Question

Assertion $\left(A\right):$ In the dissociation of $PC{l}_5$ at constant pressure and temperature addition of helium at equilibrium increases the dissociation of $PC{l}_5.$

Reason $\left(R\right):$ Helium removes $C{l}_2$ from the field of action.

• A. Both $A$ and $R$ are true and $R$ is the correct explanation of $A.$
• B. $A$ is false but $R$ is true.
• C. $A$ is true but $R$ is false.
• D. Both $A$ and $R$ are true but $R$ is not the correct explanation of $A.$

Answer 1

The correct option is C

$A$ is true but $R$ is false.

Analyzing the assertion

It is false.

$PC{l}_{5_{\left(g\right)}}\rightleftharpoons PC{l}_{3_{\left(g\right)}}+C{l}_{2_{\left(g\right)}}$

When an inert gas $\left(He\right)$ is added to the 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. i.e. forward direction. So, it will facilitate or increase the dissociation of $PC{l}_5.$ and statement given in assertion is correct.

Analyzing the reason

Helium removes $C{l}_2$ from the field of action.

It is false statement. Helium does not react with$C{l}_2,$ therefore it cannot remove $C{l}_2$ from the reaction mixture.

$A$ is true, $R$ is false

$\therefore$ Hence, correct answer is option $\left(C\right)$