7. a. A spontaneous process is one that occurs without any outside intervention.

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1 CHAPTER SIXTEEN SPONTANEITY, ENTROPY, AND FREE ENERGY Questions 7. a. A spontaneous process is one that occurs without any outside intervention. b. Entropy is a measure of disorder or randomness. c. The system is the portion of the universe in which we are interested. d. The surroundings are everything else in the universe besides the system. 8. a. Entropy increases; there is greater volume accessible to the randomly moving gas molecules which leads to an increase in disorder. b. The positional entropy doesn't change. There is no change in volume and thus no change in the numbers of positions of the molecules. The total entropy ( S univ) increases because the increase in temperature increases the energy disorder ( S ). c. Entropy decreases because the volume decreases (P and V are inversely related). 9. Living organisms need an external source of energy to carry out these processes. Green plants use the energy from sunlight to produce glucose from carbon dioxide and water by photosynthesis. In the human body, the energy released from the metabolism of glucose helps drive the synthesis of proteins. For all processes combined, S must be greater than zero (second law). univ 10. No; When using values in Appendix 4, we have specified a temperature of 25 C. Further, if gases or solutions are involved, we have specified partial pressures of 1 atm and solute concentrations of 1 molar. At other temperatures and compositions, the reaction may not be spontaneous. A negative G value means the reaction is spontaneous under standard conditions. 11. Dispersion increases the entropy of the universe since the more widely something is dispersed, the greater the disorder. We must do work to overcome this disorder. In terms of the second law, it would be more advantageous to prevent contamination of the environment than to clean it up later. As a substance disperses, we have a much larger area that must be decontaminated. surr 452

2 CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY All thermodynamic functions depend on temperature. However, H and S are the least dependent on temperature and are commonly assumed to be temperature independent. G has the strongest temperature dependence. To determine G, the temperature must be known in order to use the equation G = H - T S. 13. w max = G; When G is negative, the magnitude of G is equal to the maximum possible useful work obtainable from the process (at constant T and P). When G is positive, the magnitude of G is equal to the minimum amount of work that must be expended to make the process spontaneous. Due to waste energy (heat) in any real process, the amount of useful work obtainable from a spontaneous process is always less than w max and, for a nonspontaneous reaction, an amount of work greater than w must be applied to make the process spontaneous. max 14. The rate of a reaction is directly related to temperature. As the temperature increases, the rate of a reaction increases. Spontaneity, however, does not necessarily have a direct relationship to temperature. The temperature dependence of spontaneity depends on the signs of the values for H and S (see Table 16.5 of the text). For example, when H and S are both negative, the reaction becomes more favorable thermodynamically ( G becomes more negative) with decreasing temperature. This is just the opposite of the kinetics dependence on temperature. Other sign combinations of H and S have different spontaneity temperature dependence. Exercises Spontaneity, Entropy, and the Second Law of Thermodynamics: Free Energy 15. a, b and c; From our own experiences, salt water, colored water and rust form without any outside intervention. A bedroom, however, spontaneously gets cluttered. It takes an outside energy source to clean a bedroom. 16. c and d; It takes an outside energy source to build a house and to launch and keep a satellite in orbit. 17. We draw all of the possible arrangements of the two particles in the three levels. 2 kj x x xx 1 kj x xx x 0 kj xx x x Total E = 0 kj 1 kj 2 kj 2 kj 3 kj 4 kj The most likely total energy is 2 kj.

