AP Chemistry Kinetics — Worked Answer Explanations

Unit 5 · 12 questions explained

Below is a complete answer key for our AP Chemistry Kinetics practice questions. For each question you'll find the correct choice, a full written explanation of how to get there, and — for every wrong answer — a short note on exactly why it's tempting and where it goes wrong. Reading these straight through is one of the fastest ways to find the gaps in a unit before exam day.

Prefer to test yourself first? Take the timed Kinetics practice test and come back here to review, or head back to the Kinetics unit overview.

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Question 1 · Easy
For the reaction $A + B \to C$ , tripling $[A]$ while keeping $[B]$ constant triples the reaction rate. Tripling $[B]$ while keeping $[A]$ constant has no effect on the rate. What is the rate law?
- A
  $rate = k [A] [B]$
  Why not A: This would be second order overall; B would have a first-order dependence, but it doesn't.
- B
  $rate = k [A]$ Correct
- C
  $rate = k [A]^{2}$
  Why not C: Second order in A would cause the rate to increase by a factor of 9 when [A] triples.
- D
  $rate = k [B]$
  Why not D: Rate is independent of [B], so B should not appear in the rate law.
Explanation
Tripling [A] triples the rate → rate is first order in A (exponent = 1). Tripling [B] has no effect → rate is zero order in B (exponent = 0). Rate law: $rate = k [A]^{1} [B]^{0} = k [A]$ .

Key takeaway
Rate law exponents (orders) are determined experimentally by observing how rate changes with concentration.
Question 2 · Easy
For a first-order reaction with rate constant $k = 0.0231 mi n^{- 1}$ , what is the half-life?
- A
  $43.3 min$
  Why not A: Used $t_{1/2} = 1/ k$ instead of $ln 2/ k$ .
- B
  $30.0 min$ Correct
- C
  $0.0231 min$
  Why not C: This is just the rate constant, not the half-life.
- D
  $1.00 min$
  Why not D: Calculated $k / ln 2$ rather than $ln 2/ k$ .
Explanation
For a first-order reaction: $t_{1/2} = \frac{ln 2}{k} = \frac{0.693}{0.0231} \approx 30.0 min$ . The half-life of a first-order process is independent of concentration.

Key takeaway
First-order half-life: $t_{1/2} = \ln 2 / k$. It is constant (does not depend on initial concentration).
Question 3 · Easy
Which statement correctly describes how a catalyst increases reaction rate?
- A
  A catalyst increases the temperature of the reaction, giving molecules more kinetic energy.
  Why not A: Catalysts do not change reaction temperature; they lower activation energy.
- B
  A catalyst is consumed in the reaction, producing more product.
  Why not B: Catalysts are regenerated and not consumed in the overall reaction.
- C
  A catalyst provides an alternative pathway with a lower activation energy, increasing the fraction of successful collisions.Correct
- D
  A catalyst shifts the equilibrium toward products by raising $Δ G$ .
  Why not D: Catalysts do not change $Δ G$ , $Δ H$ , or equilibrium position; they only affect rate.
Explanation
A catalyst works by providing an alternative reaction mechanism with a lower activation energy $E_{a}$ . From the Arrhenius equation ( $k = A e^{- E_{a} / R T}$ ), reducing $E_{a}$ exponentially increases $k$ . The overall thermodynamics ( $Δ G$ , $K_{e q}$ ) are unchanged.

Key takeaway
Catalysts lower activation energy, speeding up both forward and reverse reactions without changing the equilibrium constant.
Question 4 · Easy
What distinguishes a homogeneous catalyst from a heterogeneous catalyst?
- A
  A homogeneous catalyst is consumed in the reaction, while a heterogeneous catalyst is not.
  Why not A: Neither type is consumed in the overall reaction; both are regenerated.
- B
  A homogeneous catalyst is in the same phase as the reactants; a heterogeneous catalyst is in a different phase.Correct
- C
  A homogeneous catalyst lowers activation energy; a heterogeneous catalyst does not.
  Why not C: Both types lower activation energy; the distinction is phase, not mechanism.
- D
  Homogeneous catalysts are always biological (enzymes); heterogeneous are always metals.
  Why not D: Enzymes are biological catalysts, but homogeneous catalysts are not exclusively enzymes.
Explanation
A homogeneous catalyst is dissolved in the same phase as the reactants (e.g., an acid catalyst in aqueous solution). A heterogeneous catalyst is in a different phase (e.g., solid platinum catalyst for gas-phase reactions). Both lower activation energy by providing an alternative mechanism.

