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Six original Cambridge-style questions on catalysis: how catalysts work, homogeneous and heterogeneous examples, and their effects on ΔH and equilibrium.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

Lower Ea, same ΔH.

01
Recall
[2 marks]

Distinguish between homogeneous and heterogeneous catalysis, with one example of each.

  • Homogeneous: the catalyst is in the same phase as the reactants, e.g. Fe2+(aq) in the reaction of S2O82− with I−. ✓
  • Heterogeneous: the catalyst is in a different phase, e.g. solid iron with gaseous N2 and H2 in the Haber process. ✓
02
Analysis
[3 marks]

Draw a reaction pathway diagram for an exothermic reaction with and without a catalyst. Label ΔH and both activation energies.

  • Products below reactants, ΔH arrow downwards. ✓
  • Uncatalysed curve with a high hump labelled Ea. ✓
  • Catalysed curve with a lower hump labelled Ea(cat), same start and end levels. ✓
03
Explain
[2 marks]

Use the Boltzmann distribution to explain how a catalyst increases the rate of reaction.

  • The catalyst provides a route with lower activation energy, so Ea moves to the left on the distribution. ✓
  • A greater proportion of molecules have energy ≥ Ea, so there are more successful collisions per unit time. ✓
04
Explain
[3 marks]

Describe the stages by which a catalytic converter removes carbon monoxide and nitrogen monoxide, and write the equation.

  • CO and NO are adsorbed onto the Pt/Rh surface, weakening their bonds. ✓
  • They react on the surface and the products desorb. ✓
  • 2CO + 2NO → 2CO2 + N2. ✓
05
Explain
[3 marks]

Explain, with equations, how Fe2+ ions catalyse the reaction between peroxodisulfate and iodide ions.

  • S2O82− + 2Fe2+ → 2SO42− + 2Fe3+. ✓
  • 2Fe3+ + 2I− → 2Fe2+ + I2. ✓
  • Fe2+ is regenerated; each step involves oppositely charged ions, so Ea is lower than for two repelling anions. ✓
06
Explain
[2 marks]

Explain why adding a catalyst to the Haber process does not increase the equilibrium yield of ammonia.

  • The catalyst lowers the activation energy of the forward and reverse reactions equally. ✓
  • Both rates increase by the same factor, so the position of equilibrium does not change; equilibrium is simply reached faster. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.