Question

Consider the following reaction in which dissociation of $PC{l}_5\left(g\right)$ takes place:
$PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$
At a constant pressure, an inert gas Helium is added.
Which of the following statement is correct?

• A. The dissociation of $PC{l}_5$ decreases.
• B. The dissociation of $PC{l}_5$ increases.
• C. Does not affect the degree of dissociation of $PC{l}_5$
• D. Decrease in the formation of $PC{l}_3$

Answer 1

The correct option is B The dissociation of $PC{l}_5$ increases.

For the reaction :

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

The forward reaction occurs with an increase in the number of gaseous moles from 1 to 2. Similarly, the reverse reaction occurs with a decrease in the number of gaseous moles from 2 to 1.

When inert gas is added at constant pressure, the overall volume will increase. Hence, the number of moles per unit volume (concentration) will decrease $\frac{n_g}{V}\downarrow$ or we can say that the partial pressure of the species decreases.
The equilibrium will shift in the direction in which there is an increase in the number of gaseous moles i.e $n_g\uparrow$. They increase in forward direction (from 1 to 2)

Thus, the addition of an inert gas at a constant pressure will increase the dissociation of $PC{l}_5$​ and the yield of products $PC{l}_3,C{l}_2$ increases.