3 454 CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY kj AB B A B A 1 kj AB B A A B 0 kj AB A B A B E T = 0 kj 2 kj 4 kj 1 kj 1 kj 2 kj 2 kj 3 kj 3 kj The most likely total energy is 2 kj. 19. a. H 2 at 100 C and 0.5 atm; Higher temperature and lower pressure means greater volume and hence, greater positional entropy. b. N 2 at STP has the greater volume. c. H2O(l) is more disordered than H2O(s). 20. Of the three phases (solid, liquid, and gas), solids are the most ordered and gases are the most disordered. Thus, a, b, and f (melting, sublimation, and boiling) involve an increase in the entropy of the system since going from a solid to a liquid or a solid to a gas or a liquid to a gas increases disorder. For freezing (process c), a substance goes from the more disordered liquid state to the more ordered solid state, hence, entropy decreases. Process d (mixing) involves an increase in disorder (entropy) while separation increases order (decreases the entropy of the system). So of all the processes, a, b, d, and f involve an increase in the entropy of the system. 21. a. To boil a liquid requires heat. Hence, this is an endothermic process. All endothermic processes decrease the entropy of the surroundings ( S is negative). surr b. This is an exothermic process. Heat is released when gas molecules slow down enough to form the solid. In exothermic processes, the entropy of the surroundings increases ( S is positive) a. S surr = = 7.45 kj/k = J/K surr b. S surr = = kj/k = -376 J/K 23. G = H - T S; When G is negative, the process will be spontaneous. 3 a. G = H - T S = J - (300. K)(5.0 J/K) = 24,000 J, Not spontaneous b. G = 25,000 J - (300. K)(100. J/K) = J, Spontaneous c. Without calculating G, we know this reaction will be spontaneous at all temperatures. H is negative and S is positive (-T S < 0). G will always be less than zero with these sign combinations for H and S. 4 d. G = J - (200. K)(-40. J/K) = J, Spontaneous 24. G = H - T S; A process is spontaneous when G < 0. For the following, assume H and S are

4 CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY 455 temperature independent. a. When H and S are both negative, G will be negative below a certain temperature where the favorable H term dominates. When G = 0, then H = T S. Solving for this temperature: T = = K 2 At T < K, this process will be spontaneous ( G < 0). b. When H and S are both positive, G will be negative above a certain temperature where the favorable S term dominates. T = = K 2 At T > K, this process will be spontaneous ( G < 0). c. When H is positive and S is negative, this process can never be spontaneous at any temperature since G can never be negative. d. When H is negative and S is positive, this process is spontaneous at all temperatures since G will always be negative. 25. At the boiling point, G = 0 so H = T S. S = -2 = kj/k mol = 89.3 J/K mol 26. At the boiling point, G = 0 so H = T S. T = = K 27. a. NH 3(s) NH 3(l); G = H - T S = 5650 J/mol K (28.9 J/K mol) G = 5650 J/mol J/mol = -130 J/mol Yes, NH 3 will melt since G < 0 at this temperature. b. At the melting point, G = 0 so T = = 196 K. 28. At the melting point, G = 0 so H = T S. S = = 9.57 J/K mol Chemical Reactions: Entropy Changes and Free Energy

5 456 CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY 29. a. Decrease in disorder; S (-) b. Increase in disorder; S (+) c. Decrease in disorder ( n < 0); S (-) d. Increase in disorder ( n > 0); S (+) For c and d, concentrate on the gaseous products and reactants. When there are more gaseous product molecules than gaseous reactant molecules ( n > 0), then S will be positive (disorder increases). When n is negative, then S is negative (disorder decreases). 30. a. Decrease in disorder ( n < 0); S (-) b. Decrease in disorder ( n < 0); S (-) c. Increase in disorder; S (+) d. Increase in disorder; S (+) 31. a. C graphite(s); Diamond is a more ordered structure than graphite. b. C2H5OH(g); The gaseous state is more disordered than the liquid state. c. CO 2(g); The gaseous state is more disordered than the solid state. 32. a. He (10 K); S = 0 at 0 K b. N2O; More complicated molecule c. H2O(l): The liquid state is more disordered than the solid state. 33. a. 2 H2S(g) + SO 2(g) 3 S rhombic(s) + 2 H2O(g); Since there are more molecules of reactant gases compared to product molecules of gas ( n = 2-3 < 0), then S will be negative. S = - S = [3 mol S rhombic(s) (32 J/K mol) + 2 mol H2O(g) (189 J/K mol)] S = 474 J/K J/K = -186 J/K - [2 mol H2S(g) (206 J/K mol) + 1 mol SO 2(g) (248 J/K mol)] b. 2 SO 3(g) 2 SO 2(g) + O 2(g); Since n of gases is positive ( n = 3-2), S will be positive. S = 2 mol(248 J/K mol) + 1 mol(205 J/K mol) - [2 mol(257 J/K mol)] = 187 J/K c. Fe2O 3(s) + 3 H 2(g) 2 Fe(s) + 3 H2O(g); Since n of gases = 0 ( n = 3-3), we can t easily predict if S will be positive of negative. S = 2 mol(27 J/K mol) + 3 mol(189 J/K mol) - [1 mol(90. J/K mol) + 3 mol(131 J/K mol)] S = 138 J/K 34. a. H 2(g) + 1/2 O 2(g) H2O(l); Since n of gases is negative, S will be negative. S = 1 mol H2O(l)(70. J/K mol) - [1 mol H 2(g)(131 J/K mol) + 1/2 mol O 2(g)(205 J/K mol)]

