] We insert these values into our table: \[2 NOCl_{(g)} \rightleftharpoons 2NO_{(g)}+Cl_{2(g)} \nonumber\] We sum the numbers in the \([NOCl]\) and \([NO]\) columns to obtain the final concentrations of \(NO\) and \(NOCl\): \[[NO]_f = 0.000\; M + 0.056 \;M = 0.056\; M \nonumber\] \[[NOCl]_f = 0.500\; M + (−0.056\; M) = 0.444 M \nonumber\] We can now complete the table: \[2 NOCl_{(g)} \rightleftharpoons 2NO_{(g)}+Cl_{2(g)} \nonumber\] We can now calculate the equilibrium constant for the reaction: \[K=\dfrac{[NO]^2[Cl_2]}{[NOCl]^2}=\dfrac{(0.056)^2(0.028)}{(0.444)^2}=4.5 \times 10^{−4} \nonumber\] The German chemist Fritz Haber (1868–1934; Nobel Prize in Chemistry 1918) was able to synthesize ammonia (\(NH_3\)) by reacting \(0.1248\; M \;H_2\) and \(0.0416\; M \;N_2\) at about 500°C. At equilibrium, the mixture contained 0.00272 M \(NH_3\). What is \(K\) for the reaction \[N_2+3H_2 \rightleftharpoons 2NH_3 \nonumber\] at this temperature? What is \(K_p\)? \(K = 0.105\) \(K_p = 2.61 \times 10^{-5}\) To describe how to calculate equilibrium concentrations from an equilibrium constant, we first consider a system that contains only a single product and a single reactant, the conversion of n-butane to isobutane (Equation \(\ref{Eq1}\)), for which K = 2.6 at 25°C. If we begin with a 1.00 M sample of n-butane, we can determine the concentration of n-butane and isobutane at equilibrium by constructing a table showing what is known and what needs to be calculated, just as we did in Example \(\Page {2}\). \[\text{n-butane} (g) \rightleftharpoons \text{isobutane} (g) \] The initial concentrations of the reactant and product are both known: [n-butane]i = 1.00 M and [isobutane]i = 0 M. We need to calculate the equilibrium concentrations of both n-butane and isobutane. Because it is generally difficult to calculate final concentrations directly, we focus on the change in the concentrations of the substances between the initial and the final (equilibrium) conditions. If, for example, we define the change in the concentration of isobutane (Δ[isobutane]) as +x, then the change in the concentration of n-butane is Δ[n-butane] = −x. This is because the balanced chemical equation for the reaction tells us that 1 mol of n-butane is consumed for every 1 mol of isobutane produced. We can then express the final concentrations in terms of the initial concentrations and the changes they have undergone. \[\text{n-butane}_{(g)} \rightleftharpoons \text{isobutane}_{(g)}\] Substituting the expressions for the final concentrations of n-butane and isobutane from the table into the equilibrium equation, Rearranging and solving for \(x\), \[\begin{align*} x &=2.6(1.00−x) = 2.6−2.6x \\[4pt] x+2.6x &= 2.6 \\[4pt] x&=0.72 \end{align*}\] We obtain the final concentrations by substituting this x value into the expressions for the final concentrations of n-butane and isobutane listed in the table: \[[\text{n-butane}]_f = (1.00 − x) M = (1.00 − 0.72) M = 0.28\; M\] \[[\text{isobutane}]_f = (0.00 + x) M = (0.00 + 0.72) M = 0.72\; M\] We can check the results by substituting them back into the equilibrium constant expression to see whether they give the same \(K\) that we used in the calculation: \[K=\dfrac{[\text{isobutane}]}{[\text{n-butane}]}=\left(\dfrac{0.72\; \cancel{M}}{0.28\;\cancel{M}}\right) =2.6\] This is the same \(K\) we were given, so we can be confident of our results. Example \(\Page {3}\) illustrates a common type of equilibrium problem that you are likely to encounter. The water–gas shift reaction is important in several chemical processes, such as the production of H for fuel cells. Thisreaction can be written as follows: \[H_{2(g)}+CO_{2(g)} \rightleftharpoons H_2O_{(g)}+CO_{(g)} \nonumber\] \(K = 0.106\) at 700 K. If a mixture of gases that initially contains 0.0150 M \(H_2\) and 0.0150 M \(CO_2\) is allowed to equilibrate at 700 K, what are the final concentrations of all substances present? : balanced equilibrium equation, \(K\), and initial concentrations Asked for: final concentrations : A The initial concentrations of the reactants are \([H_2]_i = [CO_2]_i = 0.0150\; M\). Just as before, we will focus on the change in the concentrations of the various substances between the initial and final states. If we define the change in the concentration of \(H_2O\) as x, then \(Δ[H_2O] = +x\). We can use the stoichiometry of the reaction to express the changes in the concentrations of the other substances in terms of x. For example, 1 mol of \(CO\) is produced for every 1 mol of \(H_2O\), so the change in the \(CO\) concentration can be expressed as \(Δ[CO] = +x\). Similarly, for every 1 mol of \(H_2O\) produced, 1 mol each of \(H_2\) and \(CO_2\) are consumed, so the change in the concentration of the reactants is \(Δ[H_2] = Δ[CO_2] = −x\). We enter the values in the following table and calculate the final concentrations. \[H_{2(g)}+CO_{2(g)} \rightleftharpoons H_2O_{(g)}+CO_{(g)} \nonumber\] We can now use the equilibrium equation and the given \(K\) to solve for \(x\): \[K=\dfrac{[H_2O,CO]}{[H_2,CO_2]}=\dfrac{(x)(x)}{(0.0150−x)(0.0150−x)}=\dfrac{x^2}{(0.0150−x)^2}=0.106 \nonumber\] We could solve this equation with the quadratic formula, but it is far easier to solve for \(x\) by recognizing that the left side of the equation is a perfect square; that is, \[\dfrac{x^2}{(0.0150−x)^2}=\left(\dfrac{x}{0.0150−x}\right)^2=0.106 \nonumber\] Taking the square root of the middle and right terms, \[\dfrac{x^2}{(0.0150−x)^2} =(0.106)^{1/2}=0.326 \nonumber\] \[x =(0.326)(0.0150)−0.326x \nonumber\] \[1.326x=0.00489 \nonumber\] \[x =0.00369=3.69 \times 10^{−3} \nonumber\] The final concentrations of all species in the reaction mixture are as follows: We can check our work by inserting the calculated values back into the equilibrium constant expression: To two significant figures, this K is the same as the value given in the problem, so our answer is confirmed. Hydrogen gas reacts with iodine vapor to give hydrogen iodide according to the following chemical equation: \(K = 54\) at 425°C. If 0.172 M \(H_2\) and \(I_2\) are injected into a reactor and maintained at 425°C until the system equilibrates, what is the final concentration of each substance in the reaction mixture? \([HI]_f = 0.270 \;M\) \([H_2]_f = [I_2]_f = 0.037\; M\) In Example \(\Page {3}\), the initial concentrations of the reactants were the same, which gave us an equation that was a perfect square and simplified our calculations. Often, however, the initial concentrations of the reactants are not the same, and/or one or more of the products may be present when the reaction starts. Under these conditions, there is usually no way to simplify the problem, and we must determine the equilibrium concentrations with other means. Such a case is described in Example \(\Page {4}\). In the water–gas shift reaction shown in Example \(\Page {3}\), a sample containing 0.632 M \(\ce{CO2}\) and 0.570 M \(\ce{H_2}\) is allowed to equilibrate at 700 K. At this temperature, \(K = 0.106\). What is the composition of the reaction mixture at equilibrium? : balanced equilibrium equation, concentrations of reactants, and \(K\) : composition of reaction mixture at equilibrium : A \([CO_2]_i = 0.632\; M\) and \([H_2]_i = 0.570\; M\). Again, x is defined as the change in the concentration of \(H_2O\): \(Δ[H_2O] = +x\). Because 1 mol of \(CO\) is produced for every 1 mol of \(H_2O\), the change in the concentration of \(CO\) is the same as the change in the concentration of H2O, so Δ[CO] = +x. Similarly, because 1 mol each of \(H_2\) and \(CO_2\) are consumed for every 1 mol of \(H_2O\) produced, \(Δ[H_2] = Δ[CO_2] = −x\). The final concentrations are the sums of the initial concentrations and the changes in concentrations at equilibrium. \[H_{2(g)}+CO_{2(g)} \rightleftharpoons H_2O_{(g)}+CO_{(g)} \nonumber\] B We can now use the equilibrium equation and the known \(K\) value to solve for \(x\): \[K=\dfrac{[H_2O,CO]}{[H_2,CO_2]}=\dfrac{x^2}{(0.570−x)(0.632−x)}=0.106 \nonumber\] In contrast to Example \(\Page {3}\), however, there is no obvious way to simplify this expression. Thus we must expand the expression and multiply both sides by the denominator: \[x^2 = 0.106(0.360 − 1.20x + x^2) \nonumber\] Collecting terms on one side of the equation, \[0.894x^2 + 0.127x − 0.0382 = 0 \nonumber\] This equation can be solved using the quadratic formula: \[ x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} = \dfrac{−0.127 \pm \sqrt{(0.127)^2−4(0.894)(−0.0382)}}{2(0.894)} \nonumber\] \[x =0.148 \text{ and } −0.290 \nonumber\] Only the answer with the positive value has any physical significance, so \(Δ[H_2O] = Δ[CO] = +0.148 M\), and \(Δ[H_2] = Δ[CO_2] = −0.148\; M\). C The final concentrations of all species in the reaction mixture are as follows: We can check our work by substituting these values into the equilibrium constant expression: \[K=\dfrac{[H_2O,CO]}{[H_2,CO_2]}=\dfrac{(0.148)^2}{(0.422)(0.484)}=0.107 \nonumber\] Because K is essentially the same as the value given in the problem, our calculations are confirmed. The exercise in Example \(\Page {1}\) showed the reaction of hydrogen and iodine vapor to form hydrogen iodide, for which \(K = 54\) at 425°C. If a sample containing 0.200 M \(H_2\) and 0.0450 M \(I_2\) is allowed to equilibrate at 425°C, what is the final concentration of each substance in the reaction mixture? \([H_I]_f = 0.0882\; M\) \([H_2]_f = 0.156\; M\) \([I_2]_f = 9.2 \times 10^{−4} M\) In many situations it is not necessary to solve a quadratic (or higher-order) equation. Most of these cases involve reactions for which the equilibrium constant is either very small (\(K ≤ 10^{−3}\)) or very large (\(K ≥ 10^3\)), which means that the change in the concentration (defined as x) is essentially negligible compared with the initial concentration of a substance. Knowing this simplifies the calculations dramatically, as illustrated in Example \(\Page {5}\). Atmospheric nitrogen and oxygen react to form nitric oxide: with \(K_p = 2.0 \times 10^{−31}\) at 25°C. What is the partial pressure of \(\ce{NO}\) in equilibrium with \(N_2\) and \(O_2\) in the atmosphere (at 1 atm) \(P_{N_2} = 0.78\; atm\) and \(P_{O_2} = 0.21\; atm\)? : balanced equilibrium equation and values of \(K_p\), \(P_{O_2}\), and \(P_{N_2}\) : partial pressure of NO : Because we are given Kp and partial pressures are reported in atmospheres, we will use partial pressures. The initial partial pressure of \(O_2\) is 0.21 atm and that of \(N_2\) is 0.78 atm. If we define the change in the partial pressure of \(NO\) as 2x, then the change in the partial pressure of \(O_2\) and of \(N_2\) is −x because 1 mol each of \(N_2\) and of \(O_2\) is consumed for every 2 mol of NO produced. Each substance has a final partial pressure equal to the sum of the initial pressure and the change in that pressure at equilibrium. Substituting these values into the equation for the equilibrium constant, \[K_p=\dfrac{(P_{NO})^2}{(P_{N_2})(P_{O_2})}=\dfrac{(2x)^2}{(0.78−x)(0.21−x)}=2.0 \times 10^{−31} \nonumber\] In principle, we could multiply out the terms in the denominator, rearrange, and solve the resulting quadratic equation. In practice, it is far easier to recognize that an equilibrium constant of this magnitude means that the extent of the reaction will be very small; therefore, the x value will be negligible compared with the initial concentrations. If this assumption is correct, then to two significant figures, (0.78 − x) = 0.78 and (0.21 − x) = 0.21. Substituting these expressions into our original equation, \[\dfrac{(2x)^2}{(0.78)(0.21)} = 2.0 \times 10^{−31 \nonumber}\] \[\dfrac{4x^2}{0.16} =2.0 \times10^{−31} \nonumber\] \[x^2=\dfrac{0.33 \times 10^{−31}}{4} \nonumber\] \[x^=9.1 \times 10^{−17} \nonumber\] Substituting this value of x into our expressions for the final partial pressures of the substances, From these calculations, we see that our initial assumption regarding x was correct: given two significant figures, \(2.0 \times 10^{−16}\) is certainly negligible compared with 0.78 and 0.21. When can we make such an assumption? As a general rule, if x is less than about 5% of the total, or \(10^{−3} > K > 10^3\), then the assumption is justified. Otherwise, we must use the quadratic formula or some other approach. The results we have obtained agree with the general observation that toxic \(NO\), an ingredient of smog, does not form from atmospheric concentrations of \(N_2\) and \(O_2\) to a substantial degree at 25°C. We can verify our results by substituting them into the original equilibrium equation: \[K_p=\dfrac{(P_{NO})^2}{(P_{N_2})(P_{O_2})}=\dfrac{(1.8 \times 10^{−16})^2}{(0.78)(0.21)}=2.0 \times 10^{−31} \nonumber\] The final \(K_p\) agrees with the value given at the beginning of this example. Under certain conditions, oxygen will react to form ozone, as shown in the following equation: \[3O_{2(g)} \rightleftharpoons 2O_{3(g)} \nonumber\] with \(K_p = 2.5 \times 10^{−59}\) at 25°C. What ozone partial pressure is in equilibrium with oxygen in the atmosphere (\(P_{O_2}=0.21\; atm\))? \(4.8 \times 10^{−31} \;atm\) Another type of problem that can be simplified by assuming that changes in concentration are negligible is one in which the equilibrium constant is very large (K ≥ 10^3). A large equilibrium constant implies that the reactants are converted almost entirely to products, so we can assume that the reaction proceeds 100% to completion. When we solve this type of problem, we view the system as equilibrating from the products side of the reaction rather than the reactants side. This approach is illustrated in Example \(\Page {6}\). The chemical equation for the reaction of hydrogen with ethylene (\(C_2H_4\)) to give ethane (\(C_2H_6\)) is as follows: \[H_{2(g)}+C_2H_{4(g)} \overset{Ni}{\rightleftharpoons} C_2H_{6(g)} \nonumber\] with \(K = 9.6 \times 10^{18}\) at 25°C. If a mixture of 0.200 M \(H_2\) and 0.155 M \(C_2H_4\) is maintained at 25°C in the presence of a powdered nickel catalyst, what is the equilibrium concentration of each substance in the mixture? : balanced chemical equation, \(K\), and initial concentrations of reactants : equilibrium concentrations : : From the magnitude of the equilibrium constant, we see that the reaction goes essentially to completion. Because the initial concentration of ethylene (0.155 M) is less than the concentration of hydrogen (0.200 M), ethylene is the limiting reactant; that is, no more than 0.155 M ethane can be formed from 0.155 M ethylene. If the reaction were to go to completion, the concentration of ethane would be 0.155 M and the concentration of ethylene would be 0 M. Because the concentration of hydrogen is greater than what is needed for complete reaction, the concentration of unreacted hydrogen in the reaction mixture would be 0.200 M − 0.155 M = 0.045 M. The equilibrium constant for the forward reaction is very large, so the equilibrium constant for the reverse reaction must be very small. The problem then is identical to that in Example \(\Page {5}\). If we define −x as the change in the ethane concentration for the reverse reaction, then the change in the ethylene and hydrogen concentrations is +x. The final equilibrium concentrations are the sums of the concentrations for the forward and reverse reactions. \[H_{2(g)}+C_2H_{4(g)} \overset{Ni}{\rightleftharpoons} C_2H_{6(g)} \nonumber\] Substituting values into the equilibrium constant expression, \[K=\dfrac{[C_2H_6]}{[H_2,C_2H_4]}=\dfrac{0.155−x}{(0.045+x)x}=9.6 \times 10^{18} \nonumber\] Once again, the magnitude of the equilibrium constant tells us that the equilibrium will lie far to the right as written, so the reverse reaction is negligible. Thus x is likely to be very small compared with either 0.155 M or 0.045 M, and the equation can be simplified [(0.045 + x) = 0.045 and (0.155 − x) = 0.155] as follows: \[K=\dfrac{0.155}{0.045x} = 9.6 \times 10^{18} \nonumber\] \[x=3.6 \times 10^{−19} \nonumber\] The small x value indicates that our assumption concerning the reverse reaction is correct, and we can therefore calculate the final concentrations by evaluating the expressions from the last line of the table: We can verify our calculations by substituting the final concentrations into the equilibrium constant expression: \[K=\dfrac{[C_2H_6]}{[H_2,C_2H_4]}=\dfrac{0.155}{(0.045)(3.6 \times 10^{−19})}=9.6 \times 10^{18} \nonumber\] This \(K\) value agrees with our initial value at the beginning of the example. Hydrogen reacts with chlorine gas to form hydrogen chloride: \[H_{2(g)}+Cl_{2(g)} \rightleftharpoons 2HCl_{(g)} \nonumber\] with \(K_p = 4.0 \times 10^{31}\) at 47°C. If a mixture of 0.257 M \(H_2\) and 0.392 M \(Cl_2\) is allowed to equilibrate at 47°C, what is the equilibrium composition of the mixture? \([H_2]_f = 4.8 \times 10^{−32}\; M\) \([Cl_2]_f = 0.135\; M\) \([HCl]_f = 0.514\; M\) Various methods can be used to solve the two fundamental types of equilibrium problems: When an equilibrium constant is calculated from equilibrium concentrations, molar concentrations or partial pressures are substituted into the equilibrium constant expression for the reaction. Equilibrium constants can be used to calculate the equilibrium concentrations of reactants and products by using the quantities or concentrations of the reactants, the stoichiometry of the balanced chemical equation for the reaction, and a tabular format to obtain the final concentrations of all species at equilibrium.

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Hydrophobic interactions describe the relations between water and (low water-soluble molecules). Hydrophobes are nonpolar molecules and usually have a long chain of carbons that do not interact with water molecules. The mixing of fat and water is a good example of this particular interaction. The common misconception is that water and fat doesn’t mix because the that are acting upon both water and fat molecules are too weak. However, this is not the case. The behavior of a fat droplet in water has more to do with the enthalpy and entropy of the reaction than its intermolecular forces. American chemist discovered that nonpolar substances like fat molecules tend to clump up together rather than distributing itself in a water medium, because this allow the fat molecules to have minimal contact with water. The image above indicates that when the hydrophobes come together, they will have less contact with water. They interact with a total of 16 water molecules before they come together and only 10 atoms after they interact. When a hydrophobe is dropped in an aqueous medium, hydrogen bonds between water molecules will be broken to make room for the hydrophobe; however, water molecules do not react with hydrophobe. This is considered an endothermic reaction, because when bonds are broken heat is put into the system. Water molecules that are distorted by the presence of the hydrophobe will make new hydrogen bonds and form an ice-like cage structure called a cage around the hydrophobe. This orientation makes the system (hydrophobe) more structured with an decrease of the total entropy of the system; therefore . The change in enthalpy (\( \Delta H \)) of the system can be negative, zero, or positive because the new hydrogen bonds can partially, completely, or over compensate for the hydrogen bonds broken by the entrance of the hydrophobe. The change in enthalpy, however, is insignificant in determining the spontaneity of the reaction (mixing of hydrophobic molecules and water) because the change in entropy ( ) is large. According to the formula with a small unknown value of and a large negative value of , the value of will turn out to be positive. A positive indicates that the mixing of the hydrophobe and water molecules is not spontaneous. The mixing hydrophobes and water molecules is not spontaneous; however, hydrophobic interactions between hydrophobes are spontaneous. When hydropobes come together and interact with each other, enthalpy increases ( \( \Delta H \) is positive) because some of hydrogen bonds that form the clathrate cage will be broken. Tearing down a portion of the clathrate cage will cause the entropy to increase ( \( \Delta S \) is positive), since forming it decreases the entropy. According to the Equation \(\ref{eq1}\) Result: negative and hence hydrophobic interactions are spontaneous. Hydrophobic interactions are relatively stronger than other weak intermolecular forces (i.e., interactions or Hydrogen bonds). The strength of Hydrophobic Interactions depend on several factors including (in order of strength of influence): Hydrophobic Interactions are important for the folding of proteins. This is important in keeping a protein stable and biologically active, because it allow to the protein to decrease in surface are and reduce the undesirable interactions with water. Besides from proteins, there are many other biological substances that rely on hydrophobic interactions for its survival and functions, like the phospholipid bilayer membranes in every cell of your body!

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The method used to extract copper from its ores depends on the nature of the ore. Sulfide ores such as chalcopyrite (\(CuFeS_2\)) are converted to copper by a different method from silicate, carbonate or sulfate ores. Chalcopyrite (also known as copper pyrites) and similar sulfide ores are the commonest ores of copper. The ores typically contain low percentages of copper and have to be concentrated before refining (e.g., via froth flotation). The concentrated ore is heated strongly with silicon dioxide (silica) and air or oxygen in a furnace or series of furnaces. An overall equation for this series of steps is: \[2CuFeS_2 + 2SiO_2 +4O_2 \rightarrow Cu_2S + 2FeSiO_3 + 3SO_2 \label{1}\] The copper(I) sulfide produced is converted to copper with a final blast of air. \[ Cu_2S + O_2 \rightarrow 2Cu + SO_2 \label{2}\] The end product of this is called - a porous brittle form of copper, about 98 - 99.5% pure. It is worthwhile spending some time sorting out what the reducing agent is in these reactions, because at first sight there does not appear to be one! Or, if you look superficially, it seems as if it might be oxygen! But that's silly! We'll start by looking at the second reaction because it is much easier to see what is happening. \[ Cu_2S + O_2 \rightarrow 2Cu + SO_2 \label{3}\] Let's look at the oxidation states of everything. That means that both the copper and the oxygen have been reduced (decrease in oxidation state). The sulfur has been oxidized (increase in oxidation state). The is therefore the sulfide ion in the copper(I) sulfide. The other reaction is more difficult to deal with, because you can't work out all of the oxidation states by following the simple rules - there are too many variables in some of the substances. You have to use some chemical knowledge as well. \[ 2CuFeS_2 + 2SiO_2 + 4O_2 \rightarrow Cu_2S + 2FeSiO_3 + 3SO_2 \label{4}\] In the CuFeS , you would have to know that the copper and iron are both in oxidation state +2, for example. You would also have to know that the oxidation state of the silicon remains unchanged at +4. So use that information to work out what has been oxidized and what reduced in this case! You should find that copper has been reduced from +2 to +1; oxygen (in the gas) has been reduced from 0 to -2 (oxygen in the SiO is unchanged); and three of the four sulfurs on the left-hand side have been oxidized from -2 to +4 (the other is unchanged). Once again, the sulfide ions are acting as the reducing agent. Copper can be extracted from non-sulfide ores by a different process involving three separate stages: When copper is made from sulfide ores by the first method above, it is impure. The blister copper is first treated to remove any remaining sulfur (trapped as bubbles of sulfur dioxide in the copper - hence "blister copper") and then cast into anodes for refining using electrolysis. The purification uses an electrolyte of copper(II) sulfate solution, impure copper anodes, and strips of high purity copper for the cathodes. The diagram shows a very simplified view of a cell. At the cathode, copper(II) ions are deposited as copper. \[ Cu^{2+}(aq) + 2e^- \rightarrow Cu(s) \label{5a}\] At the anode, copper goes into solution as copper(II) ions. \[ Cu (s) \rightarrow Cu^{2+} (aq) + 2e^- \label{6a}\] For every copper ion that is deposited at the cathode, in principle another one goes into solution at the anode. The concentration of the solution should stay the same. All that happens is that there is a transfer of copper from the anode to the cathode. The cathode gets bigger as more and more pure copper is deposited; the anode gradually disappears. In practice, it isn't quite as simple as that because of the impurities involved. Any metal in the impure anode which is below copper in the electrochemical series (reactivity series) does not go into solution as ions. It stays as a metal and falls to the bottom of the cell as an "anode sludge" together with any unreactive material left over from the ore. The anode sludge will contain valuable metals such as silver and gold. Metals above copper in the electrochemical series (like zinc) will form ions at the anode and go into solution. However, they won't get discharged at the cathode provided their concentration does not get too high. The concentration of ions like zinc will increase with time, and the concentration of the copper(II) ions in the solution will fall. For every zinc ion going into solution there will obviously be one fewer copper ion formed. (See the next note if you aren't sure about this.) The copper(II) sulfate solution has to be continuously purified to make up for this. Amongst other things copper is used for: Jim Clark ( )

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Chemistry is unique among the physical and life sciences in one very important respect. It can be manipulated extensively to man’s design. That is, molecular structures can be designed and then constructed by choosing appropriate chemical reactions. This is , which has been developed to such a degree that the economies and indeed the living standards of the industrialized nations have come to depend on it. Not everyone agrees that the present state of civilization in the industrialized nations is a way station to the millenium. But whether one agrees or not, there is no question that chemical synthesis has played an enormous role in making possible the accessories of modem life. Chemical synthesis is not a science that can be taught or learned by any well-defined set of rules. Some classify synthesis as more art than science because, as with all really creative endeavors, to be very successful requires great imagination conditioned by a wealth of background knowledge and experience. The problems of synthesis basically are problems in design and planning. Given the objective of synthesizing a specific organic compound, there always is a variety of ways that the objective can be achieved, either from the same or from different starting materials. What we hope to do here is to show how one can go about developing efficient syntheses from available starting materials. However, in planning syntheses is imperative to obtaining a good grasp of the principles and problems involved. This will be up to you; no one else can do it for you. Practice also will help greatly to convert short-term memories of organic reactions to longer-term memories through repeated review and demonstrated relevance. In almost all syntheses the is defined precisely, both as to structure and stereochemistry. Regardless of whether the synthesis is destined to be carried out on an industrial scale or on a laboratory scale, careful planning is required. The usual methodology for the planning stage involves two, not wholly independent, steps. First, one considers the various possible ways the desired can be constructed, either from smaller molecules or by changes in some existing skeleton. Second, means are considered for generation of desired on the desired carbon skeleton. In many cases, the desired functional groups can be generated as a consequence of the reactions whereby the desired skeleton itself is generated. Alternative syntheses almost always are possible and one should proceed on the notion that the first sequence one thinks of is unlikely to be the best. The choice of the best route usually is made by considering: These considerations are dealt with in the following sections and in subsequent chapters. Availability of the starting materials obviously is a limiting factor in any synthetic operation. As far as laboratory-type synthesis is concerned, “availability” means that the starting materials either may be bought “off the shelf” or may be prepared easily by standard methods from other inexpensive and available compounds. For large-scale industrial syntheses, the limiting factor usually is the cost of the starting materials, including the energy required. But in some cases the limiting factor may be problems in disposal of the byproducts. Costs will vary according to geographical location and will fluctuate widely, as with crude oil costs, so as to cause obsolescence and constant change in the chemical industry. However, it is worth remembering that the cheapest organic starting materials available are methane, ethene, ethyne, propene, butenes, benzene, and methylbenzene (toluene). Any chemical that can be prepared easily in high yield from one of these hydrocarbons is likely to be relatively inexpensive, readily available, and useful as a starting material in more involved syntheses. Among the factors considered in choosing among several possible synthetic routes is: Which gives the best yield? The definition of and its distinction from another useful term, , should be clearly understood. To help you understand, consider a specific example, the bromination of 2-methyl-propane to give -butyl bromide as the desired product. This type of reaction is carried on best with an excess of hydrocarbon to avoid polysubstitution ( ), and if we use such an excess of hydrocarbon, bromine will be the . This means simply that the amount of the desired product that could be formed is determined, or limited, by the amount of bromine used: Suppose we start with one mole of hydrocarbon and 0.2 mole of bromine and, after a specified reaction time, 0.1 mole of bromine has reacted. If only the desired product were formed, and there were no other losses of hydrocarbon or bromine, \[\% \: \text{conversion} = \frac{\text{moles of limiting reagent reacted}}{\text{moles of limiting reagent initially present}} = \frac{0.1}{0.2} \times 100 = 50\%\] If there are no losses in isolating the product or in recovering unused starting material, then \[\% \: \text{yield} = \frac{\text{moles of product}}{\text{moles of limiting reagent initially present}} = \frac{0.1}{0.1} \times 100 = 100\%\] Now suppose all of the 0.2 mole of bromine reacts, 0.08 mole of the desired product can be isolated, and 0.7 mole of hydrocarbon is recovered. Under these circumstances, the percent conversion is \(100\%\), because all of the bromine has reacted. The yield can be figured in different ways depending on which starting material one wishes to . Based on bromine (which would be logical because bromine is the more expensive reagent) the yield of -butyl bromide is \(\left( 0.08/0.2 \right) \times 100 = 40\%\). However, one also could base the yield of -butyl bromide on the unrecovered hydrocarbon, and this would be \(\left[ 0.08/\left( 1.0-0.7 \right) \right] \times 100 = 27\%\). In a multistep synthesis, the overall percent yield is the product of the fractional yields in each step times 100 and decreases rapidly with the number of steps. For this reason, a low-yield step along the way can mean practical failure for the overall sequence. Usually, the best sequence will be the one with the fewest steps. Exceptions arise when the desired product is obtained as a component of a mixture that is difficult to separate. For example, one could prepare 2-chloro-2-methylbutane in one step by direct chlorination of 2-methyl-butane ( ). But because the desired product is very difficult to separate from the other, isomeric monochlorinated products, it is desirable to use a longer sequence that may give a lower yield but avoids the separation problem. Similar separation problems would be encountered in a synthesis that gives a mixture of stereoisomers when only one isomer is desired. Again, the optimal synthesis may involve a longer sequence that would be stereospecific for the desired isomer. One way of maximizing the yield is to minimize the number of sequential steps and, whenever possible, to use rather than reactions. For example, suppose that we wish to synthesize a compound \(\ce{ABCDEF}\) by linking together \(\ce{A}\), \(\ce{B}\), \(\ce{C}\), \(\ce{D}\), \(\ce{E}\), and \(\ce{F}\). The sequential approach would involve at least five steps as follows: \[\ce{A} \overset{\ce{B}}{\rightarrow} \ce{AB} \overset{\ce{C}}{\rightarrow} \ce{ABC} \overset{\ce{D}}{\rightarrow} \ce{ABCD} \overset{\ce{E}}{\rightarrow} \ce{ABCDE} \overset{\ce{F}}{\rightarrow} \ce{ABCDEF}\] If each of these steps proceeds in \(90\%\) yield, the overall yield would be \(\left( 0.90 \right)^5 \times 100 = 59\%\). One possible parallel approach would involve synthesis of the fragments \(\ce{ABC}\) and \(\ce{DEF}\) followed by the combination of these to \(\ce{ABCDEF}\): There are still at least five reaction steps, but only three sequential steps; and if each of these proceeds in \(90\%\) yield, the overall yield would be \(\left( 0.90 \right)^3 \times 100 = 73\%\). The parallel approach is especially important in the synthesis of polymeric substances such as peptides, proteins, and nucleic acids in which many subunits have to be linked. Finally, product yields are very dependent on manipulative losses incurred in each step by isolating and purifying the synthetic intermediates. The need to minimize losses of this kind is critically important in very lengthy syntheses. and (1977)

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The term "closest packed structures" refers to the most tightly packed or space-efficient composition of crystal structures (lattices). Imagine an atom in a crystal lattice as a sphere. While cubes may easily be stacked to fill up all empty space, unfilled space will always exist in the packing of spheres. To maximize the efficiency of packing and minimize the volume of unfilled space, the spheres must be arranged as close as possible to each other. These arrangements are called . The packing of spheres can describe the solid structures of crystals. In a crystal structure, the centers of atoms, ions, or molecules lie on the lattice points. Atoms are assumed to be spherical to explain the bonding and structures of metallic crystals. These spherical particles can be packed into different arrangements. In closest packed structures, the arrangement of the spheres are densely packed in order to take up the greatest amount of space possible. When a single layer of spheres is arranged into the shape of a hexagon, gaps are left uncovered. The hole formed between three spheres is called a because it resembles a triangle. In the example below, two out of the the six trigonal holes have been highlighted green. Once the first layer of spheres is laid down, a second layer may be placed on top of it. The second layer of spheres may be placed to cover the trigonal holes from the first layer. Holes now exist between the first layer (the orange spheres) and the second (the lime spheres), but this time the holes are different. The triangular-shaped hole created over a orange sphere from the first layer is known as a . A hole from the second layer that also falls directly over a hole in the first layer is called an . In a hexagonal closest packed structure, the third layer has the same arrangement of spheres as the first layer and covers all the tetrahedral holes. Since the structure repeats itself after every two layers, the stacking for hcp may be described as "a-b-a-b-a-b." The atoms in a hexagonal closest packed structure efficiently occupy 74% of space while 26% is empty space. The arrangement in a cubic closest packing also efficiently fills up 74% of space. Similar to hexagonal closest packing, the second layer of spheres is placed on to of half of the depressions of the first layer. The third layer is completely different than that first two layers and is stacked in the depressions of the second layer, thus covering all of the octahedral holes. The spheres in the third layer are not in line with those in layer A, and the structure does not repeat until a fourth layer is added. The fourth layer is the same as the first layer, so the arrangement of layers is "a-b-c-a-b-c." A unit cell is the smallest representation of an entire crystal. All crystal lattices are built of repeating unit cells. In a unit cell, an atom's coordination number is the number of atoms it is touching. Simple Unit Cell Body-Centered Cubic Face-centered Cubic Cubic Closest Packed Hexagonal Closest Packed *For the hexagonal close-packed structure the derivation is similar. Here the unit cell consist of three primitive unit cells is a hexagonal prism containing six atoms (if the particles in the crystal are atoms). Indeed, three are the atoms in the middle layer (inside the prism); in addition, for the top and bottom layers (on the bases of the prism), the central atom is shared with the adjacent cell, and each of the six atoms at the vertices is shared with other five adjacent cells. So the total number of atoms in the cell is 3 + (1/2)×2 + (1/6)×6×2 = 6, however this results in 2 per primitive unit cell. www.quora.com/What-is-the-nu...it-cell-of-HCP

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The economies of the highly industrialized nations of the world are based in large part on energy and chemicals produced from petroleum. Although the most important and versatile intermediates for conversion of petroleum to chemicals are compounds with double or triple bonds, it also is possible to prepare many valuable substances by . In such substitutions, a hydrogen is removed from a carbon chain and another atom or group of atoms becomes attached in its place. A simple example of a substitution reaction is the formation of chloromethane and chlorine: \[ \ce{CH_4 + Cl_2 \rightarrow CH_3Cl + HCl}\] The equation for the reaction is simple, the ingredients are cheap, and the product is useful. However, if we want to decide in advance whether such a reaction is actually feasible, we have to know more. Particularly, we have to know whether the reaction proceeds in the direction it is written and, if so, whether conditions can be found under which it proceeds at a convenient rate. Obviously, if one were to mix methane and chlorine and find that, at most, only \(1 \%\) conversion to the desired product occurred and that the \(1 \%\) conversion could be achieved only after a day or so of strong heating, this reaction would be both too unfavorable and too slow for an industrial process. One way of visualizing the problems involved is with energy diagrams, which show the energy in terms of some arbitrary that is a measure of progress between the initial and final states (Figure 4-4). Diagrams such as Figure 4-4 may not be familiar to you, and a mechanical analogy may be helpful to provide better understanding of the very important ideas involved. Consider a two-level box containing a number of tennis balls. An analog to an energetically favorable reaction would be to have all of the balls on the upper level where any disturbance would cause them to roll down to the lower level under the influence of gravity, thereby losing energy. If the upper level is modified and a low fence added to hold the balls in place, it will be just as energetically favorable as when the fence is not there for the balls to be at the lower level. The difference is that the process will not occur without some major disturbance. We can say there is an to occurrence of the favorable process. Now, if we shake the box hard enough, the balls on the upper level can acquire enough energy to bounce over the barrier and drop to the lower level. The balls then can be said to acquire enough to surmount the barrier. At the molecular level, the activation energy must be acquired either by collisions between molecules as the result of their thermal motions, or from some external agency, to permit the reactants to get over the barrier and be transformed into products. We shortly will discuss this more, but first we wish to illustrate another important concept with our mechanical analogy, that of and . With gentle shaking of our two-level box, all of the balls on the upper level are expected to wind up on the lower level. There will not be enough activation to have them go from the lower to the upper level. In this circumstance, we can say that the balls are not equilibrated between the lower and upper levels. However, if we shake the box and , no matter whether we start with all of the balls on the lower or upper level, an will be set up with, on the average, most of the balls in the energetically more favorable lower level, but some in the upper level as well. To maintain a constant average fraction of the balls at each level with vigorous and continued shaking, the at which balls go from the upper to the lower level must be equal to the that they go in the opposite direction. The balls now will be between the two levels. At equilibrium, the fraction of the balls on each of the two levels is wholly independent of the height of the barrier, just as long as the activation (shaking) is sufficient to permit the balls to go ways. The diagrams of Figure 4-4 are to be interpreted in the same general way. If thermal agitation of the molecules is sufficient, then equilibrium can be expected to be established between the reactants and the products, whether the overall reaction is energetically favorable (left side of Figure 4-4) or energetically unfavorable (right side of Figure 4-4). But as with our analogy, when equilibrium is established we expect the major portion of the molecules to be in the more favorable energy state. What happens when methane is mixed with chlorine? No measurable reaction occurs when the gases are mixed and kept in the dark at room temperature. Clearly, either the reaction is energetically unfavorable or the energy barrier is high. The answer as to which becomes clear when the mixture is heated to temperatures in excess of \(300^\text{o}\) or when exposed to strong violet or ultraviolet light, whereby a rapid or even explosive reaction takes place. Therefore the reaction is energetically favorable, but the activation energy is greater than can be attained by thermal agitation alone at room temperature. Heat or light therefore must initiate a pathway for the reactants to be converted to products that has a low barrier or activation energy. Could we have predicted the results of this experiment ahead of time? First, we must recognize that there really are several questions here. Could we have decided whether the reaction was energetically favorable? That the dark reaction would be slow at room temperature? That light would cause the reaction to be fast? We consider these and some related questions in detail because they are questions and the answers to them are relevant in one way or another to the study of reactions in organic chemistry. Presumably, methane could react with chlorine to give chloromethane and hydrogen chloride, or chloromethane could react with hydrogen chloride to give methane and chlorine. If conditions were found for which both reactions proceeded at a finite rate, equilibrium finally would be established when the rates of the reactions in each direction became equal: \[ \ce{CH_4 + Cl_2 \rightleftharpoons CH_3Cl + HCl}\] At equilibrium, the relationship among the amounts of reactants and products is given by the equilibrium constant expression \[K_{eq} = \dfrac{[CH_3Cl,HCl]}{[CH_4,Cl_2]} \label{4-1}\] in which \(K_\text{eq}\) is the equilibrium constant. The quantities within the brackets of Equation \(\ref{4-1}\) denote either concentrations for liquid reactants or partial pressures for gaseous substances. If the equilibrium constant \(K_\text{eq}\) is \(1\), then on mixing equal volumes of each of the participant substances (all are gases above \(-24^\text{o}\)), reaction to the will be initially faster than reaction to the left, until equilibrium is established; at this point there will be more chloromethane and hydrogen chloride present than methane and chlorine. However, if the equilibrium constant were \(1\), the reaction initially would proceed faster to the and, at equilibrium, there would be more methane and chlorine present than chloromethane and hydrogen chloride.\(^4\) For methane chlorination, we know from experiment that the reaction goes to the right and that \(K_\text{eq}\) is much greater than unity. Naturally, it would be helpful in planning other organic preparations to be able to estimate \(K_\text{eq}\) in advance. It is a common experience to associate chemical reactions with equilibrium constants greater than one with the evolution of heat, in other words, with negative \(\Delta H^\text{0}\) values. There are, in fact, many striking examples. Formation of chloromethane and hydrogen chloride from methane and chlorine has a \(K_\text{eq}\) of \(10^{18}\) and \(\Delta H^\text{0}\) of \(-24 \: \text{kcal}\) per mole of \(CH_3Cl\) formed at \(25^\text{o}\). Combustion of hydrogen with oxygen to give water has a \(K_\text{eq}\) of \(10^{40}\) and \(\Delta H^\text{0} = -57 \: \text{kcal}\) per mole of water formed at \(25^\text{o}\). However, this correlation between \(K_\text{eq}\) and \(\Delta H^\text{0}\) is neither universal nor rigorous. Reactions are known that absorb heat (are endothermic) and yet have \(K_\text{eq} > 1\). Other reactions have large \(\Delta H^\text{0}\) values and equilibrium constants much less than \(1\). The problem is that the energy change that correlates with \(K_\text{eq}\) is not \(\Delta H^\text{0}\) but \(\Delta G^\text{0}\) (the so-called change of " ")\(^5\), and if we know \(\Delta G^\text{0}\), we can calculate \(K_\text{eq}\) by the equation \[ \Delta G^o =-2.303 RT \log_{10} K_{eq} \label{4-2}\] in which \(R\) is the gas constant and \(T\) is the absolute temperature in degrees Kelvin. For our calculations, we shall use \(R\) as \(1.987 \: \text{cal} \: \text{deg}^{-1} \: \text{mol}^{-1}\) and you should not forget to convert \(\Delta G^\text{0}\) to \(\text{cal}\). Tables of \(\Delta G^\text{0}\) values for formation of particular compounds (at various temperatures and states) from the elements are available in handbooks and the literature. With these, we can calculate equilibrium constants quite accurately. For example, handbooks give the following data, which are useful for methane chlorination: Combining these with proper regard for sign gives and \(\text{log} \: K_\text{eq} = -\left( -24.7 \times 1000 \right)/ \left(2.303 \times 1.987 \times 298.2 \right)\), so \(K_\text{eq} = 1.3 \times 10^{18}\). Unfortunately, insufficient \(\Delta G^\text{0}\) values for formation reactions are available to make this a widely applicable method for calculating \(K_\text{eq}\) values. The situation is not wholly hopeless, because there is a relationship between \(\Delta G^\text{0}\) and \(\Delta H^\text{0}\) that also involves \(T\) and another quantity, \(\Delta S^\text{0}\), the standard of the process: \[ \Delta G^o = \Delta H^o -T \Delta S^o \label{4-3}\] This equation shows that \(\Delta G^\text{0}\) and \(\Delta H^\text{0}\) are equal when \(\Delta S^\text{0}\) is zero. Therefore the sign and magnitude of \(T \Delta S^\text{0}\) determine how well \(K_\text{eq}\) correlates with \(\Delta H^\text{0}\). Now, we have to give attention to whether we can estimate \(T \Delta S^\text{0}\) values well enough to decide whether the \(\Delta H^\text{0}\) of a given reaction (calculated from bond energies or other information) will give a good or poor measure of \(\Delta G^\text{0}\). To decide whether we need to worry about \(\Delta S^\text{0}\) with regard to any particular reaction, we have to have some idea what physical meaning entropy has. To be very detailed about this subject is beyond the scope of this book, but you should try to understand the physical basis of entropy, because if you do, then you will be able to predict at least qualitatively whether \(\Delta H^\text{0}\) will be about the same or very different from \(\Delta G^\text{0}\). Essentially, the entropy of a chemical system is a measure of its or . Other things being the same, the more random the system is, the more favorable the system is. Different kinds of molecules have different degrees of translational, vibrational, and rotational freedom and, hence, different average degrees of molecular disorder or randomness. Now, if for a chemical reaction the degree of molecular disorder is different for the products than for the reactants, there will be a change in entropy and \(\Delta S^\text{0} \neq 0\). A spectacular example of the effect of molecular disorder in contributing to the difference between \(\Delta H^\text{0}\) and \(\Delta G^\text{0}\) is afforded by the formation of liquid nonane, \(C_9H_{20}\), from solid carbon and hydrogen gas at \(25^\text{o}\): \[\ce{9C(s) + 10H_2(g) \rightarrow C_910_{20}(l)}\] with \(\Delta H^o = -54.7 \, kcal\) and \(\Delta S^o = 5.0 \, kcal\). Equations \(\ref{4-2}\) and \(\ref{4-3}\) can be rearranged to calculate \(\Delta S^\text{0}\) and \(K_\text{eq}\) from \(\Delta H^\text{0}\) and \(\Delta G^\text{0}\):   and \[K_{eq} = 10^{-\Delta G^o/2.303 \,RT} = 10^{-5.900/(2.303 \times 1.987 \times 298.2)} = 4.7 \times 10^{-5}\] These \(\Delta H^\text{0}\), \(\Delta S^\text{0}\), and \(K_\text{eq}\) values can be compared to those for \(H_2 + \frac{1}{2} O_2 \longrightarrow H_2O\), for which \(\Delta H^\text{0}\) is \(-57 \: \text{kcal}\), \(\Delta S^\text{0}\) is \(8.6 \: \text{e.u.}\), and \(K_\text{eq}\) is \(10^{40}\). Obviously, there is something about the entropy change from carbon and hydrogen to nonane. The important thing is that there is a great in the constraints on the atoms on each side of the equation. In particular, hydrogen molecules in the gaseous state have great translational freedom and a high degree of disorder, the greater part of which is lost when the hydrogen atoms become attached to a chain of carbons. This makes for a large \(\Delta S^\text{0}\), which corresponds to a in \(K_\text{eq}\). The differences in constraints of the carbons are less important. Solid carbon has an ordered, rigid structure with little freedom of motion of the individual carbon atoms. These carbons are less constrained in nonane, and this would tend to make \(\Delta S^\text{0}\) more positive and \(\Delta G^\text{0}\) more negative, corresponding to an increase in \(K_\text{eq}\) (Equations \(\ref{4-2}\) and \(\ref{4-3}\)). However, this is a effect on \(\Delta S^\text{0}\) compared to the enormous difference in the degree of disorder of hydrogen between hydrogen gas and hydrogen bound to carbon in nonane. Negative entropy effects usually are observed in ring-closure reactions such as the formation of cyclohexane from 1-hexene, which occur with substantial loss of rotational freedom (disorder) about the \(C-C\) bonds: There is an even greater loss in entropy on forming cyclohexane from ethene because substantially more freedom is lost in orienting three ethene molecules to form a ring: For simple reactions, with the same number of molecules on each side of the equation, with no ring formation or other unusual changes in the constraints between the products and reactants, \(\Delta S^\text{0}\) usually is relatively small. In general, for such processes, we know from experience that \(K_\text{eq}\) \(\Delta H^\text{0}\) \(-15 \: \text{kcal}\) \(\Delta H^\text{0}\) \(+15 \: \text{kcal}\). We can use this as a "rule of thumb" to predict whether \(K_\text{eq}\) should be greater or less than unity for vapor-phase reactions involving simple molecules. Some idea of the degree of success to be expected from this rule may be inferred from the examples in Table 4-5, which also contains a further comparison of some experimental \(\Delta H^\text{0}\) values with those calculated from bond energies. Suppose \(\Delta G^\text{0}\) is positive, what hope do we have of obtaining a useful conversion to a desired product? There is no simple straightforward and general answer to this question. When the reaction is reversible the classic procedure of removing one or more of the products to prevent equilibrium from being established has many applications in organic chemistry, as will be seen later. When this approach is inapplicable, a change in reagents is necessary. Thus, iodine does not give a useful conversion with 2,2-dimethylpropane, \(1\), to give 1-iodo-2,2-dimethylpropane, \(2\), because the position of equilibrium is too far to the left (\(K_\text{eq} \cong 10^{-5}\)): Alternative routes with favorable \(\Delta G^\text{0}\) values are required. Development of ways to make indirectly, by efficient processes, what cannot be made directly is one of the most interesting and challenging activities of organic chemists. To reach an understanding of why methane and chlorine do not react in the dark, we must consider the details of the reaction occurs - that is, the . The simplest mechanism would be for a chlorine molecule to collide with a methane molecule in such a way as to have chloromethane and hydrogen chloride formed directly as a result of a breaking of the \(Cl-Cl\) and \(C-H\) bonds and making of the \(C-Cl\) and \(H-Cl\) bonds (see Figure 4-5). The failure to react indicates that there must be an energy barrier too high for this mechanism to operate. Why should this be so? First, this mechanism involves a very precisely oriented "four-center" collision between chlorine and methane that would have a low probability of occurrence (i.e., a large decrease in entropy because a precise orientation means high molecular ordering). Second, it requires pushing a chlorine molecule sufficiently deeply into a methane molecule so one of the chlorine atoms comes close enough to the carbon to form a bond and yield chloromethane. Generally, to bring nonbonded atoms to near-bonding distances (\(1.2 \: \text{A}\) to \(1.8 \: \text{A}\)) requires a large expenditure of energy, as can be seen in Figure 4-6. Interatomic repulsive forces increase rapidly at short distances, and pushing a chlorine molecule into a methane molecule to attain distances similar to the \(1.77\)-\(\text{A}\) carbon-chlorine bond distance in chloromethane would require a considerable amount of compression (see Figure 4-7). Valuable information about interatomic repulsions can be obtained with space-filling models of the CPK type ( ), which have radii scaled to correspond to actual atomic interference radii, that is, the interatomic distance at the point where curves of the type of Figure 4-6 start to rise steeply. With such models, the degree of atomic compression required to bring the nonbonded atoms to within near-bonding distance is more evident than with the ball-and-stick models. It may be noted that four-center reactions of the type postulated in Figure 4-5 are encountered only rarely. If the concerted four-center mechanism for formation of chloromethane and hydrogen chloride from chlorine and methane is discarded, all the remaining possibilities are . A slow stepwise reaction is dynamically analogous to the flow of sand through a succession of funnels with different stem diameters. The funnel with the smallest stem will be the most important bottleneck and, if its stem diameter is much smaller than the others, it alone will determine the flow rate. Generally, a multistep chemical reaction will have a slow (analogous to the funnel with the small stem) and other relatively , which may occur either before or after the slow step. A possible set of steps for the chlorination of methane follows: Reactions (1) and (2) involve dissociation of chlorine into chlorine atoms and the breaking of a \(C-H\) bond of methane to give a methyl radical and a hydrogen atom. The methyl radical, like chlorine and hydrogen atoms, has one election not involved in bonding. Atoms and radicals usually are highly reactive, so formation of chloromethane and hydrogen chloride should proceed readily by Reactions (3) and (4). The crux then will be whether Steps (1) and (2) are reasonable under the reaction conditions. In the absence of some , only collisions due to the usual thermal motions of the molecules can provide the energy needed to break the bonds. At temperatures below \(100^\text{o}\), it is very rare indeed that the thermal agitation alone can supply sufficient energy to break any significant number of bonds stronger than \(30\) to \(35 \: \text{kcal mol}^{-1}\). The \(Cl-Cl\) bond energy from Table 4-3 is \(58.1 \: \text{kcal}\), which is much too great to allow bond breaking from thermal agitation at \(25^\text{o}\) in accord with Reaction (1). For Reaction (2) it is not advisable to use the \(98.7 \: \text{kcal} \: C-H\) bond energy from Table 4-3 because this is one fourth of the energy required to break all four \(C-H\) bonds ( ). More specific are given in Table 4-5, and it will be seen that to break one \(C-H\) bond of methane requires \(104 \: \text{kcal}\) at \(25^\text{o}\), which again is too much to be gained by thermal agitation. Therefore we can conclude that Reactions (1)-(4) can not be an important mechanism for chlorination of methane at room temperature. One might ask whether dissociation into ions would provide viable mechanisms for methane chlorination. Part of the answer certainly is: Not in the vapor phase, as the following thermochemical data show: Ionic dissociation simply does not occur at ordinarily accessible temperatures by collisions between molecules in the vapor state. What is needed for formation of ions is either a highly energetic external stimulus, such as bombardment with fast-moving electrons, or an ionizing solvent that will assist ionization. Both of these processes will be discussed later. The point here is that ionic dissociation is not a viable step for the vapor-phase chlorination of methane. First, we should make clear that the light does more than provide energy merely to lift the molecules of methane and chlorine over the barrier of Figure 4-4. This is evident from the fact that very little light is needed, far less than one light photon per molecule of chloromethane produced. The light could activate either methane or chlorine, or both. However, methane is colorless and chlorine is yellow-green. This indicates that chlorine, not methane, interacts with visible light. A photon of near-ultraviolet light, such as is absorbed by chlorine gas, provides more than enough energy to split the molecule into two chlorine atoms: Once produced, a chlorine atom can remove a hydrogen atom from a methane molecule and form a methyl radical and a hydrogen chloride molecule. The bond-dissociation energies of \(CH_4\) (\(104 \: \text{kcal}\)) and \(HCl\) (\(103.1 \: \text{kcal}\)) suggest that this reaction is endothermic by about \(1 \: \text{kcal}\): Use of bond-dissociation energies gives a calculated \(\Delta H^\text{0}\) of \(-26 \: \text{kcal}\) for this reaction, which is certainly large enough, by our rule of thumb, to predict that \(K_\text{eq}\) will be greater than 1. Attack of a methyl radical on molecular chlorine is expected to require somewhat more oriented collision than for a chlorine atom reacting with methane (the chlorine molecule probably should be endwise, not sidewise, to the radical) but the interatomic repulsion probably should not be much different. The net result of \(CH_4 + Cl \cdot \longrightarrow CH_3 \cdot + HCl\) and \(CH_3 \cdot + Cl_2 \longrightarrow CH_3Cl + Cl \cdot\) is formation of chloromethane and hydrogen chloride from methane and chlorine. Notice that the chlorine atom consumed in the first step is replaced by another one in the second step. This kind of sequence of reactions is called a because, in principle, one atom can induce the reaction of an infinite number of molecules through operation of a "chain" or cycle of reactions. In our example, chlorine atoms formed by the action of light on \(Cl_2\) can induce the chlorination of methane by the : In practice, chain reactions are limited by so-called processes. In our example, chlorine atoms or methyl radicals are destroyed by reacting with one another, as shown in the following equations: Chain reactions may be considered to involve three phases: First, must occur, which for methane chlorination is activation and conversion of chlorine molecules to chlorine atoms by light. Second, steps convert reactants to products with no net consumption of atoms or radicals. The propagation reactions occur in competition with steps, which result in destruction of atoms or radicals. Putting everything together, we can write: The chain-termination reactions are expected to be exceedingly fast because atoms and radicals have electrons in unfilled shells that normally are bonding. As a result, bond formation can begin as soon as the atoms or radicals approach one another closely, without need for other bonds to begin to break. The evidence is strong that bond-forming reactions between atoms and radicals usually are , that there is almost no barrier or activation energy required, and the rates of combination are simply the rates at which encounters between radicals or atoms occur. If the rates of combination of radicals or atoms are so fast, you might well wonder how chain propagation ever could compete. Of course, competition will be possible if the propagation reactions themselves are fast, but another important consideration is the fact that the . Suppose that the concentration of \(Cl \cdot\) is \(10^{-11} \: \text{M}\) and the \(CH_4\) concentration \(1 \: \text{M}\). The probability of encounters between two \(Cl \cdot\) atoms will be proportional to \(10^{-11} \times 10^{-11}\), and between \(CH_4\) and \(Cl \cdot\) atoms it will be \(10^{-11} \times 1\). Thus, other things being the same, \(CH_4 + Cl \cdot \longrightarrow CH_3 \cdot + HCl\) (propagation) would be favored over \(2Cl \cdot \longrightarrow Cl_2\) (termination) by a factor of \(10^{11}\). Under favorable conditions, the methane-chlorination chain may go through 100 to 10,000 cycles before termination occurs by radical or atom combination. Consequently the efficiency (or ) of the reaction is very high in terms of the amount of chlorination that occurs relative to the amount of the light absorbed. The overall rates of chain reactions usually are slowed very much by substances that can combine with atoms or radicals and convert them into species incapable of participating in the chain-propagation steps. Such substances are called , or . Oxygen acts as an inhibitor in the chlorination of methane by rapidly combining with a methyl radical to form the comparatively stable (less reactive) peroxymethyl radical, \(CH_3OO \cdot\). This effectively terminates the chain: To a considerable degree, we can predict reactivities, provided we use common sense to limit our efforts to reasonable situations. In the preceding section, we argued that reactions in which atoms or radicals combine can well be expected to be extremely fast because each entity has a potentially bonding electron in an outer unfilled shell, and bringing these together to form a bond does not require that other bonds be broken: The difference between the average energy of the reactants and the energy of the transition state is called the (Figure 4-4). We expect this energy to be smaller (lower barrier) if a weak bond is being broken and a strong bond is being made. The perceptive reader will notice that we are suggesting a parallel between reaction rate and \(\Delta H^\text{0}\) because \(\Delta H^\text{0}\) depends on the difference in strengths of the bonds being broken and formed. Yet previously ( ), we pointed out that the energy barrier for a reaction need bear no relationship to how energetically feasible the reaction is, and this is indeed true for complex reactions involving many steps. But our intuitive parallel between rate and \(\Delta H^\text{0}\) usually works quite well for the rates of steps. This is borne out by experimental data on rates of removal of a hydrogen atom from methane by atoms or radicals (\(X \cdot\)), such as \(F \cdot\), \(Cl \cdot\), \(Br \cdot\), \(HO \cdot\), \(H_2N \cdot\), which generally parallel the strength of the new bond formed: Similarly, if we look at the \(H-C\) bond-dissociation energies of the hydrocarbons shown in Table 4-6, we would infer that \(Cl \cdot\) would remove a hydrogen most rapidly from the carbon forming the weakest \(C-H\) bond and, again, this is very much in accord with experience. For example, the chlorination of methylbenzene (toluene) in sunlight leads to the substitution of a methyl hydrogen rather than a ring hydrogen for the reason that the methyl \(C-H\) bonds are weaker and are attacked more rapidly than the ring \(C-H\) bonds. This can be seen explicitly in the \(\Delta H^\text{0}\) values for the chain-propagation steps calculated from the bond-dissociation energies of Table 4-6. The \(\Delta H^\text{0}\) of ring-hydrogen abstraction is unfavorable by \(+7 \: \text{kcal}\) because of the high \(C-H\) bond energy (\(110 \: \text{kcal}\)). Thus this step is not observed. It is too slow in comparison with the more favorable reaction at the methyl group even though the second propagation step is energetically favorable by \(-37 \: \text{kcal}\) and presumably would occur very rapidly. Use of bond-dissociation energies to predict relative reaction rates becomes much less valid when we try to compare different kinds of reactions. To illustrate, ethane might react with \(F \cdot\) to give fluoromethane or hydrogen fluoride: It is not a good idea to try to predict the relative rates of these two reactions on the basis of their overall \(\Delta H^\text{0}\) values because the nature of the bonds made and broken is too different. Faced with proposing a mechanism for a reaction that involves overall making or breaking of more than two bonds, the beginner almost invariably tries to concoct a process wherein, with a step, all of the right bonds break and all of the right bonds form. Such mechanisms, called , have three disadvantages. First, they are almost impossible to prove correct. Second, prediction of the relative rates of reactions involving concerted mechanisms is especially difficult. Third, concerted mechanisms have a certain sterility in that one has no control over what happens while they are taking place, except an overall control of rate by regulating concentrations, temperature, pressure, choice of solvents, and so on. To illustrate, suppose that methane chlorination appeared to proceed by way of a one-step concerted mechanism: At the instant of reaction, the reactant molecules in effect would disappear into a dark closet and later emerge as product molecules. There is no way to prove experimentally that all of the bonds were made and formed simultaneously. All one could do would be to use the most searching possible tests to probe for the existence of discrete steps. If these tests fail, the reaction still would not be concerted because other, still more searching tests might be developed later that would give a different answer. The fact is, once you accept that a particular reaction is concerted, you, in effect, accept the proposition that further work on its is futile, no matter how important you might feel that other studies would be regarding the factors affecting the reaction rate. The experienced practitioner in reaction mechanisms accepts a concerted mechanism for a reaction involving the breaking and making of more than two bonds as a last resort. He first will try to analyze the overall transformation in terms of discrete steps that are individually simple enough surely to be concerted and that also involves energetically reasonable intermediates. Such an analysis of a reaction in terms of discrete mechanistic steps offers many possibilities for experimental studies, especially in development of procedures for detecting the existence, even if highly transitory, of the proposed intermediates. We shall give many examples of the fruitfulness of this kind of approach in subsequent discussions. \(^4\)If calculations based on chemical equilibrium constants are unfamiliar to you, we suggest you study one of the general chemistry texts listed for supplemental reading at the end of Chapter 1. \(^5\)Many books and references use \(\Delta F^\text{0}\) instead of \(\Delta G^\text{0}\). The difference between standard Gibbs energy \(\Delta G^\text{0}\) and the Gibbs energy \(\Delta G\) is that \(\Delta G^\text{0}\) is defined as the value of the free energy when all of the participants are in standard states. The free energy for \(\Delta G\) for a reaction \(\text{A} + \text{B} + \cdots \longrightarrow \text{X} + \text{Y} + \cdots\) is equal to \(\Delta G^\text{0} - 2.303 RT \: \text{log} \: \frac{\left[ \text{X} \right] \left[ \text{Y} \right] \cdots}{\left[ \text{A} \right] \left[ \text{B} \right] \cdots}\) where the products, \(\left[ \text{X} \right], \left[ \text{Y} \right] \cdots\), and the reactants, \(\left[ \text{A} \right], \left[ \text{B} \right] \cdots\), do not have to be in standard states. We shall use only \(\Delta G^\text{0}\) in this book. \(^6\)The entropy unit \(\text{e.u.}\) has the dimensions calorie per degree or \(\text{cal deg}^{-1}\). and (1977)

