(a) Describe the conditions and the steps involved in the manufacture of sulphuric acid by contact process. Write the necessary reactions. (No diagram is required)

(b) Give reasons :

(i) Bond dissociation energy of ${F}_{2}$ is less than that of $C{l}_{2}.$

(ii) Nitric oxide becomes brown when released in air.

(a) Contact process : Burning sulphur in an excess of air

$S+{O}_{2}\to S{O}_{2}\left(g\right)$

or, By heating sulphide ores like pyrites in an excess of air :

$4Fe{S}_{2}+11{O}_{2}\to 2F{e}_{2}{O}_{3}+8S{O}_{2}$

In either case, an excess of air is used so that the $S{O}_{2}$ produced is already mixed with oxygen for the next stage. This is reversible reaction and the formation of $S{O}_{3}$ is exothermic in the presence of catalyst ${V}_{2}{O}_{5}$ at 720K.

This cannot be done by simply adding water to the $S{O}_{3}.$ The reaction is so uncontrollable that it creates a fog of ${H}_{2}S{O}_{4}.$ Instead, the $S{O}_{3}$ is first dissolved in conc. ${H}_{2}S{O}_{4}.$

${H}_{2}S{O}_{4}+S{O}_{3}\to {H}_{2}{S}_{2}{O}_{7}$

The product is known as fuming sulphuric acid or oleum to which water is added to get ${H}_{2}S{O}_{4}$

${H}_{2}{S}_{2}{O}_{7}+{H}_{2}O\to 2{H}_{2}S{O}_{4}$

(b) (i) Bond dissociation energy of ${F}_{2}$ is less than that of $C{l}_{2}$ because of large electron-electron repulsion among the lone pair in ${F}_{2}$ molecule.

(ii) Nitric oxide becomes brown when released in air because of the formation of $N{O}_{2}$ gas.

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