AP Chemistry Unit 7 practice questions
Try 10 multiple-choice questions from Unit 7 here. Pick an answer to see why it's right or wrong, then take the timed test above for a full AP-style set.
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?
The reverse reaction has a higher activation energy, so its rate increases more than the forward rate.
The forward reaction has a higher activation energy, so its rate increases more than the reverse rate.
The reverse reaction is endothermic, so its rate increases while the forward reaction rate decreases.
The forward reaction is exothermic, so its rate increases while the reverse reaction rate decreases.
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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?
Increased pressure increases the collision frequency for the forward reaction more than for the reverse reaction.
Increased pressure increases the collision frequency for the reverse reaction more than for the forward reaction.
Increased pressure lowers the activation energy barrier for the forward reaction more than for the reverse reaction.
Increased pressure raises the equilibrium constant value for the forward reaction more than for the reverse reaction.
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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?
Higher iron(III) concentration increases the forward reaction rate by increasing the frequency of collisions with thiocyanate.
Higher iron(III) concentration shifts equilibrium toward products by increasing collision frequency, but the solution appears darker because excess Fe³⁺ ions absorb more light.
Higher iron(III) concentration increases the forward reaction rate by increasing the concentration gradient between reactants and products.
Higher iron(III) concentration increases the forward reaction rate by providing more activated Fe³⁺ ions with sufficient energy to react with thiocyanate.
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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?
Dilution lowers ion concentrations, reducing the crystallization rate below the dissolution rate.
Dilution lowers ion concentrations, reducing the dissolution rate below the crystallization rate.
Dilution raises the solubility product, raising the dissolution rate above the crystallization rate.
Dilution raises the lattice energy, raising the crystallization rate above the dissolution rate.
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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
A particulate representation of the reaction A(g) ⇌ B(g) at equilibrium with K = 0.5 shows a vessel containing 8 atoms of A and 2 atoms of B. Based on this model, which statement correctly describes the relationship between Q and K?
Q is 0.25, which is less than K, so the reaction proceeds toward products
Q is 0.25, which is less than K, so the reaction proceeds toward reactants
Q is 4.0, which is greater than K, so the reaction proceeds toward products
Q is 4.0, which is greater than K, so the reaction proceeds toward reactants
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Answer: (A)
Calculating Q as [B]/[A] yields 2/8 = 0.25. Since 0.25 < 0.5, Q is less than K, indicating the forward reaction will proceed to form more product.
Question 6
A reaction vessel contains 2 molecules of NO₂(g) and 4 molecules of N₂O₄(g) for the reaction 2NO₂(g) ⇌ N₂O₄(g). If Kc = 4.0 at this temperature, what does the particulate data indicate about the system's status?
Q is 1.0, indicating the net reaction proceeds to form more N₂O₄
Q is 1.0, indicating the net reaction proceeds to form more NO₂
Q is 2.0, indicating the net reaction proceeds to form more N₂O₄
Q is 2.0, indicating the net reaction proceeds to form more NO₂
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Answer: (A)
Q is calculated as [N₂O₄]/[NO₂]² = 4/(2)² = 1.0. Since 1.0 < 4.0 (Q < K), the reaction proceeds to the right to form more product (N₂O₄).
Question 7
For the equilibrium PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), a particulate model shows the system volume is suddenly halved at constant temperature. How does this disturbance affect Q relative to K immediately after the volume change?
Q becomes greater than K, causing a net shift toward reactants
Q becomes greater than K, causing a net shift toward products
Q becomes less than K, causing a net shift toward reactants
Q becomes less than K, causing a net shift toward products
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Answer: (A)
Halving volume doubles all partial pressures. Qp = (2P_PCl3)(2P_Cl2) / (2P_PCl5) = 2(Q_original). Since Q_original = K, Q > K, shifting the reaction left.
Question 8
A particulate model of the reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) shows the sudden removal of O₂ molecules from an equilibrium mixture. Which statement describes the immediate effect on Q and the resulting shift?
Q increases above K, driving the net reaction toward reactants
Q increases above K, driving the net reaction toward products
Q decreases below K, driving the net reaction toward reactants
Q decreases below K, driving the net reaction toward products
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Answer: (A)
Removing a reactant decreases the denominator in the Q expression (Q = [SO₃]² / ([SO₂]²[O₂])). This causes Q to increase such that Q > K, shifting the system left.
Question 9
A particulate model depicts 1 molecule of H₂(g), 1 molecule of I₂(g), and 5 molecules of HI(g) in a 1.0 L vessel for the reaction H₂(g) + I₂(g) ⇌ 2HI(g). If K = 50, which statement correctly predicts the system's behavior?
Q is 25, so the system shifts toward products to reach equilibrium
Q is 25, so the system shifts toward reactants to reach equilibrium
Q is 5, so the system shifts toward products to reach equilibrium
Q is 5, so the system shifts toward reactants to reach equilibrium
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Answer: (A)
Q = [HI]² / ([H₂][I₂]) = 5² / (1×1) = 25. Since 25 < 50 (Q < K), the reaction must proceed forward (toward products) to increase Q until it equals K.
Question 10
For the exothermic reaction CO(g) + 2H₂(g) ⇌ CH₃OH(g), a model shows the temperature of the system is increased. Which statement describes the relationship between Q and K immediately after the temperature change?
K decreases while Q remains constant, causing a shift toward reactants
K decreases while Q remains constant, causing a shift toward products
K increases while Q remains constant, causing a shift toward reactants
K increases while Q remains constant, causing a shift toward products
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Answer: (A)
For exothermic reactions, increasing T decreases K. Q depends on concentrations, which are instantaneously unchanged. Thus Q > K, driving the shift left.
What's on the AP Chemistry Unit 7 practice test
The standard version takes about 60 minutes and follows the timing and directions of the AP exam, sized to one unit:
- 20 Multiple Choice Questions (30 minutes)
- 1 FRQ – FRQ 1 – Long Answer (15 minutes)
- 1 FRQ – FRQ 4 – Short Answer (15 minutes)
Questions are drawn from 3429 multiple-choice questions and 56 FRQs written for Unit 7, Equilibrium.
Practice AP Chemistry Unit 7 FRQs
Answer the unit's free-response questions one at a time and get a score on every part.
- FRQ 1 – Long Answer (8 questions)
- FRQ 2 – Long Answer (5 questions)
- FRQ 3 – Long Answer (5 questions)
- FRQ 4 – Short Answer (13 questions)
- FRQ 5 – Short Answer (13 questions)
- FRQ 6 – Short Answer (8 questions)
- FRQ 7 – Short Answer (4 questions)
Review Unit 7 before you start
- 7.1 Introduction to Equilibrium
- 7.2 Direction of Reversible Reactions
- 7.3 Reaction Quotient and Equilibrium Constant
- 7.4 Calculating the Equilibrium Constant
- 7.5 Magnitude of the Equilibrium Constant
- 7.6 Properties of the Equilibrium Constant
- 7.7 Calculating Equilibrium Concentrations
- 7.8 Representations of Equilibrium
- 7.9 Introduction to Le Châtelier’s Principle
- 7.10 Reaction Quotient and Le Châtelier’s Principle
- 7.11 Introduction to Solubility Equilibria
- 7.12 Common Ion Effect