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This page takes a brief look at how UV-visible absorption spectra can be used to help identify compounds and to measure the concentrations of colored solutions. It assumes that you know how these spectra arise, and know what is meant by terms such as absorbance, molar absorptivity and lambda-max. You also need to be familiar with the Beer-Lambert Law. If you have worked through the rest of this section, you will know that the wavelength of maximum absorption (lambda-max) depends on the presence of particular chromophores (light-absorbing groups) in a molecule. For example, on another page you will have come across the fact that a simple carbon-carbon double bond (for example in ethene) has a maximum absorption at 171 nm. The two conjugated double bonds in buta-1,3-diene have a maximum absorption at a longer wavelength of 217 nm. We also talked about the two peaks in the spectrum of ethanal (containing a simple carbon-oxygen double bond) at 180 and 290 nm. In carefully chosen simple cases (which is all you will get at this level), if you compared the peaks on a given UV-visible absorption spectrum with a list of known peaks, it would be fairly easy to pick out some structural features of an unknown molecule. Lists of known peaks often include molar absorptivity values as well. That might help you to be even more sure. For example (again using the simple carbon-oxygen double bond), data shows that the peak at 290 nm has a molar absorptivity of only 15, compared with the one at 180 nm of 10000. If your spectrum showed a very large peak at 180 nm, and an extremely small one at 290 nm, that just adds to your certainty. You should remember the Beer-Lambert Law: The expression on the left of the equation is known as the absorbance of the solution and is measured by a spectrometer. The equation is sometimes written in terms of that absorbance. \[A = \epsilon \, l \, c\] The symbol epsilon is the molar absorptivity of the solution. If you know the molar absorptivity of a solution at a particular wavelength, and you measure the absorbance of the solution at that wavelength, it is easy to calculate the concentration. The only other variable in the expression above is the length of the solution. That's easy to measure and, in fact, the cell containing the solution may well have been manufactured with a known length of 1 cm. For example, let's suppose you have a solution in a cell of length 1 cm. You measure the absorbance of the solution at a particular wavelength using a spectrometer. The value is 1.92. You find a value for molar absorptivity in a table of 19400 for that wavelength. Substituting those values: \[ A = \epsilon l c\] \[1.92 = 19400 \times 1 \times c\] \[c=\dfrac{1.92}{19400}\] \[= 9.90 \times 10^{-5}\; mol/l\] Notice what a very low concentration can be measured provided you are working with a substance with a very high molar absorptivity. This method, of course, depends on you having access to an accurate value of molar absorptivity. It also assumes that the Beer-Lambert Law works over the whole concentration range (not true!). It is much better to measure the concentration by plotting a calibration curve. Doing it this way you don't have to rely on a value of molar absorptivity, the reliability of the Beer-Lambert Law, or even know the dimensions of the cell containing the solution. What you do is make up a number of solutions of the compound you are investigating - each of accurately known concentration. Those concentrations should bracket the concentration you are trying to find - some less concentrated; some more concentrated. With colored solutions, this isn't a problem. You would just make up some solutions which are a bit lighter and some a bit darker in color. For each solution, you measure the absorbance at the wavelength of strongest absorption - using the same container for each one. Then you plot a graph of that absorbance against concentration. This is a calibration curve. According to the , absorbance is proportional to concentration, and so you would expect a straight line. That is true as long as the solutions are dilute, but the Law breaks down for solutions of higher concentration, and so you might get a curve under these circumstances. As long as you are working from values either side of the one you are trying to find, that isn't a problem. Having drawn a best fit line, the calibration curve will probably look something like the next diagram. (I've drawn it as a straight line because it is easier for me to draw than a curve(!), and it's what you will probably get if you are working with really dilute solutions. But if it turns out to be a curve, so be it!) Notice that no attempt has been made to force the line back through the origin. If the worked perfectly, it would pass through the origin, but you can't guarantee that it is working properly at the concentrations you are using. Now all you have to do is to measure the absorbance of the solution with the unknown concentration at the same wavelength. If, for example, it had an absorbance of 0.600, you can just read the corresponding concentration from the graph as above. Jim Clark ( )

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You may not be much interested in the way that organic chemistry developed, but if you skip to the next section without reading further, you will miss some of the flavor of a truly great achievement - of how a few highly creative chemists were able, with the aid of a few simple tools, to determine the structures of molecules, far too small and too elusive to be seen individually with the finest optical microscope, manifesting themselves only by the collective behavior of at least millions of millions at once. Try to visualize the problems confronting the organic chemist of 100 years ago. You will have no more than reasonably pure samples of organic compounds, the common laboratory chemicals of today, glassware, balances, thermometers, means of measuring densities, and a few optical instruments. You also will have a relatively embryonic theory that there are molecules in those bottles and that one compound differs from another because its molecules have different members or kinds of atoms and different arrangements of bonds. Your task will be to determine what kinds and what numbers of atoms they contain, that is, to determine their . Obviously, a compound with formula \(C_2H_6O\) and one with \(C_2H_6O_2\) are not the same compound. But suppose two compounds from different sources both are \(C_2H_6O\). To decide whether these are the you could smell them (far better to than to inhale), taste them (emphatically not recommended), see if they have the same appearance and viscosity (if liquids), or use more sophisticated criteria: boiling point, melting point, density, or refractive index. Other possibilities would be to see if they both have the same solubility in water or other solvents and whether they give the same reaction products with various reagents. Of course, all this gets a bit tough when the compounds are not pure and no good ways are available to purify them, but that is part of the job. Think about how you might proceed. In retrospect it is surprising that in less than fifty years an enormous, even if incomplete, edifice of structural organic chemistry was constructed on the basis of the results of chemical reactions without determination of a single bond distance, and with no electronic theory as a guide. Interestingly, all of the subsequent developments of the quantum mechanical theory of chemical bonds has not altered this edifice in significant ways. Indeed, for a long time, a goal of molecular quantum mechanics was simply to be able to corroborate that when an organic chemist draws a single line between two carbon atoms to show that they are bonded, he in fact knows what he is doing. And that when he draws two (or three) bonds between the carbons to indicate a double (or triple) bond, quantum mechanics supports this also as a valid idea. Furthermore, when modern tools for determining organic structures that involve actually measuring the distances between the atoms became available, these provided great convenience, but no great surprises. To be sure, a few structures turned out to be incorrect because they were based on faulty or inadequate experimental evidence. But, on the whole, the modern three-dimensional representations of molecules that accord with actual measurements of bond distances and angles are in no important respect different from the widely used three-dimensional ball-and-stick models of organic molecules, and these, in essentially their present form, date from at least as far back as E. Paterno, in 1869. How was all of this achieved? Not by any very simple process. The essence of some of the important ideas follow, but it should be clear that what actually took place was far from straightforward. A diverse group of people was involved; many firmly committed to, if not having a vested interest in, earlier working hypotheses or that had served as useful bases for earlier experimentation, but were coming apart at the seams because they could not accommodate the new facts that kept emerging. As is usual in human endeavors, espousal of new and better ideas did not come equally quickly to all those used to thinking in particular ways. To illustrate, at least one famous chemist, Berthelot, still used \(HO\) as the formula for water twenty-five years after it seemed clear that \(H_2O\) was a better choice. Before structures of molecules could be established, there had to be a means of establishing molecular formulas and for this purpose the key concept was Avogadro's hypothesis, which can be stated in the form "equal volumes of gases at the same temperature and pressure contain the same number of molecules." Avogadro's hypothesis allowed assignment of molecular weights from measurements of gas densities. Then, with analytical techniques that permit determination of the weight percentages of the various elements in a compound, it became possible to set up a self-consistent set of relative atomic weights.\(^1\) From these and the relative molecular weights, one can assign molecular formulas. For example, if one finds that a compound contains \(22.0 \%\) carbon (atomic weight \(= 12.00\)), \(4.6 \%\) hydrogen (atomic weight \(= 1.008\)), and \(73.4 \%\) bromine (atomic weight \(= 79.90\)), then the ratios of the numbers of atoms are \(\left( 22.0/12.00 \right) : \left( 4.6/1.008 \right) : \left( 73.4/79.90 \right) = 1.83:4.56:0.92\). Dividing each of the last set of numbers by the smallest (\(0.92\)) gives \(1.99:4.96:1 \cong 2:5:1\), which suggests a molecular formula of \(C_2H_5Br\) or a multiple thereof. If we know that hydrogen gas is \(H_2\) and has a molecular weight of \(2 \times 1.008 = 2.016\), we can compare the weight of a given volume of hydrogen with the weight of the same volume of our unknown in the gas phase at the same temperature and pressure. If the experimental ratio of these weights turns out to be \(54\), then the molecular weight of the unknown would be \(2.016 \times 54 = 109\) and the formula \(C_2H_5Br\) would be correct. If we assume that the molecule is held together by chemical bonds, without knowing more, we could write numerous structures such as \(H-H-H-H-H-C-C-Br, H-C-Br-H-H-C-H-H\), and so on. However, if we also know of the existence of stable \(H_2\), but not \(H_3\); of stable \(Br_2\), but not of \(Br_3\); and of stable \(CH_3Br\), \(CH_2Br_2\), \(CHBr_3\), and \(CBr_4\), but not of \(CH_4Br\), \(CHBr\), \(CBr\), and so on, a pattern of what is called emerges. It will be seen that the above formulas all are consistent if hydrogen atoms and bromine atoms form just bond (are univalent) while carbon atoms form bonds (are tetravalent). This may seem almost naively simple today, but a considerable period of doubt and uncertainty preceded the acceptance of the idea of definite valences for the elements that emerged about 1852. If we accept hydrogen and bromine as being univalent and carbon as tetravalent, we can write as a structural formula for \(C_2H_5Br\).\(^2\) However, we also might have written There is a serious problem as to whether these formulas represent the or compounds. All that was known in the early days was that every purified sample of \(C_2H_5Br\), no matter how prepared, had a boiling point of \(38^\text{o}C\) and density of \(1.460 \: \text{g} \: \text{ml}^{-1}\). Furthermore, all looked the same, all smelled the same, and all underwent the same chemical reactions. There was no evidence that \(C_2H_5Br\) was a mixture or that more than one compound of this formula could be prepared. One might conclude, therefore, that all of the structural formulas above represent a single substance even though they superficially, at least, look different. Indeed, because \(H-Br\) and \(Br-H\) are two different ways of a formula for the same substance, we suspect that the same is true for There are, though, two of these structures that could be different from one another, namely In the first of these, \(CH_3-\) is located opposite the \(Br-\) and the \(H-\)'s on the carbon with the \(Br\) also are opposite one another. In the second formula, \(CH_3-\) and \(Br-\) are located to each other as are the \(H-\)'s on the same carbon. We therefore have a problem as to whether these two different formulas also represent different compounds. A brilliant solution to the problem posed in the preceding section came in 1874 when J. H. van't Hoff proposed that all four valences of carbon are equivalent and directed to the corners of a regular tetrahedron.\(^3\) If we redraw the structures for \(C_2H_5Br\) as \(1\), we see that there is only possible arrangement and, contrary to the impression we got from our earlier structural formulas, the bromine is located with respect to each of the hydrogens on the same carbon. A convenient way of representing organic molecules in three dimensions, which shows the tetrahedral relationships of the atoms very clearly, uses the so-called ball-and-stick models. The sticks that represent the bonds or valences form the tetrahedral angles of \(109.47^\text{o}\). The tetrahedral carbon does not solve all problems without additional postulates. For example, there are two different compounds known with the formula \(C_2H_4Br_2\). These substances, which we call , can be reasonably written as However, ball-and-stick models suggest further possibilities for the second structure, for example \(3\), \(4\), and \(5\): This is a problem apparently first clearly recognized by Paterno, in 1869. We call these rotational (or conformational) isomers, because one is converted to another by rotation of the halves of the molecule with respect to one another, with the \(C-C\) bond acting as an axle. If this is not clear, you should make a ball-and-stick model and see what rotation around the \(C-C\) bond does to the relationships between the atoms on the carbons. The difficulty presented by these possibilities finally was circumvented by a brilliant suggestion by van't Hoff of "free rotation," which holds that isomers corresponding to different rotational angles, such as \(3\), \(4\), and \(5\), do not have separate stable existence, but are interconverted by rotation around the \(C-C\) bond so rapidly that they are indistinguishable from one another. Thus there is only isomer corresponding to the different possible rotational angles and a total of only isomers of formula \(C_2H_4Br_2\). As we shall see, the idea of free rotation required extensive modification some 50 years after it was first proposed, but it was an extremely important paradigm, which, as often happens, became so deeply rooted as to become essentially an article of faith for later organic chemists. Free rotation will be discussed in more detail in Chapters 5 and 27. The problem of determining whether a particular isomer of \(C_2H_4Br_2\) is could be solved today in a few minutes by spectroscopic means, as will be explained in Chapter 9. However, at the time structure theory was being developed, the structure had to be deduced on the basis of chemical reactions, which could include either how the compound was formed or what it could be converted to. A virtually unassailable proof of structure, where it is applicable, is to determine how many different products each of a given group of isomers can give. For the \(C_2H_4Br_2\) pair of isomers, will be seen to give only possibility with one compound and with the other: Therefore, if we have two bottles, one containing one \(C_2H_4Br_2\) isomer and one the other and run the substitution test, the compound that gives only one product is \(6\) and the one that gives a mixture of two products is \(7\). Further, it will be seen that the test, besides telling which isomer is \(6\) and which is \(7\), establishes the structures of the two possible \(C_2H_3Br_3\) isomers, \(8\) and \(9\). Thus only \(8\) can be formed from both of the different \(C_2H_4Br_2\) isomers whereas \(9\) is formed from only one of them. There were already many interconversion reactions of organic compounds known at the time that valence theory, structural formulas, and the concept of the tetrahedral carbon came into general use. As a result, it did not take long before much of organic chemistry could be fitted into a concordant whole. One difficult problem was posed by the structures of a group of substitution products of benzene, \(C_6H_6\), called "aromatic compounds," which for a long time defied explanation. Benzene itself had been prepared first by Michael Faraday, in 1825. An ingenious solution for the benzene structure was provided by A. Kekule, in 1866, wherein he suggested (apparently as the result of a hallucinatory perception) that the six carbons were connected in a hexagonal ring with alternating single and double carbon-to-carbon bonds, and with each carbon connected to a single hydrogen, \(10\): This concept was controversial, to say the least, mainly on two counts. Benzene did not behave as expected, as judged by the behavior of other compounds with carbon-to-carbon double bonds and also because there should be two different dibromo substitution products of benzene with the bromine on adjacent carbons (\(11\) and \(12\)) but only one such compound could be isolated. Kekule explained the second objection away by maintaining that \(11\) and \(12\) were in rapid equilibrium through concerted bond shifts, in something like the same manner as the free-rotation hypothesis mentioned previously: However, the first objection could not be dismissed so easily and quite a number of alternative structures were proposed over the ensuing years. The controversy was not really resolved until it was established that benzene is a regular planar hexagon, which means that all of its \(C-C\) bonds have the same length, in best accord with a structure written not with double, not with single, but with 1.5 bonds between the carbons, as in \(13\): This. in turn, generated a massive further theoretical controversy over just how \(13\) should be interpreted, which, for a time, even became a part of "Cold-War" politics!\(^4\) We shall examine experimental and theoretical aspects of the benzene structure in some detail later. It is interesting that more than 100 years after Kekule's proposal the final story on the benzene structure is yet to be told.\(^5\) The combination of valence theory and the substitution method as described in gives, for many compounds, quite unequivocal proofs of structure. Use of chemical transformations for proofs of structure depends on the applicability of a simple guiding principle, often called the " ." As we shall see later, many exceptions are known and care is required to keep from making serious errors. With this caution, let us see how the principle may be applied. The compound \(C_2H_5Br\) discussed in reacts slowly with water to give a product of formula \(C_2H_6O\). The normal valence of oxygen is two, and we can write two, and only two, different structures, \(19\) and \(20\), for \(C_2H_6O\): The principle of least structural change favors \(19\) as the product, because the reaction to form it is a simple replacement of bromine bonded to carbon by \(-OH\), whereas formation of \(20\) would entail a much more drastic rearrangement of bonds. The argument is really a subtle one, involving an assessment of the reasonableness of various possible reactions. On the whole, however, it works rather well and, in the specific case of the \(C_2H_6O\) isomers, is strongly supported by the fact that treatment of \(19\) with strong hydrobromic acid (\(HBr\)) converts it back to \(C_2H_5Br\). In contrast, the isomer of structure \(20\) reacts with \(HBr\) to form two molecules of \(CH_3Br\): In each case, \(C-O\) bonds are broken and \(C-Br\) bonds are formed. We could conceive of many other possible reactions of \(C_2H_6O\) with \(HBr\), for example which, as indicated by \(\nrightarrow\), does occur, but hardly can be ruled out by the principle of least structural change itself. Showing how the probability of such alternative reactions can be evaluated will be a very large part of our later discussions. The substitution method and the interconversion reactions discussed for proof of structure possibly may give you erroneous ideas about the reactions and reactivity of organic compounds. We certainly do not wish to imply that it is a simple, straightforward process to make all of the possible substitution products of a compound such as In fact, as will be shown later, direct substitution of bromine for hydrogen with compounds such as this does not occur readily, and when it does occur, the four possible substitution products indeed are formed, but in far from equal amounts because there are for substitution at the different positions. Actually, some of the substitution products are formed only in very small quantities. Fortunately, this does not destroy the validity of the substitution method but does make it more difficult to apply. If direct substitution fails, some (or all) of the possible substitution products may have to be produced by indirect means. Nonetheless, you must understand that the success of the substitution method depends on determination of the total number of possible isomers - it does depend on how the isomers are prepared. Later, you will hear a lot about compounds or reagents being "reactive" and "unreactive." You may be exasperated by the loose way that these terms are used by organic chemists to characterize how fast various chemical changes occur. Many familiar inorganic reactions, such as the neutralization of hydrochloric acid with sodium hydroxide solution, are extremely fast at ordinary temperatures. But the same is not often true of reactions of organic compounds. For example, \(C_2H_5Br\) treated in two different ways is converted to gaseous compounds, one having the formula \(C_2H_6\) and the other \(C_2H_4\). The \(C_2H_4\) compound, , reacts with bromine to give \(C_2H_4Br_2\), but the \(C_2H_6\) compound, , does not react with bromine except at high temperatures or when exposed to sunlight (or similar intense light). The reaction products then are \(HBr\) and \(C_2H_5Br\), and later, \(HBr\) and \(C_2H_4Br_2\), \(C_2H_3Br_3\), and so on. We clearly can characterize \(C_2H_4\) as "reactive" and \(C_2H_6\) as "unreactive" toward bromine. The early organic chemists also used the terms "unsaturated" and "saturated" for this behavior, and these terms are still in wide use today. But we need to distinguish between "unsaturated" and "reactive," and between "saturated" and "unreactive," because these pairs of terms are not synonymous. The equations for the reactions of ethene and ethane with bromine are different in that ethene bromine, \(C_2H_4 + Br_2 \rightarrow C_2H_4Br_2\), whereas ethane bromine, \(C_2H_6 + Br_2 \rightarrow C_2H_5Br + HBr\). You should reserve the term "unsaturated" for compounds that can, at least potentially, react by , and "saturated' for compounds that can only be expected to react by . The difference between addition and substitution became much clearer with the development of the structure theory that called for carbon to be tetravalent and hydrogen univalent. Ethene then was assigned a structure with a carbon-to-carbon bond, and ethane a structure with a carbon-to-carbon bond: Addition of bromine to ethene subsequently was formulated as breaking one of the carbon-carbon bonds of the double bond and attaching bromine to these valences. Substitution was written similarly but here bromine and a \(C-H\) bond are involved: We will see later that the way in which these reactions actually occur is much more complicated than these simple equations indicate. In fact, such equations are regarded best as chemical accounting operations. The number of bonds is shown correctly for both the reactants and the products, and there is an indication of which bonds break and which bonds are formed in the overall process. However, do not make the mistake of assuming that no other bonds are broken or made in intermediate stages of the reaction. Much of what comes later in this book will be concerned with what we know, or can find out, about the of such reactions - a reaction mechanism being the actual sequence of events by which the reactants become converted to the products. Such information is of extraordinary value in defining and understanding the range of applicability of given reactions for practical preparations of desired compounds. The distinction we have made between "unsaturated" and "reactive" is best illustrated by a definite example. Ethene is "unsaturated" (and "reactive") toward bromine, but tetrachloroethene, \(C_2Cl_4\), will not add bromine at all under the same conditions and is clearly "unreactive." But is it also "saturated"? The answer is definitely no, because if we add a small amount of aluminum bromide, \(AlBr_3\), to a mixture of tetrachloroethene and bromine, addition does occur, although sluggishly: Obviously, tetrachloroethene is "unsaturated" in the sense it can undergo addition, even if it is unreactive to bromine in the absence of aluminum bromide. The aluminum bromide functions in the addition of bromine to tetrachloroethene as a , which is something that facilitates the conversion of reactants to products. The study of the nature and uses of catalysts will concern us throughout this book. Catalysis is our principal means of controlling organic reactions to help form the product we want in the shortest possible time.   \(^1\)We will finesse here the long and important struggle of getting a truly self-consistent table of atomic weights. If you are interested in the complex history of this problem and the clear solution to it proposed by S. Cannizzaro in 1860, there are many accounts available in books on the history of chemistry. One example is J. R. Partington, , Vol. IV, Macmillan, London, 1964. Relative atomic weights now are based on \(^{12}C = 12\) (exactly). \(^2\)Formulas such as this appear to have been used first by Crum Brown, in 1864, after the originators of structural formulas, A. Kekule and A. Couper (1858), came up with rather awkward, impractical representations. It seems incredible today that even the drawing of these formulas was severely criticized for many years. The pot was kept boiling mainly by H. Kolbe, a productive German chemist with a gift for colorful invective and the advantage of a podium provided by being editor of an influential chemical journal. \(^3\)The name of J. A. Le Bel also is associated with this particular idea, but the record shows that Le Bel actually opposed the tetrahedral formulations, although, simultaneously with van't Hoff, he made a related very important contribution, as will be discussed in Chapter 5. \(^4\)The "resonance theory," to be discussed in detail in Chapters 6 and 21, was characterized in 1949 as a physically and ideologically inadmissible theory formulated by "decadent bourgeois scientists." See L. R. Graham, , Vintage Books, New York, 1973, Chapter VIII, for an interesting account of this controversy. \(^5\)Modern organic chemistry should not be regarded at all as a settled science, free of controversy. To be sure, personal attacks of the kind indulged in by Kolbe and others often are not published, but profound and indeed acrimonious differences of scientific interpretation exist and can persist for many years. and (1977)

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. Write detailed calculation down for each question involved in gambling, Microstates are specific configurations in which each particle is distinguishable. The number of ways that they can be arranged to describe the macrostate is the weight of that macrostate and used in the definition of entropy. Is the above reaction spontaneous under 1 atm and room temperature? If so, what are the driving factors? \[\ce{3NO(g) <=> N2O(g) + NO2(g)} \nonumber\] From a superficial overview of the reaction, we would conclude that the enthalpy change would be negative (i.e., \(\Delta H < 0\)) (why? more bonding in the products than the reactants). We would also expect a negative value of entropy (\(\Delta n_{gas} < 0\). It is difficult to argue which would win from just looking at the reaction and we need to do a Hess's law like approach to solve numerically using the Gibbs energies of formation from : \[ \Delta^o G_{rxn} = \Delta^o G_f \{ N_2O \} + \Delta G^o_f \{ NO_2 \} - 3 \Delta G^o_f \{ NO \}\nonumber\] For each of the following scenarios predict the system's entropy change whether it is \(\Delta S < 0\), \(\Delta S = 0\), or \(\Delta S > 0\). Determine whether the change in entropy for the processes listed down below are positive or negative: melts at 3,027 °C and has an enthalpy change of fusion of 31.0 kJ/mol. Calculate the entropy of fusion of osmium. Define the system at 1 mole of solid osmium at its melting point of 3027°C (3300 K). Imagine adding 31.0 kJ of heat infinitely slowly in such a way that the temperature remains constant, as the osmium melts. The heat is then equal to the q for the melting. Substitute the value of heat supplied and T into the equation that defines the entropy of the change of the entropy of a system. Because the change occurs at a constant temperature, T may be outside the integral sign \[\Delta S=\int\frac{dq_{rev}}{T}=\frac{1}{T}\int dq_{rev}=\frac{1}{T}q_{rev}=\frac{31.0\times10^{3}\;J\;mol^{-1}}{3300\;K}=9.39\;J\;K^{-1}\;mol^{-1}\nonumber\] Tip. The temperature must be an absolute temperature (in Kelvins, for example). Trouton's Rule is used to estimate the molar enthalpy of vaporization. What is iodine's enthalpy of vaporization if its normal boiling point is 184.4 C? Trouton's Rule: \[\Delta H_{vap}=\Delta S_{vap}T_{b}\] \[\Delta S_{vap}\approx 88\frac{J}{Kmol}\] \[T_{b}=184.4^{o}C=457.55K\] \[\Delta H_{vap}=(88\frac{J}{Kmol})(457.55K)=40264.4\frac{J}{mol}=40.26\frac{kJ}{mol}\] Enthalpy of vaporization of iodine is 40.26 kJ/mol If 0.250 mol Argon is expanded reversibly and isothermally at 400 K in its compressible oven from an initial vol­ume of 12.0 L to a final volume of 30.0 L, what will the ∆U, q, w, ∆H, and ∆S for the gas? For an isothermal process, ∆U = 0, and consequently, ∆H = 0. However, for the work: \[w = \int_{v_o}^{v_f} P \, dV \nonumber\] Where: \[P = \dfrac{nRT}{V} \nonumber\] meaning that if we put P into terms of V using PV=nRT, then we can get the work, which comes out look like: \[w= nRT \ln{\dfrac{V_f}{V_o}} \nonumber\] \[w = (0.250 \; mol) \times (8.314 \; \mathrm{\dfrac{J}{K mol}} ) \times (400 \; \mathrm{K}) \times \ln{ \dfrac{30.0 \; L}{12.0 \; L}} = 761.8 \; \mathrm{J} \nonumber\] \(PV = nRT\) is the ideal gas law. it implies that the molecules or atoms of the gas are point masses, they have no volume and undergo only elastic collision. Therefore, by the first law, the heat, q, must be be the negative of the work. Now, as for ∆S, we know: \[\Delta{S}= \dfrac{-q}{T}\nonumber\] which implies that: \[\Delta{S}= nR\ln{\dfrac{V_f}{V_o}}\nonumber\] \[\Delta{S}= (0.250 \; \mathrm{mol}) \times (8.314 \; \mathrm{\dfrac{J}{K mol}} ) \times \ln{ \dfrac{30.0 \; \mathrm{L}}{12.0 \; \mathrm{L}}} =1.90 \; \mathrm{\dfrac{J}{K}}\nonumber\] Consider this. Exactly 2 moles of ice undergoes three different processes. Find temperature X & Y and identify \(ΔS_{\text{surr}}\). Given: Since entropy changes are additive (thanks for them being state functions) and we have the entropy changes for Step I and Step III, all we need to do is calculate \(\Delta S_{II}\). However, we need to calculate temperatures first. The equation below can be used to calculate \(ΔS_{sys}\) for a temperature change. \[\Delta S = nC_{p} \ln \left(\dfrac{T_{2}}{T_{1}}\right)\] To solve for unknown temperature, the equation can be arranged in this way. \[\exp(\dfrac{\Delta S}{nC_{p}}) = \dfrac{T_{2}}{T_{1}}\] Assuming the \(C_p\) stays constant throughout all the temperature change. Initial temperature X and final temperature Y can be found using the formula below. \[T_{x}=T_{2}(\exp(\dfrac{\Delta S_{\text{I}}}{nC_{p}}))^{-1}=(273.15K)(\exp(\dfrac{12.034 J.K^{-1}}{(2.00 mol)(38 J.K^{-1}.mol^{-1})}))^{-1}=233.15K\] \[T_{y}=T_{1}(\exp(\dfrac{\Delta S_{\text{III}}}{nC_{p}}))=(273.15K)(\exp(\dfrac{136 J.K^{-1}}{(2.00 mol)(75 J.K^{-1}.mol^{-1})}))=298.15K\] ΔS must also be determined to obtain the ΔS . Since the temperature remains constant in the second step, \(ΔS = q_{rev}\). \[\Delta S_{\text{II}}= 2\,mol \times \dfrac{6007\,J/mol}{273.15\,K}=43.984\, J.K^{-1}\] Since the entire process is reversible, \[\mathrm{\Delta {S_{\text{surr}}} + \Delta {S_{\text{system}}} = 0}\nonumber\] so \[\mathrm{\Delta {S_{\text{surr}}} = - \Delta {S_{\text{system}}} }.\nonumber\] Now we add the three entropy changes: \[ \begin{align} -\Delta S_{\text{system}} &= -(\Delta S_{\text{I}} + \Delta S_{\text{II}} +\Delta S_{\text{III}}) \\[5pt] &= -(12.034 + 54.98 + 136)J.K^{-1} \\[5pt] &= -80.15\, J.K^{-1} \\[5pt] &= \Delta S_{\text{surr}} \end{align}\] Suppose 2 moles of water at standard temperature (25°C) and pressure is spontaneously evaporated by allowing it to fall onto a nickel plate maintained at 125°C. Calculate \(\Delta S\) for the water, \(\Delta S\) for the nickel plate, and \(\Delta S_{total}\) if \(C_{p\ce{(H2O)(l)}}\)= 75.4 J/(K.mol) and \(C_{p\ce{(H2O) (g)}}\) = 36.0 J/(K.mol). Take \(\Delta H_{vap}\) = 40.68 KJ/mol for water and its boiling point of 100°C. \[q_{1} = \Delta H = mc\Delta T = 2 \times 75.4 \dfrac{J}{K.mol} \times (100-25) = 11310 \; J\] \[q_{2} = 40680 \dfrac{J}{mol} \times 2 = 81.36 \times 10^{3} \; J\] \[q_{3} = \Delta H = mc\Delta T = 2 \times 36 \dfrac{J}{K.mol} \times (125-100) = 1800 \; J\] \[q_{total} = 11310 + 81.36 \times 10^{3} + 1800 = 94470 \; J\] \[\Delta S_{\ce{H2O}} = nC_{p(\ce{H2O(l)}) }) \ln{\dfrac{T_{2}}{T_{1}}} + nC_{p}(H_{2}O(g)) \ln {\dfrac{T_{2}}{T_{1}}} + n\dfrac{\Delta H}{T}\] \[ = (2\times 75.4 \times \ln{\dfrac{373}{298}}) + (2 \times 36 \times \ln {\dfrac{398}{373}}) + (2\times \dfrac{40680}{373}) = 256.6 \dfrac{J}{K}\] \[\Delta S_{iron} = \dfrac{-94470 J}{398 K} = -237.4 \dfrac{J}{K}\] \[\Delta S_{tot} = \Delta S_{\ce{H2O}} + \Delta S_{\text{iron}} = 19.2 \dfrac{J}{K} \] A 181.49 g sample of lead at 97.0 C initially is added to a coffee cup calorimeter that contains 150.0 g water which is at 24.7 C. The equilibrium temperature is 29.4 C, assuming that there is no heat lost to the calorimeter or the environment. The molar heat capacity of lead (C (Pb)) is 26.4 J K mol and that of water (C (H O)) is 75.2 J K mol . What is \(\Delta S\) for the lead sample, \(\Delta S\) for the water sample, and \(\Delta S_{total}\) for this process? Since this progress is carried out at a constant pressure, and the temperature change for this progress doesn't include a phase transition, the relationship between the \(\Delta S\) and the \(\Delta T\) of a system can be described as: \[ \Delta S = nc_{p}ln\left(\frac{T_{2}}{T_{1}}\right)\] Convert the temperatures to Kelvin: \[T_{equilibrium} = (29.4 + 273.15)K = 302.55 \; K \] \[T_{initial, \; Pb} = (97.0 + 273.15) K = 370.15 \; K \] \[T_{initial, \; H_{2}O} = (24.7 + 273.15) K = 297.85 \; K \] Calculate the amount of moles of the substance present: \[ n_{Pb} = 181.49 \; g \; Pb \times \frac{1 \; mol \; Pb}{207.2 \; g \; Pb} = 0.8759 \; mol \; Pb \] \[ n_{H_{2}O} = 150.0 \; g \; H_{2}O \times \frac{1 \; mol \; H_{2}O}{18.02 \; g \; H_{2}O} = 8.324 \; mol \; H_{2}O \] The given molar heat capacities: \[C_{p, \; Pb} = 26.4 \frac {J}{K \cdot mol} \] \[C_{p, \; H_{2}O} = 75.2 \frac {J}{K \cdot mol} \] Substitute the known variables into the equation: \( \Delta S = nc_{p}ln\left(\frac{T_{2}}{T_{1}}\right) \) \[ \Delta S_{Pb} = nc_{p,Pb}ln\left(\frac{T_{equilibrium}}{T_{inital, \; Pb}}\right)\] \[ \Delta S_{Pb} = 0.8759 \; mol \; Pb \times 26.4 \frac {J}{K \cdot mol} \times ln\left(\frac{302.55 \; K}{370.15 \; K}\right) = -4.66 \; \frac{J}{K} \] \[ \Delta S_{H_{2}O} = nc_{p,H_{2}O}ln\left(\frac{T_{equilibrium}}{T_{inital, \; H_{2}O}}\right) \] \[ \Delta S_{H_{2}O} = 8.324 \; mol \; H_{2}O \times 75.2 \frac {J}{K \cdot mol} \times ln\left(\frac{302.55 \; K}{297.85 \; K }\right) = 9.80 \; \frac{J}{K} \] \[ \Delta S_{total} = \Delta S_{Pb} + \Delta S_{H_{2}O} = (-4.66 + 9.80) \; \frac{J}{K} = 5.14 \; \frac{J}{K} \] Copper has a heat capacity of 38.5 J K mol , approximately independent of temperature between 0°C to 100°C. Calculate the enthalpy and entropy change of 5.00 moles of copper as it is cooled at atmospheric pressure from 100°C to 0°C. \[\Delta H = q=nC_{p}\Delta T\] \[= (5.00 \; mol) (38.5 \; \dfrac{J}{K.mol}) (273 \; K – 373 \; K)\nonumber\] \[\Delta H = -19250 \; J\nonumber\] \[\Delta S = nC_{p}ln(\frac{T_{2}}{T_{1}})\] \[= (38.5 \dfrac{J}{K.mol}) \ln {(\dfrac{273 \; K}{373 \; K})} (5.00 \; mol)\nonumber\] \[= -60.1 \dfrac{J}{K}\nonumber\] The react with oxygen to give the following compounds: \[\ce{2 Be (g) + O2 (g) -> 2 BeO (s) } \nonumber\] \[\ce{2 Mg (g) + O2 (g) -> 2 MgO (s) } \nonumber\] \[\ce{2 Ca (g) + O2 (g) -> 2 CaO (s) }\nonumber\] Compute Δ for each reaction, and identify a periodic trend about the entropy. All the values you need to solve this problem are in the back of the Oxtoby textbook, so just find the numbers needed and use the formula: \[\sum{S^o_{products}} -\sum{S^o_{reactants}} = ΔS^o_{rxn}\nonumber\] \[2 Be_{(g)} + O_{2(g)} \rightarrow 2 BeO_{(s)}\nonumber\] \[\mathrm{2(14.14) - (2(136.16)+205.03)= -449.07 \dfrac{J}{K.mol}}\nonumber\] \[2 Mg_{(g)} + O_{2(g)} \rightarrow 2 MgO_{(s)} \nonumber\] \[\mathrm{2(26.92) - (2(148.54)+205.03)= -448.27 \dfrac{J}{K.mol}}\nonumber\] \[2 Ca_{(g)} + O_{2(g)} \rightarrow 2 CaO_{(s)} \nonumber\] \[\mathrm{2(39.75) - (2(154.77)+205.03)= -435.07 \dfrac{J}{K.mol}}\nonumber\] The entropy of the reaction decreases as you go up the periodic table. Is the entropy change in the reaction positive, negative or zero and why? \[\ce{CH4 (g) + 2 O2 (g) -> CO2 (g) + 2 H2O (l)} \nonumber\] Hint: You don't need to actually calculate the change in entropy to determine whether the change in entropy is positive or negative. Think about it conceptually. The answer is negative because the number of gas molecules decreases. Gas molecules have more entropy than liquid molecules since they have more energy in the form of degrees of freedom of motion (translational, rotational, and vibrational). Therefore, fewer gas molecules mean less entropy. At \(25.0\text{°C}\) the reaction below is not spontaneous. \[\ce{2 H2O (g) -> 2H2 (g) + O2 (g)} \nonumber\] with \(\ce{\Delta}{G}\) = \(+228.59\ \frac{kJ}{mol}\) If the above reaction were coupled with the following nonspontaneous reaction, could it be made to proceed? Why or Why not? \[\ce{3 H2 (g) + N2 (g) <=> 2 NH3 (g)} \nonumber\] with \(\ce{\Delta}{G}\) = \(-16.48\ \frac{kJ}{mol}\) Coupling the nonspontaneous reaction \[\ce{2H_2O_{(g)} \rightleftharpoons 2H_{2\, (g)} + O_{2\, (g)}}\nonumber\] with \(\ce{\Delta}{G} = +228.59\ \frac{kJ}{mol}\) for a reaction \( \left( \ce{\Delta}{G} \right) \) is negative. \[\ce{3H_{2(g)} + N_{2(g)} \rightleftharpoons 2NH_{3(g)}}\nonumber\] with \(\ce{\Delta}{G} = -16.48\ \frac{kJ}{mol}\) yields \[\ce{2H_2O_{(g)} + \require{cancel} \cancel{3}H_{2(g)} + N_{2(g)} \rightleftharpoons \cancel{2H_{2\, (g)}} + O_{2\, (g)} + 2NH_{3(g)}}\nonumber\] with \(\ce{\Delta}{G}\) = \(+212.11\ \frac{kJ}{mol}\) Since \(\ce{\Delta}{G}\) is positive, we know the reaction is still not spontaneous after being coupled with a spontaneous reaction A reaction at constant temperature and pressure is spontaneous if \(\Delta G<0\) and nonspontaneous if \(\Delta G>0\). Over what range of temperatures is each of the following processes spontaneous? Assume that all gases are at a pressure of 1 atm. (Hint: Use to calculate \(\Delta{H}\) and \(\Delta{S}\) (assumed independent of temperature and equal to \(\Delta{H}^{\circ}\) and \(\Delta{S}^{\circ}\), respectively, and then use the definition of \(\Delta{G}\)). a. First calculate \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\)of the reaction \(\ce{6 CO2 (g) + 6 H2O(l) + light -> C6H12O6 (s) + 6 O2 (g)}\) from the data in . \[\Delta H^{\circ}=-1273.3+(6\times0)-(6\times-393.5)-(6\times-285.8)=2802.5\;\dfrac{kJ}{mol}\nonumber\] \[\Delta S^{\circ}=212.1+(6\times205.2)-(6 \times 213.8)-(6\times70)=-259.5\; \dfrac{J}{mol.K}\nonumber\] Since the problem asks for the temperature range in which the reaction is spontaneous. The changeover from spontaneity to non-spontaneity occurs at \(\Delta{G^{\circ}}=0\). To find the temperature the makes \(\Delta{G^{\circ}}=0 \; \), the relationship of \(\Delta{G^{\circ}}=\Delta{H^{\circ}}-T\;\Delta{S^{\circ}} \; \) will be used. Remember to convert \(\Delta{S^{\circ}} \; \) to \(kJ\;mol^{-1} \; \) (or \(\Delta{H^{\circ}} \; \) to \(J\;mol^{-1} \; \)) so that the units cancel out properly. \[T=\frac{\Delta H^{\circ}}{\Delta S^{\circ}}=\frac{2802.5 \;\dfrac{kJ}{mol}}{0.2595\; \dfrac{kJ}{mol.K}}=10799.6\;K\nonumber\] Because \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are both positive, the reaction is spontaneous at temperatures above 10799.6 K. Reviewer Note: The solution is incorrect here. Because \(\Delta H^{\circ} > 0\) and \(\Delta S^{\circ} < 0\), the reaction is never spontaneous.  b. Perform similar calculations for the reaction \(\ce{C3H8 (g) + 5 O2 (g) -> 3CO2 (g) + 4 H2O (l)}\) \[\Delta H^{\circ}=(3\times-393.5)+(4\times-285.83)+103.8-(5\times0)=-2220.02\;\dfrac{kJ}{mol}\nonumber\] \[\Delta S^{\circ}=(3\times28)+(4\times70)-270.3-(5\times 205.2)=-374.9\;\dfrac{J}{mol.K}\nonumber\] \[T=\frac{\Delta H^{\circ}}{\Delta S^{\circ}}=\frac{2220.02\;\dfrac{kJ}{mol}}{0.3749\;\dfrac{kJ}{mol.K}}=5922\;K\nonumber\] Since \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are both negative, the reaction is spontaneous below 5922 K. c. The reaction \(\ce{CH4 (g) + 2 O2 (g) -> CO2 (g) + 2 H2O (l)}\) \[\Delta H^{\circ}=(-393.5)+(2\times-285.8)+74.6-(2\times0)=-890.5\;\dfrac{kJ}{mol}\nonumber\] \[\Delta S^{\circ}=(28)+(2\times70)-186.3-(2\times205.2)=-242.9\; \dfrac{J}{mol.K}\nonumber\] \[T=\frac{\Delta H^{\circ}}{\Delta S^{\circ}}=\dfrac{-890.5\;\dfrac{kJ}{mol}}{-0.2429\;\dfrac{kJ}{mol.K}}=3666\;K\nonumber\] Since \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are both negative, the reaction is spontaneous below 3666 K. Over what temperatures are these reactions spontaneous (under constant pressure and temperature)? You may need to use . By calculating \(∆H_{rxn}\) and \(∆S_{rxn}\) un der standard conditions and at 25 C, we can calculate the range of temperature in which ∆G is positive and therefore spontaneous. To calculate \(∆H_{rxn}\) and \(∆S_{rxn}\), we use , in which: \[ \Delta{H_{rxn}} = \sum{nH_{products}} - \sum{nH_{reactants}} \nonumber\] \[ \Delta{S_{rxn}} = \sum{nS_{products}} - \sum{nS_{reactants}} \nonumber\] a) \[\mathrm{\Delta{H_{rxn}} = -206.0 – (0 + 0) = -206.0 \dfrac{kJ}{mol}}\nonumber\] \[\mathrm{\Delta{S_{rxn}} = 57.7 – (41.6 + 32.1) = -16 \dfrac{J}{K.mol} }\nonumber\] \[\mathrm{\Delta{G_{rxn}} = -206.0 - T(-0.016)}\nonumber\] From here, we can set T = 0 because we know that ∆G must be < 0 to be a spontaneous reaction. T<12875K b) \[\mathrm{\Delta{H_{rxn}} = (2 \times -137.2) – (0+0) = -274.4 \dfrac{kJ}{mol}}\nonumber\] \[\mathrm{\Delta{S_{rxn}} = (2 \times 86.2) – ((2 \times 33.2) + 223.1) = -117.1 \dfrac{J}{K.mol}}\nonumber\] \[\mathrm{\Delta{G_{rxn}} = -274.4- T(-0.1171)}\nonumber\] T < 2343.3K c) \[\mathrm{\Delta{H_{rxn}} = (2(-1675.7)) - (0+0) = -3351.4 \dfrac{kJ}{mol} }\nonumber\] \[\mathrm{\Delta{S_{rxn}} = (2(50.92) - (4(28.3) + (3(205.2)) = -626.96 \dfrac{J}{K.mol}}\nonumber\] \[\mathrm{\Delta{G_{rxn}} = -3351.4 - T(-0.62696)}\nonumber\] T < 5345.5K Determine if the following reaction is spontaneous at 25°C by evaluating ∆H° , ∆S° and ∆G° \[\ce{N2H4 (l) + O2 (g) -> N2 (g) + 2 H2O (l)} \nonumber\] => Spontaneous (\(\Delta{G^{\circ}_{rxn}} < 0\) ) A thermodynamic engine operates cyclically and reversibly between two temperatures reservoirs, absorbing heat from the high-temperature bath at 600 K and discarding heat to low-temperature bath at 300 K. (a) \[\mathrm{Thermodynamic \; efficiency = \dfrac{(T_1 – T_2)}{T_1}}\nonumber\] \[\mathrm{\dfrac{(600 \; K- 300 \; K)}{ (600 \; K)} \times 100 \%= 50 \% }\nonumber\] (b)\[\mathrm{\dfrac{1800 \; J}{ 50 \%}= 3600 \; J }\nonumber\] (c) \[\mathrm{-3600 \; J \times 50 \%= -1800 \; J}\nonumber\] Acetone (\(\ce{C_3H_6O}\)) is an volatile liquid with a normal boiling point of 56°C and a molar enthalpy of vaporization of 29.1 kJ mol . What is the molar entropy of vaporization of acetone under 1 atm of pressure? First we need the write the equation relating entropy and enthalpy: \[\Delta G_{vap}= \Delta H_{vap} - T\Delta S_{vap}\nonumber\] As a result of the process of normal boiling being at quilibrium \(\Delta G = 0\) so all that needs to be done is T in Kelvin and \(\Delta H_{vap}\) needs to be plugged in and then we solve for \(\Delta S_{vap}\): \[\mathrm{0 = 29.1 \; \dfrac{kJ}{mol} – (329 \; K)( \Delta{S_{vap}} )}\nonumber\] \[\mathrm{\Delta{S_{vap}} = 88.45 \dfrac{J}{K.mol}}\nonumber\] Under standard conditions and 25°C, you have this reversible process \[\ce{Sn(s, white) -> Sn(s,gray)} \nonumber\] For part a), we use the formula \[\Delta S° = \sum nS^\circ (\ce{products})-\sum nS^\circ (\ce{reactants})\nonumber\] Plugging in the values from Table T1 gives us \[\Delta S = 44.1 - 55.2 = -7.1 \frac {J}{mol \cdot K}\nonumber\] For part b), we use the formula for a reversible process \[\Delta S = \frac {\Delta H}{T}\nonumber\] \(\Delta S = \frac {-2100 \; J/mol}{298.15 \; K}\) \(\Delta S = -7.04 \frac {J}{mol \cdot K}\) Compute the \(\Delta{G_{f}^{\circ}}\) for the following reaction the reaction. The \(\Delta{G_{f}^{\circ}}\) of H SO (aq) is -537.81 kJ mol , the \(\Delta{G_{f}^{\circ}}\) of SO (g) is -300.19 kJ mol , and the \(\Delta{G_{f}^{\circ}}\) of H O (g) is -120.42 kJ mol . \[\ce{H2SO3 (aq) -> H2O (g) + SO2 (g)} \nonumber\] ∆G° for the reaction is equal to the sum of the G ° for the products minus the sum of the G ° for the reactants. \[\Delta G^{\circ}= \left(-300.19 \dfrac{kJ}{mol}+-120.42\dfrac{kJ}{mol}\right)- \left(-537.81\dfrac{kJ}{mol} \right)\nonumber\] \[\Delta G^{\circ}=117.2\dfrac{kJ}{mol}\nonumber\] Professor Nesral wants to create some water through the combustion of hydrogen. This reaction is depicted as follows: \[\ce{2 H2  (g) + O2 (g) -> 2 H2O (g)}\nonumber\] Is this reaction spontaneous? Prove it by calculating \(\Delta{G^{\circ}}\) at 298.15K. Suppose that Nesral asks his friend in El Azizia, Libya, where the temperature is around \(57^{\circ}C\) (330K), to perform the same reaction. Calculate the Gibbs energy for this value and compare it to \(\Delta{G^{\circ}}\). Entropy Values (\(S^{\circ}\); \(\dfrac{J}{mol\,K}\)) Enthalpy Values(\(H^{\circ}\); \(\dfrac{kJ}{mol}\)) From intuition, the equation shows that 3 moles of gas, 2 moles of \(\ce{H2}\) and 1 mole of \(\ce{O2}\), are reacting to form 2 moles of \(\ce{H2O}\) gas. This is a decrease in entropy, which may hint at the reaction being non spontaneous. \(\Delta{G}^{\circ}\) should still be calculated to make sure. The information provided is the entropic values of the product and reactants, and the enthalpy of water. If the \(\Delta{S}^{\circ}\) is calculated, then the \(\Delta{G}^{\circ}\) can also be found using the following equation: \(\Delta{G} = \Delta{H_{rxn}^{\circ}} – T\Delta{S_{rxn}^{\circ}}\). Where \(\Delta{H_{rxn}^{\circ}}\) is the Heat and \(\Delta{S_{rxn}^{\circ}}\) is the entropy of the reaction under standard conditions (constant pressure), and T is temperature. Knowing that \(\Delta{S_{rxn}^{\circ}}\) is a state variable, it can be calculated as the difference between the sum of the entropy values of the products and reactants multiplied by their coefficients (This is known as Hess’s Law). In other words, \(\Delta{S^{\circ}}=\sum{n_{products}\times{S_{f_{products}}^{\circ}}} - \sum{n_{reactants}\times{S_{f_{reactants}}^{\circ}}}\) From this, \(\Delta{S_{rxn}^{\circ}}\) can be calculated as such: \(\Delta{S^{\circ}} = \left(2\times188.7\dfrac{J}{molK}\right)- \left(\left[2\times130.6\dfrac{J}{molK}\right]+\left[205.6\dfrac{J}{molK}\right]\right)\) \(= -89.4\dfrac{J}{mol.K}\) Since both \(H_{2}\) and \(O_{2}\) are in their natural states, the \(H_{f}^{\circ}\) associated with them equals to zero. Since enthalpy (\(\Delta{H_{rxn}^{\circ}}\)) is also a state variable, it can be calculated with Hess's Law as well. \(\Delta{H^{\circ}} = \left(2\times-241.826\dfrac{kJ}{mol}\right)- \left(\left[0\dfrac{kJ}{mol}\right]+\left[0\dfrac{kJ}{mol}\right]\right)\) \(\Delta{H^{\circ}}= -483.652\dfrac{kJ}{mol}\) Plugging in these values, as well as the temperature, \(298.15K\), will yield \(\Delta{G}^{\circ}\). \(\Delta{G_{rxn}^{\circ}} = \Delta{H_{rxn}^{\circ}} – T\Delta{S_{rxn}^{\circ}}\). \(\Delta{G_{rxn}^{\circ}} = -483.652\dfrac{kJ}{mol} – 298.15K\left(-0.0894\dfrac{kJ}{mol.K}\right)\). \(\Delta{G_{rxn}^{\circ}}= -456.99739\dfrac{kJ}{mol}\) The \(\Delta{G_{rxn}}\) is therefore \(-456.99739\dfrac{kJ}{mol}\). This very negative value means that the reaction is DEFINITELY spontaneous. It also makes sense on a conceptual level because combustion reaction like these release a lot of heat, which means that the process if enthalpically driven! Calculating the \(\Delta{G_{rxn}}\) at a different temperature, as mentioned in the problem, has the same method: \(\Delta{G_{rxn}} = \Delta{H_{rxn}^{\circ}} – T\Delta{S_{rxn}^{\circ}}\). \(\Delta{G_{rxn}} = -483.652\dfrac{kJ}{mol} – 330K\left(-0.0894\dfrac{kJ}{mol.K}\right)\). \(\Delta{G_{rxn}}= -454.15\dfrac{kJ}{mol}\) In both situations the \(\Delta{G_{rxn}}\) is very negative, but in the hotter climate it is slightly less so. What this shows is that, at a higher temperatures the entropic factors contribute more to the gibbs energy than at standard conditions. This showcases the temperature dependence of entropy. In another reaction that released less energy, this difference may have made the reaction non-spontaneous! : Hess's Law to find \(\Delta{G_{rxn}}\). The concentration of sodium in the plasma is approximately 0.14 M. While in the fluid outside of the plasma, sodium concentration is about 0.5 M. a) The Na ions will spontaneously flow into the plasma because it flows from high to low concentration. b) \(\mathrm{\Delta{G} = nRT \ln{\frac{c_2}{c_1}} }\) where c is the destination of Na after active transport occurs while c is the original concentration of the Na ions. \(\mathrm{\Delta{G} = (5 \; mole)(8.3145 \frac{J}{K.mol})(273 \; K)\ln{\frac{0.5}{0.14}}}\) \(\mathrm{\Delta{G} = 14447.26 \; J}\)   For a hypothetical nerve cell, the sodium ion concentration is 0.015 M outside the cell and 0.00045 M inside the cell. Active transport involves using proteins and chemical energy stored in ATP to move the ions in a thermodynamically unfavorable direction. Assuming that conditions in the cell allow the hydrolysis of a single ATP molecule to give 2.05×10 J of usable enegy. How many molecules of ATP would be needed to move 0.005 moles of sodium ions using active transport at standard conditions? You can assume that the sodium ion concentrations remain constant. To answer this question, the change of free energy is needed which means the following equation is needed \[ \Delta G = - R T lnQ \nonumber\] \[ R = 8.314\; \dfrac{J}{mol.K} \; \; \; \; \; \; \; T=298K \; \; \; \; \; \; \; Q = \dfrac{[Na^{+}]_{in}}{[Na^{+}]_{out}} \nonumber\] \[\Delta G = - 8.314\; \dfrac{J}{mol.K} \times 298K \times ln(\dfrac{0.00045\; M}{0.015\; M}) \nonumber\] \[\Delta G = -2424\; \dfrac{J}{mol} \times ln(0.03) = 8688\; \dfrac{J}{mol} \nonumber\] Now that we have the the free energy, we can determine the number of ATP molecules needed \[ 8688\; \dfrac{J}{mol} \times 0.005\; mol = 43.4\; J\nonumber\] \[\dfrac {43.4\; J} {2.05 \times 10^{-18}\; J\; molecule^{-1}} = 2.119 \times 10^{19}\; \text{ ATP molecules} \nonumber\] Answer: \(2.119 \times 10^{19}\; \text{ ATP molecules}\)

https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/16%3A_Solubility_and_Precipitation_Equilibria/16.2%3A_Ionic_Equilibria_between_Solids_and_Solutions

We begin our discussion of solubility and complexation equilibria—those associated with the formation of complex ions—by developing quantitative methods for describing dissolution and precipitation reactions of ionic compounds in aqueous solution. Just as with acid–base equilibria, we can describe the concentrations of ions in equilibrium with an ionic solid using an equilibrium constant expression. When a slightly soluble ionic compound is added to water, some of it dissolves to form a solution, establishing an equilibrium between the pure solid and a solution of its ions. For the dissolution of calcium phosphate, one of the two main components of kidney stones, the equilibrium can be written as follows, with the solid salt on the left: \[Ca_3(PO_4)_{2(s)} \rightleftharpoons 3Ca^{2+}_{(aq)} + 2PO^{3−}_{4(aq)} \label{Eq1}\] As you will discover in more advanced chemistry courses, basic anions, such as S , PO , and CO , react with water to produce OH and the corresponding protonated anion. Consequently, their calculated molarities, assuming no protonation in aqueous solution, are only approximate. The equilibrium constant for the dissolution of a sparingly soluble salt is the of the salt. Because the concentration of a pure solid such as Ca (PO ) is a constant, it does not appear explicitly in the equilibrium constant expression. The equilibrium constant expression for the dissolution of calcium phosphate is therefore \[K=\dfrac{[\mathrm{Ca^{2+}}]^3[\mathrm{PO_4^{3-}}]^2}{[\mathrm{Ca_3(PO_4)_2}]} \label{Eq2a}\] \[[\mathrm{Ca_3(PO_4)_2}]K=K_{\textrm{sp}}=[\mathrm{Ca^{2+}}]^3[\mathrm{PO_4^{3-}}]^2 \label{Eq2b}\] At 25°C and pH 7.00, Ksp for calcium phosphate is 2.07 × 10 , indicating that the concentrations of Ca and PO ions in solution that are in equilibrium with solid calcium phosphate are very low. The values of for some common salts are listed in Table \(\Page {1}\), which shows that the magnitude of K varies dramatically for different compounds. Although K is not a function of pH in \(\ref{Eq2b}\), changes in pH can affect the solubility of a compound as discussed later. As with , the concentration of a pure solid does not appear explicitly in . Solubility products are determined experimentally by directly measuring either the concentration of one of the component ions or the solubility of the compound in a given amount of water. However, whereas solubility is usually expressed in terms of mass of solute per 100 mL of solvent, \(K_{sp}\), like \(K\), is defined in terms of the molar concentrations of the component ions. Calcium oxalate monohydrate [Ca(O CCO )·H O, also written as CaC O ·H O] is a sparingly soluble salt that is the other major component of kidney stones [along with Ca (PO ) ]. Its solubility in water at 25°C is 7.36 × 10 g/100 mL. Calculate its . solubility in g/100 mL We need to write the solubility product expression in terms of the concentrations of the component ions. For calcium oxalate monohydrate, the balanced dissolution equilibrium and the solubility product expression (abbreviating oxalate as ox ) are as follows: \(\mathrm{Ca(O_2CCO_2)}\cdot\mathrm{H_2O(s)}\rightleftharpoons \mathrm{Ca^{2+}(aq)}+\mathrm{^-O_2CCO_2^-(aq)}+\mathrm{H_2O(l)}\hspace{5mm}K_{\textrm{sp}}=[\mathrm{Ca^{2+}},\mathrm{ox^{2-}}]\) Neither solid calcium oxalate monohydrate nor water appears in the solubility product expression because their concentrations are essentially constant. Next we need to determine [Ca ] and [ox ] at equilibrium. We can use the mass of calcium oxalate monohydrate that dissolves in 100 mL of water to calculate the number of moles that dissolve in 100 mL of water. From this we can determine the number of moles that dissolve in 1.00 L of water. For dilute solutions, the density of the solution is nearly the same as that of water, so dissolving the salt in 1.00 L of water gives essentially 1.00 L of solution. Because each 1 mol of dissolved calcium oxalate monohydrate dissociates to produce 1 mol of calcium ions and 1 mol of oxalate ions, we can obtain the equilibrium concentrations that must be inserted into the solubility product expression. The number of moles of calcium oxalate monohydrate that dissolve in 100 mL of water is as follows: The number of moles of calcium oxalate monohydrate that dissolve in 1.00 L of the saturated solution is as follows: Because of the stoichiometry of the reaction, the concentration of Ca and ox ions are both 5.04 × 10 M. Inserting these values into the solubility product expression, \[K_{sp} = [Ca^{2+},ox^{2−}] = (5.04 \times 10^{−5})(5.04 \times10^{−5}) = 2.54 \times 10^{−9} \nonumber\] In our calculation, we have ignored the reaction of the weakly basic anion with water, which tends to make the actual solubility of many salts greater than the calculated value. One crystalline form of calcium carbonate (CaCO ) is the mineral sold as “calcite” in mineral and gem shops. The solubility of calcite in water is 0.67 mg/100 mL. Calculate its . 4.5 × 10 The reaction of weakly basic anions with \(H_2O\) tends to make the actual solubility of many salts higher than predicted. Image used with permisison from Calcite, a structural material for many organisms, is found in the teeth of sea urchins. The urchins create depressions in limestone that they can settle in by grinding the rock with their teeth. Limestone, however, also consists of calcite, so how can the urchins grind the rock without also grinding their teeth? Researchers have discovered that the teeth are shaped like needles and plates and contain magnesium. The concentration of magnesium increases toward the tip, which contributes to the hardness. Moreover, each tooth is composed of two blocks of the polycrystalline calcite matrix that are interleaved near the tip. This creates a corrugated surface that presumably increases grinding efficiency. Toolmakers are particularly interested in this approach to grinding. Tabulated values of can also be used to estimate the solubility of a salt with a procedure that is essentially the reverse of the one used in Example \(\Page {1}\). In this case, we treat the problem as a typical equilibrium problem and set up a table of initial concentrations, changes in concentration, and final concentrations ( ), remembering that the concentration of the pure solid is essentially constant. We saw that the for Ca (PO ) is 2.07 × 10 at 25°C. Calculate the aqueous solubility of Ca (PO ) in terms of the following: molar concentration and mass of salt that dissolves in 100 mL of water Although the amount of solid Ca (PO ) changes as some of it dissolves, its molar concentration does not change. We now insert the expressions for the equilibrium concentrations of the ions into the solubility product expression ( ): This is the molar solubility of calcium phosphate at 25°C. However, the molarity of the ions is 2 and 3 , which means that [PO ] = 2.28 × 10 and [Ca ] = 3.42 × 10 . \(\left(\dfrac{1.14\times10^{-7}\textrm{ mol}}{\textrm{1 L}}\right)\textrm{100 mL}\left(\dfrac{\textrm{1 L}}{\textrm{1000 mL}} \right )\left(\dfrac{310.18 \textrm{ g }\mathrm{Ca_3(PO_4)_2}}{\textrm{1 mol}}\right)=3.54\times10^{-6}\textrm{ g }\mathrm{Ca_3(PO_4)_2}\) The solubility product of silver carbonate (Ag CO ) is 8.46 × 10 at 25°C. Calculate the following: 1.28 × 10 M 3.54 mg The (by which we usually mean the ) of a solid is expressed as the concentration of the "dissolved solid" in a In the case of a simple 1:1 solid such as AgCl, this would just be the concentration of Ag or Cl in the saturated solution. But for a more complicated stoichiometry such as as silver chromate, the solubility would be only one-half of the Ag concentration. For example, let us denote the solubility of as mol L . Then for a saturated solution, we have Substituting this into Eq 5b above, \[(2S)^2 (S) = 4S^3 = 2.76 \times 10^{–12}\] \[S= \left( dfrac{K_{sp}}{4} \right)^{1/3} = (6.9 \times 10^{-13})^{1/3} = 0.88 \times 10^{-4} \label{6a}\] thus the solubility is \(8.8 \times 10^{–5}\; M\). Note that the relation between the solubility and the solubility product constant depends on the stoichiometry of the dissolution reaction. For this reason it is meaningless to compare the solubilities of two salts having the formulas A B and AB , say, on the basis of their values. It is to compare the solubilities of two salts having different formulas on the basis of their values. The solubility of CaF (molar mass 78.1) at 18°C is reported to be 1.6 mg per 100 mL of water. Calculate the value of under these conditions. moles of solute in 100 mL; = 0.0016 g / 78.1 g/mol = \(2.05 \times 10^{-5}\) mol \[\begin{align*} S &= \dfrac{2.05 \times 10^{ –5} mol}{0.100\; L} \\[4pt] &= 2.05 \times 10^{-4} M \end{align*}\] \[\begin{align*}K_{sp} &= [Ca^{2+},F^–]^2 \\[4pt] &= (S)(2S)^2 \\[4pt] &= 4 × (2.05 \times 10^{–4})^3 \\[4pt] &= 3.44 \times 10^{–11} \end{align*}\] Estimate the solubility of La(IO ) and calculate the concentration of iodate in equilibrium with solid lanthanum iodate, for which = 6.2 × 10 . The equation for the dissolution is \[\ce{La(IO_3)_3 <=> La^{3+ } + 3 IO3^{–}} \nonumber\] If the solubility is , then the equilibrium concentrations of the ions will be [La ] = and [IO ] = 3 . Then = [La ,IO ] = (3 ) = 27 27 = 6.2 × 10 , = ( ( 6.2 ÷ 27) × 10 ) = 6.92 × 10 [IO ] = 3 = 2.08 × 10 Cadmium is a highly toxic environmental pollutant that enters wastewaters associated with zinc smelting (Cd and Zn commonly occur together in ZnS ores) and in some electroplating processes. One way of controlling cadmium in effluent streams is to add sodium hydroxide, which precipitates insoluble Cd(OH) ( = 2.5E–14). If 1000 L of a certain wastewater contains Cd at a concentration of 1.6E–5 , what concentration of Cd would remain after addition of 10 L of 4 NaOH solution? As with most real-world problems, this is best approached as a series of smaller problems, making simplifying approximations as appropriate. Volume of treated water: 1000 L + 10 L = 1010 L Concentration of OH on addition to 1000 L of pure water: (4 ) × (10 L)/(1010 L) = 0.040 Initial concentration of Cd in 1010 L of water: \[(1.6 \times 10^{–5}\; M) \left( \dfrac{100}{101} \right) \approx 1.6 \times 10^{–5}\; M\] The easiest way to tackle this is to start by assuming that a stoichiometric quantity of Cd(OH) is formed — that is, of the Cd gets precipitated. Now "turn on the equilibrium" — find the concentration of Cd that can exist in a 0.04 OH solution: Substitute these values into the solubility product expression: Cd(OH) = [Cd ] [OH ] = 2.5E–14 [Cd ] = (2.5E–14) / (16E–4) = 1.6E–13 Note that the effluent will now be very alkaline: \[pH = 14 + log 0.04 = 12.6\] so in order to meet environmental standards an equivalent quantity of strong acid must be added to neutralize the water before it is released. The solubility product ( ) is used to calculate equilibrium concentrations of the ions in solution, whereas the ion product ( ) describes concentrations that are not necessarily at equilibrium. The equilibrium constant for a dissolution reaction, called the solubility product ( ), is a measure of the solubility of a compound. Whereas solubility is usually expressed in terms of mass of solute per 100 mL of solvent, is defined in terms of the molar concentrations of the component ions.

