Chemistry End of Chapter Exercises

1. What is the difference between ΔG, ΔG°, and [latex]\Delta {G}_{298}^{^\circ }[/latex] for a chemical change?
2. A reactions has [latex]\Delta {H}_{298}^{^\circ }[/latex] = 100 kJ/mol and [latex]\Delta {S}_{298}^{^\circ }=\text{250 J/mol\cdot K.}[/latex] Is the reaction spontaneous at room temperature? If not, under what temperature conditions will it become spontaneous?
3. Explain what happens as a reaction starts with ΔG < 0 (negative) and reaches the point where ΔG = 0.
4. Use the standard free energy of formation data in Standard Thermodynamic Properties for Selected Substances to determine the free energy change for each of the following reactions, which are run under standard state conditions and 25 °C. Identify each as either spontaneous or nonspontaneous at these conditions.
   1. [latex]{\text{MnO}}_{2}\left(s\right)\longrightarrow \text{Mn}\left(s\right)+{\text{O}}_{2}\left(g\right)[/latex]
   2. [latex]{\text{H}}_{2}\left(g\right)+{\text{Br}}_{2}\left(l\right)\longrightarrow \text{2HBr}\left(g\right)[/latex]
   3. [latex]\text{Cu}\left(s\right)+\text{S}\left(g\right)\longrightarrow \text{CuS}\left(s\right)[/latex]
   4. [latex]\text{2LiOH}\left(s\right)+{\text{CO}}_{2}\left(g\right)\longrightarrow {\text{Li}}_{2}{\text{CO}}_{3}\left(s\right)+{\text{H}}_{2}\text{O}\left(g\right)[/latex]
   5. [latex]{\text{CH}}_{4}\left(g\right)+{\text{O}}_{2}\left(g\right)\longrightarrow \text{C}\left(s,\text{graphite}\right)+{\text{2H}}_{2}\text{O}\left(g\right)[/latex]
   6. [latex]{\text{CS}}_{2}\left(g\right)+{\text{3Cl}}_{2}\left(g\right)\longrightarrow {\text{CCl}}_{4}\left(g\right)+{\text{S}}_{2}{\text{Cl}}_{2}\left(g\right)[/latex]
5. Use the standard free energy data in Standard Thermodynamic Properties for Selected Substances to determine the free energy change for each of the following reactions, which are run under standard state conditions and 25 °C. Identify each as either spontaneous or nonspontaneous at these conditions.
   1. [latex]\text{C}\left(s\text{, graphite}\right)+{\text{O}}_{2}\left(g\right)\longrightarrow {\text{CO}}_{2}\left(g\right)[/latex]
   2. [latex]{\text{O}}_{2}\left(g\right)+{\text{N}}_{2}\left(g\right)\longrightarrow \text{2NO}\left(g\right)[/latex]
   3. [latex]\text{2Cu}\left(s\right)+\text{S}\left(g\right)\longrightarrow {\text{Cu}}_{2}\text{S}\left(s\right)[/latex]
   4. [latex]\text{CaO}\left(s\right)+{\text{H}}_{2}\text{O}\left(l\right)\longrightarrow \text{Ca}{\left(\text{OH}\right)}_{2}\left(s\right)[/latex]
   5. [latex]{\text{Fe}}_{2}{\text{O}}_{3}\left(s\right)+\text{3CO}\left(g\right)\longrightarrow \text{2Fe}\left(s\right)+{\text{3CO}}_{2}\left(g\right)[/latex]
6. Given:
   [latex]{\text{P}}_{4}\left(s\right)+{\text{5O}}_{2}\left(g\right)\longrightarrow {\text{P}}_{4}{\text{O}}_{10}\left(s\right)\Delta {G}_{298}^{^\circ }=\text{-2697.0 kJ/mol}[/latex]
   [latex]{\text{2H}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\longrightarrow {\text{2H}}_{2}\text{O}\left(g\right)\Delta {G}_{298}^{^\circ }=\text{-457.18 kJ/mol}[/latex]
   [latex]{\text{6H}}_{2}\text{O}\left(g\right)+{\text{P}}_{4}{\text{O}}_{10}\left(g\right)\longrightarrow {\text{4H}}_{3}{\text{PO}}_{4}\left(l\right)\Delta {G}_{298}^{^\circ }=\text{-428.66 kJ/mol}[/latex]
   1. Determine the standard free energy of formation, [latex]\Delta {G}_{\text{f}}^{^\circ }[/latex], for phosphoric acid.
   2. How does your calculated result compare to the value in Standard Thermodynamic Properties for Selected Substances? Explain.
7. Is the formation of ozone (O₃(g)) from oxygen (O₂(g)) spontaneous at room temperature under standard state conditions?
8. Consider the decomposition of red mercury(II) oxide under standard state conditions.
   [latex]\text{2HgO}\left(s,\text{red}\right)\longrightarrow \text{2Hg}\left(l\right)+{\text{O}}_{2}\left(g\right)[/latex]
   1. Is the decomposition spontaneous under standard state conditions?
   2. Above what temperature does the reaction become spontaneous?
9. Among other things, an ideal fuel for the control thrusters of a space vehicle should decompose in a spontaneous exothermic reaction when exposed to the appropriate catalyst. Evaluate the following substances under standard state conditions as suitable candidates for fuels.
   1. Ammonia: [latex]{\text{2NH}}_{3}\left(g\right)\longrightarrow {\text{N}}_{2}\left(g\right)+{\text{3H}}_{2}\left(g\right)[/latex]
   2. Diborane: [latex]{\text{B}}_{2}{\text{H}}_{6}\left(g\right)\longrightarrow \text{2B}\left(g\right)+{\text{3H}}_{2}\left(g\right)[/latex]
   3. Hydrazine: [latex]{\text{N}}_{2}{\text{H}}_{4}\left(g\right)\longrightarrow {\text{N}}_{2}\left(g\right)+{\text{2H}}_{2}\left(g\right)[/latex]
   4. Hydrogen peroxide: [latex]{\text{H}}_{2}{\text{O}}_{2}\left(l\right)\longrightarrow {\text{H}}_{2}\text{O}\left(g\right)+\frac{1}{2}{\text{O}}_{2}\left(g\right)[/latex]
10. Calculate ΔG° for each of the following reactions from the equilibrium constant at the temperature given.
    1. [latex]{\text{N}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\longrightarrow \text{2NO}\left(g\right)\text{T}=2000\text{^\circ C}{K}_{p}=4.1\times {10}^{\text{-4}}[/latex]
    2. [latex]{\text{H}}_{2}\left(g\right)+{\text{I}}_{2}\left(g\right)\longrightarrow \text{2HI}\left(g\right)\text{T}=400\text{^\circ C}{K}_{p}=50.0[/latex]
    3. [latex]{\text{CO}}_{2}\left(g\right)+{\text{H}}_{2}\left(g\right)\longrightarrow \text{CO}\left(g\right)+{\text{H}}_{2}\text{O}\left(g\right)\text{T}=980\text{^\circ C}{K}_{p}=1.67[/latex]
    4. [latex]{\text{CaCO}}_{3}\left(s\right)\longrightarrow \text{CaO}\left(s\right)+{\text{CO}}_{2}\left(g\right)\text{T}=900\text{^\circ C}{K}_{p}=1.04[/latex]
    5. [latex]\text{HF}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\longrightarrow {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{F}}^{\text{-}}\left(aq\right)\text{T}=25\text{^\circ C}{K}_{p}=7.2\times {10}^{\text{-4}}[/latex]
    6. [latex]\text{AgBr}\left(s\right)\longrightarrow {\text{Ag}}^{\text{+}}\left(aq\right)+{\text{Br}}^{\text{-}}\left(aq\right)\text{T}=25\text{^\circ C}{K}_{p}=3.3\times {10}^{\text{-13}}[/latex]
11. Calculate ΔG° for each of the following reactions from the equilibrium constant at the temperature given.
    1. [latex]{\text{Cl}}_{2}\left(g\right)+{\text{Br}}_{2}\left(g\right)\longrightarrow \text{2BrCl}\left(g\right)\text{T}=25\text{^\circ C}{K}_{p}=4.7\times {10}^{\text{-2}}[/latex]
    2. [latex]{\text{2SO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\rightleftharpoons {\text{2SO}}_{3}\left(g\right)\text{T}=500\text{^\circ C}{K}_{p}=48.2[/latex]
    3. [latex]{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{2}\text{O}\left(g\right)\text{T}=60\text{^\circ C}{K}_{p}=\text{0.196 atm}[/latex]
    4. [latex]\text{CoO}\left(s\right)+\text{CO}\left(g\right)\rightleftharpoons \text{Co}\left(s\right)+{\text{CO}}_{2}\left(g\right)\text{T}=550\text{^\circ C}{K}_{p}=4.90\times {10}^{2}[/latex]
    5. [latex]{\text{CH}}_{3}{\text{NH}}_{2}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\longrightarrow {\text{CH}}_{3}{\text{NH}}_{3}{}^{\text{+}}\left(aq\right)+{\text{OH}}^{\text{-}}\left(aq\right)\text{T}=25\text{^\circ C}{K}_{p}=4.4\times {10}^{\text{-4}}[/latex]
