Hard MCAT Equilibrium Practice Questions

Mastering chemical equilibrium is a cornerstone of success on the MCAT, as it bridges the gap between general chemistry, biochemistry, and physiology. These Hard MCAT Equilibrium Practice Questions are designed to challenge your understanding of Le Châtelier’s principle, the relationship between Gibbs free energy and the equilibrium constant, and the complex calculations involving solubility products and common ion effects. By engaging with these high-level problems, you are participating in retrieval practice, the most effective way to ensure long-term retention of difficult scientific concepts.

Concept Explanation

Chemical equilibrium occurs when the rates of the forward and reverse reactions are equal, resulting in no net change in the concentrations of reactants and products over time. This state is dynamic, meaning molecules continue to react, but the macroscopic properties remain constant. The equilibrium constant, $K_{ \neq}$, is defined by the ratio of product concentrations to reactant concentrations, each raised to the power of their stoichiometric coefficients, as described by the Law of Mass Action. For a general reaction:

$$aA + bB \ \rightleftharpoons cC + dD$$

The equilibrium expression is written as:

$$K_{ \neq} = \ \frac{[C]^c [D]^d}{[A]^a [B]^b}$$

Crucially, only gaseous and aqueous species are included in this expression; pure solids and liquids are omitted because their concentrations remain effectively constant. The value of $K_{ \neq}$ is temperature-dependent. According to Le Châtelier's principle, if a system at equilibrium is stressed by changes in concentration, pressure, or temperature, the system will shift its position to counteract that stress. Furthermore, the thermodynamic stability of a system is linked to equilibrium through the equation $\Delta G^\circ = -RT \ln K_{ \neq}$. Understanding these relationships is vital for mastering medicine and the biological systems encountered on the MCAT.

Solved Examples

1. Calculating Equilibrium Concentrations: Consider the reaction $H_2(g) + I_2(g) \ \rightleftharpoons 2HI(g)$ with $K_c = 50$ at a specific temperature. If 1.0 mole of $H_2$ and 1.0 mole of $I_2$ are placed in a 1.0 L flask, what is the concentration of $HI$ at equilibrium?
   1. Set up an ICE table (Initial, Change, Equilibrium). Initial concentrations are $[H_2] = 1.0$, $[I_2] = 1.0$, and $[HI] = 0$.
   2. Let $x$ be the amount of $H_2$ reacted. At equilibrium, $[H_2] = 1.0 - x$, $[I_2] = 1.0 - x$, and $[HI] = 2x$.
   3. Substitute into the expression: $50 = \ \frac{(2x)^2}{(1.0 - x)(1.0 - x)}$.
   4. Take the square root of both sides: $\sqrt{50} \approx 7.07 = \ \frac{2x}{1.0 - x}$.
   5. Solve for $x$: $7.07 - 7.07x = 2x \Rightarrow 7.07 = 9.07x \Rightarrow x \approx 0.78$.
   6. Concentration of $HI = 2x = 1.56 \ \text{ M}$.
2. Solubility Product and Common Ion Effect: What is the molar solubility of $AgCl$ ($K_{sp} = 1.8 \ \times 10^{-10}$) in a 0.10 M solution of $NaCl$?
   1. The dissociation is $AgCl(s) \ \rightleftharpoons Ag^+(aq) + Cl^-(aq)$.
   2. Initial concentration of $Cl^-$ is 0.10 M from $NaCl$.
   3. Let $s$ be the molar solubility. Equilibrium concentrations are $[Ag^+] = s$ and $[Cl^-] = 0.10 + s$.
   4. $K_{sp} = [Ag^+][Cl^-] = (s)(0.10 + s)$.
   5. Since $K_{sp}$ is very small, assume $0.10 + s \approx 0.10$.
   6. $1.8 \ \times 10^{-10} = s(0.10) \Rightarrow s = 1.8 \ \times 10^{-9} \ \text{ M}$.
3. Thermodynamics and Equilibrium: If a reaction has a $\Delta G^\circ$ of $-5.7 \ \text{ kJ/mol}$ at 298 K, calculate $K_{ \neq}$. (Use $R = 8.314 \ \text{ J/mol}\cdot\ \text{K}$).
   1. Convert $\Delta G^\circ$ to Joules: $-5700 \ \text{ J/mol}$.
   2. Use the formula $\Delta G^\circ = -RT \ln K_{ \neq}$.
   3. $-5700 = -(8.314)(298) \ln K_{ \neq}$.
   4. $\ln K_{ \neq} = \ \frac{-5700}{-2477.57} \approx 2.3$.
   5. $K_{ \neq} = e^{2.3} \approx 10$.