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Pericyclic reactions differ from the ones we have looked at so far because they are not easily understood in Lewis acid- Lewis base terms. There is not always a clear nucleophile and electrophile in these reactions. In fact, they may appear to involve completely non-polar reactants. The classic example of a pericyclic reaction is a Diels Alder reaction. A Diels Alder reaction is a reaction between two alkenes. Normally, we think of both of these compounds as nucleophiles. It isn't easy to see why one would react with the other. It isn't easy to see how electrons would be attracted from one molecule to the other. Instead, pericyclic reactions rely on weak attractions between (or within) molecules that can lead to electronic interactions that result in new bond formation. Normally, pericyclic reactions are studied using molecular orbital calculations to map out these electronic interactions. They are also explained qualitatively using molecular orbital tools. ,

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A non-ideal solution is a solution that does not abide to the rules of an ideal solution where the interactions between the molecules are identical (or very close) to the interactions between molecules of different components. That is, there is no forces acting between the components: no Van-der-Waals nor any Coulomb forces. We assume ideal properties for dilute solutions. We use the concept of non-ideal solutions for concentrated solutions. A variety of forces act on real mixtures, making it difficult to predict the properties of such solutions. Non-ideal solutions are identified by determining the strength and specifics of the intermolecular forces between the different molecules in that particular solution. Non-ideal solutions can occur two ways: Reminder: A solvent is the major component of a mixture (i.e. water, air) while a solute is the minor component (sugar, carbon dioxide, etc...). A concrete example would be your daily cup of coffee: the coffee itself is the solvent, and anything you add (may it be sugar or cream) will be the solute. As mentioned above, non-ideal solutions are under study because their properties are not easily predictable, as forces between molecules can fluctuate over time. Non-ideal solutions cannot be defined by Raoult's law or by Henry's law, which are properties specifically unique to ideal mixtures: Since these laws assume that there are no intermolecular interactions, it is evident that they cannot be used for real mixtures, since the mathematical formulas will not hold true anymore due to the fact that the forces will have to be taken into account. However, non-ideal solutions are limited on both sides by these two laws. : Non-ideal solutions can form when forces of attraction between dissimilar molecules are than between similar molecules. At this point, a heterogeneous (non-mixing) solution may still occur, but it is not always the case.The resulting solution has a larger enthalpy of solution than pure components of the solution, causing the process to be (heat is absorbed to move the reaction forward). Putting these two components together in a mixture results in . Since dipole-dipole induced forces are not nearly as strong as the dipole-dipole interactions between acetone molecules in a pure substance, carbon disulfide-acetone mixtures are non-ideal solutions. Non-ideal solutions can also form when intermolecular forces between dissimilar molecules are than those between similar molecules. In this case, interactions between these two types of molecules release than is taken in to separate the two types of molecules. This energy is released in the form of heat, making the solution process . The of a compound corresponds to the active concentration of that particular compound. However, due to intermolecular forces we known is not the case; therefore, we introduce an , labeled \(\gamma \), as a unitless correctional factor. This coefficient takes into account the non-ideal characteristics of a mixture and it is between 0 and 1. For example, the relationship between the activity of a component and its concentration for mixtures is defined by: \[ a_1=\dfrac{C}{C_{pure}} \] While the same relationship for mixtures is defined as follows: \[ a_1=\gamma \dfrac{C}{C_{pure}} \]

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Combination bands, overtones, and Fermi resonances are used to help explain and assign peaks in vibrational spectra that do not correspond with known fundamental vibrations. Combination bands and overtones generally have lower intensities than the fundamentals, and Fermi resonance causes a spilt and shift in intensity of peaks with similar energies and identical symmetries. Hot bands will also be briefly addressed. Fundamental vibrational frequencies of a molecule corresponds to transition from v=0 to v=1. For a non-linear molecule there will by 3N-6 (where N is the number of atoms) number vibrations. The same holds true for linear molecules, however the equations 3N-5 is used, because a linear molecule has one less rotational degrees of freedom. (For a more detailed explanation see: ). Figure 1 shows a diagram for a vibrating diatomic molecule. The levels denoted by vibrational quantum numbers v represent the potenital energy for the harmonic (quadratic) oscillator. The transition \( 0 \rightarrow 1 \) is fundamental, transitions \( 0 \rightarrow n \) (n>1) are called overtones, and transitions \( 1 \rightarrow n\) (n>1) are called hot transitions (hot bands). The symmetry requirement of vibrational transisition is given by the transition moment integral, \[ \mu =\int \psi ^{\ast } \boldsymbol{\mathbf{}\mu} \psi d\tau \neq 0 \] where, \[ \boldsymbol{\mu }=\textbf{i}\mu _{x}+\textbf{j}\mu _{y}+\textbf{k}\mu _{z} \] These integrals can be separated into each component: x,y, and z. Because the ground state contains the totally symmetric representation, the coordinate x, y, or z and ψ must belong to the same representation so that the direct product will contain the totally symmetric representation. The harmonic oscillator approximation is convenient to use for diatomic molecules with quantized vibrational energy levels given by the following equation: \[ E_{v}(cm^{-1}) = \left (v + \frac{1}{2} \right) \omega_{e} \] A more accurate description of the vibrational energies is given by the anharmonic oscillator (also called Morse potential) with energy of \[ E_{v}(cm^{-1}) = \omega_{e} \left (v + \frac{1}{2} \right) - \omega_{e}x_{e} \left (v + \frac{1}{2} \right)^2 + \omega_{e}y_{e} \left (v + \frac{1}{2} \right)^3 +...\] where ω is the vibrational frequency for the r internuclear separation and ω ω x ω y . This accounts for the fact that as the higher vibrational states deviate from the perfectly parabolic shape, the level converge with increasing quantum numbers. It is because of this anharmoniticity that overtones can occur. While it may seem that the harmonic oscillator and the anharomic oscillator are closely related, this is in fact not the case. The differences in the wavefunctions lead to a breakdown of selection rules, specifically, Δv=±1 selection rule can not be applied, and higher order terms must be accounted in the energy calculations. There is only a small correction from the ground state to the first excited state for the anharmonic correction, but it becomes much larger for more highly excited states which are populated as the temperature increases. The deviation from the harmonic oscillator to the anharmonic oscillator results in expanding the energy function with additional terms and treating these terms with perturbation theory. The results in the correct vibrational energies and also relaxes the selection rules. A Δv=±1 is still most predominant, however, weaker overtones with Δv=±2, ±3,… can occur. It should be noted that a Δv=2 transition does not occur at twice the frequency of the fundamental transition, but at a lower frequency. Overtone transitions are not always observed, especially in larger molecules, because the transitions become weaker with increasing Δv. Overtones occur when a vibrational mode is excited from \(v=0\) to \(v=2\), which is called the , or v=0 to v=3, the . The fundamental transitions, \(v=±1\), are the most commonly occurring, and the probability of overtones rapid decreases as the number of quanta (\(Δv=±n\)) increases. Based on the harmonic oscillator approximation, the energy of the overtone transition would be n times larger than the energy of the fundamental transition frequency, but the anharmonic oscillator calculations show that the overtones are less than a multiple of the fundamental frequency. This is demonstrated with the vibrations of the diatomic \(\ce{HCl}\) in the gas phase: We can see from Table 1, that the anharmonic frequencies correspond much better with the observed frequencies, especially as the vibrational levels increase. If one of the symmetries is doubly degenerate in the excited state, a recursion formula is required to determine the symmetry of the v wave function, given by, \[ \chi _{v}(R) = \frac{1}{2} \left [\chi (R) \chi_{v-1}(R) + \chi (R^{v}) \right] \] Where χ (R) is the character under the operation R for the vth energy level; χ(R) is the character under R for the degenerate irreducible representation; χ (R) is the character of the (v-1)th energy level; and χ(R ) is the character of the operation R . This is demonstrated for the D point group below. Combination bands are observed when more than two or more fundamental vibrations are excited simultaneously. One reason a combination band might occur is if a fundamental vibration does not occur because of symmetry. This is comparable to in electronic transitions in which a fundamental mode can be excited and allowed as a “doubly excited state.” Combination implies addition of two frequencies, but it also possible to have a difference band where the frequencies are subtracted. To determine if two states can be excited simultaneous the transition moment integral must be evaluated with the appropriate excited state wavefunction. For example, in the transition, \[ \psi ^{_{1}}\left( 0 \right )\psi ^{_{2}}\left ( 0 \right )\psi ^{_{3}}\left ( 0 \right )\rightarrow \psi ^{_{1}}\left ( 2 \right )\psi ^{_{2}}\left ( 0 \right )\psi ^{_{3}}\left ( 1 \right ) \] the symmetry of the excited state will be the direct product of the irreducible representation for ψ (2) and ψ (1). For example, in the point group C , v has symmetry e and v has symmetry a . By performing the calculations listed above, it is determined that ψ (2) has (a + b + b ) symmetry: \[ \Gamma [\psi_{es}] = \Gamma [\psi _{1}(2)] \otimes \Gamma [\psi _{3}(1)] = (a_{1} + b_{1} + b_{2})\times a_{2} = a_{2} + b_{2} + b_{1} \] A practical use for understanding overtones and combination bands is applied to organic solvents used in spectroscopy. Most organic liquids have strong overtone and combination bands in the mid-infrared region, therefore, acetone, DMSO, or acetonitrile should only be used in very narrow spectral regions. Solvents such at CCl , CS and CDCl can be used above 1200 cm . Hot bands are observed when an already excited vibration is further excited. For example an v1 to v1' transition corresponds to a hot band in its IR spectrum. These transitions are temperature dependent, with lower signal intensity at lower temperature, and higher signal intensity at higher temperature. This is because at room temperature only the ground state is highly populated (kT ~ 200 cm ), based on the Boltzmann distribution. The Maxwell-Boltzmann distribution law states that if molecules in thermal equilibrium occupy two states of energy ε and ε , the relative populations of molecules occupying these states will be, \[ \large \dfrac{n_{j}}{n_{i}}=\dfrac{e^{-\varepsilon _{j}/RT}}{e^{-\varepsilon _{i}/RT}}=e^{-\Delta \varepsilon /RT} \] where, k is the Boltzmann constant and T is the temperature in Kelvin. In the harmonic oscillator model, hot bands are not easily distinguished from fundamental transitions because the energy levels are equally spaced. Because the spacing between energy levels in the anharmonic oscillator decrease with increasing vibrational levels, the hot bands occur at lower frequencies than the fundamentals. Also, the transition moment integrals are slightly different since the ground state will not necessarily be totally symmetric since it is not in v=0. \[ \psi ^{_{1}} \left( 0 \right )\psi ^{_{2}}\left ( 0 \right )\psi ^{_{3}}\left ( 1 \right )\rightarrow \psi ^{_{1}}\left ( 0 \right )\psi ^{_{2}}\left ( 0 \right )\psi ^{_{3}}\left ( 2 \right ) \] Fermi resonance results in the splitting of two vibrational bands that have nearly the same energy and symmetry in both IR and Raman spectroscopies. The two bands are usually a fundamental vibration and either an overtone or combination band. The wavefunctions for the two resonant vibrations mix according to the harmonic oscillator approximation, and the result is a shift in frequency and a change in intensity in the spectrum. As a result, two strong bands are observed in the spectrum, instead of the expected strong and weak bands. It is not possible to determine the contribution from each vibration because of the resulting mixed wave function. If the symmetry requirements are fulfilled and the energies of the two states are similar, mixing occurs, and the resulting modes can be described by a linear combination of the two interacting modes. The effect of this interaction is to increase the splitting between the engery levels. The splitting will be larger if the original energy difference is small and the coupling energy is large. The mixing of the two states also equalized the intensities of the vibrations which allows a weak overtone or combination band to show significant intensity from the fundamental with which it has Fermi resonance with. Because the vibrations have nearly the same frequency, the interaction will be affected if one mode undergoes a frequency shift from deuteration or a solvent effect while the other does not.The molecule most studied for this type of resonance (even what Fermi himself used to explain this phenomena), is carbon dioxide, CO . The three fundamental vibrations are v = 1337 cm , v =667 cm , v =2349 cm . The first overtone of v is v + 2v with symmetries σ and (σ + δ ), respectively, and frequencies of 1337 cm (v1) and 2(667) = 1334 cm (v ). According to group theory calculations, CO belongs to the point group D and should only have one Raman (symmetric stretching vibrations) and two IR active modes (asymmetric stretching and bending vibrations). CS is an analog to this system. Another typical example of Fermi resonance is found in the vibrational spectra of aldehydes, where the C-H bond in the CHO group interacts with the second harmonic level, 2δ(CHO), derived from the fundamental frequency of the deformation vibration of the CHO group (2*1400 cm ). The result is a Fermi doublet with branches around 2830 and 2730 cm . It is important for Fermi resonance that the vibrations connected with the two interacting levels be localized in the same part of the molecule. When bands have non-negligible widths, Fermi resonance perturbation of localized levels cannot be applied. This broadening can be the result of a number of things, such as, intermolecular interaction, shortened excited state lifetimes, or interaction of vibrational modes with phonons. In place of perturbation theory, the distribution of interacting vibrational states can be approximated as a collection of discrete level. The influence from each level can be calculated. It is useful to understand Fermi resonance because it helps assign and identify peaks within vibrational spectra (ie. IR and Raman) that may not otherwise be accounted for, however it should not be used lightly when assigning spectra.It is easy to jump to the conclusion that an unidentifiable band is the result of Fermi resonance, however this explanation may not fully account for the inconsistency and further characterization may be required for the system being investigated. It is important to assign spectra before doing the normal mode (coordinate) calculations because doing these calculations beforehand often leads to incorrect assignments of the peaks in the spectra. Q1. Given ν = 1151 cm , ν = 1361 cm , ν = 519 cm for SO , and the fact that there are 4 overtones and/or combination bands, predict the vibrational spectra and calculations. A1. Q2. What are the two main effects of Fermi resonance? A2. An overtone band can gain intensity from a nearby fundamental frequency with similiar symmetry. The energy levels of both bands are shifted away from one another. Q3. Explain the difference between a combination band and an overtone. An overtone is the result of Δv>1 from the ground state. A combination band is the result of a 2 fundamental frequencies being excited simultaneously so that the excitation is allowed by symmetry. The overtone is not subject to a symmetry requirement. Q4. Why are hot bands temperature dependent? A4. For a hot band to occur, a state other than the ground state must already be populated, and this requires >200cm to over come the thermal energy of kT (Boltzmann constant times temperature). The more heat that it put into the system, the more likely a hot band is to occur, and the stronger the signal it will produce. Q5. Show the calculations for the values in Figure 1 for both the harmonic and anharmonic oscillators. A5. Equations to use: ṽ = 2885.90v, for harmonic and ṽ = 2990.9v – 52.82v(v+1) for anharmonic. For v=3, ṽ = 2885.90(3) = 8657.7 ṽ = 2990.9(3) – 52.82(3)(3+1) = 8339.0

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The react with each other to form interhalogen compounds. The general formula of most interhalogen compounds is XY , where n = 1, 3, 5 or 7, and X is the less electronegative of the two halogens. The compounds which are formed by the union of two different halogens are called inter halogen compounds. There are never more than two types of halogen atoms in an molecule. There are of four general types: The interhalogen compounds of type AX and AX are formed between the halogen having very low electronegative difference (e.g., ClF, ClF ). The interhalogen compounds of type AX and AX are formed by larger atoms having low electronegativity with the smaller atoms having high electronegativity. This is because it is possible to fit the greater number of smaller atom around a larger one (e.g. BrF , IF ). Interhalogen are all prone to hydrolysis and ionize to give rise to polyatomic ions. The inter halogens are generally more reactive than halogens except F. This is because A-X bonds in interhalogens are weaker than the X-X bonds in dihalogen molecules. Reaction of inter halogens are similar to halogens. Hydrolysis of interhalogen compounds give halogen acid and oxy-acid. To name an Interhalogen compound, the less electronegative element is placed on to the left in formulae and naming is done straight forward. Some properties of interhalogen compounds are listed below. They are all prepared by direct combination of the elements although since in some cases more than one product is possible the conditions may vary by altering the temperature and relative proportions. For example under the same conditions difluorine reacts with dichlorine to give ClF with dibromine to give BrF but with diiodine to give IF . The structures found for the various interhalogens conform to what would be expected based on the VSEPR model. For XY the shape can be described as T-shaped with 2 lone pairs sitting in equatorial positions of a trigonal bipyramid. For XY the shape is a square pyramid with the unpaired electrons sitting in an axial position of an octahedral and XY is a pentagonal bipyramid. The interhalogens with formula XY have physical properties intermediate between those of the two parent halogens. The covalent bond between the two atoms has some ionic character, the larger element, X, becoming oxidised and having a partial positive charge. Most combinations of F, Cl, Br and I are known, but not all are stable. Chlorine trifluoride, ClF was first reported in 1931 and it is primarily used for the manufacture of uranium hexafluoride, UF as part of nuclear fuel processing and reprocessing, by the reaction: \[U + 3 ClF_3 → UF_6 + 3 ClF\] U isotope separation is difficult because the two isotopes have very nearly identical chemical properties, and can only be separated gradually using small mass differences. ( U is only 1.26% lighter than U.) A cascade of identical stages produces successively higher concentrations of U. Each stage passes a slightly more concentrated product to the next stage and returns a slightly less concentrated residue to the previous stage. There are currently two generic commercial methods employed internationally for enrichment: gaseous diffusion (referred to as first generation) and gas centrifuge (second generation) which consumes only 6% as much energy as gaseous diffusion. These both make use of the volatility of UF . ClF has been investigated as a high-performance storable oxidizer in rocket propellant systems. Handling concerns, however, prevented its use. Hypergolic means explode on contact with no need for any activator. One observer made the following comment about \(ClF_3\): It is, of course, extremely toxic, but that's the least of the problem. It is hypergolic* with every known fuel, and so rapidly hypergolic that no ignition delay has ever been measured. It is also hypergolic with such things as cloth, wood, and test engineers, not to mention asbestos, sand, and water with which it reacts explosively. It can be kept in some of the ordinary structural metals-steel, copper, aluminium, etc.-because of the formation of a thin film of insoluble metal fluoride which protects the bulk of the metal, just as the invisible coat of oxide on aluminium keeps it from burning up in the atmosphere. If, however, this coat is melted or scrubbed off, and has no chance to reform, the operator is confronted with the problem of coping with a metal-fluorine fire. For dealing with this situation, I have always recommended a good pair of running shoes." It is believed that prior to and during World War II, ClF code named N-stoff ("substance N") was being stockpiled in Germany for use as a potential incendiary weapon and poison gas. The plant was captured by the Russians in 1944, but there is no evidence that the gas was actually ever used during the war. The interhalogens of form XY have physical properties intermediate between those of the two parent halogens. The covalent bond between the two atoms has some ionic character, the less electronegative element, X, being oxidised and having a partial positive charge. Most combinations of F, Cl, Br and I are known, but not all are stable. All Interhalogens are volatile at room temperature. All are polar due to difference in their electronegativity. These are usually covalent liquids or gases due to small electronegativity difference among them. Some compounds partially ionize in solution. For example: \[2 ICl \rightarrow I^+ + ICl_2^-\] Interhalogen compounds are more reactive than normal halogens except fluorine.

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The simplest classical model of the hydrogen atom is one in which the electron moves in a circular planar orbit about the nucleus as previously discussed and as illustrated in . The angular momentum vector in this figure is shown at an angle with respect to some arbitrary axis in space. Assuming for the moment that we can somehow physically define such an axis, then in the classical model of the atom there should be an infinite number of values possible for the component of the angular momentum vector along this axis. As the angle between the axis and the vector varies continuously from 0°, through 90° to 180°, the component of along the axis would vary correspondingly from to zero to - . Thus the quantum mechanical statements regarding the angular momentum of an electron in an atom differ from the classical predictions in two startling ways. First, the magnitude of the angular momentum (the length of the vector ) is restricted to only certain values given by: \[ M = \sqrt{l(l+1)} \hbar\] with \(l= 0,1,2,... \) The magnitude of the angular momentum is quantized. Secondly, quantum mechanics states that the component of along a given axis can assume only (\(2 + 1\)) values, rather than the infinite number allowed in the classical model. In terms of the classical model this would imply that when the magnitude of is (the value when = 1), there are only three allowed values for , the angle of inclination of with respect to a chosen axis. The angle is another example of a physical quantity which in a classical system may assume any value, but which in a quantum system may take on only certain discrete values. You need not accept this result on faith. There is a simple, elegant experiment which illustrates the "quantization" of , just as a line spectrum illustrates the quantization of the energy. If we wish to measure the number of possible values which the component of the angular momentum may exhibit with respect to some axis we must first find some way in which we can physically define a direction or axis in space. To do this we make use of the magnetism exhibited by an electron in an atom. The flow of electrons through a loop of wire (an electric current) produces a magnetic field \(\Page {1}\) . At a distance from the ring of wire, large compared to the diameter of the ring, the magnetic field produced by the current appears to be the same as that obtained from a small bar magnet with a north pole and a south pole. Such a small magnet is called a magnetic dipole, i.e., two poles separated by a small distance. The electron is charged and the motion of the electron in an atom could be thought of as generating a small electric current. Associated with this current there should be a small magnetic field. The magnitude of this magnetic field is related to the angular momentum of the electron's motion in roughly the same way that the magnetic field produced by a current in a loop of wire is proportional to the strength of the current flowing in the wire. The strength of the atomic magnetic dipole is given by where: \[ \mu = \sqrt{l(l+1)} \beta_m \label{5}\] n \(\ref{5}\) we ar magnets if the electrons in these atoms possess angular momentum. In addition, the axis of the magnet lies along the axis of rotation, i.e., along the angular momentum vector. Thus the magnetism exhibited by the atoms provides an experimental means by which we may study the direction of the angular momentum vector. Thus the magnetism exhibited by the atoms provides an experimental means by which we may study the direction of the angular momentum vector. If we place the atoms in a magnetic field they will be attracted or repelled by this field, depending on whether or not the atomic magnets are aligned against or with the applied field. The applied magnetic field . By measuring the deflection of the atoms in this field we can determine the directions of their magnetic moments and hence of their angular momentum vectors with respect to this applied field. Consider an evacuated tube with a tiny opening at one end through which a stream of atoms may enter . By placing a second small hole in front of the first, inside the tube, we will obtain a narrow beam of atoms which will pass the length of the tube and strike the opposite end. If the atoms possess magnetic moments the path of the beam can be deflected by placing a magnetic field across the tube, perpendicular to the path of the atoms. The magnetic field must be one in which the lines of force diverge thereby exerting an unbalanced force on any magnetic material lying inside the field. This inhomogeneous magnetic field could be obtained through the use of N and S poles of the kind illustrated in . The direction of the magnetic field will be taken as the direction of the -axis. Let us suppose the beam consists of neutral atoms which possess units of electronic angular momentum (the angular momentum quantum number = 1). When no magnetic field is present, the beam of atoms strikes the end wall at a single in the middle of the detector. What happens when the magnetic field is present? We must assume that before the beam enters the magnetic field, the axes of the atomic magnets are randomly oriented with respect to the -axis. According to the concepts of classical mechanics, the beam should spread out along the direction of the magnetic field and produce a line rather than a point at the end of the tube . Actually, the beam is split into three distinct component beams each of equal intensity producing three spots at the end of the tube . The startling results of this experiment can be explained only if we assume that while in the magnetic field each atomic magnet could assume only one of three possible orientations with respect to the applied magnetic field . The atomic magnets which are aligned perpendicular to the direction of the field are not deflected and will follow a straight path through the tube. The atoms which are attracted upwards must have their magnetic moments oriented as shown. From the known strength of the applied inhomogeneous magnetic field and the measured distance through which the beam has been deflected upwards, we can determine that the component of the magnetic moment lying along the -axis is only in magnitude rather than the value of This latter value would result if the axis of the atomic magnet was parallel to the z-axis, i.e., the angle = 0°. Instead assumes a value such that the component of the total moment lying along the z-axis is just . Similarly the beam which is deflected downwards possesses a magnetic moment along the z-axis of - or - The classical prediction for this experiment assumes that may equal all values from 0° to 180°, and thus all values (from a maximum of ( = 0°) to 0 ( =90°) to ( = 180°)) for the component of the atomic moment along the z-axis would be possible. Instead, is found to equal only those values such that the magnetic moment along the z-axis equals + , 0 and - . The angular momentum of the electron determines the magnitude and the direction of the magnetic dipole. (Recall that the vectors for both these quantities lie along the same axis.) Thus the number of possible values which the component of the angular momentum vector may assume along a given axis must equal the number of values observed for the component of the magnetic dipole along the same axis. In the present example the values of the angular momentum component are +1( /2 ), 0 and -1( /2 ), or since = 1 in this case, + ( /2 ), 0 and - ( /2 ). In general, it is found that the number of observed values is always (2 + 1) the values being: \[ -l \hbar, (-l+1) \hbar, ... 0, ..., (l-1)\hbar, l \hbar\] for the angular momentum and \[-l \beta_m, (l-1)\beta_m ..., 0, ..., (l-1)\beta_m, l \beta_m\] for the magnetic dipole. The number governing the magnitude of the component of and , ranges from a maximum value of and decreases in steps of unity to a minimum value of - . This number is the third and final quantum number which determines the motion of an electron in a hydrogen atom. It is given the symbol and is called the magnetic quantum number. In summary, the angular momentum of an electron in the hydrogen atom is quantized and may assume only those values given by: Furthermore, it is an experimental fact that the component of the angular momentum vector along a given axis is limited to (2 + 1) different values, and that the magnitude of this component is quantized and governed by the quantum number \(m\) which may assume the values , -1, . . .,0, . . .,- . These facts are illustrated in for an electron in a orbital in which = 2. The quantum number determines the magnitude of the component of the angular momentum along a given axis in space. Therefore, it is not surprising that this same quantum number determines the axis along which the electron density is concentrated. When = 0 for a electron (regardless of the value, 2 , 3 , 4 , etc.) the electron density distribution is concentrated along the -axis (see ) implying that the classical axis of rotation must lie in the plane. Thus a electron with = 0 is most likely to be found along one axis and has a zero probability of being on the remaining two axes. The effect of the angular momentum possessed by the electron is to concentrate density along one axis. When = 1 or -1 the density distribution of a electron is concentrated in the plane with doughnut-shaped circular contours. The = 1 and -1 density distributions are identical in appearance. Classically they differ only in the direction of rotation of the electron around the z-axis; counter-clockwise for = +1 and clockwise for = -1. This explains why they have magnetic moments with their north poles in opposite directions. We can obtain density diagrams for the = +1 and -1 cases similar to the = 0 case by removing the resultant angular momentum component along the -axis. We can take combinations of the = +1 and -1 functions such that one combination is concentrated along the -axis and the other along the axis, and both are identical to the = 0 function in their appearance. Thus these functions are often labelled as , and functions rather than by their values. The value is, however, the true quantum number and we are cheating physically by labelling them , and . This would correspond to applying the field first in the direction, then in the direction and finally in the direction and trying to save up the information each time. In reality when the direction of the field is changed, all the information regarding the previous direction is lost and every atom will again align itself with one chance out of three of being in one of the possible component states with respect to the new direction. We should note that the dependence of the orbitals changes with changes in or , but the directional component changes with and only. Thus all orbitals possess spherical charge distributions and all orbitals possess dumb-bell shaped charge distributions regardless of the value of . m 0 0 1 0 -1 0 1 0 -1 2 1 0 -1 -2 summarizes the allowed combinations of quantum numbers for an electron in a hydrogen atom for the first few values of ; the corresponding name (symbol) is given for each orbital. Notice that there are orbitals for each value of , all of which belong to the same quantum level and have the same energy. There are - 1 values of for each value of and there are (2 + 1) values of for each value of . Notice also that for every increase in the value of , orbitals of the same value (same directional dependence) as found for the preceding value of are repeated. In addition, a new value of and a new shape are introduced. Thus there is a repetition in the shapes of the density distributions along with an increase in their number. We can see evidence of a periodicity in these functions (a periodic re-occurrence of a given density distribution) which we might hope to relate to the periodicity observed in the chemical and physical properties of the elements. We might store this idea in the back of our minds until later. We can summarize what we have found so far regarding the energy and distribution of an electron in a hydrogen atom thus: Some words of caution about energies and angular momentum should be added. In passing from the domain of classical mechanics to that of quantum mechanics we retain as many of the familiar words as possible. Examples are kinetic and potential energies, momentum, and angular momentum. We must, however, be on guard when we use these familiar concepts in the atomic domain. All have an altered meaning. Let us make this clear by considering these concepts for the hydrogen atom. Perhaps the most surprising point about the quantum mechanical expression for the energy is that it does not involve , the distance between the nucleus and the electron. If the system were a classical one, then we would expect to be able to write the total energy as: \[ E_n = KE = PE = \dfrac{1}{2} mv^2 - \dfrac{e^2}{r} \label{6}\] Both the and would be functions of , i.e., both would change in value as was changed (corresponding to the motion of the electron). Furthermore, the sum of the and must always yield the same value of which is to remain constant. is the potential energy diagram for the hydrogen atom and we have superimposed on it one of the possible energy levels for the atom, . Consider a classical value for at the point ". Classically, when the electron is at the point ", its is given by the value of the curve at '. The is thus equal to the length of the line ' in energy units. Thus the sum of + adds up to . When the electron is at the point ", its would equal and its would be zero. The electron would be motionless. Classically, for this value of the electron could not increase its value of beyond the point represented by ". If it did, it would be inside the "potential wall." For example, consider the point ". At this value of , the is given by the value at ' which is than and hence the must be equal to the length of the line '. But the must now be in sign so that the sum of and will still add up to . What does a negative mean? It doesn't mean anything as it never occurs in a classical system. Nor does it occur in a quantum mechanical system. It is true that quantum mechanics does predict a finite probability for the electron being inside the potential curve and indeed for all values of out to infinity. However, the quantum mechanical expression for does not allow us to determine the instantaneous values for the and . Instead, we can determine only their average values. Thus quantum mechanics does not give Equation \(\ref{6}\) but instead states only that the potential and kinetic energies may be known: \[E_n = \langle PE \rangle = \langle KE \rangle \label{7}\] The bracket denotes the fact that the energy quantity has been averaged over the complete motion (all values of ) of the electron. Why can not appear in the quantum mechanical expression for , and why can we obtain only average values for the and ? When the electron is in a given energy level its energy is precisely known; it is . The uncertainty in the value of the momentum of the electron is thus at a minimum. Under these conditions we have seen that our knowledge of the position of the electron is very uncertain and for an electron in a given energy level we can say no more about its position than that it is bound to the atom. Thus if the energy is to remain fixed and known with certainty, we cannot, because of the uncertainty principle, refer to (or measure) the electron as being at some particular distance from the nucleus with some instantaneous values for its and . Instead, we may have knowledge of these quantities only when they are averaged over all possible positions of the electron. This discussion again illustrates the pitfalls (e.g., a negative kinetic energy) which arise when a classical picture of an electron as a particle with a definite instantaneous position is taken literally. It is important to point out that the classical expressions which we write for the dependence of the potential energy on distance, - / for the hydrogen atom for example, are the expressions employed in the quantum mechanical calculation. However, only the average value of the may be calculated and this is done by calculating the value of - / at every point in space, taking into account the fraction of the total electronic charge at each point in space. The amount of charge at a given point in three-dimensional space is, of course, determined by the electron density distribution. Thus the value of for the ground state of the hydrogen atom is the electrostatic energy of interaction between a nucleus of charge +1 with the surrounding spherical distribution of negative charge. The penetration of a potential wall by the electron, into regions of negative kinetic energy, is known as "tunnelling." Classically a particle must have sufficient energy to surmount a potential barrier. In quantum mechanics, an electron may tunnel into the barrier (or through it, if it is of finite width). Tunnelling will not occur unless the barrier is of In the example of the H atom, the potential well is infinitely deep, but the energy of the electron is such that it is only a distance from the top of the well. In the example of the electron moving on a line we assumed the potential well to be infinitely deep regardless of the energy of the electron. In this case and hence must equal zero at the ends of the line and no tunnelling is possible as the potential wall is infinitely high. We can say more about the for an electron in an atom. Not only are these values constant for a given value of \( but also for any value of \( , Thus the is always positive and equal to minus one half of the . Since the total energy is negative when the electron is bound to the atom, we can interpret the stability of atoms as being due to the decrease in the when the electron is attracted by the nucleus. The question now arises as to why the electron doesn't "fall all the way" and sit right on the nucleus. When = 0, the would be equal to minus infinity, and the which is positive and thus destabilizing, would be zero. Classically this would certainly be the situation of lowest energy and thus the most stable one. The reason for the electron not collapsing onto the nucleus is a quantum mechanical one. If the electron was bound directly to the nucleus with no kinetic energy, its position and momentum would be known with certainty. This would violate Heisenberg's uncertainty principle. The uncertainty principle always operates through the kinetic energy causing it to become large and positive as the electron is confined to a smaller region of space. (Recall that in the example of an electron moving on a line, the increased as the length of the line decreased.) The smaller the region to which the electron is confined, the smaller is the uncertainty in its position. There must be a corresponding increase in the uncertainty of its momentum. This is brought about by the increase in the kinetic energy which increases the magnitude of the momentum and thus the uncertainty in its value. In other words the bound electron must always possess kinetic energy as a consequence of quantum mechanics. The and have opposite dependences on . The decreases (becomes more negative) as decreases but the increases (making the atom less stable) as decreases. A compromise is reached to make the energy as negative as possible (the atom as stable as possible) and the compromise always occurs when . A further decrease in would decrease the but only at the expense of a in the . The reverse is true for an increase in . Thus the reason the electron doesn't fall onto the nucleus may be summed up by stating that "the electron obeys quantum mechanics, and not classical mechanics."  (  /  )

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Temperature and pressure are macroscopic properties of gases. These properties are related to molecular motion, which is a microscopic phenomenon. . Kinetics means the study of motion, and in this case motions of gas molecules. At the same temperature and volume, the same numbers of moles of all gases exert the same pressure on the walls of their containers. This is known as . His theory implies that the same numbers of moles of gas have the same number of molecules. Common sense tells us that the pressure is proportional to the average kinetic energy of all the gas molecules. Avogadro's principle also implies that the kinetic energies of various gases are the same at the same temperature. The molecular masses are different from gas to gas, and if all gases have the same average kinetic energy, the average speed of a gas is unique. Based on the above assumption or theory, Boltzmann (1844-1906) and Maxwell (1831-1879) extended the theory to imply that the average kinetic energy of a gas depends on its temperature. They let \(u\) be the average or of a gas whose molar mass is . Since is the Avogadro's number, the average kinetic energy is (1/2) ( or \(\mathrm{K.E.} = \dfrac{M}{2 N}u^2 =\dfrac{3R T}{2 N}=\dfrac{3}{2}\,k T\) Note that is the mass of a single molecule. Thus, \(\begin{align} u &= \left(\dfrac{3k N T}{M}\right)^{1/2}\\ &= \left(\dfrac{3 R T}{M}\right)^{1/2} \end{align}\) where (= ) is the . Note that so evaluated is based on the average energy of gas molecules being the same, and it is called the root-mean-square speed; \(u\) is not the average speed of gas molecules. These formulas correlate temperature, pressure and kinetic energy of molecules. The distribution of gas speed has been studied by Boltzmann and Maxwell as well, but this is beyond the scope of this course. However, you may notice that at the same temperature, the average speed of hydrogen gas, \(\ce{H2}\), is 4 times more than that of oxygen, \(\ce{O2}\) in order to have the same average kinetic energy. For two gases, at the same temperature, with molecular masses and , and average speeds and , Boltzmann and Maxwell theory implies the following relationship: \[M_1 u_1^2 = M_2 u_2^2\] Thus, \[\dfrac{M_1}{M_2}=\left(\dfrac{u_2}{u_1}\right)^2\] The consequence of the above property is that the effusion rate, the root mean square speed, and the most probable speed, are all inversely proportional to the square root (SQRT) of the molar mass. Simply formulated, the Graham's law of effusion is \[\textrm{rate of effusion} = \dfrac{k}{\mathrm d^{1/2}} = \dfrac{k'}{\mathrm M^{1/2}}\] Have you noticed that helium balloons were usually deflated the next day while sizes of normal air balloons will keep at least for a few days? Small helium molecules not only effuse through the tiny holes of the balloons, they also effuse much faster through them. The theories covered here enable you to make many predictions. Apply these theories to solve the following problems. Calculate the kinetic energy of 1 mole of nitrogen molecules at 300 K. Assume nitrogen behaves as an ideal gas, then \(\begin{align*} E_{\textrm k} &= \dfrac{3}{2} R T\\ &= \mathrm{\dfrac{3}{2}\times 8.3145\: \dfrac{J}{mol\: K} \times 300\: K}\\ &= \mathrm{3742\: J / mol\: (or\: 3.74\: kJ/mol)} \end{align*}\) At 300 K, any gas that behaves like an ideal gas has the same energy per mol. Evaluate the root-mean-square speed of \(\ce{H2}\), \(\ce{He}\), \(\ce{N2}\), \(\ce{O2}\) and \(\ce{CO2}\) at 310 K (the human body temperature). Recall that \(\begin{align*} u &= \left(\dfrac{3k N T}{M}\right)^{1/2}\\ &= \left(\dfrac{3 R T}{M}\right)^{1/2}\\ &= \left(\dfrac{3 \times 8.3145 \times 310}{0.002}\right)^{1/2}\\ &= 1966\: \mathrm{m/s} \end{align*}\) Note that the molecular mass of hydrogen is 0.002 kg/mol. These units are used because the constant has been calculated using the SI units. The calculation for other gases is accomplished using their molar mass in kg. \(\begin{align*} u &= \left(\dfrac{3k N T}{M}\right)^{1/2}\\ &= \left(\dfrac{3 R T}{M}\right)^{1/2}\\ &= \dfrac{(3\times 8.3145\times 310)^{1/2}}{M^{1/2}}\\ &= \dfrac{87.9345}{M^{1/2}}\: \mathrm{m/s} \end{align*}\) The root-mean-square speeds are: Molar masses are 349 and 352 for \(\ce{^235UF6}\) and \(\ce{^238UF6}\) respectively. Using the method above, their root-mean-square speeds are 149 and 148 m/s respectively. The separation of these two isotopes of uranium was a necessity during the time of war for the US scientists. Gas diffusion was one of the methods employed for their separation. Assume air and helium molecules pass through the undetected holes in balloons with equal opportunities. If a helium balloon takes 10.0 hours to reduce its size by 5.0 %, how long will it take a nitrogen balloon to reduce its size by 5.0 %? The effusion rates are \[\textit{rate of effusion} =\dfrac{k}{\mathrm d^{1/2}}=\dfrac{k '}{\mathrm M^{1/2}} \nonumber\] Let's assume an average rate of effusion of helium to be 5/10 = 0.5, then the effusion rate of nitrogen is 0.5 * (4/28) = 0.189. The time required to effuse the same amount is thus 10*0.5/0.189 = 26.5 hr. The time required can be evaluated by \[\begin{align*} time &= \mathrm{10\times(28/4)^{1/2}\: hr}\\ &= \mathrm{26.5\: hr} \end{align*}\] Calculate the root mean square speed of \(\ce{N2}\) (molar mass = 28) at 37 °C (310 K, body temperature). R = 8.314 kg m2/(s2 mol K). Hint: 525 m/s Skill: Calculate the root mean square speed u of gas molecules. Which gas has a lowest root mean square speed: \(\ce{H2}\), \(\ce{N2}\), \(\ce{O2}\), \(\ce{CH4}\), or \(\ce{CO2}\)? carbon dioxide Know that \(\ce{CO2}\) has the highest molecular mass. Which gas has a highest root mean square speed: \(\ce{H2}\), \(\ce{N2}\), \(\ce{O2}\), \(\ce{CH4}\), or \(\ce{CO2}\)? Hint: hydrogen gas Skill: Knowing that \(\ce{H2}\) has the lowest molecular mass makes your decision easy. Assuming ideal gas behavior, calculate the kinetic energy of 1.00 mole of \(\ce{N2}\) at 37 °C (= 310 K). R = 8.314 J/(mol K); \(\ce{N2}\) molar mass = 28.0 3.87e3 J/mol Calculate the kinetic energy of any amount of any gas at any temperature. Which of the following gases has the highest effusion rate through a pinhole opening of its container (T = 300K): \(\ce{He}\) (molar mass 4), \(\ce{N2}\) (28), \(\ce{O2}\) (32), \(\ce{CO2}\) (44), \(\ce{SO2}\) (64), or \(\ce{Ar}\) (40)? helium Associate effusion rate with root mean square speed, and determine the effusion rates. For which gas in the accompanying list is the effusion rate the smallest (T = 400 K): \(\ce{NH3}\) (17), \(\ce{CO2}\) (44), \(\ce{Cl2}\) (71), \(\ce{CH4}\) (16)? Hint: chlorine gas Calculate the root mean speed for these gases. The average speed is 324 m/s for \(\ce{Cl2}\), 681 m/s for \(\ce{CH4}\). The most probable speed of \(\ce{CH4}\) (molar mass 16.0) at a given temperature is 411 m/s. What is the most probable speed of \(\ce{He}\) (molar mass 4.00) at the same temperature? 822 m/s Discussion: At the same T, the most probable speed of a gas is inversely proportional to the square-root of its molar mass.

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Electrical resistivity and conductivity is an important property for materials. Different materials have different conductivity and resistivity. Electrical conductivity is based on electrical transport properties. These can be measured with multiple techniques by using a variety of instruments. If electricity easily flows through a material, that material has high conductivity. Some materials that have high conductivity include copper and aluminum. Electrical conductivity is the measure of how easily electricity flows through a material. \[ \rho = \dfrac{1}{\sigma}\] where Since conductivity is the measure of how easily electricity flows, electrical resistivity measures how much a material resists the flow of electricity. When temperature increases, the conductivity of metals usually decreases, while the conductivity of semiconductors increases. This of course assumes that the material is homogenous, which is not always the case. You can calculate resistivity using the following equation \[\dfrac{E}{J} = ρ\] As you already read, ρ is the symbol for resistivity. \(E\) is the electric field and has units of Volts per meter (V/m). J is the current density and has units of amps per meter squared (A/m2). The electric field is calculated by dividing the Voltage by the length, l, that voltage is applied. \[E=\dfrac{V}{l}\] The current density is calculated by the equation below \[J=\dfrac{I}{A}\] I is the current and is divided by the cross sectional area, A, over which the current flows. Resistivity and resistance are two different things. Resistivity does not depend on size or shape. Resistance, however, does. You can calculate resistance with the equation below. \[ R=\dfrac{V}{I} \] R refers to resistance and is measured in Ω. \(V\) is the voltage and is measured in volts. I measures the current and its unit is amps (A). 1. E/J = ρ ---> J=E/ρ = 64V/m /12Ωm = 5.33A/m 2. E=V/l = 6V/3m = 2V/m 3. E=V/l V=IR ---> E=IR/l = 25A x 78Ω/100m = 19.5V/m 4. E/J = ρ E=V/l J=I/A ---> ρ=(V/l)/(I/A) = (150V/5m)/(62A/(24m x 5m) = 58Ωm ρ = 1/σ ---> 1/ρ = σ = 1/58Ωm 5. The material that has the greatest resistivity is the metal because as temperature increases metals are more likely to increase in resistivity and semiconductors usually decrease in resistivity as temperature increase.

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In general, the physical properties of alkenes are similar to those of alkanes. The data of Table 10-1 allow comparison of the boiling points, melting points, and densities of several alkenes with the corresponding alkanes that have the same carbon skeleton. Like the continuous-chain alkanes, the 1-alkenes form a homologous series of compounds that show regular changes in physical properties with increasing chain length. The boiling points, melting points, and densities of the simple alkynes (also included in Table 10-1) are somewhat higher than those of the corresponding alkanes or alkenes, and these properties also show regular changes as the chain length is increased. The are sufficiently different from those of alkanes in most instances to make it possible to recognize when a double bond is present. For example, in the infrared spectrum of 1-butene (Figure 10-1) the absorption band near \(1650 \: \text{cm}^{-1}\) is characteristic of the stretching vibration of the double bond. In general, the intensity and position of this band depends on the structure of the alkene; it varies with the degree of branching at the double bond, with the presence of a second unsaturated group in conjugation with the first (i.e., or ), and with the symmetry of the substitution of the double bond (see ). However, in many cases the absorption bands caused by the various modes of vibration of the alkenic \(\ce{C-H}\) bonds frequently are more useful for detecting a double bond and identifying its type than is the absorption band caused by \(\ce{C=C}\) stretch. With 1-butene, absorptions arising from the \(\ce{C-H}\) vibrations of the terminal \(\ce{=CH_2}\) group occur near \(3100 \: \text{cm}^{-1}\), \(1420 \: \text{cm}^{-1}\), and \(915 \: \text{cm}^{-1}\), and those of the \(\ce{-CH=}\) grouping near \(3020 \: \text{cm}^{-1}\), \(1420 \: \text{cm}^{-1}\), and \(1000 \: \text{cm}^{-1}\). In general, absorption bands at these frequencies are from the grouping \(\ce{-CH=CH_2}\). The bands near \(1420 \: \text{cm}^{-1}\) are due to in-plane bending, whereas those at \(915 \: \text{cm}^{-1}\) to \(1000 \: \text{cm}^{-1}\) arise from out-of-plane bending. The other intense absorptions, near \(1460 \: \text{cm}^{-1}\) and \(3000 \: \text{cm}^{-1}\), are due to \(\ce{C-H}\) vibrations of the \(\ce{CH_3CH_2}-\) group (see ). These illustrate a further point - namely, the positions of the infrared absorptions of alkyl \(\ce{C-H}\) bonds are significantly different from those of alkenic \(\ce{C-H}\) bonds. The double bonds of an alkene with no alkenic hydrogens are difficult to detect by infrared spectroscopy and in such cases Raman spectroscopy is helpful (see ). The infrared absorption of 1-butene that occurs at \(1830 \: \text{cm}^{-1}\) (Figure 10-1) falls in the region where stretching vibrations of alkene bonds usually are not observed. However, this band actually arises from an (harmonic) of the \(\ce{=CH_2}\) out-of-plane bending at \(915 \: \text{cm}^{-1}\). Such overtone absorptions come at exactly the frequency of the fundamental frequency, and whenever an absorption like this is observed that does not seem to fit with the normal fundamental vibrations, the possibility of its being an overtone should be checked. With regard to , a \(\pi\) electron of a simple alkene can be excited to a higher energy \(\left( \pi^* \right)\) state by light of wavelength \(180 \: \text{nm}\) to \(100 \: \text{nm}\). However, many other substances absorb in this region of the spectrum, including air, the quartz sample cell, and most solvents that might be used to dissolve the sample, and as a result the spectra of simple alkenes are not obtained easily with the usual ultraviolet spectrometers. When the double bond is conjugated as in or , then the wavelengths of maximum absorption shift to longer wavelengths and such absorptions are determined more easily and accurately (also see ). In proton , the chemical shifts of alkenic hydrogens are toward lower fields than those of alkane hydrogens and normally fall in the range of \(4.6\)-\(5.3 \: \text{ppm}\) relative to TMS (see and ). Spin-spin couplings of alkenic hydrogens are discussed in and . The of a monosubstituted alkyne such as ethynylbenzene, \(\ce{C_6H_5C \equiv CH}\) (Figure 10-4), has a strong band near \(3300 \: \text{cm}^{-1}\), which is characteristic of the carbon-hydrogen stretching vibration in the grouping \(\ce{\equiv C-H}\). At a lower frequency (longer wavelength) around \(2100 \: \text{cm}^{-1}\), there is a band associated with the stretching vibration of the triple bond (also see Figure 9-36). Therefore the presence of the grouping \(\ce{-C \equiv CH}\) in a molecule may be detected readily by infrared spectroscopy. However, the triple bond of a disubstituted alkyne, \(\ce{R-C \equiv C-R}\), is detected less easily because there is no \(\ce{\equiv C-H}\) absorption near \(3300 \: \text{cm}^{-1}\), and furthermore the \(\ce{C \equiv C}\) absorption sometimes is of such low intensity that it may be indiscernible. Raman spectroscopy ( ) or chemical methods must then be used to confirm the presence of a triple bond. Alkynes, like alkenes, undergo strongly only at wavelengths in the relatively inaccessible region below \(200 \: \text{nm}\). However, when a triple bond is conjugated with one or more unsaturated groups, radiation of longer wavelength is absorbed. To illustrate, ethyne absorbs at \(150 \: \text{nm}\) and \(173 \: \text{nm}\), whereas 1-buten-3-yne \(\left( \ce{CH_2=CH-C \equiv CH} \right)\) absorbs at \(219 \: \text{nm}\) and \(227.5 \: \text{nm}\). The effects of such conjugation on spectra is discussed in more detail in . The proton spectrum of ethynylbenzene is shown in Figure 10-5. The peaks near \(435 \: \text{Hz}\) and \(185 \: \text{Hz}\) correspond to resonances of the phenyl and \(\ce{\equiv C-H}\) protons, respectively. The difference in chemical shift between the two types of protons is considerably larger than between alkenic and aromatic protons come into resonance at higher magnetic fields (i.e., they are subject to more diamagnetic shielding, ) than alkenic or aromatic protons. In fact, the \(\ce{\equiv C-H}\) protons of alkynes have chemical shifts approaching those of alkyl protons. (Also see Figure 9-36.) The of alkenes and alkynes usually give distinct molecular ions; however, the fragmentation is often complex and not easily interpreted. and (1977)

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Early in their discussion of , most general chemistry texts have a Figure of the greatly increased distribution of molecular speeds at higher temperatures in gases than at moderate temperatures. When the temperature of a gas is raised (by transfer of energy from the surroundings of the system), there is a great increase in the velocity, \(v\), of many of the gas molecules (Figure \(\Page {1}\)). From 1/2mv , this means that there has also been a great increase in the translational of those faster moving molecules. Finally, we can see that an input of energy not only causes the gas molecules in the system to move faster — but also to move at very many fast speeds. (Thus, the energy in a heated system is more dispersed, spread out in being in many separate speeds rather than more localized in fewer moderate speeds.) A symbolic indication of the different distributions of the translational energy of each molecule of a gas on low to high energy levels in a 36-molecule system is in Figure \(\Page {2}\), with the lower temperature gas as Figure \(\Page {2A}\) and the higher temperature gas as Figure \(\Page {2B}\). These and later Figures in this section are symbolic because, in actuality, this small number of molecules is not enough to exhibit thermodynamic temperature. For further simplification, rotational energies that range from zero in monatomic molecules to about half the total translational energy of di- and tri-atomic molecules (and more for most polyatomic) at 300 K are not shown in the Figures. If those rotational energies were included, they would constitute a set of energy levels (corresponding to a spacing of ~10 J) each with translational energy distributions of the 36 molecules (corresponding to a spacing of ~10 J). These numbers show why translational levels, though quantized, are considered virtually continuous compared to the separation of rotational energies. The details of vibrational energy levels — two at moderate temperatures (on the ground state of which would be almost all the rotational and translational levels populated by the molecules of a symbolic or real system) — can also be postponed until physical chemistry. At this point in the first year course, depending on the instructor's preference, only a verbal description of rotational and vibrational motions and energy level spacing need be introduced. By the time in the beginning course that students reach thermodynamics, five to fifteen chapters later than kinetic theory, they can accept the concept that the total motional energies of molecules includes not just translational but also rotational and vibrational movements (that can be sketched simply below). A microstate is one of many arrangements of the molecular energies (i.e., ‘the molecules on each particular energy level') for the total energy of a system. Thus, Figure \(\Page {2A}\) is one microstate for a system with a given energy and Figure \(\Page {1B}\) is a microstate of the same system but with a greater total energy. Figure \(\Page {3A}\) (just a repeat of Figure \(\Page {2A}\), for convenience) is a different microstate than the microstate for the same system shown in Figure \(\Page {3B}\); the total energy is the same in \(\Page {3A}\) and \(\Page {3B}\), but in Figure \(\Page {3B}\) the arrangement of energies has been changed because two molecules have changed their energy levels, as indicated by the arrows. A possible scenario for that different microstate in Figure \(\Page {3}\) is that these two molecules on the second energy level collided at a glancing angle such that one gained enough energy to be on the third energy level, while the other molecule lost the same amount of energy and dropped down to the lowest energy level. In the light of that result of a single collision and the billions of collisions of molecules per second in any system at room temperature, there can be a very large number of microstates even for this system of just 36 molecules in Figures \(\Page {2}\) and \(\Page {3}\). (This is true despite the fact that not every collision would change the energy of the two molecules involved, and thus not change the numbers on a given energy level. Glancing collisions could occur with no change in the energy of either participant.) For any real system involving 6 x 10 molecules, however, the number of microstates becomes humanly incomprehensible for any system, even though we can express it in numbers, as will now be developed. The quantitative entropy change in a reversible process is given by \[ΔS = \dfrac{q_{rev}}{T}\] (Irreversible processes involving temperature or volume change or mixing can be treated by calculations from incremental steps that are reversible.) According to the Boltzmann entropy relationship, \[ΔS = k_B \ln \dfrac{\Omega_{Final}}{\Omega_{Initial}}\] where \(k_B\) is Boltzmann's constant and \(\Omega_{Final}\) or \(\Omega_{Initial}\) is the count of how many microstates correspond to the Final or Initial macrostates, respectively. The number of microstates for a system determines the number of ways in any one of which that the total energy of a macrostate can be at one instant. Thus, an increase in entropy means a greater number of microstates for the Final state than for the Initial. In turn, this means that there are more choices for the arrangement of a system's total energy at any one instant, far less possibility of localization (such as cycling back and forth between just 2 microstates), i.e., greater dispersal of the total energy of a system because of so many possibilities. An increase in entropy means a greater number of microstates for the Final state than for the Initial. In turn, this means that there are more choices for the arrangement of a system's total energy at any one instant. Some instructors may prefer “delocalization” to describe the status of the total energy of a system when there are a greater number of microstates rather than fewer, as an exact synonym for “dispersal” of energy as used here in this article for other situations in chemical thermodynamics. The advantage of uniform use of ‘dispersal' is its correct common-meaning applicability to examples ranging from motional energy becoming literally spread out in a larger volume to the cases of thermal energy transfer from hot surroundings to a cooler system, as well as to distributions of molecular energies on energy levels for either of those general cases. Students of lesser ability should be able to grasp what ‘dispersal' means in three dimensions, even though the next steps of abstraction to what it means in energy levels and numbers of microstates may result in more of a ‘feeling' than a preparation for physical chemistry that it can be for the more able. Of course, dispersal of the energy of a system in terms of microstates does mean that the energy is smeared or spread out over microstates like peanut butter on bread! All the energy of the macrostate is always in only microstate at one instant. It is the possibility that the total energy of the macrostate can be in of so many more different arrangements of that energy at the next instant — an increased probability that it could not be localized by returning to the same microstate — that amounts to a greater dispersal or spreading out of energy when there are a larger number of microstates (The numbers of microstates for chemical systems above 0 K are astounding. For any substance at a temperature about 1-4 K, there are 10 microstates ( ). For a mole of water at 273.15 K, there are 10 microstates and when it is heated to be just one degree warmer, that number is increased 10 to 10 microstates. For comparison, an estimate of the number of atoms in the entire universe is ‘only' about 10 , while a googol, considered a large number in mathematics, is `only' 10 .) Summarizing, when a substance is heated, its entropy increases because the energy acquired and that previously within it can be far more dispersed on the previous higher energy levels and on those additional high energy levels that now can be occupied. This in turn means that there are many many more possible of the molecular energies on their energy levels than before and thus, there is a great increase in accessible for the system at higher temperatures. A concise statement would be that when a system is heated, there are many more microstates accessible and this amounts to greater delocalization or dispersal of its total energy. (The common comment "heating causes or favors molecular disorder" is an anthropomorphic labeling of molecular behavior that has more flaws than utility. There is virtual chaos, so far as the distribution of energy for a system (its number of microstates) is concerned, as well as after heating at any temperature above 0 K and energy distribution is at the heart of the meaning of entropy and entropy change. ) ( ). When the volume of a gas is increased by isothermal expansion into an evacuated container, an entropy increase occurs but not for the same reason as when a gas or other substance is heated. There is no change in the quantity of energy of the system in such an expansion of the gas; \(dq\) is zero. Instead, there is a spontaneous dispersal or spreading out of that energy in space. This change in entropy can be calculated in macrothermodynamics from the equivalent \(q_{rev}\) to the work required to reversibly compress a mole of the gas back to its original volume, i.e., RT ln (V /V ), and then \[ΔS = R \ln \left(\dfrac{V_2}{V_1}\right)\] From the viewpoint of molecular thermodynamics, a few general chemistry texts use quantum mechanics to show that when a gas expands in volume, its energy levels become closer together in any small range of energy. Symbolically in Figure \(\Page {4B}\), a doubling of volume doubles the number of energy levels and increases the possibilities for energy dispersal because of these additional levels for the same molecular energies in Figure \(\Page {4A}\). Due to this increased possibility for energy dispersal — a spread over twice as many energy levels — the entropy of the system increases. Then, as could be expected from any changes in the population of energy levels for a system, there are also far greater numbers of possible arrangements of the molecular energies on those additional levels, and thus many more microstates for the system. This is the ultimate quantitative measure of an entropy increase in molecular thermodynamics (and second law of thermodynamics), any spontaneous process will result in a greater \[k_B \ln \dfrac{\Omega_{Final}}{\Omega_{Initial}}\] which is just the entropy of the reaction \(\Delta S\). When two dissimilar ideal gases mix and the volume increases, or when dissimilar liquids mix with or without a volume change, the number of energy levels that can be occupied by the molecules of each component increases. Thus, for somewhat different reasons there are similar results in this progression: additional energy levels for population by molecules (or ‘by molecular energies'), increased possibilities for motional energy dispersal on those energy levels, a far greater number of different arrangements of the molecules' energies on the energy levels, and the final result of many more accessible microstates for the system. There is an increase in entropy in the mixture. This is also the case when solutes of any type dissolve (mix) in a solvent. The entropy of the solvent increases (as does that of the solute). This phenomenon is especially important because it is the basis of colligative effects that will be discussed later.

