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 + O_{2} → 2NO_{2} 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.
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).
Interactive diagram
Explore the labelled diagram, charts and maps for this topic in full study mode.
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.
Memorize terms 3x faster
Smart flashcards show you cards right before you forget them. Perfect for definitions and key concepts.
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.