Question

In the following reversible reaction $2S{O}_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2S{O}_3\left(g\right),\Delta H=+Qcal$,
where Q is negative. Most suitable condition for a low production of $S{O}_3$ is:

• A. High temperature and high pressure
• B. High temperature and low pressure
• C. Low temperature and high pressure
• D. Low temperature and low pressure

Answer 1

The correct option is B High temperature and low pressure

Given reversible reaction $2S{O}_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2S{O}_3\left(g\right),\Delta H=+Qcal$

We need a low production of $S{O}_3$ i.e. lower amount of product.
Since $\Delta H<0$
The reaction is exothermic. For a high temperature, the reaction will proceed in backward direction, i.e. the reactant side. So, less $S{O}_3$ will be formed.

Pressure:
$\Delta n_g=2-3=-1\Delta n_g<0$
For a low pressure, the reaction will tend to go that side where number of gaseous moles increase. It increases in backward direction. So, for a low pressure more reactants are formed and less $S{O}_3$ is formed.