The big idea: At HL we don't just say what speeds a reaction up — we write an equation for the rate. For a reaction of A and B the rate equation is:
rate = k[A]m[B]n
It says the rate depends on the concentrations of the reactants, each raised to a power. Those powers are the orders of reaction, and k is the rate constant.
- the rate constant (its units depend on the overall order)
- concentrations of the reactants (mol dm⁻³)
- order with respect to A
- order with respect to B
- the overall order of reaction
Define your terms: - Order with respect to a reactant (m or n) — the power to which that reactant's concentration is raised in the rate equation. - Overall order — the sum of the individual orders, m + n. - Rate constant (k) — the proportionality constant in the rate equation; it is fixed at a given temperature (it changes only when the temperature changes). - A reactant of order 0 does not appear in the rate equation — changing its concentration has no effect on the rate.
Orders are found by EXPERIMENT — not from the equation: This is the single most-tested HL point. The orders cannot be read off the balanced (stoichiometric) equation.
For 2NO + O2 → 2NO2 you might expect order 2 in NO and 1 in O2 from the coefficients — but the only way to know the real orders is to measure how the rate responds to changing each concentration. Always determine orders from data, never from the stoichiometry.
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To find an order experimentally, change one concentration at a time (hold the others constant) and see how the initial rate responds. Doubling a concentration is the easiest test:
| Double one [reactant], hold others constant | Rate changes by | Order w.r.t. that reactant |
|---|---|---|
| rate stays the same (× 1 = 2⁰) | × 1 | 0 (zero order) — that reactant is not in the rate equation |
| rate doubles (× 2 = 2¹) | × 2 | 1 (first order) |
| rate quadruples (× 4 = 2²) | × 4 | 2 (second order) |
| rate goes up × 8 (2³) | × 8 | 3 (third order — rare) |
Each order has its own shape when you plot rate against concentration — recognising the shape is a quick way to read off the order:
Rate vs [reactant]: zero order is a flat line (rate unchanged, × 1), first order is a straight line through the origin (rate × 2 when [ ] doubles), second order curves upward (rate × 4 when [ ] doubles).
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The doubling shortcut: When you double one concentration and the rate changes by a factor F, the order is the power of 2 that gives F:
- × 1 = 2⁰ → 0 order - × 2 = 2¹ → 1st order - × 4 = 2² → 2nd order
In general, order = log(rate factor) ÷ log(concentration factor) — useful when the concentration is multiplied by 3 (×3 → factor 9 means 2nd order, since 3² = 9).
Reading the change correctly: Always compare two experiments where only the reactant you are studying changes and every other concentration is held the same. If two concentrations change at once you cannot isolate a single order — pick a different pair of experiments.
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A reaction between two reactants, A and B, was studied at constant temperature. The initial rate was measured for four mixtures:
| Experiment | [A] / mol dm⁻³ | [B] / mol dm⁻³ | Initial rate / mol dm⁻³ s⁻¹ |
|---|---|---|---|
| 1 | 0.10 | 0.10 | 2.0 × 10⁻³ |
| 2 | 0.20 | 0.10 | 4.0 × 10⁻³ |
| 3 | 0.10 | 0.20 | 8.0 × 10⁻³ |
| 4 | 0.20 | 0.20 | 1.6 × 10⁻² |
Worked example — order with respect to A
Use experiments 1 and 2 to deduce the order of reaction with respect to A.
Solution
- Pick two experiments where only [A] changes. From 1 → 2, [A] doubles (0.10 → 0.20) while [B] is held at 0.10.
- See how the rate responds:
- Doubling [A] doubles the rate (× 2 = 2¹), so the reaction is first order in A.
Final answer
First order with respect to A (m = 1).
Worked example — order with respect to B
Use experiments 1 and 3 to deduce the order of reaction with respect to B.
Solution
- Pick two experiments where only [B] changes. From 1 → 3, [B] doubles (0.10 → 0.20) while [A] is held at 0.10.
- See how the rate responds:
- Doubling [B] makes the rate × 4 (= 2²), so the reaction is second order in B.
Final answer
Second order with respect to B (n = 2).
Worked example — the rate equation and the value of k
Write the rate equation, state the overall order, and calculate the rate constant k with its units (use experiment 1).
Solution
- Formula first — combine the two orders into the rate equation:
- Overall order = sum of the orders:
- Rearrange for k:
- Substitute experiment 1 (rate = 2.0 × 10⁻³, [A] = 0.10, [B] = 0.10):
- Work out the bottom line, then divide:
- Find the units from k = rate ÷ ([A][B]²) — overall 3rd order:
Final answer
rate = k[A][B]²; overall 3rd order; k = 2.0 mol⁻² dm⁶ s⁻¹.
The units of k change with the overall order: k is not unitless — and its units are not always the same. Because rate always has units of mol dm⁻³ s⁻¹, the units of k must cancel the concentration units in the rate equation. Work them out from k = rate ÷ [ ]overall order:
| Overall order | Rate equation | Rearranged for k | Units of k |
|---|---|---|---|
| 0 | rate = k | k = rate | mol dm⁻³ s⁻¹ |
| 1 | rate = k[A] | k = rate ÷ [A] | s⁻¹ |
| 2 | rate = k[A]² | k = rate ÷ [A]² | mol⁻¹ dm³ s⁻¹ |
| 3 | rate = k[A]³ | k = rate ÷ [A]³ | mol⁻² dm⁶ s⁻¹ |
Deriving the units — every time: Never memorise the units blindly. Each step down the table divides by one more mol dm⁻³, so each step adds mol⁻¹ dm³ to the units of k:
- 1st order: rate ÷ (mol dm⁻³) → s⁻¹ - 2nd order: rate ÷ (mol dm⁻³)² → mol⁻¹ dm³ s⁻¹ - 3rd order: rate ÷ (mol dm⁻³)³ → mol⁻² dm⁶ s⁻¹
If you can derive them you will never lose the units mark.
Many reactions happen in several steps (a mechanism). The slowest step is the rate-determining step (RDS) — like the narrowest gap on a road, it sets the overall speed.
The rate equation reflects the rate-determining step: Only the species involved in (or before) the rate-determining step appear in the rate equation. This is why the orders need not match the overall stoichiometry — a reactant that is only used in a fast step after the RDS does not affect the rate, so it is order 0.
So experimentally-found orders are powerful evidence: they tell you how many of each species take part up to and including the slowest step, which lets chemists test a proposed mechanism.