Chemistry 9701 · AS & A Level · Nitrogen and sulfur

Nitrogen and sulfur — practice question

The Contact process, which is used to make sulfuric acid, was first patented in the 19th century and is still used today. The crucial stage in the overall sequence is the reversible change of sulfur dioxide into sulfur trioxide in the presence of a vanadium(V) oxide catalyst. $\text{2SO}_2(g) + \text{O}_2(g) \rightleftharpoons \text{2SO}_3(g) \qquad \Delta H = -196\ \text{kJ mol}^{-1}$
(a)[2]

One route for producing the sulfur dioxide for this reaction is to heat the sulfide ore iron pyrites, $\text{FeS}_2$, in air. Iron(III) oxide is formed as well. Write an equation for this reaction.

(b(i))[1]

Explain why the sulfur trioxide is not first mixed directly with water.

(b(ii))[2]

Write equations for the two stages involved in converting sulfur trioxide into sulfuric acid.

(c(i))[2]

Both sulfur dioxide and sulfur trioxide contain only S=O double bonds. Draw labelled diagrams to show the molecular shapes of these two molecules. $\text{SO}_2$ $\text{SO}_3$

(c(ii))[2]

For the diagrams in (i), name the shapes and suggest the bond angles. $\text{SO}_2$ shape $\text{SO}_3$ shape $\text{SO}_2$ bond angle $\text{SO}_3$ bond angle

(d(i))[4]

Using Le Chatelier’s Principle and reaction kinetics, state and explain one advantage and one disadvantage of choosing a higher temperature.

(d(ii))[1]

State the expression for the equilibrium constant, $K_p$, for sulfur trioxide formation from sulfur dioxide.

(d(iii))[5]

A flask contained 2.00 moles of sulfur dioxide and 2.00 moles of oxygen, which were left until equilibrium was reached. At equilibrium, the pressure in the flask was $2.00 \times 10^5\ \text{Pa}$ and the mixture held 1.80 moles of sulfur trioxide. Calculate $K_p$. Include the units.

Worked solution & mark scheme

This 19-mark question has a full step-by-step worked solution and mark scheme. One marking point: Balanced equation: $\mathrm{4FeS_2+11O_2\rightarrow2Fe_2O_3+8SO_2}$

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