Question

Consider the reactions
(i) $PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$
(ii) $N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$
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

Answer: (A)

Will increase the dissociation of $PC{l}_5$ as well as $N_2{O}_4$

(i) ${PC{l}_5}_{\left(g\right)}\rightleftharpoons {PC{l}_3}_{\left(g\right)}+{C{l}_2}_{\left(g\right)}$

here $\Delta n=2-1=+1$

(ii) ${N_2{O}_4}_{\left(g\right)}\rightleftharpoons {2N{O}_2}_{\left(g\right)}$

here $\Delta n=2-1=+1$

When an inert gas is added to system in equilibrium at constant pressure, then the total volume will increase. Hence, the number of moles per unit volume of various reactants and products will decrease. Hence, the equilibrium will shift towards the direction in which there is increase in number of moles of gases. Since in both reactions number of moles increase in forward direction. So for both reaction addition of inert gas at constant pressure will increase the dissociation of $PC{l}_5$ as well as $N_2{O}_4$.