When both concentrations change between two initial-rate experiments, divide the observed rate multiplier by the contribution of the reactant whose order you already know. Under the same conditions, the remaining multiplier comes from the other reactant.
The 2027 H2 Chemistry 9476 syllabus covers determining reaction orders and calculating rate constants using initial rates. The example below starts with one comparison that isolates a reactant, then uses a pair in which both concentrations change.
- Find a pair that changes one concentration.
- Determine the order for that reactant.
- Divide out its contribution when both concentrations change.
- Check the proposed rate equation against the data.
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Write the starting model
Suppose rate = k[A]ᵐ[B]ⁿ at a fixed temperature. When you divide two rates, k cancels. The rate ratio is the product of the concentration ratios raised to their respective reaction orders.
Run 1 has [A] = 0.10, [B] = 0.10 and initial rate 1.0 × 10⁻⁴. Run 2 has [A] = 0.10, [B] = 0.20 and rate 2.0 × 10⁻⁴. Run 3 has [A] = 0.20, [B] = 0.40 and rate 1.6 × 10⁻³. Concentrations are in mol dm⁻³ and rates in mol dm⁻³ s⁻¹.
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Worked example | Divide out the known factor
From run 1 to run 2, only [B] changes. It doubles and the rate doubles, so n = 1. From run 2 to run 3, both concentrations double and the rate increases eightfold.
Doubling [B] supplies a factor of two. Divide the observed factor of eight by two to obtain four. Doubling [A] therefore supplies a factor of four, so m = 2. The rate equation is rate = k[A]²[B].
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CHAPTER 3 OF 4 · Back to contents
Check another pair and calculate k
From run 1 to run 3, [A] doubles and [B] quadruples. The model predicts a rate multiplier of 2² × 4 = 16, which agrees with (1.6 × 10⁻³) ÷ (1.0 × 10⁻⁴).
From run 1, k = (1.0 × 10⁻⁴) ÷ (0.10² × 0.10) = 0.10 dm⁶ mol⁻² s⁻¹. A new run with [A] = 0.20 mol dm⁻³ and [B] = 0.10 mol dm⁻³ would have an initial rate of 4.0 × 10⁻⁴ mol dm⁻³ s⁻¹ under the same model and conditions.
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CHAPTER 4 OF 4 · Back to contents
Frequently asked questions
What if I only have the two runs in which both concentrations double?
Those two runs give m + n = 3 because the rate increases by 2³. You need an independent comparison or other information to assign the individual orders.
Where do the units of k come from?
Divide the rate units by the units of [A]²[B]. The concentration product has units of mol³ dm⁻⁹, giving dm⁶ mol⁻² s⁻¹ for k.