6 CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY 457 S = 70. J/K J/K = -164 J/K b. N 2(g) + 3 H 2(g) 2 NH 3(g); Since n of gases is negative, S will be negative. S =2(193) - [1(192) + 3(131)] = -199 J/K + - c. HCl(g) H (aq) + Cl (aq); The gaseous state dominates predictions of S. Here the gaseous state is more disordered than the ions in solution, so S will be negative. + - S = 1 mol H (0) + 1 mol Cl (57 J/K mol) - 1 mol HCl(187 J/K mol) = J/K 35. C2H 2(g) + 4 F 2(g) 2 CF 4(g) + H 2(g); S = J/K = (2 mol) J/K - [201 J/K + 4(203 J/K)], = 262 J/K mol J/K = (2 mol) - [2(28 J/K) + 3(152 J/K)], = 184 J/K mol 37. a. Srhombic S monoclinic; This phase transition is spontaneous ( G < 0) at temperatures above 95 C. G = H - T S; For G to be negative only above a certain temperature, H is positive and S is positive (see Table 16.5 of text). b. Since S is positive, S rhombic is the more ordered crystalline structure. 38. Enthalpy is not favorable, so S must provide the driving force for the change. Thus, S is positive. There is an increase in disorder, so the original enzyme has the more ordered structure. 39. a. When a bond is formed, energy is released, so H is negative. Since there are more reactant molecules of gas than product molecules of gas ( n < 0), S will be negative. b. G = H - T S; For this reaction to be spontaneous ( G < 0), the favorable enthalpy term must dominate. The reaction will be spontaneous at low temperatures where the H term dominates. 40. Since there are more product gas molecules than reactant gas molecules ( n > 0), S will be positive. From the signs of H and S, this reaction is spontaneous at all temperatures. It will cost money to heat the reaction mixture. Since there is no thermodynamic reason to do this, the purpose of the elevated temperature must be to increase the rate of the reaction, i.e., kinetic reasons. 41. a. CH 4(g) + 2 O 2(g) CO 2(g ) + 2 H2O(g) -75 kj/mol

7 458 CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY -51 kj/mol Data from Appendix 4 S 186 J/K mol H = - ; S = - H = 2 mol(-242 kj/mol) + 1 mol( kj/mol) - [1 mol(-75 kj/mol)] = -803 kj S = 2 mol(189 J/K mol) + 1 mol(214 J/K mol) - [1 mol(186 J/K mol) + 2 mol(205 J/K mol)] = -4 J/K There are two ways to get G. We can use G = H - T S (be careful of units): 3 5 G = H - T S = J - (298 K)(-4 J/K) = J = -802 kj or we can use values where G = - : G = 2 mol(- 229 kj/mol) + 1 mol(-394 kj/mol) - [1 mol(- 51 kj/mol)] G = -801 kj (Answers are the same within round-off error.) b. 6 CO 2(g) + 6 H2O(l) C6H12O 6(s) + 6 O 2(g) kj/mol S 214 J/K mol H = [6(-286) + 6(-393.5)] = 2802 kj S = 6(205) [6(214) + 6(70.)] = -262 J/K G = 2802 kj - (298 K)( kj/k) = kj c. P4O 10(s) + 6 H2O(l) 4 H3PO 4(s) (kj/mol) S (J/K mol) H = 4 mol(-1279 kj/mol) - [1 mol(-2984 kj/mol) + 6 mol(-286 kj/mol)] = -416 kj S = 4(110.) - [ (70.)] = -209 J/K G = H - T S = -416 kj - (298 K)( kj/k) = -354 kj

