The equilibrium constant $(K_{p})$ for the reaction, $PCl_{5}(g) \rightarrow PCl_{3}(g) + Cl_{2}(g)$ is 16. If the volume of the container is reduced to one half its original volume, the value of $K_{p}$ for the reaction at the same temperature will be

Question

Dr. Abubacker · Accepted Answer

(b)
$pV = nRT$
Volume become $\frac{1}{2}V$ then pressure become $2p$,
So, $2p 	imes \frac{1}{2}V = pV = nRT$
Hence, there is no effect on $K_{p}$

Q. The equilibrium constant $(K_{p})$ for the reaction, $PCl_{5}(g) \rightarrow PCl_{3}(g) + Cl_{2}(g)$ is 16. If the volume of the container is reduced to one half its original volume, the value of $K_{p}$ for the reaction at the same temperature will be

Explanation

Reviewed by Dr. Abubacker

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