Question

The equilibrium constant $\left(K_p\right)$ for the reaction $PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)$ 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:

• A. 32
• B. 64
• C. 16
• D. 4

Answer 1

Answer: (A)

32

${PCl}_5\left(g\right)\rightleftharpoons {PCl}_3\left(g\right)+{Cl}_2$

$K_P=\frac{P^{{Cl}_2}\times P^{{PCl}_3}}{P^{{PCl}_5}}$

Initially it is $K_P=16atm$.

It container volume is half of original volume according $PV=$ constant (at fixed $T$) pressure must double

$K_P=\frac{{2P}^{{PCl}_3}\times {2P}^{{Cl}_2}}{{2P}^{{PCl}_5}}=2\times 16=32$