Build a rate equation from evidence, explain what each order means and calculate a rate constant with units. Revisit concentration and ratios before attempting the data.
What the rate equation is telling you
A rate describes how quickly a concentration changes. For disappearance of a reactant, quote a positive rate using the magnitude of the negative concentration–time gradient. A typical unit is mol dm⁻³ s⁻¹. A rate equation describes how that measured rate depends on concentrations at a specified temperature.
In rate = k[A]ᵐ[B]ⁿ, square brackets mean concentration. The powers m and n are orders with respect to the individual reactants; m + n is the overall order. The constant k connects the concentration expression to the measured rate. Its numerical value and units depend on the rate equation being used.
Orders are determined experimentally. The balanced overall equation accounts for atoms but does not usually reveal the sequence of elementary steps. A reactant can therefore appear in the overall equation but be zero order in an experimentally established rate equation. Zero order means changing its concentration has no measured rate effect within the conditions investigated, not that the substance is unnecessary.
| Order in A | If [A] doubles | Rate–[A] relationship |
|---|---|---|
| 0 | Rate unchanged | Horizontal line |
| 1 | Rate doubles | Straight line through origin |
| 2 | Rate quadruples | Upward curve; rate proportional to [A]² |
Worked example: extract one dependence at a time
The table is an original illustrative experiment. Compare runs 1 and 2: only [A] doubles and the rate doubles, so A is first order. Compare runs 1 and 3: only [B] doubles and the rate quadruples, so B is second order. The overall order is three.
Write the complete relationship rate = k[A][B]². The coefficient k is essential: rate is proportional to [A][B]² but not numerically equal to it. The table is consistent with one value of k at the same temperature.
| Run | [A] | [B] | Rate / mol dm⁻³ s⁻¹ |
|---|---|---|---|
| 1 | 0.100 | 0.200 | 8.00 × 10⁻⁴ |
| 2 | 0.200 | 0.200 | 1.60 × 10⁻³ |
| 3 | 0.100 | 0.400 | 3.20 × 10⁻³ |
When two concentrations change together
If A doubles and B triples, the predicted rate factor is 2 × 3² = 18. Multiply the effects; do not add them. Conversely, if A is known to be first order, A doubles, B triples and rate increases eighteenfold, divide out the factor of two first. B is responsible for a factor of nine, so 3ⁿ = 9 and n = 2.
Ratios need not be two. A concentration factor of 1.5 giving a rate factor of 2.25 supports second order because 1.5² = 2.25. It does not support first order merely because the numbers look similar. Where data are noisy, compare the proposed relationship with every useful pair rather than forcing one rounded ratio.
Derive units instead of memorising a list
For overall order N, k has units (mol dm⁻³)^(1−N) s⁻¹ when rate is in mol dm⁻³ s⁻¹. Write the actual expression before cancelling: zero order gives mol dm⁻³ s⁻¹, first order s⁻¹, second order dm³ mol⁻¹ s⁻¹. A third-order unit is not interchangeable with a second-order one.
At fixed temperature and unchanged reaction pathway, changing a concentration changes rate through the concentration terms, not by changing k. A catalyst can change the pathway and the relevant rate constant. When checking your answer, recalculate k from another row and substitute your equation back into the data. A mismatch may expose an incorrect order, unit conversion or squared term.
Quick checks
Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official OCR A mark allocations.
Q1. For rate = k[X]²[Y]⁰, what happens if both concentrations triple?Show answer
Only X affects the rate in this expression. The factor is 3² × 3⁰ = 9. Y is still a reactant; zero order describes the observed concentration dependence.
Q2. At fixed [Y], multiplying [X] by 1.5 multiplies rate by 2.25. Deduce the order in X.Show answer
Solve 1.5ᵐ = 2.25. Since 1.5² = 2.25, the order is two. State the comparison at unchanged [Y] and temperature.
Q3. A first-order reaction has rate 6.00 × 10⁻⁵ mol dm⁻³ s⁻¹ at [A] = 0.0250 mol dm⁻³. Find k.Show answer
k = rate/[A] = 6.00 × 10⁻⁵/0.0250 = 2.40 × 10⁻³ s⁻¹. Concentration units cancel, leaving inverse seconds.
Q4. For the illustrative rate equation, [A] halves and [B] doubles. Predict the new rate relative to the original.Show answer
The factor is 0.5 × 2² = 2. The rate doubles. This changed example checks both factors rather than considering just the doubled concentration.
Q5. A student assigns rate = k[A]²[B] from 2A + B → C. Explain the problem.Show answer
The coefficients give stoichiometry, not generally kinetic orders. Experimental concentration–rate data or an appropriate elementary mechanism are needed. The proposed expression may happen to fit, but the overall equation alone does not establish it.
Sources
Sources and examiner guidance (reviewed 6 October 2026)
- OCR A H432 specification — version 3.1 — 5.1.1, printed pp. 42–43; outcomes and additional guidance, with relevant Module 1 practical skills.
- Chemrevise — OCR A 5.1.1 — Pages 1–7; secondary coverage cross-check. Lesson explanations, data and questions are original Finesse material.
- OCR H432/01 mark scheme — June 2025 — Q18(b–c); printed pp. 20–21. Question-specific evidence, not universal marking rules.
- OCR H432/01 examiner report — June 2025 — Q18(b–c); printed pp. 35–38. Read with the corresponding question context.
- OCR H432/01 question paper — June 2025 — Q18(b–c); context for the assessment references, not reproduced questions.
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