The problems are on the Gibbs free energy worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Entropy S measures the dispersal of energy and matter, and the Gibbs free energy change ΔG = ΔH − TΔS predicts spontaneity: a reaction is spontaneous when ΔG is negative. These worked answers correct the 20 problems of the AP Chemistry worksheet on Gibbs free energy, from predict the sign of ΔS to entropy of the universe: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the formula, the substitution and the result with its unit, final values in bold. Pay special attention to unit conversions between J and kJ and to the sign of each quantity, as these cost the most marks.
Level: AP Chemistry · Curriculum: AP Chem Unit 9
Problem 1 Predict the sign of ΔS
(a) positive: ions in solution are more dispersed. (b) negative: two gas molecules become one. (c) positive: a gas is produced. (d) negative: condensation.Problem 2 Rank by entropy
A solid has the least freedom, a liquid more, and a gas the most. Fe(s), then H₂O(l), then He(g), with S° of 27.3, 69.9 and 126.2 J/(mol·K).Problem 3 Entropy of water formation
ΔS° = 2(69.9) − [2(130.7) + 205.1] = 139.8 − 466.5 = −326.7 J/(mol·K). Three gas molecules become a liquid, so entropy falls.Problem 4 A simple ΔG calculation
ΔS = −0.100 kJ/K. ΔG = −40.0 − (298)(−0.100) = −40.0 + 29.8 = −10.2 kJ. ΔG is negative, so it is spontaneous.Problem 5 Endothermic but spontaneous
ΔG = 50.0 − (400)(0.150) = 50.0 − 60.0 = −10.0 kJ. It is spontaneous because the entropy term outweighs the positive ΔH.Problem 6 Four cases
1. D (ΔH negative, ΔS positive): both terms favour a negative ΔG.2. B (both positive): −TΔS outweighs ΔH only when T is large.
3. C (both negative): ΔH outweighs −TΔS only when T is small.
Problem 7 Entropy of methane combustion
1. ΔS° = [213.8 + 2(188.8)] − [186.3 + 2(205.1)] = 591.4 − 596.5 = −5.1 J/(mol·K).2. Three gas molecules of reactants give three gas molecules of products, so the number of gas particles hardly changes and ΔS° is close to zero.
Problem 8 Decomposition of limestone
1. ΔS° = 39.8 + 213.8 − 92.9 = 160.7 J/(mol·K).2. ΔG° = 178.3 − (298)(0.1607) = 130.4 kJ, non-spontaneous.
3. T = ΔH°/ΔS° = 178 300 / 160.7 = 1110 K.
Problem 9 Free energy of formation
ΔG° = ΣΔG°f(products) − ΣΔG°f(reactants) = (−603.3 − 394.4) − (−1128.8) = +131.1 kJ. This agrees with +130.4 kJ from ΔH° − TΔS°, within rounding, and the positive value shows decomposition is not spontaneous at 298 K.Problem 10 Melting of ice
1. T = 6010 / 22.0 = 273 K.2. At 263 K: ΔG = 6010 − (263)(22.0) = +224 J/mol. At 283 K: ΔG = 6010 − (283)(22.0) = −216 J/mol.
3. Melting is spontaneous at 283 K, where ΔG is negative.
Problem 11 Boiling of ethanol
1. At equilibrium ΔS = ΔH/T = 38 600 / 351 = 110 J/(mol·K).2. The liquid becomes a gas, with far greater freedom of motion.
Problem 12 K from ΔG°
ln K = −ΔG°/RT = 20 000 / (8.314 × 298) = 8.072. K = e^8.072 = 3.2 × 10³.Problem 13 ΔG° from K
ΔG° = −RT ln K = −(8.314)(298)(ln 1.0 × 10⁻⁵) = −2477.6 × (−11.513) = +28 524 J = +28.5 kJ. A K below 1 gives a positive ΔG°: reactants are favoured.Problem 14 Free energy away from standard state
1. ΔG = ΔG° + RT ln Q = 10 000 + (8.314)(298)(ln 1.0 × 10⁻³) = 10 000 − 17 115 = −7.1 kJ.2. ΔG is negative, so the forward reaction proceeds. Q is below K, so the system shifts to the right.
Problem 15 Reading a ΔG against T line
1. Slope = [20 − (−20)] / (500 − 300) = +0.20 kJ/K. Since slope = −ΔS, ΔS = −0.20 kJ/(mol·K) = −200 J/(mol·K).2. −20 = ΔH − (300)(−0.20) = ΔH + 60, so ΔH = −80 kJ.
3. T = ΔH/ΔS = −80 / −0.20 = 400 K.
Problem 16 Making sulfur trioxide
1. ΔH° = 2(−395.7) − 2(−296.8) = −197.8 kJ; ΔS° = 2(256.8) − [2(248.2) + 205.1] = −187.9 J/(mol·K).2. ΔG° = −197.8 − (298)(−0.1879) = −141.8 kJ.
3. T = −197.8 / −0.1879 = 1053 K.
Problem 17 Temperature and the equilibrium constant
1. ΔG° = −197.8 − (800)(−0.1879) = −47.5 kJ.2. ln K = 47 500 / (8.314 × 800) = 7.14, so K = 1.3 × 10³.
3. K is smaller at high T, but the rate is far greater. A catalyst and a moderate temperature compromise between yield and speed.
Problem 18 Coupled reactions
1. ΔG°(total) = 13.8 + (−30.5) = −16.7 kJ.2. ln K = 16 700 / (8.314 × 298) = 6.74; K = 8.5 × 10².
3. The sum of the free energy changes is negative, so the combined process is spontaneous even though the first step alone is not.
Problem 19 Instant cold pack
1. ΔG = 25.7 − (298)(0.1087) = 25.7 − 32.4 = −6.7 kJ/mol.2. The increase in entropy as ions disperse in water outweighs the energy absorbed (−TΔS is more negative than ΔH is positive).
3. T = 25 700 / 108.7 = 236 K, far below the freezing point of water, so dissolving is spontaneous at every temperature at which the solution exists.
Problem 20 Entropy of the universe
1. ΔS(surr) = +92 200 / 298 = 309.4 J/(mol·K).2. ΔS(univ) = 309.4 + (−198.7) = 110.7 J/(mol·K).
3. −ΔG/T = −(−33 000)/298 = 110.7 J/(mol·K), which matches within rounding, because ΔS(univ) = ΔS − ΔH/T = −ΔG/T. It is positive, so the reaction is spontaneous.
