O-LevelChemistryRates of Reaction5070/22 · Oct/Nov 2016 · Q710 Marks

In the contact process, sulfur trioxide is made by the catalytic oxidation of sulfur dioxide. In a closed container the following equilibrium is set up. 2SO2(g) + O2(g) ⇌ 2SO3(g) The reaction is exothermic.

(a)Name the catalyst used in this reaction.[1]
(b)Draw a 'dot-and-cross' diagram of an oxygen molecule. Show only the outer shell electrons.[1]
(c)(i)Describe how, and explain why, the percentage yield of sulfur trioxide changes with temperature.[3]
(c)(ii)Suggest why the reaction is carried out at 450 °C and not at 250 °C.[2]
(d)Describe how, and explain why, the position of equilibrium changes when the pressure increases.[2]
(e)Sulfur trioxide oxidises hydrogen bromide to form sulfur dioxide, bromine and water. Construct the equation for this reaction.[1]

This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2016). The mark or mark type is shown in brackets.

[1] vanadium pentoxide/vanadium(V) oxide/V2O5
[1] four bonding electrons between the two oxygen atoms AND four non-bonding electrons on each oxygen atom
[1] little/no change in % yield at low temperature
[1] yield decreases as temperature increases
[1] reaction is exothermic so equilibrium moves to the left as temperature increases/increasing temperature favours the endothermic reaction
[1] at 250 °C reaction is slow(er)/at 450 °C reaction is fast(er)
[1] idea of compromise temperature/idea of balance between lower yield and faster rate
[1] (position of) equilibrium moves to the right/(position of) equilibrium moves towards the product side
[1] more (gas) molecules on left than right of the equation/fewer (gas) molecules on right of equation than left
[1] SO3 + 2HBr → SO2 + Br2 + H2O

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