Lesson 1 of 3
Measuring and modelling rate
Use tangents, initial rates and concentration-time graphs.
A-Level 8873, revised syllabus (2026-2027)
I can use kinetic vocabulary.Syllabus
Syllabus 8873, 7(a). Use kinetic vocabulary.
I can construct and use simple rate equations.Syllabus
Syllabus 8873, 7(b). Construct and use simple rate equations.
I can explain constant first-order half-life.Syllabus
Syllabus 8873, 7(c). Explain constant first-order half-life.
Rate measures change per time
Name the quantity being followed before writing an equation.
Reaction rate tells you how quickly a reactant is used up or a product forms. A rate based on concentration is usually in mol dm-3 s-1. A reactant concentration falls, so its graph slopes down. Report the rate it disappears at as a positive number. If the equation has coefficients other than one, use them to link the rates for different substances.
For 2H2O2 → 2H2O + O2, two moles of hydrogen peroxide are used for every mole of oxygen made. So peroxide disappears twice as fast as oxygen forms. If peroxide disappears at 6.0 × 10-4 mol dm-3 s-1, oxygen forms at 3.0 × 10-4 mol dm-3 s-1 in the same volume.
| Term | Meaning |
|---|---|
| rate = k[A]m[B]n | A relationship between the stated rate and the concentrations at fixed conditions. |
| Order with respect to A: m | The exponent showing how changing [A] affects rate while other conditions remain fixed. |
| Overall order: m + n | The sum of the concentration exponents. |
| Rate constant, k | The proportionality constant for that equation at a specified temperature and catalytic conditions. |
| Half-life | Time for the amount or concentration being followed to fall to half its current value. |
| Activation energy | The minimum energy barrier for the reacting pathway. |
| Catalysis | Acceleration through an alternative pathway; the catalyst is regenerated overall. |
Worked exampleAn initial rate from a tangent
A tangent at t = 0 on a reactant concentration-time graph passes through (0 s, 0.160 mol dm-3) and (20.0 s, 0.0860 mol dm-3). Find the initial disappearance rate.
Try it first. Then show the worked stepsHide the worked steps
- Tangent slope = (0.0860 - 0.160)/(20.0 - 0) = -0.00370 mol dm-3 s-1.
- Take the positive magnitude for disappearance.
- These points are on the tangent. A chord joining distant measured points would give an average rate, not the initial rate.
Initial disappearance rate = 3.70 × 10-3 mol dm-3 s-1.
Check your understanding
For 2H2O2 -> 2H2O + O2, oxygen forms at 2.0 x 10-4 mol dm-3 s-1. How fast is H2O2 used up?
Show the answer
4.0 x 10-4 mol dm-3 s-1
Two moles of H2O2 are used for each mole of O2 formed, so H2O2 disappears twice as fast as O2 appears.
Change one concentration at a time
The rate factor reveals the exponent.
| Rate factor when [A] doubles | Order in A | Reason |
|---|---|---|
| 1 | 0 | 20 = 1 |
| 2 | 1 | 21 = 2 |
| 4 | 2 | 22 = 4 |
Compare experiments in which only one concentration changes. Keep the other concentrations, the temperature and any catalyst the same. In H1, the orders m and n are 0, 1 or 2 in simple cases. Read the orders from the evidence given. Do not copy them from the coefficients of an overall equation.
| Run | [A] / mol dm-3 | [B] / mol dm-3 | Rate / mol dm-3 s-1 |
|---|---|---|---|
| 1 | 0.0200 | 0.100 | 2.00 × 10-5 |
| 2 | 0.0400 | 0.100 | 4.00 × 10-5 |
| 3 | 0.0400 | 0.200 | 1.60 × 10-4 |
Worked exampleDerive, calculate and predict
Use the dataset to obtain a rate equation and predict the rate at [A] = 0.0300 and [B] = 0.200 mol dm-3.
Try it first. Then show the worked stepsHide the worked steps
- Runs 1 and 2: [A] doubles at fixed [B]; rate doubles, so m = 1.
- Runs 2 and 3: [B] doubles at fixed [A]; rate quadruples, so n = 2.
