Rate Laws & Reaction Order

You’ll be able to

Write a rate law and identify the rate constant k and reaction orders
Determine reaction order from initial-rate data by comparing trials
Calculate k and give its units from a determined rate law

The rate law links rate to concentration

Experiment shows the rate depends on reactant concentrations. The rate law captures that dependence: rate = k[A]ᵐ[B]ⁿ. The exponents m and n are the reaction orders — how sensitive the rate is to each reactant — and k is the rate constant, a temperature-dependent proportionality factor. The orders are found by experiment, not read off the balanced equation.

General rate law

rate = k[A]ᵐ[B]ⁿ

m is the order in A, n the order in B, and (m + n) the overall order. Orders are usually 0, 1, or 2 and come only from data.

Reading order from how rate responds

Order tells you the rate’s response to concentration. Zero order: doubling a reactant does nothing (rate ∝ [A]⁰ = 1). First order: doubling it doubles the rate (×2). Second order: doubling it quadruples the rate (×2² = 4). The method of initial rates exploits this: change one reactant at a time and watch the factor the rate changes by.

The rate constant k and its units

Once the orders are known, plug any trial back into the rate law to solve for k. Its units depend on the overall order so that the whole expression works out to M·s⁻¹: zero order is M·s⁻¹, first order is s⁻¹, second order is M⁻¹·s⁻¹, and third order is M⁻²·s⁻¹. A changing k is a clue you have the wrong orders.

Worked example

For A + B → products, initial-rate data are collected. Determine the rate law and the value of k. Trial 1: [A] = 0.10 M, [B] = 0.10 M, rate = 2.0 × 10⁻³ M·s⁻¹ Trial 2: [A] = 0.20 M, [B] = 0.10 M, rate = 4.0 × 10⁻³ M·s⁻¹ Trial 3: [A] = 0.10 M, [B] = 0.20 M, rate = 8.0 × 10⁻³ M·s⁻¹

1.Order in A: compare Trials 1→2, where [A] doubles and [B] is held constant. Rate goes 2.0 → 4.0 ×10⁻³, a factor of 2. Since 2 = 2ᵐ, m = 1 (first order in A).
2.Order in B: compare Trials 1→3, where [B] doubles and [A] is held constant. Rate goes 2.0 → 8.0 ×10⁻³, a factor of 4. Since 4 = 2ⁿ, n = 2 (second order in B).
3.Rate law: rate = k[A][B]². Overall order = 1 + 2 = 3.
4.Solve for k using Trial 1: k = rate / ([A][B]²) = (2.0 × 10⁻³) / (0.10 × 0.10²) = (2.0 × 10⁻³) / (1.0 × 10⁻³) = 2.0.
5.Units for third order: M⁻²·s⁻¹.

Answer: rate = k[A][B]², with k = 2.0 M⁻²·s⁻¹

Checkpoint

For X → products, doubling [X] from 0.20 M to 0.40 M makes the rate rise from 0.010 M·s⁻¹ to 0.040 M·s⁻¹. What is the order in X?

Tip

To find the exponent, solve (concentration factor)ᵒʳᵈᵉʳ = rate factor. Rate ×1 → order 0; ×2 → order 1; ×4 → order 2; ×8 → order 3. Always change just one reactant between the two trials you compare.

Checkpoint

A reaction has the rate law rate = k[A]²[B]. What are the correct units of k?

On the exam

The orders in a rate law come from experimental data only — never from the stoichiometric coefficients of the overall balanced equation. Reading exponents off the coefficients is the single most common kinetics mistake on the exam.

Answer the 2 checkpoints as you read.