AP Chemistry Thermodynamics FRQ — Practice Questions & Scoring Guide 2026
Thermodynamics is one of the four highest-frequency FRQ topics on AP Chemistry alongside equilibrium, kinetics, and electrochemistry. It appears on virtually every exam. This guide covers the key concepts, how AP readers score thermodynamics FRQs, and practice questions with complete solutions.
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What AP Chem Thermodynamics FRQs Test
AP Chemistry thermodynamics FRQs draw from Unit 6 (Thermodynamics). Most-tested concepts:
| Concept | Frequency |
|---|---|
| ΔH from bond enthalpies or Hess's law | Every year |
| Predicting sign of ΔS | Every year |
| ΔG = ΔH − TΔS | Every year |
| Spontaneity and conditions for spontaneous reaction | Every year |
| Relationship between ΔG° and equilibrium constant K | Every 1–2 years |
| Calorimetry (q = mcΔT) | Every 2–3 years |
Practice FRQ 1: Hess's Law
Problem: Calculate ΔH° for the reaction:
$$\text{C}(s) + 2\text{H}_2(g) \rightarrow \text{CH}_4(g)$$
Given:
- $\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l)$, $\Delta H_1° = -890$ kJ
- $\text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g)$, $\Delta H_2° = -393$ kJ
- $\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(l)$, $\Delta H_3° = -286$ kJ
Solution:
Target reaction: $\text{C}(s) + 2\text{H}_2(g) \rightarrow \text{CH}_4(g)$
Strategy: Reverse reaction 1 (to get CH₄ as product), keep reaction 2 as-is, multiply reaction 3 by 2.
Reverse reaction 1: $$\text{CO}_2(g) + 2\text{H}_2\text{O}(l) \rightarrow \text{CH}_4(g) + 2\text{O}_2(g), \quad \Delta H = +890 \text{ kJ}$$
Reaction 2 (unchanged): $$\text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g), \quad \Delta H = -393 \text{ kJ}$$
Reaction 3 × 2: $$2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(l), \quad \Delta H = -572 \text{ kJ}$$
Add all three: CO₂, 2H₂O, and O₂ cancel.
$$\Delta H° = +890 + (-393) + (-572) = -75 \text{ kJ}$$
AP Reader note: Show each step with the reversed/multiplied equations written out. Do not just combine numbers — show the cancellation.
Practice FRQ 2: Predicting ΔS
Problem: For each reaction, predict whether ΔS° is positive, negative, or near zero. Justify your answer.
(a) $\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)$
(b) $\text{NaCl}(s) \rightarrow \text{Na}^+(aq) + \text{Cl}^-(aq)$
(c) $\text{H}_2\text{O}(g) \rightarrow \text{H}_2\text{O}(l)$
Solution:
(a) ΔS° < 0 (negative) 4 moles of gas → 2 moles of gas. Fewer gas molecules means less disorder. Entropy decreases.
(b) ΔS° > 0 (positive) Ordered solid → dispersed ions in solution. Dissolution increases the freedom of motion of ions. Entropy increases.
(c) ΔS° < 0 (negative) Gas → liquid. Gas phase has much higher entropy than liquid. Entropy decreases significantly.
Rules for predicting ΔS sign:
- More gas moles on products → ΔS > 0
- Fewer gas moles on products → ΔS < 0
- Solid → liquid or aqueous → ΔS > 0
- Gas → liquid or solid → ΔS < 0
- More particles on products → ΔS > 0
Practice FRQ 3: Gibbs Free Energy and Spontaneity
Problem: A reaction has ΔH° = −120 kJ/mol and ΔS° = −200 J/(mol·K).
(a) Calculate ΔG° at 25°C (298 K). Is the reaction spontaneous? (b) At what temperature does the reaction become non-spontaneous?
Solution:
(a) $\Delta G° = \Delta H° - T\Delta S°$
Convert ΔS to kJ: $-200 \text{ J/mol·K} = -0.200 \text{ kJ/mol·K}$
$$\Delta G° = -120 - (298)(-0.200) = -120 + 59.6 = -60.4 \text{ kJ/mol}$$
ΔG° < 0 → spontaneous at 25°C.
(b) Reaction becomes non-spontaneous when ΔG° > 0, i.e., when ΔH° − TΔS° > 0.
$$T > \frac{\Delta H°}{\Delta S°} = \frac{-120 \text{ kJ/mol}}{-0.200 \text{ kJ/mol·K}} = 600 \text{ K}$$
Above 600 K, the reaction is non-spontaneous. (Both ΔH and ΔS are negative — entropy term dominates at high T.)
Practice FRQ 4: ΔG° and the Equilibrium Constant
Problem: For a reaction at 298 K, ΔG° = −17.1 kJ/mol. Calculate the equilibrium constant K.
Solution:
Relationship: $\Delta G° = -RT\ln K$
$$-17100 = -(8.314)(298)\ln K$$
$$\ln K = \frac{17100}{8.314 \times 298} = \frac{17100}{2477.6} = 6.90$$
$$K = e^{6.90} = 992 \approx 1.0 \times 10^3$$
Since K >> 1, the reaction strongly favors products at equilibrium — consistent with ΔG° < 0.
Important: Use ΔG° (standard conditions), not ΔG, for this relationship. ΔG = ΔG° + RT ln Q is a different equation for non-standard conditions.
Spontaneity Summary Table
This table is essential for AP Chem thermodynamics FRQs:
| ΔH | ΔS | Spontaneous? |
|---|---|---|
| − | + | Always spontaneous (ΔG always −) |
| + | − | Never spontaneous (ΔG always +) |
| − | − | Spontaneous at low T only |
| + | + | Spontaneous at high T only |
When asked "under what conditions is the reaction spontaneous," use this table directly.
AP Chem Thermodynamics FRQ Scoring Tips
Units are a common point loss. ΔH is in kJ/mol, ΔS is in J/(mol·K). Before using ΔG = ΔH − TΔS, convert ΔS to kJ/(mol·K) or ΔH to J/mol. Mixing units is the #1 arithmetic error on thermodynamics FRQs.
Justify ΔS sign qualitatively. AP readers want more than "more products." Say: "More moles of gas on the product side means greater molecular disorder, so ΔS > 0."
When using ΔG° = −RT ln K: R = 8.314 J/(mol·K), and ΔG° must be in joules (not kJ) unless you use R = 0.008314 kJ/(mol·K).
Hess's Law: Always write out the modified reactions before adding. Partial credit is available at each step — showing your work is worth points even if the final answer is wrong.