    6. [latex]{\text{PbI}}_{2}\left(s\right)\longrightarrow {\text{Pb}}^{2+}\left(aq\right)+{\text{2I}}^{\text{-}}\left(aq\right)\text{T}=25\text{^\circ C}{K}_{p}=8.7\times {10}^{\text{-9}}[/latex]
12. Calculate the equilibrium constant at 25 °C for each of the following reactions from the value of ΔG° given.
    1. [latex]{\text{O}}_{2}\left(g\right)+{\text{2F}}_{2}\left(g\right)\longrightarrow {\text{2OF}}_{2}\left(g\right)\Delta G\text{^\circ }=\text{-9.2 kJ}[/latex]
    2. [latex]{\text{I}}_{2}\left(s\right)+{\text{Br}}_{2}\left(l\right)\longrightarrow \text{2IBr}\left(g\right)\Delta G\text{^\circ }=\text{7.3 kJ}[/latex]
    3. [latex]\text{2LiOH}\left(s\right)+{\text{CO}}_{2}\left(g\right)\longrightarrow {\text{Li}}_{2}{\text{CO}}_{3}\left(s\right)+{\text{H}}_{2}\text{O}\left(g\right)\Delta G\text{^\circ }=\text{-79 kJ}[/latex]
    4. [latex]{\text{N}}_{2}{\text{O}}_{3}\left(g\right)\longrightarrow \text{NO}\left(g\right)+{\text{NO}}_{2}\left(g\right)\Delta G\text{^\circ }=\text{-1.6 kJ}[/latex]
    5. [latex]{\text{SnCl}}_{4}\left(l\right)\longrightarrow {\text{SnCl}}_{4}\left(l\right)\Delta G\text{^\circ }=\text{8.0 kJ}[/latex]
13. Calculate the equilibrium constant at 25 °C for each of the following reactions from the value of ΔG° given.
    1. [latex]{\text{I}}_{2}\left(s\right)+{\text{Cl}}_{2}\left(g\right)\longrightarrow \text{2ICl}\left(g\right)\Delta G\text{^\circ }=\text{-10.88 kJ}[/latex]
    2. [latex]{\text{H}}_{2}\left(g\right)+{\text{I}}_{2}\left(s\right)\longrightarrow \text{2HI}\left(g\right)\Delta G\text{^\circ }=\text{3.4 kJ}[/latex]
    3. [latex]{\text{CS}}_{2}\left(g\right)+{\text{3Cl}}_{2}\left(g\right)\longrightarrow {\text{CCl}}_{4}\left(g\right)+{\text{S}}_{2}{\text{Cl}}_{2}\left(g\right)\Delta G\text{^\circ }=\text{-39 kJ}[/latex]
    4. [latex]{\text{2SO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\longrightarrow {\text{2SO}}_{3}\left(g\right)\Delta G\text{^\circ }=\text{-141.82 kJ}[/latex]
    5. [latex]{\text{CS}}_{2}\left(g\right)\longrightarrow {\text{CS}}_{2}\left(l\right)\Delta G\text{^\circ }=\text{-1.88 kJ}[/latex]
14. Calculate the equilibrium constant at the temperature given.
    1. [latex]{\text{O}}_{2}\left(g\right)+{\text{2F}}_{2}\left(g\right)\longrightarrow {\text{2F}}_{2}\text{O}\left(g\right)\left(\text{T}=100\text{^\circ C}\right)[/latex]
    2. [latex]{\text{I}}_{2}\left(s\right)+{\text{Br}}_{2}\left(l\right)\longrightarrow \text{2IBr}\left(g\right)\left(\text{T}=0.0\text{^\circ C}\right)[/latex]
    3. [latex]\text{2LiOH}\left(s\right)+{\text{CO}}_{2}\left(g\right)\longrightarrow {\text{Li}}_{2}{\text{CO}}_{3}\left(s\right)+{\text{H}}_{2}\text{O}\left(g\right)\left(\text{T}=575\text{^\circ C}\right)[/latex]
    4. [latex]{\text{N}}_{2}{\text{O}}_{3}\left(g\right)\longrightarrow \text{NO}\left(g\right)+{\text{NO}}_{2}\left(g\right)\left(\text{T}=-10.0\text{^\circ C}\right)[/latex]
    5. [latex]{\text{SnCl}}_{4}\left(l\right)\longrightarrow {\text{SnCl}}_{4}\left(g\right)\left(\text{T}=200\text{^\circ C}\right)[/latex]
15. Calculate the equilibrium constant at the temperature given.
    1. [latex]{\text{I}}_{2}\left(s\right)+{\text{Cl}}_{2}\left(g\right)\longrightarrow \text{2ICl}\left(g\right)\left(\text{T}=100\text{^\circ C}\right)[/latex]
    2. [latex]{\text{H}}_{2}\left(g\right)+{\text{I}}_{2}\left(s\right)\longrightarrow \text{2HI}\left(g\right)\left(\text{T}=0.0\text{^\circ C}\right)[/latex]
    3. [latex]{\text{CS}}_{2}\left(g\right)+{\text{3Cl}}_{2}\left(g\right)\longrightarrow {\text{CCl}}_{4}\left(g\right)+{\text{S}}_{2}{\text{Cl}}_{2}\left(g\right)\left(\text{T}=125\text{^\circ C}\right)[/latex]
    4. [latex]{\text{2SO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\longrightarrow {\text{2SO}}_{3}\left(g\right)\left(\text{T}=675\text{^\circ C}\right)[/latex]
    5. [latex]{\text{CS}}_{2}\left(g\right)\longrightarrow {\text{CS}}_{2}\left(l\right)\left(\text{T}=90\text{^\circ C}\right)[/latex]
16. Consider the following reaction at 298 K:
    [latex]{\text{N}}_{2}{\text{O}}_{4}\left(g\right)\rightleftharpoons {\text{2NO}}_{2}\left(g\right){K}_{P}=0.142[/latex]
    What is the standard free energy change at this temperature? Describe what happens to the initial system, where the reactants and products are in standard states, as it approaches equilibrium.
17. Determine the normal boiling point (in kelvin) of dichloroethane, CH₂Cl₂. Find the actual boiling point using the Internet or some other source, and calculate the percent error in the temperature. Explain the differences, if any, between the two values.
18. Under what conditions is [latex]{\text{N}}_{2}{\text{O}}_{3}\left(g\right)\longrightarrow \text{NO}\left(g\right)+{\text{NO}}_{2}\left(g\right)[/latex] spontaneous?
19. At room temperature, the equilibrium constant (K_w) for the self-ionization of water is 1.00 [latex]\times[/latex] 10⁻¹⁴. Using this information, calculate the standard free energy change for the aqueous reaction of hydrogen ion with hydroxide ion to produce water. (Hint: The reaction is the reverse of the self-ionization reaction.)
20. Hydrogen sulfide is a pollutant found in natural gas. Following its removal, it is converted to sulfur by the reaction [latex]{\text{2H}}_{2}\text{S}\left(g\right)+{\text{SO}}_{2}\left(g\right)\rightleftharpoons \frac{3}{8}{\text{S}}_{8}\left(s,\text{rhombic}\right)+{\text{2H}}_{2}\text{O}\left(l\right)[/latex]. What is the equilibrium constant for this reaction? Is the reaction endothermic or exothermic?
21. Consider the decomposition of CaCO₃(s) into CaO(s) and CO₂(g). What is the equilibrium partial pressure of CO₂ at room temperature?
22. In the laboratory, hydrogen chloride (HCl(g)) and ammonia (NH₃(g)) often escape from bottles of their solutions and react to form the ammonium chloride (NH₄Cl(s)), the white glaze often seen on glassware. Assuming that the number of moles of each gas that escapes into the room is the same, what is the maximum partial pressure of HCl and NH₃ in the laboratory at room temperature? (Hint: The partial pressures will be equal and are at their maximum value when at equilibrium.)
23. Benzene can be prepared from acetylene. [latex]{\text{3C}}_{2}{\text{H}}_{2}\left(g\right)\rightleftharpoons {\text{C}}_{6}{\text{H}}_{6}\left(g\right)[/latex]. Determine the equilibrium constant at 25 °C and at 850 °C. Is the reaction spontaneous at either of these temperatures? Why is all acetylene not found as benzene?
24. Carbon dioxide decomposes into CO and O₂ at elevated temperatures. What is the equilibrium partial pressure of oxygen in a sample at 1000 °C for which the initial pressure of CO₂ was 1.15 atm?
25. Carbon tetrachloride, an important industrial solvent, is prepared by the chlorination of methane at 850 K.
    [latex]{\text{CH}}_{4}\left(g\right)+{\text{4Cl}}_{2}\left(g\right)\longrightarrow {\text{CCl}}_{4}\left(g\right)+\text{4HCl}\left(g\right)[/latex]
    What is the equilibrium constant for the reaction at 850 K? Would the reaction vessel need to be heated or cooled to keep the temperature of the reaction constant?
26. Acetic acid, CH₃CO₂H, can form a dimer, (CH₃CO₂H)₂, in the gas phase.
    [latex]{\text{2CH}}_{3}{\text{CO}}_{2}\text{H}\left(g\right)\longrightarrow {\left({\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right)}_{2}\left(g\right)[/latex]
    The dimer is held together by two hydrogen bonds with a total strength of 66.5 kJ per mole of dimer.
    
