AP Chemistry / Topic
Thermodynamics
Unit 6 (Thermochemistry) covers energy transfer as heat, endothermic and exothermic processes, calorimetry, enthalpy of reaction, Hess's law, formation enthalpies, and bond-energy estimates.
What the exam asks
- Distinguish endothermic and exothermic changes and draw their energy diagrams
- Run q = mcΔT calorimetry calculations, including metal-in-water thermal equilibrium problems
- Compute reaction enthalpy from Hess's law, formation enthalpies, or bond energies
- Relate heating/cooling curves and phase changes to heat flow
- Track energy in particle terms: bond breaking absorbs, bond forming releases
Key formulas and rules
q = m·c·ΔT; heat lost = −heat gained
q = n·ΔH(fusion or vaporization), temperature constant during the change
ΔH(target) = Σ ΔH(steps); reverse a step ⇒ flip the sign
ΔH°rxn = Σ ΔHf°(products) − Σ ΔHf°(reactants)
ΔH ≈ Σ bonds broken − Σ bonds formed
exothermic: ΔH < 0, surroundings warm; endothermic: ΔH > 0
Question bank breakdown
31
1
8 easy · 13 medium · 10 hard
Skills covered
Every question in the bank comes with a per-choice explanation, 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.
Given the following standard enthalpies of formation: ΔH°f [CO₂(g)] = -393.5 kJ/mol ΔH°f [H₂O(l)] = -285.8 kJ/mol ΔH°f [C₂H₆(g)] = -84.7 kJ/mol ΔH°f [O₂(g)] = 0 kJ/mol What is ΔH°rxn for the reaction as written? 2 C₂H₆(g) + 7 O₂(g) → 4 CO₂(g) + 6 H₂O(l)
- A. -2856.6 kJ
- B. -3119.4 kJ
- C. -1559.7 kJ
- D. +3119.4 kJ
Show answer and explanation
Answer: B. Correct. Using Hess's law: ΔH°rxn = Σ[ΔH°f(products)] - Σ[ΔH°f(reactants)]. Products: 4 mol CO₂ × (-393.5) + 6 mol H₂O × (-285.8) = -1574.0 + (-1714.8) = -3288.8 kJ. Reactants: 2 mol C₂H₆ × (-84.7) + 7 mol O₂ × (0) = -169.4 + 0 = -169.4 kJ. ΔH°rxn = -3288.8 - (-169.4) = -3288.8 + 169.4 = -3119.4 kJ. The large negative value confirms this is highly exothermic, as expected for combustion. ✓
Study tip
In mixing calorimetry, the mass in q = mcΔT is the water (or total solution) mass, not the solute's — and both objects end at the same final temperature. Set q(hot) = −q(cold) and solve.