You are viewing a free preview of this lesson.
Subscribe to unlock all 10 lessons in this course and every other course on LearningBro.
Understanding rates of reaction is a core part of the AQA GCSE Combined Science Trilogy (8464) chemistry specification. In this lesson you will learn how to measure the rate of a chemical reaction using different experimental methods, how to calculate the mean rate, and how to interpret graphical data including drawing tangents for instantaneous rate.
The rate of reaction measures how quickly reactants are used up or how quickly products are formed during a chemical reaction.
| Reaction Example | Approximate Timescale | Rate Description |
|---|---|---|
| Explosion of hydrogen and oxygen | Fractions of a second | Very fast |
| Burning magnesium ribbon | A few seconds | Fast |
| Marble chips reacting with dilute acid | A few minutes | Moderate |
| Rusting of iron | Days to months | Slow |
| Weathering of limestone | Years to centuries | Very slow |
Exam Tip (AQA 8464): The specification uses the term "rate of reaction" rather than "speed of reaction." Always use the correct terminology in your written answers.
The mean rate of reaction is calculated using:
$$\text{mean rate of reaction} = \frac{\text{quantity of reactant used or product formed}}{\text{time taken}}$$
The quantity measured depends on the experimental method used:
| Measurement | Units | Method |
|---|---|---|
| Volume of gas produced | cm³ | Gas syringe or collection over water |
| Mass lost | g | Measuring decrease in mass on a balance |
| Time for a precipitate to obscure a mark | s | Disappearing cross experiment |
| Colour change | s | Timing a colour change (colorimeter) |
The units of rate depend on the quantities involved:
| Quantity Unit | Time Unit | Rate Unit |
|---|---|---|
| cm³ | s | cm³/s |
| g | s | g/s |
| mol | s | mol/s |
| cm³ | min | cm³/min |
Exam Tip: Always include units with your calculated rate. A numerical answer without units will lose marks.
Question: In an experiment, 60 cm³ of gas was collected in 150 seconds. Calculate the mean rate of reaction.
Solution:
$$\text{mean rate} = \frac{\text{volume of gas}}{\text{time taken}}$$
$$\text{mean rate} = \frac{60\text{ cm}^3}{150\text{ s}}$$
$$\text{mean rate} = 0.40\text{ cm}^3\text{/s}$$
Question: A reaction caused the mass of a flask to decrease by 0.90 g over 3 minutes. Calculate the mean rate of reaction in g/s.
Solution:
First convert minutes to seconds: 3 min = 180 s
$$\text{mean rate} = \frac{0.90\text{ g}}{180\text{ s}} = 0.005\text{ g/s}$$
Common Mistake: Forgetting to convert minutes to seconds when the question asks for a rate in g/s or cm³/s. Always check the units required in the question.
A gas syringe collects the gas produced and measures its volume at regular intervals. This is accurate for gases that do not dissolve in water.
Place the reaction flask on a balance. As gas escapes, the mass decreases. Record mass at regular time intervals.
A cross is drawn on paper beneath a conical flask. When the reaction produces a precipitate, the solution becomes opaque. The time taken for the cross to disappear is recorded.
| Method | Advantages | Disadvantages |
|---|---|---|
| Gas syringe | Accurate volume readings; easy to plot | Gas can leak if syringe sticks |
| Mass loss | Continuous data; simple equipment | Small mass changes hard to detect; drafts affect readings |
| Disappearing cross | Quick and simple | Subjective endpoint; only gives one time reading |
A typical graph of product formed (y-axis) against time (x-axis) shows:
graph TD
A["Start: reactant concentration highest"] --> B["Steep curve — fast rate"]
B --> C["Curve becomes less steep — rate slows"]
C --> D["Curve levels off — reaction complete"]
D --> E["Flat line — no more product formed"]
| Feature | Meaning |
|---|---|
| Steep gradient at start | Rate is fastest when reactant concentration is highest |
| Decreasing gradient | Rate slows as reactants are used up |
| Horizontal line | Reaction has finished — limiting reactant used up |
| Final y-value | Total amount of product formed |
The mean rate gives the average over the whole reaction. The instantaneous rate at a particular time is found by drawing a tangent to the curve at that point.
$$\text{gradient} = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1}$$
A graph shows volume of gas (cm³) against time (s). A tangent drawn at t = 40 s passes through (10, 8) and (70, 56).
$$\text{gradient} = \frac{56 - 8}{70 - 10} = \frac{48}{60} = 0.80\text{ cm}^3\text{/s}$$
The instantaneous rate at 40 seconds is 0.80 cm³/s.
Exam Tip: When drawing a tangent in an exam, make it as long as possible across the graph. Longer lines give more accurate gradient calculations because small errors in reading coordinates have less impact.
Exam Tip: Rate calculation questions appear frequently on AQA 8464 papers. Practise with different unit combinations and always show your working clearly.