    At 25 °C, the equilibrium constant for the dimerization is 1.3 [latex]\times[/latex] 10³ (pressure in atm). What is ΔS° for the reaction?
27. Nitric acid, HNO₃, can be prepared by the following sequence of reactions:
    [latex]{\text{4NH}}_{3}\left(g\right)+{\text{5O}}_{2}\left(g\right)\longrightarrow \text{4NO}\left(g\right)+{\text{6H}}_{2}\text{O}\left(g\right)[/latex]
    [latex]\text{2NO}\left(g\right)+{\text{O}}_{2}\left(g\right)\longrightarrow {\text{2NO}}_{2}\left(g\right)[/latex]
    [latex]{\text{3NO}}_{2}\left(g\right)+{\text{H}}_{2}\text{O}\left(l\right)\longrightarrow {\text{2HNO}}_{3}\left(l\right)+\text{NO}\left(g\right)[/latex]
    How much heat is evolved when 1 mol of NH₃(g) is converted to HNO₃(l)? Assume standard states at 25 °C.
28. Determine ΔG for the following reactions.
    1. Antimony pentachloride decomposes at 448 °C. The reaction is: [latex]{\text{SbCl}}_{5}\left(g\right)\longrightarrow {\text{SbCl}}_{3}\left(g\right)+{\text{Cl}}_{2}\left(g\right)[/latex]
       An equilibrium mixture in a 5.00 L flask at 448 °C contains 3.85 g of SbCl₅, 9.14 g of SbCl₃, and 2.84 g of Cl₂.
    2. Chlorine molecules dissociate according to this reaction: [latex]{\text{Cl}}_{2}\left(g\right)\longrightarrow 2\text{Cl}\left(g\right)[/latex]
       1.00% of Cl₂ molecules dissociate at 975 K and a pressure of 1.00 atm.
29. Given that the [latex]\Delta {G}_{\text{f}}^{^\circ }[/latex] for Pb²⁺(aq) and Cl⁻(aq) is −24.3 kJ/mole and −131.2 kJ/mole respectively, determine the solubility product, K_sp, for PbCl₂(s).
30. Determine the standard free energy change, [latex]\Delta {G}_{\text{f}}^{^\circ }[/latex], for the formation of S²⁻(aq) given that the [latex]\Delta {G}_{\text{f}}^{^\circ }[/latex] for Ag⁺(aq) and Ag₂S(s) are 77.1 k/mole and −39.5 kJ/mole respectively, and the solubility product for Ag₂S(s) is 8 [latex]\times[/latex] 10⁻⁵¹.
31. Determine the standard enthalpy change, entropy change, and free energy change for the conversion of diamond to graphite. Discuss the spontaneity of the conversion with respect to the enthalpy and entropy changes. Explain why diamond spontaneously changing into graphite is not observed.
32. The evaporation of one mole of water at 298 K has a standard free energy change of 8.58 kJ.
    [latex]{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{2}\text{O}\left(g\right)\Delta {G}_{298}^{^\circ }=\text{8.58 kJ}[/latex]
    1. Is the evaporation of water under standard thermodynamic conditions spontaneous?
    2. Determine the equilibrium constant, K_P, for this physical process.
    3. By calculating ∆G, determine if the evaporation of water at 298 K is spontaneous when the partial pressure of water, [latex]{P}_{{\text{H}}_{2}\text{O}}[/latex], is 0.011 atm.
    4. If the evaporation of water were always nonspontaneous at room temperature, wet laundry would never dry when placed outside. In order for laundry to dry, what must be the value of [latex]{P}_{{\text{H}}_{2}\text{O}}[/latex] in the air?
33. In glycolysis, the reaction of glucose (Glu) to form glucose-6-phosphate (G6P) requires ATP to be present as described by the following equation: [latex]\text{Glu}+\text{ATP}\longrightarrow \text{G6P}+\text{ADP}\Delta {G}_{298}^{^\circ }=\text{-17 kJ}[/latex]
    In this process, ATP becomes ADP summarized by the following equation: [latex]\text{ATP}\longrightarrow \text{ADP}\Delta {G}_{298}^{^\circ }=\text{-30 kJ}[/latex]
    Determine the standard free energy change for the following reaction, and explain why ATP is necessary to drive this process: [latex]\text{Glu}\longrightarrow \text{G6P}\Delta {G}_{298}^{^\circ }=?[/latex]
34. One of the important reactions in the biochemical pathway glycolysis is the reaction of glucose-6-phosphate (G6P) to form fructose-6-phosphate (F6P): [latex]\text{G6P}\rightleftharpoons \text{F6P}\Delta {G}_{298}^{^\circ }=\text{1.7 kJ}[/latex]
    1. Is the reaction spontaneous or nonspontaneous under standard thermodynamic conditions?
    2. Standard thermodynamic conditions imply the concentrations of G6P and F6P to be 1 M, however, in a typical cell, they are not even close to these values. Calculate ΔG when the concentrations of G6P and F6P are 120 μM and 28 μM respectively, and discuss the spontaneity of the forward reaction under these conditions. Assume the temperature is 37 °C.
35. Without doing a numerical calculation, determine which of the following will reduce the free energy change for the reaction, that is, make it less positive or more negative, when the temperature is increased. Explain.
    1. [latex]{\text{N}}_{2}\left(g\right)+{\text{3H}}_{2}\left(g\right)\longrightarrow {\text{2NH}}_{3}\left(g\right)[/latex]
    2. [latex]\text{HCl}\left(g\right)+{\text{NH}}_{3}\left(g\right)\longrightarrow {\text{NH}}_{4}\text{Cl}\left(s\right)[/latex]
    3. [latex]{\left({\text{NH}}_{4}\right)}_{2}{\text{Cr}}_{2}{\text{O}}_{7}\left(s\right)\longrightarrow {\text{Cr}}_{2}{\text{O}}_{3}\left(s\right)+{\text{4H}}_{2}\text{O}\left(g\right)+{\text{N}}_{2}\left(g\right)[/latex]
    4. [latex]\text{2Fe}\left(s\right)+{\text{3O}}_{2}\left(g\right)\longrightarrow {\text{Fe}}_{2}{\text{O}}_{3}\left(s\right)[/latex]
36. When ammonium chloride is added to water and stirred, it dissolves spontaneously and the resulting solution feels cold. Without doing any calculations, deduce the signs of ΔG, ΔH, and ΔS for this process, and justify your choices.
37. An important source of copper is from the copper ore, chalcocite, a form of copper(I) sulfide. When heated, the Cu₂S decomposes to form copper and sulfur described by the following equation: [latex]{\text{Cu}}_{2}\text{S}\left(s\right)\longrightarrow \text{Cu}\left(s\right)+\text{S}\left(s\right)[/latex]
    1. Determine [latex]\Delta {G}_{298}^{^\circ }[/latex] for the decomposition of Cu₂S(s).
    2. The reaction of sulfur with oxygen yields sulfur dioxide as the only product. Write an equation that describes this reaction, and determine [latex]\Delta {G}_{298}^{^\circ }[/latex] for the process.
    3. The production of copper from chalcocite is performed by roasting the Cu₂S in air to produce the Cu. By combining the equations from Parts (a) and (b), write the equation that describes the roasting of the chalcocite, and explain why coupling these reactions together makes for a more efficient process for the production of the copper.
38. What happens to [latex]\Delta {G}_{298}^{^\circ }[/latex] (becomes more negative or more positive) for the following chemical reactions when the partial pressure of oxygen is increased?
    1. [latex]\text{S}\left(s\right)+{\text{O}}_{2}\left(g\right)\longrightarrow {\text{SO}}_{2}\left(g\right)[/latex]
    2. [latex]{\text{2SO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\longrightarrow {\text{SO}}_{3}\left(g\right)[/latex]
    3. [latex]\text{HgO}\left(s\right)\longrightarrow \text{Hg}\left(l\right)+{\text{O}}_{2}\left(g\right)[/latex]

