UNIT 2 CHEMISTRY · PEARSON AAQ · D1.2.5
Rate Equations
FREE CHEMISTRY REVISION · TOPIC 37
Rate Equations
Pearson BTEC Level 3 AAQ Applied Science
Unit 2: Principles and Applications of Chemistry · D1.2.5
What you will learn
Use initial-rate data to determine reaction orders, write r = k[A]m[B]n, calculate k and derive its units. Work from one changing concentration at a time before handling experiments where both change. The data below are original synthetic experimental-style teaching datasets, not measurements claimed from a laboratory.
1. What a rate equation tells you
A rate equation describes how the measured reaction rate depends on concentrations under specified conditions. Square brackets mean concentration, usually in mol dm⁻³. In r = k[A]m[B]n, m is the order with respect to A, n is the order with respect to B, and m + n is the overall order. k is the rate constant.
Orders are determined experimentally. Do not copy the coefficients of a balanced equation into the powers unless a particular elementary step and its kinetic interpretation have been established. A balanced equation describes overall stoichiometry, not usually the rate law.
For a fixed reaction pathway, temperature and catalyst conditions, k is constant across experiments even when initial concentrations change. Temperature or a catalyst can change k. The value and units also depend on the rate and concentration units used. Here r is in mol dm⁻³ s⁻¹. Keep the same definition of rate in every row; do not mix disappearance rates of one species with formation rates of another without accounting for stoichiometry.
2. Zero-, first- and second-order behaviour
| Order in A | Dependence with other factors fixed | Double [A] | Triple [A] |
|---|---|---|---|
| 0 | [A]⁰ = 1; rate independent of [A] | ×1 | ×1 |
| 1 | Rate ∝ [A] | ×2 | ×3 |
| 2 | Rate ∝ [A]² | ×4 | ×9 |
Zero order in A means changing [A] has no effect over the tested range. It does not mean a zero rate or that A is absent from the reaction. A second-order reaction can be second order in A or first order in each of two reactants. Always name the reactant and state the overall order separately.
These are rate-versus-concentration graphs, not concentration-versus-time graphs. For r = k[A] with all other factors absent or absorbed into a fixed effective constant, the first-order straight-line gradient is k. A zero-order horizontal line has height k; a second-order rate-versus-[A] graph is curved.
3. A reliable initial-rates method
- Measure initial rates in separate runs before appreciable reactant consumption. Control temperature, catalyst and the measurement method.
- Compare two runs where only [A] changes. Divide their rates and concentrations.
- Use rate ratio = (concentration ratio)order. Decide which power fits the data.
- Repeat for B using a pair with [A] constant. Add orders to get the overall order.
- Rearrange k = r/([A]m[B]n). Substitute one full row with consistent units.
- Check k in another row and derive its units from the final equation.
For appropriate non-unit ratios, order = log(rate ratio)/log(concentration ratio). The logarithm base can be any consistent base. Do not divide by log(1); if concentration has not changed, that pair cannot determine its order. Real data have uncertainty, so a ratio near 4 when concentration doubles may support second order rather than an exactly calculated non-integer.
4. Dataset A: build the complete rate equation
All five runs use the same temperature and catalyst conditions. Rates have been measured consistently at the start of each run. Values are intentionally idealised so that the arithmetic can be checked exactly.
| Run | Initial [A] / mol dm⁻³ | Initial [B] / mol dm⁻³ | Initial r / mol dm⁻³ s⁻¹ |
|---|---|---|---|
| 1 | 0.100 | 0.200 | 1.00 × 10⁻³ |
| 2 | 0.200 | 0.200 | 4.00 × 10⁻³ |
| 3 | 0.100 | 0.400 | 2.00 × 10⁻³ |
| 4 | 0.300 | 0.200 | 9.00 × 10⁻³ |
| 5 | 0.200 | 0.400 | 8.00 × 10⁻³ |
Worked calculation 1 — determine m and n
Compare runs 1 and 2: [B] is fixed, [A] doubles and r increases by 4.00 × 10⁻³ / 1.00 × 10⁻³ = 4. Thus 2m = 4 and m = 2. Compare runs 1 and 3: [A] is fixed, [B] doubles and r doubles. Thus 2n = 2 and n = 1. The rate equation is r = k[A]²[B], overall order 3. Run 4 independently checks the A order: tripling A multiplies rate by 9.
Worked calculation 2 — calculate k and its units
Using run 1: k = (1.00 × 10⁻³)/(0.100² × 0.200) = 0.00100/0.00200 = 0.500 dm⁶ mol⁻² s⁻¹. Units: (mol dm⁻³ s⁻¹)/(mol dm⁻³)³ = mol⁻² dm⁶ s⁻¹. Recheck run 5: 0.00800/(0.200² × 0.400) = 0.00800/0.0160 = 0.500, the same k.
Worked calculation 3 — predict a new rate
At [A] = 0.150 mol dm⁻³ and [B] = 0.300 mol dm⁻³, with the same conditions, r = 0.500 × 0.150² × 0.300 = 0.003375 mol dm⁻³ s⁻¹ = 3.38 × 10⁻³ mol dm⁻³ s⁻¹ to three significant figures. Check the units: (dm⁶ mol⁻² s⁻¹)(mol² dm⁻⁶)(mol dm⁻³) = mol dm⁻³ s⁻¹.
