AP - Advanced Placement Kinetics and Reaction Rates 1

Question 1

A chemist plots [A] vs. t, ln[A] vs. t, and 1/[A] vs. t for a decomposition reaction. Only the plot of 1/[A] vs. t produces a straight line with a positive slope. Which integrated rate law and reaction order apply?

Accepted Answer

Second order; 1/[A] = 1/[A]₀ + kt

Answer

A linear 1/[A] vs. t graph is the diagnostic signature of a second-order reaction. The integrated second-order rate law is 1/[A] = 1/[A]₀ + kt, giving a straight line whose slope equals k. A linear ln[A] vs. t graph would indicate first order, and a linear [A] vs. t graph would indicate zero order.

Question 2

For a second-order reaction A → products, the half-life is 40.0 s when [A]₀ = 0.500 M. What is the rate constant k, and what will the half-life be when [A]₀ is halved to 0.250 M?

Accepted Answer

k = 0.0500 M⁻¹s⁻¹; t₁/₂ = 80.0 s

Answer

For a second-order reaction, t₁/₂ = 1/(k[A]₀). Using the first condition: 40.0 = 1/(k × 0.500), so k = 0.0500 M⁻¹s⁻¹. When [A]₀ is halved to 0.250 M: t₁/₂ = 1/(0.0500 × 0.250) = 80.0 s. Unlike first-order reactions, second-order half-lives depend on initial concentration and double when [A]₀ is halved.

Question 3

A reaction follows the rate law: rate = k[A][B]². What are the correct units of the rate constant k if concentration is expressed in M and time in seconds?

Accepted Answer

M⁻²s⁻¹

Answer

The overall reaction order is 1 + 2 = 3. Since rate has units of M·s⁻¹ and [A][B]² has units of M³, k must have units of (M·s⁻¹)/M³ = M⁻²s⁻¹. In general, for an nth-order reaction k has units of M^(1−n)·s⁻¹.

Question 4

A catalyst is added to a reaction at equilibrium. Which statement best describes its effect on the reaction's thermodynamics and kinetics?

Accepted Answer

It lowers Ea for both the forward and reverse reactions without altering ΔG° or the equilibrium constant

Answer

A catalyst works by providing an alternative, lower-energy reaction pathway, reducing Ea for both the forward and reverse directions equally. Because ΔG° (and therefore Keq) depends only on the difference in energy between reactants and products—which the catalyst does not change—the equilibrium position is unaffected. The catalyst speeds up the approach to equilibrium but does not shift it.

Question 5

Consider the two-step mechanism below:
  Step 1 (fast equilibrium): A ⇌ B    (equilibrium constant K₁)
  Step 2 (slow):            B + C → D  (rate constant k₂)
What is the rate law for the overall reaction in terms of the original reactants A and C?

Accepted Answer

rate = K₁k₂[A][C]

Answer

Because Step 2 is rate-determining, rate = k₂[B][C]. However, B is a reactive intermediate that cannot appear in the final rate law. Using the fast pre-equilibrium in Step 1, K₁ = [B]/[A], so [B] = K₁[A]. Substituting gives rate = K₁k₂[A][C], expressed entirely in terms of observable reactants.

AP Practice Test

AP Practice Test

AP - Advanced Placement Kinetics and Reaction Rates 1