Lesson 1 of 5
Rates, orders and rate constants
Read changing concentrations and compare controlled initial-rate experiments.
A-Level 9476 (2026-2027)
I can explain the central kinetic terms.Syllabus
Syllabus 9476, 8(a). Explain the central kinetic terms.
I can construct and use simple rate equations.Syllabus
Syllabus 9476, 8(b). Construct and use simple rate equations.
I can calculate a rate constant from initial-rate data.Syllabus
Syllabus 9476, 8(d). Calculate a rate constant from initial-rate data.
A rate measures change per unit time
A rate law is experimental evidence about speed, not a rearranged balanced equation.
Rate is how fast a concentration changes with time. For a reactant, write the rate as -Δ[A]/Δt, so the number comes out positive. For a product, write +Δ[P]/Δt. A tangent to a concentration-time curve gives the rate at one instant. A chord between two points gives the average rate over that interval. Always say which substance you measured. The coefficients in the equation link the rates for different substances.
A rate equation has the form rate = k[A]m[B]n. Here m is the order with respect to A, and n is the order with respect to B. Their sum, m + n, is the overall order. The rate constant k stays the same for a reaction at a fixed temperature, with the same catalyst. When you change a concentration, the rate changes through the concentration terms. k itself does not change.
| 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. |
Watch out for this
A student says: "The equation is 2H2O2 -> 2H2O + O2, so the reaction must be second order in H2O2." What is wrong with this?
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A coefficient tells you how much reacts overall. An order tells you how the measured rate depends on a concentration. The two match only within a single elementary step. You usually cannot read orders from the overall equation of a reaction that has several steps.
Check your understanding
On a concentration-time graph for a reactant, how do you find the rate at one instant?
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Draw a tangent at that time and take minus its gradient
The tangent gives the instantaneous rate. The reactant's concentration falls, so its gradient is negative; rate = -d[A]/dt is reported as a positive number.
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 exampleDetermine the complete rate equation
Find the orders, k and the predicted initial rate when [A] = 0.150 and [B] = 0.300 mol dm-3.
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- 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.
Try it step by step
Work it out in steps: Order in A, then Order in B, then Rate constant k.
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Problem 1.
Order in A: Experiments 1 and 2: [A] doubles at fixed [B] and the rate doubles, so the order in A is 1.
Order in B: Experiments 2 and 3: [B] doubles at fixed [A] and the rate quadruples, so the order in B is 2.
Rate constant k: Rate = k[A][B]2, so k = 0.0010 / (0.10 x 0.102) = 1.00 dm6 mol-2 s-1. The overall order is 3, which sets the units.
Problem 2.
Order in A: Experiments 1 and 2: [A] doubles and the rate rises four times, so the order in A is 2.
Order in B: Experiments 2 and 3: [B] triples and the rate does not change, so the order in B is 0.
Rate constant k: Rate = k[A]2, so k = 0.00030 / 0.0502 = 0.120 dm3 mol-1 s-1.
Problem 3.
Order in A: Experiments 1 and 2: [A] triples and the rate triples, so the order in A is 1.
Order in B: Experiments 2 and 3: [B] doubles and the rate doubles, so the order in B is 1.
Rate constant k: Rate = k[A][B], so k = 0.00040 / (0.10 x 0.20) = 0.0200 dm3 mol-1 s-1.
Sometimes two concentrations change at once. First use another pair of experiments to find one order. Then divide out the effect of that order. For example, suppose doubling both A and B makes the rate eight times faster, and B is first order. Doubling B alone would double the rate. The remaining factor of four must come from A, so A is second order. Finally, check your rate equation against every row of the table, not only the pair you used.
| 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 |