Rate Data That Don't Fit a Simple Order
A kinetics team studies the reaction A + B → products at 300 K and collects initial-rate data in a sealed reactor:
| Trial | [A]₀ (M) | [B]₀ (M) | Initial rate (M/s) |
|---|---|---|---|
| 1 | 0.10 | 0.10 | 2.0 × 10⁻⁴ |
| 2 | 0.20 | 0.10 | 8.0 × 10⁻⁴ |
| 3 | 0.10 | 0.20 | 2.0 × 10⁻⁴ |
| 4 | 0.10 | 0.10 | 2.0 × 10⁻⁴ (with 0.05 M added product P) |
A junior analyst concludes the reaction is second order in A and first order in B (overall third order) because 'doubling A quadruples the rate and B must matter too.' The team lead is skeptical of the B conclusion and of the analyst's proposed mechanism.
1. Determine the rate law from Trials 1–3. Is the analyst correct about the order in B? Explain the reasoning from the data.
Read the worked answer
Order in A: Compare Trials 1 and 2 (B fixed). [A] doubles; rate goes from 2.0×10⁻⁴ to 8.0×10⁻⁴, a factor of 4. So 2ⁿ = 4 → n = 2. Second order in A.
Order in B: Compare Trials 1 and 3 (A fixed). [B] doubles; rate stays 2.0×10⁻⁴, a factor of 1. So 2ᵐ = 1 → m = 0. The reaction is zero order in B.
Rate law: rate = k[A]², with k = 2.0×10⁻⁴ / (0.10)² = 2.0×10⁻² M⁻¹s⁻¹.
The analyst was wrong about B. Zero order in B means B does not appear in the rate-determining step - it may be involved in a fast step after the slow step, or its surface/saturation is already saturated. The 'B must matter too' reasoning is intuition, not data.
2. Trial 4 shows the rate unchanged when 0.05 M product P is added. What does this tell you, and how does it constrain a possible mechanism?
Read the worked answer
No rate change on adding product means the reaction is not product-inhibited and P does not appear in the rate law - consistent with an irreversible rate-determining step that does not involve P. This rules out mechanisms where product binds the catalyst or competes for the active site (which would lower the rate).
A mechanism consistent with rate = k[A]²: a slow bimolecular step 2A → intermediate, followed by fast reaction with B. The slow step's rate = k[A]² matches the data, and B enters only after the slow step, explaining zero order in B. The lead's skepticism was warranted: the data demand a mechanism where B is not in the rate-determining step.
3. The team wants to double the rate without changing temperature or concentrations. Based on the rate law, which single change is most effective, and by what factor?
Read the worked answer
Since rate = k[A]², only [A] affects the rate (B has no effect). Doubling [A] multiplies the rate by 2² = 4 - a fourfold increase. Doubling [B] does nothing (factor of 1). So increase [A] by a factor of 2.
If instead they wanted exactly 2× rate, they would increase [A] by √2 ≈ 1.41×. The key judgment: because the reaction is zero order in B, effort spent on B is wasted; the second-order dependence on A makes A the lever. Temperature would raise k, but the problem fixes T.