Medium MCAT Kinetics Practice Questions

Mastering chemical kinetics is a fundamental requirement for success on the Chemical and Physical Foundations of Biological Systems section of the MCAT. This guide provides Medium MCAT Kinetics Practice Questions designed to bridge the gap between basic definitions and complex passage-based analysis. By engaging with these problems, you will refine your ability to interpret rate laws, understand activation energy, and analyze reaction mechanisms.

Concept Explanation

MCAT kinetics focuses on the study of reaction rates, the factors affecting these rates, and the microscopic pathways reactions follow to reach completion. Unlike thermodynamics, which determines the spontaneity and equilibrium position of a reaction, kinetics describes the speed at which a reaction occurs. The core of this topic involves the Rate Law, generally expressed as $\text{rate} = k[A]^m[B]^n$, where $k$ is the rate constant and the exponents represent the reaction order with respect to each reactant.

Key variables influencing kinetics include reactant concentration, temperature, the presence of a catalyst, and the medium of the reaction. According to Collision Theory, for a reaction to occur, molecules must collide with sufficient energy (the activation energy, $E_a$) and proper orientation. The Arrhenius Equation, $k = Ae^{-E_a/RT}$, mathematically links the rate constant to temperature and activation energy. Understanding these relationships is vital when using retrieval practice for medical students to memorize and apply formulas under timed conditions.

Reaction Orders and Graphs

• Zero-order: Rate is independent of concentration. A plot of $[A]$ vs. time is linear.
• First-order: Rate is proportional to one reactant. A plot of $\ln[A]$ vs. time is linear.
• Second-order: Rate is proportional to the square of a reactant or the product of two. A plot of $1/[A]$ vs. time is linear.

Solved Examples

Example 1: Determining Rate Law from Initial Rates
Given the reaction $A + B \rightarrow C$, find the rate law based on the following data:
Exp 1: $[A]=0.1, [B]=0.1, \text{Rate}=2.0 \times 10^{-3}$
Exp 2: $[A]=0.2, [B]=0.1, \text{Rate}=8.0 \times 10^{-3}$
Exp 3: $[A]=0.1, [B]=0.2, \text{Rate}=2.0 \times 10^{-3}$

1. Compare Exp 1 and 2: $[A]$ doubles, rate quadruples ($2^2=4$). Therefore, the reaction is second-order with respect to $A$.
2. Compare Exp 1 and 3: $[B]$ doubles, rate remains constant ($2^0=1$). Therefore, the reaction is zero-order with respect to $B$.
3. Combine for the rate law: $\text{rate} = k[A]^2$.

Example 2: Calculating Activation Energy
If a reaction's rate constant doubles when the temperature increases from 300 K to 310 K, what is the approximate activation energy? (Use $R = 8.314 \text{ J/mol}\cdot \text{K}$)

1. Use the logarithmic form of the Arrhenius equation: $\ln(\frac{k_2}{k_1}) = \frac{E_a}{R} (\frac{1}{T_1} - \frac{1}{T_2})$.
2. Substitute values: $\ln(2) = \frac{E_a}{8.314} (\frac{1}{300} - \frac{1}{310})$.
3. Calculate: $0.693 = \frac{E_a}{8.314} (0.00333 - 0.00322)$.
4. Solve for $E_a$: $E_a \approx 52,000 \text{ J/mol}$ or $52 \text{ kJ/mol}$.

Example 3: Reaction Mechanisms
Consider the mechanism:
Step 1: $NO_2 + NO_2 \rightarrow NO_3 + NO$ (Slow)
Step 2: $NO_3 + CO \rightarrow NO_2 + CO_2$ (Fast)
What is the predicted rate law?

1. Identify the rate-determining step (RDS), which is the slowest step.
2. The rate law is derived from the stoichiometry of the RDS reactants.
3. Result: $\text{rate} = k[NO_2]^2$.

Practice Questions

1. A reaction is found to be second-order overall. If the concentration of the single reactant is tripled, by what factor does the reaction rate change?

2. In a first-order reaction, the half-life is 20 minutes. If the initial concentration is 0.8 M, what will be the concentration after 60 minutes?

3. Consider the reaction $2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g)$. If the rate of disappearance of $N_2O_5$ is $1.2 \times 10^{-2} \text{ M/s}$, what is the rate of appearance of $NO_2$?

4. A catalyst is added to a reaction at equilibrium. Which of the following will change? (A) The equilibrium constant $K_{ \neq}$, (B) The enthalpy of reaction $\Delta H$, (C) The rate of the forward reaction, or (D) The position of equilibrium.

5. A graph of $\ln[A]$ versus time yields a straight line with a slope of $-0.045 \text{ s}^{-1}$. What is the order of the reaction and the value of the rate constant $k$?