Key takeaway
Homogeneous catalyst: same phase as reactants. Heterogeneous catalyst: different phase (usually solid catalyst with gas or liquid reactants).
Question 5 · Medium
The rate of decomposition of $N_{2} O_{5}$ is measured at two temperatures:

Temperature Rate constant $k$
$25° C$ $3.0 \times 1 0^{- 5} s^{- 1}$
$45° C$ $1.2 \times 1 0^{- 4} s^{- 1}$

Which statement about this data is correct?
- A
  The reaction rate doubles for every $10° C$ increase because that is a chemistry rule.
  Why not A: The rule of thumb is approximate; this data shows a 4-fold increase over 20°C, not a strict doubling per 10°C.
- B
  The rate constant increases by a factor of 4 over a 20°C range, consistent with Arrhenius behavior.Correct
- C
  The rate constant doubles when temperature is doubled, so the reaction is second order in temperature.
  Why not C: Temperature is not doubled here (25 to 45°C is an increase of 20°, not doubling); also, order refers to concentration dependence, not temperature.
- D
  The rate constant is independent of temperature for a first-order reaction.
  Why not D: All rate constants increase with temperature regardless of reaction order.
Explanation
$k_{45} / k_{25} = (1.2 \times 1 0^{- 4}) / (3.0 \times 1 0^{- 5}) = 4.0$ . This 4-fold increase over 20°C is typical Arrhenius behavior: higher temperature provides more molecules with enough energy to overcome the activation energy barrier ( $k = A e^{- E_{a} / R T}$ ).

Key takeaway
Rate constants increase with temperature (Arrhenius equation). A ~2–4× increase per 10°C rise is common.
Question 6 · Medium
A proposed mechanism for a reaction is:

Step 1 (slow): $N O_{2} + N O_{2} \to N O_{3} + NO$

Step 2 (fast): $N O_{3} + CO \to N O_{2} + C O_{2}$

What is the overall reaction and the predicted rate law?
- A
  Overall: $N O_{2} + CO \to NO + C O_{2}$ ; rate $= k [N O_{2}] [CO]$
  Why not A: The rate law is determined by the slow step only, and CO is not in the slow step.
- B
  Overall: $N O_{2} + CO \to NO + C O_{2}$ ; rate $= k [N O_{2}]^{2}$ Correct
- C
  Overall: $2 N O_{2} + CO \to N O_{3} + NO + C O_{2}$ ; rate $= k [N O_{2}]^{2} [CO]$
  Why not C: NO $_{3}$ is an intermediate (produced and consumed), not a final product.
- D
  Overall: $N O_{2} + CO \to NO + C O_{2}$ ; rate $= k [N O_{3}] [CO]$
  Why not D: NO $_{3}$ is an intermediate; rate laws cannot contain intermediates.
Explanation
Add the two steps: NO $_{2}$ + NO $_{2}$ + NO $_{3}$ + CO → NO $_{3}$ + NO + NO $_{2}$ + CO $_{2}$ . Cancel intermediates (NO $_{3}$ and one NO $_{2}$ ): overall = $N O_{2} + CO \to NO + C O_{2}$ . Rate law comes from the slow (rate-determining) step: $rate = k [N O_{2}]^{2}$ .

Key takeaway
The rate-determining step (slowest step) determines the rate law; intermediates cannot appear in the final rate law.
Question 7 · Medium
A reaction has the energy profile shown below (described): the reactants are at a higher energy than the products, and there is a single energy maximum (transition state) between them. The activation energy for the forward reaction is $80 kJ/mol$ and for the reverse reaction is $120 kJ/mol$ . What is $Δ H$ for the forward reaction?
- A
  $- 40 kJ/mol$ Correct
- B
  $+ 40 kJ/mol$
  Why not B: Wrong sign: products are at lower energy, so the forward reaction is exothermic.
- C
  $+ 200 kJ/mol$
  Why not C: Added the two activation energies rather than taking their difference.
- D
  $- 120 kJ/mol$
  Why not D: Used only the reverse activation energy for $Δ H$ .
Explanation
$Δ H = E_{a, forward} - E_{a, reverse} = 80 - 120 = - 40 kJ/mol$ .

The negative sign confirms the products are at lower energy (exothermic). This relationship holds when there is a single transition state.

Key takeaway
$\Delta H = E_{a,\text{forward}} - E_{a,\text{reverse}}$. If $E_{a,\text{rev}} > E_{a,\text{fwd}}$, the reaction is exothermic.
Question 8 · Medium
For the reaction $2 N O_{2} (g) \to 2 NO (g) + O_{2} (g)$ , experimental data shows that doubling $[N O_{2}]$ increases the initial rate by a factor of 4. What is the rate law and overall reaction order?
- A
  $rate = k [N O_{2}]$ ; first order overall
  Why not A: First order would only double the rate when [NO $_{2}$ ] doubles.
- B
  $rate = k [N O_{2}]^{2}$ ; second order overallCorrect
- C
  $rate = k [N O_{2}]^{2} [NO] [O_{2}]$ ; fourth order overall
  Why not C: Products do not appear in a rate law determined from initial rate experiments.
- D
  $rate = k [N O_{2}]^{1/2}$ ; half order overall
  Why not D: Half-order would increase the rate by $2$ ≈ 1.41 when [NO $_{2}$ ] doubles.
Explanation
Doubling [NO $_{2}$ ] causes a $2^{n} = 4$ factor increase in rate. Solving: $2^{n} = 4 \Rightarrow n = 2$ . Rate law: $rate = k [N O_{2}]^{2}$ , which is second order overall.