Selected Answers

2. [latex]\Delta {G}_{298}^{^\circ }=\Delta {H}_{298}^{^\circ }-T\Delta {S}_{298}^{^\circ }[/latex]

[latex]\Delta {G}_{298}^{^\circ }=100 - 298.15\times 250\frac{\text{J}}{\text{mol\cdot K}}\left(\frac{\text{1 kJ}}{\text{1000 J}}\right)=\text{25.5 kJ/mol}[/latex]

The [latex]\Delta {G}_{298}^{^\circ }>0[/latex], so the reaction is nonspontaneous at room temperature.

ΔG = ΔH − TΔS

[latex]0=100-T\left[250\frac{\text{J}}{\text{mol\cdot K}}\left(\frac{\text{1 kJ}}{\text{1000 J}}\right)\right][/latex]

T = 400 K

Above 400 K, ΔG will become negative, and the reaction will become spontaneous.

4. (a) [latex]\Delta {G}_{298}^{^\circ }=\sum \nu \Delta {G}_{\text{f}}^{^\circ }\left(\text{products}\right)-\sum \nu \Delta {G}_{\text{f}}^{^\circ }\left(\text{reactants}\right)[/latex]

[latex]\Delta {G}_{298}^{^\circ }=\left[\text{1 mol}\left(\text{0 kJ/mol}\right)+\text{1 mol}\left(\text{0 kJ/mol}\right)\right]-\left[\text{1 mol}\left(-\text{465.1 kJ/mol}\right)\right]=465.1[/latex] nonspontaneous

(b) [latex]\Delta {G}_{298}^{^\circ }=\sum \nu \Delta {G}_{\text{f}}^{^\circ }\left(\text{products}\right)-\sum \nu \Delta {G}_{\text{f}}^{^\circ }\left(\text{reactants}\right)[/latex]
[latex]\Delta {G}_{298}^{^\circ }=\left[\text{2 mol}\left(-\text{53.43 kJ/mol}\right)\right]-\left[\text{1 mol}\left(\text{0 kJ/mol}\right)+\text{1 mol}\left(\text{0 kJ/mol}\right)\right]=\text{-106.86 kJ}[/latex] spontaneous

(c) [latex]\Delta {G}_{298}^{^\circ }=\sum \nu \Delta {G}_{\text{f}}^{^\circ }\left(\text{products}\right)-\sum \nu \Delta {G}_{\text{f}}^{^\circ }\left(\text{reactants}\right)[/latex]

[latex]\Delta {G}_{298}^{^\circ }=\left[\text{1 mol}\left(-\text{53.6 kJ/mol}\right)\right]-\left[\text{1 mol}\left(\text{0 kJ/mol}\right)+\text{1 mol}\left(\text{0 kJ/mol}\right)\right]=\text{-53.6 kJ}[/latex] spontaneous

(d) [latex]\Delta {G}_{298}^{^\circ }=\sum \nu \Delta {G}_{\text{f}}^{^\circ }\left(\text{products}\right)-\sum \nu \Delta {G}_{\text{f}}^{^\circ }\left(\text{reactants}\right)[/latex]

