3.2.6Stoichiometry
Practical- Effect of Catalysts on Rate of Reaction — Cambridge IGCSE Chemistry 0620 Revision Notes & Topical Questions
5 min read 5 parts Lesson 20 of 28 Unit quiz
Part 1 of 5Aim and Set-Up
Aim
- To compare how different solid catalysts affect the rate of decomposition of hydrogen peroxide solution.
Equation
- 2H₂O₂(aq) → 2H₂O(l) + O₂(g), with the oxygen gas collected over water.
Catalyst choices
- Manganese(IV) oxide, MnO₂, is a common catalyst,
- and other solid oxides such as iron(III) oxide or copper(II) oxide can also be compared.
Control variables
- Volume and concentration of hydrogen peroxide, mass of catalyst and temperature must be the same in each test.
Safety
- Hydrogen peroxide solution is an oxidising agent that can irritate skin and eyes, so eye protection must be worn.
Part 2 of 5Method
1
Add a fixed volume of hydrogen peroxide solution to a conical flask and connect it to a measuring cylinder inverted over water.
2
Add a fixed mass of the chosen solid catalyst to the flask and close the bung quickly so that the timing starts with the reaction.
3
Record the volume of oxygen collected at regular intervals or after a fixed time.
4
Use a clean, dry flask for each catalyst so that the results are not contaminated.
5
Run a repeat test without any catalyst, then compare the volumes or the gradients of the volume–time graphs.
Part 3 of 5Interpreting Results
Interpreting Results
Gradient
- A steeper gradient on a volume–time graph shows a faster rate of oxygen production.
Control test
- A repeat without catalyst shows whether the catalyst, and not something else, caused the faster rate.
Not used up
- Filter and dry the catalyst afterwards; its mass remains the same, showing that it was not consumed.
Explanation
- Catalysts give an alternative pathway with lower activation energy, so more collisions are successful.
| Variable | Type | How it is controlled |
|---|---|---|
| Catalyst | Independent | Change the solid used |
| Volume of oxygen in a fixed time | Dependent | Measure with the gas collection apparatus |
| Hydrogen peroxide volume and concentration | Control | Keep the same in each test |
| Temperature | Control | Keep the same throughout |
Part 4 of 5Worked Example
Question
- In one test 40 cm³ of oxygen is collected in 20 s with a catalyst, and 10 cm³ in 20 s without one.
- Calculate the rate of each reaction and state how many times faster the catalysed reaction is.
- Rate with catalyst = 40 cm³ ÷ 20 s = 2.0 cm³/s
- Rate without catalyst = 10 cm³ ÷ 20 s = 0.50 cm³/s
- Ratio = 2.0 ÷ 0.50 = 4
Answer 2.0 cm³/s with the catalyst and 0.50 cm³/s without, so the catalysed reaction is 4 times faster.
Part 5 of 5Exam Checklist & Tips
3.2.6 – Key Points for the Exam
- Catalyst A substance that increases the rate of a reaction and is not used up by it.
- Decomposition A reaction in which a single compound breaks down into two or more simpler substances.
- Control variable A factor kept the same in every test so that the comparison is fair.
- Oxidising agent A substance that gains electrons or oxygen, or causes another substance to lose electrons.
- Gradient of a volume–time graph The steepness of the line, which gives the rate of gas production.
Common Mistakes
- Changing the volume or concentration of hydrogen peroxide between tests, so the results cannot be compared fairly.
- Leaving out the repeat test without catalyst, so the catalyst cannot be shown to be the cause of the change.
- Stating that the catalyst produces the oxygen; the oxygen comes from the hydrogen peroxide.
- Comparing total volumes collected at different times rather than volumes collected in the same fixed time.
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