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The rate equation is the empirical mathematical statement that links the rate of a chemical reaction to the concentrations of the species involved. For a generic reaction it takes the form Rate = k[A]ᵐ[B]ⁿ…, in which the exponents m and n are the orders of reaction with respect to each reactant and k is the rate constant. A point that catches almost every new A2 student: the orders cannot be read off the balanced chemical equation. They are determined experimentally, by varying concentrations and measuring how the initial rate responds. The reason is that the rate equation describes only what happens at the rate-determining step — the slowest step of the mechanism — not the overall stoichiometry. In this lesson we will define the rate equation rigorously, work through how to deduce orders from initial-rate tables and from half-life data, derive the units of k for each overall order, sketch the diagnostic concentration-time and rate-concentration graphs, and signpost how rate equations are then used (in lesson 3) to deduce reaction mechanisms.
Spec mapping (AQA 7405): This lesson maps to §3.1.9 of the AQA A-Level Chemistry specification (rate equations). It builds on lesson 0 of this course (collision theory and the qualitative factors that affect rate, §3.1.5), prepares the way for lesson 2 (the Arrhenius equation — how k depends on T, §3.1.9.2), and underpins lesson 3 (using rate equations to deduce the rate-determining step and propose mechanisms, §3.1.9.3). The experimental measurement of rates and orders is the anchor for Required Practicals 3 (iodine-thiosulfate clock), 6 (rate by gas-volume collection) and 7 (rate by colorimetry), developed in lesson 7 of this course. Refer to the official AQA specification document for the exact wording of each section.
Assessment objectives: Definitions of "rate equation", "order with respect to a reactant", "overall order", and "rate constant" are AO1 recall items. Deducing orders from initial-rate data and computing k with correct units is bread-and-butter AO2 — these calculations appear on essentially every Paper 1 in some form. Sketching and interpreting concentration-time and rate-concentration graphs for orders 0, 1 and 2, and deducing the order from half-life data, are AO3 reasoning tasks that frequently feature in multi-mark structured questions.
For a reaction with reactants A, B, C … the rate equation has the form:
Rate = k[A]ᵐ[B]ⁿ[C]ᵖ …
where:
Key Definition: The order of reaction with respect to a particular reactant is the power to which the concentration of that reactant is raised in the experimentally-determined rate equation. The overall order is the sum of the individual orders. Orders are empirical quantities — they must be determined by experiment and cannot in general be predicted from the balanced equation. A reactant that does not appear in the rate equation is said to be zero order with respect to the reaction (its concentration has no effect on the rate).
Key Point: A common stumbling block is the assumption that orders match stoichiometric coefficients. They do not. For example, the reaction 2NO + 2H₂ → N₂ + 2H₂O has a balanced equation whose experimentally-determined rate law is second-order in NO and first-order in H₂ (Rate = k[NO]²[H₂]) — even though both reactants have a coefficient of 2 in the overall equation. The match between order and coefficient is coincidental for NO and broken for H₂. This is because the rate equation reflects only the rate-determining step of the mechanism, not the overall balanced equation.
Because Rate always has units mol dm⁻³ s⁻¹ and concentrations have units mol dm⁻³, the units of k must adjust to balance the dimensional algebra. Work it out by substituting units into the rate equation and solving for k.
| Overall order | Rate equation example | Units of k |
|---|---|---|
| 0 | Rate = k | mol dm⁻³ s⁻¹ |
| 1 | Rate = k[A] | s⁻¹ |
| 2 | Rate = k[A]² or k[A][B] | mol⁻¹ dm³ s⁻¹ (equivalently dm³ mol⁻¹ s⁻¹) |
| 3 | Rate = k[A]²[B] or k[A][B][C] | mol⁻² dm⁶ s⁻¹ (equivalently dm⁶ mol⁻² s⁻¹) |
For Rate = k[A]²: k = Rate / [A]² = (mol dm⁻³ s⁻¹) / (mol dm⁻³)² = (mol dm⁻³ s⁻¹) / (mol² dm⁻⁶) = mol⁻¹ dm³ s⁻¹.