[latex]\Delta {G}_{298}^{^\circ }=\left[\text{1 mol}\left(-\text{1132.19 kJ/mol}\right)+\text{1 mol}\left(-\text{228.59 kJ/mol}\right)\right]-\left[\text{2 mol}\left(-\text{441.5 kJ/mol}\right)+\text{1 mol}\left(-\text{394.36 kJ/mol}\right)\right]=\text{-83.4 kJ}[/latex] spontaneous

(e) [latex]\Delta {G}_{298}^{^\circ }=\sum \nu \Delta {G}_{\text{f}}^{^\circ }\left(\text{products}\right)-\sum \nu \Delta {G}_{\text{f}}^{^\circ }\left(\text{reactants}\right)[/latex]
[latex]\Delta {G}_{298}^{^\circ }=\left[\text{1 mol}\left(\text{0 kJ/mol}\right)+\text{2 mol}\left(-\text{228.59 kJ/mol}\right)\right]-\left[\text{1 mol}\left(-\text{50.5 kJ/mol}\right)+\text{1 mol}\left(\text{0 kJ/mol}\right)\right]=\text{-406.7 kJ}[/latex] spontaneous

(f) [latex]\Delta {G}_{298}^{^\circ }=\sum \nu \Delta {G}_{\text{f}}^{^\circ }\left(\text{products}\right)-\sum \nu \Delta {G}_{\text{f}}^{^\circ }\left(\text{reactants}\right)[/latex]

[latex]\Delta {G}_{298}^{^\circ }=\left[\text{1 mol}\left(-\text{62.5 kJ/mol}\right)+\text{1 mol}\left(-\text{29.25 kJ/mol}\right)\right]-\left[\text{1 mol}\left(\text{66.8 kJ/mol}\right)+\text{3 mol}\left(\text{0 kJ/mol}\right)\right]=\text{-30.0 kJ}[/latex] spontaneous

6. (a) The standard free energy of formation is the standard free energy change for [latex]\frac{1}{4}{\text{P}}_{4}\left(s\right)+\frac{3}{2}{\text{H}}_{2}\left(g\right)+{\text{2O}}_{2}\left(g\right)\longrightarrow {\text{H}}_{3}{\text{PO}}_{4}\left(l\right)[/latex]. We can use a Hess’s law-like approach. Note that adding the first reaction plus three times the second reaction plus the third reaction gives, after canceling terms:

[latex]{\text{P}}_{4}\left(s\right)+{\text{6H}}_{2}\left(g\right)+{\text{8O}}_{2}\left(g\right)\longrightarrow {\text{4H}}_{3}{\text{PO}}_{4}\left(l\right)[/latex]

[latex]\Delta {G}_{\text{rxn}}^{^\circ }=\left[\left(-2697.0\right)+3\left(-457.18\right)+\left(-428.66\right)\right]\text{kJ/mol}=\text{-4497.2 kJ/mol}[/latex]

Dividing this result by four gives the equation of interest, and it gives −1124.3 kJ/mol for the standard free energy change.

(b) The calculation agrees with the value in Standard Thermodynamic Properties for Selected Substances because free energy is a state function (just like the enthalpy and entropy), so its change depends only on the initial and final states, not the path between them.

8. (a) Using the data in Standard Thermodynamic Properties for Selected Substances, determine [latex]\Delta {G}_{298}^{^\circ }:[/latex]

[latex]\Delta {G}_{298}^{^\circ }=2\Delta {G}_{\text{f}}^{^\circ }\left(\text{Hg}\left(l\right)\right)+\Delta {G}_{\text{f}}^{^\circ }\left({\text{O}}_{2}\left(g\right)\right)-2\Delta {G}_{\text{f}}^{^\circ }\left(\text{HgO}\left(s,\text{red}\right)\right)=\left\{2\left(0\right)+0 - 2\left(-58.5\right)\right\}\text{kJ/mol}=\text{117.0 kJ/mol}[/latex]

From its value at 298.15 K, the reaction is nonspontaneous

(b) requires the ratio of the standard enthalpy change to the standard entropy change:

[latex]\Delta {H}_{298}^{^\circ }=2\Delta {H}_{\text{f}}^{^\circ }\left(\text{Hg}\left(l\right)\right)+\Delta {H}_{\text{f}}^{^\circ }\left({\text{O}}_{2}\left(g\right)\right)-2\Delta {H}_{\text{f}}^{^\circ }\left(\text{HgO}\left(s,\text{red}\right)\right)[/latex]

= [2(0) + 0 − 2(−90.83)] kJ/mol = 181.66 kJ/mol

[latex]\Delta {S}_{298}^{^\circ }=2{S}_{298}^{^\circ }\left(\text{Hg}\left(l\right)\right)+{S}_{298}^{^\circ }\left({\text{O}}_{2}\left(g\right)\right)-2{S}_{298}^{^\circ }\left(\text{HgO}\left(s,\text{red}\right)\right)[/latex]

= [2(75.9) + 205.0 − 2(70.29)] J/K·mol = 216.42 J/K·mol

[latex]T=\frac{\Delta {H}_{298}^{^\circ }}{\Delta {S}_{298}^{^\circ }}=\frac{181.66\times {10}^{3}\text{J/mol}}{\text{216.42 J/K\cdot mol}}=\text{839 K}=566\text{^\circ C}[/latex]

Above 566 °C the process is spontaneous.

10. (a) ΔG° = −RT ln K_P = −(8.314 J K⁻¹)(2273.15 K)(ln 4.1 [latex]\times[/latex] 10⁻⁴) = 147 kJ = 1.5 [latex]\times[/latex] 10² kJ

(b) ΔG° = −(8.314 J K⁻¹)(673.15 K)(ln 50.0) = −21,893 J = 21.9 kJ

(c) ΔG° = −(8.314 J K⁻¹)(1253.15 K)(ln 1.67) = −5.34 kJ

(d) ΔG° = −(8.314 J K⁻¹)(1173.15 K)(ln 1.04) = −0.383 kJ

(e) ΔG° = −(8.314 J K⁻¹)(298.15 K)(ln 7.2 [latex]\times[/latex] 10⁻⁴) = 17,937 J = 18 kJ

(f) ΔG° = −(8.314 J K⁻¹)(298.15 K)(ln 3.3 [latex]\times[/latex] 10⁻¹³) = 71,240 J = 71 kJ

12. Equilibrium constants are calculated from [latex]\text{ln}K=\frac{-\Delta G\text{^\circ }}{RT}[/latex]. Note that K is a function of T and thus changes as T changes.