8 CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY 459 d. HCl(g) + NH 3(g) NH4Cl(s) (kj/mol) S (J/K mol) H = [-92-46] = -176 kj; S = 96 - [ ] = -284 J/K G = H - T S = -176 kj - (298 K)( kj/k) = -91 kj 42. G = kj - (298 K)( kj/k) = kj G = 0 = H - T S, T = = K G is negative below K where the favorable H term dominates. 43. CH 4(g) + CO 2(g) CH3CO2H(l) H = [-75 + (-393.5)] = -16 kj; S = [ ] = J/K G = H - T S = -16 kj - (298 K)( kj/k) = 56 kj This reaction is spontaneous only at temperatures below T = H / S = 67 K (where the favorable H term will dominate, giving a negative G value). This is not practical. Substances will be in condensed phases, and rates will be very slow at this extremely low temperature. CH3OH(g) + CO(g) CH3CO2H(l) H = [ (-201)] = -173 kj; S = [ ] = -278 J/K G = -173 kj - (298 K)( kj/k) = -90. kj This reaction also has a favorable enthalpy and an unfavorable entropy term. This reaction is spontaneous at temperatures below T = H / S = 622 K. The reaction of CH3OH and CO will be preferred. It is spontaneous at high enough temperatures that the rates of reaction should be reasonable. 44. C2H 4(g) + H2O(g) CH3CH2OH(l) H = (52-242) = -88 kj; S = ( ) = -247 J/K When G = 0, H = T S, T = = 360 K Since the signs of H and S are both negative, this reaction will be spontaneous at temperatures

9 460 CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY below 360 K (where the favorable H term will dominate). C2H 6(g) + H2O(g) CH3CH2OH(l) + H 2(g) H = ( ) = 49 kj; S = ( ) = -127 J/K This reaction can never be spontaneous because of the signs of H and S. Thus, the reaction C2H 4(g) + H2O(g) C2H5OH(l) would be preferred. 45. SO 3(g) SO 2(g) + 1/2 O 2(g) G = -1/2 (-142 kj) S(s) + 3/2 O (g) SO (g) G = -371 kj 2 3 S(s) + O 2(g) SO 2(g) G = 71 kj kj = kj C(s) + 6 O 2(g) 6 CO 2(g) G = 6(-394 kj) 3 H 2(g) + 3/2 O 2(g) 3 H2O(l) G = 3(-237 kj) 6 CO (g) + 3 H O(l) C H (l) + 15/2 O (g) G = -1/2 (-6399 kj) C(s) + 3 H 2(g) CH(l) 6 6 G = 125 kj 47. G =, -374 kj = kj -, = -731 kj/mol kj = 8(-394 kj) + 10(-237 kj) - = -16 kj/mol 49. G = G = 1 mol(-2698 kj/mol) + 6 mol(-237 kj/mol) - [4 mol(13 kj/mol) + 8 mol(0)] = kj 50. a. G = 2(-270. kj) - 2(-502 kj) = 464 kj b. Since G is positive, this reaction is not spontaneous at standard conditions at 298 K. c. G = H - T S, H = G + T S = 464 kj K(0.179 kj/k) = 517 kj We need to solve for the temperature when G = 0: G = 0 = H - T S, T = = 2890 K This reaction will be spontaneous at standard conditions ( G < 0) at T > 2890 K, where the favorable entropy term will dominate. Free Energy: Pressure Dependence and Equilibrium 51. G = G + RT ln Q; For this reaction: G = G + RT ln

