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Answers · Reversible reactions and equilibrium

Reversible reactions and equilibrium — return questions key

Check meaning, not exact wording. These are teaching points, not official Cambridge marks.

1. XR1 · Return

Model answer. hydrated copper(II) sulfate ⇌ anhydrous copper(II) sulfate + water. Heating drives the water out, the forward reaction, and the blue crystals turn white; adding water reverses it.

A full answer includes

  • hydrated copper(II) sulfate ⇌ anhydrous copper(II) sulfate + water
  • Heating drives the forward reaction (blue to white).

Also accept

  • Hydrated ⇌ anhydrous + water; heating drives it forward

Watch for

  • “Adding water needs heat.” — Adding water gives out heat: it is heating the hydrated crystals that takes heat in.

2. XR2 · Return

Model answer. (a) Water turns the blue anhydrous cobalt(II) chloride pink, so the liquid contains water (not necessarily pure water). (b) The pink compound is hydrated cobalt(II) chloride, CoCl2·6H2O. (c) Gentle heating drives the water out again, the forward reaction, and the strip turns blue.

A full answer includes

  • (a) The liquid contains water.
  • (b) Hydrated cobalt(II) chloride.
  • (c) Heat it gently (dry it in a warm oven): the water is driven out.

Also accept

  • Contains water; hydrated cobalt(II) chloride; heat it

Watch for

  • “If the test turns blue (or pink), the liquid is pure water.” — The test shows that water is present, not that nothing else is dissolved in it. Check the boiling point (100 °C) or the melting point (0 °C) for purity.

3. XR3 · Return

Model answer. 2SO2(g) + O2(g) ⇌ 2SO3(g), at about 450 °C and 200 kPa (2 atm), with a vanadium(V) oxide catalyst.

A full answer includes

  • 2SO2(g) + O2(g) ⇌ 2SO3(g).
  • 450 °C, 200 kPa (2 atm), vanadium(V) oxide catalyst.

Also accept

  • 2SO2 + O2 ⇌ 2SO3; 450 °C, 2 atm, V2O5

Watch for

  • “The Contact process uses an iron catalyst.” — Iron is the Haber catalyst. The Contact process uses vanadium(V) oxide.

4. XR4 · Return

Model answer. (a) The forward reaction is endothermic, so the reverse is exothermic. Cooling favours the exothermic direction: the position moves to the left, and there is less C. (b) A catalyst speeds up both directions equally, so the amount of C is unchanged; equilibrium is reached sooner.

A full answer includes

  • (a) Less C: cooling favours the exothermic reverse reaction.
  • (b) No change: equilibrium is reached sooner.

Also accept

  • Less C; unchanged (sooner)

Watch for

  • “A catalyst gives more product at equilibrium.” — A catalyst speeds up the forward and the reverse reactions equally. Equilibrium comes sooner, but the position does not move.