AP Chemistry / Topic
Applications of Thermodynamics
Unit 9 covers entropy, Gibbs free energy and spontaneity, the link between ΔG° and K, and electrochemistry: galvanic and electrolytic cells, standard cell potential, and Faraday's-law calculations.
What the exam asks
- Predict the sign of ΔS° from states of matter and gas mole counts
- Use ΔG = ΔH − TΔS to decide when a process is thermodynamically favored
- Relate ΔG° to K (favored means K > 1) and to cell potential
- Analyze galvanic cells: anode/cathode, electron flow, salt bridge, and E°cell
- Contrast electrolytic cells and compute deposited mass with Faraday's law
- Explain why favored reactions can still be slow (kinetic control)
Key formulas and rules
ΔG° = ΔH° − TΔS°; favored when ΔG° < 0
ΔH<0, ΔS>0: always favored; ΔH>0, ΔS<0: never; otherwise temperature decides
ΔG° = −RT ln K
E°cell = E°cathode − E°anode; favored when E°cell > 0; ΔG° = −nFE°
moles e− = I·t / 96,485; then use mole ratios to get metal deposited
RED CAT / AN OX - reduction at cathode, oxidation at anode (both cell types)
Question bank breakdown
43
2
13 easy · 22 medium · 8 hard
Skills covered
Every question in the bank comes with a worked explanation, and most add a per-choice breakdown so you learn why each wrong answer is wrong - not just the key.
Sample question
From the free tier of the ScoreMint AP Chemistry bank - try it, then check the answer.
For which of the following reactions is ΔS° expected to be the most positive (greatest increase in entropy)?
- A. 2 H₂(g) + O₂(g) → 2 H₂O(l)
- B. CaCO₃(s) → CaO(s) + CO₂(g)
- C. N₂(g) + 3 H₂(g) → 2 NH₃(g)
- D. AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)
Show answer and explanation
Answer: B. Correct. This reaction goes from 1 mol solid → 1 mol solid + 1 mol gas. A gas is produced from a solid, which represents a large increase in entropy (gases have much greater entropy than solids due to greater molecular freedom). The increase from 0 mol gas on the left to 1 mol gas on the right means ΔS° is strongly positive. ✓
Study tip
Keep units honest in ΔG = ΔH − TΔS: ΔH is usually kJ/mol while ΔS° is J/(mol·K). Convert ΔS to kJ before subtracting - the exam plants answer choices that punish skipping this step.