Chemical Kinetics and Reaction Mechanisms
What this domain tests
Kinetics questions test how concentration and temperature control rate: deducing rate orders from initial-rate data, using integrated rate laws and half-lives, and connecting a proposed mechanism to the observed rate law through the rate-determining step.
Core rules
- Rate = k[A]^m[B]^n, where m and n are experimental orders, not the equation coefficients. The overall order is m + n.
- Unit of k: mol^(1−order) dm^(3(order−1)) s⁻¹. First order is s⁻¹, zero order mol dm⁻³ s⁻¹.
- First order: ln[A] = ln[A]₀ − kt, and t½ = 0.693/k, independent of concentration. Second order: 1/[A] = 1/[A]₀ + kt. Zero order: [A] = [A]₀ − kt.
- Arrhenius: k = Ae^(−Ea/RT); ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂). A small Ea means a fast reaction and little temperature sensitivity.
- The rate-determining step is the slow step; intermediates may not appear in the final rate law and must be eliminated by substitution.
Worked example
Doubling [A] with [B] constant doubles the rate, so the order in A is 1. Doubling [B] with [A] constant quadruples the rate, so the order in B is 2. The rate law is rate = k[A][B]² and the reaction is overall third order, with k in dm⁶ mol⁻² s⁻¹. If the mechanism began with a slow step A + 2B → products, that would be consistent, but a fast pre-equilibrium would need substitution of the intermediate concentration.
Traps that cost marks
- Reading orders from the stoichiometric coefficients instead of the data.
- Using the first-order half-life for a second-order reaction - here t½ = 1/(k[A]₀) and it lengthens as the reaction proceeds.
- Forgetting that catalysts lower Ea without changing ΔH or K.
- Confusing rate with rate constant: dilution lowers rate but leaves k unchanged if T is fixed.