---
title: "AP Chemistry Unit 7 Practice Test"
description: "A timed AP Chemistry Unit 7 (Equilibrium) practice test in the AP format, drawn from 3,429 multiple-choice questions and 56 FRQs written for the unit. The standard version takes about 60 minutes."
canonical: "https://fiveable.me/ap-chem/unit-test/7"
type: "unit-test"
subject: "AP Chemistry"
unit: "Unit 7"
---

# AP Chemistry Unit 7 Practice Test

## Overview

A timed AP Chemistry Unit 7 (Equilibrium) practice test in the AP format, drawn from 3,429 multiple-choice questions and 56 FRQs written for the unit. The standard version takes about 60 minutes.

## Test Format

- 20 Multiple Choice Questions (30 minutes)
- 1 FRQ – FRQ 1 – Long Answer (15 minutes)
- 1 FRQ – FRQ 4 – Short Answer (15 minutes)

## Unit FRQs by Type

- [FRQ 1 – Long Answer](/ap-chem/frqs/frq-1?unit=7): 8 questions
- [FRQ 2 – Long Answer](/ap-chem/frqs/frq-2?unit=7): 5 questions
- [FRQ 3 – Long Answer](/ap-chem/frqs/frq-3?unit=7): 5 questions
- [FRQ 4 – Short Answer](/ap-chem/frqs/frq-4?unit=7): 13 questions
- [FRQ 5 – Short Answer](/ap-chem/frqs/frq-5?unit=7): 13 questions
- [FRQ 6 – Short Answer](/ap-chem/frqs/frq-6?unit=7): 8 questions
- [FRQ 7 – Short Answer](/ap-chem/frqs/frq-7?unit=7): 4 questions

## Practice Questions

### Question 1

A student observes that heating a sealed glass tube containing an equilibrium mixture of NO₂(g) (brown) and N₂O₄(g) (colorless) causes the mixture to turn darker brown. Given that the forward reaction 2NO₂(g) ⇌ N₂O₄(g) is exothermic, which explanation best justifies this observation using particulate-level reasoning?

- (A) The reverse reaction has a higher activation energy, so its rate increases more than the forward rate.
- (B) The forward reaction has a higher activation energy, so its rate increases more than the reverse rate.
- (C) The reverse reaction is endothermic, so its rate increases while the forward reaction rate decreases.
- (D) The forward reaction is exothermic, so its rate increases while the reverse reaction rate decreases.

**Answer: (A)** Connects the endothermic nature of the reverse reaction to kinetics. Because the reverse reaction is endothermic, it has a higher activation energy; therefore, an increase in temperature increases the fraction of successful collisions for the reverse step more significantly than for the forward step, causing a net shift toward reactants.

### Question 2

A student claims that decreasing the volume of a reaction vessel containing the equilibrium system 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) will increase the amount of SO₃(g) produced. Which of the following provides the best particulate-level justification for this claim?

- (A) Increased pressure increases the collision frequency for the forward reaction more than for the reverse reaction.
- (B) Increased pressure increases the collision frequency for the reverse reaction more than for the forward reaction.
- (C) Increased pressure lowers the activation energy barrier for the forward reaction more than for the reverse reaction.
- (D) Increased pressure raises the equilibrium constant value for the forward reaction more than for the reverse reaction.

**Answer: (A)** Connects partial pressure changes to collision theory. Decreasing volume increases the partial pressures of all gases; since the forward reaction involves 3 moles of gas colliding compared to 2 moles for the reverse, the frequency of effective collisions increases more drastically for the forward direction.

### Question 3

A student adds a solution of Fe(NO₃)₃ to an equilibrium mixture of Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq) and observes that the solution turns a darker red. Which reasoning best justifies this observation based on particulate-level interactions?

- (A) Higher iron(III) concentration increases the forward reaction rate by increasing the frequency of collisions with thiocyanate.
- (B) Higher iron(III) concentration shifts equilibrium toward products by increasing collision frequency, but the solution appears darker because excess Fe³⁺ ions absorb more light.
- (C) Higher iron(III) concentration increases the forward reaction rate by increasing the concentration gradient between reactants and products.
- (D) Higher iron(III) concentration increases the forward reaction rate by providing more activated Fe³⁺ ions with sufficient energy to react with thiocyanate.

**Answer: (A)** Links concentration to collision theory. Adding reactant ions increases the number of Fe³⁺ particles available to collide with SCN⁻ ions, thereby increasing the rate of the forward reaction relative to the reverse reaction until a new equilibrium is established.

### Question 4

A student adds distilled water to a beaker containing a saturated solution of Mg(OH)₂ in contact with excess solid Mg(OH)₂ and observes that more solid dissolves. Which explanation best justifies this observation using particulate-level reasoning?

- (A) Dilution lowers ion concentrations, reducing the crystallization rate below the dissolution rate.
- (B) Dilution lowers ion concentrations, reducing the dissolution rate below the crystallization rate.
- (C) Dilution raises the solubility product, raising the dissolution rate above the crystallization rate.
- (D) Dilution raises the lattice energy, raising the crystallization rate above the dissolution rate.