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The Equilibrium Constant is . It simply signifies the ratio between the forward and reversal rates of a chemical reaction at equilibrium. There are a few different Equilibrium Constants such as: For the sake of brevity, let us stick to the former, \(K_c\) to follow this Module. Generally, we follow this equation when dealing with the calculation of the equilibrium constant, \(K_c\): \(K_c = \dfrac{Products}{Reactants}\) : We only include the concentrations of substances. include those of or ! This guide assumes you know how to setup and ICE table. If you do not, please refer to this excellent guide: Using . : Remember that we are obtaining the equilibrium constant K, NOT the Q. Q deals with using the initial values from the row of the ICE table. Take this general, easy to follow example, where A/B/C are different chemicals and M is molarity (mols/liter). : Calculate the equilibrium constant for a reaction between .100 M A and .300 M B, given that the product, C has a concentration of .200 M and that these values are those when the reaction is in a state of . Here is the chemical equation and assume it is balanced. \(A_{(aq)} + B_{(aq)} \rightleftharpoons 3C_{(aq)}\) : Given that the concentrations at equilibrium, we can assume that the ICE table (omit IC as they are not needed here) ends up looking like: Based on the steps mentioned above: \(K_c = \dfrac{Products}{Reactants}\) Thus: \(K_c = \dfrac{(0.200)^3}{(0.100)(0.300)}\) Notice that the concentration of C (0.200) is raised to the power of 3, since the number 3 is the coefficient of C. 4. Following basic order of operations of mathematics, first calculating exponential values then multiplying out the products and the reactants, and finally dividing them, we obtain our answer as follows: \(K_c = 0.267\) Therefore, we can state that the ratio of the forward and reversal rates of this reaction is 0.267

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We will consider here the reverse process of photosynthesis, namely how carbohydrates, especially glucose, are converted to energy by being broken down into carbon dioxide and water. A general summary of the several stages involved is shown in Figure 20-8. Initially, the storage fuels or foodstuffs (fats, carbohydrates, and proteins) are hydrolyzed into smaller components (fatty acids and glycerol, glucose and other simple sugars, and amino acids). In the next stage, these simple fuels are degraded further to two-carbon fragments that are delivered as the \(\ce{CH_3C=O}\) group (ethanoyl, or acetyl) in the form of the thioester of coenzyme A, \(\ce{CH_3COS}\) . The structure of this compound and the manner in which fatty acids are degraded has been considered in , and amino acid metabolism is discussed briefly in . This section is concerned mainly with the pathway by which glucose is metabolized by the process known as . In the conversion of glucose to \(\ce{CH_3COS}\) , two carbons are oxidized to carbon dioxide with consumption of two oxygen molecules: \[\ce{C_6H_{12}O_6} + 2 \textbf{CoA} \ce{SH} + \left[ 2 \ce{O_2} \right] \rightarrow 2 \ce{CH_3COS} \textbf{CoA} + 4 \ce{H_2O} + 2 \ce{CO_2} \tag{20-5}\] For further oxidation to occur, the \(\ce{CH_3COS}\) must enter the next stage of metabolism, whereby the \(\ce{CH_3C=O}\) group is converted to \(\ce{CO_2}\) and \(\ce{H_2O}\). This stage is known variously as the citric acid cycle, the tricarboxylic acid cycle, or the Krebs cycle, in honor of H. A. Krebs (Nobel Prize, 1953), who first recognized its cyclic nature in 1937. We can write an equation for the process as if it involved oxygen: \[2 \ce{CH_3COS} \textbf{CoA} + \left[ 4 \ce{O_2} \right] \rightarrow 4 \ce{CO_2} + 2 \ce{H_2O} + 2 \textbf{CoA} \ce{SH} \tag{20-6}\] Notice that combination of the reactions of Equations 20-5 and 20-6, glycolysis plus the citric acid cycle, oxidizes glucose completely to \(\ce{CO_2}\) and \(\ce{H_2O}\): \[\ce{C_6H_{12}O_6} + \left[ 6 \ce{O_2} \right] \rightarrow 6 \ce{CO_2} + 6 \ce{H_2O} \tag{20-7}\] But, as you will see, none of the steps uses molecular oxygen directly. Hence there must be a stage in metabolism whereby molecular oxygen is linked to production of oxidizing agents that are consumed in glycolysis and in the citric acid cycle. The coupling of oxygen into the metabolism of carbohydrates is an extremely complex process involving transport of the oxygen to the cells by an oxygen carrier such as hemoglobin, myoglobin, or hemocyanin. This is followed by a series of reactions, among which \(\ce{NADH}\) is converted to \(\ce{NAD}^\oplus\) with associated formation of three moles of ATP from three moles of ADP and inorganic phosphate. Another electron-carrier is flavin adenine dinucleotide (\(\ce{FAD}\); ), which is reduced to \(\ce{FADH_2}\) with an associated production of two moles of ATP from two moles of ADP. These processes are known as and can be expressed by the equations: Oxidative phosphorylation resembles photophosphorylation, discussed in , in that electron transport in photosynthesis also is coupled with ATP formation. By suitably juggling Equations 20-7 through 20-9, we find that the metabolic oxidation of one mole of glucose is achieved by ten moles of \(\ce{NAD}^\oplus\) and two moles of \(\ce{FAD}\): The overall result is production of 36 moles of ATP from ADP and phosphate per mole of glucose oxidized to \(\ce{CO_2}\) and \(\ce{H_2O}\). Of these, 34 ATPs are produced according to Equation 20-10 and, as we shall see, two more come from glycolysis. Glycolysis is the sequence of steps that converts glucose into two \(\ce{C_3}\) fragments with the production of ATP. The \(\ce{C_3}\) product of interest here is 2-oxopropanoate (pyruvate): There are features in this conversion that closely resemble the dark reactions of photosynthesis, which build a \(\ce{C_6}\) chain (fructose) from \(\ce{C_3}\) chains ( ). For example, the reactants are either phosphate esters or mixed anhydrides, and the phosphorylating agent is ATP: \[\ce{ROH} + \text{ATP} \rightarrow \ce{RO-PO_3^2-} + \text{ADP} + \ce{H}^\oplus\] Furthermore, rearrangements occur that interconvert an aldose and ketose, and the cleavage of a \(\ce{C_6}\) chain into two \(\ce{C_3}\) chains is achieved by a reverse aldol condensation: Also, oxidation of an aldehyde to an acid is accomplished with \(\ce{NAD}^\oplus\). There is a related reaction in photosynthesis ( ) that accomplishes the reduction of an acid to an aldehyde and is specific for \(\ce{NADPH}\), not \(\ce{NADH}\): First, glucose is phosphorylated to glucose 6-phosphate with ATP. Then an aldose \(\rightleftharpoons\) ketose rearrangement converts glucose 6-phosphate into fructose 6-phosphate. A second phosphorylation with ATP gives fructose 1,6-diphosphate: At this stage the enzyme aldolase catalyzes the aldol cleavage of fructose 1,6-diphosphate. One product is glyceraldehyde 3-phosphate and the other is 1,3-dihydroxypropanone phosphate. Another ketose \(\rightleftharpoons\) aldose equilibrium converts the propanone into the glyceraldehyde derivative: The next step oxidizes glyceraldehyde 3-phosphate with \(\ce{NAD}^\oplus\) in the presence of phosphate with the formation of 1,3-diphosphoglycerate: The mixed anhydride of phosphoric acid and glyceric acid then is used to convert ADP to ATP and form 3-phosphoglycerate. Thereafter the sequence differs from that in photosynthesis. The next few steps accomplish the formation of pyruvate by transfer of the phosphoryl group from \(\ce{C_3}\) to \(\ce{C_2}\) followed by dehydration to phosphoenolpyruvate. Phosphoenolpyruvate is an effective phosphorylating agent that converts ADP to ATP and forms pyruvate: The net reaction at this point produces more ATP than is consumed in the phosphorylation of glucose and fructose. What happens thereafter depends on the organism. With yeast and certain other microorganisms, pyruvate is decarboxylated and reduced to ethanol. The end result of glycolysis in this instance is . In higher organisms, pyruvate can be stored temporarily as a reduction product (lactate) or it can be oxidized further to give \(\ce{CH_3COS}\) and \(\ce{CO_2}\). The \(\ce{CH_3COS}\) then enters the citric acid cycle to be oxidized to \(\ce{CO_2}\) and \(\ce{H_2O}\), as discussed in the next section: Glycolysis to the pyruvate or lactate stage liberates heat, which can help keep the organism warm and produce ATP from ADP for future conversion into energy. However, glycolysis does not directly involve oxygen and does not liberate \(\ce{CO_2}\), as we might expect from the overall process of the metabolic conversion of glucose to carbon dioxide and water (Equation 20-10). The liberation of \(\ce{CO_2}\) occurs subsequent to pyruvate formation in a process called variously, the citric acid cycle, the Krebs cycle, or the tricarboxylic acid (TCA) cycle. The initial step, which is not really part of the cycle, is conversion of pyruvate to ethanoyl (acetyl ): To achieve the oxidation of acetyl on a continuing basis, intermediates consumed in certain steps must be regenerated in others. Thus we have a situation similar to that in the Calvin cycle ( ), whereby the first stage of the cycle achieved the desired reaction (\(\ce{CO_2}\) formation) and the second stage is designed to regenerate intermediates necessary to perpetuate the cycle. The entry point is the reaction between acetyl and a four-carbon unit, 2-oxobutanedioic acid. An aldol-type addition of the \(\ce{CH_3CO}\) group to this \(\ce{C_4}\) keto acid extends the chain to a branched \(\ce{C_6}\) acid (as citric acid): Dehydration-rehydration of citrate converts it to isocitrate: From here, oxidation of the hydroxyl function with \(\ce{NAD}^\oplus\) gives a keto acid, which loses \(\ce{CO_2}\) readily ( ) and affords 2-oxopentanedioate: We now have a \(\ce{C_5}\) keto acid that can be oxidized in the same way as the \(\ce{C_3}\) keto acid, pyruvic acid, to give a butanedioyl : Two molecules of \(\ce{CO_2}\) now have been produced and the remaining part of the citric acid cycle is concerned with regeneration of the for forming acetyl from 2-oxopropanoate, and also with regenerating the 2-oxobutanedioate, which is the precursor of citrate. The steps involved are The hydrolysis of the acyl in the first step is used for energy storage by conversion of guanosine diphosphate (GDP) to guanosine triphosphate (GTP): The hydration of the -butenedioate ( ) and the final oxidation reaction ( ) have been discussed previously. There is an alternative route, called the , by which glucose enters the glycolytic sequence to pyruvate. This route achieves the oxidative decarboxylation of glucose to give ribose, as the 5-phosphate ester. The essential steps are The net result is that three pentoses are converted into two molecules of fructose and one of glyceraldehyde \(\left( 3 \ce{C_5} \rightarrow 2 \ce{C_6} + \ce{C_3} \right)\). The relationship of the pentose-phosphate pathway to glycolysis is shown in Figure 20-11. The steps involved in the pentose shunt are readily reversible, but there are several steps in glycolysis that are not. These are the phosphorylation steps (Figure 20-9). Yet, there has to be a return route from pyruvate to glucose. This route is called and, in animals, takes place in the liver. We shall not discuss the steps in gluconeogenesis except to indicate again that they are not all the reverse of glycolysis. For comparison, the steps that differ are indicated in Figure 20-9 by dashed lines. Why is lactate formed from pyruvate in the metabolism of glucose? Pyruvate \( + \: \ce{NADH} + \ce{H}^\oplus \rightarrow\) lactate \(+ \: \ce{NAD}^\oplus\) is a dead-end path, but it does furnish the \(\ce{NAD}^\oplus\) needed for glycolysis in active muscle. This route for forming \(\ce{NAD}^\oplus\) is important, because in circumstances of physical exertion, the rate of production of \(\ce{NAD}^\oplus\) from oxidative phosphorylation may be slower than the demand for \(\ce{NAD}^\oplus\), in which case a temporary supply is available from the pyruvate \(\rightarrow\) lactate reduction. The lactate so formed builds up in muscle tissue under conditions of physical exertion and is apt to cause muscles to "cramp". The excess lactate so formed ultimately is removed by being converted back to pyruvate by oxidation with \(\ce{NAD}^\oplus\). The beauty of the metabolic cycle through pyruvate, shown in summary in Figure 20-11, is the way it can be tapped at various points according to whether the organism requires ATP (from glycolysis), \(\ce{NADH}\) (from pentose shunt), or \(\ce{NAD}^\oplus\) (from the lactate siding). and (1977)

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In the quest to identify a property that may reliably predict the spontaneity of a process, we have identified a very promising candidate: entropy. Processes that involve an increase in entropy (\(ΔS_{sys} > 0\)) are very often spontaneous; however, examples to the contrary are plentiful. By expanding consideration of entropy changes to include , we may reach a significant conclusion regarding the relation between this property and spontaneity. In thermodynamic models, the system and surroundings comprise everything, that is, the universe, and so the following is true: \[ΔS_\ce{univ}=ΔS_\ce{sys}+ΔS_\ce{surr} \label{\(\Page {1}\)}\] To illustrate this relation, consider again the process of heat flow between two objects, one identified as the system and the other as the surroundings. There are three possibilities for such a process: These results lead to a profound statement regarding the relation between entropy and spontaneity known as the : A summary of these three relations is provided in Table \(\Page {1}\). , \[ΔS_\ce{univ} > 0.\] For many realistic applications, the surroundings are vast in comparison to the system. In such cases, the heat gained or lost by the surroundings as a result of some process represents a very small, nearly infinitesimal, fraction of its total thermal energy. For example, combustion of a fuel in air involves transfer of heat from a system (the fuel and oxygen molecules undergoing reaction) to surroundings that are infinitely more massive (the earth’s atmosphere). As a result, \(q_{surr}\) is a good approximation of , and the second law may be stated as the following: \[ \begin{align} ΔS_\ce{univ} &=ΔS_\ce{sys}+ΔS_\ce{surr} \\[4pt] &=ΔS_\ce{sys}+\dfrac{q_\ce{surr}}{T} \label{4} \end{align}\] We may use this equation to predict the spontaneity of a process as illustrated in Example \(\Page {1}\). The entropy change for the process \[\ce{H2O}(s)⟶\ce{H2O}(l) \nonumber\] is 22.1 J/K and requires that the surroundings transfer 6.00 kJ of heat to the system. Is the process spontaneous at −10.00 °C? Is it spontaneous at +10.00 °C? We can assess the spontaneity of the process by calculating the entropy change of the universe. If Δ is positive, then the process is spontaneous. At both temperatures, Δ = 22.1 J/K and = −6.00 kJ. At −10.00 °C (263.15 K), the following is true: \[\begin{align*} ΔS_\ce{univ}&=ΔS_\ce{sys}+ΔS_\ce{surr} \\[4pt] &=ΔS_\ce{sys}+\dfrac{q_\ce{surr}}{T} \\[4pt] &=\mathrm{22.1\: J/K+\dfrac{−6.00×10^3\:J}{263.15\: K}} \\[4pt] &=−0.7\:J/K \end{align*} \] < 0, so melting is nonspontaneous ( spontaneous) at −10.0 °C. At 10.00 °C (283.15 K), the following is true: \[\begin{align*} ΔS_\ce{univ} &=ΔS_\ce{sys}+\dfrac{q_\ce{surr}}{T} \\[4pt] &= 22.1 \:J/K + \dfrac{−6.00×10^3\:J}{283.15\: K} \\[4pt] &=+0.9\: J/K \end{align*}\] \(\Delta S_{univ} > 0\), so melting spontaneous at 10.00 °C. Using this information, determine if liquid water will spontaneously freeze at the same temperatures. What can you say about the values of \(\Delta S_{univ}\)? Entropy is a state function, and freezing is the opposite of melting. At −10.00 °C spontaneous, +0.7 J/K; at +10.00 °C nonspontaneous, −0.9 J/K. The second law of thermodynamics states that a process increases the entropy of the universe, \(S_{univ} > 0\). If \(ΔS_{univ} < 0\), the process is , and if \(ΔS_{univ} = 0\), the system is at equilibrium.   ).

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The atoms, molecules, or ions that compose a chemical system can undergo several types of molecular motion, including translation, rotation, and vibration (Figure \(\Page {1}\)). 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 ). This system may be described by a single microstate, as its purity, perfect crystallinity and complete lack of motion (at least classically, quantum mechanics argues for constant motion) means there is but one possible location for each identical atom or molecule comprising the crystal (\(\Omega = 1\)). According to the Boltzmann equation, the entropy of this system is zero. \[\begin{align*} S&=k\ln \Omega \\[4pt] &= k\ln(1) \\[4pt] &=0 \label{\(\Page {5}\)} \end{align*}\] 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 entropy of a pure, perfect crystalline substance at 0 K is zero. 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 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. One way of calculating \(ΔS\) for a reaction is to use tabulated values of the standard molar entropy (\(S^o\)), which is the entropy of 1 mol of a substance under standard pressure (1 bar). Often the standard molar entropy is given at 298 K and is often demarked as \(ΔS^o_{298}\). The units of \(S^o\) 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 \(S = 0\)) and 298 K ( and ). As shown in Table \(\Page {1}\), for substances with approximately the same molar mass and number of atoms, \(S^o\) values fall in the order \[S^o(\text{gas}) \gg S^o(\text{liquid}) > S^o(\text{solid}).\] For instance, \(S^o\) for liquid water is 70.0 J/(mol•K), whereas \(S^o\) for water vapor is 188.8 J/(mol•K). Likewise, \(S^o\) is 260.7 J/(mol•K) for gaseous \(\ce{I2}\) and 116.1 J/(mol•K) for solid \(\ce{I2}\). This order makes qualitative sense based on the kinds and extents of motion available to atoms and molecules in the three phases (Figure \(\Page {1}\)). The correlation between physical state and absolute entropy is illustrated in Figure \(\Page {2}\), which is a generalized plot of the entropy of a substance versus temperature. The of a substance at any temperature above 0 K must be determined by calculating the increments of heat \(q\) required to bring the substance from 0 K to the temperature of interest, and then summing the ratios \(q/T\). Two kinds of experimental measurements are needed: \[ S_{0 \rightarrow T} = \int _{0}^{T} \dfrac{C_p}{T} dt \label{eq20}\] Because the heat capacity is itself slightly temperature dependent, the most precise determinations of absolute entropies require that the functional dependence of \(C\) on \(T\) be used in the integral in Equation \ref{eq20}, i.e.,: \[ S_{0 \rightarrow T} = \int _{0}^{T} \dfrac{C_p(T)}{T} dt. \label{eq21}\] When this is not known, one can take a series of heat capacity measurements over narrow temperature increments \(ΔT\) and measure the area under each section of the curve. The area under each section of the plot represents the entropy change associated with heating the substance through an interval \(ΔT\). To this must be added the enthalpies of melting, vaporization, and of any solid-solid phase changes. Values of \(C_p\) for temperatures near zero are not measured directly, but can be estimated from quantum theory. The cumulative areas from 0 K to any given temperature (Figure \(\Page {3}\)) are then plotted as a function of \(T\), and any phase-change entropies such as \[S_{vap} = \dfrac{H_{vap}}{T_b}\] are added to obtain the absolute entropy at temperature \(T\). As shown in Figure \(\Page {2}\) above, the entropy of a substance increases with temperature, and it does so for two reasons: We can make careful calorimetric measurements to determine the temperature dependence of a substance’s entropy and to derive absolute entropy values under specific conditions. are given the label \(S^o_{298}\) for values determined for one mole of substance at a pressure of 1 bar and a temperature of 298 K. The for any process may be computed from the standard entropies of its reactant and product species like the following: \[ΔS^o=\sum νS^o_{298}(\ce{products})−\sum νS^o_{298}(\ce{reactants}) \label{\(\Page {6}\)}\] Here, \(ν\) represents stoichiometric coefficients in the balanced equation representing the process. For example, \(ΔS^o\) for the following reaction at room temperature is computed as the following: \[ΔS^o=[xS^o_{298}(\ce{C})+yS^o_{298}(\ce{D})]−[mS^o_{298}(\ce{A})+nS^o_{298}(\ce{B})] \label{\(\Page {8}\)}\] Table \(\Page {1}\) lists some standard entropies at 298.15 K. You can find additional standard entropies in and A closer examination of Table \(\Page {1}\) also reveals that substances with similar molecular structures tend to have similar \(S^o\) 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 (\(S^o = 2.4 \,J/(mol•K)\)). In contrast, graphite, the softer, less rigid allotrope of carbon, has a higher \(S^o\) (5.7 J/(mol•K)) due to more (microstates) 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 \(S^o\) values for CH OH(l) and CH CH OH(l). Finally, substances with strong hydrogen bonds have lower values of \(S^o\), which reflects a more ordered structure. Entropy increases with softer, less rigid solids, solids that contain larger atoms, and solids with complex molecular structures. To calculate \(ΔS^o\) for a chemical reaction from standard molar entropies, we use the familiar “products minus reactants” rule, in which the absolute entropy of each reactant and product is multiplied by its stoichiometric coefficient in the balanced chemical equation. Example \(\Page {1}\) illustrates this procedure for the combustion of the liquid hydrocarbon isooctane (\(\ce{C8H18}\); 2,2,4-trimethylpentane). Use the data in Table \(\Page {1}\) to calculate \(ΔS^o\) for the reaction of liquid isooctane with \(\ce{O2(g)}\) to give \(\ce{CO2(g)}\) and \(\ce{H2O(g)}\) at 298 K. : standard molar entropies, reactants, and products : ΔS° : Write the balanced chemical equation for the reaction and identify the appropriate quantities in Table \(\Page {1}\). Subtract the sum of the absolute entropies of the reactants from the sum of the absolute entropies of the products, each multiplied by their appropriate stoichiometric coefficients, to obtain \(ΔS^o\) for the reaction. : The balanced chemical equation for the complete combustion of isooctane (\(\ce{C8H18}\)) is as follows: \[\ce{C8H18(l) + 25/2 O2(g) -> 8CO2(g) + 9H2O(g)} \nonumber\] We calculate \(ΔS^o\) for the reaction using the “products minus reactants” rule, where m and n are the stoichiometric coefficients of each product and each reactant: \[\begin{align*}\Delta S^o_{\textrm{rxn}}&=\sum mS^o(\textrm{products})-\sum nS^o(\textrm{reactants}) \\[4pt] &=[8S^o(\mathrm{CO_2})+9S^o(\mathrm{H_2O})]-[S^o(\mathrm{C_8H_{18}})+\dfrac{25}{2}S^o(\mathrm{O_2})] \\[4pt] &=\left \{ [8\textrm{ mol }\mathrm{CO_2}\times213.8\;\mathrm{J/(mol\cdot K)}]+[9\textrm{ mol }\mathrm{H_2O}\times188.8\;\mathrm{J/(mol\cdot K)}] \right \} \\[4pt] & \,\,\, -\left \{[1\textrm{ mol }\mathrm{C_8H_{18}}\times329.3\;\mathrm{J/(mol\cdot K)}]+\left [\dfrac{25}{2}\textrm{ mol }\mathrm{O_2}\times205.2\textrm{ J}/(\mathrm{mol\cdot K})\right ] \right \} \\[4pt] &=515.3\;\mathrm{J/K}\end{align*}\] \(ΔS^o\) is positive, as expected for a combustion reaction in which one large hydrocarbon molecule is converted to many molecules of gaseous products. Use the data in Table \(\Page {1}\) to calculate \(ΔS^o\) for the reaction of \(\ce{H2(g)}\) with liquid benzene (\(\ce{C6H6}\)) to give cyclohexane (\(\ce{C6H12}\)) at 298 K. -361.1 J/K Calculate the standard entropy change for the following process at 298 K: \[\ce{H2O}(g)⟶\ce{H2O}(l)\nonumber\] The value of the standard entropy change at room temperature, \(ΔS^o_{298}\), is the difference between the standard entropy of the product, H O( ), and the standard entropy of the reactant, H O( ). \[\begin{align*} ΔS^o_{298} &=S^o_{298}(\ce{H2O (l)})−S^o_{298}(\ce{H2O(g)})\nonumber \\[4pt] &= (70.0\: J\:mol^{−1}K^{−1})−(188.8\: Jmol^{−1}K^{−1})\nonumber \\[4pt] &=−118.8\:J\:mol^{−1}K^{−1} \end{align*}\] The value for \(ΔS^o_{298}\) is negative, as expected for this phase transition (condensation), which the previous section discussed. Calculate the standard entropy change for the following process at 298 K: \[\ce{H2}(g)+\ce{C2H4}(g)⟶\ce{C2H6}(g)\nonumber\] −120.6 J mol K Calculate the standard entropy change for the combustion of methanol, CH OH at 298 K: \[\ce{2CH3OH}(l)+\ce{3O2}(g)⟶\ce{2CO2}(g)+\ce{4H2O}(l)\nonumber\] The value of the standard entropy change is equal to the difference between the standard entropies of the products and the entropies of the reactants scaled by their stoichiometric coefficients. The standard entropy of formations are found in Table \(\Page {1}\). \[\begin{align*} ΔS^o &=ΔS^o_{298} \\[4pt] &= ∑νS^o_{298}(\ce{products})−∑νS^o_{298} (\ce{reactants}) \\[4pt] & = 2S^o_{298}(\ce{CO2}(g))+4S^o_{298}(\ce{H2O}(l))]−[2S^o_{298}(\ce{CH3OH}(l))+3S^o_{298}(\ce{O2}(g))]\nonumber \\[4pt] &= [(2 \times 213.8) + (4×70.0)]−[ (2 \times 126.8) + (3 \times 205.03) ]\nonumber \\[4pt] &= −161.6 \:J/mol⋅K\nonumber \end{align*} \] Calculate the standard entropy change for the following reaction at 298 K: \[\ce{Ca(OH)2}(s)⟶\ce{CaO}(s)+\ce{H2O}(l)\nonumber\] 24.7 J/mol•K Energy values, as you know, are all relative, and must be defined on a scale that is completely arbitrary; there is no such thing as the absolute energy of a substance, so we can arbitrarily define the enthalpy or internal energy of an element in its most stable form at 298 K and 1 atm pressure as zero. The same is true of the entropy; since entropy is a measure of the “dilution” of thermal energy, it follows that the less thermal energy available to spread through a system (that is, the lower the temperature), the smaller will be its entropy. In other words, as the absolute temperature of a substance approaches zero, so does its entropy. This principle is the basis of the , which states that the entropy of a perfectly-ordered solid at 0 K is zero. 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. The second law of thermodynamics states that a spontaneous process increases the entropy of the universe, > 0. If Δ < 0, the process is nonspontaneous, and if Δ = 0, the system is at equilibrium. The third law of thermodynamics establishes the zero for entropy as that of a perfect, pure crystalline solid at 0 K. With only one possible microstate, the entropy is zero. We may compute the standard entropy change for a process by using standard entropy values for the reactants and products involved in the process.   ). )

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To what extent will quantum mechanics permit us to pinpoint the position of an electron when it is bound to an atom? We can obtain an order of magnitude answer to this question by applying the uncertainty principle \[\Delta x \Delta p \ge \dfrac{\hbar}{2}\] \[ \approx h\] to estimate \(\Delta x\), which represents the minimum uncertainty in our knowledge of the position of the electron. The momentum of an electron in an atom is of the order of magnitude of \(9 \times 10^{-19}\; g\, cm/sec\). The in the momentum \(Delta p\) must necessarily be of the same order of magnitude. Thus \[ \Delta x =\dfrac{ 7 \times 10^{-27}}{9 \times 10^{-19}} \approx 10^{-8} \,cm\] The uncertainty in the position of the electron is of the same order of magnitude as the diameter of the atom itself. As long as the electron is bound to the atom, we will not be able to say much more about its position than that it is in the atom. Certainly all models of the atom which describe the electron as a particle following a definite trajectory or orbit must be discarded. We can obtain an energy and one or more wavefunctions for every value of \(n\), the principal quantum number, by solving Schrödinger's equation for the hydrogen atom. A knowledge of the wavefunctions, or probability amplitudes \(\psi_n\), allows us to calculate the probability distributions for the electron in any given quantum level. When = 1, the wave function and the derived probability function are independent of direction and depend only on the distance between the electron and the nucleus. In Figure \(\Page {1}\), we plot both \(\psi_1\) and \(P_1\) versus \(r\), showing the variation in these functions as the electron is moved further and further from the nucleus in any one direction. (These and all succeeding graphs are plotted in terms of the atomic unit of length, = 0.529 10 cm.) Two interpretations can again be given to the \(P_1\) curve. An experiment designed to detect the position of the electron with an uncertainty much less than the diameter of the atom itself (using light of short wavelength) will, if repeated a large number of times, result in Figure \(\Page {1}\) for \(P_1\). That is, the electron will be detected close to the nucleus most frequently and the probability of observing it at some distance from the nucleus will decrease rapidly with increasing \(r\). The atom will be ionized in making each of these observations because the energy of the photons with a wavelength much less than 10 cm will be greater than \(K\), the amount of energy required to ionize the hydrogen atom. If light with a wavelength comparable to the diameter of the atom is employed in the experiment, then the electron will not be excited but our knowledge of its position will be correspondingly less precise. In these experiments, in which the electron's energy is not changed, the electron will appear to be "smeared out" and we may interpret \(P_1\) as giving the fraction of the total electronic charge to be found in every small volume element of space. (Recall that the addition of the value of for every small volume element over all space adds up to unity, i.e., one electron and one electronic charge.) When the electron is in a definite energy level we shall refer to the \(P_n\) distributions as , since they describe the manner in which the total electronic charge is distributed in space. The electron density is expressed in terms of the number of electronic charges per unit volume of space, / . The volume is usually expressed in atomic units of length cubed, and one atomic unit of electron density is then / . To give an idea of the order of magnitude of an atomic density unit, 1 au of charge density / = 6.7 electronic charges per cubic Ångstrom. That is, a cube with a length of \(0.52917 \times 10^{-8}\; cm\), if uniformly filled with an electronic charge density of 1 au, would contain 6.7 electronic charges. \(P_1\) may be represented in another manner. Rather than considering the amount of electronic charge in one particular small element of space, we may determine the total amount of charge lying within a thin spherical shell of space. Since the distribution is independent of direction, consider adding up all the charge density which lies within a volume of space bounded by an inner sphere of radius and an outer concentric sphere with a radius only infinitesimally greater, say \(r + Dr\). The area of the inner sphere is 4 and the thickness of the shell is Thus the volume of the shell is \(4\pi r^2 \Delta r\) and the product of this volume and the charge density ( ), which is the charge or number of electrons per unit volume, is therefore the total amount of electronic charge lying between the spheres of radius \(r\) and \(r + Dr\). The product \(4pr^2P_n\) is given a special name, the radial distribution function, which we shall label \(Q_n(r)\). The reader may wonder why the volume of the shell is not taken as: \[ \dfrac{4}{3} \pi \left[ (r + \Delta r)^3 -r^3 \right]\] the difference in volume between two concentric spheres. When this expression for the volume is expanded, we obtain \[\dfrac{4}{3} \pi \left(3r^2 \Delta r + 3r \Delta r^2 + \Delta r^3\right)\] and for very small values of \(\Delta r\) the \(3r \Delta r^2\) and \(\Delta r^3\) terms are negligible in comparison with \(3r^2\Delta r\). Thus for small values of \(\Delta r\), the two expressions for the volume of the shell approach one another in value and when \(\Delta r\) represents an infinitesimal small increment in \(r\) they are identical. The radial distribution function is plotted in Figure \(\Page {2}\) for the ground state of the hydrogen atom. 4 Figure \(\Page {1}\): e \(n = 1\) leve We could also represent the distribution of negative charge in the hydrogen atom in the manner used previously for the electron confined to move on a plane (Figure \(\Page {1}\)), by displaying the charge density in a plane by means of a contour map. Imagine a plane through the atom including the nucleus. The density is calculated at every point in this plane. All points having the same value for the electron density in this plane are joined by a contour line . Since the electron density depends only on , the distance from the nucleus, and not on the direction in space, the contours will be circular. A contour map is useful as it indicates the "shape" of the density distribution. This completes the description of the most stable state of the hydrogen atom, the state for which \( = 1\). Before proceeding with a discussion of the excited states of the hydrogen atom we must introduce a new term. When the energy of the electron is increased to another of the allowed values, corresponding to a new value for \(n\), \(y_n\) and \(P_n\) change as well. The wavefunctions \(y_n\) for the hydrogen atom are given a special name, , because they play such an important role in all of our future discussions of the electronic structure of atoms. In general the word orbital is the name given to a wavefunction which determines the motion of a single electron. If the one-electron wave function is for an atomic system, it is called an atomic orbital. Do not confuse the word orbital with the classical word and notion of an orbit. First, an orbit implies the knowledge of a definite trajectory or path for a particle through space which in itself is not possible for an electron. Secondly, an orbital, like the wave function, has no physical reality but is a mathematical function which when squared gives the physically measurable electron density distribution. For every value of the energy , for the hydrogen atom, there is a degeneracy equal to \(n^2\). Therefore, for = 1, there is but one atomic orbital and one electron density distribution. However, for = 2, there are four different atomic orbitals and four different electron density distributions, all of which possess the same value for the energy, Thus for all values of the principal quantum number there are different ways in which the electronic charge may be distributed in three-dimensional space and still possess the same value for the energy. For every value of the principal quantum number, of the possible atomic orbitals is independent of direction and gives a spherical electron density distribution which can be represented by circular contours as has been exemplified above for the case of = 1. The other atomic orbitals for a given value of exhibit a directional dependence and predict density distributions which are not spherical but are concentrated in planes or along certain axes. The angular dependence of the atomic orbitals for the hydrogen atom and the shapes of the contours of the corresponding electron density distributions are intimately connected with the angular momentum possessed by the electron. The physical quantity known as angular momentum plays a dominant role in the understanding of the electronic structure of atoms. To gain a physical picture and feeling for the angular momentum it is necessary to consider a model system from the classical point of view. The simplest classical model of the hydrogen atom is one in which the electron moves in a circular orbit with a constant speed or angular velocity (Figure \(\Page {4}\)). Just as the ordinary momentum \(m\vec{v}\) plays a dominant role in the analysis of straight line or linear motion, so angular momentum plays the central role in the analysis of a system with circular motion as found in the model of the hydrogen atom. In Figure \(\Page {4}\), is the mass of the electron, is the linear velocity (the velocity the electron would possess if it continued moving at a tangent to the orbit as indicated in the figure) and is the radius of the orbit. The linear velocity is a vector since it possesses at any instant both a magnitude and a direction in space. Obviously, as the electron rotates in the orbit the direction of \(\vec{v}\) is constantly changing, and thus the linear momentum \(m\vec{v}\) is constant for the circular motion. This is so even though the speed of the electron (i.e, the magnitude of \(\vec{v}\) which is denoted by \(|\vec{v}|\)) remains unchanged. According to Newton's second law, a force must be acting on the electron if its momentum changes with time. This is the force which prevents the electron from flying on tangent to its orbit. In an atom the attractive force which contains the electron is the electrostatic force of attraction between the nucleus and the electron, directed along the radius at right angles to the direction of the electron's motion. The angular momentum, like the linear momentum, is a vector and is defined as follows: \[|\vec{M}| = m \nu r\] or \(\vec{M}\) \vec{M}\) The important point of the above discussion is that both the angular momentum and the energy of an atom remain constant if the atom is left undisturbed. Any physical quantity which is constant in a classical system is both conserved and quantized in a quantum mechanical system. Thus both the energy and the angular momentum are quantized for an atom. Any physical quantity which is constant in a classical system is both conserved and quantized in a quantum mechanical system. There is a quantum number, denoted by \(l\), which governs the magnitude of the angular momentum, just as the quantum number \(n\) determines the energy. The of the angular momentum may assume only those values given by: \[ M = \sqrt{l(l+1)} \hbar\] with \(l = 0, 1, 2, 3, ... n-1\) Furthermore, the value of limits the maximum value of the angular momentum as the value of cannot be greater than - 1. For the state = 1 discussed above, may have the value of zero only. When = 2, may equal 0 or 1, and for = 3, = 0 or 1 or 2, etc. When = 0, it is evident from Equation \(\ref{4}\) that the angular momentum of the electron is zero. The atomic orbitals which describe these states of zero angular momentum are called orbitals. The orbitals are distinguished from one another by stating the value of , the principal quantum number. They are referred to as the 1 , 2 , 3 , etc., atomic orbitals. The preceding discussion referred to the 1 orbital since for the ground state of the hydrogen atom = 1 and = 0. This orbital, and all orbitals in general, predict spherical density distributions for the electron as exemplified by Figure \(\Page {2}\) for the 1s density. Figure \(\Page {5}\) shows the radial distribution functions \( ( )\) which apply when the electron is in a 2 or 3 orbital to illustrate how the character of the density distributions change as the value of \(n\) is increased. It is common usage to refer to an electron as being "in" an orbital even though an orbital is, but a mathematical function with no physical reality. To say an electron is in a particular orbital is meant to imply that the electron is in the quantum state which is described by that orbital. For example, when the electron is in the 2 orbital the hydrogen atom is in a state for which = 2 and = 0. Comparing these results with those for the 1 orbital in Figure \(\Page {2}\) we see that as \( increases the average value of \( increases. This agrees with the fact that the energy of the electron also increases as \( increases. The increased energy results in the electron being on the average pulled further away from the attractive force of the nucleus. As in the simple example of an electron moving on a line, nodes (values of \(r\) for which the electron density is zero) appear in the probability distributions. The number of nodes increases with increasing energy and equals \(n - 1\). When the electron possesses angular momentum the density distributions are no longer spherical. In fact for each value of , the electron density distribution assumes a characteristic shape Figure \(\Page {6}\). When = 1, the orbitals are called orbitals. In this case the orbital and its electron density are concentrated along a line (axis) in space. The 2 orbital or wave function is positive in value on one side and negative in value on the other side of a plane which is perpendicular to the axis of the orbital and passes through the nucleus. The orbital has a node in this plane, and consequently an electron in a 2 orbital does not place any electronic charge density at the nucleus. The electron density of a 1 orbital, on the other hand, is a maximum at the nucleus. The same diagram for the 2 density distribution is obtained for any plane which contains this axis. Thus in three dimensions the electron density would appear to be concentrated in two lobes, one on each side of the nucleus, each lobe being circular in cross section . When = 2, the orbitals are called orbitals andFigure \(\Page {7}\) shows the contours in a plane for a 3 orbital and its density distribution. Notice that the density is again zero at the nucleus and that there are now two nodes in the orbital and in its density distribution. As a final example, shows the contours of the orbital and electron density distribution obtained for a 4 atomic orbital which occurs when = 4 and = 3. The point to notice is that as the angular momentum of the electron increases, the density distribution becomes increasingly concentrated along an axis or in a plane in space. Only electrons in orbitals with zero angular momentum give spherical density distributions and in addition place charge density at the position of the nucleus. There seems to be neither rhyme nor reason for the naming of the states corresponding to the different values of \( ( for = 0, 1, 2, 3). This set of labels had its origin in the early work of experimental atomic spectroscopy. The letter stood for sharp, for principal, for diffuse and for fundamental in characterizing spectral lines. From the letter onwards the naming of the orbitals is alphabetical \(l = 4,5,6 \rightarrow g,h,i, ....\). We have not as yet accounted for the full degeneracy of the hydrogen atom orbitals which we stated earlier to be for every value of . For example, when = 2, there are four distinct atomic orbitals. The remaining degeneracy is again determined by the angular momentum of the system. Since angular momentum like linear momentum is a vector quantity, we may refer to the component of the angular momentum vector which lies along some chosen axis. For reasons we shall investigate, the number of values a particular component can assume for a given value of is (2 + 1). Thus when = 0, there is no angular momentum and there is but a single orbital, an orbital. When = 1, there are three possible values for the component (2 1 + 1) of the total angular momentum which are physically distinguishable from one another. There are, therefore, three orbitals. Similarly there are five orbitals, (2 2+1), seven orbitals, (2 3 +1), etc. All of the orbitals with the same value of and , the three 2 orbitals for example, are similar but differ in their spatial orientations. To gain a better understanding of this final element of degeneracy, we must consider in more detail what quantum mechanics predicts concerning the angular momentum of an electron in an atom.  (  /  )

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The 1,2-dienes, which have cumulated double bonds, commonly are called allenes. The simplest example is 1,2-propadiene, Allenes of the type \(\ce{RR'C=C=CRR'}\) are molecules and can exist in two stereoisomeric forms, one being the mirror image of the other and neither being superimposable on the other (i.e., enantiomers, Figure 13-5). Verification of the chirality of such allenes (originally proposed by van’t Hoff in 1875) was slow in coming and was preceded by many unsuccessful attempts to resolve suitably substituted allenes into their enantiomers. The first successful resolutions were achieved in 1935 for the enantiomers of two compounds \(1\) and \(2\). This was a classic achievement because it dispelled the suspicion prevalent at the time that rotation about the bonds of the cumulated diene system was free enough to preclude the isolation of configurationally stable enantiomers. The chirality observed in this kind of substituted allene is a consequence of dissymmetry resulting from restricted rotation about the double bonds, not because of a tetrahedral atom carrying four different groups. Restricted rotation occurs in many other kinds of compounds and a few examples are shown in Table 13-3, which includes -cycloalkenes ( ), cycloalkylidenes, spiranes, and -substituted biphenyl compounds. To have enantiomers, the structure must not have a plane or center of symmetry ( ). The chirality of biphenyls results from restricted rotation about a bond imposed by the bulky nature of ortho substituents. Models will help you visualize the degree of difficulty of having the substituents pass by one another. If \(\ce{X} = \ce{H}\) and \(\ce{Y} = \ce{F}\) (Table 13-3), the enantiomers are not stable at room temperature; if \(\ce{X} = \ce{H}\) and \(\ce{Y} = \ce{Br}\), they are marginally stable; if \(\ce{X} = \ce{H}\) and \(\ce{Y} = \ce{I}\), the rate of loss of optical activity is about 700 times slower than with \(\ce{Y} = \ce{Br}\). This is in keeping with the fact that the atomic size of the halogens increases in the order \(\ce{F} < \ce{Br} < \ce{I}\). In a cumulated triene, or any cumulated polyene with an number of double bonds, the atoms connected to the terminal carbons lie in the same plane, just as they do in an ordinary alkene. Van’t Hoff pointed out that suitably substituted cumulated polyenes of this type should then have cis and trans forms: Like the resolution of allenes, the separate existence of cis and trans isomers of cumulated trienes was not verified until many years after van’t Hoff’s original predictions, but a separation finally was achieved, in 1954, by R. Kuhn and K. Scholler for compounds \(3\) and \(4\): There are relatively few cis-trans forms of 1,2,3-alkatrienes known. They appear to interconvert readily on mild heating, which suggests that one of the double bonds has a lower rotational barrier than is normal for an alkene double bond. The properties of allenes are similar to those of alkenes, although the pure compounds often are difficult to prepare and are not indefinitely stable. Allenes undergo many of the usual double-bond reactions, being readily hydrogenated, adding bromine, and being oxidized with potassium permanganate solution. The hydration of allenes resembles the hydration of alkynes in giving initially an unstable enol that rapidly rearranges to a ketone: Allenes are not as stable as dienes with conjugated or isolated double bonds. The heats of hydrogenation (Table 11-2) indicate that the order of stability is conjugated dienes \(>\) isolated dienes \(>\) cumulated dienes. The relative instability of allenes probably reflects extra strain as the result of one carbon atom forming two double bonds. 1,2-Propadiene is slightly more strained than that of propyne. It is not surprising then that 1,2-propadiene isomerizes to propyne. This isomerization occurs under the influence of strongly basic substances such as sodium amide in liquid ammonia or potassium hydroxide in ethyl alcohol: Indeed, one of the difficulties associated with syntheses of allenes and alkynes (which often are carried out in the presence of strong bases) is the concurrent formation of isomerization products. The basic catalyst in the isomerization of 1,2-butadiene to butynes acts by removing an alkenic proton from the hydrocarbon. Two different anions can be formed, each of which is stabilized by electron delocalization involving the adjacent multiple bond. Either anion can react with the solvent by proton transfer to form the starting material or an alkyne. At equilibrium the most stable product, which is 2-butyne, predominates [1-butyne \(\left( g \right) \rightleftharpoons\) 2-butyne \(\left( g \right)\), \(\Delta G^0 = -4.0 \: \text{kcal mol}^{-1}\)]: and (1977)

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The Rayleigh-Jeans Radiation Law was a useful, but not completely successful attempt at establishing the functional form of the spectra of thermal radiation. The energy density \(u_ν\) per unit frequency interval at a frequency \(ν\) is, according to the The Rayleigh-Jeans Radiation, \[u_ν = \dfrac{8πν^2kT}{c^2} \nonumber \] where \(k\) is Boltzmann's constant, \(T\) is the absolute temperature of the radiating body, and \(c\) is the speed of light in a vacuum. This formula fits the empirical measurements for low frequencies, but fails increasingly for higher frequencies. The failure of the formula to match the new data was called the . The significance of this inadequate so-called law is that it provides an asymptotic condition which other proposed formulas, such as Planck's, need to satisfy. It gives a value to an otherwise arbitrary constant in Planck's thermal radiation formula. Consider a cube of edge length \(L\) in which radiation is being reflected and re-reflected off its walls. Standing waves occur for radiation of a wavelength \(λ\) only if an integral number of half-wave cycles fit into an interval in the cube. For radiation parallel to an edge of the cube this requires \[\dfrac{L}{λ/2} = m \nonumber \] where \(m\) is an integer or, equivalently \[λ = \dfrac{2L}{m} \nonumber \] Between two end points there can be two standing waves, one for each polarization. In the following the matter of polarization will be ignored until the end of the analysis and there the number of waves will be doubled to take into account the matter of polarization. Since the frequency \(\nu\) is equal to \(c/λ\), where \(c\) is the speed of light \[ν = \dfrac{cm}{2L} \nonumber \] It is convenient to work with the quantity \(q\), known as the , which is defined as \[q = \dfrac{2π}{λ} \nonumber \] and hence \[q = \dfrac{2πν}{c} \nonumber \] In terms of the relationship for the cube, \[q = \dfrac{2πm}{2L} = π\left(\dfrac{m}{L}\right) \nonumber \] and hence \[q^2 = π^2\left(\dfrac{m}{L}\right)^2 \nonumber \] Another convenient term is the radian frequency \(ω=2πν\). From this it follows that \(q=ω/c\). If \(m_X\), \(m_Y\), \(m_Z\) denote the integers for the three different directions in the cube then the condition for a standing wave in the cube is that \[q^2 = π^2\left[ \left(\dfrac{m_X}{L}\right)^2 + \left(\dfrac{m_Y}{L}\right)^2 + \left(\dfrac{m_Z}{L}\right)^2\right] \nonumber \] which reduces to \[m_X^2 + m_Y^2 + m_Z^2 = \dfrac{4L^2ν^2}{c^2} \nonumber \] Now the problem is to find the number of nonnegative combinations of (\(m_X\), \(m_Y\), \(m_Z\)) that fit between a sphere of radius \(R\) and and one of radius \(R+dR\). First the number of combinations ignoring the nonnegativity requirement can be determined. The volume of a spherical shell of inner radius \(R\) and outer radius \(R+dR\) is given by \[dV = 4πR^2\,dR \nonumber \] If \[R = \sqrt{m_X^2+m_Y^2+m_Z^2} \nonumber \] then \[R = \sqrt{\dfrac{4L^2ν^2}{c^2}} = \dfrac{2Lν}{c} \nonumber \] and hence \[dR=\dfrac{2L\,dν}{c}. \nonumber \] This means that \[\begin{align*} dV &= 4π\left(\dfrac{2Lν}{c}\right)^2 \left(\dfrac{2L}{c}\right)\,dν \\[4pt] &= 32π \left(\dfrac{L^3ν^2}{c^3}\right) dν \end{align*}\] Now the nonnegativity require for the combinations (\(m_X\), \(m_Y\), \(m_Z\)) must be taken into account. For the two dimensional case the nonnegative combinations are approximately those in one quadrant of circle. The approximation arises from the matter of the combinations on the boundaries of the nonnegative quadrant. For the three dimensional case the nonnegative combinations constitute approximately one octant of the total. Thus the number \(dN\) for the nonnegative combinations of (\(m_X\), \(m_Y\), \(m_Z\)) in this volume is equal to \((1/8)dV\) and hence \[dN = 4πν^2\left(\dfrac{L^3}{c^3}\right)\,dν \nonumber \] The average kinetic energy per degree of freedom is \(½kT\), where \(k\) is Boltzmann's constant. For harmonic oscillators there is an equality between kinetic and potential energy so the average energy per degree of freedom is \(kT\). This means that the average radiation energy \(E\) per unit frequency is given by \[\dfrac{dE}{dν} = kT\left(\dfrac{dN}{dν}\right) = 4πkT\left(\dfrac{L^3}{c^3}\right)ν^2 \nonumber \] and the average energy , \(u_ν\), is given by \[\dfrac{du_ν}{dν} = \left(\dfrac{1}{L^3}\right)\left(\dfrac{dE}{dν}\right) = \dfrac{4πkTν^2}{c^3} \nonumber \] The previous only considered one direction of polarization for the radiation. If the two directions of polarization are taken into account a factor of 2 must be included in the above formula; i.e., \[\dfrac{du_ν}{dν} = \dfrac{8πkTν^2}{c^3} \nonumber \] This is the Raleigh-Jeans Law of Radiation and holds empirically as the frequency goes to zero.

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A method for estimating electronegativity was developed by Robert Mulliken (1896–1986; Nobel Prize in Chemistry 1966) who noticed that elements with large first ionization energies tend to have very negative electron affinities and gain electrons in chemical reactions. Conversely, elements with small first ionization energies tend to have slightly negative (or even positive) electron affinities and lose electrons in chemical reactions. Mulliken recognized that an atom’s tendency to gain or lose electrons could therefore be described quantitatively by the average of the values of its first ionization energy and the absolute value of its electron affinity. Robert S. Mulliken proposed that the arithmetic mean of the first (\(E_{I_1}\)) and the (\(E_{ea}\)) should be a measure of the tendency of an atom to attract electrons. As this definition is not dependent on an arbitrary relative scale, it has also been termed Using our definition of electron affinity, we can write Mulliken’s original expression for electronegativity as follows: \[ \chi = \dfrac{|E_{I_1} + E_{ea}|}{2} \label{1}\] Elements with a large first ionization energy and a very negative electron affinity have a large positive value in the numerator of Equation \(\ref{1}\), so their electronegativity is high. Elements with a small first ionization energy and a small electron affinity have a small positive value for the numerator in Equation \(\ref{1}\), so they have a low electronegativity. However, it is more usual to use a linear transformation to transform these absolute values into values that resemble the more familiar . For ionization energies and electron affinities in electronvolts: \[\chi_{Mulliken} = 0.187 (E_{I_1}+E_{ea})+0.17 \label{2}\] and for energies in kJ/mol, \[\chi_{Mulliken} = (1.97 \times 10^{-3})(E_{I_1}+E_{ea})+0.19 \label{3}\] The Mulliken electronegativity can only be calculated for an element for which the electron affinity is known, fifty-seven elements as of 2006. The Mulliken electronegativity of an atom is sometimes said to be the negative of the chemical potential. By inserting the energetic definitions of the ionization potential and electron affinity into the Mulliken electronegativity, it is possible to show that the Mulliken chemical potential is a finite difference approximation of the electronic energy with respect to the number of electrons., i.e., \[\mu_{Mulliken}= -\chi_{Mulliken} = -\dfrac{E_{I_1} + E_{ea}}{2} \label{4}\] All electronegativity scales give essentially the same results for one element relative to another. Even though the Mulliken scale is based on the properties of individual and the Pauling scale is based on the properties of atoms in , they both apparently measure the same basic property of an element. In the following discussion, we will focus on the relationship between electronegativity and the tendency of to form positive or negative ions. We will therefore be implicitly using the Mulliken definition of electronegativity. Because of the parallels between the Mulliken and Pauling definitions, however, the conclusions are likely to apply to atoms in molecules as well. Despite being developed from a very different set of principles than , which is based on bond dissociation energies, there is a good correlation between Mullikin and Pauling Electronegativities for the atoms, as shown in the plot below. Although Pauling electronegativities are usually what are found in textbooks, the Mullikin electronegativity more intuitively corresponds to the "ability of an atom to draw electrons toward itself in bonding," and is probably a better indicator of that property. However, because of the good correlation between the two scales, using the Pauling scale is sufficient for most purposes.

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Chemists frequently need to know the equilibrium constant for a reaction that has not been previously studied. In such cases, the desired reaction can often be written as the sum of other reactions for which the equilibrium constants are known. The equilibrium constant for the unknown reaction can then be calculated from the tabulated values for the other reactions. To illustrate this procedure, let’s consider the reaction of \(N_2\) with \(O_2\) to give \(NO_2\). This reaction is an important source of the \(NO_2\) that gives urban smog its typical brown color. The reaction normally occurs in two distinct steps. In the first reaction (1), \(N_2\) reacts with \(O_2\) at the high temperatures inside an internal combustion engine to give \(NO\). The released \(NO\) then reacts with additional \(O_2\) to give \(NO_2\) (2). The equilibrium constant for each reaction at 100°C is also given. Summing reactions (1) and (2) gives the overall reaction of \(N_2\) with \(O_2\): The equilibrium constant expressions for the reactions are as follows: \[K_1=\dfrac{[NO]^2}{[N_2,O_2]}\;\;\; K_2=\dfrac{[NO_2]^2}{[NO]^2[O_2]}\;\;\; K_3=\dfrac{[NO_2]^2}{[N_2,O_2]^2}\] What is the relationship between \(K_1\), \(K_2\), and \(K_3\), all at 100°C? The expression for \(K_1\) has \([NO]^2\) in the numerator, the expression for \(K_2\) has \([NO]^2\) in the denominator, and \([NO]^2\) does not appear in the expression for \(K_3\). Multiplying \(K_1\) by \(K_2\) and canceling the \([NO]^2\) terms, \[ K_1K_2=\dfrac{\cancel{[NO]^2}}{[N_2,O_2]} \times \dfrac{[NO_2]^2}{\cancel{[NO]^2}[O_2]}=\dfrac{[NO_2]^2}{[N_2,O_2]^2}=K_3\] Thus the product of the equilibrium constant expressions for \(K_1\) and \(K_2\) is the same as the equilibrium constant expression for \(K_3\): \[K_3 = K_1K_2 = (2.0 \times 10^{−25})(6.4 \times 10^9) = 1.3 \times 10^{−15}\] The equilibrium constant for a reaction that is the sum of two or more reactions is equal to the product of the equilibrium constants for the individual reactions. In contrast, recall that according to , \(ΔH\) for the sum of two or more reactions is the sum of the ΔH values for the individual reactions. To determine \(K\) for a reaction that is the sum of two or more reactions, add the reactions but multiply the equilibrium constants. The following reactions occur at 1200°C: Calculate the equilibrium constant for the following reaction at the same temperature. : two balanced equilibrium equations, values of \(K\), and an equilibrium equation for the overall reaction : equilibrium constant for the overall reaction : Arrange the equations so that their sum produces the overall equation. If an equation had to be reversed, invert the value of \(K\) for that equation. Calculate \(K\) for the overall equation by multiplying the equilibrium constants for the individual equations. : The key to solving this problem is to recognize that reaction 3 is the sum of reactions 1 and 2: \[CO_{(g)}+ \cancel{3H_{2(g)}} \rightleftharpoons \cancel{CH_{4(g)}} + H_2O_{(g)}\] \[\cancel{CH_{4(g)}} +2H_2S_{(g)} \rightleftharpoons CS_{2(g)} + \cancel{3H_{2(g)}} + H_{2(g)}\] \[ CO_{(g)} + 3H_{2(g)} \rightleftharpoons CS_{2(g)}+H_2O_{(g)}+H_{2(g)}\] The values for \(K_1\) and \(K_2\) are given, so it is straightforward to calculate \(K_3\): \[K_3 = K_1K_2 = (9.17 \times 10^{−2})(3.3 \times 10^4) = 3.03 \times 10^3\] In the first of two steps in the industrial synthesis of sulfuric acid, elemental sulfur reacts with oxygen to produce sulfur dioxide. In the second step, sulfur dioxide reacts with additional oxygen to form sulfur trioxide. The reaction for each step is shown, as is the value of the corresponding equilibrium constant at 25°C. Calculate the equilibrium constant for the overall reaction at this same temperature. \(K_3 = 1.1 \times 10^{66}\) An equilibrium system that contains products and reactants in a single phase is a homogeneous equilibrium; a system whose reactants, products, or both are in more than one phase is a heterogeneous equilibrium. When a reaction can be expressed as the sum of two or more reactions, its equilibrium constant is equal to the product of the equilibrium constants for the individual reactions.

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For two molecules to react, they must first come into contact with each other. This contact can be considered a "collision." The more mobile the molecules are, the more likely they are to collide. In addition, the closer the molecules are together, the more likely they are to collide. In the following drawings, the molecules are closer together in the picture on the right than they are in the picture on the left. The molecules are more likely to collide and react in the picture on the right. The two figures above might be described in terms of population density. Both drawings appear to offer the same amount of space, but are inhabited by different amounts of molecules. The difference is much like the difference between human population densities in various locations around the world. Some places, such as Mexico City or Tokyo, are very crowded; they have high population densities. Some places, such as the Australian Outback or the Canadian Arctic, have low population densities. ,

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It should be stressed that there . All that is properly defined are differences in enthalpy ΔH and these are only defined for For example for the process of: We could write ΔH = ∫ C . dT from T=253K to T=273K = H(273) -H(253). But before moving beyond the melting point first a different process needs to take place, that of This gives us Δ H = H - H (both at 273K!). When we heat the liquid water further to say +20 C we would have to integrate over the heat capacity of the liquid. We could write \[ΔH_3 = \int C_{P,water} dT \nonumber \] from T=273K to T=293K = H(293)-H(273). The total change in enthalpy between -20 and +20 would be the sum of the three enthalpy changes. \[ΔH_{total process} = ΔH_1 + Δ_{fus}H + ΔH_3 \nonumber \] Of course we could consider doing the same calculation for any temperature between -20 and +20 and summarize all our results in a graph. The three processes can thus schematically be shown in Figure 19.10.1 . Notice that the slopes (i.e. the heat capacities!) before and after the melting point differ. The slope for the liquid is a little steeper because the liquid has more degrees of freedom and therefore the heat capacity of the liquid tends to be higher than of the solid. In the figure the enthlapy curves are shown as straight lines. This would be the case if the heat capacities are over the temperature interval. Although C is typically a 'slow' or 'weak' function of temperature it usually does change a bit, which means that the straight lines for H become curves. Although C is typically a 'slow' or 'weak' function of temperature and is well approximated as a constant. Notice that for process two, the temperature is constant, that means that \(ΔT\) or \(dT\) is zero, but ΔH is finite, consequently ΔH/ΔT is infinitly large. Taking the limit for ΔT going to zero, we get a derivative: \[ \left( \dfrac{\partial H}{ \partial T} \right)_p = C_p \nonumber \] This derivative, the heat capacity must undergo a singularity: the slope is infinitely large (i.e, the \(H\) curve goes straight up). When there are more phase transitions, more discontinuities in \(H\) and singularities in \(C_p\) result (Figure 19.10.1 ). Note that \(H(T)-H(0)\), not \(H(T)\) is plotted to avoid the question what the absolute enthalpy is. There is technique that allows us to measure the heat capacity as a function of temperature fairly directly. It is called Differential Scanning Calorimetry (DSC). You put a sample in a little pan and put the pan plus an empty reference pan in the calorimeter. The instrument heats up both pans with a constant heating rate. Both pans get hotter by , but the heat capacity of the filled pan is obviously bigger. This means that the heat flow into the sample pan must be a bit bigger than into the empty one. This differential heat flow induces a tiny temperature difference ΔT between the two pans that can be measured. This temperature difference is proportional to the heat flow difference which is proportional to the heat capacity difference. \[ΔT ~ ΔΦ ~ Δ_{between pans}C_p = C_p^{sample} \,(if the pans cancel) \nonumber \] However, there are number of serious broadening issues with the technique. If you melt something you will never get to see the infinite singularity of the heat capacity. Instead it broadens out into a peak. If you integrate the peak you get the Δ H and the onset is calibrated to give you the melting point. It is even possible to heat the sample with a rate that fluctuates with a little sine wave. This "Modulated DSC" version can even give you the (small) difference in \(C_p\) before and after the melting event.