10 CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY 461 G = 1 mol(52 kj/mol) + 1 mol(0) - [1 mol(87 kj/mol) + 1 mol(163 kj/mol)] = -198 kj G = -198 kj + (298 K) ln G = -198 kj kj = -188 kj 52. G = 3(0) + 2(-229) - [2(-34) + 1(-300.)] = -90. kj G = G + RT ln = -90. kj + kj G = -90. kj kj = -50. kj 53. G = G + RT ln Q = G + RT ln G = 1 mol(98 kj/mol) - 2 mol(52 kj/mol) = -6 kj a. These are standard conditions, so G = G since Q = 1 and ln Q = 0. Since G is negative, then the forward reaction is spontaneous. The reaction shifts right to reach equilibrium. 3 b. G = J J/K mol (298 K) ln 3 3 G = J J = 0 Since G = 0, this reaction is at equilibrium (no shift). 3 c. G = J J/K mol (298 K) ln G = J J = J = 1 10 J Since G is positive, the reverse reaction is spontaneous, so the reaction shifts to the left to reach equilibrium. 54. H = 2 = 2(-46) = -92 kj; G = 2 = 2(-17) = -34 kj S = 2(193 J/K) - [192 J/K + 3(131 J/K)] = -199 J/K; G = -RT ln K K = exp = e = Note: When determining exponents, we will round off after the calculation is complete. This helps eliminate excessive round-off error.

11 462 CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY a. G = G + RT ln = -34 kj + G = -34 kj - 33 kj = -67 kj b. G = -34 kj + G = -34 kj kj = -68 kj At 25.0 C: G = H T S = J/mol (298.2 K)( J/K mol) 3 = J/mol G = RT ln K, ln K = = = 2.166, K = e = At C: G = J/mol (373.2 K)( J/K mol) = J/mol ln K = = 2.540, K = e = Note: When determining exponents, we will round off after the calculation is complete. This helps eliminate excessive round-off error. 56. a. H = 2 mol(-92 kj/mol) - [1 mol(0) + 1 mol(0)] = -184 kj S = 2 mol(187 J/K mol) - [1 mol(131 J/K mol) + 1 mol(223 J/K mol)] = 20. J/K 3 5 G = H - T S = J K (20. J/K) = J = kj G = -RT ln K, ln K = = , K = e = b. These are standard conditions, so G = G = kj. Since G is negative, the forward reaction is spontaneous, so the reaction shifts right to reach equilibrium a. G = -RT ln K = - (298 K) ln ( ) = J = 79.9 kj/mol b. = -RT ln K = - (313 K) ln ( ) = J = 81.1 kj/mol 58. a. (kj/mol) S (J/K mol) NH 3(g) O 2(g) NO(g) H O(g)

12 CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY 463 NO 2(g) HNO 3(l) H O(l) NH 3(g) + 5 O 2(g) 4 NO(g) + 6 H2O(g) H = 6(-242) + 4(90.) - [4(-46)] = -908 kj S = 4(211) + 6(189) - [4(193) + 5(205)] = 181 J/K G = -908 kj K (0.181 kj/k) = -962 kj G = - RT ln K, ln K = = ln K = log K, log K = 168, K = 10 (an extremely large value) 2 NO(g) + O 2(g) 2 NO 2(g) H = 2(34) -[2(90.)] = -112 kj; S = 2(240.) - [2(211) + (205)] = -147 J/K G = -112 kj - (298 K)( kj/k) = -68 kj K = exp = e = Note: When determining exponents, we will round off after the calculation is complete. 3 NO 2(g) + H2O(l) 2 HNO 3(l) + NO(g) H = 2(-174) + (90.) - [3(34) + (-286)] = -74 kj S = 2(156) + (211) - [3(240.) + (70.)] = -267 J/K G = -74 kj - (298 K)( kj/k) = 6 kj K = exp = e = 9 10 b. G = -RT ln K; T = 825 C = ( )K = 1098 K; We must determine G at 1098 K. = H - T S = -908 kj - (1098 K)(0.181 kj/k) = kj K = exp = e = c. There is no thermodynamic reason for the elevated temperature since H is negative and S is positive. Thus, the purpose of the high temperature must be to increase the rate of the reaction.

