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The rate of a chemical reaction describes how quickly reactants are converted into products. Understanding rate is essential for both GCSE chemistry and for real-world applications — from industrial manufacturing to cooking to the reactions happening inside your body. This lesson introduces the concept of rate and how to measure it.
The rate of reaction is a measure of how quickly a reaction takes place. It can be defined as:
rate of reaction=time takenamount of reactant used or product formed
Rate can be measured in terms of:
There are several experimental methods for measuring rate. The choice depends on the reaction:
If the reaction produces a gas, the mass of the reaction mixture decreases as the gas escapes. By placing the reaction vessel on a balance and recording the mass at regular intervals, you can track the rate.
| Time (s) | Mass of flask and contents (g) | Mass lost (g) |
|---|---|---|
| 0 | 150.00 | 0.00 |
| 30 | 149.56 | 0.44 |
| 60 | 149.20 | 0.80 |
| 90 | 148.92 | 1.08 |
| 120 | 148.76 | 1.24 |
| 150 | 148.68 | 1.32 |
| 180 | 148.68 | 1.32 |
The reaction is complete when the mass stops changing (here, at ~150 s).
If the reaction produces a gas, the volume of gas can be collected in a gas syringe and recorded at regular time intervals.
Some reactions produce a visible change — a colour change, a precipitate forming, or a solution becoming opaque. The time taken for this change can be measured (as in the sodium thiosulfate "disappearing cross" experiment).
graph TD
A["Methods for Measuring Rate"] --> B["Mass loss on balance\n(gas escapes)"]
A --> C["Gas syringe\n(collect gas volume)"]
A --> D["Colour change / precipitate\n(time to reach endpoint)"]
B --> E["Plot mass lost vs time"]
C --> F["Plot volume vs time"]
D --> G["Use 1/time as\nmeasure of rate"]
A volume vs time or mass lost vs time graph gives you detailed information about the rate:
Question: In an experiment, 48 cm³ of gas was collected in the first 60 seconds. Calculate the average rate of reaction.
rate = volume / time = 48 / 60 = 0.80 cm³/s
Question: A graph of mass lost against time shows that 1.2 g of gas was lost in the first 40 seconds. What is the average rate of reaction over this period?
rate = mass lost / time = 1.2 / 40 = 0.030 g/s
Question: Two experiments produce the following volume-time graphs. Experiment A reaches 60 cm³ in 80 s. Experiment B reaches 60 cm³ in 40 s. Both produce the same final volume. Compare the rates.
Answer: Both reactions produce the same total volume of gas, meaning the same amount of product is formed (same amount of reactant was used). However, Experiment B reaches completion in half the time, so its average rate is double that of Experiment A. The gradient of B's graph is steeper than A's, especially in the early stages, confirming the faster rate.
Since both experiments produce the same final volume, the amounts of reactants used must be the same — the difference is the conditions (e.g., temperature, concentration, or surface area).
The rate of a chemical reaction can be changed by altering:
| Factor | Effect on rate | Why (brief) |
|---|---|---|
| Temperature | Increase T → increase rate | Particles move faster → more frequent and energetic collisions |
| Concentration | Increase conc → increase rate | More particles per volume → more frequent collisions |
| Surface area | Increase SA → increase rate | More exposed particles → more collisions |
| Catalyst | Add catalyst → increase rate | Provides alternative pathway with lower activation energy |
| Pressure (gases) | Increase P → increase rate | Particles closer together → more frequent collisions |
Each of these factors is explained in detail in subsequent lessons using collision theory.
Question: Explain how the volume-time graph would differ if the same reaction was carried out at a higher temperature, using the same amounts of reactants.
Answer: At a higher temperature:
Common GCSE exam mistake: Saying that a higher temperature produces "more product." If the amounts of reactants are the same, the same amount of product is formed — it is just formed faster. The final plateau on the graph is the same height; only the time to reach it changes.
Question: From a volume-time graph, a tangent is drawn to the curve at t = 20 s. The tangent line passes through the points (10 s, 15 cm³) and (30 s, 45 cm³). Calculate the instantaneous rate of reaction at t = 20 s.
Answer: The gradient of the tangent gives the instantaneous rate:
rate = change in volume / change in time = (45 − 15) / (30 − 10) = 30 / 20 = 1.5 cm³/s
This is the rate at exactly t = 20 s. Compare this to the average rate over the full experiment — the instantaneous rate at 20 s is higher than the average because the reaction is still relatively fast at this early stage.
| Method | Best for | Advantages | Disadvantages |
|---|---|---|---|
| Mass loss on balance | Reactions producing gas | Continuous data; simple setup | Gas must not be too light (H₂ hard to detect); draughts affect balance |
| Gas syringe | Reactions producing gas | Accurate volume measurement; sealed system | Gas syringe can stick; limited capacity (~100 cm³) |
| Colour change / precipitate | Reactions with visible change | Simple; no specialist equipment | Subjective endpoint; only gives one data point (time to endpoint) |
| pH probe + datalogger | Acid-base reactions | Continuous, quantitative, objective data | Requires specialist equipment |
In the exam, you may be asked to evaluate which method is most appropriate for a given reaction. Always consider whether the reaction produces a gas, changes colour, or involves acids/alkalis.