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This lesson covers how changing the surface area of a solid reactant affects the rate of reaction, with collision theory explanations, graphs and examples as required by the Edexcel GCSE Combined Science specification (1SC0).
When a solid reactant is broken into smaller pieces, its surface area increases. More of the solid's particles are exposed at the surface and are available to collide with particles in the solution (or gas). This means there are more collisions per second, so the rate of reaction increases.
graph LR
subgraph "Large piece (low SA)"
LP1["Only outer particles<br/>exposed"]
LP2["Fewer collisions<br/>per second"]
LP3["Slower rate"]
LP1 --> LP2 --> LP3
end
subgraph "Powder (high SA)"
PP1["Many more particles<br/>exposed at the surface"]
PP2["Many more collisions<br/>per second"]
PP3["Faster rate"]
PP1 --> PP2 --> PP3
end
style LP3 fill:#c0392b,color:#fff
style PP3 fill:#27ae60,color:#fff
Exam Tip: Increasing surface area does not change the energy of collisions — it only increases the frequency of collisions. The activation energy is unchanged, and the proportion of energetic collisions stays the same.
| Form of solid | Relative surface area | Relative rate |
|---|---|---|
| Large lump | Low | Slow |
| Small chips | Medium | Moderate |
| Fine powder | Very high | Fast |
| Feature | Powder (high SA) | Lumps (low SA) |
|---|---|---|
| Initial gradient | Steep (fast rate) | Shallow (slow rate) |
| Total product | Same (if same mass of reactant used) | Same |
| Time to complete | Shorter | Longer |
Key graph features:
A student reacts 5 g of calcium carbonate with excess dilute hydrochloric acid. In experiment A, the calcium carbonate is in large lumps. In experiment B, the same mass is ground to a fine powder.
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
| Experiment | Form | Time to collect 100 cm³ of CO₂ (s) | Mean rate (cm³/s) |
|---|---|---|---|
| A | Large lumps | 250 | 100 ÷ 250 = 0.40 |
| B | Fine powder | 60 | 100 ÷ 60 = 1.67 |
The powder reacted about 4 times faster than the large lumps.
Exam Tip: Always state that the mass of solid must be kept the same for a fair test. Changing the surface area without changing the mass means breaking the same amount of solid into smaller pieces.
In a surface area investigation, you must control:
| Variable | How it is controlled |
|---|---|
| Mass of solid | Use the same mass each time (weigh on a balance) |
| Volume and concentration of acid | Use a measuring cylinder; same acid each time |
| Temperature | Carry out all experiments at room temperature |
| Type of reactants | Same chemicals in every experiment |
The independent variable is the surface area (size of pieces), and the dependent variable is the time taken (or volume of gas at set intervals).
The relationship between particle size and surface area can be illustrated by imagining cutting a cube:
| Starting cube | Number of pieces | Total surface area |
|---|---|---|
| 1 large cube (1 cm sides) | 1 | 6 cm² |
| Cut into 8 cubes (0.5 cm sides) | 8 | 12 cm² |
| Cut into 64 cubes (0.25 cm sides) | 64 | 24 cm² |
Each time you halve the side length, the total surface area doubles (while the total mass and volume stay the same).
Question: A sugar cube of side 1 cm is ground into 1 000 tiny cubes each of side 0.1 cm. Calculate the surface area before and after, and the ratio.
Before: 6 × (1)² = 6 cm². After (per cube): 6 × (0.1)² = 0.06 cm². Total: 1 000 × 0.06 = 60 cm². Ratio: 60 ÷ 6 = 10× increase in surface area.
Exam Tip: Halving the side length of each piece doubles the total surface area. Cutting to one-tenth the side length increases total surface area ten-fold — which is why fine powders react so fast.
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