(a) [latex]\text{ln}K=-\left[\frac{-9.2\times {10}^{3}\text{J}}{\left({\text{8.314 J mol}}^{\text{-1}}{\text{K}}^{\text{-1}}\times \text{298.15 K}\right)}\right]=3.71[/latex], K = 41

(b) [latex]\text{ln}K=\frac{-7300\text{J}}{\left(8.314\text{J}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-1}}\times 298.15\text{K}\right)}=-2.945[/latex], K = 0.053

(c) [latex]\text{ln}K=-\left[\frac{-79\times {10}^{3}\text{J}}{\left({\text{8.314 J mol}}^{\text{-1}}{\text{K}}^{\text{-1}}\times \text{298.15 K}\right)}\right]=31.870[/latex], K = 6.9 [latex]\times[/latex] 10¹³

(d) [latex]\text{ln}K=-\left[\frac{-1.6\times {10}^{3}\text{J}}{\left({\text{8.314 J mol}}^{\text{-1}}{\text{K}}^{\text{-1}}\times \text{298.15 K}\right)}\right]=0.645[/latex], K = 1.9

(e) [latex]\text{ln}K=\frac{\text{-8}\times {10}^{3}\text{J}}{\left(8.314\text{J}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-1}}\times 298.15\text{K}\right)}=-3.227[/latex], K = 0.04

14. In each of the following, the value of ΔG is not given at the temperature of the reaction. Therefore, we must calculate ΔG from the values ΔH° and ΔS and then calculate ΔG from the relation ΔG = ΔH° − TΔS°.

(a) [latex]\begin{array}{l}\\ \Delta H\text{^\circ }=2\Delta {H}_{{\text{f}}_{{\text{OF}}_{2}\left(g\right)}}^{^\circ }-\Delta {H}_{{\text{f}}_{{\text{O}}_{2}\left(g\right)}}^{^\circ }-\Delta {H}_{{\text{f}}_{{\text{F}}_{2}\left(g\right)}}^{^\circ }=\text{2 mol}\left(-22{\text{kJ mol}}^{\text{-1}}\right)-0 - 2\left(0\right)=\text{-44 kJ}=-4.4\times {10}^{4}\text{J}\\ \Delta S\text{^\circ }=2\Delta {S}_{{\text{OF}}_{2}\left(g\right)}^{^\circ }-\Delta {S}_{{\text{O}}_{2}\left(g\right)}^{^\circ }-2\Delta {S}_{{\text{F}}_{2}\left(g\right)}^{^\circ }=2\left(247.3\right)-205.03 - 2\left(202.7\right)=-115.83\text{J}\\ \Delta G=\Delta H\text{^\circ }-T\Delta S\text{^\circ }=\text{-44,000 J}-373.15\left(-115.83\text{J}\right)=\text{-778 J}\\ \text{ln}K=\frac{\Delta G}{-RT}=\frac{-778}{-8.314\times 373.15}=0.2507,K=1.29;\end{array}[/latex]

(b) [latex]\begin{array}{l}\Delta H\text{^\circ }=2\Delta {H}_{{\text{f}}_{\text{IBr}\left(g\right)}}^{^\circ }-\Delta {H}_{{\text{f}}_{{\text{I}}_{2}\left(s\right)}}^{^\circ }-\Delta {H}_{{\text{f}}_{{\text{Br}}_{2}\left(l\right)}}^{^\circ }=2\left(\text{40,800 J}\right)=\text{81,600 J}\\ \Delta S\text{^\circ }=2\Delta {S}_{\text{IBr}\left(g\right)}^{^\circ }-\Delta {S}_{{\text{I}}_{2}\left(s\right)}^{^\circ }-2\Delta {S}_{{\text{Br}}_{2}\left(l\right)}^{^\circ }=2\left(258.66\right)-116.14 - 152.23=248.95\\ \Delta G=81,600 - 273.15\left(248.95\right)=1.36\times {10}^{4}\text{J}\\ \text{ln}K=\frac{13,600}{-8.314\left(273.15\right)}=-5.989,K=2.51\times {10}^{-3};\end{array}[/latex]

(c) [latex]\begin{array}{l}\Delta H\text{^\circ }=2\Delta {H}_{{\text{f}}_{{\text{Li}}_{2}{\text{CO}}_{3}\left(s\right)}}^{^\circ }+\Delta {H}_{{\text{f}}_{{\text{H}}_{2}\text{O}\left(g\right)}}^{^\circ }-2\Delta {H}_{{\text{f}}_{\text{LiOH}\left(g\right)}}^{^\circ }-\Delta {H}_{{\text{f}}_{\text{C}{\text{O}}_{2}\left(g\right)}}^{^\circ }=-1215.6 - 241.82 - 2\left(-487.23\right)-\left(-393.51\right)=\text{-44 kJ}\\ \Delta S\text{^\circ }=\Delta {S}_{{\text{f}}_{{\text{Li}}_{2}{\text{CO}}_{3}\left(s\right)}}^{^\circ }+\Delta {S}_{{\text{f}}_{{\text{H}}_{2}\text{O}\left(g\right)}}^{^\circ }-2\left(\Delta {S}_{{\text{f}}_{\text{LiOH}\left(g\right)}}^{^\circ }\right)-\Delta {S}_{{\text{f}}_{{\text{CO}}_{2}\left(g\right)}}^{^\circ }=90.4+188.71 - 2\left(50.2\right)-213.6=-34.9{\text{J K}}^{\text{-1}}\\ \Delta G=\Delta H\text{^\circ }-T\Delta S\text{^\circ },T=575\text{^\circ C}+273.15=\text{848.15 K}\\ \Delta G=-89,400 - 848.15\left(-34.9\right)=-5.9800\times {10}^{-4}\\ \text{ln}K=\frac{\Delta G}{-RT}=\frac{-5.98\times {10}^{4}}{-8.314\times 848.15}=8.4816,K=4.83\times {10}^{3};\end{array}[/latex]

(d) [latex]\begin{array}{l}\Delta H\text{^\circ }=\Delta {H}_{{\text{f}}_{\text{NO}\left(g\right)}}^{^\circ }+\Delta {H}_{{\text{f}}_{{\text{NO}}_{2}\left(g\right)}}^{^\circ }-\Delta {H}_{{\text{f}}_{{\text{N}}_{2}\text{O}\left(g\right)}}^{^\circ }=90.25+33.2 - 83.72=\text{39.73 kJ}\\ \Delta S\text{^\circ }=\Delta {S}_{\text{NO}\left(g\right)}^{^\circ }+\Delta {S}_{{\text{f}}_{{\text{NO}}_{2}\left(g\right)}}^{^\circ }-\Delta {S}_{{\text{N}}_{2}{\text{O}}_{3}\left(g\right)}^{^\circ }=210.65+239.9 - 312.2={\text{138.35 J K}}^{-1}\\ \Delta G=\Delta H\text{^\circ }-T\Delta S\text{^\circ }=39,730 - 263.15\left(138.35\right)=3323\\ \text{ln}K=\frac{\Delta G}{-RT}=\frac{3323}{-8.314\times 263.15}=-1.519,K=0.219;\end{array}[/latex]

(e) [latex]\begin{array}{l}\Delta H\text{^\circ }=2\Delta {H}_{{\text{f}}_{{\text{SnCl}}_{4}\left(g\right)}}^{^\circ }-\Delta {H}_{{\text{f}}_{{\text{SnCl}}_{4}\left(l\right)}}^{^\circ }=-471.5{\text{kJ mol}}^{\text{-1}}-\left(-511.2{\text{kJ mol}}^{\text{-1}}\right)={\text{39.7 kJ mol}}^{-1}\\ \Delta S\text{^\circ }=\Delta {S}_{{\text{SnCl}}_{4}\left(g\right)}^{^\circ }-{\Delta }_{S}^{{\text{SnCl}}_{4}\left(l\right)}=-366{\text{J K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \Delta G=\Delta H\text{^\circ }-T\Delta S\text{^\circ }=39,700 - 473.15\left(107\right)=\text{-10.927 kJ}\\ \text{ln}K=\frac{-10,927}{-8.314\times 473.15}=2.778,K=16.1\end{array}[/latex]

16. The standard free energy change is [latex]\Delta {G}_{298}^{^\circ }=\text{-}RT\text{ln}K=\text{4.84 kJ/mol}[/latex].