13 464 CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY 59. K = = = = -RT ln K = J/K mol (800. K) ln ( ) = J/mol = -71 kj/mol SO 2(g) + O 2(g) 2 SO 3(g); G = 2(-371 kj) - [2(-300. kj)] = -142 kj G = -RT ln K, ln K = = = , K = e = K = = =, = atm From the negative value of G, this reaction is spontaneous at standard conditions. Since there are more molecules of reactant gases than product gases, S will be negative (unfavorable). Therefore, this reaction must be exothermic ( H < 0). When H and S are both negative, the reaction will be spontaneous at relatively low temperatures where the favorable H term dominates. 61. The equation ln K = is in the form of a straight line equation (y = mx + b). A graph of ln K vs. 1/T will yield a straight line with slope = m = - H /R and a y-intercept = b = S /R. From the plot: slope = = = K K = - H /R, H = K J/K mol = J/mol y-intercept = 40. = S /R, S = J/K mol = 330 J/K mol As seen here, when H is positive, the slope of the ln K vs. 1/T plot is negative. When H is negative, as in an exothermic process, the slope of the ln K vs. 1/T plot will be positive (slope = - H /R). 62. A graph of ln K vs. 1/T will yield a straight line with slope equal to - H /R and y-intercept equal to S /R. -1 Temp ( C) T(K) 1000/T (K ) K ln K w w

14 CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY The straight line equation (from a calculator) is: ln K = Slope = K =, H = -( K J/K mol) = J/mol y-intercept = =, S = J/K mol = J/K mol Additional Exercises 63. It appears the sum of the two processes has no net change. This is not so. By the second law of thermodynamics, S univ must have increased even though it looks as if we have gone through a cyclic process. 64. When an ionic solid dissolves, one would expect the disorder of the system to increase, so S sys is positive. Since temperature increased as the solid dissolved, this is an exothermic process and S surr is positive ( S surr = - H/T). Since the solid did dissolve, the dissolving process is spontaneous, so S is positive (as it must be when S and S are both positive). univ sys. surr 65. The introduction of mistakes is an effect of entropy. The purpose of redundant information is to provide a control to check the "correctness" of the transmitted information. 66. At boiling point, G = 0 so S = ; For methane: S = = 73.2 J/mol K For hexane: S = = 84.5 J/mol K V = = = 9.19 L; V = = R(342 K) = 28.1 L met hex Hexane has the larger molar volume at the boiling point, so hexane should have the larger entropy.

15 466 CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY As the volume of a gas increases, positional disorder increases. 67. S (monoclinic) S (rhombic); H = = kj; S = = J/K At the conversion temperature: G = 0 so H = T S ; T = = 370 K a. G = -RT ln K = -( J/K mol)(298 K) ln 0.090, G = J/mol = 6.0 kj/mol b. HSOSH + ClSOSCl 2 HSOSCl On each side of the reaction, there are 2 HSO bonds and 2 OSCl bonds. Both sides have the same number and type of bonds. Thus, H H 0. c. G = H - T S, S = = -20. J/K d. For H2O(g), = -242 kj/mol and S = 189 J/K mol H = 0 = 2 - [1 mol(-242 kj/mol) + 1 mol(80.3 J/K/mol)], = -81 kj/mol -20. J/K = 2 - [1 mol(189 J/K mol) + 1 mol(266.1 J/K mol)], = 218 J/K mol 4 e. Assuming H and S are T independent: = 0 - (500. K)(-20. J/K) = J G = -RT ln K, K = exp = exp = e = f. G = G + RT ln ; From part a, G = 6.0 kj/mol. We should express all P s in atm. However, because we perform the pressure conversion the same number of times in the numerator and denominator, the factors of 760 torr/atm will all cancel. Thus, we can use the pressures in units of torr. G = 6.0 kj/mol + ln = = -14 kj/mol 69. HgbO 2 Hgb + O2 G = -(-70 kj) Hgb + CO HgbCO G = -80 kj HgbO 2 + CO HgbCO + O2 G = -10 kj G = -RTln K, K = exp = exp = Ba(NO 3) 2(s) Ba (aq) + 2 NO 3(aq) K = K sp; G = (-109) - (-797) = 18 kj G = -RT ln K sp, ln K sp = = -7.26, K sp = e =