Key takeaway
If doubling concentration causes rate to increase by $2^n$, the reaction is $n$th order in that reactant.
Question 9 · Medium
Which of the following best describes the orientation and energy requirements for an effective collision in collision theory?
- A
  Any collision produces a reaction; increasing collision frequency always increases rate.
  Why not A: Most collisions do not lead to reaction — molecules must collide with sufficient energy and correct orientation.
- B
  Only collisions with energy $\geq E_{a}$ AND correct geometric orientation lead to reaction.Correct
- C
  Orientation does not matter; only the kinetic energy of the collision is important.
  Why not C: Orientation is critical for breaking and forming specific bonds.
- D
  Energy is irrelevant; rate depends only on the frequency of correctly oriented collisions.
  Why not D: Activation energy is an energy threshold; low-energy collisions cannot break the necessary bonds.
Explanation
Collision theory: for a collision to be effective, (1) molecules must have kinetic energy $\geq E_{a}$ (activation energy threshold), AND (2) they must approach with the correct spatial orientation so that reactive sites interact. Both conditions must be met simultaneously.

Key takeaway
Collision theory: reaction rate depends on collision frequency, collision energy (≥ E_a), and proper molecular orientation.
Question 10 · Hard
An Arrhenius plot of $ln k$ vs $1/ T$ gives a straight line with slope $- 8000 K$ . What is the activation energy $E_{a}$ ? ( $R = 8.314 J/ (mol \cdot K)$ )
- A
  $9.63 kJ/mol$
  Why not A: Divided slope by R without multiplying by R correctly.
- B
  $66.5 kJ/mol$ Correct
- C
  $8000 kJ/mol$
  Why not C: Used the slope value directly without dividing by R.
- D
  $960 kJ/mol$
  Why not D: Used incorrect units for R or made a factor-of-1000 error.
Explanation
Arrhenius equation: $ln k = ln A - \frac{E _{a}}{R} \cdot \frac{1}{T}$ . The slope $= - E_{a} / R$ , so:

$E_{a} = - slope \times R = 8000 K \times 8.314 J/ (mol \cdot K) = 66, 512 J/mol \approx 66.5 kJ/mol$

Key takeaway
Arrhenius plot: slope = $-E_a/R$. $E_a = -\text{slope} \times R$.
Question 11 · Hard
For the second-order integrated rate law $\frac{1}{[ A ] _{t}} = k t + \frac{1}{[ A ] _{0}}$ , if $[A]_{0} = 0.500 M$ and $k = 0.0400 M^{- 1} \cdot s^{- 1}$ , what is $[A]$ after $100 s$ ?
- A
  $0.167 M$ Correct
- B
  $0.250 M$
  Why not B: Assumed first-order behavior and took one half-life.
- C
  $0.100 M$
  Why not C: Arithmetic error: used $1/ [A]_{0} = 2$ but forgot to add correctly.
- D
  $0.333 M$
  Why not D: Subtracted $k t$ from $[A]_{0}$ directly instead of using the reciprocal form.
Explanation
$\frac{1}{[ A ] _{t}} = (0.0400) (100) + \frac{1}{0.500} = 4.00 + 2.00 = 6.00 M^{- 1}$

$[A]_{t} = \frac{1}{6.00} \approx 0.167 M$

Key takeaway
Second-order integrated rate law: $1/[A]_t = kt + 1/[A]_0$. A plot of $1/[A]$ vs $t$ is linear for a second-order reaction.
Question 12 · Hard
The concentration of a reactant A in a first-order reaction decreases from $0.800 M$ to $0.100 M$ over $90 min$ . What is the half-life of the reaction?
- A
  $11.25 min$
  Why not A: Divided total time by the number of half-lives (8) rather than (3).
- B
  $30 min$ Correct
- C
  $45 min$
  Why not C: Used half the total time; the concentration dropped 8-fold, not 4-fold.
- D
  $90 min$
  Why not D: Assumed one half-life spans the full 90 min, but 0.800 to 0.100 is three half-lives.
Explanation
Ratio: $\frac{0.800}{0.100} = 8 = 2^{3}$ , so 3 half-lives elapsed in 90 min.

$t_{1/2} = \frac{90}{3} = 30 min$

Key takeaway
Determine how many half-lives fit the concentration change ($[A]_0/[A]_t = 2^n$), then $t_{1/2} = t_{\text{total}}/n$.

Temperature	Rate constant $k$
$25° C$	$3.0 \times 1 0^{- 5} s^{- 1}$
$45° C$	$1.2 \times 1 0^{- 4} s^{- 1}$