AP Chemistry / Topic
Equilibrium
Unit 7 covers reversible reactions and the equilibrium constant: writing K expressions, reaction quotient Q, ICE tables, Le Châtelier's principle, and solubility equilibria with Ksp.
What the exam asks
- Write Kc and Kp expressions, leaving out pure solids and liquids
- Compare Q to K to predict which direction a system shifts
- Solve ICE-table problems for equilibrium concentrations
- Predict shifts from concentration, pressure/volume, and temperature changes (Le Châtelier)
- Work Ksp problems: molar solubility, common-ion effect, and whether a precipitate forms
Key formulas and rules
K = [products]^coeff / [reactants]^coeff (aq and gas species only)
Q < K: shifts right; Q > K: shifts left; Q = K: at equilibrium
reverse: 1/K; multiply by n: K^n; add reactions: multiply K's
only temperature changes the value of K; heat acts like a reactant (endo) or product (exo)
existing ion concentration lowers solubility
Question bank breakdown
44
2
8 easy · 20 medium · 16 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.
For the reaction N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g), the equilibrium constant Kc = 0.50 at a certain temperature. If the equilibrium concentrations are [N₂] = 1.0 M and [H₂] = 2.0 M, what is [NH₃] at equilibrium?
- A. 1.0 M
- B. 2.0 M
- C. 4.0 M
- D. 0.50 M
Show answer and explanation
Answer: B. Correct. The equilibrium expression is Kc = [NH₃]² / ([N₂][H₂]³). Substitute known values: 0.50 = [NH₃]² / [(1.0)(2.0)³]. 0.50 = [NH₃]² / (1.0 × 8.0) = [NH₃]² / 8.0. Solve: [NH₃]² = 0.50 × 8.0 = 4.0. [NH₃] = √4.0 = 2.0 M. Check: Kc = (2.0)² / [(1.0)(2.0)³] = 4.0 / 8.0 = 0.50 ✓.
Study tip
Le Châtelier answers earn points only with a Q-versus-K argument: say which concentration changed, what that does to Q, and which way the system must shift to make Q equal K again.