Sulphuric acid is manufactured by the Contact process.
(a) Give two balanced equations to obtain $S{O}_2$ in this process.
(b) Give the conditions for the oxidation of $S{O}_2$.
(c) Why is $H_{2}S{O}_4$ not obtained by directly reacting $S{O}_3$ with water?
(d) Name the chemical used to dissolve $S{O}_3$ and name also the product formed.
(e) Give all the main reactions to this process.
[5 MARKS]

Each option: 1 Mark

(a) Sulphur dioxide is produced by burning Louisiana sulphur ($99.5\%$ pure) in purified air.
$S+O_2\to S{O}_2$
It is also obtained by roasting Iron pyrites $\left(Fe{S}_2\right)$
$4Fe{S}_2+11O_2\to 2Fe2O_3+8S{O}_2$

(b) Conditions for oxidation of $S{O}_2$. The reaction of conversion of sulphur dioxide sulphur trioxide is :
$\underset{2vol.}{2S{O}_2}+\underset{1vol.}{O_2}\underset{{450}^{\circ }C}{\stackrel{V_2{O}_5}{\rightleftharpoons }}\underset{2vol.}{2S{O}_3}+198kJ$

Clearly, the reaction is reversible; exothermic and proceeds with a decrease in volume. Therefore, according to Le-Chatelier's principle, the favourable conditions for maximum yield of sulphur trioxide are :
(i) Low temperature (about ${410}^{\circ }$C to ${450}^{\circ }$C)
(ii) High pressure (1-2 atm)
(iii) Purity of the gases
(iv) Concentration of gases (Excess of oxygen)
(v) Suitable catalyst (platinum or vanadium pentaoxide)

(c) Since sulphur trioxide does not dissolve in water satisfactorily and it gives a lot of heat and forms misty droplets of sulphuric acid, therefore, sulphuric acid is not obtained by directly, reacting $S{O}_3$ with water.
(d) The chemical used to dissolve $S{O}_3$ is conc. sulphuric acid and the product fanned is oleum or pyrosulphuric acid.
$S{O}_3+H_{2}S{O}_4\to \underset{Oleum}{H_2{S}_2{O}_7}$

(e) The main reactions taking place in the manufacture of sulphuric acid by Contact process are as follows:
$S+O_2\to S{O}_2$
$4Fe{S}_2+11O_2\to 2F{e}_2{O}_3+8S{O}_2$
$2S{O}_2+O_2\underset{400-{450}^{\circ }C}{\stackrel{V_2{O}_5/Pt}{\rightleftharpoons }}2S{O}_3$