Question

At constant pressure, the presence of inert gases:
(A) Reduces the dissociation of $PC{l_5}$
(B) Increases the association of $PC{l_5}$
(C) Does not affect the degree of dissociation of $PC{l_5}$
(D) Steps up the formation of $PC{l_5}$

Answer 1

Hint: The effect of inert gases would be determined by Le Chatelier’s Principle. The reaction is more likely to move forward where the number of moles is high .

Complete answer: The above reaction would be :
$PC{l_5}\overset {} \leftrightarrows PC{l_3} + C{l_2}$
The above reaction is in equilibrium . So any changes made in physical and chemical factors will tend to shift the reaction or show some changes.
If the reaction moves backward, more of $PC{l_5}$ will be formed. If the reaction moves forward more of the $PC{l_5}$ will dissociate.
We will have to see what changes would the presence of inert gases make .
Any changes made in volume, pressure , catalyst will make some changes. These changes will be in accordance with Le Chatelier’s Principle . It states that : “If any changes are made in an equilibrium system in any of the conditions , then the system corresponds in a way to counteract the change.”
At constant pressure , the presence of inert gases will shift the reaction which has a higher number of moles . Clearly , we can see the number of moles on the product side is more . Hence, the forward reaction will be favourable . More of the $PC{l_5}$ will dissociate now .

Hence , the correct option is (A).

Note: This is the case when the system is at constant pressure. If however the pressure is varying the explanation and answer would differ . Like for eg – at constant volume, there would be no changes.