Unit 9 Explanation 5

Unit 9: Applications of Thermodynamics

📝 Past AP Questions (MCQ + FRQ)

Multiple Choice Questions

MCQ 1: The standard free energy change (ΔG°) for a reaction is +12.4 kJ/mol. What can you conclude?

1. K > 1, products favored
2. K = 1, system at equilibrium
3. K < 1, reactants favored
4. Reaction is spontaneous

✅ Correct Answer: C ΔG° > 0 means non-spontaneous, so K < 1.

MCQ 2: Which of the following processes results in a positive entropy change (ΔS)?

1. Condensation of steam
2. Freezing of water
3. Decomposition of H₂O₂ into gases
4. Formation of solid salt from aqueous ions

✅ Correct Answer: C Gas formation increases entropy — more randomness!

MCQ 3: A student constructs a galvanic cell and measures E°cell = +0.76 V. What is ΔG°?

• A. –147.4 kJ/mol
• B. +147.4 kJ/mol
• C. –76.0 kJ/mol
• D. Cannot be determined without temperature

Use n = 2, F = 96,485 C/mol

✅ Correct Answer: A ΔG° = –nFE° = –(2)(96485)(0.76) ≈ –147.4 kJ

Free Response Excerpts

FRQ 1: (Based on ATP Hydrolysis)
ATP hydrolysis has a ΔG° of –30.5 kJ/mol. A reaction that requires +18.0 kJ/mol is coupled with it.

1. (a) Show using calculations that the overall reaction is spontaneous.
2. (b) Explain why biological systems rely on coupling instead of direct heat use.

✅ Strategy:

• (a) ΔG_total = –30.5 + 18.0 = –12.5 kJ/mol → spontaneous ✅
• (b) Direct heat would denature proteins or damage cells. Coupling allows targeted energy use.

FRQ 2: (ΔG and K)
Given: ΔG° = –34.6 kJ/mol at 298 K

1. (a) Calculate the value of K using ΔG° = –RT ln K
2. (b) Interpret the meaning of this K value in terms of equilibrium position.

✅ Strategy:

• Use R = 8.314 J/mol·K → ΔG in J: –34600 J
• ln K = –ΔG / RT = 34600 / (8.314×298) ≈ 13.9 → K ≈ e^13.9 ≈ 1.1×10⁶
• ✅ Large K → strongly product-favored at equilibrium