When reactants and products are in their standard states (1 bar or 1 atm), Q = 1. As the reaction proceeds toward equilibrium, the reaction shifts left (the amount of products drops while the amount of reactants increases): Q < 1, and [latex]\Delta {G}_{298}[/latex] becomes less positive as it approaches zero. At equilibrium, Q = K, and ΔG = 0.

18. This reaction will become spontaneous when ΔG goes from positive to negative, or at the point where ΔG = 0. [latex]\Delta G\text{^\circ }=\Delta H\text{^\circ }-T\Delta S\text{^\circ }T=\frac{\Delta H}{\Delta S}[/latex]

Calculate ΔH° and ΔS°:

[latex]\begin{array}{l}\Delta H\text{^\circ }=\Delta {H}_{\text{NO}\left(g\right)}^{^\circ }+\Delta {H}_{{\text{NO}}_{2}\left(g\right)}^{^\circ }-\Delta {H}_{{\text{N}}_{2}{\text{O}}_{3}\left(g\right)}^{^\circ }=\left(90.25+33.2 - 83.72\right)\text{kJ}=\text{39.73 kJ}\\ \Delta S\text{^\circ }={S}_{\text{NO}\left(g\right)}^{^\circ }+{S}_{{\text{NO}}_{2}\left(g\right)}^{^\circ }-{S}_{{\text{N}}_{2}{\text{O}}_{3}\left(g\right)}^{^\circ }=\left(210.65+239.9 - 312.2\right){\text{J K}}^{\text{-1}}=138.35{\text{J K}}^{\text{-1}}\\ \Delta G\text{^\circ }=\Delta H\text{^\circ }-T\Delta S\text{^\circ }=\text{39.73 kJ}-\left(\text{298.15 K}\right)\left({\text{0.13835 kJ K}}^{\text{-1}}\right)=\text{-1.519 kJ}\end{array}[/latex]

Thus this reaction is spontaneous at 298 K because ΔG is negative at this temperature, and because TΔS° term is negative, ΔG will become progressively more negative as the temperature increases above 298 K. Thus the temperature at which ΔG first becomes negative must lie at a temperature lower than 298 K. We can find this temperature by substituting into [latex]T=\frac{\Delta H}{\Delta S}[/latex].

[latex]T=\frac{\Delta H}{\Delta S}=\frac{\text{39,700 J}}{{\text{138.35 J K}}^{\text{-1}}}=\text{287 K}[/latex]

The reaction will be spontaneous at temperatures greater than 287 K.

20. Find ΔG° for the reaction, and from ΔG° find K.

[latex]\begin{array}{l}\Delta G\text{^\circ }=3\Delta {G}_{\text{S}\left(s\right)}^{^\circ }+2\Delta {G}_{{\text{H}}_{2}\text{O}\left(g\right)}^{^\circ }-2\Delta {G}_{{\text{H}}_{2}\text{S}\left(g\right)}^{^\circ }-\Delta {G}_{{\text{SO}}_{2}\left(s\right)}^{^\circ }=3\left(0\right)+2\left(-228.59\right)-2\left(-33.6\right)-\left(-300.19\right)=\text{-89.79 kJ}\\ \Delta G\text{^\circ }=\text{-}RT\text{ln}K=-2.303RT\text{log}K\\ \\ K=\text{antilog}\left(\frac{\text{-\Delta }G}{2.303RT}\right)=\left[\frac{89,790}{2.303\left(8.314\right)\left(298.15\right)}\right]=\text{antilog}15.729=5.35\times {10}^{15}\end{array}[/latex]

The process is exothermic.

22. The equilibrium may be written as

[latex]{\text{NH}}_{4}\text{Cl}\left(s\right)\rightleftharpoons \text{HCl}\left(g\right)+{\text{NH}}_{3}\left(g\right){K}_{P}={P}_{{\text{NH}}_{3}}{P}_{\text{HCl}}={x}^{2}[/latex]

to help simplify the mathematics. Use the standard free energy change to determine the equilibrium constant for the reaction: [latex]\Delta {G}_{298}^{^\circ }=\text{91.1 kJ/mol}[/latex], K_P = 1.1 [latex]\times[/latex] 10⁻¹⁶. At equilibrium, the partial pressures are equal so that [latex]x=\sqrt{{K}_{P}}=1.0\times {10}^{\text{-8}}\text{atm}[/latex]. This is the maximum pressure of the gases under the stated conditions.

24. The reaction is [latex]{\text{2CO}}_{2}\left(g\right)\longrightarrow \text{2CO}\left(g\right)+{\text{O}}_{2}\left(g\right)[/latex].

ΔH° = (2 mol)(−110.52 kJ mol⁻¹) + (1 mol)(0) − (2 mol)(−393.51 kJ mol⁻¹) = 565.98 kJ;

ΔS° = (2 mol)(197.56 J K⁻¹ mol⁻¹) + (1 mol)(205.03 J K⁻¹ mol⁻¹) − (2 mol)(−393.51 kJ mol⁻¹) = 172.95 J K⁻¹;

ΔG° = (565,980 J) − (1273 K)(172.95 J K⁻¹) = 345,810 J;

[latex]\begin{array}{l}\\ \\ \text{log}K=\frac{\text{345,810 J}}{\left(-2.303\right)\left(8.314{\text{J K}}^{\text{-1}}\right)\left(1273K\right)}=-14.1975\\ \\ \\ K=6.494\times {10}^{-15}=\frac{{\left({P}_{\text{CO}}\right)}^{2}\left({P}_{{\text{O}}_{2}}\right)}{\left({P}_{{\text{CO}}_{2}}\right)}=\frac{{\left(2x\right)}^{2}\left(x\right)}{{\left(1.15 - 2x\right)}^{2}}\approx \frac{{\left(2x\right)}^{2}\left(x\right)}{{\left(1.15\right)}^{2}}\\ 4{x}^{3}=8.59\times {10}^{-15}\\ x=1.29\times {10}^{-5}\text{atm}={P}_{{\text{O}}_{2}}\end{array}[/latex]

26. The equilibrium constant allows the calculation of [latex]\Delta {G}_{298}^{^\circ }[/latex].

[latex]\begin{array}{l}\Delta {G}_{298}^{^\circ }=\text{-}RT\text{ln}K=-2.303RT\text{log}K=-2.303\left(8.314{\text{J K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)\left(298.2K\right)\left(\text{log}1.3\times {10}^{3}\right)\\ =-17.78{\text{kJ mol}}^{\text{-1}}=1.8{\text{kJ mol}}^{\text{-1}}\left(\text{2 significant figures}\right)\end{array}[/latex]

The strength of the bond (66.5 kJ) means that it requires 66.5 kJ to pull 1 mol of bonds apart. In other words, [latex]\Delta {H}_{298}^{^\circ }=\text{-66.5 kJ}[/latex]. The values of [latex]\Delta {G}_{298}^{^\circ }[/latex] and [latex]\Delta {H}_{298}^{^\circ }[/latex] allow us to calculate ΔS° by use of the equation [latex]\begin{array}{l}\Delta {G}_{298}^{^\circ }=\Delta {H}_{298}^{^\circ }-T\Delta {H}_{298}^{^\circ }\\ \\ \Delta S\text{^\circ }=-\frac{\Delta {G}_{298}^{^\circ }-\Delta {H}_{298}^{^\circ }}{T}=-\frac{\left(\text{-17,780 J}\right)-\left(\text{-66,500 J}\right)}{\text{298.2 K}}=-163{\text{J K}}^{\text{-1}}=\text{-0.16 kJ}\end{array}[/latex]

28. (a) Find the equilibrium constant by first calculating the number of moles of each component.

