Question

The conditions favourable for the reaction:
$2S{O}_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2S{O}_3\left(g\right);△H^o=-198kJ$ are :

• A. low temperature, high pressure
• B. any value of T and P
• C. low temperature and low pressure
• D. high temperature and high pressure

Answer 1

The correct option is A low temperature, high pressure

According to Le-chatelier's principle, if the temperature of the system is changed (increased or decreased) the equilibrium shifts in opposite direction in order to neutralise the effect of change in temperature.
$2S{O}_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2S{O}_3\left(g\right);△H^o=-198kJ$
Since the enthalpy is negative, the reaction is exothermic.
If the temperature of the reaction mixture is decreased, the equilibirum will get shifted to the right (towards product) in order to neutralise the effect of decrease in temperature.
$2S{O}_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2S{O}_3\left(g\right)$
$\text{Change in moles =Product moles-reactant moles}=2-3=-1$
Moles are decreased.
According to "Le-chatellier's principle",
increase in pressure shifts the equilibrium in the direction of decreasing gaseous moles towards the formation of sulfur trioxide.
So, the favourable condition for the reaction is low temperature and high pressure.