5.2 Rate Law

Method of Initial Rates

Select two experiments to derive the order for each reactant. Bar charts show rate and concentration ratios before the full rate law and k are assembled.

5 ReactionsOrder DeterminationCalculate kRate Law
Topic 5.2

Introduction to Rate Law

Represent experimental data with a consistent rate law expression.

Experimental methods monitor the amounts of reactants or products over time — spectrophotometry via Beer's law (3.13) is the standard technique in AP labs, because absorbance is directly proportional to concentration.

The rate law expresses the rate as proportional to reactant concentrations raised to powers:

rate = k[A]m[B]n

The power on each reactant is the order with respect to that reactant, and the sum of the powers is the overall order.

The proportionality constant k is the rate constant. Two facts about it are heavily tested: its value is temperature dependent, and its units reflect the overall reaction order. For an overall order x, the units of k are M1−x·s⁻¹:

  • Zeroth order → M·s⁻¹
  • First order → s⁻¹
  • Second order → M⁻¹·s⁻¹

This means you can often deduce the overall order just from the units of k.

The method of initial rates is the standard way to find the orders. Compare two trials in which only one concentration changes:

  • Double [A], rate unchanged → zeroth order in A
  • Double [A], rate doubles → first order in A
  • Double [A], rate quadruples → second order in A

Algebraically, rate₂/rate₁ = ([A]₂/[A]₁)m, so m = log(rate ratio)/log(concentration ratio).

The critical conceptual point: orders are determined experimentally and generally do not equal the stoichiometric coefficients. They match the coefficients only for an elementary step (5.4).

Key points

  • Orders come from experiment, never from the balanced equation’s coefficients.
  • The units of k tell you the overall order.
  • k depends on temperature (and on the presence of a catalyst) but not on concentration.
  • A species absent from the rate law does not affect the rate — that is a mechanistic clue (5.8).

Equations

  • not on the sheetThe general rate law. Not on the sheet — orders must come from data.
    • rate constant (units depend on overall order)
    • experimentally determined orders

Common mistakes

  • Using coefficients as orders. Only valid for an elementary step.
  • Omitting the units of k. AP rubrics award a point for them.
  • Thinking k changes with concentration. It does not — only with temperature or a catalyst.
  • Including a catalyst or solvent in the rate law by default. A catalyst may appear, but only if the data say so.

Worked example

Determine the rate law and the value of k, with units, from these initial-rate data for A + B → products. Trial 1: [A] = 0.10 M, [B] = 0.10 M, rate = 2.0 × 10⁻³ M/s Trial 2: [A] = 0.20 M, [B] = 0.10 M, rate = 4.0 × 10⁻³ M/s Trial 3: [A] = 0.10 M, [B] = 0.20 M, rate = 8.0 × 10⁻³ M/s

Order in A — compare trials 1 and 2 ([B] constant):
[A] doubles, rate doubles (2.0 → 4.0 × 10⁻³). Rate ratio 2 = 2^m → m = 1 (first order in A).

Order in B — compare trials 1 and 3 ([A] constant):
[B] doubles, rate quadruples (2.0 → 8.0 × 10⁻³). Rate ratio 4 = 2ⁿ → n = 2 (second order in B).

Rate law: rate = k[A][B]² (overall order 3)

Solve for k using trial 1:
2.0 × 10⁻³ = k(0.10)(0.10)² = k(1.0 × 10⁻³)
k = 2.0 M⁻²s⁻¹

Units check: for overall order 3, k has units M1−3s⁻¹ = M⁻²s⁻¹ ✓

Full notes for topic 5.2 →