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The idea of oxidation states is not normally applied to organic compounds, but it can be useful to do so. When we do, we can gain some insight into certain reactions of organic molecules. For example,carbon dioxide, CO , can be thought of as having carbon in an oxidized state. If we apply the usual oxidation state rule, carbon dioxide is overall neutral and contains two oxygens, each with 2- charge. To counter that charge, carbon must be in oxidation state 4+. On the other hand, methane, CH , can be thought of as having carbon in a reduced state. If we apply the usual oxidation state rule here, methane is overall neutral but contains four protons. That means the carbon must be in a 4- oxidation state. Of course, the carbon does not behave as if it has a minus four charge. But we will see that this sort of exercise can be useful for book-keeping purposes. Assign the formal oxidation state to carbon in the following molecules. a) methanol, CH OH b) formaldehyde or methanal, CH O c) carbonate, CO d) hydrogen cyanide, HCN e) ethane, CH CH f) ethene, CH CH g) ethyne, CHCH The general trend here is that the more bonds there are to oxygen, the more oxidized is carbon. The more bonds there are to hydrogen, the more reduced is carbon. Adding a hydrogen nucleophile to a carbonyl electrophile is routinely referred to as a reduction. For example, adding sodium borohydride to methanal would result in reduction to form methanol. Of course, a hydride is really a proton plus two electrons. We could write an equation for the reduction of methanal that looks a lot like the redox reactions we see in a table of standard reduction potentials. \[ CH_2O + H^- + H^+ \rightarrow CH_3OH \] or \[ CH_2O + 2e^- + 2H^+ \rightarrow CH_3OH \] It stands to reason that the opposite reaction, the conversion of methanol to methanal, is a two electron oxidation. \[ CH_3OH \rightarrow CH_2O + 2e^- + 2H^+ \] We know how to accomplish the reduction of methanal, at least on paper. We just add a complex metal hydride, such as lithium aluminum hydride or sodium borohydride, to the carbonyl compound. After an acidic aqueous workup to remove all the lithium and aluminum compounds, we get methanol. For practical reasons, methanol may be difficult to isolate this way, but that's the general idea of the reaction. How do we accomplish the reverse reaction? One way would be to provide a hydride acceptor in the reaction, so that we could catch hydride as it comes off the methanol. The most well-known such entity is NAD+, of course. There are biological oxidations that employ NAD+ for this reason. More generally, the reaction can be accomplished in a number of ways, on paper, by separating out the two tasks involved. We need something to accept the two protons: that's a base. We need something to accept the two electrons: that's an oxidizing agent. For the latter task, there are a number of high oxidation state transition metal compounds that are quite willing to accept two electrons. One of the most widely employed is Cr(VI), which accepts two electrons to become Cr(IV). A number of other methods are available, having been developed partly to avoid the toxicity of chromium salts, but let's look at the chromium case as an example. A simple chromium(VI) compound is chromium trioxide. A simple base is pyridine. If we took these two reagents together with benzyl alcohol in a solvent such as dichloromethane, what would happen? OK, you might not want to try this, because chromium trioxide has an alarming capacity to cause spontaneous combustion in organic compounds, but we can do it on paper. Is chromium trioxide a nucleophile or an electrophile? That Cr(VI) is pretty electrophilic, surely. So what part of the benzyl alcohol is nucleophilic? The oxygen atom. When we mix these things, the oxygen atom would likely coordinate to the chromium. When the oxygen atom coordinates to the chromium, the oxygen gets a positive formal charge. It is now motivated to lose a proton. That's what the pyridine is for. Now we have accomplished one of the goals of the reaction. We have removed a proton from benzyl alcohol. We have one more proton and two electrons left. The second proton will have to come from the carbon attached to the oxygen; that's the place where we need to form a carbonyl. But wait a minute. You can't take two protons off the same molecule, can you? And certainly not from two atoms that are right next to each other. Doesn't that generate an unstable dianion? Not this time. The chromium is there to accept two electrons. We won't generate an anion at all, as far as the benzyl alcohol is concerned. It is oxidized to benzaldehyde. A completely different outcome to this reaction would be obtained in aqueous solution because of the equilibrium that exists between a carbonyl and the geminal diol (or hydrate) in water. Instead of obtaining an aldehyde, a carboxylic acid would be obtained via a second reduction. Provide a mechanism for this reaction. ,

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Hydrogen addition to multiple bonds is catalyzed by certain complex metal salts . This may be described as catalysis and, compared to heterogeneous catalysis, is a relatively new development in the area of hydrogenation reactions. Rhodium and ruthenium salts appear to be generally useful catalysts: At present, homogeneous catalysis for routine hydrogenation reactions offers little advantage over the convenience and simplicity of heterogeneous catalysis. Suprafacial addition of hydrogen is observed with both types of catalytic systems. However, greater selectivity can be achieved with homogeneous catalysts because they appear to be more sensitive to steric hindrance and are less likely to cause rearrangement, dissociation, and hydrogenation of other bonds (e.g., \(\ce{-NO_2}\) and ). The most thoroughly investigated homogeneous hydrogenation catalyst is the four-coordinated rhodium complex \(\ce{Rh} \left[ \ce{(C_6H_5)_3P} \right]_3 \ce{Cl}\). This catalyst is called after its discoverer, G. Wilkinson. In 1973, the Nobel Prize in chemistry was awarded jointly to Wilkinson and E. O. H. Fischer for their respective contributions to the field of . As you will see in this and later chapters, compounds with carbon-metal bonds (organometallic compounds) are extremely useful reagents, reactive intermediates, or catalysts in organic reactions. To a very large extent, the work of Fischer and Wilkinson created the current interest and developments in the field of transition-metal organic chemistry, which will be discussed in . and (1977)

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Linus Pauling described electronegativity as “the power of an atom in a molecule to attract electrons to itself.” Basically, the electronegativity of an atom is a relative value of that atom's ability to attract election density toward itself when it bonds to another atom. The higher the electronegativity of an element, the more that atom will attempt to pull electrons towards itself and away from any atom it bonds to. The main properties of an atom dictate it's electronegativity are it's atomic number as well as its atomic radius. The trend for electronegativity is to increase as you move from left to right and bottom to top across the periodic table. This means that the most electronegative atom is Fluorine and the least electronegative is Francium. There are a few different 'types' of electronegativity which differ only in their definitions and the system by which they assign values for electronegativity. For example, defines electronegativity as the "the average of the and of an atom." As we will see, this definition differs slightly from Pauling's definition of electronegativity. Linus Pauling was the original scientist to describe the phenomena of electronegativity. The best way to describe his method is to look at a hypothetical molecule that we will call XY. By comparing the measured X-Y bond energy with the theoretical X-Y bond energy (computed as the average of the X-X bond energy and the Y-Y bond energy), we can describe the relative affinities of these two atoms with respect to each other. \[Δ\text{Bond Energies} = (X-Y)_{measured} – (X-Y)_{expected} \nonumber\] If the electonegativities of X and Y are the same, then we would expect the measured bond energy to equal the theoretical (expected) bond energy and therefore the Δ bond energies would be zero. If the electronegativities of these atoms are not the same, we would see a polar molecule where one atom would start to pull electron density toward itself, causing it to become partially negative. By doing some careful experiments and calculations, Pauling came up with a slightly more sophisticated equation for the relative electronegativities of two atoms in a molecule: \[EN(X) - EN(Y) = 0.102 \sqrt{Δ}.\nonumber\] In that equation, the factor 0.102 is simply a conversion factor between kJ and eV to keep the units consistent with bond energies. By assigning a value of 4.0 to Fluorine (the most electronegative element), Pauling was able to set up relative values for all of the elements. This was when he first noticed the trend that the electronegativity of an atom was determined by it's position on the periodic table and that the electronegativity tended to increase as you moved left to right and bottom to top along the table. The range of values for Pauling's scale of electronegativity ranges from Fluorine (most electronegative = 4.0) to Francium (least electronegative = 0.7). Furthermore, if the electronegativity difference between two atoms is very large, then the bond type tends to be more ionic, however if the difference in electronegativity is small then it is a nonpolar covalent bond. Explain the difference between Electronegativity and Electron Affinity Predict the order or increasing electronegativity from the following elements

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We now introduce two concepts useful in describing heat flow and temperature change. The (\(C\)) of a body of matter is the quantity of heat (\(q\)) it absorbs or releases when it experiences a temperature change (\(ΔT\)) of 1 degree Celsius (or equivalently, 1 kelvin) \[C=\dfrac{q}{ΔT} \label{12.3.1} \] Heat capacity is determined by both the type and amount of substance that absorbs or releases heat. It is therefore an —its value is proportional to the amount of the substance. For example, consider the heat capacities of two cast iron frying pans. The heat capacity of the large pan is five times greater than that of the small pan because, although both are made of the same material, the mass of the large pan is five times greater than the mass of the small pan. More mass means more atoms are present in the larger pan, so it takes more energy to make all of those atoms vibrate faster. The heat capacity of the small cast iron frying pan is found by observing that it takes 18,140 J of energy to raise the temperature of the pan by 50.0 °C \[C_{\text{small pan}}=\dfrac{18,140\, J}{50.0\, °C} =363\; J/°C \label{12.3.2} \nonumber\] The larger cast iron frying pan, while made of the same substance, requires 90,700 J of energy to raise its temperature by 50.0 °C. The larger pan has a (proportionally) larger heat capacity because the larger amount of material requires a (proportionally) larger amount of energy to yield the same temperature change: \[C_{\text{large pan}}=\dfrac{90,700\, J}{50.0\,°C}=1814\, J/°C \label{12.3.3} \nonumber\] The (\(c\)) of a substance, commonly called its , is the quantity of heat required to raise the temperature of 1 gram of a substance by 1 degree Celsius (or 1 kelvin): \[c = \dfrac{q}{m\Delta T} \label{12.3.4} \] Specific heat capacity depends only on the kind of substance absorbing or releasing heat. It is an intensive property—the type, but not the amount, of the substance is all that matters. For example, the small cast iron frying pan has a mass of 808 g. The specific heat of iron (the material used to make the pan) is therefore: \[c_{iron}=\dfrac{18,140\; J}{(808\; g)(50.0\;°C)} = 0.449\; J/g\; °C \label{12.3.5} \nonumber\] The large frying pan has a mass of 4040 g. Using the data for this pan, we can also calculate the specific heat of iron: \[c_{iron}=\dfrac{90,700 J}{(4,040\; g)(50.0\;°C)}=0.449\; J/g\; °C \label{12.3.6} \nonumber\] Although the large pan is more massive than the small pan, since both are made of the same material, they both yield the same value for specific heat (for the material of construction, iron). Note that specific heat is measured in units of energy per temperature per mass and is an intensive property, being derived from a ratio of two extensive properties (heat and mass). The molar heat capacity, also an intensive property, is the heat capacity per mole of a particular substance and has units of J/mol °C (Figure \(\Page {1}\)). The heat capacity of an object depends on both its and its . For example, doubling the mass of an object doubles its heat capacity. Consequently, the amount of substance must be indicated when the heat capacity of the substance is reported. The is the amount of energy needed to increase the temperature of 1 mol of a substance by 1°C; the units of are thus J/(mol•°C). The (\(c_s\)) is the amount of energy needed to increase the temperature of 1 g of a substance by 1°C; its units are thus J/(g•°C). We can relate the quantity of a substance, the amount of heat transferred, its heat capacity, and the temperature change either via moles (Equation \(\ref{12.3.7}\)) or mass (Equation \(\ref{12.3.8}\)): \[q = nc_p ΔT \label{12.3.7}\] e \[q = mc_s ΔT \label{12.3.8}\] Both Equations \ref{12.3.7} and \ref{12.3.8} are under constant pressure (which matters) and both show that we know the amount of a substance and its specific heat (for mass) or molar heat capcity (for moles), we can determine the amount of heat, \(q\), entering or leaving the substance by measuring the temperature change before and after the heat is gained or lost. The specific heats of some common substances are given in Table \(\Page {1}\). Note that the specific heat values of most solids are less than 1 J/(g•°C), whereas those of most liquids are about 2 J/(g•°C). Water in its solid and liquid states is an exception. The heat capacity of ice is twice as high as that of most solids; the heat capacity of liquid water, 4.184 J/(g•°C), is one of the highest known. The specific heat of a substance varies somewhat with temperature. However, this variation is usually small enough that we will treat specific heat as constant over the range of temperatures that will be considered in this chapter. Specific heats of some common substances are listed in Table \(\Page {1}\). The value of \(C\) is intrinsically a positive number, but \(ΔT\) and \(q\) can be either positive or negative, and they both must have the same sign. If \(ΔT\) and \(q\) are positive, then . If \(ΔT\) and \(q\) are negative, then . If a substance gains thermal energy, its temperature increases, its final temperature is higher than its initial temperature, then \(ΔT>0 \) and \(q\) is positive. If a substance loses thermal energy, its temperature decreases, the final temperature is lower than the initial temperature, so \(ΔT<0 \) and \(q\) is negative. A flask containing \(8.0 \times 10^2\; g\) of water is heated, and the temperature of the water increases from \(21\, °C\) to \(85\, °C\). How much heat did the water absorb? To answer this question, consider these factors: The specific heat of water is 4.184 J/g °C (Table \(\Page {1}\)), so to heat 1 g of water by 1 °C requires 4.184 J. We note that since 4.184 J is required to heat 1 g of water by 1 °C, we will need 800 times as much to heat 800 g of water by 1 °C. Finally, we observe that since 4.184 J are required to heat 1 g of water by 1 °C, we will need 64 times as much to heat it by 64 °C (that is, from 21 °C to 85 °C). This can be summarized using Equation \ref{12.3.8}: \[\begin{align*} q&=mc_sΔT \nonumber \\[4pt] &= m c_s (T_\ce{final}−T_\ce{initial}) \\[4pt] &=\mathrm{(4.184\:J/\cancel{g}°C)×(800\:\cancel{g})×(85−21)°C} \\[4pt] &=\mathrm{(4.184\:J/\cancel{g}°\cancel{C})×(800\:\cancel{g})×(64)°\cancel{C}} \\[4pt] &=\mathrm{210,000\: J(=210\: kJ)} \nonumber \end{align*} \] Because the temperature increased, the water absorbed heat and \(q\) is positive. How much heat, in joules, must be added to a \(5.00 \times 10^2 \;g\) iron skillet to increase its temperature from 25 °C to 250 °C? The specific heat of iron is 0.451 J/g °C. \(5.07 \times 10^4\; J\) Note that the relationship between heat, specific heat, mass, and temperature change can be used to determine any of these quantities (not just heat) if the other three are known or can be deduced. A piece of unknown metal weighs 348 g. When the metal piece absorbs 6.64 kJ of heat, its temperature increases from 22.4 °C to 43.6 °C. Determine the specific heat of this metal (which might provide a clue to its identity). Since mass, heat, and temperature change are known for this metal, we can determine its specific heat using Equation \ref{12.3.8}: \[\begin{align*} q&=m c_s \Delta T &=m c_s (T_{final}−T_{initial}) \end{align*}\] Substituting the known values: \[6,640\; J=(348\; g) c_s (43.6 − 22.4)\; °C \nonumber\] Solving: \[c=\dfrac{6,640\; J}{(348\; g)(21.2°C)} =0.900\; J/g\; °C \nonumber\] Comparing this value with the values in Table \(\Page {1}\), this value matches the specific heat of aluminum, which suggests that the unknown metal may be aluminum. A piece of unknown metal weighs 217 g. When the metal piece absorbs 1.43 kJ of heat, its temperature increases from 24.5 °C to 39.1 °C. Determine the specific heat of this metal, and predict its identity. \(c = 0.45 \;J/g \;°C\); the metal is likely to be iron from checking Table \(\Page {1}\). A home solar energy storage unit uses 400 L of water for storing thermal energy. On a sunny day, the initial temperature of the water is 22.0°C. During the course of the day, the temperature of the water rises to 38.0°C as it circulates through the water wall. How much energy has been stored in the water? (The density of water at 22.0°C is 0.998 g/mL.) volume and density of water and initial and final temperatures amount of energy stored The mass of water is \[ mass \; of \; H_{2}O=400 \; \cancel{L}\left ( \dfrac{1000 \; \cancel{mL}}{1 \; \cancel{L}} \right ) \left ( \dfrac{0.998 \; g}{1 \; \cancel{mL}} \right ) = 3.99\times 10^{5}g\; H_{2}O \nonumber \] The temperature change (Δ ) is 38.0°C − 22.0°C = +16.0°C. From Table \(\Page {1}\), the specific heat of water is 4.184 J/(g•°C). From Equation \ref{12.3.8}, the heat absorbed by the water is thus \[ q=mc_s\Delta T=\left ( 3.99 \times 10^{5} \; \cancel{g} \right )\left ( \dfrac{4.184 \; J}{\cancel{g}\cdot \bcancel{^{o}C}} \right ) \left ( 16.0 \; \bcancel{^{o}C} \right ) = 2.67 \times 10^{7}J = 2.67 \times 10^{4}kJ \nonumber \] Both and Δ are positive, consistent with the fact that the water has absorbed energy. Some solar energy devices used in homes circulate air over a bed of rocks that absorb thermal energy from the sun. If a house uses a solar heating system that contains 2500 kg of sandstone rocks, what amount of energy is stored if the temperature of the rocks increases from 20.0°C to 34.5°C during the day? Assume that the specific heat of sandstone is the same as that of quartz (SiO ) in Table \(\Page {1}\). \(2.7 × 10^4\, kJ \) Even though the mass of sandstone is more than six times the mass of the water in Example \(\Page {1}\), the amount of thermal energy stored is the same to two significant figures. When two objects at different temperatures are placed in contact, heat flows from the warmer object to the cooler one until the temperature of both objects is the same. The law of conservation of energy says that the total energy cannot change during this process: \[q_{cold} + q_{hot} = 0 \label{12.3.9}\] The equation implies that the amount of heat that flows a warmer object is the same as the amount of heat that flows a cooler object. Because the direction of heat flow is opposite for the two objects, the sign of the heat flow values must be opposite: \[q_{cold} = −q_{hot} \label{12.3.10}\] Thus heat is conserved in any such process, consistent with the law of conservation of energy. The amount of heat lost by a warmer object equals the amount of heat gained by a cooler object. Substituting for \(q\) from Equation \(\ref{12.3.8}\) gives \[ \left [ mc_s \Delta T \right ] _{cold} + \left [ mc_s \Delta T \right ] _{hot}=0 \label{12.3.11} \nonumber \] which can be rearranged to give \[ \left [ mc_s \Delta T \right ] _{cold} = - \left [ mc_s \Delta T \right ] _{hot} \label{12.3.12} \] When two objects initially at different temperatures are placed in contact, we can use Equation \(\ref{12.3.12}\) to calculate the final temperature if we know the chemical composition and mass of the objects. If a 30.0 g piece of copper pipe at 80.0°C is placed in 100.0 g of water at 27.0°C, what is the final temperature? Assume that no heat is transferred to the surroundings. mass and initial temperature of two objects final temperature Using Equation \(\ref{12.3.12}\) and writing \(ΔT= T_{final} − T_{initial}\) for both the copper and the water, substitute the appropriate values of \(m\), \(c_s\), and \(T_{initial}\) into the equation and solve for \(T_{final}\). \[ \left [ mc_s \left (T_{final} - T_{initial} \right ) \right ] _{Cu} + \left [ mc_s \left (T_{final} - T_{initial} \right ) \right ] _{H_{2}O} =0 \nonumber \] Substituting the data provided in the problem and Table \(\Page {1}\) gives \[\begin{align*} \left (30 \; g \right ) (0.385 \; J/ (g °C) ) (T_{final} - 80°C) + (100\;g) (4.184 \; J/ (g °C) ) (T_{final} - 27.0°C ) &= 0 \nonumber \\[4pt] T_{final}\left ( 11.6 \; J/ ^{o}C \right ) -924 \; J + T_{final}\left ( 418.4 \; J/ ^{o}C \right ) -11,300 \; J &= 0 \\[4pt] T_{final}\left ( 430 \; J/\left ( g\cdot ^{o}C \right ) \right ) &= 12,224 \; J \nonumber \\[4pt] T_{final} &= 28.4 \; ^{o}C \end{align*} \] If a 14.0 g chunk of gold at 20.0°C is dropped into 25.0 g of water at 80.0°C, what is the final temperature if no heat is transferred to the surroundings? 80.0°C A 28.0 g chunk of aluminum is dropped into 100.0 g of water with an initial temperature of 20.0°C. If the final temperature of the water is 24.0°C, what was the initial temperature of the aluminum? (Assume that no heat is transferred to the surroundings.) 90.6°C One technique we can use to measure the amount of heat involved in a chemical or physical process is known as . Calorimetry is used to measure amounts of heat transferred to or from a substance. To do so, the heat is exchanged with a calibrated object (calorimeter). The change in temperature of the measuring part of the calorimeter is converted into the amount of heat (since the previous calibration was used to establish its heat capacity). The measurement of heat transfer using this approach requires the definition of a (the substance or substances undergoing the chemical or physical change) and its (the other components of the measurement apparatus that serve to either provide heat to the system or absorb heat from the system). Knowledge of the heat capacity of the surroundings, and careful measurements of the masses of the system and surroundings and their temperatures before and after the process allows one to calculate the heat transferred as described in this section. A is a device used to measure the amount of heat involved in a chemical or physical process. The thermal energy change accompanying a chemical reaction is responsible for the change in temperature that takes place in a calorimeter. If the reaction releases heat ( < 0), then heat is absorbed by the calorimeter ( > 0) and its temperature increases. Conversely, if the reaction absorbs heat ( > 0), then heat is transferred from the calorimeter to the system ( < 0) and the temperature of the calorimeter decreases. In both cases, . The heat capacity of the calorimeter or of the reaction mixture may be used to calculate the amount of heat released or absorbed by the chemical reaction. The amount of heat released or absorbed per gram or mole of reactant can then be calculated from the mass of the reactants. Because \(ΔH\) is defined as the heat flow at constant pressure, measurements made using a constant-pressure calorimeter (a device used to measure enthalpy changes in chemical processes at constant pressure) give \(ΔH\) values directly. This device is particularly well suited to studying reactions carried out in solution at a constant atmospheric pressure. A “student” version, called a (Figure \(\Page {3}\)), is often encountered in general chemistry laboratories. Commercial calorimeters operate on the same principle, but they can be used with smaller volumes of solution, have better thermal insulation, and can detect a change in temperature as small as several millionths of a degree (10 °C). Before we practice calorimetry problems involving chemical reactions, consider a simpler example that illustrates the core idea behind calorimetry. Suppose we initially have a high-temperature substance, such as a hot piece of metal (M), and a low-temperature substance, such as cool water (W). If we place the metal in the water, heat will flow from M to W. The temperature of M will decrease, and the temperature of W will increase, until the two substances have the same temperature—that is, when they reach thermal equilibrium. If this occurs in a calorimeter, ideally all of this heat transfer occurs between the two substances, with no heat gained or lost by either the calorimeter or the calorimeter’s surroundings. Under these ideal circumstances, the net heat change is zero: \[q_\mathrm{\,substance\: M} + q_\mathrm{\,substance\: W}=0 \label{12.3.13}\] This relationship can be rearranged to show that the heat gained by substance M is equal to the heat lost by substance W: \[q_\mathrm{\,substance\: M}=-q_\mathrm{\,substance\: W} \label{12.3.14}\] The magnitude of the heat (change) is therefore the same for both substances, and the negative sign merely shows that \(q_{substance\; M}\) and \(q_{substance\; W}\) are opposite in direction of heat flow (gain or loss) but does not indicate the arithmetic sign of either value (that is determined by whether the matter in question gains or loses heat, per definition). In the specific situation described, \(q_{substance\, M}\) is a negative value and is positive, since heat is transferred from M to W. A 360-g piece of rebar (a steel rod used for reinforcing concrete) is dropped into 425 mL of water at 24.0 °C. The final temperature of the water was measured as 42.7 °C. Calculate the initial temperature of the piece of rebar. Assume the specific heat of steel is approximately the same as that for iron ( ), and that all heat transfer occurs between the rebar and the water (there is no heat exchange with the surroundings). The temperature of the water increases from 24.0 °C to 42.7 °C, so the water absorbs heat. That heat came from the piece of rebar, which initially was at a higher temperature. Assuming that all heat transfer was between the rebar and the water, with no heat “lost” to the surroundings, then heat given off by rebar = −heat taken in by water, or: Since we know how heat is related to other measurable quantities, we have: Letting f = final and i = initial, in expanded form, this becomes: \[ c_\ce{rebar}×m_\ce{rebar}×(T_\mathrm{f,rebar}−T_\mathrm{i,rebar})=−c_\ce{water}×m_\ce{water}×(T_\mathrm{f,water}−T_\mathrm{i,water}) \nonumber\] The density of water is 1.0 g/mL, so 425 mL of water = 425 g. Noting that the final temperature of both the rebar and water is 42.7 °C, substituting known values yields: \[ \mathrm{(0.449\:J/g\: °C)(360g)(42.7°C−\mathit T_\mathrm{i,rebar})=-(4.184\:J/g\: °C)(425\:g)(42.7°C−24.0°C)} \nonumber\] \[\mathrm{\mathit T_{i,rebar}=\dfrac{(4.184\:J/g\: °C)(425\:g)(42.7°C−24.0°C)}{(0.449\:J/g\: °C)(360\:g)}+42.7°C} \nonumber\] Solving this gives \(T_{i,rebar}\)= 248 °C, so the initial temperature of the rebar was 248 °C. A 248-g piece of copper is dropped into 390 mL of water at 22.6 °C. The final temperature of the water was measured as 39.9 °C. Calculate the initial temperature of the piece of copper. Assume that all heat transfer occurs between the copper and the water. The initial temperature of the copper was 335.6 °C. A 248-g piece of copper initially at 314 °C is dropped into 390 mL of water initially at 22.6 °C. Assuming that all heat transfer occurs between the copper and the water, calculate the final temperature. The final temperature (reached by both copper and water) is 38.8 °C. This method can also be used to determine other quantities, such as the specific heat of an unknown metal. A 59.7 g piece of metal that had been submerged in boiling water was quickly transferred into 60.0 mL of water initially at 22.0 °C. The final temperature is 28.5 °C. Use these data to determine the specific heat of the metal. Use this result to identify the metal. Assuming perfect heat transfer, heat given off by metal = −heat taken in by water, or: In expanded form, this is: \[c_\ce{metal}×m_\ce{metal}×(T_\mathrm{f,metal}−T_\mathrm{i, metal})=−c_\ce{water}×m_\ce{water}×(T_\mathrm{f,water}−T_\mathrm{i,water}) \nonumber\] Noting that since the metal was submerged in boiling water, its initial temperature was 100.0 °C; and that for water, 60.0 mL = 60.0 g; we have: \[\mathrm{(\mathit c_{metal})(59.7\:g)(28.5°C−100.0°C)=−(4.18\:J/g\: °C)(60.0\:g)(28.5°C−22.0°C)} \nonumber\] Solving this: \[\mathrm{\mathit c_{metal}=\dfrac{−(4.184\:J/g\: °C)(60.0\:g)(6.5°C)}{(59.7\:g)(−71.5°C)}=0.38\:J/g\: °C} \nonumber \] Comparing this with values in , our experimental specific heat is closest to the value for copper (0.39 J/g °C), so we identify the metal as copper. A 92.9-g piece of a silver/gray metal is heated to 178.0 °C, and then quickly transferred into 75.0 mL of water initially at 24.0 °C. After 5 minutes, both the metal and the water have reached the same temperature: 29.7 °C. Determine the specific heat and the identity of the metal. (Note: You should find that the specific heat is close to that of two different metals. Explain how you can confidently determine the identity of the metal). \(c_{metal}= 0.13 \;J/g\; °C\) This specific heat is close to that of either gold or lead. It would be difficult to determine which metal this was based solely on the numerical values. However, the observation that the metal is silver/gray in addition to the value for the specific heat indicates that the metal is lead. When we use calorimetry to determine the heat involved in a chemical reaction, the same principles we have been discussing apply. The amount of heat absorbed by the calorimeter is often small enough that we can neglect it (though not for highly accurate measurements, as discussed later), and the calorimeter minimizes energy exchange with the surroundings. Because energy is neither created nor destroyed during a chemical reaction, there is no overall energy change during the reaction. The heat produced or consumed in the reaction (the “system”), , plus the heat absorbed or lost by the solution (the “surroundings”), , must add up to zero: This means that the amount of heat produced or consumed in the reaction equals the amount of heat absorbed or lost by the solution: \[q_\ce{reaction}=−q_\ce{solution} \label{12.3.16}\] This concept lies at the heart of all calorimetry problems and calculations. Because the heat released or absorbed at constant pressure is equal to Δ , the relationship between heat and Δ is \[ \Delta H_{rxn}=q_{rxn}=-q_{calorimater}=-mc_s \Delta T \label{12.3.17} \] The use of a constant-pressure calorimeter is illustrated in Example \(\Page {7}\). When 5.03 g of solid potassium hydroxide are dissolved in 100.0 mL of distilled water in a coffee-cup calorimeter, the temperature of the liquid increases from 23.0°C to 34.7°C. The density of water in this temperature range averages 0.9969 g/cm . What is \(ΔH_{soln}\) (in kilojoules per mole)? Assume that the calorimeter absorbs a negligible amount of heat and, because of the large volume of water, the specific heat of the solution is the same as the specific heat of pure water. mass of substance, volume of solvent, and initial and final temperatures Δ To calculate Δ , we must first determine the amount of heat released in the calorimetry experiment. The mass of the solution is \[ \left (100.0 \; \cancel{mL}\; \ce{H2O} \right ) \left ( 0.9969 \; g/ \cancel{mL} \right )+ 5.03 \; g \; KOH=104.72 \; g \nonumber\] The temperature change is (34.7°C − 23.0°C) = +11.7°C. Because the solution is not very concentrated (approximately 0.9 M), we assume that the specific heat of the solution is the same as that of water. The heat flow that accompanies dissolution is thus \[ \begin{align*} q_{calorimater} &= mc_s \Delta T \nonumber \\[4pt] &= \left ( 104.72 \; \cancel{g} \right ) \left ( \dfrac{4.184 \; J}{\cancel{g}\cdot \bcancel{^{o}C}} \right )\left ( 11.7 \; \bcancel{^{o}C} \right ) \nonumber \\[4pt] &= 5130 \; J \\[4pt] &=5.13 \; kJ \end{align*} \] The temperature of the solution increased because heat was absorbed by the solution ( > 0). Where did this heat come from? It was released by KOH dissolving in water. From Equation \ref{12.3.1}, we see that \[ΔH_{rxn} = −q_{calorimeter} = −5.13\, kJ \nonumber\] This experiment tells us that dissolving 5.03 g of \(\ce{KOH}\) in water is accompanied by the of 5.13 kJ of energy. Because the temperature of the solution increased, the dissolution of KOH in water must be exothermic. The last step is to use the molar mass of \(\ce{KOH}\) to calculate \(ΔH_{soln}\) - the heat associated when dissolving 1 mol of \(\ce{KOH}\): \[ \begin{align*} \Delta H_{soln} &= \left ( \dfrac{5.13 \; kJ}{5.03 \; \cancel{g}} \right )\left ( \dfrac{56.11 \; \cancel{g}}{1 \; mol} \right ) \nonumber \\[4pt] &= -57.2 \; kJ/mol\end{align*} \] A coffee-cup calorimeter contains 50.0 mL of distilled water at 22.7°C. Solid ammonium bromide (3.14 g) is added and the solution is stirred, giving a final temperature of 20.3°C. Using the same assumptions as in Example \(\Page {7}\), find \(ΔH_{soln}\) for NH Br (in kilojoules per mole). 16.6 kJ/mol Constant-pressure calorimeters are not very well suited for studying reactions in which one or more of the reactants is a gas, such as a combustion reaction. The enthalpy changes that accompany combustion reactions are therefore measured using a constant-volume calorimeter, such as the shown schematically in Figure \(\Page {4}\)). The reactant is placed in a steel cup inside a steel vessel with a fixed volume (the “bomb”). The bomb is then sealed, filled with excess oxygen gas, and placed inside an insulated container that holds a known amount of water. Because combustion reactions are exothermic, the temperature of the bath and the calorimeter increases during combustion. If the heat capacity of the bomb and the mass of water are known, the heat released can be calculated. Because the volume of the system (the inside of the bomb) is fixed, the combustion reaction occurs under conditions in which the volume, but not the pressure, is constant. The heat released by a reaction carried out at constant volume is identical to the change in (\(ΔU\)) rather than the enthalpy change (Δ ); ΔU is related to Δ by an expression that depends on the change in the number of moles of gas during the reaction. The difference between the heat flow measured at constant volume and the enthalpy change is usually quite small, however (on the order of a few percent). Assuming that \(ΔU < ΔH\), the relationship between the measured temperature change and Δ is given in Equation \ref{12.3.18}, where is the total heat capacity of the steel bomb and the water surrounding it \[ \Delta H_{comb} < q_{comb} = q_{calorimater} = C_{bomb} \Delta T \label{12.3.18}\] To measure the heat capacity of the calorimeter, we first burn a carefully weighed mass of a standard compound whose enthalpy of combustion is accurately known. Benzoic acid (C H CO H) is often used for this purpose because it is a crystalline solid that can be obtained in high purity. The combustion of benzoic acid in a bomb calorimeter releases 26.38 kJ of heat per gram (i.e., its Δ = −26.38 kJ/g). This value and the measured increase in temperature of the calorimeter can be used to determine . The use of a bomb calorimeter to measure the Δ of a substance is illustrated in Example \(\Page {8}\). The combustion of 0.579 g of benzoic acid in a bomb calorimeter caused a 2.08°C increase in the temperature of the calorimeter. The chamber was then emptied and recharged with 1.732 g of glucose and excess oxygen. Ignition of the glucose resulted in a temperature increase of 3.64°C. What is the Δ of glucose? mass and Δ for combustion of standard and sample Δ of glucose The first step is to use Equation \ref{12.3.1} and the information obtained from the combustion of benzoic acid to calculate . We are given Δ , and we can calculate \(q_{comb}\) from the mass of benzoic acid: \[ q_{comb} = \left ( 0.579 \; \cancel{g} \right )\left ( -26.38 \; kJ/\cancel{g} \right ) = - 15.3 \; kJ \nonumber \] From Equation \ref{12.3.1}, \[ -C_{bomb} = \dfrac{q_{comb}}{\Delta T} = \dfrac{-15.3 \; kJ}{2.08 \; ^{o}C} =- 7.34 \; kJ/^{o}C \nonumber \] According to the strategy, we can now use the heat capacity of the bomb to calculate the amount of heat released during the combustion of glucose: \[ q_{comb}=-C_{bomb}\Delta T = \left ( -7.34 \; kJ/^{o}C \right )\left ( 3.64 \; ^{o}C \right )=- 26.7 \; kJ \nonumber\] Because the combustion of 1.732 g of glucose released 26.7 kJ of energy, the Δ of glucose is \[ \Delta H_{comb}=\left ( \dfrac{-26.7 \; kJ}{1.732 \; \cancel{g}} \right )\left ( \dfrac{180.16 \; \cancel{g}}{mol} \right )=-2780 \; kJ/mol =2.78 \times 10^{3} \; kJ/mol \nonumber\] This result is in good agreement (< 1% error) with the value of \(ΔH_{comb} = −2803\, kJ/mol\) that calculated using enthalpies of formation. When 2.123 g of benzoic acid is ignited in a bomb calorimeter, a temperature increase of 4.75°C is observed. When 1.932 g of methylhydrazine (CH NHNH ) is ignited in the same calorimeter, the temperature increase is 4.64°C. Calculate the Δ of methylhydrazine, the fuel used in the maneuvering jets of the US space shuttle. −1.30 × 10 kJ/mol Calorimetry measures enthalpy changes during chemical processes, where the magnitude of the temperature change depends on the amount of heat released or absorbed and on the heat capacity of the system. is the set of techniques used to measure enthalpy changes during chemical processes. It uses devices called , which measure the change in temperature when a chemical reaction is carried out. The magnitude of the temperature change depends on the amount of heat released or absorbed and on the heat capacity of the system. The of an object is the amount of energy needed to raise its temperature by 1°C; its units are joules per degree Celsius. The (\(c_s\)) of a substance is the amount of energy needed to raise the temperature of 1 g of the substance by 1°C, and the (\(c_p\)) is the amount of energy needed to raise the temperature of 1 mol of a substance by 1°C. Liquid water has one of the highest specific heats known. Heat flow measurements can be made with either a , which gives \(ΔH\) values directly, or a , which operates at constant volume and is particularly useful for measuring enthalpies of combustion. Thermal energy itself cannot be measured easily, but the temperature change caused by the flow of thermal energy between objects or substances can be measured. describes a set of techniques employed to measure enthalpy changes in chemical processes using devices called . To have any meaning, the quantity that is actually measured in a calorimetric experiment, the change in the temperature of the device, must be related to the heat evolved or consumed in a chemical reaction. We begin this section by explaining how the flow of thermal energy affects the temperature of an object.   ).

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An organic compound commonly is said to be "reduced" if reaction leads to an increase in its hydrogen content or a decrease in its oxygen content. The compound would be "oxidized" if the reverse change took place: This is a very unsatisfactory definition because many oxidation-reduction or reactions do not involve changes in hydrogen or oxygen content, as the following example illustrates: Redox reactions are better defined in terms of the concept of electron transfer. Thus . This simple definition can be used to identify oxidation or reduction processes at carbon in terms of a scale of oxidation states for carbon based on the electronegativities of the atoms attached to carbon. The idea is to find out whether in a given reaction carbon becomes more, or less, electron-rich. We will use the following somewhat arbitrary rules: To illustrate, the oxidation state of carbon in four representative examples is determined as follows: Using this approach, we can construct a carbon oxidation scale, as in Table 11-1. Any reaction that increases the degree of oxidation of carbon corresponds to a loss of electrons (oxidation), and a reaction that decreases the oxidation level corresponds to a gain of electrons (reduction). Two examples follow: We recommend this scheme of oxidation states only as an aid to identify and balance redox reactions. Also, the terminology "redox" should not be confused with the mechanism of a reaction, as there is no connection between them. A moment's reflection also will show that virtually all reactions theoretically can be regarded as redox reactions, because in almost every reaction the reacting atoms experience some change in their electronic environments. Traditionally, however, reactions are described as redox reactions of carbon only when there is a change in the oxidation state of the carbon atoms involved. An indication of just how arbitrary this is can be seen by the example of addition of water to ethene. This reaction usually is not regarded as an oxidation-reduction reaction because there is no change in the oxidation state of the ethene carbons, despite the fact that, by our rules, one carbon is oxidized and the other reduced: \[\overset{-2}{\ce{CH_2}} = \overset{-2}{\ce{CH_2}} + \ce{H_2O} \rightarrow \overset{-3}{\ce{CH_3}} \overset{-1}{\ce{CH_2}} \ce{OH}\] Apart from indicating when oxidation or reduction occurs, the oxidation scales is useful in balancing redox equations. For example, consider the following oxidation of ethenylbenzene (styrene) with potassium permanganate: To determine how many moles of permanganate ion are required to oxidize one mole of styrene in this reaction, first determine the net change in oxidation state of the reacting carbons: Second, determine the net change in oxidation state of manganese for \(\ce{MnO_4^-} \rightarrow \ce{MnO_2}\): Therefore we need three moles of styrene for every eight moles of permanganate: To get the overall atom and electrical balance for Equation 11-1, the requisite amounts of \(\ce{H_2O}\) must be added, but the 3:8 ratio will remain unchanged: Because \(\ce{KOH}\) reacts in a nonoxidative way with carboxylic acids to form carboxylate salts \(\left( \ce{RCO_2H} + \ce{KOH} \rightarrow \ce{RCO_2K} + \ce{H_2O} \right), the final equation is and (1977)

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This page explains how to use an infra-red spectrum to identify the presence of a few simple bonds in organic compounds. Ethanoic acid has the structure: You will see that it contains the following bonds: The carbon-carbon bond has absorptions which occur over a wide range of wavenumbers in the fingerprint region - that makes it very difficult to pick out on an infra-red spectrum. The carbon-oxygen single bond also has an absorbtion in the fingerprint region, varying between 1000 and 1300 cm depending on the molecule it is in. You have to be very wary about picking out a particular trough as being due to a C-O bond. The other bonds in ethanoic acid have easily recognized absorptions outside the fingerprint region. The infrared spectrum for ethanoic acid looks like this: The possible absorption due to the C-O single bond is queried because it lies in the fingerprint region. You couldn't be sure that this trough wasn't caused by something else. The O-H bond in an alcohol absorbs at a higher wavenumber than it does in an acid - somewhere between 3230 - 3550 cm . In fact this absorption would be at a higher number still if the alcohol isn't hydrogen bonded - for example, in the gas state. All the infra-red spectra on this page are from liquids - so that possibility will never apply. Notice the absorption due to the C-H bonds just under 3000 cm , and also the troughs between 1000 and 1100 cm - one of which will be due to the C-O bond. This time the O-H absorption is missing completely. Don't confuse it with the C-H trough fractionally less than 3000 cm . The presence of the C=O double bond is seen at about 1740 cm . The C-O single bond is the absorption at about 1240 cm . Whether or not you could pick that out would depend on the detail given by the table of data which you get in your exam, because C-O single bonds vary anywhere between 1000 and 1300 cm depending on what sort of compound they are in. Some tables of data fine it down, so that they will tell you that an absorption from 1230 - 1250 is the C-O bond in an ethanoate. You will find that this is very similar to the infra-red spectrum for ethyl ethanoate, an ester. Again, there is no trough due to the O-H bond, and again there is a marked absorption at about 1700 cm due to the C=O. Confusingly, there are also absorptions which look as if they might be due to C-O single bonds - which, of course, aren't present in propanone. This reinforces the care you have to take in trying to identify any absorptions in the fingerprint region. will have similar infra-red spectra to ketones. This is interesting because it contains two different sorts of O-H bond - the one in the acid and the simple "alcohol" type in the chain attached to the -COOH group. The O-H bond in the acid group absorbs between 2500 and 3300, the one in the chain between 3230 and 3550 cm . Taken together, that gives this immense trough covering the whole range from 2500 to 3550 cm . Lost in that trough as well will be absorptions due to the C-H bonds. Notice also the presence of the strong C=O absorption at about 1730 cm . Primary amines contain the -NH group, and so have N-H bonds. These absorb somewhere between 3100 and 3500 cm . That double trough (typical of primary amines) can be seen clearly on the spectrum to the left of the C-H absorptions. Jim Clark ( )

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This page explores how you write electronic structures for atoms using s, p, and d notation. It assumes that you know about simple atomic orbitals - at least as far as the way they are named, and their relative energies. Hydrogen has its only electron in the 1s orbital - 1s , and at helium the first level is completely full - 1s . Now we need to start filling the second level, and hence start the second period. Lithium's electron goes into the 2s orbital because that has a lower energy than the 2p orbitals. Lithium has an electronic structure of 1s 2s . Beryllium adds a second electron to this same level - 1s 2s . Now the 2p levels start to fill. These levels all have the same energy, and so the electrons go in singly at first. The next electrons to go in will have to pair up with those already there. You can see that it is going to get progressively tedious to write the full electronic structures of atoms as the number of electrons increases. There are two ways around this, and you must be familiar with both.vShortcut 1: All the various p electrons can be lumped together. For example, fluorine could be written as 1s 2s 2p , and neon as 1s 2s 2p . This is what is normally done if the electrons are in an inner layer. If the electrons are in the bonding level (those on the outside of the atom), they are sometimes written in shorthand, sometimes in full. Don't worry about this. Be prepared to meet either version, but if you are asked for the electronic structure of something in an exam, write it out in full showing all the p , p and p orbitals in the outer level separately. For example, although we have not yet met the electronic structure of chlorine, you could write it as 1s 2s 2p 3s 3p 3p 3p . Notice that the 2p electrons are all lumped together whereas the 3p ones are shown in full. The logic is that the 3p electrons will be involved in bonding because they are on the outside of the atom, whereas the 2p electrons are buried deep in the atom and are not really of any interest. Shortcut 2: You can lump all the inner electrons together using, for example, the symbol [Ne]. In this context, [Ne] means the electronic structure of neon - in other words: 1s 2s 2p 2p 2p You wouldn't do this with helium because it takes longer to write [He] than it does 1s . On this basis the structure of chlorine would be written [Ne]3s 3p 3p 3p . At neon, all the second level orbitals are full, and so after this we have to start the third period with sodium. The pattern of filling is now exactly the same as in the previous period, except that everything is now happening at the 3-level. At this point the 3-level orbitals are not all full - the 3d levels have not been used yet. But if you refer back to the energies of the orbitals, you will see that the next lowest energy orbital is the 4s - so that fills next. There is strong evidence for this in the similarities in the chemistry of elements like sodium (1s 2s 2p 3s ) and potassium (1s 2s 2p 3s 3p 4s ). The outer electron governs their properties and that electron is in the same sort of orbital in both of the elements. That wouldn't be true if the outer electron in potassium was 3d . The elements in Group 1 of the Periodic Table all have an outer electronic structure of ns (where n is a number between 2 and 7). All Group 2 elements have an outer electronic structure of ns . Elements in Groups 1 and 2 are described as s-block elements. Elements from Group 3 (the boron group) across to the noble gases all have their outer electrons in p orbitals. These are then described as p-block elements. We are working out the electronic structures of the atoms using the ("building up") Principle. So far we have got to calcium with a structure of 1s 2s 2p 3s 3p 4s . The 4s level is now full, and the structures of the next atoms show electrons gradually filling up the 3d level. These are known as d-block elements. Once the 3d orbitals have filled up, the next electrons go into the 4p orbitals as you would expect. d-block elements are elements in which the last electron to be added to the atom using the Aufbau Principle is in a d orbital. The first series of these contains the elements from scandium to zinc, which are called or transition metals. Technically, the terms "transition element" and "d-block element" do not quite have the same meaning, but it doesn't matter in the present context. d electrons are almost always described as, for example, d or d - and not written as separate orbitals. Remember that there are five d orbitals, and that the electrons will inhabit them singly as far as possible. Up to 5 electrons will occupy orbitals on their own. After that they will have to pair up. d means d means Notice in what follows that all the 3-level orbitals are written together - with the 4s electrons written at the end of the electronic structure. However, chromium breaks the sequence. In chromium, the electrons in the 3d and 4s orbitals rearrange so that there is one electron in each orbital. It would be convenient if the sequence was tidy - but it is not! And at zinc the process of filling the d orbitals is complete. The next orbitals to be used are the 4p, and these fill in exactly the same way as the 2p or 3p. We are back now with the p-block elements from gallium to krypton. Bromine, for example, is 1s 2s 2p 3s 3p 3d 4s 4p 4p 4p . First work out the number of outer electrons. This is quite likely all you will be asked to do anyway. The number of outer electrons is the same as the group number. (The noble gases are a bit of a problem here, because they are normally called group 0 rather then group 8. Helium has 2 outer electrons; the rest have 8.) All elements in group 3, for example, have 3 electrons in their outer level. Fit these electrons into s and p orbitals as necessary. Which level orbitals? Count the periods in the Periodic Table (not forgetting the one with H and He in it). Iodine is in group 7 and so has 7 outer electrons. It is in the fifth period and so its electrons will be in 5s and 5p orbitals. Iodine has the outer structure 5s 5p 5p 5p . What about the inner electrons if you need to work them out as well? The 1, 2 and 3 levels will all be full, and so will the 4s, 4p and 4d. That gives the full structure: 1s 2s 2p 3s 3p 3d 4s 4p 4d 5s 5p 5p 5p . Once finished, count all the electrons to make sure that they come to the same as the atomic number. Don't forget to make this check - it is easy to miss an orbital out when it gets this complicated. Barium is in group 2 and so has 2 outer electrons. It is in the sixth period. Barium has the outer structure 6s . Including all the inner levels: 1s 2s 2p 3s 3p 3d 4s 4p 4d 5s 5p 6s . It would be easy to include 5d as well by mistake, but the d level always fills after the next s level - so 5d fills after 6s just as 3d fills after 4s. As long as you counted the number of electrons you could easily spot this mistake because you would have 10 too many. Steps in writing the electronic structure of an element from hydrogen to krypton:

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Glucose is by far the most abundant monosaccharide; it occurs free in fruits, plants, honey, in the blood of animals, and combined in many glycosides, disaccharides, and polysaccharides. The structure and properties of glucose will be considered in greater detail than those of the other monosaccharides, not only because of its importance, but because much of what can be said about glucose also can be said about the other monosaccharides. Glucose is an aldohexose, which means that it is a six-carbon sugar with a terminal aldehyde group, shown by \(1\): The carbons labeled with an asterisk in \(1\) are chiral; thus there are \(2^4\), or sixteen, possible configurational isomers. All are known - some occur naturally and the others have been synthesized (see Table 20-1). The problem of identifying glucose as a particular one of the sixteen possibilities was solved by Emil Fischer during the latter part of the nineteenth century, for which he was awarded the Nobel Prize in chemistry in 1902. The configurations he deduced for each of the chiral carbons, \(\ce{C_2}\)-\(\ce{C_5}\), are shown in the projection formula \(2\).\(^2\) Although Fischer was aware that natural glucose could be the enantiomer of Structure \(2\), his original guess at the configuration proved to be correct and the configuration at \(\ce{C_5}\) is the same as the configuration of the simplest "sugar" \(D\)-\(\left( + \right)\)-glyceraldehyde, \(3\), (see ). Therefore natural glucose is specifically \(D\)-glucose: The complete logic of Fischer's procedures for determination of the configuration of glucose is too involved to be explained here in detail. What you will be unable to fully appreciate is the great difficulties of working with carbohydrates - that is, their considerable solubility in water, instability to strong oxidizing agents and acidic or basic reagents, reluctance to crystallize, and their tendency to decompose rather than give sharp melting points. Fortunately for Fischer, many different pentoses and hexoses already were available from the efforts of earlier investigators, and the principles of optical isomerism were well understood as the result of the work of van't Hoff. Two of the key ideas used by Fischer can be illustrated best with aldotetroses because they have only two chiral carbons and far fewer possible structures to consider. Writing the four possibilities as the aldehyde rather than hemiacetal structures, we have \(4\)-\(7\). Of these, \(4\) and \(5\) constitute a pair of enantiomers, as do \(6\) and \(7\). These pairs can be identified by careful oxidation of the terminal groups to give the corresponding tartaric (2,3-dihydroxybutanedioic) acids. Oxidation of both \(4\) and \(5\) gives -tartaric acid, whereas oxidation of \(6\) and \(7\) gives, respectively, \(\left( + \right)\) and \(\left( - \right)\) tartaric acids: It should be clear from this that the configurations of \(6\) and \(7\) are established by being related to the respective tartaric acids. However, we have no way of telling which of the tetroses represented by \(4\) and \(5\) is \(D\) and which is \(L\), because, on oxidation, they give the same achiral tartaric acid. What we need to do is relate one or the other of the chiral carbons of these tetroses to the corresponding carbon of either \(6\) or \(7\). One way that this can be done is by the , whereby the chain length is reduced by one carbon by removing the aldehyde carbon: As applied to \(4\), \(5\), \(6\), and \(7\), the Wohl degradation forms enantiomers of glyceraldehyde: Here we see that \(4\) and \(6\) give the same enantiomer, \(D\)-glyceraldehyde, and therefore have the same configuration of their highest-numbered asymmetric carbon. In contrast, \(5\) and \(7\) give \(L\)-glyceraldehyde and thus must be similarly related. By this kind of procedure, the configurations of \(4\) to \(7\) can be established unequivocally, although, as you might imagine, it is far easier to do on paper than in the laboratory. Knowing the configurations of the tetroses aids in establishing the configurations of the pentoses. Thus, \(4\), by ,\(^3\) can be converted to a mixture of two aldopentoses, \(8\) and \(9\), by a carbon at the aldehyde end of the molecule. The configurations of the two carbons at the lower end of the starting materials remain unchanged, but two diastereomeric aldopentoses are formed in the syntheses because a new chiral center is created: Products \(8\) and \(9\) present a new configurational problem, but a less difficult one than before, because the configurations of two of the three chiral centers already are known. Controlled oxidation of \(8\) and \(9\) will give diastereomeric 2,3,4-trihydroxypentanedioic acids, \(10\) and \(11\), respectively: Of these, \(11\) is ( ), whereas \(10\) is . Therefore, by simply determining which oxidation product is optically active, and hence chiral, we can assign the configurations of \(8\) and \(9\). Direct comparison of these synthetic aldopentoses with the naturally occurring compounds then could be used as proof of the structure of natural aldopentoses. By this reasoning \(8\) turns out to be \(D\)-arabinose and \(9\) is \(D\)-ribose. Some of the key reactions in carbohydrate chemistry involve oxidation of aldoses to carboxylic acids. There is a simple nomenclature system for these acids. In abbreviated notation, the products of oxidation at \(\ce{C_1}\), \(\ce{C_6}\), or both are called: The carboxylic acids derived from glucose are therefore gluconic acid, glucuronic acid, and glucaric acid. Although glucose has some of the properties expected of an aldehyde, it lacks others. For example, it forms certain carbonyl derivatives (e.g., oxime and cyanohydrin), and can be reduced to the hexahydroxyhexane (sorbitol), and oxidized with bromine to gluconic acid (a monocarboxylic acid). (With nitric acid, oxidation proceeds further to give the dicarboxylic acid, \(D\)-glucaric acid.) Glucose also will reduce Fehling's solution \(\left[ \ce{Cu} (II) \rightarrow \ce{Cu} (I) \right]\) and Tollen's reagent \(\left[ \ce{Ag} (I) \rightarrow \ce{Ag} (0) \right]\) and, for this reason, is classified as a .\(^4\) However, it fails to give a hydrogen sulfite addition compound and, although it will react with amines \(\left( \ce{RNH_2} \right)\), the products are not the expected Schiff's bases of the type \(\ce{-C=NR}\). Furthermore, glucose forms two monomethyl derivatives (called methyl \(\alpha\)-\(D\)-glucoside and methyl \(\beta\)-\(D\)-glucoside) under conditions that normally convert an aldehyde to a dimethyl acetal: All of these reactions can be explained on the bases that the carbonyl group is not free in glucose but is combined with one of the hydroxyl groups, which turns out to be the one at \(\ce{C_5}\), to form a hemiacetal, \(12\) or \(13\). Why are two hemiacetals possible? Because a new asymmetric center is created at \(\ce{C_1}\) by hemiacetal formation, and this leads to diastereomeric forms of \(D\)-glucose called \(\alpha\)-\(D\)-glucose and \(\beta\)-\(D\)-glucose. In general, carbohydrate stereoisomers that differ only in configuration at the hemiacetal carbon are called : Although formulas \(12\) and \(13\) show the at each of the chiral centers, they do not provide the crucial information for understanding the properties of glucose with respect to the . Conversion of a projection formula such as \(12\) or \(13\) to a three-dimensional representation is not at all a trivial task. We have indicated the procedure for doing in before ( and ). The result of these procedures applied to \(12\) and \(13\) are the so-called Haworth projection formulas, \(14\) and \(15\), and the sawhorse conformations, \(16\) and \(17\): You should be able to satisfy yourself that the configuration at \(\ce{C_5}\) is the same in both Fischer and Haworth representations. This amounts to asking if \(18\) and \(19\) represent the same configurations: They do, but if you have trouble visualizing this, it will be very helpful to use a ball-and-stick model to see that \(18\) and \(19\) are different representations of the same configuration. If you do not have models, remember that if transposition of any three groups converts one projection into the other, the formulas are identical. Thus \(18\) and \(19\) have the same configuration because \(18\) becomes \(19\) by transposition of \(\ce{C_4}\) with \(\ce{CH_2OH}\), then \(\ce{C_4}\) with \(\ce{H}\). X-ray studies of crystalline \(\alpha\)- and \(\beta\)-\(D\)-glucose show that these molecules have their atoms arranged in space as correspond to \(16\) and \(17\). This is what we would expect from our studies of cyclohexane conformations ( to ), because for the \(\beta\) form, of the substituents on the oxacyclohexane ring are in equatorial positions, and for the \(\alpha\) form, all except the hydroxyl at the carbon \(\left( \ce{C_1} \right)\) are equatorial. Although the crystalline forms of \(\alpha\)- and \(\beta\)-\(D\)-glucose are quite stable, in solution each form slowly changes into an equilibrium mixture of both. The process can be observed as a decrease in the optical rotation of the \(\alpha\) anomer \(\left( +112^\text{o} \right)\) or an increase for the \(\beta\) anomer \(\left( +18.7^\text{o} \right)\) to the equilibrium value of \(52.5^\text{o}\). The phenomenon is known as and commonly is observed for reducing sugars. Both acids and bases catalyze mutarotation; the mechanism, Equation 20-1, is virtually the same as described for acid- and base-catalyzed hemiacetal and hemiketal equilibria of aldehydes and ketones (see ): At equilibrium, about \(64\%\) of the \(\beta\) anomer and \(36\%\) of the \(\alpha\) anomer are present. The amount of the free aldehyde form at equilibrium is very small (about 0.024 mole percent in neutral solution). Preponderance of the \(\beta\) anomer is attributed to the fact that glucose exists in solution in the chair conformation with the large \(\ce{-CH_2OH}\) group equatorial. In this conformation, the hydroxyl substituent at \(\ce{C_1}\) is equatorial in the \(\beta\) anomer and axial in the \(\alpha\) anomer; hence the \(\beta\) anomer is slightly more stable. When glucose is in the aldehyde form, the hydroxyl at \(\ce{C_4}\) also could add to the aldehyde carbonyl to produce a hemiacetal with a five-membered ring. This does not occur to a significant degree with glucose because the hemiacetal with the six-membered ring and many equatorial groups is more stable. With other sugars, mixtures of five- and six-membered hemiacetal or ketal rings and their respective anomers are produced in water solution. Carbon-13 nmr spectra ( ) provide an especially powerful tool for studying the anomeric forms of sugars. For example, with glucose the resonances of \(\ce{C_1}\), \(\ce{C_3}\), and \(\ce{C_5}\) of the \(\alpha\) anomer are seen in Figure 20-3 to be shifted substantially upfield relative to those of the \(\beta\) anomer because of the axial substituent effect ( ). In the presence of dilute base, \(D\)-glucose rearranges to a mixture containing the anomers of \(D\)-glucose \(\left( \sim 64\% \right)\), \(D\)-fructose \(\left( \sim 31\% \right)\), and \(D\)-mannose \(\left( 3\% \right)\). This interconversion undoubtedly involves enolization of the hexoses by way of a common enediol intermediate according to Equation 20-2: The rearrangement is of interest because the corresponding enzymatic interconversion of aldoses and ketoses is an important part of the biosynthetic, photosynthetic, and metabolic pathways, as we shall see in . Although the biochemical rearrangement also may proceed by way of enediol intermediates, it is highly stereospecific and yields only of two possible stereoisomeric aldoses. For example, glucose, but not mannose, can be enzymatically interconverted with fructose as the 6-phosphate ester derivative: \(^3\)The steps of the Kiliani-Fischer synthesis are: \(^4\)In general, reducing sugars are hemiacetals or hemiketals and the nonreducing sugars are acetals or ketals. The difference is between the presence of the structural elements \(\ce{C-O-C-O-H}\) and \(\ce{C-O-C-O-C}\). and (1977)

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Van der Waals forces' is a general term used to define the attraction of intermolecular forces between molecules. There are two kinds of Van der Waals forces: weak and stronger . The chance that an electron of an atom is in a certain area in the electron cloud at a specific time is called the "electron charge density." Since there is no way of knowing exactly where the electron is located and since they do not all stay in the same area 100 percent of the time, if the electrons all go to the same area at once, a dipole is formed momentarily. Even if a molecule is nonpolar, this displacement of electrons causes a nonpolar molecule to become polar for a moment. Since the molecule is polar, this means that all the electrons are concentrated at one end and the molecule is partially negatively charged on that end. This negative end makes the surrounding molecules have an instantaneous dipole also, attracting the surrounding molecules' positive ends. This process is known as the Force of attraction. The ability of a molecule to become polar and displace its electrons is known as the molecule's " ." The more electrons a molecule contains, the higher its ability to become polar. Polarizability increases in the periodic table from the top of a group to the bottom and from right to left within periods. This is because the higher the molecular mass, the more electrons an atom has. With more electrons, the outer electrons are easily displaced because the inner electrons shield the nucleus' positive charge from the outer electrons which would normally keep them close to the nucleus. When the molecules become polar, the melting and boiling points are raised because it takes more heat and energy to break these bonds. Therefore, the greater the mass, the more electrons present, and the more electrons present, the higher the melting and boiling points of these substances. London dispersion forces are stronger in those molecules that are not compact, but long chains of elements. This is because it is easier to displace the electrons because the forces of attraction between the electrons and protons in the nucleus are weaker. The more readily displacement of electrons means the molecule is also more “polarizable.” These forces are similar to London Dispersion forces, but they occur in molecules that are permanently polar versus momentarily polar. In this type of intermolecular interaction, a polar molecule such as water or H O attracts the positive end of another polar molecule with its negative end of its dipole. The attraction between these two molecules is the dipole-dipole force. Van der Waals equation is required for special cases, such as non-ideal (real) gases, which is used to calculate an actual value. The equation consist of: \[ \left (P+\frac{n^2a}{V^2} \right ) \left (V-nb \right )=nRT \tag{1}\] The V in the formula refers to the volume of gas, in moles . The intermolecular forces of attraction is incorporated into the equation with the \( \frac{n^2a}{V^2} \) term where is a specific value of a particular gas. P represents the pressure measured, which is expected to be lower than in usual cases. The variable expresses the , which accounts for the volume of gas molecules and is also a value of a particular gas. R is a known constant, 0.08206 L atm mol K , and T stands for temperature. Unlike most equations used for the calculation of real, or ideal, gases, van der Waals equation takes into account, and corrects for, the volume of participating molecules and the intermolecular forces of attraction.

