Question

Sulphur trioxide is not directly dissolved in water to form sulphuric acid because:
A. $\text{S}{{\text{O}}_{3}}$ does not react with water to form acid
B. $\text{S}{{\text{O}}_{3}}$ gets oxidised to ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$ when dissolved in water
C. it results in the formation of dense fog of sulphuric acid which is difficult to condense
D. sulphur trioxide is insoluble in water due to its covalent nature

Answer 1

Contact process is the industrial method of producing sulphuric acid. It is also called lead-chamber process. Sulphur dioxide and oxygen combines to form sulphur trioxide, which reacts with water to form sulphuric acid. See the process to know the reason.

Complete step by step answer:
Let us see the contact process to see why $\text{S}{{\text{O}}_{3}}$ does not react with water. The process is divided into five stages:
(1) Sulphur and oxygen $\left( {{\text{O}}_{2}} \right)$ combines to form sulphur dioxide.
$\text{S}+{{\text{O}}_{2}}\left( \text{g} \right)\to \text{S}{{\text{O}}_{2}}\left( \text{g} \right)$

(2) Add an excess of oxygen to sulphur dioxide $\left( \text{S}{{\text{O}}_{2}} \right)$ in the presence of the catalyst $\left( {{\text{V}}_{2}}{{\text{O}}_{5}} \right)$ vanadium pentoxide at 1-2 atm and ${{450}^{\text{o}}}\text{C}$.
Sulphur trioxide $\left( \text{S}{{\text{O}}_{3}} \right)$ will be formed. $\text{2S}{{\text{O}}_{2}}\left( \text{g} \right)+{{\text{O}}_{2}}\left( \text{g} \right)\to 2\text{S}{{\text{O}}_{3}}\left( \text{g} \right)$.

(3) Sulphur trioxide $\left( \text{S}{{\text{O}}_{3}} \right)$ which is formed is added to sulphuric acid $\left( {{\text{H}}_{2}}\text{S}{{\text{O}}_{4}} \right)$ which forms oleum $\left( {{\text{H}}_{2}}{{\text{S}}_{2}}{{\text{O}}_{7}} \right)$.
Hot sulphur trioxide passes through heat exchanger and is dissolved in concentrated ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$ in the absorption tower to form oleum.
The reaction is ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}\left( \text{l} \right)+\text{S}{{\text{O}}_{3}}\left( \text{g} \right)\to {{\text{H}}_{2}}{{\text{S}}_{2}}{{\text{O}}_{7}}\left( \text{l} \right)$.
Directly dissolving $\left( \text{S}{{\text{O}}_{3}} \right)$ in water is impractical and not possible due to the highly exothermic nature of the reaction. The reaction gives acidic vapours or mists and the formation of fog of sulphuric acid which is difficult to condense instead of a liquid.
Reaction between water and sulphur trioxide forms a ‘mist’ of sulphuric acid which can be difficult to store and handle.

(4) Then, oleum is added to water to form concentrated sulphuric acid. The reaction is ${{\text{H}}_{2}}{{\text{S}}_{2}}{{\text{O}}_{7}}\left( \text{l} \right)+{{\text{H}}_{2}}\text{O}\left( \text{l} \right)\to 2{{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}\left( \text{l} \right)$.
The correct answer of this question is option ‘c’, sulphur trioxide is not directly dissolved in water to form sulphuric acid because it results in the formation of dense fog of sulphuric acid which is difficult to condense.

Note: The process of formation of sulphuric acid is an exothermic reaction, so the temperature should be as low as possible. The yield of sulphuric acid is maximum at about ${{410}^{\text{o}}}-{{450}^{\text{o}}}\text{C}$. Too lowering of a temperature will also lower the formation rate to an uneconomical level.