1. Average, initial and instantaneous rates
- Average rate: total change ÷ total time over an interval.
- Instantaneous rate: the rate at one moment, found from the gradient of the tangent to the curve at that time.
- Initial rate: the instantaneous rate at t = 0.
As a reactant is used up, its concentration falls, so collisions become less frequent and the curve flattens. For most reactions the initial rate is the fastest, but not always: in autocatalysed reactions a product speeds up the reaction, so the rate can rise at first.
2. Tangents from a concentration–time graph
| t / s | 0 | 20 | 40 | 60 | 80 | 100 | 120 |
|---|---|---|---|---|---|---|---|
| [A] / mol dm⁻³ | 0.400 | 0.300 | 0.230 | 0.180 | 0.145 | 0.120 | 0.104 |
Tips: draw a large triangle using two points on the tangent line far apart, not two points on the curve. Read values carefully from the axes and include units.
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Concentration–time curve with tangents
Labels to include:
- x-axis: Time / s; y-axis: [A] / mol dm⁻³
- Curve falling from 0.400 at t = 0 and flattening
- Tangent at t = 0 through (0, 0.400) and (40, 0.180)
- Tangent at t = 40 s through (0, 0.326) and (100, 0.086)
- Large gradient triangle drawn on the t = 40 s tangent with Δ[A] and Δt labelled
Each tangent just touches the curve at one point. The t = 0 tangent is steeper than the t = 40 s tangent, showing the rate falls as A is used up. All the numbers needed are given in the worked examples above.
3. Gas volume and mass loss
Example reaction: CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)
Gas syringe
- Connect the flask to a gas syringe with an airtight bung and delivery tube.
- Start timing as soon as the reactants are mixed (e.g. by tipping a small tube of acid inside the sealed flask).
- Record the volume at regular intervals. Keep temperature and pressure constant, since gas volume depends on both.
Mass loss
- Stand the flask on a balance with a loose cotton-wool plug in the neck. CO2 escapes, but acid spray is held back. Never use an airtight stopper here.
- Record the mass at regular intervals; the loss is the mass of CO2.
- Limitations: unsuitable for gases that are very soluble in water (some stays dissolved) or have very low mass (e.g. H2, where the loss is tiny compared with balance resolution).
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Gas syringe and mass-loss set-ups
Labels to include:
- Left: conical flask with reactants, airtight bung, delivery tube, gas syringe on a clamp, scale on syringe
- Right: conical flask on a top-pan balance, cotton-wool plug in neck (not airtight)
- Stopwatch/timer
Left set-up collects the gas; right set-up lets the gas escape and records the falling mass.
4. What the plateau tells you
For a reaction that proceeds to completion, the curve levels off when the limiting reactant is used up, and the plateau height shows the final amount of product, not how fast it was made. A plateau on its own does not prove completion: it can also mean the reaction has reached equilibrium or become extremely slow.
With the same moles of limiting reactant, a higher temperature, a powder or a catalyst makes the curve steeper but, for a complete, irreversible reaction with the gas measured at the same temperature and pressure, it reaches the same plateau.
Average rate from a gas curve: if 60 cm3 is collected in the first 20 s, the average rate = 60 ÷ 20 = 3.0 cm3 s−1.
5. Changing amounts and volumes
- Doubling the concentration of the excess reactant usually makes the curve steeper but leaves the plateau the same. The final amount doubles only if the moles of the limiting reactant double.
- Doubling all solution volumes at the same concentrations (for a reaction happening throughout the solution) leaves the rate in mol dm−3 s−1 the same, but doubles the total moles reacting per second, so the gas collected per second and the final volume both double.
So “more volume means faster” and “same concentration means the same graph” are both too simple; say which quantity you mean.
6. Required Practical 3: thiosulfate and acid
The sulfur forms a pale yellow precipitate that gradually hides a cross drawn on paper under the flask.
Method (investigating temperature)
- Measure fixed volumes of sodium thiosulfate solution and of dilute hydrochloric acid into separate tubes.
- Stand both in a water bath at the target temperature until they reach it. Repeat at about five different temperatures.
- Pour the thiosulfate into a conical flask on the cross; add the acid, swirl in the same way each time and start the timer when the acid is added.
- Measure the actual temperature of the mixture (not just the bath setting), ideally at the start and end, and use the mean.
- Stop the timer when the cross can no longer be seen, viewed from the same position each time.
- Keep constant: volumes and concentrations, total depth (same flask), the same cross, lighting and viewing point.
- Repeat at each temperature and take a mean of concordant times. Only reject a result if there is a reason to think it is anomalous.
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RP3 apparatus: disappearing cross
Labels to include:
- Conical flask containing reaction mixture
- Paper with a black cross under the flask
- Eye viewing from above (same position each run)
- Thermometer in the mixture
- Water bath with tubes of thiosulfate and acid
- Stopwatch
The flask stands on the cross. As sulfur forms, the mixture clouds until the cross disappears from view above.