For Rate = k[A][B] (also overall second-order): k = Rate / ([A][B]) = (mol dm⁻³ s⁻¹) / (mol dm⁻³ × mol dm⁻³) = mol⁻¹ dm³ s⁻¹. Same units — the overall order determines the units of k, not the form of the rate equation.
Exam Tip: A common AQA mark on rate-equation calculations is "1 mark for correct value of k, 1 mark for correct units". If you give the numerical value without units, you lose half the mark. If asked to "calculate k including units", always derive the units from first principles by substituting units into Rate = k[A]ᵐ[B]ⁿ — do not guess from a table.
The initial-rate method is the standard experimental approach: run the reaction several times, varying the initial concentration of one reactant at a time while holding the others constant, and measure the initial rate in each run. The ratio of rates tells you the order.
The logic: if [A] is multiplied by a factor f and the rate changes by a factor fᵐ, then the order with respect to A is m. So:
The reaction 2A + B → C was studied at 298 K. The initial concentrations and initial rates are tabulated:
| Experiment | [A]₀ / mol dm⁻³ | [B]₀ / mol dm⁻³ | Initial rate / mol dm⁻³ s⁻¹ |
|---|---|---|---|
| 1 | 0.10 | 0.10 | 2.0 × 10⁻⁴ |
| 2 | 0.20 | 0.10 | 8.0 × 10⁻⁴ |
| 3 | 0.10 | 0.20 | 4.0 × 10⁻⁴ |
| 4 | 0.30 | 0.30 | ? |
Step 1 — Order with respect to A: Compare experiments 1 and 2. [B] is held constant. [A] doubles (0.10 → 0.20). Rate changes from 2.0 × 10⁻⁴ to 8.0 × 10⁻⁴, a factor of 4. Since 2ᵐ = 4 → m = 2.
Step 2 — Order with respect to B: Compare experiments 1 and 3. [A] is held constant. [B] doubles (0.10 → 0.20). Rate changes from 2.0 × 10⁻⁴ to 4.0 × 10⁻⁴, a factor of 2. Since 2ⁿ = 2 → n = 1.
Step 3 — Rate equation: Rate = k[A]²[B]. Overall order = 2 + 1 = 3 (third order).
Step 4 — Calculate k: Use any row. Take experiment 1: k = Rate / ([A]²[B]) = (2.0 × 10⁻⁴) / (0.10² × 0.10) = (2.0 × 10⁻⁴) / (1.0 × 10⁻³) = 0.20.
Step 5 — Units of k: Overall third order, so k has units mol⁻² dm⁶ s⁻¹. k = 0.20 mol⁻² dm⁶ s⁻¹ (or equivalently 0.20 dm⁶ mol⁻² s⁻¹).
Step 6 — Predict the rate of experiment 4: Rate = 0.20 × (0.30)² × 0.30 = 0.20 × 0.09 × 0.30 = 5.4 × 10⁻³ mol dm⁻³ s⁻¹.
The gas-phase reaction 2NO(g) + 2H₂(g) → N₂(g) + 2H₂O(g) was studied at 1100 K with the following data:
| Experiment | [NO]₀ / mol dm⁻³ | [H₂]₀ / mol dm⁻³ | Initial rate / mol dm⁻³ s⁻¹ |
|---|---|---|---|
| 1 | 5.0 × 10⁻³ | 2.0 × 10⁻³ | 1.25 × 10⁻⁵ |
| 2 | 1.0 × 10⁻² | 2.0 × 10⁻³ | 5.00 × 10⁻⁵ |
| 3 | 5.0 × 10⁻³ | 4.0 × 10⁻³ | 2.50 × 10⁻⁵ |
Order in NO: Experiments 1 → 2: [NO] doubles, [H₂] constant. Rate goes 1.25 → 5.00 (×4). So 2ᵐ = 4 → m = 2 (second order in NO).
Order in H₂: Experiments 1 → 3: [H₂] doubles, [NO] constant. Rate goes 1.25 → 2.50 (×2). So 2ⁿ = 2 → n = 1 (first order in H₂).
