Manganese dioxide (MnO₂) is used as a catalyst in the decomposition of hydrogen peroxide solution to produce water and oxygen. A student carries out the reaction at room temperature and observes rapid gas production. The student then repeats the experiment using the same mass of manganese dioxide but ground into a fine powder instead of small lumps. Explain why the rate of reaction is faster when the manganese dioxide is in powdered form, and explain why the total volume of oxygen collected at the end of both experiments is the same.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
Hydrogen peroxide decomposes according to the equation: 2H₂O₂(aq) → 2H₂O(l) + O₂(g). Manganese dioxide acts as a catalyst in this reaction.
Model answer (5 marks)
Powdering MnO₂ increases its surface area.
The larger surface area exposes more active sites to H₂O₂, so more molecules collide successfully with the catalyst surface per unit time.
This higher frequency of successful collisions raises the reaction rate, giving rapid gas evolution.
The catalyst is not consumed; the same mass of MnO₂ is used in both runs.
Because the amount of H₂O₂ is unchanged, the same number of moles of H₂O₂ decomposes in each experiment.
From 2 H₂O₂ → 2 H₂O + O₂, 2 mol H₂O₂ give 1 mol O₂, so the total moles – and therefore the volume – of O₂ produced is identical in both cases.
The larger surface area exposes more active sites to H₂O₂, so more molecules collide successfully with the catalyst surface per unit time.
This higher frequency of successful collisions raises the reaction rate, giving rapid gas evolution.
The catalyst is not consumed; the same mass of MnO₂ is used in both runs.
Because the amount of H₂O₂ is unchanged, the same number of moles of H₂O₂ decomposes in each experiment.
From 2 H₂O₂ → 2 H₂O + O₂, 2 mol H₂O₂ give 1 mol O₂, so the total moles – and therefore the volume – of O₂ produced is identical in both cases.
Examiner tips
- Mention surface area first, then link to active sites and collision frequency; finish with stoichiometry for volume equality.
- Use the exact reaction equation to justify the 2:1 molar ratio of H₂O₂ to O₂.
Common mistakes
- Saying the catalyst is ‘reactive’ instead of ‘active sites’; "catalyst is used up"; ignoring the 2:1 stoichiometry when explaining volume equality.
Mark scheme (5 marks)
- Powdering the manganese dioxide increases the surface area (of the catalyst / of the solid)
- A greater surface area means more active sites / more of the catalyst is exposed to the hydrogen peroxide molecules
- This increases the frequency of successful collisions (between hydrogen peroxide molecules and the catalyst surface), so the rate increases
- The catalyst is not used up / the amount of hydrogen peroxide (reactant) is the same in both experiments
- Therefore the same amount of hydrogen peroxide is decomposed / the same number of moles of reactant produces the same number of moles of oxygen, so the total volume of oxygen is the same
Key terms in this question
catalyst · rate of reaction · decomposition
Related
- All Edexcel A-Level Chemistry (9CH0) revision notes →
- How to answer a "Explain" question →
- Decode the mark scheme abbreviations →
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