AP Chemistry Unit 7: Equilibrium

Explore AP Chemistry Unit 7 questions on reversible reactions, equilibrium expressions, Q and K, concentration changes, Le Châtelier shifts, Ksp, and solubility.

Covered syllabus points

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

[Maximum number: 3]

When heated, calcium carbonate decomposes according to the equation above. In a study of the decomposition of calcium carbonate, a student added a 50.0 g sample of powdered $\mathrm{CaCO}_{3}(s)$ to a 1.00 L rigid container. The student sealed the container, pumped out all the gases, then heated the container in an oven at 1100 K. As the container was heated, the total pressure of the $\mathrm{CO}_{2}(g)$ in the container was measured over time. The data are plotted in the graph below.

Figure for Question 7 Equilibrium question 1 — AP Chemistry

The student repeated the experiment, but this time the student used a 100.0 g sample of powdered $\mathrm{CaCO}_{3}(s)$ . In this experiment, the final pressure in the container was 1.04 atm, which was the same final pressure as in the first experiment.

(a)

The student claimed that the final pressure in the container in each experiment became constant because all of the $\mathrm{CaCO}_{3}(s)$ had decomposed. Based on the data in the experiments, do you agree with this claim? Explain.

[ 1 ]

(b)

After 20 minutes some $\mathrm{CO}_{2}(g)$ was injected into the container, initially raising the pressure to 1.5 atm. Would the final pressure inside the container be less than, greater than, or equal to 1.04 atm? Explain your reasoning.

[ 1 ]

(c)

Are there sufficient data obtained in the experiments to determine the value of the equilibrium constant, $K_{p}$ , for the decomposition of $\mathrm{CaCO}_{3}(s)$ at 1100 K? Justify your answer.

[ 1 ]

[Maximum number: 1]

Hydrogen gas reacts with iodine gas at constant temperature in a sealed rigid container. The gases are allowed to reach equilibrium according to the equation above. Which of the following best describes what will happen to the reaction immediately after additional iodine gas is added to the system?

The rates of both the forward and reverse reactions decrease.

The rates of both the forward and reverse reactions do not change.

The rate of the forward reaction becomes greater than the rate of the reverse reaction.

The rate of the forward reaction becomes less than the rate of the reverse reaction.

Table for Question 7 Equilibrium question 2 — AP Chemistry

Figure for Question 7 Equilibrium question 2 — AP Chemistry

[Maximum number: 3]

Answer the following questions about magnesium.

(a)

The dissolution of magnesium hydroxide is represented by the following equation.

\mathrm{Mg}(\mathrm{OH})_{2}(s) \rightleftarrows \mathrm{Mg}^{2+}(a q)+2 \mathrm{OH}^{-}(a q) \quad K_{s p}=5.61 \times 10^{-12}

[ 3 ]

(i)

Write the expression for the solubility product constant, $K_{s p}$ .

[ 1 ]

E (i) For the correct expression:

$K_{s p}=\left[\mathrm{Mg}^{2+}\right]\left[\mathrm{OH}^{-}\right]^{2}$

(ii)

After the two solutions are combined in beaker 3 as described in part D, but before any

reaction takes place, $\left[\mathrm{OH}^{-}\right]=1.65 \times 10^{-4} M$ . Using your answer to part D, calculate the

value of the reaction quotient, Q.

[ 1 ]

(ii) For the correct calculated value, consistent with part D and part E (i):

$Q=\left[\mathrm{Mg}^{2+}\right]\left[\mathrm{OH}^{-}\right]^{2}=\left(7.62 \times 10^{-4}\right)\left(1.65 \times 10^{-4}\right)^{2}=2.07 \times 10^{-11}$

(iii)

Using the reaction quotient, Q, predict whether a precipitate should form as the mixture

in beaker 3 approaches equilibrium. Justify your answer.

[ 1 ]

(iii) For the correct prediction and justification, consistent with part E (ii).

Examples of acceptable responses may include the following:

- $Q>K_{s p}$ , so a precipitate will form.

- The concentration of the ions in solution (represented by Q) is greater than that of a saturated solution, so a precipitate will form.

[Maximum number: 2]

In the gas phase, $\mathrm{AlCl}_{3}$ is a molecular substance. A reaction of gaseous $\mathrm{AlCl}_{3}$ at high temperature is represented by the following balanced equation.

\text { Reaction 1: } \mathrm{AlCl}_{3}(g) \rightarrow \mathrm{Al}(g)+3 \mathrm{Cl}(g) \quad \Delta H_{1}^{\circ}=\text { ? }

(a)

Write the expression for the equilibrium constant, $K_{p}$ , for this reaction.

A particle-level diagram of an equilibrium mixture of $\mathrm{AlCl}_{3}(g)$ and $\mathrm{Al}_{2} \mathrm{Cl}_{6}(g)$ at 400°C in a 25 L closed container is shown.

[ 1 ]

For the correct answer: 1 point

K_{p}=\frac{P_{\mathrm{Al}_{2} \mathrm{Cl}_{6}}}{\left(P_{\mathrm{AlCl}_{3}}\right)^{2}}

(b)

Using the particle-level diagram, calculate the value of $K_{p}$ for the reaction if the total pressure in the container is 22.1 atm.

[ 1 ]

For the correct calculated value, consistent with part (e): 1 point

K_{p}=\frac{\chi_{\mathrm{Al}_{2} \mathrm{Cl}_{6}}\left(P_{\text {total }}\right)}{\left(\chi_{\mathrm{AlCl}_{3}}\left(P_{\text {total }}\right)\right)^{2}}=\frac{\frac{3}{10}(22.1)}{\left(\frac{7}{10}(22.1)\right)^{2}}=0.0277

Total for question 210 points

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Covered syllabus points

7 Equilibrium question 1

Question (a)

Question (b)

Question (c)

7 Equilibrium question 2

7 Equilibrium question 3

Question (a)

Question (i)

Question (ii)

Question (iii)

7 Equilibrium question 4

Question (a)

Question (b)