7. Processing RP3 results
Each run stops at the same amount of sulfur (the same cloudiness, for the same depth and volume). So 1/t is proportional to the rate. It is a relative rate in s−1, not a rate in mol dm−3 s−1, and t is not the time to completion.
| T / °C | t / s | 1/t / s⁻¹ | 1000/t / s⁻¹ |
|---|---|---|---|
| 20 | 64 | 0.0156 | 15.6 |
| 30 | 38 | 0.0263 | 26.3 |
| 40 | 22 | 0.0455 | 45.5 |
| 50 | 13 | 0.0769 | 76.9 |
The rate at 50 °C compared with 20 °C = 64 ÷ 13 = 4.9 times faster.
Plot 1/t (or 1000/t, for easier numbers) against temperature and draw a smooth best-fit curve. Don't assume it must be a straight line or that the rate always doubles every 10 °C. To go back from a value read off a 1000/t graph, use t = 1000 ÷ value: reading 35 gives t = 1000 ÷ 35 = 28.6 s.
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Graph of 1000/t against temperature
Labels to include:
- x-axis: Temperature / °C (20–50)
- y-axis: 1000/t / s⁻¹ (0–80)
- Four plotted points from the table
- Smooth curve of best fit, rising more steeply at higher temperature
- Read-off line at 1000/t = 35 down to the temperature axis
The curve rises with temperature and gets steeper. A horizontal read-off at 35 meets the curve between 30 °C and 40 °C.
8. Errors, improvements and safety
| Problem | Improvement |
|---|---|
| Judging when the cross disappears is subjective | Same observer and viewpoint; or a colorimeter/light sensor timing to a fixed transmission |
| Reaction time and human reaction delay are significant for fast runs | Use lower concentrations or temperatures so times are longer; a data logger |
| Mixture temperature drifts from bath temperature | Pre-heat both reagents; measure the mixture's temperature; keep the flask in the bath |
| Random scatter | Repeat and average concordant results (repeats do not fix systematic errors) |
Safety: wear eye protection; use dilute acid and small volumes; SO2 is toxic and irritating, so work in a well-ventilated room and avoid breathing the fumes; dispose of the mixture promptly as directed. Follow your school's risk assessment.
Quick checks
These are Finesse practice questions. The step-by-step answers are indicative worked solutions, not official AQA mark allocations.
Q1. 45 cm3 of gas is collected in the first 30 s. Calculate the average rate of gas production.Show answer
Step 1: rate = volume ÷ time.
Step 2: 45 ÷ 30 = 1.5 cm3 s−1.
Q2. A tangent to a reactant concentration–time curve passes through (10 s, 0.50 mol dm−3) and (50 s, 0.10 mol dm−3). Find the rate at the point of contact.Show answer
Step 1: gradient = (0.10 − 0.50) ÷ (50 − 10) = −0.40 ÷ 40 = −1.0 × 10−2 mol dm−3 s−1.
Step 2: rate of disappearance = 1.0 × 10−2 mol dm−3 s−1.
Q3. 0.120 g of Mg (Ar 24.3) reacts with excess HCl: Mg + 2HCl → MgCl2 + H2. Predict the final gas volume (24.0 dm3 mol−1), and say whether mass loss would be a good method.Show answer
Step 1: n(Mg) = 0.120 ÷ 24.3 = 4.938 × 10−3 mol = n(H2).
Step 2: volume = 4.938 × 10−3 × 24 000 = 119 cm3.
Step 3: mass of H2 = 4.938 × 10−3 × 2.0 ≈ 0.0099 g, which is too small to measure well on a typical balance, so a gas syringe is better.
Q4. In RP3 a run takes 25 s. Find 1/t. Separately, a value of 50 is read from a 1000/t graph: what time does it correspond to?Show answer
Step 1: 1/t = 1 ÷ 25 = 0.040 s−1.
Step 2: t = 1000 ÷ 50 = 20 s.
Q5. A student sets the water bath to 40 °C but measures only the bath, and pours room-temperature acid into the warmed thiosulfate. Explain the effect and an improvement.Show answer
The mixture is cooler than 40 °C, so the reaction is slower than it should be at 40 °C, and the recorded temperature is too high for the time measured.
Improvement: warm both solutions in the bath first and measure the actual temperature of the mixture.
Sources
Sources and examiner guidance (reviewed 1 October 2026)
- Chemrevise — AQA 1.5 Reaction kinetics revision guide (N. Goalby) — Checklist for rate measurement and RP3; the guide's volume/concentration point is restated here with quantities separated.
- AQA 7405 specification — 3.1.5 Kinetics (incl. Required Practical 3) — 3.1.5.1 Collision theory (rate from concentration–time graphs); Required Practical 3.
- AQA 7404/2 mark scheme, June 2023 — Q03.2: steeper gradient = faster production.
- AQA 7404/2 examiner report, June 2023 — Q03.2: graph interpretation.
- AQA 7404/2 mark scheme, June 2022 — Q01.3–Q01.5: actual temperature, smooth best fit, converting 1000/t.
- AQA 7405/3 mark scheme, June 2019 — Q01.4: method, measurements, use of results.
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