Rate equation: Rate = k[NO]²[H₂]. Overall order = 3.
k from experiment 1: k = (1.25 × 10⁻⁵) / ((5.0 × 10⁻³)² × (2.0 × 10⁻³)) = (1.25 × 10⁻⁵) / (2.5 × 10⁻⁵ × 2.0 × 10⁻³) = (1.25 × 10⁻⁵) / (5.0 × 10⁻⁸) = 250 mol⁻² dm⁶ s⁻¹.
Note that the orders (2 and 1) do not match the stoichiometric coefficients (both 2). This is empirical evidence that the mechanism is multi-step (see lesson 3).
The half-life t½ is the time taken for the concentration of a reactant to fall to half its initial value. The way half-life depends on initial concentration is itself a fingerprint of the order:
| Order in A | t½ behaviour | Mathematical form |
|---|---|---|
| 0 | t½ decreases linearly as [A]₀ decreases (proportional to [A]₀) | t½ = [A]₀ / 2k |
| 1 | t½ is constant — independent of [A]₀ | t½ = ln 2 / k = 0.693 / k |
| 2 | t½ increases as [A]₀ decreases (proportional to 1/[A]₀) | t½ = 1 / (k[A]₀) |
The constant half-life of a first-order reaction is the single most useful diagnostic in undergraduate kinetics. If you measure successive half-lives from a concentration-time graph and they are all equal within experimental error, the reaction is first order in the reactant being followed. Radioactive decay (a classic first-order process) is the textbook example: a 5730-year half-life of carbon-14 means a sample that starts at 100 % takes 5730 years to fall to 50 %, then another 5730 years to fall to 25 %, then another 5730 years to fall to 12.5 %, regardless of starting mass.
A first-order reaction is followed colorimetrically. The concentration of A halves from 0.80 to 0.40 mol dm⁻³ in 120 s. From 0.40 to 0.20 mol dm⁻³ takes a further 120 s, and from 0.20 to 0.10 mol dm⁻³ another 120 s. The constancy of the half-life confirms first-order kinetics. Calculate k:
k = 0.693 / t½ = 0.693 / 120 = 5.78 × 10⁻³ s⁻¹.
A reactant decays such that t½ = 200 s starting from [A]₀ = 0.40 mol dm⁻³, but t½ = 400 s starting from [A]₀ = 0.20 mol dm⁻³. Since halving the initial concentration doubles the half-life (t½ ∝ 1/[A]₀), the reaction is second order in A.
A concentration-time graph plots [A] (y-axis) against time (x-axis) for a single experiment in which one reactant is followed continuously.
To deduce the order from a concentration-time graph at A-Level: measure two or three successive half-lives. If equal → first order. If lengthening → second order. If shortening (or if the graph is a straight line) → zero order.
A rate-concentration graph plots the rate of reaction (y-axis) against the concentration of a reactant (x-axis). Each data point comes from a separate experiment (or from tangents to a concentration-time graph at different times).
If the graph is a straight line through the origin, the order is 1. If horizontal, order 0. If curving upward through the origin, order 2.
Consider the acid-catalysed iodination of propanone:
CH₃COCH₃(aq) + I₂(aq) → CH₃COCH₂I(aq) + HI(aq) (catalysed by H⁺)
The experimentally-determined rate equation is:
Rate = k[CH₃COCH₃][H⁺]
Two observations of immense pedagogical value:
These two observations together tell you a great deal about the mechanism — a topic developed properly in lesson 3.
The reaction H₂O₂ + 2I⁻ + 2H⁺ → 2H₂O + I₂ can be followed using the iodine clock method: a small fixed quantity of thiosulfate is added, which reacts instantly with any I₂ produced, regenerating I⁻ and keeping [I₂] effectively zero. Starch indicator is also present. When the thiosulfate runs out, free I₂ accumulates and the starch turns blue-black at a stopwatch-measurable instant. The time t taken for the blue-black colour to appear is inversely proportional to the initial rate (since the same small fixed quantity of I₂ is produced in each run).
Typical experimental finding: Rate = k[H₂O₂][I⁻] — first order in each (overall second order). H⁺ does not appear in the rate equation under these conditions, meaning H⁺ is involved only in a fast step after the rate-determining one. This experimental procedure underpins Required Practical 6 (the iodine-thiosulfate clock) — see lesson 7 for the full protocol, hazard assessment, and graphical analysis.
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