Worked calculation 4 — both concentrations change
Relative to run 1, [A] is multiplied by 1.50 and [B] is halved. Rate factor = 1.50² × 0.500 = 1.125. New rate = (1.00 × 10⁻³) × 1.125 = 1.125 × 10⁻³ mol dm⁻³ s⁻¹, or 1.13 × 10⁻³ to three significant figures. Direct substitution with [A] = 0.150 and [B] = 0.100 gives the same result.
Worked calculation 5 — find an unknown concentration
For r = 6.00 × 10⁻³ mol dm⁻³ s⁻¹ and [B] = 0.300 mol dm⁻³, [A]² = r/(k[B]) = 0.00600/(0.500 × 0.300) = 0.0400 mol² dm⁻⁶. Taking the positive square root gives [A] = 0.200 mol dm⁻³. A negative concentration is not physically valid. Recheck: 0.500 × 0.200² × 0.300 = 0.00600.
Worked calculation 6 — use non-doubling data
Comparing runs 1 and 4 at constant B, the A ratio is 0.300/0.100 = 3 and the rate ratio is 9. Therefore m = log(9)/log(3) = 2. A third A value confirms the square dependence; do not assume only doubling experiments can identify an order.
5. Dataset B: a zero-order reactant
| Run | [A] / mol dm⁻³ | [B] / mol dm⁻³ | r / mol dm⁻³ s⁻¹ |
|---|---|---|---|
| 1 | 0.100 | 0.100 | 2.00 × 10⁻³ |
| 2 | 0.100 | 0.400 | 8.00 × 10⁻³ |
| 3 | 0.200 | 0.100 | 2.00 × 10⁻³ |
Worked calculation 7 — zero order and first order together
Runs 1 and 3 double A at fixed B, with no rate change: 2m = 1, so m = 0. Runs 1 and 2 quadruple B at fixed A, and the rate quadruples: 4n = 4, so n = 1. Thus r = k[A]⁰[B] = k[B], overall order 1. k = 0.00200/0.100 = 0.0200 s⁻¹. Run 2 checks this: 0.00800/0.400 = 0.0200 s⁻¹.
This conclusion is supported over the tested concentration range. It does not imply that the reaction can continue indefinitely when A has been consumed.
6. Dataset C: second order in one reactant
| Run | [A] / mol dm⁻³ | r / mol dm⁻³ s⁻¹ |
|---|---|---|
| 1 | 0.0500 | 3.00 × 10⁻⁴ |
| 2 | 0.100 | 1.20 × 10⁻³ |
| 3 | 0.150 | 2.70 × 10⁻³ |
Worked calculation 8 — a second-order rate constant
Doubling A multiplies rate by 4; tripling A multiplies it by 9. Therefore r = k[A]². k = 0.000300/0.0500² = 0.120 dm³ mol⁻¹ s⁻¹. Run 3: 0.00270/0.150² = 0.120. Units are (mol dm⁻³ s⁻¹)/(mol² dm⁻⁶) = dm³ mol⁻¹ s⁻¹.
Worked calculation 9 — recognise a fully zero-order law
A separate reaction has r = k[A]⁰ and a measured rate of 3.00 × 10⁻³ mol dm⁻³ s⁻¹ over the tested range. Hence k = 3.00 × 10⁻³ mol dm⁻³ s⁻¹. No concentration factor remains. k has the same units as rate, not s⁻¹.
7. Derive units instead of guessing
Let overall order p = m + n and concentration unit C = mol dm⁻³. Since r has units C s⁻¹, k has units C1−p s⁻¹. This dimensional rule assumes all concentration factors use the same units and the rate uses concentration per second.
| Overall order | Units of k | Example law |
|---|---|---|
| 0 | mol dm⁻³ s⁻¹ | r = k |
| 1 | s⁻¹ | r = k[A] |
| 2 | dm³ mol⁻¹ s⁻¹ | r = k[A][B] or k[A]² |
| 3 | dm⁶ mol⁻² s⁻¹ | r = k[A]²[B] |
If times are in minutes, the time unit is min⁻¹ until converted. Do not compare numerical k values from different overall orders as if their sizes alone showed which reaction was faster: their dimensions differ, and rate depends on concentration.
Common misconceptions
- Orders equal equation coefficients: find orders from experiments.
- Second order means every concentration is squared: the total order is the sum of the powers.
- Zero order means no reaction: it means no rate dependence on that concentration in the tested range.
- Change two variables and assign all the rate change to one: isolate one effect or use an already known order.
- k always has units s⁻¹: that is the first-order case with concentration/time rate units.
- k increases when concentration increases: at fixed conditions, the equation accounts for that change through the concentration factors.
Exam tips
- Name the pair of runs and the concentration held constant.
- Write the rate ratio and concentration ratio before identifying the power.
- Show the final rate equation and overall order explicitly.
- Calculate k using one complete row; verify it against another.
- Derive and simplify units from the equation you actually found.
- Keep extra digits until the final answer and include units.
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