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Exothermic and Endothermic Reactions: Question 4

Syllabus 5.1

Multiple choice Extended 1 mark

Early rocket engineers used concentrated hydrogen peroxide as a "monopropellant" for small emergency thrusters: passing it over a solid catalyst makes it decompose rapidly, producing a jet of hot steam and oxygen gas that provides thrust.

2H2O2(l)→2H2O(g)+O2(g)2\text{H}_2\text{O}_2(l) \rightarrow 2\text{H}_2\text{O}(g) + \text{O}_2(g)

This decomposition is strongly exothermic.

Which statement correctly explains this, in terms of bond breaking and bond making?

Choose an answer to check it, then compare with the worked solution below.
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Worked solution

Step 1: Recall the bond-energy rule

Bond breaking is an endothermic process (it absorbs energy), and bond making is an exothermic process (it releases energy). The overall enthalpy change of a reaction depends on the balance between these two energy changes:

ΔH=(energy absorbed breaking bonds)−(energy released forming bonds)\Delta H = (\text{energy absorbed breaking bonds}) - (\text{energy released forming bonds})

Step 2: Apply this to the decomposition

For the decomposition to be exothermic overall, ΔH\Delta H must be negative. This happens when less energy is needed to break the bonds in the hydrogen peroxide reactant than is released when the new bonds in water and oxygen are formed.

Why the other options are wrong

  • Option B describes the reverse balance, which would make the reaction endothermic (ΔH\Delta H positive), not exothermic.
  • Option C reverses the fundamental rule: bond breaking absorbs energy, and bond making releases it, not the other way round.
  • Option D is incorrect because bonds are always broken and formed in a chemical reaction; the catalyst only lowers the activation energy needed and speeds up the reaction, without providing or storing any energy itself.

Final answers

  • The decomposition is exothermic because less energy is absorbed breaking bonds in the reactant than is released forming bonds in the products, option A\boxed{\text{option A}}.