# A student adds a small lump of manganese(IV) oxide to a solution of hydrogen peroxide. Bubbles of oxygen gas are produced rapidly. The student then removes the manganese(IV) oxide, dries it, and finds its mass is unchanged. Explain how the manganese(IV) oxide increases the rate of decomposition of hydrogen peroxide, and justify why its mass is unchanged at the end of the reaction.

> Pearson Edexcel International GCSE Chemistry (4CH1) — 3.2 Rates of reaction · Explain · 4 marks

> Hydrogen peroxide decomposes slowly at room temperature: H₂O₂(aq) → H₂O(l) + ½O₂(g). In the presence of manganese(IV) oxide the reaction is much faster.

## Mark scheme (4 marks)

1. Manganese(IV) oxide acts as a catalyst
2. The catalyst provides a different (alternative) reaction pathway with a lower activation energy
3. A lower activation energy means a greater proportion of colliding particles have sufficient energy to react, so there are more successful collisions per unit time / the rate of successful collisions increases
4. The catalyst is not used up during the reaction, so its mass remains unchanged at the end

## Related

- [Revision notes for Pearson Edexcel International GCSE Chemistry (4CH1)](https://www.gradenine.co.uk/learn)
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