Question

Sulphur trioxide gas is an important precursor to manufacture sulphuric acid. It is produced by the catalytic oxidation of $S{O}_2$ gas. In the contact process, once we obtain $S{O}_3\left(g\right)$, Sulphuric acid is obtained by the hydrolysis of $S{O}_3\left(g\right)$. True or False?

• A. True
• B. False

Answer 1

The correct option is B

False

In the contact process catalytic oxidation happens as

$2S{O}_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2S{O}_3\left(g\right):\Delta H=-197kJ.mo{l}^{-1}$

The catalyst used is $V_2{O}_5$ – Vanadium pentoxide.

Now we know that $S{O}_3$ is the anhydride of $H_{2}S{O}_4$. But in the actual contact process, the $S{O}_3$ produced is cooled and absorbed in concentrated Sulphuric acid in a controlled manner. This yields oleum $H_2{S}_2{O}_7$:

$H_{2}S{O}_4\left(l\right)+S{O}_3\left(g\right)\to H_2{S}_2{O}_7\left(l\right)$

This has two advantages:

1. We can avoid the violently exothermic hydrolysis reaction of SO₃. Controlling this exothermic reaction becomes impractical
2. Further, the hydrolysis produces a mist of Sulphuric acid vapour, which could also be avoided

Oleum is diluted with water to produce Sulphuric acid. Could you try the reaction?

$H_2{S}_2{O}_7\left(l\right)+H_{2}O\left(l\right)\to 2H_{2}S{O}_4\left(l\right)$