Question

For the following equilibria in a close container:

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

At a constant temperature, the volume of the container is halved suddenly. Which of the following is correct for $K_p$ and $\alpha$?

• A. None of the $K_p$ and $\alpha$ changes
• B. Both $K_p$ and $\alpha$ changes
• C. $\alpha$ - changes but $K_p$ does not
• D. $K_p$ changes but $\alpha$ does not

Answer 1

Answer: (C)

$PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$
Let initially no. of moles $x$ - -
At equilibrium (volume V) $x(1-\alpha )$ $x\alpha$ $x\alpha$

$K_c=\frac{x^2{\alpha }^2}{x(1-\alpha )V}$, also $K_{\rho }=K_c(RT{)}^{\Delta n_9}$

On reducing volume to $\frac{V}{2}$ initial concentration of $PC{l}_5$ doubles and its degree of dissociation is increased but $K_p$ does not change, as it is characteristic constant for a reaction at constant temperature.

Hence, option D is correct