The questions are on the Gibbs free energy test. Take it under real conditions before reading the mark scheme.
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. This mark scheme gives the model answers and the marks for the 8 questions of the AP Chemistry test on Gibbs free energy, from entropy basics to comparing entropies, with the equivalent answers to accept.
This is the Gibbs free energy test answers for AP Chemistry, with the marks for each answer. First sit the test without notes, then mark it here and revise with the lesson.
Each model answer is followed by a marking guide. Allow follow-through marks when a later step correctly uses an earlier wrong value, and accept answers within rounding.
Level: AP Chemistry · Curriculum: AP Chem Unit 9
Gibbs free energy test answers: answers and marks
Question 1 Entropy basics (2 marks)
Entropy is a measure of the dispersal of energy and matter, or the number of microstates available. Its unit is J/(mol·K). Evaporation has positive ΔS because a gas is more disordered than a liquid.Marking: 1 mark for the definition and unit, 1 mark for the sign with reason.
Question 2 Entropy of a decomposition (3 marks)
ΔS° = [2(82.6) + 3(205.1)] − 2(143.1) = [165.2 + 615.3] − 286.2 = +494.3 J/(mol·K).Marking: 1 mark for the method, 1 for products sum, 1 for the final value.
Question 3 Gibbs calculation (3 marks)
- ΔG = −120 − (298)(−0.250) = −120 + 74.5 = −45.5 kJ, so spontaneous.
- T = ΔH/ΔS = −120 / −0.250 = 480 K.
Marking: (a) 1 mark for ΔG and 1 for the conclusion; (b) 1 mark.
Question 4 K from ΔG° (3 marks)
ln K = 15 000 / (8.314 × 298) = 6.054. K = 426 (about 4.3 × 10²). K is much greater than 1, so the equilibrium mixture contains mostly products.Marking: 1 mark for ln K, 1 for K, 1 for the interpretation.
Question 5 Sign combinations (2 marks)
ΔG = ΔH − TΔS. At low T the positive ΔH dominates, so ΔG is positive. As T increases the −TΔS term becomes more negative and eventually outweighs ΔH, so ΔG becomes negative above T = ΔH/ΔS.Marking: 1 mark for the low-T case, 1 for the high-T case.
Question 6 Reading a graph (3 marks)
- ΔH is the intercept at T = 0: +60 kJ.
- ΔG = 0 at 300 K.
- ΔS = ΔH/T = 60 / 300 = 0.20 kJ/(mol·K) = +200 J/(mol·K).
Marking: 1 mark for each part.
Question 7 Non-standard conditions (2 marks)
ΔG = ΔG° + RT ln Q = −5000 + (8.314)(298)(ln 50) = −5000 + 9692 = +4692 J (+4.7 kJ). ΔG is positive, so the reverse reaction proceeds (Q is above K).Marking: 1 mark for ΔG, 1 for the direction.
Question 8 Comparing entropies (2 marks)
(a) H₂O(g): gas particles have far more freedom. (b) C₃H₈(g): the larger molecule has more ways to vibrate and rotate.Marking: 1 mark for each part with reason.