Practice Questions

1. A reaction vessel contains 2.0 atm of $NO_2$ and 1.0 atm of $N_2O_4$. The reaction is $2NO_2(g) \ \rightleftharpoons N_2O_4(g)$ with $K_p = 0.5$. In which direction will the reaction shift to reach equilibrium?

2. The decomposition of calcium carbonate is endothermic: $CaCO_3(s) \ \rightleftharpoons CaO(s) + CO_2(g)$. If the temperature of a sealed vessel at equilibrium is increased, what happens to the mass of $CaCO_3(s)$ and the partial pressure of $CO_2$?

3. Calculate the pH of a saturated solution of $Mg(OH)_2$, given that the $K_{sp} = 1.8 \ \times 10^{-11}$.

4. For the reaction $A(g) + B(g) \ \rightleftharpoons 2C(g)$, the equilibrium constant $K_c$ is 100. If a 1.0 L container is filled with 1.0 M of A, 1.0 M of B, and 5.0 M of C, determine if the system is at equilibrium. If not, which way does it shift?

5. A buffer solution is prepared with 0.5 M acetic acid ($K_a = 1.8 \ \times 10^{-5}$) and 0.5 M sodium acetate. If 0.1 moles of $NaOH$ is added to 1.0 L of this buffer, what is the final pH?

6. Consider the Haber process: $N_2(g) + 3H_2(g) \ \rightleftharpoons 2NH_3(g)$ ($\Delta H < 0$). Which of the following changes would increase the yield of $NH_3$: increasing volume, adding a catalyst, or decreasing temperature?

7. The molar solubility of a salt with the formula $MX_2$ is $1.0 \ \times 10^{-4} \ \text{ M}$. Calculate the value of the solubility product constant, $K_{sp}$.

8. In a 0.1 M solution of a weak acid $HA$, the acid is 1% dissociated. What is the $K_a$ of the acid?

9. A reaction has $K_{ \neq} = 0.01$ at 300 K and $K_{ \neq} = 100$ at 400 K. Is the reaction exothermic or endothermic?

10. If the pressure of a system containing $PCl_5(g) \ \rightleftharpoons PCl_3(g) + Cl_2(g)$ is increased by decreasing the volume, how does the concentration of $PCl_5$ change?

Answers & Explanations

1. Towards Reactants: Calculate the reaction quotient $Q_p = \ \frac{P_{N_2O_4}}{P_{NO_2}^2} = \ \frac{1.0}{(2.0)^2} = 0.25$. Since $Q_p < K_p$ (0.25 < 0.5), the reaction actually shifts towards the products (right). Correction: The calculation shows Q < K, so it shifts right.

2. Mass decreases, Pressure increases: Since the reaction is endothermic, adding heat (increasing temperature) shifts the equilibrium to the right. This consumes $CaCO_3(s)$ (decreasing its mass) and produces more $CO_2(g)$ (increasing the partial pressure).

3. pH = 10.52: $K_{sp} = [Mg^{2+}][OH^-]^2$. Let $s = [Mg^{2+}]$, then $[OH^-] = 2s$. $K_{sp} = s(2s)^2 = 4s^3$. $1.8 \ \times 10^{-11} = 4s^3 \Rightarrow s = 1.65 \ \times 10^{-4} \ \text{ M}$. $[OH^-] = 2s = 3.3 \ \times 10^{-4} \ \text{ M}$. $pOH = -\log(3.3 \ \times 10^{-4}) \approx 3.48$. $pH = 14 - 3.48 = 10.52$.

