Question

Consider the reactions
(i) $PC{l_5}(g) \rightleftharpoons PC{l_3}(g) + C{l_2}(g)$
(ii) ${N_2}{O_4}(g) \rightleftharpoons 2N{O_2}(g)$
The addition of an inert gas at constant pressure:
A. Will increase the dissociation of $PC{l_5}$ as well as ${N_2}{O_4}$
B. Will reduce the dissociation of $PC{l_5}$ as well as ${N_2}{O_4}$
C. Will increase the dissociation of $PC{l_5}$ and step up the formation of $N{O_2}$ .
D. Will not disturb the equilibrium of the reactions.

Answer 1

The effect of inert gas on the reaction at constant pressure will depend on the principle of Le Chatelier’s. The dissociation of the molecule gives rise to formation of smaller molecules.

Complete step by step answer:
The two reactions are at equilibrium means any change in the parameters of the reaction will affect the equilibrium of the reaction. The parameters of a reaction are the pressure, temperature volume or catalyst.
According to Le Chatelier’s principle any change in pressure, temperature, concentration or catalyst will affect the equilibrium of a reaction and the equilibrium shifts to that side of the reaction so as to counteract the change.
In both the reactions the change in number of moles is
(i) $PC{l_5}(g) \rightleftharpoons PC{l_3}(g) + C{l_2}(g)$, here number of moles, $\Delta n = 2 - 1 = + 1$
(ii) ${N_2}{O_4}(g) \rightleftharpoons 2N{O_2}(g)$​, here number of moles, $\Delta n = 2 - 1 = + 1$
The addition of an inert gas to the system at equilibrium at constant pressure will change the total volume of the system. The total volume of the system increases by the addition. This results in a decrease in the number of moles per unit volume of the reactants and products.
As the number of moles of the system the reaction proceeds in that direction in which the number of moles increases. Thus the equilibrium shifts towards the direction having greater number of moles of the gas. As the product in the forward direction has a greater number of moles so both the reaction proceeds in the forward direction.
Hence the addition of inert gas at constant pressure will increase the dissociation of $PC{l_5}$ as well as ${N_2}{O_4}$, i.e. option A is the correct answer.

Note:
The inert gases would affect the equilibrium of the system differently at constant volume. At constant volume the pressure of the system increases as the number of moles of the system increases.