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AP Chemistry Practice Quiz: Calculating the Equilibrium Constant

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 7 questions to check your progress.

Question 1 of 7

Consider the following reaction at equilibrium: H₂(g) + I₂(g) ⇌ 2HI(g). An experimental measurement finds the following equilibrium concentrations: [H₂] = 0.10 M, [I₂] = 0.10 M, and [HI] = 0.80 M. Based on these observations, what is the value of the equilibrium constant, Kc?

All Questions (7)

Consider the following reaction at equilibrium: H₂(g) + I₂(g) ⇌ 2HI(g). An experimental measurement finds the following equilibrium concentrations: [H₂] = 0.10 M, [I₂] = 0.10 M, and [HI] = 0.80 M. Based on these observations, what is the value of the equilibrium constant, Kc?

A) 8.0

B) 16

C) 64

D) 0.016

Correct Answer: C

The equilibrium constant expression for this reaction is Kc = [HI]² / ([H₂][I₂]). Plugging in the experimentally measured concentrations gives: Kc = (0.80)² / (0.10 * 0.10) = 0.64 / 0.010 = 64.

For the reversible reaction N₂O₄(g) ⇌ 2NO₂(g), a system reaches equilibrium. Experimental measurements of the partial pressures are P(N₂O₄) = 0.50 atm and P(NO₂) = 1.0 atm. What is the value of the equilibrium constant, Kp, for this reaction?

A) 0.50

B) 1.0

C) 2.0

D) 4.0

Correct Answer: C

The expression for Kp is the partial pressure of the products raised to their stoichiometric coefficients divided by that of the reactants. Kp = P(NO₂)² / P(N₂O₄). Substituting the given equilibrium pressures: Kp = (1.0 atm)² / 0.50 atm = 1.0 / 0.50 = 2.0.

The synthesis of sulfur trioxide proceeds according to the equation: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g). At equilibrium, the measured concentrations are [SO₂] = 2.0 M, [O₂] = 1.0 M, and [SO₃] = 4.0 M. What is the value of Kc?

A) 0.50

B) 1.0

C) 2.0

D) 4.0

Correct Answer: D

The equilibrium constant expression is Kc = [SO₃]² / ([SO₂]²[O₂]). Using the provided equilibrium concentrations: Kc = (4.0)² / ((2.0)² * 1.0) = 16 / (4.0 * 1.0) = 16 / 4.0 = 4.0.

For the decomposition of phosphorus pentachloride, PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), the following equilibrium concentrations were observed in an experiment: [PCl₅] = 0.80 M, [PCl₃] = 0.20 M, and [Cl₂] = 0.20 M. Calculate the value of Kc.

A) 20

B) 5.0

C) 0.20

D) 0.050

Correct Answer: D

The equilibrium expression is Kc = ([PCl₃][Cl₂]) / [PCl₅]. Substituting the measured values: Kc = (0.20 * 0.20) / 0.80 = 0.040 / 0.80 = 0.050. Option A represents the inverse of the correct calculation.

The Haber process for ammonia synthesis is represented by N₂(g) + 3H₂(g) ⇌ 2NH₃(g). In an equilibrium mixture, the partial pressures were measured to be P(N₂) = 2.0 atm, P(H₂) = 1.0 atm, and P(NH₃) = 2.0 atm. What is the value of Kp for this reaction under these conditions?

A) 0.50

B) 1.0

C) 2.0

D) 4.0

Correct Answer: C

The expression for Kp is P(NH₃)² / (P(N₂) * P(H₂)³). Plugging in the experimental values: Kp = (2.0)² / (2.0 * (1.0)³) = 4.0 / (2.0 * 1.0) = 2.0. Common mistakes include forgetting the exponent for H₂ (which would lead to answer D) or inverting the expression (which would lead to answer A).

Consider the decomposition of calcium carbonate: CaCO₃(s) ⇌ CaO(s) + CO₂(g). An experiment is run in a closed container, and at equilibrium, the partial pressure of CO₂(g) is measured to be 0.25 atm. Solid CaCO₃ and CaO are both present. What is the value of Kp?

A) 0.25

B) 0.50

C) 1.0

D) 4.0

Correct Answer: A

The equilibrium constant expression for a heterogeneous equilibrium does not include pure solids or liquids. Therefore, Kp = P(CO₂). Based on the experimental measurement, Kp = 0.25.

For the reaction 4HCl(g) + O₂(g) ⇌ 2H₂O(g) + 2Cl₂(g), an equilibrium mixture was analyzed and found to contain [HCl] = 0.20 M, [O₂] = 0.10 M, [H₂O] = 0.40 M, and [Cl₂] = 0.40 M. Based on these experimental data, what is the calculated value of Kc?

A) 8.0

B) 40

C) 160

D) 400

Correct Answer: C

The equilibrium constant expression is Kc = ([H₂O]²[Cl₂]²) / ([HCl]⁴[O₂]). Substituting the given concentrations: Kc = ((0.40)² * (0.40)²) / ((0.20)⁴ * 0.10) = (0.16 * 0.16) / (0.0016 * 0.10) = 0.0256 / 0.00016 = 160.