4. Shifts Right: $Q_c = \ \frac{[C]^2}{[A][B]} = \ \frac{5.0^2}{1.0 \times 1.0} = 25$. Since $Q_c < K_c$ (25 < 100), the reaction shifts toward the products (right) to increase the numerator and reach equilibrium.

5. pH = 4.92: Initially, $pH = pK_a + \log(\ \frac{0.5}{0.5}) = 4.74$. Adding 0.1 M $OH^-$ reacts with the acid: $[Acid] = 0.5 - 0.1 = 0.4 \ \text{ M}$; $[Base] = 0.5 + 0.1 = 0.6 \ \text{ M}$. New $pH = 4.74 + \log(\ \frac{0.6}{0.4}) = 4.74 + 0.18 = 4.92$.

6. Decreasing temperature: The reaction is exothermic ($\Delta H < 0$). Decreasing temperature shifts the equilibrium toward the product side to generate heat. Increasing volume shifts toward the side with more moles of gas (reactants). A catalyst only speeds up the rate, not the yield.

7. $4.0 \ \times 10^{-12}$: For $MX_2 \ \rightleftharpoons M^{2+} + 2X^-$, $K_{sp} = [M^{2+}][X^-]^2 = (s)(2s)^2 = 4s^3$. $K_{sp} = 4(1.0 \ \times 10^{-4})^3 = 4.0 \ \times 10^{-12}$.

8. $1.0 \ \times 10^{-5}$: 1% of 0.1 M is $0.001 \ \text{ M}$. $[H^+] = [A^-] = 0.001$. $[HA] = 0.1 - 0.001 \approx 0.1$. $K_a = \ \frac{(0.001)^2}{0.1} = 1.0 \ \times 10^{-5}$.

9. Endothermic: As temperature increases, $K_{ \neq}$ increases. This means the reaction shifts right with added heat, which is characteristic of an endothermic process.

10. Concentration increases: Increasing pressure shifts the equilibrium to the side with fewer moles of gas. The reactant side has 1 mole, and the product side has 2 moles. The shift occurs toward the reactant (left), increasing the concentration of $PCl_5$.

1. Which of the following species is NOT included in the equilibrium constant expression?

• AH2O (l)
• BCO2 (g)
• CH+ (aq)
• DNH3 (aq)

Frequently Asked Questions

What is the difference between Kc and Kp?

$K_c$ is the equilibrium constant expressed in terms of molar concentrations, while $K_p$ uses the partial pressures of gaseous reactants and products. They are related by the equation $K_p = K_c(RT)^{\Delta n}$, where $\Delta n$ is the change in the number of moles of gas.

How does temperature affect the equilibrium constant?

Temperature is the only factor that can change the numerical value of $K_{ \neq}$. For endothermic reactions, increasing temperature increases $K_{ \neq}$, while for exothermic reactions, increasing temperature decreases $K_{ \neq}$.

Why are solids and liquids omitted from Keq?

Pure solids and liquids have activities defined as 1 in thermodynamics because their concentrations (density divided by molar mass) do not change significantly during a reaction. Including them would not affect the ratio of the equilibrium expression.

What is the common ion effect?

The common ion effect is a decrease in the solubility of an ionic compound when a soluble salt containing one of its constituent ions is added to the solution. This is a direct application of Le Châtelier's principle shifting the equilibrium toward the solid precipitate.

Can Keq be negative?

No, the equilibrium constant $K_{ \neq}$ cannot be negative because it is a ratio of concentrations or pressures, which are always non-negative values. A very small $K_{ \neq}$ (close to zero) simply indicates that the reaction favors the reactants.

How can I use retrieval practice for MCAT chemistry?

You can use retrieval practice vs practice tests by attempting to write out equilibrium expressions and Le Châtelier shifts from memory before checking your notes. This strengthens the neural pathways associated with these complex chemical relationships.