Question

The equilibrium which remains unaffected by the change in pressure of the reactants is
A.${{N}_{2}}(g)+{{O}_{2}}(g)\rightleftharpoons 2NO(g)$
B.$2S{{O}_{2}}(g)+{{O}_{2}}(g)\rightleftharpoons 2S{{O}_{3}}(g)$
C.$2{{O}_{3}}(g)\rightleftharpoons 3{{O}_{2}}(g)$
D.$2N{{O}_{2}}(g)\rightleftharpoons {{N}_{2}}{{O}_{4}}(g)$

Answer 1

Hint: According to Le-Chatelier’s principle, increasing the pressure of the system causes the reaction equilibrium to shift towards the side with fewer moles or vice versa. So, the difference in the number of moles between the gaseous products and reactants must be either greater than or less than zero.

Formula Used:Consider a general equilibrium: $R(g)\rightleftharpoons P(g)$
$\Delta {{n}_{g}}={{n}_{P}}(g)-{{n}_{R}}(g)$
${{n}_{P}}(g)=$Number of moles in gaseous product molecules
${{n}_{R}}(g)=$Number of moles in gaseous reactant molecules

Complete answer:When the pressure of any reaction system changes at equilibrium, That chemical reaction is no longer at equilibrium. In this condition, the system will shift the equilibrium either in a forward or backward direction to neutralize the changes.
The equilibrium is not always disturbed upon changing the pressure of the overall system. It is only disturbed when there is a change in the partial pressure of any constituents or all the constituents (reactants and products) in the equilibrium in which $\Delta {{n}_{g}}\ne 0$.
In this problem, we have four reactions at equilibrium. Let us check where $\Delta {{n}_{g}}=0$because only in that case equilibrium remains unaffected by changes in pressure.
In (A)${{N}_{2}}(g)+{{O}_{2}}(g)\rightleftharpoons 2NO(g)$
$\Delta {{n}_{g}}=2-(1+1)=0$
As the value $\Delta {{n}_{g}}$is zero, hence equilibrium remains unaffected by the change in pressure.
In (B)$2S{{O}_{2}}(g)+{{O}_{2}}(g)\rightleftharpoons 2S{{O}_{3}}(g)$
$\Delta {{n}_{g}}=2-(2+1)=-1$
Hence this equilibrium will be affected by the change in pressure.
In (C)$2{{O}_{3}}(g)\rightleftharpoons 3{{O}_{2}}(g)$
$\Delta {{n}_{g}}=3-2=1$
Here also equilibrium will be affected by pressure changes.
In (D)$2N{{O}_{2}}(g)\rightleftharpoons {{N}_{2}}{{O}_{4}}(g)$
$\Delta {{n}_{g}}=1-2=-1$
Finally in (D) equilibrium is also affected by increasing pressure since $\Delta {{n}_{g}}\ne 0$.

Thus, option (A) is correct.

Note: When a non-reacting inert gas is added to the system at constant volume, the pressure of the whole system is increased. However, the individual partial pressure of gasses present in that reaction system is not changed as the volume is not changed. Thereby no effect in the equilibrium of adding a non-reacting inert gas to the system.