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A monatomic gas has three degrees of freedom per molecule, all of them translational: A polyatomic gas, including diatomic molecules, has other levels that you can 'stuff' energy into. Polyatomic molecules can rotate and vibrate, and if enough energy is available you could also excite the electrons involved in the \(σ\) and \(π\) bonds. A reasonable approximation of the partition function of the molecule would become: \[q_\text{tot}(V,T) = q_\text{trans}q_\text{vib}q_\text{rot}q_\text{elec} \] The partition function of a molecular idea gas is then: \[ Q(N,V,T) = \frac{(q_\text{trans}q_\text{vib}q_\text{rot}q_\text{elec})^N}{N!} \] The total energy of the molecule is then the sum of the translation, vibrational, rotational, and electronic energies: \[E_{tot} = E_\text{trans}E_\text{vib}E_\text{rot}E_\text{elec} \] We will only scratch the surface of the additional degrees of freedom and their partition functions. At room temperature the system is usually in its ground electronic state. This means that the \(q_\text{elec} = 1\). Usually we do not have to worry about these degrees of freedom. If we do, there are usually just a few levels to worry about. This includes their . If there is a single state at a certain energy (\(g=1\)), two states (\(g=2\)), or more states, we must multiply the Boltzmann factor by this degeneracy number. If there are more than one state to worry about, we could follow the same procedure as we did for the translational states:

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To review the general characteristics of phase equilibria, let us consider a familiar system. Suppose that we have a transparent but very strong cylinder, sealed with a frictionless piston, within which we have trapped a quantity of pure liquid water at some high pressure. We can fix the pressure of the liquid water at any value we choose by applying an appropriate force to the piston. Suppose that we hold the temperature constant and force the volume to increase by withdrawing the piston in very small increments. Because pure water is not compressed easily, we find initially that the pressure of the water decreases and does so in very large increments. However, after some small increase in the volume, we find that imposing a further volume increase changes the system’s behavior abruptly. The system undergoes a profound change. What was formerly pure liquid becomes a mixture of liquid and gas. As we impose still further volume increases, the pressure of the system remains constant, additional liquid passes from the liquid to the gas phase, and we find that we must supply substantial amounts of heat in order to keep the temperature of the system constant. If we continue to force volume increases in this manner, vaporization continues until all of the liquid evaporates. If we impose a decrease in the volume of the two-phase system, we see the process reverse. The pressure of the system remains constant, some of the gas condenses to liquid, and the system gives up heat to the surroundings. For any given temperature, these conversions are precisely balanced at some particular pressure, and these conditions characterize a state of liquid–vapor equilibrium. At any given pressure, the equilibrium temperature is called the of the liquid. The equilibrium pressure and temperature completely specify the state of the system, except for the exact amounts of liquid and gaseous water present. If we begin with this system in a state of liquid–vapor equilibrium, we can increase the amount of vapor by imposing a small volume increase. Conversely, we can decrease the amount of vapor by imposing a very small volume decrease. At the equilibrium temperature and pressure, changing the imposed volume by an arbitrarily small amount (from \(V\) to \(V\pm dV\)) is sufficient to reverse the direction of the change that occurs in the system. . Evidently, there is a close connection between reversible processes and equilibrium states. In this description, the reversible, constant-temperature vaporization of water is driven by arbitrarily small volume changes. The system responds to these imposed volume changes so as to maintain a constant equilibrium vapor pressure at the specified temperature. We say that the reversible process “takes place at constant pressure and temperature.” We can also describe this process as being driven by arbitrarily small changes in the applied pressure: If the applied pressure exceeds the equilibrium vapor pressure by an arbitrarily small increment, \(dP>0\), condensation occurs; if the applied pressure is less than the equilibrium vapor pressure by an arbitrarily small increment, \(dP<0\), vaporization occurs. To describe this tersely, we introduce a figure of speech and say that the reversible process occurs “while the system pressure and the applied pressure are equal.” Literally, of course, there can be no change when these pressures are equal. To cause water to vaporize at a constant temperature and pressure, we must add heat energy to the system. This heat is called the or the , and it must be supplied from some entity in the surroundings. When water vapor condenses, this latent heat must be removed from the system and taken up by the surroundings. (The enthalpy change for vaporizing one mole of a substance is usually denoted \({\Delta }_{vap}H\). It varies with temperature and pressure. Tables usually give experimental values of the equilibrium boiling temperature at a pressure of 1 bar or 1 atm; then they give the enthalpy of vaporization at this temperature and pressure. We discuss the enthalpy function in .) Four conditions are sufficient to exactly specify either the initial or the final state: the number of moles of liquid, the number of moles of gas, the pressure, and the temperature. The change is a conversion of some liquid to gas, or . We can represent this change as a transition from an initial state to a final state where \(n^o_{liquid}\) and \(n^o_{gas}\) are the initial numbers of moles of liquid and gas, respectively, and \(\delta n\) is the incremental number of moles vaporized: \[\left(P,\ T,n^o_{liquid},n^o_{gas}\right)\to \left(P,\ T,n^o_{liquid}-\delta n,n^o_{gas}+\delta n\right)\] The initial pressure and temperature are the same as the final pressure and temperature. Effecting this change requires that a quantity of heat, \(\left({\Delta }_{vap}H\right)\delta n\), be added to the system, without changing the temperature of the system. This introduces another requirement that a reversible process must satisfy. If the reversibly vaporizing water is to take up an arbitrarily small amount of heat, the system must be in contact with surroundings that are hotter than the system. The temperature difference between the system and its surroundings must be arbitrarily small, because we can describe exactly the same process as being driven by contacting the system, at temperature \(T\), with surroundings at temperature \(\hat{T}+\delta \hat{T}\). If we keep the applied pressure constant at the temperature-\(T\) equilibrium vapor pressure, the system volume increases. We can reverse the direction of change by changing the temperature of the surroundings from \(\hat{T}+\delta \hat{T}\) to \(\hat{T}-\delta \hat{T}\). If the process is to satisfy our criterion for reversibility, the difference between these two temperatures must be arbitrarily small. To describe this requirement tersely, we again introduce a figure of speech and say that the reversible process occurs “while the system temperature and the surroundings temperature are equal.” If we repeat the water-in-cylinder experiment with the temperature held constant at a slightly different value, we get similar results. There is again a pressure at which the process of converting liquid to vapor is at equilibrium. At this temperature and pressure, both liquid and gaseous water can be present in the system, and, so long as no heat is added or removed from the system, the amount of each remains constant. When we hold the pressure of the system constant at the equilibrium value and supply a quantity of heat to the system, a quantity of liquid is again converted to gaseous water. (The quantity of heat required to convert one mole of liquid to gaseous water, phase equilibria is slightly different from the quantity required in the previous experiment. This is what we mean when we say that the enthalpy of vaporization varies with temperature.) This experiment can be repeated for many temperatures. So long as the temperature is in the range \(\mathrm{273.16} Below \(273.16\ \mathrm{K}\), an equilibrium system containing only liquid and gaseous water cannot exist. At high pressures, a two-phase equilibrium system contains solid and liquid; at sufficiently low pressures, it contains solid and gas. Above \(647.1\ \mathrm{K}\), the distinction between liquid and gaseous water vanishes. The water exists as a single dense phase. This is the . Above the critical temperature, there is a single fluid phase at any pressure. If we keep the pressure constant and remove heat from a quantity of liquid water, the temperature decreases until we eventually reach a temperature at which the water begins to freeze to ice. At this point, water and ice are in equilibrium. Further removal of heat does not decrease the temperature of the water–ice system; rather, the temperature remains constant and additional water freezes into ice. Only when all of the liquid has frozen does further removal of heat cause a further decrease in the temperature of the system. When we repeat this experiment at a series of temperatures, we find a continuous line of pressure–temperature points that are liquid–ice equilibrium points. As sketched in Figure 4, the liquid–ice equilibrium line intersects the liquid–vapor equilibrium line. At this intersection, liquid water, ice, and water vapor are all in equilibrium with one another. There is only one such point. It is called the of water. The or of water is the temperature at which solid and liquid water are in equilibrium at one atmosphere in the presence of air. The water contains dissolved air. The triple point occurs in a pure-water system; it is the temperature and pressure at which gaseous, liquid, and solid water are in equilibrium. By definition, the triple point temperature is 273.16 K. Experimentally, the pressure at the triple point is 611 Pa. Experimentally, the melting point is 273.15 K. To freeze a liquid, we must remove heat. To fuse (melt) the same amount of the solid at the same temperature and pressure, we must add the same amount of heat. This heat is called the or the . The enthalpy of fusion for one mole of a substance is usually denoted \({\Delta }_{fus}H\). It varies slightly with temperature and pressure. Tables usually give experimental values of the equilibrium melting temperature at a pressure of 1 bar or 1 atm; then they give the enthalpy of fusion at this temperature and pressure. At low pressures and temperatures, ice is in equilibrium with gaseous water. A continuous line of pressure–temperature points represents the conditions under which the system contains only ice and water vapor. As the temperature increases, the ice–vapor equilibrium line ends at the triple point. The conversion of a solid directly into its vapor is called sublimation. To sublime a solid to its vapor requires the addition of heat. This heat is called the or the . The enthalpy of sublimation for one mole of a substance is usually denoted \({\Delta }_{sub}H\). It varies slightly with temperature and pressure.

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Spectrophotometry is a method to measure how much a chemical substance absorbs light by measuring the intensity of light as a beam of light passes through sample solution. The basic principle is that each compound absorbs or transmits light over a certain range of wavelength. This measurement can also be used to measure the amount of a known chemical substance. Spectrophotometry is one of the most useful methods of quantitative analysis in various fields such as chemistry, physics, biochemistry, material and chemical engineering and clinical applications. Every chemical compound absorbs, transmits, or reflects light (electromagnetic radiation) over a certain range of wavelength. Spectrophotometry is a measurement of how much a chemical substance absorbs or transmits. Spectrophotometry is widely used for quantitative analysis in various areas (e.g., chemistry, physics, biology, biochemistry, material and chemical engineering, clinical applications, industrial applications, etc). Any application that deals with chemical substances or materials can use this technique. In biochemistry, for example, it is used to determine enzyme-catalyzed reactions. In clinical applications, it is used to examine blood or tissues for clinical diagnosis. There are also several variations of the spectrophotometry such as atomic absorption spectrophotometry and atomic emission spectrophotometry. A spectrophotometer is an instrument that measures the amount of photons (the intensity of light) absorbed after it passes through sample solution. With the spectrophotometer, the amount of a known chemical substance (concentrations) can also be determined by measuring the intensity of light detected. Depending on the range of wavelength of light source, it can be classified into two different types: In visible spectrophotometry, the absorption or the transmission of a certain substance can be determined by the observed color. For instance, a solution sample that absorbs light over all visible ranges (i.e., transmits none of visible wavelengths) appears black in theory. On the other hand, if all visible wavelengths are transmitted (i.e., absorbs nothing), the solution sample appears white. If a solution sample absorbs red light (~700 nm), it appears green because green is the complementary color of red. Visible spectrophotometers, in practice, use a prism to narrow down a certain range of wavelength (to filter out other wavelengths) so that the particular beam of light is passed through a solution sample. Figure 1 illustrates the basic structure of spectrophotometers. It consists of a light source, a collimator, a monochromator, a wavelength selector, a cuvette for sample solution, a photoelectric detector, and a digital display or a meter. Detailed mechanism is described below. Figure 2 shows a sample spectrophotometer (Model: Spectronic 20D). A spectrophotometer, in general, consists of two devices; a spectrometer and a photometer. A spectrometer is a device that produces, typically disperses and measures light. A photometer indicates the photoelectric detector that measures the intensity of light. You need a spectrometer to produce a variety of wavelengths because different compounds absorb best at different wavelengths. For example, p-nitrophenol (acid form) has the maximum absorbance at approximately 320 nm and p-nitrophenolate (basic form) absorb best at 400nm, as shown in Figure 3. Looking at the graph that measures absorbance and wavelength, an isosbestic point can also be observed. An is the wavelength in which the absorbance of two or more species are the same. The appearance of an isosbestic point in a reaction demonstrates that an intermediate is NOT required to form a product from a reactant. Figure 4 shows an example of an isosbestic point. Referring back to Figure 1 (and Figure 5), the amount of photons that goes through the cuvette and into the detector is dependent on the length of the cuvette and the concentration of the sample. Once you know the intensity of light after it passes through the cuvette, you can relate it to transmittance (T). Transmittance is the fraction of light that passes through the sample. This can be calculated using the equation: \(Transmittance (T) = \dfrac{I_t}{I_o}\) Where I is the light intensity after the beam of light passes through the cuvette and I is the light intensity before the beam of light passes through the cuvette. Transmittance is related to absorption by the expression: \(Absorbance (A) = - log(T) = - log(\dfrac{I_t}{I_o})\) Where absorbance stands for the amount of photons that is absorbed. With the amount of absorbance known from the above equation, you can determine the unknown concentration of the sample by using Beer-Lambert Law. Figure 5 illustrates transmittance of light through a sample. The length \(l\) is used for Beer-Lambert Law described below. (also known as Beer's Law) states that there is a linear relationship between the absorbance and the concentration of a sample. For this reason, Beer's Law can be applied when there is a linear relationship. Beer's Law is written as: \(A = \epsilon{lc}\) where The molar extinction coefficient is given as a constant and varies for each molecule. Since absorbance does not carry any units, the units for \(\epsilon\) must cancel out the units of length and concentration. As a result, \(\epsilon\) has the units: L·mol ·cm . The path length is measured in centimeters. Because a standard spectrometer uses a cuvette that is 1 cm in width, \(l\) is always assumed to equal 1 cm. Since absorption, \(\epsilon\), and path length are known, we can calculate the concentration \(c\) of the sample. has a maximum absorbance of 275 nm. \(\epsilon_{275} = 8400 M^{-1} cm^{-1}\) and the path length is 1 cm. Using a spectrophotometer, you find the that \(A_{275}= 0.70\). What is the concentration of guanosine? To solve this problem, you must use Beer's Law. \[A = \epsilon lc \nonumber \] 0.70 = (8400 M cm )(1 cm)(\(c\)) Next, divide both side by [(8400 M cm )(1 cm)] \(c\) = 8.33x10 mol/L There is a substance in a solution (4 g/liter). The length of cuvette is 2 cm and only 50% of the certain light beam is transmitted. What is the absorption coefficient? Using Beer-Lambert Law, we can compute the absorption coefficient. Thus, \(- \log \left(\dfrac{I_t}{I_o} \right) = - \log(\dfrac{0.5}{1.0}) = A = {8} \epsilon\) Then we obtain that \(\epsilon\) = 0.0376 In example 2 above, how much is the beam of light is transmitted when 8 g/liter ? Since we know \(\epsilon\), we can calculate the transmission using Beer-Lambert Law. Thus, \(\log(1) - \log(I_t) = 0 - \log(I_t)\) = 0.0376 x 8 x 2 = 0.6016 \(\log(I_t)\) = -0.6016 Therefore, \(I_t\) = 0.2503 = 25% In example 2 above, what is the molar absorption coefficient if the molecular weight is 100? It can simply obtained by multiplying the absorption coefficient by the molecular weight. Thus, \(\epsilon\) = 0.0376 x 100 = 3.76 L·mol ·cm The absorption coefficient of a glycogen-iodine complex is 0.20 at light of 450 nm. What is the concentration when the transmission is 40 % in a cuvette of 2 cm? It can also be solved using Beer-Lambert Law. Therefore, \[- \log(I_t) = - \log(0.4) = 0.20 \times c \times 2 \nonumber \] Then \(c\) = 0.9948

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Cadmium sulfide (CdS) exists in two natural forms: greenockite and hawleyite, which differ in their crystal structure. Greenockite forms hexagonal crystals with the wurtzite structure, hawleyite has the sphalerite (zinc blende) structure. Cadmium sulfide is a direct band gap semiconductor with E = 2.42 eV at room temperature. CdS is used in optoelectronics (photosensitive and photovoltaic devices). One simple use is as a photoresistor whose electrical resistance changes with incident light levels. Mixed with zinc sulfide, cadmium sulfide acts as a phosphor with long afterglow. Cadmium sulfide was used as a pigment in paints as far back as 1819. Synthetic cadmium sulfide pigments are valued for their good thermal stability in many polymers, for example in engineering plastics. By adding selenium it is possible to obtain colors ranging from a greenish yellow to red violet.

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Now it is time to turn our attention to the ways in which the three-dimensional structure of organic molecules affects their stability, reactivity, and the ways in which they interact with one another and with solvent molecules. To begin with, we will consider compounds that are composed of \(\mathrm{sp}^{3}\) hybridized (i.e. tetrahedral) carbons which are inherently three-dimensional since each of the four bonds points toward the vertex of a tetrahedron (a three-dimensional as opposed to a planar or two-dimensional organization). It is important to remember and understand this aspect of organic chemistry, since we tend to represent organic compounds in ever more simplified (and abstract) two-dimensional diagrams. The Lewis structure of a molecule is a \(2\mathrm{-D}\) (that is, flat) cartoon that contains a great deal of information (if you know how to look for it). As we move to even more abstracted representations of molecular structures it is easy to forget about the wealth of information encoded within a structural diagram. This includes both how many carbons and hydrogens are present and the three-dimensional relationships between the parts of the molecule. While there are ways to show the three-dimensional nature of \(\mathrm{sp}^{3}\) hybridized carbon (such as through the use of wedge-dash structures à), they can be quite cumbersome when it comes time represent complex molecules and we don’t often use them unless we want to specifically address the three-dimensional arrangement of atoms within the molecule. The other idea that gets lost in a static two-dimensional representation is that the bonds between \(\mathrm{sp}^{3}\) hybridized carbon atoms are sigma (\(\sigma\)) bonds; the two parts of the molecule linked by a \(\sigma\) bond can rotate relative to each other, without disrupting the overlap between the bonding orbitals. At room temperatures the carbons in most σ bonds are rotating freely and rapidly with respect to each other, although we have to portray them in a fixed orientation when we draw them.

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To deal with the complexity of the vibrational motion in polyatomic molecules, we need to utilize the three important concepts listed as the title of this section. By a spatial degree of freedom, we mean an independent direction of motion. A single atom has three spatial degrees of freedom because it can move in three independent or orthogonal directions in space, i.e. along the x, y, or z-axes of a Cartesian coordinate system. Motion in any other direction results from combining velocity components along two or three of these directions. Two atoms have six spatial degrees of freedom because each atom can move in any of these three directions independently. Equivalently, we also can say one atom has three spatial degrees of freedom because we need to specify the values of three coordinates \((x_1, y_1, z_1)\) to locate the atom. Two atoms have six spatial degrees of freedom because we need to specify the values of six coordinates, \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\), to locate two atoms in space. In general, to locate N atoms in space, we need to specify 3N coordinates, so a molecule comprised of N atoms has 3N spatial degrees of freedom. Identify the number of spatial degrees of freedom for the following molecules: \(Cl_2\), \(CO_2\), \(H_2O\), \(CH_4\), \(C_2H_2\), \(C_2H_4\), \(C_6H_6\). The motion of the atomic nuclei in a molecule is not as simple as translating each of the nuclei independently along the x, y, and z axes because the nuclei, which are positively charged, are coupled together by the electrostatic interactions with the electrons, which are negatively charged. The electrons between two nuclei effectively attract them to each other, forming a chemical bond. Consider the case of a diatomic molecule, which has six degrees of freedom. The motion of the atoms is constrained by the bond. If one atom moves, a force will be exerted on the other atom because of the bond. The situation is like two balls coupled together by a spring. There are still six degrees of freedom, but the motion of atom 1 along x, y, and z is not independent of the motion of atom 2 along x, y, and z because the atoms are bound together. It therefore is not very useful to use the six Cartesian coordinates, \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\), to describe the six degrees of freedom because the two atoms are coupled together. We need new coordinates that are independent of each other and yet account for the coupled motion of the two atoms. These new coordinates are called normal coordinates, and the motion described by a normal coordinate is called a normal mode. A normal coordinate is a linear combination of Cartesian displacement coordinates. A linear combination is a sum of terms with constant weighting coefficients multiplying each term. The coefficients can be imaginary or any positive or negative number including +1 and -1. For example, the point or vector r = (1, 2, 3) in three-dimensional space can be written as a linear combination of unit vectors. \[r = 1 \bar {x} + 2 \bar {y} + 3 \bar {z} \label {6-1}\] A Cartesian displacement coordinate gives the displacement in a particular direction of an atom from its equilibrium position. The equilibrium positions of all the atoms are those points where no forces are acting on any of the atoms. Usually the displacements from equilibrium are considered to be small. For illustration, the Cartesian displacement coordinates for HCl are defined in Table \(\Page {1}\), and they are illustrated in Figure \(\Page {1}\). Table 6. Cartesian displacement coordinates for HCl.* \[q_1 = X_{H} - X^e_H\] \[q_2 = y_{H} - y^e_H\] \[q_3 = z_H - z^e_H\] \[q_4 = x_{Cl} - x^e_{Cl}\] \[q_5 = y_{Cl} - y^e_{Cl}\] \[q_6 = z_{Cl} - z^e_{Cl}\] *The superscript e designates the coordinate value at the equilibrium position. Note that the position of one atom can be written as a vector \(r_1\) where \(r_1 = (x_1, y_1, z_1)\), and the positions of two atoms can be written as two vectors \(r_1\) and \(r_2\) or as a generalized vector that contains all six components \(r = (x_1, y_1, z_1, x_2, y_2, z_2)\). Similarly the six Cartesian displacement coordinates can be written as such a generalized vector \(q = (q_1, q_2, q_3, q_4, q_5, q_6)\). For a diatomic molecule it is easy to find the linear combinations of the Cartesian displacement coordinates that form the normal coordinates and describe the normal modes. Just take sums and differences of the Cartesian displacement coordinates. Refer to Table \(\Page {1}\) and Figure \(\Page {4}\). for the definition of the q's. The combination q1 + q4 corresponds to translation of the entire molecule in the x direction; call this normal coordinate Tx. Similarly we can define Ty = q2 + q5 and Tz = q3 + q6 as translations in the y and z directions, respectively. Now we have three normal coordinates that account for three of the degrees of freedom, the three translations of the entire molecule. What do we do about the remaining three degrees of freedom? Here let's use a simple rule for doing creative science: if one thing works, try something similar and examine the result. In this case, if adding quantities works, try subtracting them. Examine the combination q2 - q5. This combination means that H is displaced in one direction and Cl is displaced in the opposite direction. Because of the bond, the two atoms cannot move completely apart, so this small displacement of each atom from equilibrium is the beginning of a rotation about the z-axis. Call this normal coordinate Rz. Similarly define Ry = q3 - q6 to be rotation about the y-axis. We now have found two rotational normal coordinates corresponding to two rotational degrees of freedom. The remaining combination, q1 - q4, corresponds to the atoms moving toward each other along the x-axis. This motion is the beginning of a vibration, i.e. the oscillation of the atoms back and forth along the x-axis about their equilibrium positions, and accounts for the remaining sixth degree of freedom. We use Q for the vibrational normal coordinate. \[Q = q_1 - q_4 \label {6-2}\] To summarize: a normal coordinate is a linear combination of atomic Cartesian displacement coordinates that describes the coupled motion of all the atoms that comprise a molecule. A normal mode is the coupled motion of all the atoms described by a normal coordinate. While diatomic molecules have only one normal vibrational mode and hence one normal vibrational coordinate, polyatomic molecules have many. Draw and label six diagrams, each similar to Figure \(\Page {1}\), to show the 3 translational, 2 rotational and 1 vibrational normal coordinates of a diatomic molecule. Vibrational normal modes have several distinguishing characteristics. Examine the animations for the normal modes of benzene shown i identify and make a list of these characteristics. Use a molecular modeling program to calculate and visualize the normal modes of another molecule. The list of distinguishing characteristics of normal modes that you compiled in Exercise \(\Page {4}\) should include the following four properties. If not, reexamine the animations to confirm that these characteristics are present. For the example of HCl, see Table \(\Page {1}\), the first property, stated mathematically, means \[q_1 = A_1 \sin (\omega t) \text {and} q_4 = A_4 \sin (\omega t) \label {6-3}\] The maximum displacements or amplitudes are given by A1 and A4, and the frequency of oscillation (in radians per second) is ω for both displacement coordinates involved in the normal vibrational mode of HCl. Substitution of Equations (6-3) for the displacement coordinates into the expression determined above for the vibrational normal coordinate, Equation (6-2), yields \[ Q = q_1 - q_4 = A_1 \sin (\omega t) - A_4 \sin (\omega t) \label {6-4}\] This time-dependent expression describes the coupled motions of the hydrogen and chlorine atoms in a vibration of the HCl molecule. In general for a polyatomic molecule, the magnitude of each atom's displacement in a vibrational normal mode may be different, and some can be zero. If an amplitude, the A, for some atom in some direction is zero, it means that atom does not move in that direction in that normal mode. In different normal modes, the displacements of the atoms are different, and the frequencies of the motion generally are different. If two or more vibrational modes have the same vibrational frequency, these modes are called degenerate. You probably noticed in Exercise \(\Page {4}\) that the atoms reached extreme points in their motion at the same time but that they were not all moving in the same direction at the same time. These characteristics are described by the second and third properties from the list above. For the case of HCl, the two atoms always move in exactly opposite directions during the vibration. Mathematically, the negative sign in Equation that we developed for the normal coordinate, Q, accounts for this relationship. This timing with respect to the direction of motion is called the phasing of the atoms. In a normal mode, the atoms move with a constant phase relationship to each other. The phase relationship is represented by a phase angle φ in the argument of the sine function that describes the time oscillation, sin(ωt + φ). The angle is called a phase angle because it shifts the sine function on the time axis. We can illustrate this phase relationship for HCl. Use the trigonometric identity \[- \sin {\theta} = \sin (\theta + 180^0) \label {6-5}\] in Equation \ref{6-4} to obtain \[Q = A_1 \sin (\omega t ) + A_4 \sin (\omega t + 180^0) \label {6-6}\] to see that that the phase angle for this case is 180 . The phase angle \(\varphi\) accounts for the fact that the H atom and the Cl atom reach their maximum displacements in the positive x-direction, +A1 and +A4, at different times. Generally in a normal mode the phase angle \(\varphi\) is 0o or 180o. If \(\varphi\) = 0o for both atoms, the atoms move together, and they are said to be in-phase. For the vibration of a diatomic molecule such as HCl, the phase angle for one atom is \(\varphi\) = 0o, and the phase angle for the other atom is \(\varphi\) = 180o. The atoms therefore move in opposite directions any time, and the atoms are said to be 180o out-of-phase. When \(\varphi\) is 180o, two atoms reach the extreme points in their motion at the same time, but one is in the positive direction and the other is in the negative direction. Phase relationships can be seen by watching a marching band. All the players are executing the same marching motion at the same frequency, but a few may be ahead or behind the rest. You might say, "They are out-of-step." You also could say, "They are out-of-phase." To illustrate the fourth property for HCl, recall that the center of mass for a diatomic molecule is defined as the point where the following equation is satisfied. \[m_H d_H = m_{Cl} d_{Cl} \label {6-7}\] The masses of the atoms are given by \(m_H\) and \(m_Cl\), and \(d_H\) and \(d_{Cl}\) are the distances of these atoms from the center of mass. Find the distances, dH and dCl, of the H and Cl atoms from the center of mass in HCl given that the bond length is 0.13 nm. In general for a diatomic molecule, AB, what determines the ratio dA/dB, and which atom moves the greater distance in the vibration? In general, to satisfy the center of mass condition, a light atom is located further from the center of mass than a heavy atom. To keep the center of mass fixed during a vibration, the amplitude of motion of an atom must depend inversely on its mass. In other words, a light atom is located further from the center of mass and moves a longer distance in a vibration than a heavy atom. Find the ratio of A1 to A4 from Equ s the HCl center of mass stationary during a vibration. Find values for A1 and A4 that satisfy the condition \[A12 + A42= 1.\] For a vibrating HCl molecule, use the four properties of a normal vibrational mode, listed previously, to sketch a graph showing the position of the H atom (plot 1) and the position of the Cl atom (plot 2) as a function of time. Both plots should be on the same scale. Hint: place x on the vertical axis and time on the horizontal axis. In general, a molecule with 3N spatial degrees of freedom has 3 translational normal modes (along each of the three axes), 3 rotational normal modes (around each of the three axes), and 3N-6 (the remaining number) different vibrational normal modes of motion. A linear molecule, as we have just seen, only has two rotational modes, so there are 3N-5 vibrational normal modes. Rotational motion about the internuclear axis in a linear molecule is not one of the 3N spatial degrees of freedom derived from the translation of atoms in three-dimensional space. Rather, such motion corresponds to other degrees of freedom, rotational motion of the electrons and a spinning motion of the nuclei. Indeed, the electronic wavefunction for a linear molecule is characterized by some angular momentum (rotation) about this axis, and nuclei have a property called spin. Identify the number of translational, rotational, and vibrational normal modes for the following molecules: \(Cl_2, CO_2, H_2O, CH_4, C_2H_2, C_2H_4, C_2H_6, C_6H_6\). Using your intuition, draw diagrams similar to the ones in Exercise \(\Page {3}\) to show the normal modes of \(H_2O\) and \(C_2H_4\). It is difficult to identify the normal modes of triatomic and larger molecules by intuition. A mathematical analysis is essential. It is easier to see the normal modes if you use a molecular modeling program like Spartan or Gaussian to generate and display the normal modes. You probably found in trying to complete Exercise \(\Page {7}\) that it is difficult to identify the normal modes and normal coordinates of triatomic and large molecules by intuition. A mathematical analysis is essential. A general analysis based on the Lagrangian formulation of classical mechanics is described separately.

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The second law of thermodynamics can be formulated in many ways, but in one way or another they are all related to the fact that the state function entropy, \(S\), tends to increase over time in isolated systems. For a long time, people have looked at the entire universe as an example of an isolated system and concluded that its entropy must be steadily increasing until \(\delta S_{universe}\) becomes zero. As we will see below, the second law has important consequences for the question of how we can use heat to do useful work. Of late, cosmologists like the late Richard Hawkins have begun to question the assumption that the entropy of the universe is steadily increasing. The key problem is the role that gravity and relativity play in creating black holes. Let's compare two expansions from \(V_1\) to \(V_2\) for an ideal gas, both are isothermal. The first is an irreversible one, where we pull a peg an let the piston move against vacuum: The second one is a reversible isothermal expansion from \(V_1\) to \(V_2\) (and \(P_1\) to \(P_2\)) that we have examined before. In both cases, the is no change in internal energy since \(T\) does not change. During the irreversible expansion, however, there is also no volume work because the piston is expanding against a vacuum and the following integral: \[\int -P_{ext}dV = 0 \nonumber \] integrates to zero. The piston has nothing to perform work against until it slams into the right hand wall. At this point \(V=V_2\) and then \(dV\) becomes zero. This is not true for the reversible isothermal expansion as the external pressure must always equal the internal pressure. No energy and no work means no heat! Clearly the zero heat is irreversible heat (\(q_{irr} = 0\)) and this makes it hard to calculate the entropy of this spontaneous process. But then this process ends in the same final state as the reversible expansion from \(V_1\) to \(V_2\). We know that \(dU\) is still zero, but now \(δw_{rev} = -δq_{rev}\) is nonzero. We calculated its value before: \[q_{rev} = nRT \ln \left(\dfrac{V_2}{V_1} \right) \label{Vacuum} \] The Claussius definition of entropy change can be used to find \(\Delta S\) (under constant temperature). \[\Delta S = \dfrac{q_{rev}}{T} \label{Claussius}\] Substituting Equation \(\ref{Vacuum}\) into Equation \ref{Claussius} results in  \[\Delta S = nR \ln \left(\dfrac{V_2}{V_1} \right) \nonumber \] As \(S\) is a state function this equation also holds for the irreversible expansion against vacuum. Always calculate the entropy difference between two points along a path. For the irreversible expansion into vacuum we see that \[\begin{align*}\Delta S_\text{total} &= \Delta S_\text{sys} + \Delta S_\text{surr} \\[4pt] &= nR\ln \left( \dfrac{V_2}{V_1}\right) + 0 \\[4pt] &= nR\ln \left( \dfrac{V_2}{V_1}\right) \end{align*} \] For the reversible expansion, heat is transferred to the system while the system does work on the surroundings in order to keep the process isothermal: \[\begin{align*}\Delta S_\text{sys} &= nR\ln \left( \dfrac{V_2}{V_1}\right) \end{align*}\] The entropy change for the surrounding is the opposite of the system: \[\begin{align*}\Delta S_\text{surr} &= -nR\ln \left( \dfrac{V_2}{V_1}\right) \end{align*}\] This is because the amount of heat transferred to the system is the same as the heat transferred from the surroundings and this process is reversible so the system and surroundings are at the same temperature (equilibrium). Heat is related to entropy by the following equation: \[ dS = \frac{\delta q}{T} \] Therefore, the total entropy change for the reversible process is zero: \[\begin{align*}\Delta S_\text{total} &= \Delta S_\text{sys} + \Delta S_\text{surr} \\[4pt] &= nR\ln \left( \dfrac{V_2}{V_1}\right)- nR\ln \left( \dfrac{V_2}{V_1}\right) \\[4pt] &=0 \end{align*}\] Consider two ideal gases at same pressure separated by a thin wall that is punctured. Both gases behave as if the other one is not there and again we get a spontaneous process, mixing in this case. If the pressure is the same the number of moles of each gas should be proportional to the original volumes, \(V_A\) and \(V_B\), and the total number of moles to the total volume \(V_{tot}\). For gas A we can write: \[\Delta S_A = n_A R \ln \dfrac{V_{tot}}{V_A} = n_A R \ln \dfrac{n_{tot}}{n_A} \nonumber \] and similarly for gas B we can write: \[\Delta S_B = n_B R \ln \dfrac{V_{tot}}{V_B} = n_B R \ln \dfrac{n_{tot}}{n_B} \nonumber \] The total entropy change is therefore the sum of constituent entropy changes: \[ \Delta S = \Delta S_A + \Delta S_B \nonumber \] and the entropy change total per mole of gas is: \[ \dfrac{\Delta S}{n_{tot}} =R \dfrac{\left[n_B \ln \dfrac{n_{tot}}{n_B}+ n_A \ln \dfrac{n_{tot}}{n_A} \right ]}{n_{tot}} \label{EqTot} \] Equation \(\ref{EqTot}\) can be simplified using mole fractions: \[\chi_A = \dfrac{n_A}{n_{tot}} \nonumber \] and the mathematical relationship of logarithms that: \[\ln \left( \dfrac{x}{y} \right)= - \ln \left( \dfrac{y}{x} \right) \nonumber \] to: \[\Delta \bar{S} = -R \left [\chi_A\ln \chi_A +\chi_B \ln \chi_B \right] \label{Molar Entropy} \] In the case of mixing of more than two gases, Equation \(\ref{Molar Entropy}\) can be expressed as: \[\Delta \bar{S} = -R \sum \chi_i\ln \chi_i \label{Sum Entropy} \] This entropy expressed in Equations \(\ref{Molar Entropy}\) and \(\ref{Sum Entropy}\) is known as the ; its existence is the major reason why there is such a thing as diffusion and mixing when gases, and also solutions (even solid ones), are brought into contact with each other.

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Some of the most important reactions of amines are brought about by nitrous acid \(\left( \ce{HONO} \right)\). The character of the products depends very much on whether the amine is primary, secondary, or tertiary. In fact, nitrous acid is a useful reagent to determine whether a particular amine is primary, secondary, or tertiary. With primary amines nitrous acid results in evolution of nitrogen gas; with secondary amines insoluble yellow liquids or solid \(\ce{N}\)-nitroso compounds, \(\ce{R_2N-N=O}\), separate; tertiary alkanamines dissolve in and react with nitrous acid solutions without evolution of nitrogen, usually to give complex products: Nitrous acid is unstable and always is prepared as needed, usually by mixing a solution of sodium nitrite, \(\ce{NaNO_2}\), with a strong acid at \(0^\text{o}\). These conditions provide a source of \(^\oplus \ce{NO}\), which is transferred readily to the nucleophilic nitrogen of the amine: With this common key step, why do amines react differently with nitrous acid depending on their degree of substitution? The answer can bee seen from the reactions that are most easily possible for the \(\ce{-} \overset{\oplus}{\ce{N}} \ce{-NO}\) intermediate. Clearly, if there is a hydrogen on the positive nitrogen, it can be lost as a proton and a \(\ce{N}\)-nitrosamine formed: With a amine, the reaction stops here, with formation of \(\ce{R_2N-NO}\), and because these substances are weak bases, they are in dilute aqueous acids. They are characteristically yellow or orange-yellow solids or oils. A amine\(\cdot \overset{\oplus}{\ce{N}} \ce{O}\) complex, \(\ce{R_3} \overset{\oplus}{\ce{N}} \ce{-NO}\), cannot lose a proton from nitrogen, but instead may lose a proton from carbon and go on to form complex products. With a amine, the initially formed \(\ce{N}\)-nitrosamine can undergo a proton shift by a sequence analogous to interconversion of a ketone to an enol. The product is called a : Some diazoic acids form salts that are quite stable, but the acids themselves usually decompose rapidly to diazonium ions: Diazonium salts can be regarded as combinations of carbocations \(\ce{R}^\oplus\) with \(\ce{N_2}\) and, because of the considerable stability of nitrogen in the form of \(\ce{N_2}\), we would expect diazonium salts to decompose readily with evolution of nitrogen and formation of carbocations. This expectation is realized, and diazonium salts normally decompose in this manner in water solution. The aliphatic diazonium ions decompose so rapidly that their presence can only be inferred from the fact that the products are typically those of reactions of carbocations: With propanamine, loss of nitrogen from the diazonium ion gives the very poorly stabilized propyl cation, which then undergoes a variety of reactions that are consistent with the carbocation reactions discussed previously (see and ): The isopropyl cation formed by rearrangement undergoes substitution and elimination like the propyl cation. About half of the products arise from isopropyl cations. There is one exceptional reaction of the propyl cation that involves and formation of about \(10\%\) of cyclopropane: Clearly, the plethora of products to be expected, particularly those resulting from rearrangement, prevents the reaction of the simple primary amines with nitrous acid from having any substantial synthetic utility. Unlike primary alkylamines, primary arenamines react with nitrous acid at \(0^\text{o}\) to give diazonium salts that, in most cases, are stable enough to be isolated as crystalline \(\ce{BF_4^-}\) salts. Other salts can be isolated, but some of these, such as benzenediazonium chloride, in the solid state may decompose with explosive violence. The reason for the greater stability of arenediazonium salts compared with alkanediazonium salts appears to be related to the difficulty of forming aryl carbocations ( ). Even the gain in energy associated with having nitrogen as the leaving group is not sufficient to make aryl cations form readily, although solvolysis of arenediazonium ions in water does proceed by an \(S_\text{N}1\) mechanism: This reaction has general utility for replacement of aromatic amino groups by hydroxyl groups. In contrast to the behavior of alkylamines, no rearrangements occur. Generally, diazonium salts from arenamines are much more useful intermediates than diazonium salts from alkanamines. In fact, arenediazonium salts provide the only substances that undergo nucleophilic substitution reactions on the aromatic ring under mild conditions, without the necessity of having activating groups, such as nitro or cyano, in the ortho or para position. The most important reactions of this type include the replacement of the diazonium group by nucleophiles such as \(\ce{Cl}^\ominus\), \(\ce{Br}^\ominus\), \(\ce{I}^\ominus\), \(\ce{CN}^\ominus\), \(\ce{NO_2^-}\), and these reactions lead to the formation of aryl halogen, cyano, and nitro compounds. Most of these reactions require cuprous ions, \(\ce{Cu}\)(I), as catalysts. The method is known as the . The following examples illustrate how a primary arenamine can be converted to a variety of different groups by way of its diazonium salt: Aryl fluorides also may be prepared from arenamines by way of diazonium salts if the procedure is slightly modified. The amine is diazotized with nitrous acid in the usual way; then fluoroboric acid or a fluoroborate salt is added, which usually causes precipitation of a sparingly soluble diazonium fluoroborate. The salt is collected and thoroughly dried, then carefully heated to the decomposition point - the products being an aryl fluoride, nitrogen, and boron trifluoride: \[\ce{C_6H_5} \overset{\oplus}{\ce{N}} \: \overset{\ominus}{\ce{BF_4}} \overset{\text{heat}}{\longrightarrow} \ce{C_6H_5F} + \ce{N_2} + \ce{BF_3}\] This reaction is known as the . An example (which gives a better than usual yield) follows: Later in the chapter we shall see that amines can be prepared by the reduction of nitro compounds, which permits the following sequence of reactions: \[\ce{ArH} \overset{\ce{HNO_3}}{\longrightarrow} \ce{ArNO_2} \overset{\left[ \ce{H} \right]}{\longrightarrow} \ce{ArNH_2} \overset{\ce{HONO}}{\longrightarrow} \ce{ArN_2^+} \underset{-\ce{N_2}}{\overset{\ce{CuX}}{\longrightarrow}} \ce{ArX}\] This sequence is especially useful to introduce groups or produce orientations of substituents that may not be possible by direct substitution. The Sandmeyer group of reactions is an example of the production of (see ): This mechanism is supported by the fact that \(\ce{Cu}\)(II) is important in the formation of \(\ce{C_6H_5X}\). If the concentration of \(\ce{Cu}\)(II) is kept very low so as to slow down conversion of \(\ce{C_6H_5} \cdot\) to \(\ce{C_6H_5^+}\), and a compound with a reactive double bond is present, then products are formed by attack of \(\ce{C_6H_5} \cdot\) on the double bond. This is called the : \[\ce{C_6H_5} \overset{\oplus}{\ce{N_2}} \: \overset{\ominus}{\ce{X}} + \ce{CH_2=CHCN} \overset{\ce{Cu} \left( \text{I} \right)}{\longrightarrow} \ce{C_6H_5CH_2-CHXCN} + \ce{N_2}\] Iodide ion appears to be a good enough reducing agent to form \(\ce{C_6H_5} \cdot\) without the intervention of \(\ce{Cu}\)(I); considerable \(\ce{I_2}\) usually is formed in the reaction: \[\begin{align} \ce{C_6H_5} \overset{\oplus}{\ce{N_2}} + \overset{\ominus}{\ce{I}} &\rightarrow \ce{C_6H_5} \cdot + \ce{N_2} + \ce{I} \cdot \\ \ce{C_6H_5} \cdot + \ce{I} \cdot &\rightarrow \ce{C_6H_5I} \\ 2 \ce{I} \cdot &\rightarrow \ce{I_2} \end{align}\] Secondary arenamines react with nitrous acid to form \(\ce{N}\)-nitroso compounds while tertiary arenamines undergo electrophilic substitution with \(\ce{NO}^\oplus\) if they have an unsubstituted position: Not all reactions of diazonium ions involve cleavage of the \(\ce{C-N}\) bond. An important group of reactions of arenediazonium ions involves aromatic substitution by the diazonium ion acting as an agent to yield azo compounds, \(\ce{Ar-N=N-Ar}\): This reaction is highly sensitive to the nature of the substituent \(\ce{X}\), and coupling to benzene derivatives normally occurs only when \(\ce{X}\) is a strongly electron-donating group such as \(\ce{-O}^\ominus\), \(\ce{-N(CH_3)_2}\), and \(\ce{-OH}\). However, coupling with \(\ce{X} = \ce{-OCH_3}\) may take place with particularly active diazonium ions. Diazo coupling has considerable technical value, because the azo compounds that are produced are highly colored. Many are used as fabric dyes and for other coloring purposes. A typical example of diazo coupling is formation of 4-\(\ce{N}\),\(\ce{N}\)-dimethylaminoazobenzene from benzenediazonium chloride and \(\ce{N}\),\(\ce{N}\)-dimethylbenzenamine: The product once was used to color edible fats and therefore was known as "Butter Yellow", but its use to color food is prohibited because it is reported to be a potent liver carcinogen for rats. The pH used for diazo coupling of amines is very important in determining the nature of the products. Under near-neutral conditions the diazonium ion may attack the of the arenamine rather than a ring carbon. In this event a diazoamino compound, a triazene, \(\ce{-N=N-N}-\), is formed: The reaction is readily reversed if the pH is lowered sufficiently. As you see from this brief discussion of arenediazonium salts, their chemistry is complex. It is inappropriate to discuss all of their many reactions here, but a summary of the most important types of reactions is given in Table 23-4. A secondary arenamine behaves like a secondary alkanamine in reacting with nitrous acid to give an \(\ce{N}\)-nitrosamine. However, when treated with an acid the \(\ce{N}\)-nitrosamine rearranges: This is one example of a group of formally related rearrangements in which a substituent, \(\ce{Y}\), attached to the nitrogen of a benzenamine derivative migrates to the ortho or para positions of the aromatic ring under the influence of acid: Rearrangement occurs most readily when \(\ce{Y}\) is a strongly electron-attracting group and the \(\ce{N-Y}\) bond that is broken is not as strong as the \(\ce{C-Y}\) bond that is formed. A few of the many examples of this type of reaction follow: and (1977)

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The carbonyl group, \(\ce{-C=O}\), is a structural feature of many different types of compounds. It is present in carbon dioxide and in methanal, which represent respectively the high and low extremes in the level of oxidation of a carbonyl carbon: In between, there are carbonyl compounds ranging from aldehydes and ketones to carboxylic acids and their derivatives (esters, amides, anhydrides, and acyl halides). The naming of these compounds is described in Sections 7-4 to 7-7. At the upper end of the oxidation scale, along with \(\ce{CO_2}\), are the carbonic acid derivatives such as carbonic esters, amides, halides, and carbonate salts, and isocyanates: In this and succeeding chapters we describe the chemistry of these compounds with the intent of emphasizing the similarities that exist between them. The differences turn out to be more in degree than in kind. Even so, it is convenient to discuss aldehydes and ketones separately from carboxylic acids and, following some general observations about the carbonyl group, this chapter mainly is concerned with aldehydes and ketones. Apart from \(\ce{CO_2}\) and metal carbonates, the most abundant carbonyl compounds of natural origin are carboxylic esters and amides. These occur as fats and lipids, which are esters of long-chain alkanoic acids, and as proteins, which are polyamides of natural amino acids. The same structural features are found in certain synthetic polymers, in particular the polyesters (e.g., Dacron) and the polyamides (e.g., nylon 6): Compared to carboxylic and carbonic acid derivatives, the less highly oxidized carbonyl compounds such as aldehydes and ketones are not so widely-spread in nature. That is not to say that they are unimportant. To the contrary, aldehydes and ketones are of great importance both in biological chemistry and in synthetic organic chemistry. However, the high reactivity of the carbonyl group in these compounds enables them to function more as intermediates in metabolism or in synthesis than as end products. This fact will become evident as we discuss the chemistry of aldehydes and ketones. Especially important are the reactions of carbonyl groups, and this chapter is mostly concerned with this kind of reaction of aldehydes and ketones. and (1977)

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The noble gases (Group 18) are located in the far right of the and were previously referred to as the "inert gases" due to the fact that their filled valence shells (octets) make them extremely nonreactive. Noble gases are odorless, colorless, nonflammable, and monotonic gases that have low chemical reactivity. The full valence electron shells of these atoms make noble gases extremely stable and unlikely to form chemical bonds because they have little tendency to gain or lose electrons. Although noble gases do not normally react with other elements to form compounds, there are some exceptions. Xe may form compounds with fluoride and oxide. \(XeF_​2\) \(XeF_4\) \(XeF_6\) Two other short-lived xenon compounds with an oxidation state of +8, XeO F and XeO F , are produced in the reaction of xenon tetroxide with xenon hexafluoride. Radon difluoride (RnF ) is one of the few reported compounds of radon. Radon reacts readily with fluorine to form a solid compound, but this decomposes on attempted vaporization and its exact composition is uncertain. The usefulness of radon compounds is limited because of the noble gas's radioactivity. The longest-lived isotope, Ra, has a half-life of only 3.82 days.

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Many important molecules have alternating single and double bonds (are conjugated), but have atoms that are more (or less) electron-attracting than carbon. An example is propenal (acrolein), \(18\): With such molecules we need to take into account the fact that the \(\pi\) electrons will be attracted to oxygen from carbon, because oxygen is more electronegative than carbon. With the VB method we can do this by considering ionic electron-pairing schemes, \(18c\) and \(18d\), along with the dienelike structures, \(18a\) and \(18b\). The hybrid, \(18e\), is drawn to reflect the expected relative contributions of the various forms, with \(18a\) being most important. Ionic structures such as \(19a\) and \(19b\) need not be considered for propenal because carbon is much less electron-attracting than oxygen: Analysis of the electronic configuration resulting from the MO calculations accords generally with the VB hybrid \(18e\). An especially important type of carbocation is represented by the 2-propenyl electron-pairing schemes, \(21a\) and \(21b\), which correspond to the hybrid \(21c\). Because \(21a\) and \(21b\) are equivalent and no other single low-energy structure is possible, a sizable delocalization energy is expected. Evidence for this delocalization energy of \(21c\) is available from the comparative ease of reactions involving formation of carbocation intermediates. An example is in \(S_\text{N}1\) ionizations of alkenyl and alkyl halides. The ionization \(\ce{CH_2=CHCH_2Br} \rightarrow \ce{CH_2=CHCH_2^+} + \ce{Br^+}\) proceeds than \(\ce{CH_3CH_2CH_2Br} \rightarrow \ce{CH_3CH_2CH_2^+} + \ce{Br^+}\) (for which no \(\pi\)-electron delocalization is possible). MO treatment of the 2-propenyl cation begins with the atomic-orbital model \(22\): Any \(\pi\) electrons will be delocalized through the orbitals of \(22\), but it is not so easy to be confident that when two electrons are placed into the lowest molecular orbital the resulting electron distribution will be the same as \(21c\) with half of the positive charge on \(\ce{C_1}\) and half on \(\ce{C_3}\). The complete calculation gives the result shown in Figure 21-9. Here the lowest-energy molecular orbital has a higher proportion of the \(p\) orbital of \(\ce{C_2}\) mixed in than the \(p\) orbitals of \(\ce{C_1}\) and \(\ce{C_3}\) - in fact, just the right amount to have \(\ce{C_2}\) neutral and \(\ce{C_1}\) and \(\ce{C_3}\) each with \(\frac{1}{2}^\oplus\) when this MO is filled with two paired electrons. The delocalization energy calculated for the cation is \(\left( 2 \alpha + 2.82 \beta \right) - \left( 2 \alpha + 2 \beta \right) = 0.82 \beta\) or about \(16 \: \text{kcal}\) if \(\beta\) is taken to be \(19 \: \text{kcal}\). Thus in every respect the simple VB and MO methods give the same representation of the 2-propenyl carbocation. You will notice that the 2-propenyl radical and the 2-propenyl carbanion can be formulated by the same set of \(\pi\) molecular orbitals (Figure 21-9) used for the carbocation by putting one or two electrons into the nonbonding MO. The delocalization energies calculated for the radical and anion are the same as for the cation. Thus \(\left( 3 \alpha + 2.82 \beta \right) - \left( 3 \alpha + 2 \beta \right) = 0.82 \beta\) for the radical and \(\left( 4 \alpha + 2.82 \beta \right) - \left( 4 \alpha + 2 \beta \right) = 0.82 \beta\) for the anion. covers qualitative explanations of how the VB method is used to account for the lower-energy (longer-wavelength) radiation required for electron excitation of conjugated polyenes compared to nonconjugated polyenes. Thus 1,3-butadiene has a \(\lambda_\text{max}\) for ultraviolet light at \(217 \: \text{nm}\), whereas 1,5-hexadiene has a corresponding \(\lambda_\text{max}\) at \(185 \: \text{nm}\). We will now consider how the MO approach can be used to understand these differences in excitation energy. The \(\pi\)-energy levels and electronic configurations for delocalized and localized 1,3-butadiene are shown in Figure 21-10 (also see ). Because the double bonds are so far apart, the \(\pi\)-electron system of 1,5-hexadiene by the simple MO approach is identical with that of localized 1,3-butadiene. The calculated energy change for the lowest-energy \(\pi \rightarrow \pi^*\) transition is \(\left( \alpha - 0.62 \beta \right) - \left( \alpha + 0.6 \beta \right) = -1.24 \beta\) for 1,3-butadiene and \(\left( \alpha - \beta \right) - \left( \alpha + \beta \right) = -2 \beta\) for 1,5-hexadiene. In each case the energy of the electron in the orbital (the ) is subtracted from the energy that an electron would have in the orbital (the ). Other transitions are possible, as of an electron from the lowest occupied orbital of energy \(\alpha + 1.62 \beta\) to the highest unoccupied orbital of energy \(\alpha - 1.62 \beta\), but these would have far greater energies. Qualitatively, the \(\pi \rightarrow \pi^*\) transition energy is predicted to be substantially less for 1,3-butadiene than for 1,5-hexadiene. However, any attempt at a quantitative correlation is suspect, because the lowest energy \(\pi \rightarrow \pi^*\) transition calculated for 1,3-butadiene is \(-1.24 \beta\) and, if \(\beta\) is \(19 \: \text{kcal}\) (see ), \(\lambda_\text{max}\) from Equation 9-2 should be \(1214 \: \text{nm}\) instead of the observed \(217 \: \text{nm}\). and (1977)

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Unlike hydrogenic atoms, the wavefunctions satisfying Schrödinger's equation for multi-electron atoms cannot be solved analytically. Instead, various techniques are used for giving approximate solutions to the wave functions. The wavefunctions of multi-electron atoms can be considered, as a first approximation, to be built up of components, where the combined wavefunction for an atom with k electrons is of the form: \(\Psi = \Psi(1) \; \Psi(2) \;...\; \Psi(k)\) Here, \(\Psi(1)\), \(\Psi(2)\), up to \(\Psi(k)\) represent wave functions for the first, second, up to k electrons. Each \(\Psi(i)\) is considered to be in the form of a wave function for the single electron of the hydrogenic atom subject to the Pauli Exclusion Principle and after making adjustments to account for . The Pauli Exclusion Principle allows at most two electrons in any one orbital. This is explained by postulating an additional quantum number for electron spin, which can have values of +1/2 or -1/2. The Dirac theory of quantum mechanics, applied to electron orbitals, more naturally explains this because the theory goes beyond the assumptions of the Schrödinger equation by also accounting for the relativistic behavior of orbiting electrons. Shielding occurs because other electrons that are closer to the nucleus shield an electron from the attractive force of the nucleus. Adjusting the hydrogenic orbitals can be done by reducing the value of Z to Z . The amount of the reduction depends on which inner orbitals are occupied and how much the orbital being calculated is able to penetrate the shielding. As the amount of shielding increases, Z becomes smaller, and the energy levels of the orbital increase. Conversely, as the amount of penetration increases, the shielding is reduced, Z becomes bigger, and the energy level of the orbital decreases. For example, when the 1s orbital is fully occupied by two electrons, the third electron could occupy any of 2s or 2p orbitals, which would have the same energy level in a hydrogenic atom. However, as illustrated in the diagrams below, the 2s orbital concentrates more of its probability near the nucleus than the 2p orbital does. As a result, the 2s orbital penetrates the 1s orbital shielding more than the 2p orbital does. Hence, a third electron will occupy the 2s orbital where it has a slightly lower energy than in the 2p orbital. The values of Z can be calculated by various techniques. Slater's rule is a relatively simple ad hoc method of estimating Z for values of up to 4, that is for electrons in the s, p, d, or f orbitals.