6. For the reaction $A + B \rightarrow C$, the rate law is $\text{rate} = k[A][B]^2$. If the volume of the reaction vessel is halved at constant temperature, how does the rate change?

7. The decomposition of a drug follows zero-order kinetics with $k = 0.02 \text{ M/year}$. If a pharmacy stocks a 1.0 M solution, how many years will it take for the concentration to reach 0.8 M?

8. Which of the following statements best describes the effect of increasing temperature on an endothermic reaction?

9. A reaction mechanism has two steps. Step 1: $A + B \rightleftharpoons C$ (fast equilibrium). Step 2: $C + D \rightarrow E$ (slow). Derive the rate law in terms of reactants $A, B,$ and $D$.

10. If the activation energy of a forward reaction is $75 \text{ kJ/mol}$ and the $\Delta H$ of the reaction is $-25 \text{ kJ/mol}$, what is the activation energy of the reverse reaction?

Answers & Explanations

1. Answer: 9. For a second-order reaction $\text{rate} = k[A]^2$. If $[A]$ is tripled ($3[A]$), the new rate is $k(3[A])^2 = 9k[A]^2$. The rate increases by a factor of 9.
2. Answer: 0.1 M. After 60 minutes (3 half-lives), the concentration is halved three times: $0.8 \rightarrow 0.4 \rightarrow 0.2 \rightarrow 0.1$.
3. Answer: $2.4 \times 10^{-2} \text{ M/s}$. According to stoichiometry, for every 2 moles of $N_2O_5$ consumed, 4 moles of $NO_2$ are produced. The rate of $NO_2$ appearance is twice the rate of $N_2O_5$ disappearance: $2 \times (1.2 \times 10^{-2}) = 2.4 \times 10^{-2}$.
4. Answer: (C) The rate of the forward reaction. Catalysts lower activation energy, increasing both forward and reverse rates equally. They do not affect thermodynamic properties like $K_{ \neq}$, $\Delta H$, or the equilibrium position. For more on how to optimize your study of these concepts, check out our guide on retrieval practice vs practice tests.
5. Answer: First-order; $k = 0.045 \text{ s}^{-1}$. A linear plot of $\ln[A]$ vs. time is characteristic of first-order kinetics. The slope of this line is $-k$, so $k = 0.045$.
6. Answer: Increases by a factor of 8. Halving the volume doubles the concentration of all gaseous or aqueous reactants. The new rate is $k(2[A])(2[B])^2 = k(2)(4)[A][B]^2 = 8 \times \text{original rate}$.
7. Answer: 10 years. For zero-order: $[A]_t = -kt + [A]_0$. Substituting: $0.8 = -(0.02)t + 1.0$. Solving for $t$: $0.02t = 0.2$, so $t = 10$.
8. Answer: Both the rate constant and the rate increase. Increasing temperature increases the kinetic energy of molecules, leading to more frequent and energetic collisions, regardless of whether the reaction is endothermic or exothermic.
9. Answer: $\text{rate} = k'[A][B][D]$. The slow step is the RDS: $\text{rate} = k_2[C][D]$. From the fast equilibrium: $k_1[A][B] = k_{-1}[C]$, so $[C] = \frac{k_1}{k_{-1}}[A][B]$. Substituting $[C]$ into the RDS rate law gives $\text{rate} = k_2(\frac{k_1}{k_{-1}})[A][B][D]$.
10. Answer: 100 kJ/mol. The activation energy of the reverse reaction is $E_{a, \text{fwd}} - \Delta H$. Since $\Delta H$ is negative (exothermic), $E_{a, \text{rev}} = 75 - (-25) = 100 \text{ kJ/mol}$.

Quick Quiz

Frequently Asked Questions

What is the difference between reaction order and molecularity?

Reaction order is an experimentally determined value relating concentration to rate, while molecularity refers to the number of molecules colliding in a single elementary step. Order can be fractional or zero, but molecularity must be a whole number.

How does temperature affect the rate constant?

According to the Arrhenius equation, increasing temperature increases the rate constant $k$ exponentially. This is because a higher fraction of molecules possess the kinetic energy required to surpass the activation energy barrier.

Can the rate law be determined from the balanced chemical equation?

No, the rate law cannot be determined from the overall balanced equation; it must be determined experimentally or from the slowest step of a known mechanism. For more on effective study habits for these complex topics, see our guide on retrieval practice vs highlighting notes.

What is a reaction intermediate?

A reaction intermediate is a species that is produced in one step of a mechanism and consumed in a subsequent step. Unlike transition states, intermediates have finite lifetimes and can sometimes be isolated or detected.

How does a catalyst affect the equilibrium constant?

A catalyst has no effect on the equilibrium constant $K_{ \neq}$ or the final concentrations of reactants and products. It only increases the speed at which the system reaches that equilibrium state by lowering the activation energy for both directions.

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