Chemical Kinetics -- Practice Problems

Chemical Kinetics — Practice Problems

10 MCQ questions covering rate laws, integrated rate laws, reaction mechanisms, the Arrhenius equation, and catalysis. Select an option to check your answer and view the explanation.

Worked Examples

Example 1: Determining Reaction Order from Data

Problem: The decomposition of $\text{N}_2\text{O}_5$ in $\text{CCl}_4$ solution at 45°C gives:

Time (s)	$[\text{N}_2\text{O}_5]$ (mol/L)
0	1.00
120	0.88
240	0.78
360	0.69
480	0.61

Determine the order and rate constant.

Solution:

Step 1: Test first-order (plot $\ln[A]$ vs $t$ )

$t$ (s)	$\ln[\text{N}_2\text{O}_5]$
0	0.000
120	-0.128
240	-0.249
360	-0.371
480	-0.494

The differences are approximately constant: $\Delta \ln[A] \approx -0.123$ per 120 s.

Step 2: Calculate $k$

$k = -\frac{\Delta \ln[A]}{\Delta t} = \frac{0.123}{120} = 1.02 \times 10^{-3} \text{ s}^{-1}$

Step 3: Verify with half-life

$t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{1.02 \times 10^{-3}} = 679 \text{ s} \approx 11.3 \text{ min}$

From the data, $[\text{N}_2\text{O}_5]$ drops from 1.00 to 0.50 in approximately 680 s. ✓

Key insight: Always test which integrated rate law gives a straight line. First-order: $\ln[A]$ vs $t$ . Second-order: $1/[A]$ vs $t$ . Zero-order: $[A]$ vs $t$ .

Example 2: Michaelis-Menten Kinetics

Problem: An enzyme-catalyzed reaction has $V_{\max} = 100$ μmol/min and $K_M = 5.0$ mM. Calculate the reaction velocity at $[S] = 2.5$ mM, 5.0 mM, and 25 mM.

Solution:

Using the Michaelis-Menten equation:

$v = \frac{V_{\max}[S]}{K_M + [S]}$

At $[S] = 2.5$ mM:

$v = \frac{100 \times 2.5}{5.0 + 2.5} = \frac{250}{7.5} = 33.3 \text{ μmol/min}$

At $[S] = 5.0$ mM (= $K_M$ ):

$v = \frac{100 \times 5.0}{5.0 + 5.0} = \frac{500}{10} = 50.0 \text{ μmol/min} = \frac{V_{\max}}{2}$

At $[S] = 25$ mM (5 × $K_M$ ):

$v = \frac{100 \times 25}{5.0 + 25} = \frac{2500}{30} = 83.3 \text{ μmol/min}$

Key insight: When $[S] = K_M$ , the velocity is exactly half of $V_{\max}$ . At $[S] \gg K_M$ , the enzyme approaches saturation ( $v \approx V_{\max}$ ). At $[S] \ll K_M$ , the reaction is first-order in substrate ( $v \approx V_{\max}[S]/K_M$ ).

Example 3: Arrhenius — Temperature Dependence

Problem: A reaction has $k_1 = 3.2 \times 10^{-4}$ s⁻¹ at 300 K and $k_2 = 1.5 \times 10^{-3}$ s⁻¹ at 320 K. Calculate the activation energy.

Solution:

Using the two-point Arrhenius equation:

$\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$

$\ln\frac{1.5 \times 10^{-3}}{3.2 \times 10^{-4}} = \frac{E_a}{8.314}\left(\frac{1}{300} - \frac{1}{320}\right)$

$\ln(4.688) = \frac{E_a}{8.314}(0.003333 - 0.003125)$

$1.545 = \frac{E_a}{8.314}(0.000208)$

$E_a = \frac{1.545 \times 8.314}{0.000208} = 61,800 \text{ J/mol} = 61.8 \text{ kJ/mol}$

Key insight: A 20 K temperature increase caused the rate to increase by a factor of ~4.7. This is consistent with the rule of thumb that rates roughly double for every 10 K increase near room temperature for activation energies around 50-60 kJ/mol.

Rate Laws and Integrated Rate Laws

Q1. For a first-order reaction, the integrated rate law is $\\ln[A] = -kt + \\ln[A]_0$ . What is the half-life in terms of the rate constant $k$ ?

A. $t_\{1/2\} = \\ln 2 / k = 0.693/k$ B. $t_\{1/2\} = [A]_0 / 2k$ C. $t_\{1/2\} = 1/k[A]_0$ D. $t_\{1/2\} = 1/k$ ’,

Answer: A

Answer: A — At half-life: $[A] = [A]_0/2$ , so $\\ln([A]_0/2) = -kt_\{1/2\} + \\ln[A]_0$ , giving $-\\ln 2 = -kt_\{1/2\}$ , thus $t_\{1/2\} = \\ln 2/k$ . For a first-order reaction, the half-life is independent of the initial concentration. For zero-order: $t_\{1/2\} = [A]_0/2k$ (depends on $[A]_0$ ). For second-order: $t_\{1/2\} = 1/k[A]_0$ .

