Topic 1 of 5
Rates, orders and rate constants
Read changing concentrations and compare controlled initial-rate experiments.
A-Level 9476 (2026-2027)
A rate measures change per unit time
A rate law is experimental evidence about speed, not a rearranged balanced equation.
A reaction rate can be measured as the change in a reactant or product concentration per unit time. Reactant disappearance rates are reported positively: -Δ[A]/Δt; product appearance rates use +Δ[P]/Δt. A tangent to a concentration-time curve gives the instantaneous rate, while a chord gives an average rate over the interval. State which species is being measured because stoichiometric coefficients connect their different disappearance and appearance rates.
For rate = k[A]m[B]n, m and n are the orders with respect to A and B; m + n is the overall order. The rate constant k is constant for a given reaction under specified temperature and catalytic conditions. Changing concentrations changes the rate through the concentration terms, not k, within the conditions where that rate law applies.
| Order in A | Dependence on [A] | Effect of doubling [A] |
|---|---|---|
| Zero | The [A] factor is 1. | No change in rate. |
| First | Rate is proportional to [A]. | Rate doubles. |
| Second | Rate is proportional to [A] squared. | Rate becomes four times as large. |
Compare experiments that isolate one concentration
Divide the rate equations so the constant and unchanged factors cancel.
| Experiment | [A] / mol dm-3 | [B] / mol dm-3 | Initial rate / mol dm-3 s-1 |
|---|---|---|---|
| 1 | 0.100 | 0.100 | 2.00 × 10-4 |
| 2 | 0.200 | 0.100 | 8.00 × 10-4 |
| 3 | 0.200 | 0.200 | 1.60 × 10-3 |
Worked example
Determine the complete rate equation
Find the orders, k and the predicted initial rate when [A] = 0.150 and [B] = 0.300 mol dm-3.
- Compare experiments 1 and 2: [A] doubles at fixed [B], while rate quadruples. Therefore m = 2.
- Compare experiments 2 and 3: [B] doubles at fixed [A], while rate doubles. Therefore n = 1.
- k = (2.00 × 10-4)/(0.1002 × 0.100) = 0.200.
- Overall order is 3. Units of k = (mol dm-3 s-1)/(mol dm-3)3 = dm6 mol-2 s-1.
- New rate = 0.200(0.150)2(0.300) = 0.00135 mol dm-3 s-1.
Rate = k[A]2[B], with k = 0.200 dm6 mol-2 s-1; predicted rate = 1.35 × 10-3 mol dm-3 s-1.
If both concentrations change, first use another pair to find one order, then divide out its known contribution. For example, if doubling both A and B raises the rate eightfold and B is first order, the remaining factor of four comes from A, which is second order. Check the final equation against every row, not just the pair used to derive it.
| Overall order | Units of k |
|---|---|
| 0 | mol dm-3 s-1 |
| 1 | s-1 |
| 2 | dm3 mol-1 s-1 |
| 3 | dm6 mol-2 s-1 |