16 CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY HF(aq) H (aq) + F (aq); G = G + RT ln -4 4 G = -RT ln K = -( J/K mol) (298 K) ln ( ) = J/mol 4 a. The concentrations are all at standard conditions, so G = G = J/mol (since Q = 1.0 and ln Q = 0). Since G is positive, then the reaction shifts left to reach equilibrium. 4 b. G = J/mol + ( J/K mol) (298 K) ln 4 4 G = J/mol J/mol = 0 Since G = 0, then the reaction is at equilibrium (no shift). 4 4 c. G = (298) ln = J/mol; shifts right d. G = (298) ln = = 0; at equilibrium 4 3 e. G = (298) ln = 2 10 J/mol; shifts left K (blood) K (muscle) G = 0; G = RT ln ; G = w max 3 G = (310. K) ln, G = J/mol = 8.8 kj/mol At least 8.8 kj of work must be applied. Other ions will have to be transported in order to maintain electroneutrality. Either anions must be + transported into the cells, or cations (Na ) in the cell must be transported to the blood. The latter is + + what happens: [Na ] in blood is greater than [Na ] in cells as a result of this pumping. 73. a. G = - RT ln K, K = exp (- G /RT) = exp = b. C6H12O 6(s) + 6 O 2(g) 6 CO 2(g) + 6 H2O(l) G = 6 mol(-394 kj/mol) + 6 mol(-237 kj/mol) - 1 mol(-911 kj/mol) = kj = ; 94.3 molecules ATP/molecule glucose This is an overstatement. The assumption that all of the free energy goes into this reaction is false. Actually, only 38 moles of ATP are produced by the metabolism of one mole of glucose.

17 468 CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY 74. a. G = -RT ln K ln K = = -5.65, K = e = b. Glutamic acid + NH Glutamine + H O G = 14 kj ATP + H O ADP + H PO G = kj Glutamic acid + ATP + NH Glutamine + ADP + H PO G = = -17 kj ln K = = 6.86, K = e = S is more favorable for reaction two than for reaction one, resulting in K 2 > K 1. In reaction one, seven particles in solution are forming one particle. In reaction two, four particles form one particle, which results in a smaller decrease in disorder than for reaction one. 76. G = -RT ln K; When K =1.00, G = 0 since ln 1.00 = 0. G = 0 = H - T S, H = T S H = 3(-242 kj) - [-826 kj] = 100. kj; S = 2(27 J/K) + 3(189 J/K) - [90. J/K + 3(131 J/K)] = 138 J/K H = T S, T = = 725 K Challenge Problems O 2(g) 2 O 3(g); H = 2(143 kj) = 286 kj; G = 2(163 kj) = 326 kj ln K = = , K = e = We need the value of K at 230. K. From Section 16.8 of the text: ln K = For two sets of K and T: ln K 1 = ; ln K 2 = Subtracting the first expression from the second: ln K 2 - ln K 1 = Let K 2 = , T 2 = 298; K 1 = K 230, T 1 = 230. K; H = J ln = 34.13