• molar masses: SbCl₅ = 299.11 g mol⁻¹
• SbCl₃ = 228.11 g mol⁻¹
• Cl₂ = 70.9054 g mol⁻¹

[latex]\begin{array}{l}\\ {\text{mol SbCl}}_{5}=\frac{\text{3.85 g}}{299.01g{\text{mol}}^{\text{-1}}}=\text{0.0129 mol}\\ {\text{mol SbCl}}_{3}=\frac{\text{9.14 g}}{228.11{\text{g mol}}^{\text{-1}}}=\text{0.0401 mol}\\ {\text{mol Cl}}_{2}=\frac{\text{2.84 g}}{70.9054{\text{g mol}}^{\text{-1}}}=\text{0.0401 mol}\end{array}[/latex]

[latex]K=\frac{\left[{\text{SbCl}}_{3}\right]\left[C{l}_{2}\right]}{\left[{\text{SbCl}}_{5}\right]}=\frac{\left(\frac{\text{0.0401 mol}}{\text{5.00 L}}\right)\left(\frac{\text{0.0401 mol}}{\text{5.00 L}}\right)}{\left(\frac{\text{0.0129 mol}}{\text{5.00 L}}\right)}=0.0249{\text{mol L}}^{\text{-1}}[/latex]
At 448 °C = 721.15 K, [latex]\Delta G^\circ =\text{-}RT\text{ln}K=\text{-}\left(8.314{\text{J K}}^{\text{-1}}\right)\text{(721.15 K)(ln 0.0249)}=\text{-2.21}\times {10}^{4}\text{J}=\text{-}\text{22.1 kJ}[/latex]

(b) Determine the value of K and then determine the value of ΔG° from the value of K. Molar concentrations are required to find K. The concentration of Cl₂ at 1.00 atm and 975 K is determined from the ideal gas law, PV = nRT

[latex]\frac{n}{V}=\frac{P}{RT}=\frac{\text{1.00 atm}}{0.08206\frac{\text{L atm}}{\text{K mol}}\left(\text{975 K}\right)}=0.0125M[/latex]

Since the Cl₂ gas is 1.00% dissociated, the concentration of Cl₂ is 0.9900 [latex]\times[/latex] 0.0125 M = 0.0124 M. The concentration of Cl₂ is 2 [latex]\times[/latex] 0.0100 [latex]\times[/latex] 0.0125 = 0.000250 M (we have to use the factor of two since each dissociated Cl₂ molecule produces two Cl particles).

[latex]K=\frac{{\left[\text{Cl}\right]}^{2}}{\left[{\text{Cl}}_{2}\right]}=\frac{{0.00250}^{2}}{0.0125}=0.000500[/latex]
Using this result, ΔG° = −RT ln K = −8.314 J/mol K [latex]\times[/latex] 975 K ln (0.000500) = 61,600 J/mol = 61.6 kJ/mol.

30. The reaction of interest is: [latex]{\text{Ag}}_{2}\text{S}\left(s\right)\rightleftharpoons {\text{2Ag}}^{\text{+}}\left(aq\right)+{\text{S}}^{2-}\left(aq\right){K}_{\text{sp}}=8\times {10}^{-51}[/latex]

[latex]\begin{array}{l}\Delta {G}_{298}^{^\circ }=\text{-}RT\text{ln}K\\ \Delta {G}_{298}^{^\circ }=-8.314\times \text{298.15 ln}8\times {10}^{-51}\\ \Delta {G}_{298}^{^\circ }=\text{286 kJ}\end{array}[/latex]

[latex]\Delta {G}_{298}^{^\circ }=\sum \nu \Delta {G}_{\text{f}}^{^\circ }\left(\text{products}\right)-\sum \nu \Delta {G}_{\text{f}}^{^\circ }\left(\text{reactants}\right)[/latex]

[latex]\begin{array}{l}\\ 286=\left[\text{1 mol}\left(\Delta {G}_{\text{f}}^{^\circ }{S}^{2-}\text{kJ/mol}\right)+\text{2 mol}\left(\text{77.1 kJ/mol}\right)\right]-\left[\text{1 mol}\left(\text{-39.5 kJ/mol}\right)\right]\\ \Delta {G}_{\text{f}}^{^\circ }{S}^{2-}=\text{90 kJ/mol}\end{array}[/latex]

32. (a) Under standard thermodynamic conditions, the temperature is 298 K, and the pressure of water vapor would be 1 atm (or 1 bar). Under these conditions, the evaporation is nonspontaneous as indicated by [latex]\Delta {G}_{298}^{^\circ }>0[/latex]

(b) [latex]\Delta {G}_{298}^{^\circ }=\text{-}RT\text{ln}K[/latex], 8.58 [latex]\times[/latex] 10³ = −8.314 [latex]\times[/latex] 2981.15 ln K, K_p = 0.031

(c) [latex]Q={P}_{{\text{H}}_{2}\text{O}}=\text{0.011 atm}[/latex], [latex]\Delta G=\Delta {G}_{298}^{^\circ }+RT\text{ln}Q[/latex], ΔG = 8.58 [latex]\times[/latex] 10³ + (8.314 [latex]\times[/latex] 298 [latex]\times[/latex] ln 0.011) = −2.6 kJ, Under these conditions, the evaporation of water is spontaneous

(d) [latex]{P}_{{\text{H}}_{2}\text{O}}[/latex] must always be less than K_p or less than 0.031 atm. 0.031 atm represents air saturated with water vapor at 25 °C, or 100% humidity.

34. (a) Nonspontaneous as [latex]\Delta {G}_{298}^{^\circ }>0[/latex]

(b) [latex]\Delta {G}_{298}^{^\circ }=\text{-}RT\text{ln}K[/latex], [latex]\Delta G=1.7\times {10}^{3}+\left(8.314\times 335\times \text{ln}\frac{28}{128}\right)=\text{-2.5 kJ}[/latex]. The forward reaction to produce F6P is spontaneous under these conditions.

36. ΔG is negative as the process is spontaneous. ΔH is positive as with the solution becoming cold, the dissolving must be endothermic. ΔS must be positive as this drives the process, and it is expected for the dissolution of any soluble ionic compound.

38. (a) Increasing [latex]{P}_{{\text{O}}_{2}}[/latex] will shift the equilibrium toward the products, which increases the value of K. [latex]\Delta {G}_{298}^{^\circ }[/latex] therefore becomes more negative.

(b) Increasing [latex]{P}_{{\text{O}}_{2}}[/latex] will shift the equilibrium toward the products, which increases the value of K. [latex]\Delta {G}_{298}^{^\circ }[/latex] therefore becomes more negative.

(c) Increasing [latex]{P}_{{\text{O}}_{2}}[/latex] will shift the equilibrium the reactants, which decreases the value of K. [latex]\Delta {G}_{298}^{^\circ }[/latex] therefore becomes more positive.