Topic 1 of 3
Measuring and modelling rate
Use tangents, initial rates and concentration-time graphs.
A-Level 8873, revised syllabus (2026-2027)
Rate measures change per time
Name the quantity being followed before writing an equation.
Reaction rate measures how quickly a reactant is consumed or a product appears. A concentration-based rate is commonly in mol dm-3 s-1. Reactant concentration falls, so its graph has a negative slope; report its disappearance rate as the positive magnitude. If the equation has coefficients other than one, rates for different species need their stoichiometric factors.
For 2H2O2 → 2H2O + O2, hydrogen peroxide disappears twice as fast in molar terms as oxygen forms. If peroxide disappearance is 6.0 × 10-4 mol dm-3 s-1, oxygen formation is 3.0 × 10-4 mol dm-3 s-1 for the same volume convention.
| 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 example
An 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.
- 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.
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 |
Use experiments with all other reactant concentrations, temperature and catalyst conditions held constant. The H1 scope uses m and n equal to 0, 1 or 2 in simple single-step cases. Rate orders are read from the stated evidence; do not copy coefficients from an overall equation without justification.
| 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 example
Derive, 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.
- 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.
Check your understandingDoes zero order in A mean no A is consumed?Think it through, then reveal the answer
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: the same amount is lost per second. Its successive halving intervals become shorter. A first-order graph curves because rate falls with concentration; its half-life is constant even when measured from a later starting concentration.
| Time / s | 0 | 30 | 60 | 90 |
|---|---|---|---|---|
| [A] / mol dm-3 | 0.160 | 0.0800 | 0.0400 | 0.0200 |
Worked example
Half-life is independent of the chosen starting concentration
Show how the table supports first order.
- 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.
In real data, plot a sensible curve through measurements and read several halving intervals. Keep temperature controlled; warming during the run changes k and can distort the shape. An anomalous measurement needs a reasoned treatment, not automatic deletion to make the curve look ideal.