AP Chemistry Unit 7 review: Equilibrium

• Treating equilibrium as stopped reactions or equal concentrations. The rates are equal; reactant and product concentrations may differ. CED topic 7.1.
• Ignoring a very large K when choosing an approximation. A product-favored reaction may be nearly complete, so an assumption of tiny conversion can fail. CED printed p. 125 highlights this problem.
• Including a pure solid or pure liquid in a standard equilibrium expression. Its activity is fixed in the model. CED topic 7.3.
• Assuming pressure always shifts a gas equilibrium. Compare gaseous stoichiometric totals and explain how Q changes; equal totals give no compression-driven shift in the ideal model. CED topics 7.9–7.10.

1. Question 1 · 7.4

   For A(g) ⇌ 2B(g), equilibrium concentrations are [A] = 0.200 M and [B] = 0.300 M. What is Kc?

   • A0.667
   • B1.50
   • C0.450
   • D2.25
   Show the answer

   C. Kc = [B]²/[A] = (0.300)²/0.200 = 0.450. The coefficient 2 becomes an exponent, not a factor multiplying [B].

2. Question 2 · 7.10

   For A ⇌ B, K = 4.0. A mixture has [A] = 0.50 M and [B] = 1.0 M. Which change occurs as equilibrium is approached?

   • ANet conversion of A to B
   • BNet conversion of B to A
   • CNo net reaction because Q = K
   • DK decreases until it equals Q
   Show the answer

   A. Q = [B]/[A] = 1.0/0.50 = 2.0. Since Q < K, the forward reaction increases the ratio toward 4.0. At fixed temperature, K does not adjust to the initial mixture.

3. Question 3 · 7.6

   The reaction A ⇌ B has K = 9.0. What is K for 2B ⇌ 2A at the same temperature?

   • A18
   • B81
   • C1/18
   • D1/81
   Show the answer

   D. Reversing gives 1/9. Doubling all coefficients squares that value: (1/9)² = 1/81. Doubling a reaction does not merely double K.

4. Question 4 · 7.7

   Initially [A] = 0.600 M and [B] = 0 in A ⇌ B at constant volume. If K = 2.00, what is [B] at equilibrium?

   • A0.200 M
   • B0.400 M
   • C0.300 M
   • D0.600 M
   Show the answer

   B. Let x M of A convert to B. Then K = x/(0.600 − x) = 2.00. Solving x = 1.200 − 2x gives x = 0.400 M, leaving [A] = 0.200 M.

5. Question 5 · 7.11

   For MX₂(s) ⇌ M²⁺ + 2X⁻, Ksp = 1.08 × 10⁻¹⁰. What is the molar solubility in pure water?

   • A1.04 × 10⁻⁵ M
   • B4.76 × 10⁻⁴ M
   • C3.00 × 10⁻⁴ M
   • D1.08 × 10⁻¹⁰ M
   Show the answer

   C. If solubility is s, [M²⁺] = s and [X⁻] = 2s. Ksp = s(2s)² = 4s³. Thus s³ = 2.70 × 10⁻¹¹ and s = 3.00 × 10⁻⁴ M.

6. Question 6 · 7.12

   AgCl has Ksp = 1.8 × 10⁻¹⁰. In 0.020 M NaCl, neglecting added chloride from dissolution, what is [Ag⁺] at saturation?

   • A9.0 × 10⁻⁹ M
   • B3.6 × 10⁻¹² M
   • C1.3 × 10⁻⁵ M
   • D9.0 × 10⁻⁷ M
   Show the answer

   A. Ksp = [Ag⁺][Cl⁻], so [Ag⁺] = 1.8 × 10⁻¹⁰/0.020 = 9.0 × 10⁻⁹ M. This is much smaller than 0.020 M, consistent with neglecting the extra chloride.

For each disturbance, write the expression for Q and mark which term changes immediately. Compare that new Q with unchanged K unless temperature changes. Only then state the net reaction direction.

For an equilibrium calculation, define x and write the stoichiometric changes before solving. Substitute the answer into the original K expression and check that no concentration is negative. For solubility, derive every ion concentration from the dissolution equation rather than setting all of them equal to s.

Ashore's Chemistry course has a short lesson for every topic in this unit, each ending with a few questions on what it taught.

• AP Chemistry: all 9 units, exam format and scores
• AP Chemistry Unit 1: Atomic Structure and Properties
• AP Chemistry Unit 2: Compound Structure and Properties
• AP Chemistry Unit 3: Properties of Substances and Mixtures
• AP Chemistry Unit 4: Chemical Reactions
• AP Chemistry Unit 5: Kinetics
• AP Chemistry Unit 6: Thermochemistry
• AP Chemistry Unit 8: Acids and Bases
• AP Chemistry Unit 9: Thermodynamics and Electrochemistry