Rate Law, a Badly Chosen Mechanism, and the Activation Energy
The reaction 2 A(g) + B(g) → C(g) is studied at 300 K by measuring initial rates from the appearance of C.
- Trial 1: [A]₀ = 0.100 M, [B]₀ = 0.100 M, rate = 2.0 × 10⁻⁴ M s⁻¹
- Trial 2: [A]₀ = 0.200 M, [B]₀ = 0.100 M, rate = 8.0 × 10⁻⁴ M s⁻¹
- Trial 3: [A]₀ = 0.100 M, [B]₀ = 0.200 M, rate = 2.0 × 10⁻⁴ M s⁻¹
A proposed mechanism is: Step 1: A + A ⇌ A₂ (fast equilibrium) Step 2: A₂ + B → C (slow)
A separate run at 320 K with [A]₀ = [B]₀ = 0.100 M gives an initial rate of 8.0 × 10⁻⁴ M s⁻¹.
1. Determine the experimental rate law and the value of k with correct units, showing how each trial comparison supports your conclusion.
Read the worked answer
Order in A: compare Trials 1 and 2. [B]₀ is unchanged; [A]₀ doubles (0.100 → 0.200 M) and the rate goes from 2.0 × 10⁻⁴ to 8.0 × 10⁻⁴ M s⁻¹, a factor of 4 = 2². So the reaction is second order in A.
Order in B: compare Trials 1 and 3. [A]₀ is unchanged; [B]₀ doubles and the rate stays at 2.0 × 10⁻⁴ M s⁻¹. So the reaction is zero order in B.
Rate law: rate = k[A]²[B]⁰ = k[A]².
Using Trial 1: k = rate/[A]² = (2.0 × 10⁻⁴ M s⁻¹)/(0.100 M)² = 2.0 × 10⁻² M⁻¹ s⁻¹. The units M⁻¹ s⁻¹ confirm an overall second-order reaction. The zero order in B is chemically meaningful: B's concentration does not affect the rate, which means B cannot appear in or before the rate-determining step.
2. Show that the proposed mechanism predicts a rate law inconsistent with the data, identify the flaw, and propose a modified mechanism that is consistent with the experimental rate law.
Read the worked answer
From the fast equilibrium, K = [A₂]/[A]², so [A₂] = K[A]². The slow step gives rate = k₂[A₂][B] = k₂K[A]²[B]. That is first order in B, but the data demand zero order in B - so the proposed mechanism is wrong.
The flaw is the choice of rate-determining step. A species that appears in the slow step (or before it in the mechanism) will generally affect the rate. Since changing [B] has no effect, B must be consumed in a fast step that occurs after the rate-determining step. A consistent mechanism is:
Step 1: A + A → A₂ (slow, rate = k₁[A]²) Step 2: A₂ + B → C (fast)
This gives rate = k₁[A]², matching the data, and the overall stoichiometry 2 A + B → C. Because Step 2 is fast, once A₂ forms it is consumed essentially as soon as B is available, and B's concentration never enters the rate law.
3. Use the 300 K and 320 K data to calculate the activation energy. Then state whether adding a catalyst would change Ea, k, ΔH, or the equilibrium constant, and justify each.
Read the worked answer
At 300 K, k₁ = (2.0 × 10⁻⁴)/(0.100)² = 0.020 M⁻¹ s⁻¹. At 320 K, k₂ = (8.0 × 10⁻⁴)/(0.100)² = 0.080 M⁻¹ s⁻¹.
ln(k₂/k₁) = ln(4.00) = 1.386 = −(Ea/R)(1/T₂ − 1/T₁) 1/320 − 1/300 = 0.003125 − 0.003333 = −2.083 × 10⁻⁴ K⁻¹ Ea = (1.386)(8.314 J mol⁻¹ K⁻¹)/(2.083 × 10⁻⁴ K⁻¹) ≈ 5.5 × 10⁴ J/mol ≈ 55 kJ/mol.
A catalyst provides an alternative pathway with a lower Ea, so k increases (both forward and reverse) and the reaction is faster at the same temperature. It does not change ΔH, because enthalpy depends only on the initial and final states, and it does not change the equilibrium constant, because K is fixed by ΔG° = −RT ln K, which is also state-function based. The catalyst speeds the approach to equilibrium but cannot shift the position or the equilibrium constant.