**Answer: (A)** Connects dilution to the rates of opposing processes. Dilution decreases the concentration of dissolved ions, which reduces the frequency of collisions required for crystallization (reverse reaction), while the rate of dissolution (forward reaction) remains constant for the solid, causing a net shift toward dissolution.

### Question 5

The energy profile for an elementary step shows the forward activation energy is 20 kJ/mol and the reverse activation energy is 80 kJ/mol. Which statement links these barriers to the equilibrium constant $$K$$?

- (A) $$K > 1$$ because the forward activation energy is smaller than the reverse activation energy.
- (B) $$K < 1$$ because the forward activation energy is smaller than the reverse activation energy.
- (C) $$K > 1$$ because the forward activation energy is larger than the reverse activation energy.
- (D) $$K < 1$$ because the forward activation energy is larger than the reverse activation energy.

**Answer: (A)** Connects the lower forward barrier to a higher forward rate constant relative to the reverse, resulting in an equilibrium constant greater than 1.

### Question 6

A proposed mechanism for the reaction $$2O_3 \rightarrow 3O_2$$ involves a fast reversible step $$O_3 \rightleftharpoons O_2 + O$$ followed by a slow step $$O + O_3 \rightarrow 2O_2$$. Based on the pre-equilibrium approximation, how is the concentration of the intermediate $$O$$ expressed in terms of the stable reactants and products?

- (A) $$[O] = K_{eq} \frac{[O_3]}{[O_2]}$$
- (B) $$[O] = K_{eq} [O_3][O_2]$$
- (C) $$[O] = k [O_3]^2$$
- (D) $$[O] = \frac{k_1}{k_2} [O_3]$$

**Answer: (A)** Analyzing the fast equilibrium step $$O_3 \rightleftharpoons O_2 + O$$ allows for the derivation of the intermediate's concentration. At equilibrium, $$Rate_{fwd} = Rate_{rev}$$, so $$k_1[O_3] = k_{-1}[O_2][O]$$. Solving for $$[O]$$ yields $$K_{eq}[O_3]/[O_2]$$.

### Question 7

Consider a mechanism where Step 1 is $$A + B \rightleftharpoons C$$ (fast, equilibrium constant $$K$$) and Step 2 is $$C + A \rightarrow D$$ (slow, rate constant $$k_2$$). Which equation correctly maps the particulate rate constants to the macroscopic observed rate constant $$k_{obs}$$ for the overall rate law $$Rate = k_{obs}[A]^2[B]$$?

- (A) $$k_{obs} = k_2 K$$
- (B) $$k_{obs} = k_2 / K$$
- (C) $$k_{obs} = K / k_2$$
- (D) $$k_{obs} = k_2 + K$$

**Answer: (A)** Deriving the rate law involves substituting $$[C] = K[A][B]$$ into the slow step rate $$Rate = k_2[C][A]$$. This yields $$Rate = k_2(K[A][B])[A] = k_2 K [A]^2[B]$$, meaning the observed constant is the product $$k_2 K$$.

### Question 8

The reaction $$2NO + Cl_2 \rightarrow 2NOCl$$ follows a mechanism with a fast equilibrium step $$NO + Cl_2 \rightleftharpoons NOCl_2$$ followed by a slow step $$NOCl_2 + NO \rightarrow 2NOCl$$. Which experimental rate law supports this mechanism?

- (A) $$Rate = k[NO]^2[Cl_2]$$, because the intermediate $$NOCl_2$$ concentration depends on $$[NO]$$ and $$[Cl_2]$$
- (B) $$Rate = k[NO][Cl_2]$$, because the first step is the rate-determining step involving $$NO$$ and $$Cl_2$$
- (C) $$Rate = k[NOCl_2][NO]$$, because the slow step involves a collision between $$NOCl_2$$ and $$NO$$
- (D) $$Rate = k[NO]^2[Cl_2]^2$$, because the overall reaction order must match the stoichiometric coefficients

**Answer: (A)** Deriving the rate law involves substituting the intermediate $$NOCl_2$$. The fast equilibrium establishes $$[NOCl_2] = K[NO][Cl_2]$$, and the slow step rate is $$k[NOCl_2][NO]$$, combining to give $$k'[NO]^2[Cl_2]$$.

### Question 9

The decomposition of ozone follows the mechanism below:
Step 1: $$O_3 \rightleftharpoons O_2 + O$$ (fast equilibrium)
Step 2: $$O + O_3 \rightarrow 2O_2$$ (slow)
Which rate law is consistent with this mechanism?

