Hydrogen peroxide solution decomposes slowly at room temperature. When manganese(IV) oxide powder is added, the decomposition is much faster. Explain how the manganese(IV) oxide increases the rate of decomposition, and why the same mass of manganese(IV) oxide can be recovered at the end of the reaction.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
Model answer (5 marks)
Manganese(IV) oxide is a catalyst. It provides an alternative reaction pathway with a lower activation energy. Consequently, a greater proportion of the hydrogen peroxide molecules have sufficient energy to react in each collision. The decomposition therefore proceeds faster. The catalyst is not consumed in the reaction, so the same mass of manganese(IV) oxide can be recovered after the reaction.
Examiner tips
- Use the word ‘catalyst’ first, then explain lower activation energy and increased rate.
- Show the chain: catalyst → lower Ea → more effective collisions → faster rate.
- Mention that the catalyst is unchanged to justify recovery.
Common mistakes
- Saying the oxide is a reagent rather than a catalyst.
- Forgetting to explain why the catalyst is not consumed.
- Using vague terms like ‘helps’ instead of ‘lowers activation energy’.
Mark scheme (5 marks)
- Manganese(IV) oxide acts as a catalyst
- A catalyst provides an alternative (reaction) pathway with a lower activation energy
- More particles (collisions) have energy greater than or equal to the (lower) activation energy
- So the rate of reaction increases / decomposition is faster
- The catalyst is not used up / not consumed in the reaction, so the same mass remains at the end
Related
- All AQA A-Level Chemistry (7405) revision notes →
- How to answer a "Explain" question →
- Decode the mark scheme abbreviations →