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Another example of a synthesis problem makes use of the cycloaddition reactions discussed here. Consider the synthesis of bicyclo[2.2.1]heptane, \(9\), from compounds with fewer carbons. Whenever a ring has to be constructed, you should consider the possibility of cycloaddition reactions, especially [4 + 2] cycloaddition by the . A first glance at \(9\), written in the usual sawhorse-perspective formula, might lead to overlooking the possibility of constructing the skeleton by [4 + 2] addition, because the compound seems only to be made up of five-membered rings. If the structure is rewritten as \(10\), the six-membered ring stands out much more clearly: If we now try to divide the six-membered ring into [2] and [4] fragments, we find that there are only two ways this can be done: The left division corresponds to a simple [4 + 2] cycloaddition, whereas the right division corresponds to a complex reaction involving formation of three ring bonds at once. Actual Diels-Alder reactions require diene and dienophile starting materials, and two possibilities, using 1,3-cyclopentadiene as the diene and ethene or ethyne as dienophile, follow: Either of the products can be converted to bicyclo[2.2.1]heptane by hydrogenation (Table 13-5): Neither ethene nor ethyne is a very good dienophile but [4 + 2] cycloadditions of either with 1,3-cyclopentadiene go well at temperatures of \(160\)-\(180^\text{o}\) because 1,3-cyclopentadiene is a very reactive diene. Achieving the overall result of addition of ethene or ethyne to a less reactive diene could necessitate a synthetic sequence wherein one of the reactive dienophiles listed in Table 13-1 is used to introduce the desired two carbons, and the activating groups are subsequently removed. An example follows: Reactions that can be used to remove a \(\ce{-CO_2H}\) group will be discussed in Chapter 18. and (1977)

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Band Theory was developed with some help from the knowledge gained during the quantum revolution in science. In 1928, Felix Bloch had the idea to take the quantum theory and apply it to solids. In 1927, Walter Heitler and Fritz London discovered bands- very closely spaced orbitals with not much difference in energy. In this image, orbitals are represented by the black horizontal lines, and they are being filled with an increasing number of electrons as their amount increases. Eventually, as more orbitals are added, the space in between them decreases to hardly anything, and as a result, a band is formed where the orbitals have been filled. Different metals will produce different combinations of filled and half filled bands. Sodium's bands are shown with the rectangles. Filled bands are colored in blue. As you can see, bands may overlap each other (the bands are shown askew to be able to tell the difference between different bands). The lowest unoccupied band is called the conduction band, and the highest occupied band is called the valence band. Bands will follow a trend as you go across a period: The probability of finding an electron in the conduction band is shown by the equation: \[ P= \dfrac{1}{e^{ \Delta E/RT}+1} \] The ∆E in the equation stands for the change in energy or energy gap. t stands for the temperature, and R is a bonding constant. That equation and this table below show how the bigger difference in energy is, or gap, between the valence band and the conduction band, the less likely electrons are to be found in the conduction band. This is because they cannot be excited enough to make the jump up to the conduction band. Metals are conductors. There is no band gap between their valence and conduction bands, since they overlap. There is a continuous availability of electrons in these closely spaced orbitals. In insulators, the band gap between the valence band the the conduction band is so large that electrons cannot make the energy jump from the valence band to the conduction band. Semiconductors have a small energy gap between the valence band and the conduction band. Electrons can make the jump up to the conduction band, but not with the same ease as they do in conductors. There are two different kinds of semiconductors: and . An intrinsic semiconductor is a semiconductor in its pure state. For every electron that jumps into the conduction band, the missing electron will generate a hole that can move freely in the valence band. The number of holes will equal the number of electrons that have jumped. In extrinsic semiconductors, the band gap is controlled by purposefully adding small impurities to the material. This process is called . Doping, or adding impurities to the lattice can change the electrical conductivity of the lattice and therefore vary the efficiency of the semiconductor. In extrinsic semiconductors, the number of holes will not equal the number of electrons jumped. There are two different kinds of extrinsic semiconductors, p-type (positive charge doped) and n-type (negative charge doped). Jim Clark ( )

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Among the organic nitrogen compounds having nitrogen above the oxidation level of ammonia are a wide variety of substances with \(\ce{N-N}\) bonds. We shall mention only a very few of the more important of these substances: hydrazines, azo and diazo compounds, and azides. Organic hydrazines or diazanes are substitution products of \(\ce{NH_2-NH_2}\) and have many properties similar to those of amines in being basic and forming acyl derivatives as well as undergoing alkylation and condensations with carbonyl compounds ( ). Unsymmetrical hydrazines can be prepared by careful reduction of \(\ce{N}\)-nitrosamines. 1,1-Dimethyldiazane is prepared in this way for use as a rocket fuel: Aromatic hydrazines are best prepared by reduction of aromatic diazonium salts (Table 23-4). Hydrazines of the type \(\ce{R-NH-NH-R}\) are easily oxidized to the corresponding azo compounds, \(\ce{R-N=N-R}\). With nitrous acid, monosubstituted hydrazines are converted to azides: \[\ce{R-NH-NH_2} + \ce{HONO} \rightarrow \ce{R-N=} \overset{\oplus}{\ce{N}} \ce{=} \overset{\ominus}{\ce{N}} + 2 \ce{H_2O}\] Azo or diazo compounds possess the \(\ce{-N=N}-\) grouping. Aliphatic azo compounds of the type \(\ce{R-N=N-H}\) appear to be highly unstable and decompose to \(\ce{R-H}\) and nitrogen. Derivatives of the type \(\ce{R-N=N-R}\) are much more stable and can be prepared as mentioned above by oxidation of the corresponding hydrazines. Aromatic azo compounds are available in considerable profusion from diazo coupling reactions ( ) and are of commercial importance as dyes and coloring materials. A prime characteristic of azo compounds is their tendency to decompose into organic free radicals and liberate nitrogen: \[\ce{R-N=N-R} \rightarrow 2 \ce{R} \cdot + \ce{N_2}\] The ease of these reactions is usually a fairly reliable guide to the stabilities of the free radicals that result. For instance, it is found that dimethyldiazene (azomethane, \(\ce{CH_3N=NCH_3}\)) is stable to about \(400^\text{o}\), and diphenyldiazene (azobenzene, \(\ce{C_6H_5N=NC_6H_5}\)) also is resistant to thermal decomposition; but, when the azo compound decomposes to radicals that have extra stability because of delocalization of the odd electron, the decomposition temperature is greatly reduced. Thus the azo compound, \(2\), decomposes to radicals at moderate temperatures (\(60^\text{o}\) to \(100^\text{o}\)), and for this reason is a very useful agent for generating radicals, such as those required for the initiation of polymerization of ethenyl compounds: The parent of the diazo compounds, diazomethane, \(\ce{CH_2=} \overset{\oplus}{\ce{N}} \ce{=} \overset{\ominus}{\ce{N}}\), has been mentioned before in connection with ylide reactions for ring enlargement ( ) and the preparation of methyl esters from acids (Table 18-7). It is one of the most versatile and useful reagents in organic chemistry, despite the fact that it is highly toxic, dangerously explosive, and cannot be stored without decomposition. Diazomethane is an intensely yellow gas, bp \(-23^\text{o}\), which customarily is prepared and used in diethyl ether or dichloromethane solution. It can be synthesized in a number of ways, the most useful of which employs the action of base on an \(\ce{N}\)-nitroso-\(\ce{N}\)-methylamide: As a methylating agent of reasonably acidic substances, diazomethane has nearly ideal properties. It can be used in organic solvents; reacts very rapidly without need for a catalyst (except with alcohols, which do require an acid catalyst); the coproduct is nitrogen which offers no separation problem; it gives essentially quantitative yields; and it act as as its own indicator to show when reaction is complete. With enols, it gives \(\ce{O}\)-alkylation: Besides being a methylating agent, diazomethane also is a source of \(\colon \ce{CH_3}\) when irradiated with light. The carbene formed in this way is highly reactive and even will react with the electrons of a carbon-hydrogen bond to "insert" the carbon of the carbene between carbon and hydrogen. This transforms \(\ce{-C-H}\) to \(\ce{-C-CH_3}\): This \(\colon \ce{CH_3}\) species is one of the most reactive reagents known in organic chemistry. Diazomethane undergoes a wealth of other unusual reactions. Besides those already mentioned are the following two examples: (\(\ce{COCl} \rightarrow \ce{-CH_2CO_2H}\), ) ([2 + 3] cycloaddition) The Arndt-Eistert synthesis is useful for converting an acid to the next higher member of the series. Pyrazolines are important intermediates for the preparation of cyclopropanes: Diazomethane originally was believed to possess the three-membered 1,2-diazacyclopropene ring structure, but this concept was disproved by electron-diffraction studies, which showed the linear structure to be correct: Recently, a variety of authentic 1,2-diazacyclopropenes (sometimes called ) have been prepared, and these have been found to have very different properties from the diazoalkanes. The simple 1,2-diazacyclopropenes are colorless and do not react with dilute acids, bases, or even bromine. The syntheses of these substances are relatively simple. One of several possible routes follows: Organic azides can be prepared from hydrazines and nitrous acid ( ) and by the reaction of sodium azide with acyl halides or with alkyl halides having good \(S_\text{N}2\) reactivity: \[\ce{RBr} + \ce{N_3^-} \underset{\ce{CH_3OH}}{\overset{S_\text{N}2}{\longrightarrow}} \ce{R-N=} \overset{+}{\ce{N}} \ce{=} \overset{-}{\ce{N}} + \ce{Br^-}\] The lower-molecular-weight organic azides often are unpredictably explosive and are best handled in solution. The use of acyl azides in the preparation of amines by the has been discussed previously ( ). Alkyl azides can be reduced readily by lithium aluminum hydride to amines and, if a pure primary amine is desired, the sequence halide \(\rightarrow\) azide \(\rightarrow\) amine may give as good or better results than does the Gabriel synthesis ( ). and (1977)

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For a review of determining concentration of solutions, click the following link and select CoreChem: Solution Concentrations Mercury is a liquid metal which is found naturally in ores, typically combined with other elements. It is commonly used in electrical products, including fluorescent light bulbs, dry cell batteries, switches, and various control equipment. Mercury in drinking water is a possible health concern. Those who drink water with high levels of mercury over several years are at risk of experiencing kidney damage. The Environmental Protection Agency (EPA) in the USA has established particular guidelines regarding contaminants such as mercury in the drinking water: maximum contaminant level goals (MCLG) and the maximum contaminant level (MCL). The MCLG is the highest level of protection based on the best available data in order to prevent potential health issues. The MCL is the enforceable standard set by EPA; it is often the same as the MCLG, but sometimes it is a little less stringent due to cost restriants or limits of public water systems to detect, treat, or remove contaminants. For mercury, both the MCLG and the MCL are 0.002 mg/L. Remember that molarity, or molar concentration, can be expressed in a variety of ways, including moles per liter. Expressed in this way, and using the conversion 1 g = 1000 mg and the molar mass of mercury, allows us to solve this problem: (0.002 mg / L) x (1 g / 1000 mg ) x (1 mol Hg / 200.59 g)= 9.97 x 10 mol / L In example one above, the units for the final answer were expressed in moles per liter. Using the above information, grams can be converted into moles and cubic meters can be converted into liters: 51.70 g Hg x ( 1 mol Hg / 200.59 g Hg ) = 2.58 x 10 mol Hg 2.35 x 10 m x ( 1000 L / 1 m ) = 2.35 x 10 L The concentration in moles per liter can be found by dividing results of the two calculations above: Molarity = moles / liters = 2.58 x 10 mol Hg / 2.35 x 10 L = 1.10 x 10 M This concentration is an acceptable level for mercury because it is less than the maximum contaminant level of 9.97 x 10 mol / L set by the EPA. 1. 92.56 grams of mercury have contaminated a drinking water source with a volume of 7.5 x 10 m . Does this mercury concentration fall below the EPA's maximum contaminant level? 2. Suppose the EPA revises its maximum contaminant level goal to 0.0015 mg/L. What would this concentration be expressed in moles per liter? 1. 7.5 x 10 m x ( 1000 L / 1 m ) = 7.5 x 10 L 92.56 g Hg x ( 1 mol Hg / 200.59 g Hg )= 4.61 x 10 mol Hg Concentration = moles / liters = 4.61 x 10 mol Hg / 7.5 x 10 L = 6.15 x 10 M Yes, this concentration falls below the EPA's maximum contaminant level. 2. 0.0015 mg / L x ( 1 g / 1000 mg ) x ( 1 mol / 200.59 g )= 7.48 x 10 mol/L Reference

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The general character of alkene polymerization by radical and ionic mechanisms was discussed briefly in . The same principles apply to the polymerization of alkadienes, with the added feature that there are additional ways of linking the monomer units. The polymer chain may grow by either 1,2 or 1,4 addition to the monomer. With 1,3-butadiene, for example, In 1,2 addition, a chiral carbon (marked with \(^*\)) is created as each molecule of the monomer adds to the growing chain radical. The physical properties of the polymer greatly depend on whether these carbons have the same or different configurations, as we will show in greater detail in . However, in polymer nomenclature, an polymer is one with essentially all chiral carbons having the same configuration, whereas an polymer has a ordering of the chiral carbons with different configurations. For polymerization of 1,3-butadiene by 1,4 addition, there are no chiral carbons, but there is the possibility of cis-trans isomerism: A polymer made of identical repeating units is called a . If the units are nonidentical, as when different monomers are polymerized, the product is called a . Many of the polymers formed from conjugated dienes are elastic and are used to manufacture synthetic rubbers. The raw polymers usually are tacky and of little direct use, except as adhesives and cements. They are transformed into materials with greater elasticity and strength by , in which the polymer is heated with sulfur and various other substances called , with the result that the polymer chains become cross-linked to one another by carbon-sulfur and carbon-carbon bonds. Some of the cross-linking appears to occur by addition to the double bonds, but the amount of sulfur added generally is insufficient to saturate the polymer. With large proportions of sulfur, hard rubber is formed such as is used in storage-battery cases. Because of the many double bonds present, diene rubbers are sensitive to air oxidation unless are added to inhibit oxidation. The more important dienes for the manufacture of synthetic rubbers are 1,3-butadiene, 2-chloro-1,3-butadiene (chloroprene), and 2-methyl-1,3-butadiene (isoprene): Several rubbers that have desirable properties of elasticity, flexibility, abrasive resistance, and resistance to chemicals are listed in Table 13-2. The homogeneity of these polymers depends greatly on the way in which they are prepared, particularly on the polymerization catalyst employed. A synthetic rubber that is virtually identical to natural Hevea rubber is made from 2-methyl-1,3-butadiene (isoprene) using finely divided lithium metal or transition-metal catalysts; the product is formed almost exclusively by cis 1,4 addition\(^3\): Polymerization of 2-methylpropene in the presence of small amounts of 2-methyl-1,3-butadiene (isoprene) gives a copolymer with enough double bonds to permit cross-linking of the polymer chains through vulcanization. The product is a hard-wearing, chemically resistant rubber called “butyl rubber.” It is highly impermeable to air and is used widely for inner tubes for tires. \(^3\)Synthetic rubber has provided severe competition for natural rubber and, for many years, it seemed as though rubber plantations eventually would become extinct. However, rising petroleum prices and higher 2-methyl-1,3-butadiene costs coupled with methods developed for greatly increasing the output of rubber latex per tree, and the fact that natural rubber has superior properties in radial automobile tires, have reversed the trend and rubber plantations currently are being expanded. and (1977)

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A multipole expansion is a mathematical series representing a function that depends on angles—usually the two angles on a sphere. These series are useful because they can often be truncated, meaning that only the first few terms need to be retained for a good approximation to the original function. Multipole expansions are very frequently used in the study of electromagnetic and gravitational fields, where the fields at distant points are given in terms of sources in a small region. The multipole expansion with angles is often combined with an expansion in radius. Such a combination gives an expansion describing a function throughout three-dimensional space. The multipole expansion is expressed as a sum of terms with progressively finer angular features. For example, the initial term—called the zeroth, or monopole, moment—is a constant, independent of angle. The following term—the first, or dipole, moment—varies once from positive to negative around the sphere. Higher-order terms (like the quadrupole and octupole) vary more quickly with angles. A multipole moment usually involves powers (or inverse powers) of the distance to the origin, as well as some angular dependence. Consider an arbitrary charge distribution \( \rho (\mathbf {r} ')\). We wish to find the electrostatic potential due to this charge distribution at a given point \( \mathbf {r} \). We assume that this point is at a large distance from the charge distribution, that is if \( \mathbf {r} '\) varies over the charge distribution, then \( \mathbf {r} \gg \mathbf {r} '\). Now, the Coulomb potential for an arbitrary charge distribution is given by \[ V(\mathbf {r} ) ={\dfrac {1}{4\pi \epsilon _{0}}}\int _{V'}{\dfrac {\rho (\mathbf {r} ')}{|\mathbf {r} -\mathbf {r'} |}}dV' \label{eq2} \] Here, \[ \begin{align} | \mathbf {r} -\mathbf {r'} |&= \sqrt{ |r^{2}-2\mathbf {r} \cdot \mathbf {r} '+r'^{2}|} \\[5pt] &=r \sqrt{ \left|1-2 \dfrac {\hat {\mathbf {r}} \cdot \mathbf {r} '}{r} +\left( \dfrac {r'}{r} \right)^2 \right|} \end{align}\] where \[ \hat {\mathbf {r} } = \mathbf {r} /r \] Thus, using the fact that \({ \mathbf {r} }\) is much larger than \( \mathbf {r} '\), we can write \[ \dfrac {1}{|\mathbf {r} -\mathbf {r'} |} = \dfrac {1}{r} \dfrac {1}{ \sqrt{\left|1-2 \dfrac {\hat {\mathbf {r} }\cdot \mathbf {r} '}{r} +\left(\dfrac {r'}{r}\right)^2 \right|}} \label{eq6}\] and using the binomial expansion (see below), \[ \dfrac {1} {\sqrt{ \left|1-2 \dfrac {\hat {\mathbf {r}} \cdot \mathbf {r} '}{r} + \left( \dfrac {r'}{r} \right)^{2} \right|} } =1-{\dfrac {\hat {\mathbf {r} }\cdot \mathbf {r} '}{r}} + {\dfrac {1}{2r^{2}}}\left(r'^{2}-3({\hat {\mathbf {r} }}\cdot \mathbf {r} '\right)^{2})+O\left({\dfrac {r'}{r}}\right)^{3} \label{eq10}\] (we neglect the third and higher order terms). The can be used to expand specific functions into an infinite series: \[\begin{align} (1+ x )^s &= \sum_{n=0}^{\infty} \dfrac{s!}{n! (s-n)!} x^n \\[5pt] &= 1 + \dfrac{s}{1!} x + \dfrac{s(s-1)}{2!} x^2 + \dfrac{s(s-1)(s-2)}{3!} x^3 + \ldots \end{align}\] Equation \ref{eq6} can be rewritten as \[ \dfrac {1}{|\mathbf {r} -\mathbf {r'} |} = \dfrac {1}{r} \dfrac {1}{ \sqrt{1 + \epsilon } } \label{eq6B}\] where \[\epsilon = -2 \dfrac {\hat {\mathbf {r} }\cdot \mathbf {r} '}{r} +\left(\dfrac {r'}{r}\right)^2 \label{eq6c}\] Applying the Binomial Theorem to Equation \ref{eq6B} (with \(s = -1/2\)) results in \[ \dfrac {1}{|\mathbf {r} -\mathbf {r'} |} = \dfrac {1}{r} \left( 1 - \dfrac{1}{2} \epsilon + \dfrac{3}{8} \epsilon^2 - \dfrac{5}{16} \epsilon^3 + \ldots \right) \label{eq6d}\] Equation \ref{eq10} originates from substituting Equation \ref{eq6c} into Equation \ref{eq6d}. Inserting Equation \ref{eq10} into Equation \ref{eq2} shows that the potential can be written as \[ V(\mathbf {r} )={\dfrac {1}{4\pi \epsilon _{0}r}}\int _{V'}\rho (\mathbf {r} ')\left(1-{\dfrac {\hat {\mathbf {r} }\cdot \mathbf {r} '}{r}}+{\dfrac {1}{2r^{2}}}\left(3({\hat {\mathbf {r} }}\cdot \mathbf {r} '\right)^{2}-r'^{2})+O\left({\dfrac {r'}{r}}\right)^{3}\right)dV' \] We write this as \[ V(\mathbf {r} )=V_{\text{mon}}(\mathbf {r} )+V_{\text{dip}}(\mathbf {r} )+V_{\text{quad}}(\mathbf {r} )+\ldots \label{expand}\] The first (the zeroth-order) term in the expansion is called the monopole moment, the second (the first-order) term is called the dipole moment, the third (the second-order) is called the quadrupole moment, the fourth (third-order) term is called the octupole moment, and the fifth (fourth-order) term is called the hexadecapole moment. Given the limitation of Greek numeral prefixes, terms of higher order are conventionally named by adding "-pole" to the number of poles—e.g., 32-pole (i.e., dotriacontapole) and 64-pole (hexacontatetrapole). These moments can be expanded thusly \[ \begin{align} V_{\text{mon}}(\mathbf {r} ) &={\dfrac {1}{4\pi \epsilon _{0}r}}\int _{V'}\rho (\mathbf {r} ')dV' \\[5pt] V_{\text{dip}}(\mathbf {r} ) &=-{\dfrac {1}{4\pi \epsilon _{0}r^{2}}}\int _{V'}\rho (\mathbf {r} ')\left({\hat {\mathbf {r} }}\cdot \mathbf {r} '\right)dV' \\[5pt] V_{\text{quad}}(\mathbf {r} ) &={\dfrac {1}{8\pi \epsilon _{0}r^{3}}}\int _{V'}\rho (\mathbf {r} ')\left(3\left({\hat {\mathbf {r} }}\cdot \mathbf {r} '\right)^{2}-r'^{2}\right)dV' \end{align}\] and so on. In principle, a multipole expansion provides an exact description of the potential and generally converges under two conditions: In the first (more common) case, the coefficients of the series expansion are called exterior multipole moments or simply multipole moments whereas, in the second case, they are called interior multipole moments. Observe that \[ V_{mon}(\mathbf {r}) =\dfrac {1}{4\pi \epsilon _0r}\int _{V'}\rho (\mathbf {r} ')dV' = \dfrac{q}{ 4\pi \epsilon _0 r}\] is a scalar, (actually the total charge in the distribution) and is called the . This term indicates point charge electrical potential with charge \(q\). If a charge distribution has a net total charge, it will tend to look like a monopole (point charge) from large distances. We can write \[ V_{\text{dip}}(\mathbf {r} )=-{\dfrac {\hat {\mathbf {r} }}{4\pi \epsilon _{0}r^{2}}}\cdot \int _{V'}\rho (\mathbf {r} ')\mathbf {r} 'dV' \] The vector \[ \mathbf {p} =\int _{V'}\rho (\mathbf {r} ')\mathbf {r} 'dV' \] is called the electric dipole. And its magnitude is called the dipole moment of the charge distribution. This terms indicates the linear charge distribution geometry of a dipole electrical potential. Let \( \hat {\mathbf {r} }\) and \( \mathbf {r} '\) be expressed in Cartesian coordinates as \( (r_{1},r_{2},r_{3})\) and \( (x_{1},x_{2},x_{3})\). Then, \( ({\hat {\mathbf {r} }}\cdot \mathbf {r} ')^{2}=(r_{i}x_{i})^{2}=r_{i}r_{j}x_{i}x_{j}\). We define a dyad to be the tensor \( {\hat {\mathbf {r} }}{\hat {\mathbf {r} }}\) given by \[ \left({\hat {\mathbf {r} }}{\hat {\mathbf {r} }}\right)_{ij}=r_{i}r_{j} \] Define the as \[ T=\int _{V'} \rho (\mathbf {r} ') \left(3 (\mathbf {r} '\mathbf {r} ')-\mathbf {I} r'^{2}\right)dV' \] Then, we can write \( V_{\text{qua}}\) as the tensor contraction \[ V_{\text{qua}}(\mathbf {r} )=- \dfrac {\hat {\mathbf {r}} \hat {\mathbf {r}}} {4\pi \epsilon _{0}r^{3}} ::T\] this term indicates the three dimensional distribution of a quadruple electrical potential.

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In any introductory chemistry course you will have come across the electronic structures of hydrogen and carbon drawn as: The circles show energy levels - representing increasing distances from the nucleus. You could straighten the circles out and draw the electronic structure as a simple energy diagram. Orbits and orbitals sound similar, but they have quite different meanings. It is essential that you understand the difference between them. To plot a path for something you need to know exactly where the object is and be able to work out exactly where it's going to be an instant later. You can't do this for electrons. The says that you cannot know with certainty both where an electron is and where it's going next. That makes it impossible to plot an orbit for an electron around a nucleus. Is this a big problem? No. If something is impossible, you have to accept it and find a way around it. Suppose you had a single hydrogen atom and at a particular instant plotted the position of the one electron. Soon afterwards, you do the same thing, and find that it is in a new position. You have no idea how it got from the first place to the second. You keep on doing this over and over again, and gradually build up a sort of 3D map of the places that the electron is likely to be found. In the hydrogen case, the electron can be found anywhere within a spherical space surrounding the nucleus. The diagram shows a through this spherical space. 95% of the time (or any other percentage you choose), the electron will be found within a fairly easily defined region of space quite close to the nucleus. Such a region of space is called an You can think of an orbital as being the region of space in which the electron lives. What is the electron doing in the orbital? We don't know, we can't know, and so we just ignore the problem! All you can say is that if an electron is in a particular orbital it will have a particular definable energy. Each orbital has a name. The orbital occupied by the hydrogen electron is called a The represents the fact that the orbital is in the energy level closest to the nucleus. The tells you about the shape of the orbital. s orbitals are spherically symmetric around the nucleus - in each case, like a hollow ball made of rather chunky material with the nucleus at its center. The orbital on the left is a This is similar to a 1s orbital except that the region where there is the greatest chance of finding the electron is further from the nucleus - this is an orbital at the second energy level. If you look carefully, you will notice that there is another region of slightly higher electron density (where the dots are thicker) nearer the nucleus. ("Electron density" is another way of talking about how likely you are to find an electron at a particular place.) 2s (and 3s, 4s, etc) electrons spend some of their time closer to the nucleus than you might expect. The effect of this is to slightly reduce the energy of electrons in s orbitals. The nearer the nucleus the electrons get, the lower their energy. 3s, 4s (etc) orbitals get progressively further from the nucleus. Not all electrons inhabit s orbitals (in fact, very few electrons live in s orbitals). At the first energy level, the only orbital available to electrons is the 1s orbital, but at the second level, as well as a 2s orbital, there are also orbitals called A p orbital is rather like 2 identical balloons tied together at the nucleus. The diagram on the right is a cross-section through that 3-dimensional region of space. Once again, the orbital shows where there is a 95% chance of finding a particular electron. Unlike an s orbital, a p orbital points in a particular direction - the one drawn points up and down the page. At any one energy level it is possible to have three absolutely equivalent p orbitals pointing mutually at right angles to each other. These are arbitrarily given the symbols and . This is simply for convenience - what you might think of as the x, y or z direction changes constantly as the atom tumbles in space. The p orbitals at the second energy level are called 2p , 2p and 2p . There are similar orbitals at subsequent levels - 3p , 3p , 3p , 4p , 4p , 4p and so on. All levels except for the first level have p orbitals. At the higher levels the lobes get more elongated, with the most likely place to find the electron more distant from the nucleus. Because for the moment we are only interested in the electronic structures of hydrogen and carbon, we do not need to concern ourselves with what happens beyond the second energy level. Remember: Orbitals can be represented as boxes with the electrons in them shown as arrows. Often an up-arrow and a down-arrow are used to show that the electrons are in some way different. A 1s orbital holding 2 electrons would be drawn as shown on the right, but it can be written even more quickly as 1s . This is read as "one s two" - not as "one s squared". You mustn't confuse the two numbers in this notation: Electrons fill low energy orbitals (closer to the nucleus) before they fill higher energy ones. Where there is a choice between orbitals of equal energy, they fill the orbitals singly as far as possible. The diagram (not to scale) summarizes the energies of the various orbitals in the first and second levels. Notice that the 2s orbital has a slightly lower energy than the 2p orbitals. That means that the 2s orbital will fill with electrons before the 2p orbitals. All the 2p orbitals have exactly the same energy. Carbon has six electrons. Two of them will be found in the 1s orbital close to the nucleus. The next two will go into the 2s orbital. The remaining ones will be in two separate 2p orbitals. This is because the p orbitals all have the same energy and the electrons prefer to be on their own if that's the case. The electronic structure of carbon is normally written 1s 2s 2p 2p 0 Jim Clark ( )

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Precipitation reactions occur when cations and anions in aqueous solution combine to form an insoluble ionic solid called a . Whether or not such a reaction occurs can be determined by using the for common ionic solids. Because not all aqueous reactions form precipitates, one must consult the solubility rules before determining the state of the products and writing a The ability to predict these reactions allows scientists to determine which ions are present in a solution, and allows industries to form chemicals by extracting components from these reactions. Precipitates are insoluble ionic solid products of a reaction, formed when certain cations and anions combine in an aqueous solution. The determining factors of the formation of a precipitate can vary. Some reactions depend on temperature, such as solutions used for buffers, whereas others are dependent only on solution concentration. The solids produced in precipitate reactions are crystalline solids, and can be suspended throughout the liquid or fall to the bottom of the solution. The remaining fluid is called (or just the supernate). The two components of the mixture (precipitate and supernate) can be separated by various methods, such as filtration, centrifuging, or decanting. Figure \(\Page {1}\): Above is a diagram of the formation of a precipitate in solution. (Public Domain; ) The use of solubility rules require an understanding of the way that ions react. Most precipitation reactions are single replacement reactions or double replacement reactions. A double replacement reaction occurs when two ionic reactants dissociate and bond with the respective anion or cation from the other reactant. The ions replace each other based on their charges as either a cation or an anion. This can be thought of as a double displacement reaction where the partners "switching; that is, the two reactants each "lose" their partner and form a bond with a different partner: A double replacement reaction is specifically classified as a when the chemical equation in question occurs in aqueous solution and one of the of the products formed is insoluble. An example of a precipitation reaction is given below: \[\ce{CdSO4(aq) + K2S (aq) \rightarrow CdS (s) + K2SO4(aq)}\] Both reactants are aqueous and one product is solid. Because the reactants are ionic and aqueous, they dissociate and are therefore . However, there are six solubility guidelines used to predict which molecules are insoluble in water. These molecules form a solid precipitate in solution. Whether or not a reaction forms a precipitate is dictated by the solubility rules. These rules provide guidelines that tell which ions form solids and which remain in their ionic form in aqueous solution. The rules are to be followed from the top down, meaning that if something is insoluble (or soluble) due to rule 1, it has precedence over a higher-numbered rule. If the rules state that an ion is soluble, then it remains in its aqueous ion form. If an ion is insoluble based on the solubility rules, then it forms a solid with an ion from the other reactant. If all the ions in a reaction are shown to be soluble, then no precipitation reaction occurs. To understand the definition of a , recall the equation for the double replacement reaction. Because this particular reaction is a precipitation reaction, states of matter can be assigned to each variable pair: \[\color{blue}{A}\color{red}{B}\color{black} (aq) + \color{blue}{C}\color{red}{D}\color{black} (aq) → \color{blue}{A}\color{red}{D}\color{black} (aq) \color{black}+ \color{blue}{C}\color{red}{B}\color{black} (s) \] The first step to writing a net ionic equation is to separate the soluble (aqueous) reactants and products into their respective cations and anions. Precipitates do not dissociate in water, so the solid should not be separated. The resulting equation looks like that below: \[\color{blue}{A}^+ \color{black} (aq) + \color{red}{B}\color{black}^- (aq) + \color{blue}{C}\color{black}^+ (aq) + \color{red}{D}^-\color{black} (aq) → \color{blue}{A}^+\color{black} (aq) + \color{red}{D}^-\color{black} (aq) + \color{blue}{C}\color{red}{B}\color{black} (s) \] In the equation above, and ions are present on both sides of the equation. These are called because they remain unchanged throughout the reaction Since they go through the equation unchanged, they can be eliminated to show the : \[ \color{red}{B}\color{black}^- (aq) + \color{blue}{C}\color{black}^+ (aq) → + \color{blue}{C}\color{red}{B}\color{black} (s) \] The net ionic equation only shows the precipitation reaction. A net ionic equation must be balanced on both sides not only in terms of atoms of elements, but also in terms of electric charge. Precipitation reactions are usually represented solely by net ionic equations. If all products are aqueous, a net ionic equation cannot be written because all ions are canceled out as spectator ions. Therefore, The of a salt is the product of the concentrations of the ions in solution raised to the same powers as in the solubility product expression. It is analogous to the reaction quotient ( ) discussed for gaseous equilibria. Whereas describes equilibrium concentrations, the ion product describes concentrations that are not necessarily equilibrium concentrations. An ion product can in principle have positive value, depending on the concentrations of the ions involved. Only in the special case when its value is identical with does it become the solubility product. A solution in which this is the case is said to be . Thus when \[[\ce{Ag^{+}}]^2 [\ce{CrO4^{2-}}] = 2.76 \times 10^{-12}\] at the temperature and pressure at which this value \(K_{sp}\) of applies, we say that the "solution is saturated in silver chromate". The ion product \(Q\) is analogous to the reaction quotient \(Q\) for gaseous equilibria. As summarized in , there are three possible conditions for an aqueous solution of an ionic solid: The process of calculating the value of the ion product and comparing it with the magnitude of the solubility product is a straightforward way to determine whether a solution is unsaturated, saturated, or supersaturated. More important, the ion product tells chemists whether a precipitate will form when solutions of two soluble salts are mixed. Barium sulfate is used in medical imaging of the gastrointestinal tract. Its solubility product is \(1.08 \times 10^{−10}\) at 25°C, so it is ideally suited for this purpose because of its low solubility when a “barium milkshake” is consumed by a patient. The pathway of the sparingly soluble salt can be easily monitored by x-rays. Will barium sulfate precipitate if 10.0 mL of 0.0020 M Na SO is added to 100 mL of 3.2 × 10 M BaCl ? Recall that \(\ce{NaCl}\) is highly soluble in water. and volumes and concentrations of reactants whether precipitate will form The only slightly soluble salt that can be formed when these two solutions are mixed is \(\ce{BaSO4}\) because \(\ce{NaCl}\) is highly soluble. The equation for the precipitation of \(\ce{BaSO4}\) is as follows: \[\ce{BaSO4(s) <=> Ba^{2+} (aq) + SO^{2−}4(aq)} \nonumber\] The solubility product expression is as follows: To solve this problem, we must first calculate the ion product: \[Q = [\ce{Ba^{2+}},\ce{SO4^{2−}}] \nonumber\] using the concentrations of the ions that are present after the solutions are mixed and before any reaction occurs. The concentration of Ba when the solutions are mixed is the total number of moles of Ba in the original 100 mL of \(\ce{BaCl2}\) solution divided by the final volume (100 mL + 10.0 mL = 110 mL): \[ \begin{align*} \textrm{moles Ba}^{2+}=\textrm{100 mL}\left(\dfrac{\textrm{1 L}}{\textrm{1000 mL}}\right)\left(\dfrac{3.2\times10^{-4}\textrm{ mol}}{\textrm{1 L}} \right )=3.2\times10^{-5}\textrm{ mol Ba}^{2+} \\[4pt] [\mathrm{Ba^{2+}}]=\left(\dfrac{3.2\times10^{-5}\textrm{ mol Ba}^{2+}}{\textrm{110 mL}}\right)\left(\dfrac{\textrm{1000 mL}}{\textrm{1 L}}\right)=2.9\times10^{-4}\textrm{ M Ba}^{2+} \end{align*}\] Similarly, the concentration of SO after mixing is the total number of moles of SO in the original 10.0 mL of Na SO solution divided by the final volume (110 mL): \[ \begin{align*} \textrm{moles SO}_4^{2-} &=\textrm{10.0 mL}\left(\dfrac{\textrm{1 L}}{\textrm{1000 mL}}\right)\left(\dfrac{\textrm{0.0020 mol}}{\textrm{1 L}}\right)=2.0\times10^{-5}\textrm{ mol SO}_4^{2-} \\[4pt] [\mathrm{SO_4^{2-}}] &=\left(\dfrac{2.0\times10^{-5}\textrm{ mol SO}_4^{2-}}{\textrm{110 mL}} \right )\left(\dfrac{\textrm{1000 mL}}{\textrm{1 L}}\right)=1.8\times10^{-4}\textrm{ M SO}_4^{2-} \end{align*}\] We can now calculate \( : We now compare \(Q\) with the \(K_{sp}\). If > , then \(\ce{BaSO4}\) will precipitate, but if < , it will not. Because > , we predict that \(\ce{BaSO4}\) will precipitate when the two solutions are mixed. In fact, \(\ce{BaSO4}\) will continue to precipitate until the system reaches equilibrium, which occurs when \[[\ce{Ba^{2+}},\ce{SO4^{2−}}] = K_{sp} = 1.08 \times 10^{−10}. \nonumber\] The solubility product of calcium fluoride (\(\ce{CaF2}\)) is \(3.45 \times 10^{−11}\). If 2.0 mL of a 0.10 M solution of \(\ce{NaF}\) is added to 128 mL of a \(2.0 \times 10^{−5}\,M\) solution of \(\ce{Ca(NO3)2}\), will \(\ce{CaF2}\) precipitate? Yes, since \(Q_{sp} = 4.7 \times 10^{−11} > K_{sp}\). This is a condition for solubility equilibrium, but it is not by itself . True chemical equilibrium can only occur when all components are simultaneously present. A solubility system can be in equilibrium only when some of the solid is in contact with a saturated solution of its ions. Failure to appreciate this is a very common cause of errors in solving solubility problems. If the ion product is smaller than the solubility product, the system is not in equilibrium and no solid can be present. Such a solution is said to be . A solution is one in which the ion product exceeds the solubility product. A supersaturated solution is not at equilibrium, and no solid can ordinarily be present in such a solution. If some of the solid is added, the excess ions precipitate out and until solubility equilibrium is achieved. How to know the saturation status of a solution? Just comparing the ion product with the solubility product . as shown in Table \(\Page {1}\). For example, \[\ce{Ag2CrO4(s) <=> 2 Ag^{+} + CrO_4^{2–}} \label{4ba}\] a solution in which  < (i.e.,  / > 1) is undersaturated (blue shading) and the no solid will be present. The combinations of [Ag ] and [CrO ] that correspond to a saturated solution (and thus to equilibrium) are limited to those described by the curved line. The pink area to the right of this curve represents a supersaturated solution. For some substances, formation of a solid or crystallization does not occur automatically whenever a solution is saturated. These substances have a tendency to form oversaturated solutions. For example, syrup and honey are oversaturated sugar solutions, containing other substances such as citric acids. For oversatureated solutions, is greater than . When a seed crystal is provided or formed, a precipitate will form immediately due to equilibrium of requiring \(Q_{sp}\) to approach \(K_{sp}\). For example Sodium acetate trihydrate, \(\ce{NaCH3COO\cdot 3H2O}\), when heated to 370 K will become a liquid and stays as a liquid when cooled to room temperature or even below 273 K ( \(\Page {1}\)). As soon as a seed crystal is present, crystallization occurs rapidly. In such a process, heat is released since this is an exothermic process \(\Delta H < 0\). A sample of groundwater that has percolated through a layer of gypsum (\(\ce{CaSO4}\)) with \(K_{sp} = 4.9 \times 10^{–5} = 10^{–4.3}\)) is found to have be \(8.4 \times 10^{–5}\; M\) in Ca and \(7.2 \times 10^{–5}\; M\) in SO . What is the equilibrium state of this solution with respect to gypsum? The ion product \[Q_s = (8.4 \times 10^{–5})(7.2 \times 10^{-5}) = 6.0 \times 10^{–4} \nonumber\] exceeds \(K_{sp}\), so the ratio \(K_{sp} /Q_{sp} > 1\) and the solution is in \(\ce{CaSO_4}\). The (by which we usually mean the ) of a solid is expressed as the concentration of the "dissolved solid" in a In the case of a simple 1:1 solid such as AgCl, this would just be the concentration of Ag or Cl in the saturated solution. But for a more complicated stoichiometry such as as silver chromate, the solubility would be only one-half of the Ag concentration. For example, let us denote the solubility of as mol L . Then for a saturated solution, we have Substituting this into Eq 5b above, \[(2S)^2 (S) = 4S^3 = 2.76 \times 10^{–12}\] \[S= \left( \dfrac{K_{sp}}{4} \right)^{1/3} = (6.9 \times 10^{-13})^{1/3} = 0.88 \times 10^{-4} \label{6a}\] thus the solubility is \(8.8 \times 10^{–5}\; M\). Note that the relation between the solubility and the solubility product constant depends on the stoichiometry of the dissolution reaction. For this reason it is meaningless to compare the solubilities of two salts having the formulas A B and AB , say, on the basis of their values. It is to compare the solubilities of two salts having different formulas on the basis of their values. The solubility of CaF (molar mass 78.1) at 18°C is reported to be 1.6 mg per 100 mL of water. Calculate the value of under these conditions. moles of solute in 100 mL; = 0.0016 g / 78.1 g/mol = \(2.05 \times 10^{-5}\) mol \[S = \dfrac{2.05 \times 10^{ –5} mol}{0.100\; L} = 2.05 \times 10^{-4} M\] \[K_{sp}= [Ca^{2+},F^–]^2 = (S)(2S)^2 = 4 × (2.05 \times 10^{–4})^3 = 3.44 \times 10^{–11}\] Estimate the solubility of La(IO ) and calculate the concentration of iodate in equilibrium with solid lanthanum iodate, for which = 6.2 × 10 . The equation for the dissolution is \[La(IO_3)_3 \rightleftharpoons La^{3+ }+ 3 IO_3^–\] If the solubility is , then the equilibrium concentrations of the ions will be [La ] = and [IO ] = 3 . Then = [La ,IO ] = (3 ) = 27 27 = 6.2 × 10 , = ( ( 6.2 ÷ 27) × 10 ) = 6.92 × 10 [IO ] = 3 = 2.08 × 10 Cadmium is a highly toxic environmental pollutant that enters wastewaters associated with zinc smelting (Cd and Zn commonly occur together in ZnS ores) and in some electroplating processes. One way of controlling cadmium in effluent streams is to add sodium hydroxide, which precipitates insoluble Cd(OH) ( = 2.5E–14). If 1000 L of a certain wastewater contains Cd at a concentration of 1.6E–5 , what concentration of Cd would remain after addition of 10 L of 4 NaOH solution? As with most real-world problems, this is best approached as a series of smaller problems, making simplifying approximations as appropriate. Volume of treated water: 1000 L + 10 L = 1010 L Concentration of OH on addition to 1000 L of pure water: (4 ) × (10 L)/(1010 L) = 0.040 Initial concentration of Cd in 1010 L of water: \[(1.6 \times 10^{–5}\; M) \left( \dfrac{100}{101} \right) \approx 1.6 \times 10^{–5}\; M\] The easiest way to tackle this is to start by assuming that a stoichiometric quantity of Cd(OH) is formed — that is, of the Cd gets precipitated. Now "turn on the equilibrium" — find the concentration of Cd that can exist in a 0.04 OH solution: Substitute these values into the solubility product expression: Cd(OH) = [Cd ] [OH ] = 2.5E–14 [Cd ] = (2.5E–14) / (16E–4) = 1.6E–13 Note that the effluent will now be very alkaline: \[pH = 14 + \log 0.04 = 12.6\] so in order to meet environmental standards an equivalent quantity of strong acid must be added to neutralize the water before it is released. It has long been known that the solubility of a sparingly soluble ionic substance is markedly decreased in a solution of another ionic compound when the two substances have an ion in common. This is just what would be expected on the basis of the Le Chatelier Principle; whenever the process \[CaF_{2(s)} \rightleftharpoons Ca^{2+} + 2 F^– \label{7}\] is in equilibrium, addition of more fluoride ion (in the form of highly soluble NaF) will shift the composition to the left, reducing the concentration of Ca , and thus effectively reducing the solubility of the solid. We can express this quantitatively by noting that the \[[Ca^{2+},F^–]^2 = 1.7 \times 10^{–10} \label{8}\] must always hold, even if some of the ionic species involved come from sources other than . For example, if some quantity of fluoride ion is added to a solution initially in equilibrium with solid CaF , we have so that \[K_{sp} = [Ca^{2+}, F^–]^2 = S (2S + x)^2 . \label{9a}\] \[K_{sp} ≈ S x^2 \] \[S = \dfrac{K_{sp}}{x^2} \label{9b}\] The plots shown below illustrate the common ion effect for silver chromate as the chromate ion concentration is increased by addition of a soluble chromate such as Na CrO . What's different about the plot on the right? If you look carefully at the scales, you will see that this one is plotted logarithmically (that is, in powers of 10.) Notice how a much wider a range of values can display on a logarithmic plot. The point of showing this pair of plots is to illustrate the great utility of log-concentration plots in equilibrium calculations in which simple approximations (such as that made in Equation \(\ref{9b}\)) can yield straight-lines within the range of values for which the approximation is valid. Calculate the solubility of strontium sulfate ( = 2.8 × 10 ) in In pure water, = [Sr ,SO ] = = √ = (2.8 × 10 ) = 5.3 × 10 (b) In 0.10 mol L Na SO , we have = [Sr ,SO ] = × (0.10 + ) = 2.8 × 10 Because is negligible compared to 0.10 M, we make the approximation = [Sr ,SO ] ≈ × (0.10 M) = 2.8 × 10 so ≈ (2.8 × 10 ) / 0.10M = 2.8 × 10 — which is roughly 100 times smaller than the result from . The common ion effect usually decreases the solubility of a sparingly soluble salt. Calculate the solubility of calcium phosphate [Ca (PO ) ] in 0.20 M CaCl . concentration of CaCl solution solubility of Ca (PO ) in CaCl solution The balanced equilibrium equation is given in the following table. If we let equal the solubility of Ca (PO ) in moles per liter, then the change in [Ca ] is once again +3 , and the change in [PO ] is +2 . We can insert these values into the ICE table. \[Ca_3(PO_4)_{2(s)} \rightleftharpoons 3Ca^{2+} (aq) + 2PO^{3−}_{4(aq)}\] The expression is as follows: Because Ca (PO ) is a sparingly soluble salt, we can reasonably expect that << 0.20. Thus (0.20 + 3 ) M is approximately 0.20 M, which simplifies the expression as follows: This value is the solubility of Ca (PO ) in 0.20 M CaCl at 25°C. It is approximately nine orders of magnitude less than its solubility in pure water, as we would expect based on Le Chatelier’s principle. With one exception, this example is identical to Example \(\Page {2}\)—here the initial [Ca ] was 0.20 M rather than 0. Calculate the solubility of silver carbonate in a 0.25 M solution of sodium carbonate. The solubility of silver carbonate in pure water is 8.45 × 10 at 25°C. 2.9 × 10 M (versus 1.3 × 10 M in pure water) Precipitation reactions are useful in determining whether a certain element is present in a solution. If a precipitate is formed when a chemical reacts with lead, for example, the presence of lead in water sources could be tested by adding the chemical and monitoring for precipitate formation. In addition, precipitation reactions can be used to extract elements, such as magnesium from seawater. Precipitation reactions even occur in the human body between antibodies and antigens; however, the environment in which this occurs is still being studied. Complete the double replacement reaction and then reduce it to the net ionic equation. \[NaOH (aq) + MgCl_{2 \;(aq)} \rightarrow \nonumber \] First, of this reaction using knowledge of double replacement reactions (remember the cations and anions “switch partners”). \[2NaOH (aq) + MgCl_{2\;(aq)} \rightarrow 2NaCl + Mg(OH)_2 \nonumber\] Second, to determine if the products are soluble. Group 1 cations (\(Na^+\)) and chlorides are soluble from rules 1 and 3 respectively, so \(NaCl\) will be soluble in water. However, rule 6 states that hydroxides are insoluble, and thus \(Mg(OH)_2\) will form a precipitate. The resulting equation is the following: \[2NaOH(aq) + MgCl_{2\;(aq)} \rightarrow 2NaCl (aq) + Mg(OH)_{2\;(s)} \nonumber\] Third, into their ionic forms, as they would exist in an aqueous solution. Be sure to balance both the electrical charge and the number of atoms: \[2Na^+ (aq) + 2OH^- (aq) + Mg^{2+} (aq) + 2Cl^- (aq) \rightarrow Mg(OH)_{2\;(s)} + 2Na^+ (aq) + 2Cl^- (aq) \nonumber\] Lastly, (the ions that occur on both sides of the equation unchanged). In this case, they are the sodium and chlorine ions. The final is: \[Mg^{2+} (aq) + 2OH^- (aq) \rightarrow Mg(OH)_{2(s)} \nonumber\] Complete the double replacement reaction and then reduce it to the net ionic equation. \[CoCl_{2\;(aq)} + Na_2SO_{4\;(aq)} \rightarrow \nonumber\] The predicted products of this reaction are \(CoSO_4\) and \(NaCl\). From the solubility rules, \(CoSO_4\) is soluble because rule 4 states that sulfates (\(SO_4^{2-}\)) are soluble. Similarly, we find that \(NaCl\) is soluble based on rules 1 and 3. After balancing, the resulting equation is as follows: \[CoCl_{2\;(aq)} + Na_2SO_{4\;(aq)} \rightarrow CoSO_{4\;(aq)} + 2 NaCl (aq) \nonumber\] Separate the species into their ionic forms, as they would exist in an aqueous solution. Balance the charge and the atoms. Cancel out all spectator ions (those that appear as ions on both sides of the equation.): → This particular example is important because all of the reactants and the products are aqueous, meaning they cancel out of the net ionic equation. There is no solid precipitate formed; therefore, no precipitation reaction occurs. Write the net ionic equation for the potentially double displacement reactions. Make sure to include the states of matter and balance the equations. a. Regardless of physical state, the products of this reaction are \(Fe(OH)_3\) and \(NaNO_3\). The solubility rules predict that \(NaNO_3\) is soluble because all nitrates are soluble (rule 2). However, \(Fe(OH)_3\) is insoluble, because hydroxides are insoluble (rule 6) and \(Fe\) is not one of the cations which results in an exception. After dissociation, the ionic equation is as follows: \[Fe^{3+} (aq) + NO^-_{3\;(aq)} + Na^+ (aq) + 3OH^- (aq) \rightarrow Fe(OH)_{3\;(s)} + Na^+ (aq) + NO^-_{3\;(aq)} \nonumber\] Canceling out spectator ions leaves the net ionic equation: \[Fe^{3+} (aq) + OH^- (aq) \rightarrow Fe(OH)_{\;3(s)} \nonumber\] b. From the double replacement reaction, the products are \(AlCl_3\) and \(BaSO_4\). \(AlCl_3\) is soluble because it contains a chloride (rule 3); however, \(BaSO_4\) is insoluble: it contains a sulfate, but the \(Ba^{2+}\) ion causes it to be insoluble because it is one of the cations that causes an exception to rule 4. The ionic equation is (after balancing): \[2Al^{3+} (aq) + 6Cl^- (aq) + 3Ba^{2+} (aq) + 3SO^{2-}_{4\;(aq)} \rightarrow 2 Al^{3+} (aq) +6Cl^- (aq) + 3BaSO_{4\;(s)} \nonumber\] Canceling out spectator ions leaves the following net ionic equation: \[Ba^{2+} (aq) + SO^{2-}_{4\;(aq)} \rightarrow BaSO_{4\;(s)} \nonumber\] c. From the double replacement reaction, the products \(HNO_3\) and \(ZnI_2\) are formed. Looking at the solubility rules, \(HNO_3\) is soluble because it contains nitrate (rule 2), and \(ZnI_2\) is soluble because iodides are soluble (rule 3). This means that both the products are aqueous (i.e. dissociate in water), and thus no precipitation reaction occurs. d. The products of this double replacement reaction are \(Ca_3(PO_4)_2\) and \(NaCl\). Rule 1 states that \(NaCl\) is soluble, and according to solubility rule 6, \(Ca_3(PO_4)_2\) is insoluble. The ionic equation is: \[Ca^{2+} (aq) + Cl^- (aq) + Na^+ (aq) + PO^{3-}_{4\;(aq)} \rightarrow Ca_3(PO_4)_{2\;(s)} + Na^+ (aq) + Cl^- (aq) \nonumber \] After canceling out spectator ions, the net ionic equation is given below: \[Ca^{2+} (aq) + PO^{3-}_{4\;(aq)} \rightarrow Ca_3(PO_4)_{2\;(s)} \nonumber\] e. The first product of this reaction, \(PbSO_4\), is soluble according to rule 4 because it is a sulfate. The second product, \(KNO_3\), is also soluble because it contains nitrate (rule 2). Therefore, no precipitation reaction occurs. In contrast to \(K_{sp}\), the ion product (\(Q_{sp}\)) describes concentrations that are not necessarily equilibrium concentrations. Comparing and enables us to determine whether a precipitate will form when solutions of two soluble salts are mixed. Adding a common cation or common anion to a solution of a sparingly soluble salt shifts the solubility equilibrium in the direction predicted by Le Chatelier’s principle. The solubility of the salt is almost always decreased by the presence of a common ion. )

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In measuring optical rotation, plane-polarized light travels down a long tube containing the sample. If it is a liquid, the sample may be placed in the tube as a pure liquid (its is sometimes called a neat sample). Usually, the sample is dissolved in a solvent and the resulting solution is placed in the tube. There are important factors affecting the outcome of the experiment. \[[\alpha] = \dfrac{\alpha}{c l}\] A pure sample of the naturally-occurring, chiral compound A (0.250 g) is dissolved in acetone (2.0 mL) and the solution is placed in a 0.5 dm cell. Three polarimetry readings are recorded with the sample: 0.775 , 0.806 , 0.682 . TBA A pure sample of the (+) enantiomer of compound B shows [a] = 32 . What would be the observed a if a solution of the sample was made by dissolving 0.150 g in 1.0 mL of dichloromethane and was then placed in a 0.5 dm cell? TBA ,

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If you're observant and pay attention to nutrition labels on foods, you may have noticed labels like the one here, where the fats don't seem to add up. If one 12 g serving of Crisco contains 3 g of saturated fat, 0g of trans fat, 6 g of polyunsaturated fat, and 2.5 g of monounsaturated fat what happened to the missing 0.5 g of fat? 3 g + 0 g + 6 g + 2.5 g = 11.5 g! In many cases there's a bigger disparity than this. The fats don't add up because the weight of glycerol is not included in the separately listed components. fatty acids are now recognized as a major dietary risk factor for cardiovascular diseases, and the US FDA has revised food labeling requirements to include trans fats. Are companies pulling the wool over our eyes? In order to understand what's going on, we need to look into the nature of vegetable fats and oils, which are . Vegetable fats and oils are all triglycerides, which contain a glycerol ( ) three carbon "backbone" with 3 long chain attached through ester linkages, as in the figure below. The actual shape is shown in the Jmol model, which can be rotated with the mouse. Triglycerides are called "fats" when they're solids or semisolids, and "oils" when they're liquids. The long chain fatty acids may be with hydrogen atoms, in which case they have all single bonds like the top fatty acid in the Figure (which is palmitic acid). If they have fewer hydrogen atoms, they are and have double bonds like the middle fatty acid in the Figure (which is oleic acid). The bottom fatty acid is , with multiple double bonds (it is linolenic acid). Various cooking oils have As of 2010, Crisco consists of a blend of soybean oil, fully hydrogenated cottonseed oil, and partially hydrogenated soybean and cottonseed oils. Each of these oils is a complex mixture of triglycerides, all with different substituents. (We've discussed the benefits and drawbacks of saturated and unsaturated oils ). , so how can the actual amount of saturated and unsaturated fatty acids be reliably reported? The tryglycerides have to be decomposed into their component fatty acids, and the total amount of each kind (saturated, monounsaturated, and polyunsaturated) reported separately. But this leaves out the glycerol that results from the decomposition, as shown in the equation below for the triglyceride containing 2 palmitic acid (P = C H O ) and 1 linolenic acid (L = C H O ) substituents. The triglyceride can be abbreviated "GPPL" for glycerol (G) with 2 palmitic acid and 1 linolenic acid (L) substituents: \[\ce{(C16H32O2)}\textbf{CH}_\textbf{2}\textbf{CH} \ce{(C16H32O2)}\textbf{CH}_\textbf{2}\ce{-(C18H30O2) + 3 H2O} \xrightarrow{\ce{NaOH}} \ce{1 C3H6O3 + 2 C16H32O2 + 1 C18H30O2} \nonumber \] This is called a reaction, because water causes the decomposition. Sodium hydroxide (NaOH) written above the arrow indicates that NaOH is a catalyst, and isn't consumed or integrated into the products of the chemical reaction. This balanced chemical equation can tell us where the missing mass from the nutrition label has gone. The equation not only tells how many molecules of each kind are involved in a reaction, it also indicates the of each substance that is involved, so it will allow us to find out how much water is consumed and how much glycerol is produced, and those will explain the "missing" masses on the nutrition label. The equation says that 1 GPPL can react with 3 H O to give 1 G , 2 P and 1 L . It also says that 1 GPPL would react with 3 H O yielding 1 G, 2 P, and 1 L. The balanced equation does more than this, though. It also tells us that 2 × 1 = 2 mol GPPL will react with 2 × 3 = 6 mol H O, to form 2 × 1 = 2 mol G, and that ½ × 1 = 0.5 mol GPPL requires only ½ × 3 = 1.5 mol H O. In other words, the equation indicates that exactly 3 mol H O must react 1 mol GPPL consumed, and for every 1 mol GPPL is consumed, 1 mol G, 2 mol P, and 1 mol L will be produced. For the purpose of calculating how much H O is require to react with a certain amount of GPPL, the significant information contained in Eq. (1) is the \[\dfrac {\text {3 mol H}_{\text{2}}{\text{O}}} {\text {1 mol GPPL}} \nonumber \] We shall call such a ratio derived from a balanced chemical equation a and give it the symbol . Thus, for Eq. (1), \(\text{S}\left( \frac{\text{H}_{\text{2}}\text{O}}{\text{GPPL}} \right)~=~\frac{\text{3 mol H}_{\text{2}}\text{O}}{\text{1 mol GPPL}\text{)}}\) The word comes from the Greek words , “element,“ and , “measure.“ Hence the stoichiometric ratio measures one element (or compound) against another. Derive all possible stoichiometric ratios from Eq. (1) Any ratio of amounts of substance given by coefficients in the equation may be used: \[\text{S}\left( \frac{\text{GPPL}}{\text{G}} \right)=\frac{\text{1 mol GPPL}}{\text{1 mol G}}\] \[\text{S}\left( \frac{\text{L}}{\text{P}} \right)=\frac{\text{1 mol L}}{\text{2 mol P}}\] \[\text{S}\left( \frac{\text{GPPL}}{\text{P}} \right)=\frac{\text{1 mol GPPL}}{\text{2 mol P}}\] \[\text{S}\left( \frac{\text{L}}{\text{H}_{\text{2}}\text{O}} \right)=\frac{\text{1 mol L}}{\text{3 mol H}_{\text{2}}\text{O}}\] \[\text{S}\left( \frac{\text{P}}{\text{H}_{\text{2}}\text{O}} \right)=\frac{\text{2 mol P}}{\text{3 mol H}_{\text{2}}\text{O}}\] \[\text{S}\left( \frac{\text{G}}{\text{H}_{\text{2}}\text{O}} \right)=\frac{\text{1 mol G}}{\text{3 mol H}_{\text{2}}\text{O}}\] . Using Eq. (1) as an example, this means that the ratio of the amount of H O consumed to the amount of GPPL consumed must be the stoichiometric ratio S(H O/GPPL): \(\frac{n_{\text{H}_{\text{2}}\text{O}\text{ consumed}}}{n_{\text{GPPL}\text{ consumed}}}\)\(=\text{S}\left( \frac{\text{H}_{\text{2}}\text{O}}{\text{GPPL}} \right)=\frac{\text{3 mol H}_{\text{2}}\text{O}}{\text{1 mol GPPL}}\) Similarly, the ratio of the amount of G produced to the amount of GPPL consumed must be S(G/GPPL): \[\frac{n_{\text{G produced}}}{n_{\text{GPPL}\text{ consumed}}}~=~\text{S}\left( \frac{\text{G}}{\text{GPPL}} \right)=\frac{\text{1 mol G}}{\text{1 mol GPPL}}\] In general we can say that \[\text{Stoichiometric ratio }\left( \frac{\text{X}}{\text{Y}} \right)=\frac{\text{amount of X consumed or produced}}{\text{amount of Y consumed or produced}}\text{ (3}\text{a)}\] or, in symbols, \[\text{S}\left( \frac{\text{X}}{\text{Y}} \right)=\frac{n_{\text{X consumed or produced}}}{n_{\text{Y consumed or produced}}}\text{ (3}\text{b)}\] Note that in the word Eq. (3a) and the symbolic Eq. (3b), X and Y may represent reactant or product in the balanced chemical equation from which the stoichiometric ratio was derived. No matter how much of each reactant we have, the amounts of reactants and the amounts of products will be in appropriate stoichiometric ratios. Find the amount of glycerol produced when 3.68 mol H O is consumed according to Eq. (1). (C H O ) (C H O ) -(C H O ) + 3 H O \(\xrightarrow{\text{NaOH}}\)1 C H O + 2 C H O + 1 C H O The amount of glycerol produced must be in the stoichiometric ratio S(G/H O)to the amount of water consumed: \[\text{S}\left( \frac{\text{G}}{\text{H}_{\text{2}}\text{O}} \right)=\frac{n_{\text{G produced}}}{n_{\text{H}_{\text{2}}\text{O consumed}}}\] Multiplying both sides , by we have \[n_{\text{G}\text{O produced}}=n_{\text{H}_{\text{2}}\text{O}\text{ consumed}}\times \text{S}\left( \frac{\text{G}}{\text{H}_{\text{2}}\text{O}} \right)=\text{3}\text{.68 mol H}_{\text{2}}\text{O}\times \frac{\text{1 mol G}}{\text{3 mol H}_{\text{2}}\text{O}}=\text{1}\text{0.23 mol H}\] This is a typical illustration of the use of a stoichiometric ratio as a conversion factor. Example 2 is analogous to , where density was employed as a conversion factor between mass and volume. Example 2 is also analogous to Examples 2.4 and 2.6, in which the Avogadro constant and molar mass were used as conversion factors. As in these previous cases, there is no need to memorize or do algebraic manipulations with Eq. (3) when using the stoichiometric ratio. Simply remember that the coefficients in a balanced chemical equation give stoichiometric ratios, and that the proper choice results in cancellation of units. In road-map form \(\text{amount of X consumed or produced}\overset{\begin{smallmatrix} \text{stoichiometric} \\ \text{ ratio X/Y} \end{smallmatrix}}{\longleftrightarrow}\text{amount of Y consumed or produced}\) or symbolically. \(n_{\text{X consumed or produced}}\text{ }\overset{S\text{(X/Y)}}{\longleftrightarrow}\text{ }n_{\text{Y consumed or produced}}\) When using stoichiometric ratios, be sure you indicate moles . You can only cancel moles of the same substance. In other words, 1 mol NH cancels 1 mol NH but does not cancel 1 mol H O The next example shows that stoichiometric ratios are also useful in problems involving the mass of a reactant or product. The problem asks that we calculate the mass of glycerol produced. As we learned in , the molar mass can be used to convert from the amount of glycerol to the mass of glycerol. Therefore this problem in effect is asking that we calculate the amount of glycerol produced from the amount of GPPL consumed. This is the same problem as in Example 2. It requires the stoichiometric ratio \[\text{S}\left( \frac{\text{G}}{\text{GPPL}} \right)=\frac{\text{1 mol G}}{\text{1 mol GPPL}}\] The glycerol, G, produced is then \[n_{\text{G produced}}=n_{\text{GPPL consumed}}\text{ }\!\!\times\!\!\text{ conversion factor}=\text{3}\text{.84 mol GPPL}\times \frac{\text{1 mol G}}{\text{1 mol GPPL}}=\text{3}\text{.84 mol G}\] The of glycerol, (G = C H O ) is \(\text{m}_{\text{G}}~=~\text{3}\text{.84 mol G}\times \frac{\text{92}\text{.1 g G}}{\text{1 mol G}}=\text{354 g G}\) With practice this kind of problem can be solved in one step by concentrating on the units. The appropriate stoichiometric ratio will convert moles of O to moles of SO and the molar mass will convert moles of SO to grams of SO . A schematic road map for the one-step calculation can be written as \(n_{\text{G}} ~ \xrightarrow{S\text{(G}\text{/GPPL}\text{)}}~ n_{\text{G}}~\xrightarrow{M_{\text{G}}} ~ m_{\text{G}}\) Thus \(\text{m}_{\text{G}}=\text{3}\text{.84 mol GPPL}\times ~ \frac{\text{1 mol G}}{\text{1 mol GPPL}} ~ \times ~ \frac{\text{92}\text{.1 g}}{\text{1 mol G}}=\text{354 g}\) These calculations can be organized as a table, with entries below the respective reactants and products in the chemical equation. You may verify the additional calculations. b. When we calculate the masses of all reactants and products, we get the results shown in the table. In this case, the total mass of fat is 3185 g. It is a mixed saturated (palmitic acid) and polyunsaturated (linolenic acid) fat, that is degraded to 1969 g of saturated fat and 1069 g of polyunsaturated fat. We assume that no trans fat was produced, and no monounsaturated fatty acids are present. So the label would have numbers proportional to: The components add up to 3038 g, which is less than the total fat because the mass of glycerol (354 g) and water (207.6) are not accounted for. Show that the calculation of the mass of water required to react with 3.68 mol of GPPL in the example above is correct. Symbolically \[n_{\text{GPPL}} ~ \xrightarrow{S\text{(H}_{\text{2}}\text{O}\text{/GPPL}\text{)}} ~ n_{\text{H}_{\text{2}}\text{O}}~\xrightarrow{M_{\text{H}_{\text{2}}\text{O}}} ~ m_{\text{H}_{\text{2}}\text{O}}\] \[3.84 \text{mol GPPL}~\times~\frac{\text{3 mol H}_{\text{2}}\text{O}}{\text{1 mol GPPL}} ~ \times ~ \frac{\text{18}\text{.02 g}}{\text{1 mol H}_{\text{2}}\text{O}}=\text{208 g }\]

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Aldehdyes are oxidized easily by moist silver oxide or by potassium permanganate solution to the corresponding acids. The mechanism of the permanganate oxidation has some resemblance to the chromic acid oxidation of alcohols (Section 15-6B): Many aldehydes are oxidized easily by atmospheric oxygen in a radical-chain mechanism. Oxidation of benzenecarbaldehyde to benzenecarboxylic acid has been studied particularly well and involves formation of a peroxy acid as an intermediate. Reaction is initiated by a radical \(\ce{R} \cdot\) which breaks the relatively weak aldehyde \(\ce{C-H}\) bond \(\left( 86 \: \text{kcal} \right)\). The benzenecarbonyl radical, \(\ce{C_6H_5} \overset{\cdot}{\ce{C}} \ce{O}\), then propagates a chain reaction. The peroxy acid formed then reacts with benzenecarbaldehyde to give two molecules of carboxylic acid: The oxidation of benzenecarbaldehyde with peroxybenzenecarboxylic acid (Equation 16-8) is an example of a reaction of wide applicability in which aldehydes are oxidized to carboxylic acids, and ketones are oxidized to esters. The reaction, which is known as the , has synthetic utility, particularly for the oxidation of ketones to esters because ketones normally are difficult to oxidize without degrading the structure to smaller fragments. Two examples of the Baeyer-Villiger reaction follow: The mechanism of the Baeyer-Villiger oxidation has been studied extensively and is of interest because it involves a rearrangement step in which a substituent group \(\left( \ce{R} \right)\) moves from carbon to oxygen. The reaction sequence is shown in Equations 16-9 through 16-11: In the first step, Equation 16-9, the peroxy acid adds to the carbonyl group. The adduct has several oxygen atoms on which protons can reside, and there will be rapid shifts of protons between these oxygens. However, at some stage the structure will be appropriate to allow elimination of a molecule of carboxylic acid, \(\ce{R'CO_2H}\), Equation 16-10. The resulting intermediate has an electron-deficient oxygen atom with only six valence electrons. As with carbocations and borane complexes (Sections 8-9B, 15-5E, 11-6E, and 16-9D,G), a neighboring \(\ce{R}\) group can move over with its bonding electron-pair to the electron-deficient (oxygen) atom, Equation 16-11. You will notice that for aldehydes, the aldehyde hydrogen migrates in preference to the alkyl or aryl group. In the other examples given, a cycloalkyl migrates in preference to a methyl group, and aryl in preference to methyl. and (1977)