Q2. A plot of $\\ln[A]$ versus time yields a straight line with slope $-0.0231\\,\\text\{s\}^\{-1\}$ . What is the order of the reaction and the rate constant?

A. First order; $k = 0.0231\\,\\text\{s\}^\{-1\}$ B. Zero order; $k = -0.0231\\,\\text\{s\}^\{-1\}$ C. Second order; $k = 0.0231\\,\\text\{s\}^\{-1\}$ D. First order; $k = 2.31\\,\\text\{s\}^\{-1\}$ ’,

Answer: A

Answer: A — A straight line of $\\ln[A]$ vs. time indicates a first-order reaction, since $\\ln[A] = -kt + \\ln[A]_0$ is linear with slope $-k$ . Here $k = 0.0231$ s $^\{-1\}$ and $t_\{1/2\} = 0.693/0.0231 = 30.0$ s. For second-order, a plot of $1/[A]$ vs. time would be linear. For zero-order, $[A]$ vs. time is linear.

Q3. For the reaction $2\\text\{NO\}_2 \\to \\text\{NO\} + \\text\{O\}_2$ , the experimental rate law is $v = k[\\text\{NO\}_2]^2$ . This suggests the reaction is:

A. Second order overall and likely bimolecular (but the stoichiometric coefficient does not necessarily equal the order) B. First order because the stoichiometry is 2:1:1 C. Termolecular because three molecules are involved D. Zero order with respect to $\\text\{NO\}_2$ ’,

Answer: A

Answer: A — The overall order is 2 (second order in $\\text\{NO\}_2$ ). The rate law is experimentally determined and does not necessarily match the stoichiometric coefficients. A second-order rate law suggests a bimolecular rate-determining step, but the actual mechanism might involve multiple steps with intermediates.

Q4. For a reaction with the experimental rate law $v = k[A]^2[B]$ , if the concentration of $A$ is doubled and $B$ is tripled, by what factor does the rate increase?

A. $2 \times 3 = 6

Answer: B

Answer: B — The rate is proportional to $[A]^2[B]$ . Doubling $[A]$ gives a factor of $2^2 = 4$ , and tripling $[B]$ gives a factor of $3$ . The combined effect is $4 \\times 3 = 12$ . The new rate is 12 times the original rate. This is a third-order reaction overall (second order in A, first in B).

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Reaction Mechanisms and Catalysis

Q5. For a reaction mechanism with a slow rate-determining step followed by fast steps, what is the relationship between the rate law and the rate-determining step?

A. The rate law reflects the molecularity of the rate-determining step (for an elementary step) B. The rate law always matches the overall stoichiometry C. The rate law has no relationship to any individual step D. The rate law is determined by the fastest step’,

Answer: A

Answer: A — For an elementary (single molecular event) rate-determining step, the rate law directly reflects its molecularity. For example, if the RDS is $\\text\{A\} + \\text\{B\} \\to \\text\{products\}$ (bimolecular), the rate law is $v = k[A][B]$ . If the RDS involves intermediates, the steady-state approximation or pre-equilibrium approximation may be needed to express the rate law in terms of reactants.

Q6. The steady-state approximation assumes that the concentration of reactive intermediates remains approximately constant during the reaction. Mathematically, this means:

A. $d[\\text\{intermediate\}]/dt \\approx 0$ B. $[\\text\{intermediate\}] = 0$ at all times C. $d[\\text\{intermediate\}]/dt = k[\\text\{reactants\}]$ D. $[\\text\{intermediate\}] = [\\text\{products\}]$ ’,

Answer: A

Answer: A — The steady-state approximation sets $d[\\text\{I\}]/dt = 0$ , meaning the rate of formation equals the rate of consumption of the intermediate. This allows expressing $[\\text\{I\}]$ in terms of reactant concentrations without solving the full differential equations. It is valid when the intermediate is highly reactive (consumed rapidly after formation), so its concentration remains low and nearly constant.

Q7. The Arrhenius equation $k = A e^\{-E_a/RT\}$ relates the rate constant to temperature. What happens to the rate constant if the temperature increases from 300 K to 310 K for a reaction with $E_a = 50$ kJ/mol?

A. The rate constant approximately doubles ( $k_\{310\}/k_\{300\} \\approx e^\{50\\,000 \\cdot 10 / (8.314 \\cdot 300 \\cdot 310)\} \\approx 2.0$ ) B. The rate constant increases by a factor of 10 C. The rate constant decreases D. The rate constant is unchanged’,

Answer: A

Answer: A — Using the two-point Arrhenius form: $\\ln(k_2/k_1) = (E_a/R)(1/T_1 - 1/T_2) = (50000/8.314)(1/300 - 1/310) = 6014 \\times 0.0001074 = 0.645$ . So $k_2/k_1 = e^\{0.645\} \\approx 1.91$ , approximately doubling. The common rule of thumb is that the rate roughly doubles for every 10 K temperature increase near room temperature for typical activation energies.