- (A) $$Rate = k[O_3]^2[O_2]^{-1}$$, derived by substituting $$[O]$$ from the equilibrium expression
- (B) $$Rate = k[O_3][O_2]$$, derived by assuming the first step determines the overall rate
- (C) $$Rate = k[O_3]^2$$, derived by omitting the product $$O_2$$ from the equilibrium expression
- (D) $$Rate = k[O][O_3]$$, derived by using the reactants directly from the rate-determining step

**Answer: (A)** Solving for the intermediate $$[O]$$ from Step 1 gives $$[O] = K[O_3]/[O_2]$$. Substituting this into the Step 2 rate law $$Rate = k[O][O_3]$$ results in an inverse dependence on $$[O_2]$$.

### Question 10

For the reaction $$H_2 + I_2 \rightarrow 2HI$$, the proposed mechanism involves a fast equilibrium $$I_2 \rightleftharpoons 2I$$ followed by a slow step $$H_2 + 2I \rightarrow 2HI$$. Why is the observed rate law $$Rate = k[H_2][I_2]$$ consistent with this mechanism?

- (A) The concentration of the intermediate $$I$$ is related to $$[I_2]$$ by the equilibrium constant $$K = [I]^2/[I_2]$$
- (B) The concentration of the intermediate $$I$$ is related to $$[I_2]$$ by the rate constant $$k = [I]/[I_2]$$
- (C) The rate-determining step involves $$H_2$$ and $$I_2$$ directly colliding, ignoring the intermediate $$I$$
- (D) The fast equilibrium step dictates the rate law because it occurs before the rate-determining step

**Answer: (A)** Justifies the substitution of the intermediate. The slow step depends on $$[I]^2$$, and since the fast equilibrium establishes $$[I]^2 = K[I_2]$$, the rate law becomes first order in $$I_2$$, matching the experimental observation.

## Review the Unit

- [7.1 Introduction to Equilibrium](/ap-chem/unit-7/intro-equilibrium/study-guide/dzIPBIOsEPKoTL4VKEH2)
- [7.2 Direction of Reversible Reactions](/ap-chem/unit-7/direction-reversible-reactions/study-guide/QKsYOmy4YBiELg6YUHOv)
- [7.3 Reaction Quotient and Equilibrium Constant](/ap-chem/unit-7/reaction-quotient-equilibrium-constant/study-guide/Y12LZEr3OWIAqV3gErat)
- [7.4 Calculating the Equilibrium Constant](/ap-chem/unit-7/calculating-equilibrium-constant/study-guide/WMmdcS9qD8yHDQwepK31)
- [7.5 Magnitude of the Equilibrium Constant](/ap-chem/unit-7/magnitude-equilibrium-constant/study-guide/dvXT7PLceyYd2QH8KiV4)
- [7.6 Properties of the Equilibrium Constant](/ap-chem/unit-7/properties-equilibrium-constant/study-guide/PGLxCyabBaR6axW5E18w)
- [7.7 Calculating Equilibrium Concentrations](/ap-chem/unit-7/calculating-equilibrium-concentrations/study-guide/ou9xNlxg758Auz6D3WIO)
- [7.8 Representations of Equilibrium](/ap-chem/unit-7/representations-equilibrium/study-guide/wLQChBkGSKiEP5xvlXB8)
- [7.9 Introduction to Le Châtelier’s Principle](/ap-chem/unit-7/intro-le-chateliers-principle/study-guide/ST8UE6kcdhi0hkA5bSkN)
- [7.10 Reaction Quotient and Le Châtelier’s Principle](/ap-chem/unit-7/reaction-quotient-le-chateliers-principle/study-guide/JFx1InPfZCZ9SugPKDCE)
- [7.11 Introduction to Solubility Equilibria](/ap-chem/unit-7/intro-solubility-equilibria/study-guide/lkqnvEmZIxrvjX1sogng)
- [7.12 Common Ion Effect](/ap-chem/unit-7/common-ion-effect/study-guide/z1tr3Mg2Afz6kj9yQGtc)

## Other Unit Tests

- [Unit 1: Atomic Structure & Properties](/ap-chem/unit-test/1)
- [Unit 2: Compound Structure and Properties](/ap-chem/unit-test/2)
- [Unit 3: Properties of Substances and Mixtures](/ap-chem/unit-test/3)
- [Unit 4: Chemical Reactions](/ap-chem/unit-test/4)
- [Unit 5: Kinetics](/ap-chem/unit-test/5)
- [Unit 6: Thermochemistry](/ap-chem/unit-test/6)
- [Unit 8: Acids & Bases](/ap-chem/unit-test/8)
- [Unit 9: Thermodynamics and Electrochemistry](/ap-chem/unit-test/9)

## About This Document

Canonical Fiveable pages are available as Markdown at the same path plus `.md`.

- [llms.txt](https://fiveable.me/llms.txt): index of Fiveable's sections and URL patterns
- [llms-full.txt](https://fiveable.me/llms-full.txt): complete subject and unit listing
- [MCP server](https://fiveable.me/mcp): call Fiveable as tools instead of fetching pages (`https://fiveable.me/api/mcp`)
- [MCP server for AP teachers](https://fiveable.me/mcp/teachers): a teacher's classes, assignments and AP-rubric grading (`https://fiveable.me/api/mcp/teacher`)