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Most alkenes react readily with ozone \(\left( \ce{O_3} \right)\), even at low temperatures, to yield cyclic peroxidic derivatives known as . For example, These substances, like most compounds with peroxide \(\left( \ce{O-O} \right)\) bonds, may explode violently and unpredictably. Therefore ozonizations must be carried out with appropriate caution. The general importance of these reactions derives not from the ozonides, which usually are not isolated, but from their subsequent products. The ozonides can be converted by hydrolysis with water and reduction, with hydrogen (palladium catalyst) or with zinc and acid, to carbonyl compounds that can be isolated and identified. For example, 2-butene gives ethanal on ozonization, provided the ozonide is destroyed with water and a reducing agent which is effective for hydrogen peroxide: An alternative procedure for decomposing ozonides from di- or trisubstituted alkenes is to treat them with methanol \(\left( \ce{CH_3OH} \right)\). The use of this reagent results in the formation of an aldehyde or ketone and a carboxylic acid: The overall ozonization reaction sequence provides an excellent means for locating the positions of double bonds in alkenes. The potentialities of the method may be illustrated by the difference in reaction products from the 1- and 2-butenes: Ozonization of alkenes has been studied extensively for many years, but there is still disagreement about the mechanism (or mechanisms) involved because some alkenes react with ozone to give oxidation products other than ozonides. It is clear that the ozonide is not formed directly, but by way of an unstable intermediate called a . the molozonide then either isomerizes to the "normal" ozonide or participates in other oxidation reactions. Although the structure of normal ozonides has been established beyond question, that of the molozonide, which is very unstable even at \(-100^\text{o}\), is much less certain. The simplest and most widely accepted mechanism involves formation of a molozonide by a direct of ozone to the double bond.\(^1\) Isomerization of the molozonide appears to occur by a fragmentation-recombination reaction, as shown in Equations 11-7 and 11-8: Several oxidizing reagents react with alkenes under mild conditions to give, as the overall result, addition of hydrogen peroxide as \(\ce{HO-OH}\). Of particular importance are alkaline permanganate \(\left( \ce{MnO_4^-} \right)\) and osmium tetroxide \(\left( \ce{OsO_4} \right)\), both of which react in an initial step by a suprafacial cycloaddition mechanism like that postulated for ozone. Each of these reagents produces -1,2-dihydroxy compounds (diols) with cycloalkenes: Osmium tetroxide is superior to permanganate in giving good yields of diol, but its use is restricted because it is a very costly and very toxic reagent. Alkenes can be oxidized with peroxycarboxylic acids, \(\ce{RCO_3H}\), to give oxacyclopropanes (oxiranes, epoxides), which are three-membered cyclic ethers: The reaction, known as , is valuable because the oxacyclopropane ring is cleaved easily, thereby providing a route to the introduction of many kinds of functional groups. In fact, oxidation of alkenes with peroxymethanoic acid (peroxyformic acid), prepared by mixing methanoic acid and hydrogen peroxide, usually does not stop at the oxacyclopropane stage, but leads to ring-opening and the subsequent formation of a diol: This is an alternative scheme for the hydroxylation of alkenes (see ). However, the overall stereochemistry is opposite to that in permanganate hydroxylation. For instance, cyclopentene gives -1,2-cylcopentanediol. First the oxirane forms by suprafacial addition and then undergoes ring opening to give the trans product: The ring opening is a type of \(S_\text{N}2\) reaction. Methanoic acid is sufficiently acidic to protonate the ring oxygen, which makes it a better leaving group, thus facilitating nucleophilic attack by water. The nucleophile always attacks from the side remote from the leaving group: The peroxyacids that are used in the formation of oxacyclopropanes include peroxyethanoic \(\left( \ce{CH_3CO_3H} \right)\), peroxybenzoic \( \left( \ce{C_6H_5CO_3H} \right)\), and trifluoroperoxyethanoic \(\left( \ce{CF_3CO_3H} \right)\) acids. A particularly useful peroxyacid is 3-chloroperoxybenzoic acid, because it is relatively stable and is handled easily as the crystalline solid. The most reactive reagent is trifluoroperoxyethanoic acid, which suggests that the peroxyacid behaves as an electrophile (the electronegativity of fluorine makes the \(\ce{CF_3}\) group strongly electron-attracting). The overall reaction can be viewed as a , in which the proton on oxygen is transferred to the neighboring carbonyl oxygen more or less simultaneously with formation of the three-membered ring: A reaction of immense industrial importance is the formation of oxacyclopropane itself (most often called ethylene oxide) by oxidation fo ethene with oxygen over a silver oxide catalyst at \(300^\text{o}\): Oxacyclopropane is used for many purposes, but probably the most important reaction is ring opening with water to give 1,2-ethanediol (ethylene glycol, bp \(197^\text{o}\)). This diol, mixed with water, is employed widely in automotive cooling systems to provide both a higher boiling and lower freezing coolant than water alone: Propene and higher alkenes are not efficiently epoxidized by oxygen and \(\ce{Ag_2O}\) in the same way as ethene because of competing attack at other than the double-bond carbons. \(^1\)The ozone structure shown here with single electrons having paired spins on the terminal oxygens accords both with the best available quantum mechanical calculations and the low dipole moment of ozone, which is not consonant with the conventional \(\ce{O=} \overset{\oplus}{\ce{O}} - \overset{\ominus}{\ce{O}}\) structure. See W. A. Goddard III, T. H. Dunning, Jr., W. J. Hunt, and P. J. Hay, , 368 (1973). and (1977)

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Heat and work are both forms of transferring energy, and under the right circumstance, one form may be transformed into the other. However, the second law of thermodynamics puts a limitation on this. To go from work to heat is called dissipation and there is no limitation on this at all. In fact it was through dissipation (by friction) that we discovered that heat and work were both forms of energy. There is, however, a limitation on converting heat to work. Let's consider a circular, reversible path of an ideal gas on a PV diagram: This cycle forms the 4-stage Carnot cycle heat engine. A heat engine converts heat energy into work. The cycle consists of: The total four-step process produces work because \(w_{hot} > w_{cold}\). The work is the integral under the upper isotherm minus the one under the lower curve, i.e.the surface . Sadi Carnot was a French engineer at the beginning of the 19th century. He considered a cyclic process involving a cylinder filled with gas. This cycle the Carnot cycle contributed greatly to the development of thermodynamics and the improvement of the steam engine. Carnot demonstrated that the cold temperature on the right is as important as the heat source on the left in defining the possible efficiency of a heat engine Of course we spend good money on the fuel to start the cycle by heating things up. So how much work do we get for the heat we put in? In other words, we want to know how efficient our heat engine is. The efficiency, \(\eta\) of a heat engine is: \[\eta=\frac{|w_\text{cycle}|}{q_h}=\frac{q_h+q_c}{q_h}=1+\frac{q_c}{q_h} \nonumber \] To get the work of the cycle, we can make use of internal energy as a state function. As the path is circular the circular integrals for \(U\) is zero: \[\begin{align*} \oint{dU} &= \Delta U_{\text{cycle}} \\[4pt] &=\sum{\Delta U_i} \\[4pt] &=w_1+q_h+w_2+w_3+q_c+w_4 \\[4pt] &=0 \end{align*} \] Rearranging: \[\begin{align*} q_h + q_c &=-w_1-w_2-w_3-w_4 \\[4pt] &=-w_\text{cycle} \end{align*} \] An ideal engine would take \(q_h\rightarrow q_c\) with 100% efficiency. The work of the cycle will be equivalent to the heat transfer. For ideal gases: Finding an expression for \(w_\text{cycle}\): \[\begin{split} w_\text{cycle} &= -RT_h\ln{\left(\frac{V_B}{V_A}\right)}+{\bar{C}}_V\left(T_c-T_h\right)-RT_c\ln{\left(\frac{V_D}{V_C}\right)}+{\bar{C}}_V\left(T_h-T_c\right) \\ &= -RT_h\ln{\left(\frac{V_B}{V_A}\right)}-RT_c\ln{\left(\frac{V_D}{V_C}\right)}= -R(T_h-T_c)\ln{\left(\frac{V_B}{V_A}\right)} \end{split} \nonumber \] We have an expression for work, so we can evaluate the efficiency, \(\eta\). The efficiency of the Carnot engine is: \[\eta=\frac{|w_\text{cycle}|}{q_h}=\frac{R\left(T_h-T_c\right)\ln{\left(\frac{V_B}{V_A}\right)}}{RT_h\ln{\left(\frac{V_B}{V_A}\right)}}=\frac{T_h-T_c}{T_h}=1-\frac{T_c}{T_h} \nonumber \] Paths (2) and (4) are adiabats, so we can also use entropy, \(S\), to get the same solution: \[\oint{dS}= \frac{q_h}{T_h}+\frac{q_c}{T_c}=0 \nonumber \] Therefore: \[ \dfrac{q_c}{q_h} = -\dfrac{T_c}{T_h} \nonumber \] And we get that: \[η= 1+ \dfrac{q_c}{q_h} = 1-\dfrac{T_c}{T_h} \label{eff} \] As you see we can only get full efficiency if \(T_{cold}\) is 0 K, which is never (i.e., we always waste energy). Another implication is that if \(T_c = T_h\) then can be obtained, no matter how much energy is available in the from of heat. Or in other words, if one dissipates work into heat isothermally, of it can be retrieved. Equation \(\ref{eff}\) is not very forgiving at all. Just imaging that you have a heat source of say 400 K (a superheated pool of water, e.g. a geyser) and you are dumping in the river at room temperature 300 K. The best efficiency you'll ever get is: \[η= 1-\dfrac{300}{400}= 24\% \nonumber \] Sadly, you'd be dumping three quarters of your energy as heat in the river (and that is best case scenario as there are always more losses, e.g. due to friction). The arrow saying \(Q_C\) in Figure 20.7.3 should then be three times as fat as the one that says \(w\). So getting our work from heat is hard and always less than 100% successful. The other way around should be easy. After all, we can dissipate work into heat freely even under isothermal conditions! What happens if we let the heat engine run backwards? Consider reversing all the flows in the above diagram. Obviously we must to make the cycle run in reverse. The heat will now flow from cold to hot, say from your cold garden into your nice and warm apartment. The amount of heat you get in your humble abode will be the sum of all the work (say 100 Joules) you dissipate the heat you pumped out of the garden (say 300 Joules). Thus if you are willing to pay for the energy you dissipate (100 Joules), you may well end up with a total of 400 Joules of heat in your apartment! Obviously if it is heat you after this is a better deal than just dissipating the work in your apartment (by burning some oil). Then you'd only get 100 J for your precious buck. Dissipating it as electrical heating is even worse because you would Refrigerators are also heat pumps. They heat the kitchen by pumping heat from its innards to the kitchen. If I keep the door open, however, all it does is dissipate precious electrical work, because the pumped heat will flow back into its innards and spoil the milk. Let's consider another type of heat engine, the Stirling engine. The Stirling engine uses a circular reversible path in an ideal PV diagram: The path consists of four steps: Notice that this gray area if \(T_{h}= T_{c}\). Obviously how cold the cold side is of great importance! The amount of work is also equal to the difference in the heat picked up at high temperature \(q_h\) and dumped at low temperature \(q_{c}\). The isochoric heats cancel. The problem is that \(q_{c}\) is only zero if the cold temperature is 0 K. That means that we can never get all the heat we pick up at high temperatures to come out as work.

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Silver nitrate, AgNO , is the least expensive silver salt and is relatively stable to light. It easily dissolves in water (2150 g/L at 20 °C). As the nitrate can be easily replaced by other ligands AgNO is a versatile starting point for the synthesis of other silver compounds. Silver nitrate can be prepared by dissolving silver in with nitric acid: 3 Ag + 4 HNO 3 AgNO + NO + 2 H O When a sheet of copper is put into a silver nitrate solution, the silver nitrate reacts with copper to form hairlike crystals of metallic silver and a blue solution of copper nitrate: 2 AgNO + Cu Cu(NO ) + 2 Ag AgNO reacts with solutions of halide ions to give a precipitate of AgX (X = Cl, Br, I), which are used in photographic films. When heated, silver nitrate decomposes into metallic silver, oxygen and nitrogen oxide: 2 AgNO 2 Ag + O + 2 NO Silver salts have antimicrobial properties and are commonly used to disinfect drinking water. When diluted silver nitrate is braught into contact with skin, the skin becomes brown/black after a short time due to elementary silver which is introduced into the skin according to the following reaction: AgNO + H (from the skin) Ag + HNO Concentrated solutions of AgNO will cause burns due to the same reaction.

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A rearrangement is a reaction in which one molecule undergoes bonding changes, with the transfer of one atom or group from one position in the molecule to another. Proton tautomerism is a kind of rearrangement. A proton is removed from one site in the molecule and put back in a different site nearby. Tautomerism generally requires a couple of proton transfer steps in a row. A proton is removed from one site and then a proton is placed on the new site. (In another variation, a proton is added at one site and then a proton is removed from the old site.) However, rearrangements often involve the concerted transfer of a group from one site to another within the molecule. The group loses its bond to one site and gains its bond to the other site at the same time. A Cope rearrangement happens that way. Figure PR2.1. A Cope rearrangement At first it may not seem like much has happened in this reaction. The two pi bonds have changed position, however, and so has one of the sigma bonds. That means a total of six electrons have moved (two electrons per bond). Figure PR2.2. A diagram of electron movement in a Cope rearrangement. It does not matter which directions you draw the arrows moving in figure PR2.2. They could be shown going clockwise or counterclockwise. There is no electrophile or nucleophile. However, the arrows help with "electron book-keeping." The number of electrons is significant, however. That patterns is reminiscent of benzene. This reaction may be related to the unusual stability of benzene. The transition state for this reaction is considered to be somewhat like benzene. Halfway between one structure and the other, the electrons are delocalized around the ring of atoms. Figure PR2.3. The transition state in a Cope rearrangement. A Cope rearrangement can be considered to occur via a rearrangement of overlap between a group of orbitals around this ring. Two orbitals forming a sigma bond tilt away from each other while two orbitals that are pi bonding tilt toward each other. Figure PR2.4. Orbital rearrangement in a Cope rearrangement Now there are p orbitals parallel to each other on the left, able to form new pi bonds. Many concerted rearrangements can be thought of in terms of these orbital reorganizations. A Claisen rearrangement is very similar to a Cope rearrangement, but oxygen is involved. ,

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. State the balanced equation representing the dissolution of potassium chromate and write its solubility product expression. \[ \mathrm{K_2CrO_4 \rightleftharpoons 2K^+ + CrO_4^{2-} }\nonumber \] \[ \mathrm{K_{sp} = [K^+]^2 [CrO_4^{2-}] }\nonumber \] Consider Mercury(I) bromide. Estimate the concentration of \(Hg_2^{2+}\) and \(Br^-\) given the \(K_{sp}\) as 5.6 x 10 . \[\ce{Hg2Br2 (s) <=> Hg2^{2+} (aq) + 2Br^- (aq)}\] \[K_{sp}=5.6\cdot 10^{-23}=[Hg_{2}^{2+},Br^{-}]^{2}=x(2x)^{2}\] \[\mathrm{x = 2.41 \times 10^{-8} = [Hg_2^{2+}] } \nonumber \] \[\mathrm{[Br^-] = 4.82 \times 10^{-8}}\nonumber \] The solubility product constant of Calcium chlorate (\(Ca(ClO_3)_2\)) of water is 7.1 x 10 at 25 C. How many grams of \(Ca(ClO_3)_2\) were dissolved in 750 mL? First, write out the solubility product constant expression for \(\ce{Ca(ClO3)2}\) \[\ce{ Ca(ClO3)2 (s) <=> Ca^{2+} (aq) + 2 ClO3^- (aq)} \nonumber \] \[\mathrm{K_{sp} = [Ca^{2+},ClO_3^-]^2}\nonumber \] \(Ca(ClO_3)_2\) does not matter because it is a solid \[\mathrm{K_{sp}= [2x]^2[x]}\nonumber \] \[\mathrm{7.1 \times 10^{-7} = 4x^3}\nonumber \] \[\mathrm{x = 0.0056 \; M}\nonumber \] The number of moles of \(Ca^{2-}\) is the same as the number of moles of \(Ca(ClO_3)_2\) because matter cannot be created or destroyed. \[\mathrm{\dfrac{0.0056 \: moles}{L} \times 0.750 \: L = 0.0042 \; moles \; Ca(ClO_3)_2}\nonumber \] Finally, the mass can be calculated by multiplying the number of moles with the molar mass: \[\mathrm{0.0042 \: moles Ca(ClO_3)_2 \times (\dfrac{206.98 \: g}{1 \; mole} ) = 0.872 \: g }\nonumber \] At 25°C, water dissolves 0.8108g of PbCl per liter, calculate the K of PbCl at 25°C. \[PbCl_{2}(s)\rightleftharpoons Pb^{2+}(aq)+2Cl^{-}(aq)\nonumber \] \[K_{sp}=[Pb^{2+}]\cdot [Cl^{-}]^{2}\nonumber \] \[n(PbCl_{2})=\dfrac{0.8108g}{278.1g/mol}=2.9155 \times 10^{-3}mol=n(Pb^{2+})=0.5n(Cl^{-})\nonumber \] \[K_{sp}=\dfrac{2.9155 \times 10^{-3}mol}{1L}\cdot (\dfrac{5.831 \times 10^{-3}mol}{1L})^{2}=9.91 \times 10^{-8}\nonumber \] 0.986635g of \(\ce{BaCO_3}\) is dissolved in 1.00dm water at 80 C. the solution was then cooled down to 25 C. Given that the Ksp of \(\ce{BaCO_3}\) in water is 2.58x10 at 25 C, determine with an explanation whether or not a precipitate will form at 25 C. x moles of BaCO dissolves into x moles of Ba and x moles CO due to equal stoichiometric coefficients (since solution is 1.00L, the moles also equal to molarities). Calculate the moles of BaCO then square the result to find Qsp, then compare Qsp to Ksp at 25 C. \[\mathrm{moles \: of \; BaCO_3 = 0.986635 \; g \times (137.327 \dfrac{g}{mol} + 12 \dfrac{g}{mol} + 3 \times 16 \dfrac{g}{mol}) = 5.0 \times 10^{-3} \; mol }\nonumber \] \[\mathrm{Q_{sp} = (5.0 \times 10^{-3} \; M)^2 = 2.5 \times 10^{-5} }\nonumber \] \[\mathrm{2.5 \times 10^{-5} > 2.58 \times 10^{-9}}\nonumber \] \[\mathrm{Q_{sp} > K_{sp} }\nonumber \] Therefore, precipitate forms. Suppose that you take a solution of 500.0-mL of 0.001234 M \(FeF_2\) and mix it well with a 500.0-mL solution of 0.003142 M \(KOH\) at 25ºC. Given that \(\mathrm{K_{sp} = 4.87 \times 10^{-17} }\) is for \(Fe(OH)_2\), determine whether or not a precipitate will be formed from the mixture. Before starting any calculations it is important to write down the equation for precipitation of the Fe(OH) \[\ce{Fe(OH)2}(s) \rightleftharpoons \ce{Fe^2+}(aq)+\ce{2OH-}(aq)\nonumber \] State the solubility product expression and set it equal to the solubility product constant: \[ K_{\ce sp}={[Fe^2+,OH-]^2}=4.87\times10^{-17}\nonumber \] Now that the essential basic steps are out of the way, find the new concentrations of both \(\ce{Fe^{2+} }\) Fe and OH by first finding the number of moles in the original unmixed solutions and then dividing that number by the total new volume in liters. \[\textrm{moles Fe}^{2+}=\textrm{500} \cancel{mL}\;\left(\dfrac{\textrm{1} \cancel{L}\;}{\textrm{1000} \cancel{mL}\;}\right)\left(\dfrac{0.001234\textrm{ mol}}{\textrm{1} \cancel{L}\;}\right )=6.17\times10^{-4}\textrm{ mol Fe}^{2+}\nonumber \] \[\textrm[{Fe}^{2+}]=\left(\dfrac{6.17\times10^{-4}\textrm{ mol Fe}^{2+}}{\textrm{1000} \cancel{mL}\;}\right)\left(\dfrac{\textrm{1000} \cancel{ mL}\;}{\textrm{1 L}} \right )=6.17\times10^{-4}\textrm{ M Fe}^{2+}\nonumber \] \[\textrm{moles OH-}=\textrm{500} \cancel{mL}\;\left(\dfrac{\textrm{1} \cancel{L}\;}{\textrm{1000} \cancel{mL}\;}\right)\left(\dfrac{0.003142\textrm{ mol}}{\textrm{1} \cancel{L}\;}\right )=1.57\times10^{-3}\textrm{ mol OH-}\nonumber \] \[\textrm[{OH-}]=\left(\dfrac{1.57\times10^{-3}\textrm{ mol OH-}}{\textrm{1000} \cancel{mL}\;}\right)\left(\dfrac{\textrm{1000} \cancel{ mL}\;}{\textrm{1 L}} \right )=1.57\times10^{-3}\textrm{ M OH-}\nonumber \] Now that the new concentrations have been found they can be substituted in the Q equation and solved for. \[ Q_{\ce sp}={[6.17\times10^{-4},1.57\times10^{-3}]^2}=1.52\times10^{-9}\nonumber \] As shown above the \(\mathrm{Q_{sp} > K_{sp}}\), therefore a precipitate does in fact form. \[2AgNO_{3(aq)}+Na_2CO_{3(aq)} \rightleftharpoons Ag_{2}CO_{3(s)}+2NaNO_{3(aq)}\] \[mole_{AgNO_{3}}=mole_{Ag^{+}}=(0.003 \; L)(0.5 \; M)=1.5\times 10^{-3}\ mol\] \[mole_{Na_2CO_{3}}=mole_{CO_{3}^{2-}}=(0.006 \; L)(0.1 \; M)=6\times 10^{-4}\ mol\] \[Total\ Volume = 3 \; mL + 6 \; mL = 9 \; mL = 0.009 \; L\] \[M_{CO_{3}^{2-}}=\dfrac{6\times 10^{-4}mol}{0.009 \; L}=0.0667 \; M\] \[M_{Ag^{+}}=\dfrac{1.5\times 10^{-3}\ mol}{0.009 \; L}=0.1667 \; M\] \[2Ag^{+} +CO_{3}^{2-} \rightarrow Ag_{2}CO_{3(s)}\] \[K_{sp}=8.1\times 10^{-12}=[CO_{3}^{-},Ag^{+}]^{2}\] \[8.1\times 10^{-12}=[0.0667-x ,0.1667-2x]^{2}\] \[x=0.06669999M\] \[[Ag^{+}]=0.1667-2x=0.03336M\] \[[CO_{3}^{2-}]=0.0667-x=7.30\times 10^{-9}\] Your nemesis has prepared a perfectly saturated solution of CaF . To sabotage their work, you add 10 M NaF to their solution. How many grams of CaF fall out of solution if they had prepared 500 mL. \(\mathrm{CaF_2 :K_{sp} = 4.0 \times 10^{-11}} \) First, calculate the molar solubility of CaF from it's K with the expression: \[K_{sp} = (s)(2s)^2= 4s^{3}\nonumber \] \[4.0 \times 10^{-11} = 4s^3\nonumber \] \[s = 2.15 \times 10^{-4}\nonumber \] Then, create an ICE table for the addition of F as NaF: Now, it's reasonable to assume that 10>>x and s, so our expression simplifies to: \(K_{sp} = 10^{2}(s-x)\) Solving for x yields the moles/L which fall out, so multiplying x by 0.5 L (500 mL) yields the correct answer. \[K_{sp} = 4.0 \times 10^{-11} = 10^2(s-x)\nonumber \] \[x = 0.000215 \; M\nonumber \] \[0.5 \; L (0.000215 \; M) (78.07 \; \dfrac{g}{mol}) = 0.00839 \; g\nonumber \] Given copper(II) hydroxide, \(\ce{Cu(OH)_2}\). The concentration of \(\ce{Cu^{2+}}\) and \(\ce{OH^{-}}\) at equilibrium in 25°C water is 1.765 x 10 M and 3.530 x 10 M respectively. \[Cu(OH)_{2(s)}\rightleftharpoons Cu_{(aq)}^{2+}+2OH_{(aq)}^{-}\] 1) Since the solubilities given are already at equilibrium, K is found as such: \[K_{sp}= [Cu^{2+},OH^{-}]^{2}=(1.765 \times 10^{-7}M)(3.530 \times 10^{-7}M)^{2}=2.2 \times 10^{-20}\] 2) Apply the ice table result to the K found in 1), we get \[K_{sp}=2.2\times 10^{-20} =[Cu^{2+},OH^{-}]^{2}=(z)(0.100+2z)^{2}\] Z can be found either by plugging in quadratic formula or finding intersection in a graphing calculator. \[K_{sp}=2.2\times 10^{-20} =(z)(0.100+2z)^{2}\] \[\mathrm{z = 2.2 \times 10^{-18} \; M}\nonumber \] b) due to the common ion effect, \(PbCO_3\) reduces the molar solubility of Barium Carbonate. Determine how will the solubility of the following compounds change (increase, decrease, or unchanged) if the neutral solution is made to be more acidic. Explain your answer in one sentence. As the acidity of a solution increases, the solubility of salts consist of conjugate bases of weak acids will increases. 18-crown-6, or \(\ce{C_12H_24O_6}\), is an organic compound that can bind to group 1 metal ions by wrapping around them when in aqueous solutions. Due to its particular size, it can fit the \(K^+\) ion better than any other. This physical property can be seen in the equilibrium constants for 18-crown-6 and various alkali metals: \(\ce{Na+_{(aq)} + C_12H_24_O_{6(aq)} \rightarrow Na-crown+_{(aq)}}\) K = 6.6 \(\ce{K+_{(aq)} + C_12H_24_O_{6(aq)} \rightarrow K-crown+_{(aq)}}\) K = 111.6 \(\ce{Rb+_{(aq)} + C_12H_24_O_{6(aq)} \rightarrow Rb-crown+_{(aq)}}\) K = 36 If an aqueous solution is made that is 0.006 M in both 18-crown-6 and Rb , what is the concentration of unbound Rb at equilibrium? If the same solution is made, but with K instead of Rb ions, what is the concentration of unbound K+ ions at equilibrium? First, construct an ICE table. \[\ce{Rb+_{aq} + C_12H_24O_{6(aq)} \rightleftharpoons Rb-crown+_{aq}}\nonumber \] Then, use the K value of the reaction to solve for x. Use x to solve for the concentration of the desired ion. \( K = \dfrac{[Rb-crown+]}{[Rb+,C_{12}H_{24}O_6]} \) \( 36 = \dfrac{[x]}{[0.006-x,0.006-x]} \) \( x = 9.2954 \times 10^{-4} \) \( [Rb^+] = 0.0507 M\) Repeat this process for the other reaction. \(\ce{K+_{aq} + C_{12}H_{24}O_{6(aq)} \rightleftharpoons K-crown^+_{aq}}\) \( K = \dfrac{[K-crown^+]}{[K^+,C_{12}H_{24}O_6]} \) \( 111.6 = \dfrac{[x]}{[0.006-x,0.006-x]} \) \( x = 0.0019 \) \( [K^+] = 0.0041 M\) Is it easier to dissolve CuBr into 1M of NaBr solution than to dissolve it into pure water? Does the answer change if when the concentration of NaBr is 0.5M? common ion effect Le Chatelier's principle 2. If the concentration is changed to 0.5M, it is still harder to dissolve CuBr in NaBr solution, but less difficult.   \( \ce{ NH_{4}F} \) can be classified as either acidic or basic depending upon which species is present in a greater concentration ( \( \text{H}^{+} \) or \( \text{OH}^{-} \) ) in the solution when \( \ce{NH_{4}F} \) is added. \( \ce{ NH^{+}_{4(aq)} + F^{-}_{(aq)} + H_2O_{(l)} \rightleftharpoons NH_{3(aq)} + HF_{(aq)} + H_{3}O^{+}_{(aq)} + OH^{-}_{(aq)}} \) \( \ce{ K_a = 5.6 \times 10^{-10}} \) for \( \ce{NH^{+}_{4(aq)} + H_{2}O_{(1)} \rightleftharpoons NH_{3(aq)} + H_{3}O^{+}_{(aq)}} \) \( \ce{ K_b = 1.6 \times 10^{-11}} \) for \( \ce{ F^{-}_{(aq)} + H_{2}O_{(1)} \rightleftharpoons HF_{(aq)} + OH^{-}_{(aq)}} \) \[ [Fe(H_2O)_{6}]^{3+}_{(aq)} + H_2O_{(aq)} \rightleftharpoons [Fe(H_2O)_5OH]^{2+}_{(aq)} + H_3O^{+}_{(aq)} \nonumber \] From the ICE table: \[K_a = \dfrac{[H_3O^{3+},Fe(H_2O)_5OH^{2+}]}{[Fe(H_2O)_6^{3+}]}\nonumber \] \[7.7 \times 10^{-3} = K_{a} = \dfrac{\left[x\right]\left[x\right]}{\left[0.4-x\right]}\nonumber \] By rearranging the equation: \[\mathrm{x^{2}+0.0077x-0.00308=0}\nonumber \] and using the quadratic, \[\mathrm{\dfrac{-0.0077\pm{\sqrt {0.0077^{2}-4\left(1\right)\left(-0.00308\right)}}}{2}}\nonumber \] \[\mathrm{x=0.05178}\nonumber \] \[\mathrm{x = 0.1048}\nonumber \] \[\mathrm{x = 0.000154}\nonumber \] \[\mathrm{[H_3O^+] = 0.1048 + 0.000154 \; M = 0.105 \; M}\nonumber \] An aqueous solution at 25 °C is 0.10 M in both Ba and Ca ions. One wants to separate the two ions by taking advantage of the different solubility of BaCO and CaCO . \(BaCO_3 \; K_{sp} = 2.58 \times 10^{-9} \; M \) \(CaCO_3 \; K_{sp} = 3.36 \times 10^{-9} \; M \) What is the highest possible CO concentration that allows only one salt to present at equilibrium? Which ion is present in the solid, Ba or Ca ? \[\mathrm{BaCO_{3(s)} \leftrightharpoons Ba^{2+}_{(aq)} + CO_{3(aq)}^{2-}}\nonumber \] \[\mathrm{[CO_3^{2-}] < \dfrac{K_{sp}}{[Ba^{2+}]} = \dfrac{2.58 \times 10^{-9}}{0.1 \; M} = 2.58 \times 10^{-8} \; M \; (Using \: the \; formula \; for \; K_{sp})}\nonumber \] \[\mathrm{CaCO_{3(s)} \rightleftharpoons Ca_{(aq)}^{2+} + CO_{3(aq)}^{2-} }\nonumber \] \[\mathrm{[CO_3^{2-}] < \dfrac{K_{sp}}{[Ca^{2+}]} = \dfrac{3.36 \times 10^{-9}}{0.1 \; M} = 3.36 \times 10^{-8} \; M}\nonumber \] Therefore, the highest possible \(CO_3^{2-}\) concentration that allows only one salt to present at equilibrium is \(\mathrm{2.58 \times 10^{-8} \; M}\). \(Ba^{2+}\) is present in the solid. A solution was created by dissolving 1.31 kg of Ba(NO ) into 1 L of 0.50 M Ca(NO ) . Determine which molecule is more soluble. Then determine the concentration of SO ion is needed to precipitate all of one molecule but leave the other molecule completely unreacted, use . First we must determine the mass action equation for dissolving the salts: \[\mathrm{K_{sp} = [Ba^+,SO_4^-] = 1.08 \times 10^{-10} } \nonumber \] \[\mathrm{K_{sp} = [Ca^+,SO_4^-] = 4.93 \times 10^{-5} } \nonumber \]   We will use CaSO as the salt because it is the most soluble. For Ca to to remain in solution, its reaction quotient must remain smaller than then its K : \[\mathrm{Q = [Ca^+,SO_4^-] < 4.93 \times 10^{-5}}\nonumber \] Now we just need to impute the concentration of Ca and solve for the concentration of SO ion: \[\mathrm{[0.5 \; M,SO_4^-] < 4.93 \times 10^{-5}}\nonumber \] \[\mathrm{[SO_4^-] < 9.86 \times 10^{-5} \; M}\nonumber \] So as long as the concentration of SO ion is smaller than 9.86 x 10 M, no CaSO will precipitate Let's say you have a solution that is saturated with \(HF\) at a concentration of \([HF] = 0.20 M\). Part a) Since you have the equation \(XF(s) + H_2O(l) \leftrightharpoons X(aq) + F^-(aq) + OH^-(aq)\), the equilibrium constant K will be equal to \(K = [X] \; [F^-] \; [OH^-] = 2.5 \times 10^{-16}\) We now replace these variables above with known derived equations \(K = [X] \left(\dfrac {[K_a] [HF]}{[H_3O^+]} \right) \left(\dfrac {[K_w]}{[H_3O^+]}\right) = 2.5 \times 10^{-16}\) Plug in known values and solve for \([H_3O^+]\) \([H_3O^+] = \sqrt{\dfrac {(0.20) \; (6.6 \times 10^{-4}) \; (0.20) \; (1.0 \times 10^{-14})}{2.5 \times 10^{-16}}}\) \([H_3O^+] = 0.0325 \; M\) Now we solve for pH and see that \[pH = -log(0.0325) = 1.50\nonumber \] Part b) For part b), we solve the question in a very similar manner as used in part a) with the only differences being a few different variables we plug in. Now, we are solving for our concentration of some element \(G\) using the value of \(H_3O^+\) we found above. Given the reaction \(GF(s) + H_2O(l) \leftrightharpoons G(aq) + F^-(aq) + OH^-(aq)\), the equilibirum constant K will be equal to \(K = [G] \; [F^-] \; [OH^-] = 4.90 \times 10^{-22}\) We now replace these variables above with known derived equations and this time, solve for \([G]\) using the \([H_3O^+]\) value found above \(K = [G] \; (\dfrac {[K_a] [HF]}{[H_3O^+]}) \; (\dfrac {[K_w]}{[H_3O^+]}) = 4.9 \times 10^{-22}\) \([G] = 3.92 \times 10^{-7} \; M\) Because NaCl and AgNO are strong electrolytes, they dissolve in water by dissociation. NaNO precipitates according to the net ionic equation \[Na^{+}_{(aq)}+NO^{-}_{3(aq)}\rightarrow NaNO_{3(s)}\nonumber \] First calculate the moles of each ion present. \[n_{Na^{+}}=0.2L\times 0.5M=0.1 mol Na^{+}\nonumber \] \[n_{NO_{3}^{-}}=0.8L\times 0.1M= 0.08 mol NO_{3}^{-}\nonumber \] Assuming that NaNO is completely insoluble, the reaction continues until the entire 0.08 mol of NO is consumed, which is the limiting reactant, creating 0.08 mol NaNO . \[0.08 \; mol \; NaNO_{3}(84.99\dfrac{g}{mol})=6.80 \; g \; NaNO_{3}\nonumber \] Siobhan has a \(1L\) solution with containing \(0.050M\) of \(CO_{3\, (aq)}^{2-}\) and \(0.100M\) of \(F_{(aq)}^{-}\). She titrates the solution with a 0.100 M titrand of Iron Iodide, \(FeI_{2}\). The \(K_{sp}\) expressions are: \(FeCO_{3\, (s)} \rightleftharpoons CO_{3\, (aq)}^{2-} + Fe_{(aq)}^{2+}\) \(K_{sp}\)= \(3.13\times{10}^{-11}\) \(FeF_{2\, (s)} \rightleftharpoons 2 F_{(aq)}^{-} + Fe_{(aq)}^{2+}\) \(K_{sp}\)= \(2.36\times{10}^{-6}\) At what volume of iron iodide will the precipitate \(FeCO_{3\, (s)}\) appear? How about \(FeF_{2\, (s)}\)? This problem primarily focuses on expanding on how the solubility constant \(K_{sp}\) can predict which solution will titrate first. When Iron Iodide is dripped into the solution, it disassociates into Iron(II) and Iodide Ions. \(FeI_{2} \rightarrow Fe^{2+}_{(aq)} + 2I^-_{aq}\) This facilitates a "common ion effect", where the ions from the dissolution of the titrand Iron Iodide contribute to shifting the concentration of the other two reactions: \(FeCO_{3\, (s)} \rightleftharpoons CO_{3\, (aq)}^{2-} + Fe_{(aq)}^{2+}\) \(FeF_{2\, (s)} \rightleftharpoons 2 F_{(aq)}^{-} + Fe_{(aq)}^{2+}\) As seen here, the introduction of \(Fe^{2+}\) ions will mean there is a larger iron ion concentration, which shifts both of the above reactions to the left. This concept that explains this phenomena is colloquially known as Le Chatlier's principle. However, in order to precisely determine at what point precipitate will form, the \(K_{sp}\) values need to be taken into consideration. Recall that the K constant is defined in terms of either concentrations or partial pressures. As the species in this problem are only aqueous solutes and not gaseous compounds, the resulting \(K_{sp}\) will be defined by concentrations as depicted below: \(FeCO_{3\, (s)} \rightleftharpoons CO_{3\, (aq)}^{2-} + Fe_{(aq)}^{2+}\) \(K_{sp}= \dfrac{\left[CO_{3}^{2-}\right]\left[Fe^{2+}\right]}{1} = 3.13\times{10}^{-11}\) \(FeF_{2\, (s)} \rightleftharpoons 2 F_{(aq)}^{-} + Fe_{(aq)}^{2+}\) \(K_{sp}= \dfrac{\left[F^{-}\right]^{2}\left[Fe^{2+}\right]}{1} = 2.36\times{10}^{-6}\) Notice that each \(K_{sp}\) is divided by 1. This is to represent the solid reactants. Also, based on the information above, recognize that, at equilibrium, the product of the concentrations of the two ions should equal to thier \(K_{sp}\), and since the problem provided both \(\left[CO^{2+}_{3}\right]\) and \(\left[F^{-}\right]\). By substituting these concentrations into the respective \(K_{sp}\) formulas, the equations can be algebraically manipulated to provide the \(\left[Fe^{2+}\right]\) at equilibrium. This process is seen below for \(FeF_{2\, s}\). The process is basically the same to finding \(\left[Fe^{2+}\right]\) for \(\left[CO^{2+}_{3}\right]\), but there is no need to square root the value since there is no stoichiometric coefficients for that particular equation. In contrast, \(F^{-}\) has a stoichiometric coefficient of 2). Since only initial concentrations were given, we can determine at which point \(Q_{sp} > K_{sp}\) \(Q_{sp}= \dfrac{\left[F^{-}\right]^{2}\left[Fe^{2+}\right]}{1} = 2.36\times{10}^{-6}\) \(Q_{sp}= \left(0.05M\right)^{2}\left[Fe^{2+}\right] = 2.36\times{10}^{-6}\) \(\left[Fe^{2+}\right] = \dfrac{2.36\times{10}^{-6}}{\left(0.100M^{2}\right)}= 2.36\times10^{-4}M\) Therefore, \(\left[Fe^{2+}\right]\) for \(FeF_{2}= 2.36\times10^{-4}M\) And \(\left[Fe^{2+}\right]\) for \(FeCO_{3}=6.26\times10^{-10} M\), using the same method. What this information tells us is the calculated \(\left[Fe^{2+}\right]\) is the concentration of \(Fe^{2+}\) that one would expect to be in equilibrium at the given concentration of the other ion. Therefore, if \(\left[Fe^{2+}\right]\) is higher than this value, then the resulting reaction quotient will be larger than \(K_{sp}\), causing the reaction to shift towards the product and consequently causing the precipitate to form. From the values, it is clear that \(FeCO_{3\, (s)}\) will end up precipitating first, and then \(FeF_{2\, s}\). Finding the exact volume that the precipitate for each reaction will form is relatively straightforward. Based on the information overviewed before, after the \(\left[Fe^{2+}\right]\) reaches the calculated point, precipitate will form. Therefore, to find the volume of titrant needed, just set up the following equation: \( molarity_{titrant}\times{V_{titrant}} = mol_{titrant}= mol_{Fe^{2+} to precipitate}\) \(0.100M\times{V_{titrant}}= \left(6.26\times10^{-10}M\right)\left(1L\right)\) \(0.100M\times{V_{titrant}}= 6.26\times10^{-10}mol\) \(V_{titrant}= \dfrac{6.26\times10^{10}mol}{0.100M}\) \(V_{titrant}=6.26\times10^{-9}L\) for \(FeCO_{3\, (s)}\) precipitate to form. Via the same process, \(V_{titrant}=0.00236L\) for \(FeF_{2\, (s)}\) to form. : Use \(K_{sp}\) to determine \(Q_{sp}\) concentration \(\left[Fe^{2+}\right]\). Divide by molarity of titrant to get desired volume for precipitate to form. Complex ion plays an important role in the level of solubility. For example: \[Hg^{2+}_{(aq)} + 2I^-_{(aq)} \rightleftharpoons HgI_{2 (s)}\nonumber \] This reaction results in the formation of solid HgI . However, when solid HgI reacts in the same process: \[HgI_{2(s)} + I^-_{(aq)} \rightleftharpoons HgI^-_{3(aq)}\nonumber \] The solubility increases greatly and the reaction results in an aqueous product. Explain this phenomenon. The first reaction produces a solid precipitate and the second reaction is soluble because complex ions are created in the reaction. \(HgI_3^-\) is a complex ion, which means it has Hg ion in the center that is surrounded by iodine molecules that are able to act as Lewis bases and attract the protons of the Lewis acid in the reaction. These properties are brought out as a result of the reaction and result in an aqueous product. Why does the inclusion of \(NH_3\) increase the solubility of \(CuCl_2\) in an aqueous solution? The solubility of Cu increases when NH present because NH forms a complex ion with Cu . The corresponding the chemical reaction for is as follows, \[\mathrm{CuCl_{2(s)} \rightleftharpoons Cu^{2+}_{(aq)} + 2Cl_{(aq)}^- + 4NH_{3(aq)} \rightleftharpoons [Cu(NH_3)_4]^{2+}}\nonumber \] The formation of the complex ion pulls equilibrium towards the right which allows more CuCl to dissolve.  

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The wavefunctions that describe electrons in atoms and molecules are called orbitals. An orbital is a wavefunction for a single electron. When we say an electron is in orbital \(n\), we mean that it is described by a particular wavefunction \(Ψ_n\) and has energy \(E_n\). All the properties of this electron can be calculated from \(Ψ_n\) as described in Chapter 3. We will now use the particle-in-a-box model to explain the absorption spectra of the cyanine dyes. When an atom or molecule absorbs a photon, the atom or molecule goes from one energy level, designated by quantum number \(n_i\), to a higher energy level, designated by \(n_f\). We can also say that the molecule goes from one electronic state to another. This change is called a transition. Sometimes it is said that an electron goes from one orbital to another in a transition, but this statement is not general. It is valid for a particle-in-a-box, but not for real atoms and molecules, which are more complicated than the simple particle-in-a-box model. The energy of the photon absorbed (\(E_{photon} = h\nu\)) matches the difference in the energy between the two states involved in the transition (\(ΔE_{states}\)). In general, the observed frequency or wavelength for a transition is calculated from the change in energy using the following equalities, \[\Delta E_{states} = E_f - E_i = E_{photon} = h \nu = hc \bar {\nu} \label {4-18}\] Then, for the specific case of the particle-in-a-box, \[ E_{photon} = \Delta E_{states} = E_f - E_i = \frac {(n_{f}^2 - n_{i}^2) h^2}{8mL^2} \label {4-19}\] where \(n_f\) is the quantum number associated with the final state and \(n_i\) is the quantum number for the initial state. A negative value for \(E_{photon}\) means the photon is emitted as a result of the transition in states; a positive value means the photon is absorbed. Generally the transition energy, \(E_{photon}\) or \(ΔE_{states}\), is taken to correspond to the peak in the absorption spectrum. When high accuracy is needed for the electronic transition energy, the spectral line shape must be analyzed to account for rotational and vibrational motion as well as effects due to the solvent or environment. Contributions of rotational and vibrational motion to an absorption spectrum will be discussed in later chapters. In a cyanine dye molecule that has three carbon atoms in the chain, there are six \(π\)-electrons. When light is absorbed, one of these electrons increases its energy by an amount \(hν\) and jumps to a higher energy level. In order to use Equation \ref{4-18}, we need to know which energy levels are involved. We assign the electrons to the lowest energy levels to create the ground-state lowest-energy electron configuration. We could put all six electrons in the \(n = 1\) level, or we could put one electron in each of \(n = 1\) through \(n = 6\), or we could pair the electrons in \(n = 1\) through \(n = 3\), etc. The Pauli Exclusion Principle says that each spatial wavefunction can describe, at most, two electrons, or in other words, that each energy level can have only two electrons assigned to it. Spatial refers to our 3-dimensional space, and a spatial wavefunction depends upon the spatial coordinates x, y, or z. We will discuss the Pauli Exclusion Principle more fully later, but you probably have encountered it in other courses. Rather than appeal to the to assign the electrons to the energy levels, let’s try an empirical approach and discover the Pauli Exclusion Principle as a result. Assign the electrons to the energy levels in different ways until you find an assignment that agrees with experiment. When there is an even number of electrons, the lowest-energy transition is the energy difference between the highest occupied level (HOMO) and the lowest unoccupied level (LUMO). HOMO designates the highest-energy occupied molecular orbital, and LUMO designates the lowest-energy unoccupied molecular orbital. The term orbital refers to the wavefunction or energy level for one electron. All other transitions have a higher energy. For the case with all the electrons in the first energy level, the lowest-energy transition energy would be \(hν = E_2 – E_1\). With one electron in each of the first six levels, \(hν = E_7 – E_6\), and with the electrons paired, \(hν = E_4 – E_3\). Draw energy level diagrams indicating the HOMO, the LUMO, the electrons and the lowest energy transition for each of the three cases mentioned in the preceding paragraph. For the three ways of assigning the 6 electrons to the energy levels in Exercise \(\Page {17}\), calculate the peak absorption wavelength λ for a cyanine dye molecule with 3 carbon atoms in the chain using a value for L of 0.849 nm, which is obtained by estimating bond lengths. Which wavelength agrees most closely with the experimental value of 309 nm for this molecule? It turns out that the assignment that gives a reasonable wavelength for the absorption of a cyanine dye with 6 \(\pi\) electrons is \(hν = E_4 – E_3\) as you concluded from Exercise \(\Page {2}\). In this way we have “discovered” the Pauli Exclusion Principle, electrons should be paired in the same energy level whenever possible, and we accept it for now because it agrees with the experimental observations of the cyanine dye spectra. In molecules with an odd number of electrons, it is possible to have transitions between the doubly occupied molecular orbitals and the singly occupied molecular orbital as well as from the singly occupied orbital to an unoccupied orbital.

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The synthesis of element 114 was reported in January of 1999 by scientists from the Joint Institute for Nuclear Research in Dubna (near Moscow) and Lawrence Livermore National Laboratory (in California). In an experiment lasting more than 40 days Russian scientists bombarded a film of Pu-244 supplied by Livermore scientists with a beam of Ca-48. One atom of element 114 was detected with a half-life of more than 30 seconds. This is about 100,000 times as long as the previously longest-lived isotope of 112 produced (element 113 has yet to be made) and bolsters the arguments of theorists who envision an "island of stability" in the super-heavy elements. In May of 2012 the IUPAC approved the name "Flerovium" (symbol Fl), in honor of the Flerov Laboratory of Nuclear Reactions where superheavy elements are synthesized.

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Antimony and its compounds have been known for centuries. Scientific study of the element began during the early 17th century, much of the important work being done by Nicolas Lemery. The name of the element comes from the Greek anti + monos for "not alone", while the modern symbol is rooted in the Latin-derived name of the common ore, stibnite. Antimony is a hard, brittle metalloid which is alloyed with other metals to increase hardness. It is also used in some semi-conductor devices. The recovery of elemental antimony parallels that of arsenic: the sulfide ore (stibnite) is roasted in air and then heated with carbon. Antimony trisulfide, \(Sb_2S_3\), is a sulfide mineral commonly called or . Antimony trisulfide exists as a gray/black crystalline solid (orthorombic crystals) and an amorphous red-orange powder. It turns black due to oxidation by air. Antimony trisulfide is the most important source for antimony. It is insoluble in water and melts at 550°C. The chemical symbol of antimony (Sb) is derived from stibnite. Amorphous (red to yellow-orange) antimony trisulfide can be prepared by treating an antimony trichloride solution with hydrogen sulfide: \[ 2 SbCl_3 + 3 H_2S \rightarrow Sb_2S_3 + 6 HCl\] When melting antimony trisulfide with iron at approx. 600°C the following reaction yields elementary antimony: \[Sb_2S_3 + 3 Fe \rightarrow Sb + 3 FeS\] \(Sb_2S_3\) is used as a pigment, in pyrotechnics (glitter and fountain mixtures) and on safety matches. In combination with antimony oxides it is also used as a yellow pigment in glass and porcelain. Antimony trisulfide photoconductors are used in vidicons for CCTV.

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The collisional cross section is an "effective area" that quantifies the likelihood of a scattering event when an incident species strikes a target species. In a a hard object approximation, the cross section is the area of the conventional geometric cross section. The collisional cross sections typically denoted σ and measured in units of area. Atoms and molecules can move around in space and bump into each other. If certain conditions of the collision are met, a chemical reaction occurs and a product forms. Sometimes, however, particles may get extremely close to each other but do not strike. We can use the collisional cross section to find how large the distance between two particles must be in order for a collision to occur. A few assumptions must be made: The collisional cross section is defined as the area around a particle in which the center of another particle must be in order for a collision to occur. In the image below, the area within the black circle is molecule A's collisional cross section. That area can change depending on the size of the two particles involved. The collisional cross section \(\sigma_{AA}\) between molecule \(A\) and molecule \(A\) can be calculated using the following equation: \[ \sigma_{AB} = \pi{(r_A + r_B)^2} \nonumber \] Collision occurs when the distance between the center of the two reactant molecules is less than the sum of the radii of these molecules, as shown in Figure \(\Page {2}\). The collisional cross section describes the area around a single reactant. For a collisional reaction to occur, the center of one reactant must be within the collisional cross section of a corresponding reactant. It is first assumed that all particles, whether it be an atom or a molecule, are hard spheres. Particle A must come in contact with Particle B in order for a collision to occur. Particle B can approach particle A from any direction; thus, consider a circle with radius \(r\): \[Area \; of \; a \; Circle = \pi{r}^2 \nonumber \] Assume that the two particles involved in the collision are the same in size and have the same radius. The furthest distance the two centers can be and still have a collision is \(2r\). Substitute \(r\) with \(2r\): \[ \sigma_{AA}= \pi{(2r)}^2 \label{SameEq} \] We will define this area as the collisional cross section. Anytime the center of another particle is within this area, there will be parts of the two particles that will overlap, touch, and cause a collision. Consider the following reaction: \(\ce{H + H -> H_2}\) The radius of hydrogen is \(5.3 \times 10^{-11}\; m\). What is the collisional cross section for this reaction? Use Equation \ref{SameEq} with the given atomic radius for hydrogen atoms: \[\text{Collisional Cross Section}= \pi{(2r)}^2= \pi{[(2)(5.3 \times 10^{-11}\,m)]}^2= 3.53 \times 10^{-20} m^2 \nonumber \] Although the collisional cross section of a particle can be calculated, it is usually not used on its own (Table \(\Page {1}\)). Instead, it is a component of more complex theories such as and . Now what if the particles were of different size and different radii? The \(2r\) term in Equation \(\ref{SameEq}\) is really the sum of the radius of each molecule (i.e., \(2r= r + r\)). However, if the colliding molecules have differing sizes (e.g., \(r_A\) and \(r_B\)) for the radius of particle A and B, respectively, then \(2r\) in Equation \(\ref{SameEq}\) is substituted with \(r_A+r_B\): \[ \sigma_{AB} = \pi{(r_A + r_B)}^2 \nonumber \] What is the collisional cross section for this reaction? \[ H + F \rightarrow HF \nonumber \] The radius of fluorine atom is 4.2 x 10 m. \[\text{ Collisional Cross-Section}= \pi{(r_A+r_B)}^2= \pi{[(5.3 \times 10^{-11})+(4.3 \times 10^{-11})]}^2= 2.90 \times 10^{-20} \,m^2 \nonumber \] \( H_2 + O_2 \rightarrow H_2O \). The radius of oxygen is 4.8 x 10 m. What is the collisional cross section for this reaction? (Hint: Assume that the radius of a molecule is just the sum of its atoms.) The radius of H is 2(r )=1.06 x 10 m. The radius of O is 2(r )=9.6 x 10 m. \[Collisional \; Cross \; Section= \pi{(r_A+r_B)}^2= \pi{[(1.06 \times 10^{-10})+(9.6 \times 10^{-11})]}^2= 1.28 \times 10^{-19} \nonumber \] \( N_2 + O_2 \rightarrow N_2O \). The radius of nitrogen is 5.6 x 10 m. If the distance between the two centers is 2.00 x 10 m, is there a collision between the two molecules? Yes, there is a collision. The radius of N is 2(r )=1.12 x 10 m. The radius of O is 2(r )=9.6 x 10 m. \[Collisional \; Cross \; Section= \pi{(r_A+r_B)}^2= \pi{[(1.12 \times 10^{-10})+(9.6 \times 10^{-11})]}^2= 1.36 \times 10^{-19} \nonumber \] 1.36 x 10 m is the furthest the two molecules can be and still get a collision. Since 2.00 x 10 m is larger than that distance, the center of one molecule is not in the collisional cross section of the other molecule. Therefore, no collision occurs. The molecules are too far apart. \( F + F \rightarrow F_2 \) The center of the two atoms are 3.50 x 10 m apart from each other. How much closer do the centers have to be in order for a collision to occur? \[Collisional \; Cross \; Section= \pi{(2r)}^2= \pi{[(2)(4.2 \times 10^{-11})]}^2= 2.22 \times 10^{-20} \nonumber \] Because the molecules are 3.50 x 10 m apart, the center of one F is not in the Collisional Cross Section of the other F. They must be closer. \[(3.50 \times 10^{-20})-(2.22 \times 10^{-20})=1.28 \times 10^{-20} \nonumber \] The molecules must be at least 1.28 x 10 m closer.​

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Why do reactions take place at different rates? Why do some happen quickly, and others proceed very slowly? Why might the same reaction proceed at different rates under different conditions? There are a number of factors that influence reaction rates, but this article focuses on the activation barrier. An activation barrier is a sort of energetic hurdle that a reaction must bypass. Some reactions have higher hurdles and some have lower hurdles. It is easier to overcome lower hurdles, so reactions with low activation barriers can proceed more quickly than ones with higher activation barriers: A reaction can be exergonic overall, but still have an activation barrier at the beginning. Even if the system decreases in energy by the end of the reaction, it generally experiences an initial increase in energy. This situation is similar to investing in a business. A business generally requires a financial investment to get started. If the business is successful, it will eventually make products and pay money back to the investors. If the business is unable to make back its initial investment, it may fail. Reactions require an initial investment of energy. This energy may come from surrounding molecules or the environment in general. If the reaction is successful, it will proceed to make products and it will emit energy back to its surroundings. All reactions must overcome activation barriers in order to occur. The activation barrier is the sum of the energy that must be expended to get the reaction going. An activation barrier is often pictured as a hill the reactants must climb over during the reaction. Once, there, it can slide down the other side of the hill to become products. At the top of the hill, the molecule exists in what is called the "transition state." At the transition state, the structure is somewhere between its original form and the structure of the products. The type of diagram shown above is sometimes called a "reaction progress diagram." It shows energy changes in the system as a reaction proceeds. One or more activation barriers may exist along the reaction pathways, due to various elementary steps in the reaction. In order to understand more concretely the terms "reaction progress" and "transition state," consider a real reaction. Suppose a nucleophile, such as an acetylide ion, donates its electrons to an electrophilic carbonyl. The π bond breaks and an alkoxide ion is formed. "Reaction progress" refers to how far the reaction has proceeded. The transition state refers specifically to the highest energy point on the pathway from reactants to products. It refers to the structure at that point, and the energy associated with that structure. In the following diagram, the term "reaction progress" has been replaced by an illustration that matches the status of the reaction with the corresponding point in the energy curve. The structure in the square brackets is the transition state, corresponding to the maximum of the curve. The "double dagger" symbol indicates a transition state structure. The transition state is not a true chemical structure. It does not necessarily obey the rules of Lewis structures, because some new bonds have started to form and some old bonds have started to break; partial bonds have no place in a Lewis structure. Physically, the transition state structure cannot be isolated. Because it sits at the top of an energy curve, the transition state tends to convert into something else. A change in either direction will lower its energy. The tendency is to proceed to lowest energy if possible. As soon as the transition state forms, it will either slide back into the original starting materials or slip forward into the final products. ,

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The Jahn-Teller effect is a geometric distortion of a non-linear molecular system that reduces its symmetry and energy. This distortion is typically observed among where the two axial bonds can be shorter or longer than those of the equatorial bonds. This effect can also be observed in . This effect is dependent on the electronic state of the system. In 1937, Hermann Jahn and Edward Teller postulated a theorem stating that "stability and degeneracy are not possible simultaneously unless the molecule is a linear one," in regards to its electronic state.[1] This leads to a break in degeneracy which stabilizes the molecule and by consequence, reduces its symmetry. Since 1937, the theorem has been revised which Housecroft and Sharpe have eloquently phrased as "any non-linear molecular system in a degenerate electronic state will be unstable and will undergo distortion to form a system of lower symmetry and lower energy, thereby removing the degeneracy."[2] This is most commonly observed with transition metal octahedral complexes, however, it can be observed in tetrahedral compounds as well. For a given octahedral complex, the five d atomic orbitals are split into two degenerate sets when constructing a molecular orbital diagram. These are represented by the sets' symmetry labels: \(t_{2g}\) (\(d_{xz}\), \(d_{yz}\), \(d_{xy}\)) and \(e_g\) (\(d_{z^2}\) and \(d_{x^2−y^2}\)). When a molecule possesses a degenerate electronic ground state, it will distort to remove the degeneracy and form a lower energy (and by consequence, lower symmetry) system. The octahedral complex will either elongate or compress the \(z\) ligand bonds as shown in Figure \(\Page {1}\) below: When an octahedral complex exhibits elongation, the axial bonds are longer than the equatorial bonds. For a compression, it is the reverse; the equatorial bonds are longer than the axial bonds. Elongation and compression effects are dictated by the amount of overlap between the metal and ligand orbitals. Thus, this distortion varies greatly depending on the type of metal and ligands. In general, the stronger the metal-ligand orbital interactions are, the greater the chance for a Jahn-Teller effect to be observed. Elongation Jahn-Teller distortions occur when the degeneracy is broken by the stabilization (lowering in energy) of the d orbitals with a component, while the orbitals without a component are destabilized (higher in energy) as shown in Figure \(\Page {2}\) below: This is due to the \(d_{xy}\) and \(d_{x^2−y^2}\) orbitals having greater overlap with the ligand orbitals, resulting in the orbitals being higher in energy. Since the \(d_{x^2−y^2}\) orbital is antibonding, it is expected to increase in energy due to elongation. The \(d_{xy}\) orbital is still nonbonding, but is destabilized due to the interactions. Jahn-Teller elongations are well-documented for copper(II) octahedral compounds. A classic example is that of copper(II) fluoride as shown in Figure \(\Page {3}\). Notice that the two axial bonds are both elongated and the four shorter equatorial bonds are the same length as each other. According the theorem, the orbital degeneracy is eliminated by distortion, making the molecule more stable based on the model presented in Figure \(\Page {2}\). Compression Jahn-Teller distortions occur when the degeneracy is broken by the stabilization (lowering in energy) of the d orbitals a component, while the orbitals with a component are destabilized (higher in energy) as shown in Figure \(\Page {4}\) below: This is due to the -component d orbitals having greater overlap with the ligand orbitals, resulting in the orbitals being higher in energy. For Jahn-Teller effects to occur in transition metals there must be degeneracy in either the t or e orbitals. The electronic states of octahedral complexes depend on the number of d-electrons and the splitting energy, \(\Delta\). When \(\Delta\) is large and is greater than the energy required to pair electrons, electrons pair in t before occupying e . On the other hand, when \(Delta\) small and is less than the pairing energy, electrons will occupy e before pairing in t . The \(\Delta\) of an octahedral complex is dictated by the chemical environment (ligand identity), and the identity and charge of the metal ion. If there electron configurations for any d-electron count is different depending on \(\Delta\), the configuration with more paired electrons is called while the one with more unpaired electrons is called . The electron configurations diagrams for d1 through d10 with large and small \(\delta\) are illustrated in the figures below. Notice that the electron configurations for d , d , d , d , d , and d are the same no matter what the magnitude of \(\Delta\). Low spin and high spin configurations exist only for the electron counts d , d , d , and d . Figure \(\Page {5}\) (below) shows the various electronic configurations for octahedral complexes with large \(\Delta\), including the low-spin configurations of d , d , d , and d : The figure illustrates the electron configurations in the case of large \(\Delta\). The electron configurations highlighted in red (d , low spin d , d , and d ) exhibit Jahn-Teller distortions. On the other hand d , d , low spin d , low spin d , low spin d , and d , would be expected to exhibit Jhan-Teller distortion. These electronic configurations correspond to a variety of transition metals. Some common examples include Cr , Co , and Ni . Figure \(\Page {6}\) (below) shows the various electronic configurations for octahedral complexes with small \(\Delta\), including the high-spin configurations of d , d , d , and d :: The figure illustrates the electron configurations in the case of small \(\Delta\). The electron configurations highlighted in red (d , high spin d , d , and d ) exhibit Jahn-Teller distortions. In general, degenerate electronic states occupying the \(e_g\) orbital set tend to show stronger Jahn-Teller effects. This is primarily caused by the occupation of these high energy orbitals. Since the system is more stable with a lower energy configuration, the degeneracy of the e set is broken, the symmetry is reduced, and occupations at lower energy orbitals occur. Jahn-Teller distortions can be observed using a variety of spectroscopic techniques. In UV-VIS absorption spectroscopy, distortion causes splitting of bands in the spectrum due to a reduction in symmetry (O to D ). Consider a hypothetical molecule with octahedral symmetry showing a single absorption band. If the molecule were to undergo Jahn-Teller distortion, the number of bands would increase as shown in Figure \(\Page {7}\) below: A similar phenomenon can be seen with IR and Raman vibrational spectroscopy. The number of vibrational modes for a molecule can be calculated using the 3n - 6 rule (or 3n - 5 for linear geometry) rule. If a molecule exhibits an O symmetry point group, it will have fewer bands than that of a Jahn-Teller distorted molecule with D symmetry. Thus, one could observe Jahn-Teller effects through either IR or Raman techniques. The Jahn-Teller Theorem predicts that distortions should occur for , including degeneracy of the t level, however distortions in bond lengths are much more distinctive when the degenerate electrons are in the e level.