18 CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY = e = , K 230 = K 230 = =, = atm The volume occupied by one molecule of ozone is: 17, V = L Equilibrium is probably not maintained under these conditions. When only two ozone molecules are 17 in a volume of L, the reaction is not at equilibrium. Under these conditions, Q > K and the reaction shifts left. But with only 2 ozone molecules in this huge volume, it is extremely unlikely they will collide with each other. In these conditions, the concentration of ozone is not large enough to maintain equilibrium. 78. Arrangement I and V: S = k ln W; W = 1; S = k ln 1 = Arrangement II and IV: W = 4; S = k ln 4 = J/K ln 4, S = J/K -23 Arrangement III: W = 6; S = k ln 6 = J/K 79. a. From the plot, the activation energy of the reverse reaction is E a + (- G ) = E a - G ( G is a negative number as drawn in the diagram). k = A exp and k = A exp, f r If the A factors are equal: = exp From G = -RT ln K, K = exp ; Since K and are both equal to the same expression, then K =. b. A catalyst will lower the activation energy for both the forward and reverse reaction (but not change G ). Therefore, a catalyst must increase the rate of both the forward and reverse reactions. 80. At equilibrium: -10 = atm

19 470 CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY -10 The pressure of H 2 decreased from 1.00 atm to atm. Essentially all of the H 2 and Br 2 has reacted. Therefore, P HBr = 2.00 atm since there is a 2:1 mol ratio between HBr and H 2 in the balanced equation. Since we began with equal moles of H 2 and Br 2, then we will have equal moles of H 2 and -10 Br at equilibrium. Therefore, = atm K = = Assumptions good G = -RT ln K = -( J/K mol)(298 K) ln ( ) = J/mol S = = 20 J/K mol 81. a. G = = 11, = 2722 J K = exp = exp = b. Since Q = 1.00 > K, reaction shifts left. Let x = atm of B(g) which reacts to reach equilibrium. A(g) B(g) Initial 1.00 atm 1.00 atm Equil x x = 0.333, x = x, x = 0.50 atm P = = 0.50 atm; P = = 1.50 atm B A c. G = G + RT ln Q = G + RT ln (P /P ) B A G = 2722 J + (8.3145)(298) ln (0.50/1.50) = 2722 J J = 0 (carrying extra sig. figs.) 82. From Exercise 16.61, ln K =. For K at two temperatures, T 1 and T 2, the equation can be manipulated to give (see Exercise 16.77): ln = ( mol/j) ( H ), H = J/mol For K = 8.84 at T = 25 C: ln 8.84 = = -37, S = -310 J/K mol

20 CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY We get the same value for S using K = at T = 348 K data. G = -RT ln K. When K = 1.00, then G = 0 since ln 1.00 = 0. G = 0 = H - T S. Assuming H and S do not depend on temperature: H = T S, T = = 310 K 83. K p = ; To insure Ag2CO 3 from decomposing, should be greater than K p. From Exercise 16.61, ln K =. For two conditions of K and T, the equation is: -3 Let T 1 = 25 C = 298 K, K 1 = torr; T 2 = 110. C = 383 K, K 2 =? = 7.1, = e = , K 2 = 7.5 torr To prevent decomposition of Ag2CO 3, the partial pressure of CO 2 should be greater than 7.5 torr. o 84. From the problem, = We need the pure vapor pressures (P ) in order to calculate the vapor pressure of the solution. Using the thermodynamic data: C6H 6(l) C6H 6(g) K = at 25 C = kj/mol kj/mol = 5.16 kj/mol G = -RT ln K, ln K = = K = = e = atm For CCl 4: = kj/mol - ( kj/mol) = 4.62 kj/mol K = = exp = atm

21 472 CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY = (0.125 atm) = atm; = (0.155 atm) = atm = 0.446; = = Use the thermodynamic data to calculate the boiling point of the solvent. At boiling point, G = 0 = H - T S, H = T S, T = = K T = Kbm, (355.4 K K) = 2.5 K kg/mol (m), m = = 0.84 mol/kg mass solvent = 150. ml = kg mass solute = kg solvent = 15.7 g = 16 g solute