- rate = k[A][B]2. From run 1, k = (2.00 × 10-5)/(0.0200 × 0.1002) = 0.100.
- Units of k = (mol dm-3 s-1)/(mol dm-3)3 = dm6 mol-2 s-1.
- New rate = 0.100 × 0.0300 × 0.2002 = 1.20 × 10-4 mol dm-3 s-1.
rate = k[A][B]2; k = 0.100 dm6 mol-2 s-1; predicted rate = 1.20 × 10-4 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.
Show the answer
Problem 1.
Order in A: Runs 1 and 2: [A] doubles and the rate rises four times, so the order in A is 2.
Order in B: Runs 2 and 3: [B] doubles and the rate does not change, so the order in B is 0.
Rate constant k: rate = k[A]2, so k = 0.000040 / 0.0202 = 0.100 dm3 mol-1 s-1.
Problem 2.
Order in A: Runs 1 and 2: [A] triples and the rate triples, so the order in A is 1.
Order in B: Runs 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 = 0.00050 / (0.10 x 0.10) = 0.0500 dm3 mol-1 s-1.
Problem 3.
Order in A: Runs 1 and 2: [A] doubles and the rate doubles, so the order in A is 1.
Order in B: Runs 2 and 3: [B] doubles and the rate rises four times, so the order in B is 2.
Rate constant k: rate = k[A][B]2, so k = 0.00020 / (0.050 x 0.0502) = 1.60 dm6 mol-2 s-1.
Check your understanding
A reaction is zero order in A. Does that mean no A is used up?
Show the answer
No. A is still used up; the rate just does not depend on [A] under these conditions.
Order describes how rate depends on concentration in the rate equation. A zero-order reactant still reacts. Its concentration falls at a constant rate.
Recognise zero and first order from the whole curve
Constant slope differs from constant halving time.
Two concentration-time patterns
Both illustrative curves start at 0.160 mol dm-3 at constant temperature. Zero-order concentration falls linearly to zero at 120 seconds. The first-order curve halves every 30 seconds and progressively flattens.
A zero-order concentration-time graph is a straight line while reactant remains, because the same amount is used up every second. Each halving then takes less time than the one before. A first-order graph curves, because the rate falls as the concentration falls. Its half-life stays the same, even if you start measuring from a later point.
| Time / s | 0 | 30 | 60 | 90 |
|---|---|---|---|---|
| [A] / mol dm-3 | 0.160 | 0.0800 | 0.0400 | 0.0200 |
Worked exampleHalf-life is independent of the chosen starting concentration
Show how the table supports first order.
Try it first. Then show the worked stepsHide the worked steps
- 0.160 to 0.0800 takes 30 s.
- 0.0800 to 0.0400 also takes 30 s, despite the lower starting concentration.
- 0.0400 to 0.0200 again takes 30 s.
The constant 30 s halving interval supports first order. A falling curve alone is insufficient evidence: many rate laws produce falling curves.
Try it step by step
Work it out in steps: Number of half-lives, then Concentration remaining, then Percentage reacted.
Show the answer
Problem 1.
Number of half-lives: 90.0 / 30.0 = 3 half-lives.
Concentration remaining: 0.320 -> 0.160 -> 0.0800 -> 0.0400 mol dm-3.
Percentage reacted: 1/8 remains, so 7/8 has reacted: 87.5%.
Problem 2.
Number of half-lives: 10.0 / 5.00 = 2 half-lives.
Concentration remaining: 0.600 -> 0.300 -> 0.150 mol dm-3.
Percentage reacted: 1/4 remains, so 75.0% has reacted.
Problem 3.
Number of half-lives: 160 / 40.0 = 4 half-lives.
Concentration remaining: 0.480 -> 0.240 -> 0.120 -> 0.0600 -> 0.0300 mol dm-3.
Percentage reacted: 1/16 remains, so 93.75% has reacted, 93.8% to 3 significant figures.
With real data, draw a smooth curve through the points and read several halving times. Keep the temperature steady. If the mixture warms during the run, k changes and the shape of the curve is distorted. Treat an odd point with a reason. Do not delete it just to make the curve look ideal.