Q8. In enzyme kinetics, the Michaelis-Menten equation $v = \\frac\{V_\{\\max\}[S]\}\{K_M + [S]\}$ gives the reaction velocity. What does $K_M$ represent?

A. The substrate concentration at which the velocity is half of $V_\{\\max\}$ B. The maximum velocity of the enzyme-catalyzed reaction C. The equilibrium constant for enzyme-substrate binding D. The minimum substrate concentration for the reaction to proceed’,

Answer: A

Answer: A — When $[S] = K_M$ : $v = V_\{\\max\}[S]/(K_M + [S]) = V_\{\\max\}/2$ . $K_M$ equals $(k_\{-1\} + k_2)/k_1$ for the mechanism $\\text\{E\} + \\text\{S\} \\underset\{k_\{-1\}\}\{\\overset\{k_1\}\{\\rightleftharpoons\}\} \\text\{ES\} \\xrightarrow\{k_2\} \\text\{E\} + \\text\{P\}$ . When $k_2 \\ll k_\{-1\}$ , $K_M \\approx K_d$ (dissociation constant). A Lineweaver-Burk plot ( $1/v$ vs $1/[S]$ ) yields $K_M$ from the $x$ -intercept.

Q9. A catalyst increases the rate of a chemical reaction. Which of the following is true about the catalyzed vs. uncatalyzed reaction?

A. The catalyst lowers the activation energy but does not change $\\Delta H$ or $\\Delta G^\\circ$ ; it provides an alternative pathway B. The catalyst changes the equilibrium constant C. The catalyst increases the activation energy to slow the reverse reaction D. The catalyst is consumed in the reaction’,

Answer: A

Answer: A — A catalyst provides an alternative reaction pathway with lower activation energy $E_a$ . It accelerates both forward and reverse reactions equally, so the equilibrium constant $K$ (and thus $\\Delta G^\\circ$ ) is unchanged. $\\Delta H$ is also unchanged since the answer varies based on only on initial and final states. A true catalyst is not consumed in the reaction.

Q10. For a consecutive reaction $\\text\{A\} \\xrightarrow\{k_1\} \\text\{B\} \\xrightarrow\{k_2\} \\text\{C\}$ , under what condition does the concentration of intermediate B reach a maximum?

A. When $k_1[A] = k_2[B]$ , i.e., the rate of formation of B equals the rate of consumption of B B. When $[A] = [B] = [C]$ C. When $k_1 = k_2$ D. When $[A] = 0$ ’,

Answer: A

Answer: A — B is formed from A ( $d[B]/dt = k_1[A] - k_2[B]$ ) and consumed to give C. B reaches maximum concentration when $d[B]/dt = 0$ , i.e., $k_1[A] = k_2[B]$ . Before this point, B accumulates ( $k_1[A] > k_2[B]$ ); after, B is consumed faster than formed ( $k_2[B] > k_1[A]$ ). The time of maximum B depends on both rate constants.

Intuition

Chemical kinetics is about timing, not just possibility: Thermodynamics tells you whether a reaction can happen; kinetics tells you how fast it actually happens. Think of it like a ball sitting at the top of a hill — thermodynamics says it can roll down, but kinetics tells you whether it rolls now or in a million years.

Why it matters: Understanding reaction rates lets you control industrial processes, design drugs that metabolise at the right speed, and predict how quickly pollutants break down in the environment.

The key insight: The rate law is always determined experimentally — you can never predict reaction order from the balanced equation alone.

Common Mistakes

Confusing reaction order with stoichiometric coefficients: The rate law $\text{rate} = k[A]^m[B]^n$ has orders $m$ and $n$ determined experimentally, not from the balanced equation. A reaction $2A \to B$ is not necessarily second order in $A$ — it could be first order, zero order, or any order. Always determine orders from experimental data.

Using the wrong integrated rate law: Each reaction order has a different integrated rate law: zero order ( $[A] = [A]_0 - kt$ ), first order ( $\ln[A] = \ln[A]_0 - kt$ ), second order ( $1/[A] = 1/[A]_0 + kt$ ). Using the wrong one gives incorrect rate constants and half-lives. Check which plot gives a straight line to identify the order.

Confusing the Arrhenius equation activation energy with the overall reaction energy: The activation energy $E_a$ is the energy barrier for the rate-determining step, not the overall $\Delta H$ or $\Delta G$ of the reaction. A reaction can be highly exothermic ( $\Delta H \ll 0$ ) but slow (high $E_a$ ). Do not confuse thermodynamic favourability with kinetic speed.

Cross-References

Chemical Kinetics: Detailed notes on rate laws, reaction mechanisms, and activation energy.
Thermodynamics: Covers Gibbs free energy and spontaneity that complement kinetic analysis.
Practice Physical Chemistry: Interactive practice problems covering quantum chemistry and statistical mechanics.

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