Question

When $PC{l}_5$ is heated in a closed container at $575K$, the total pressure at equilibrium is found to be $1atm$ and partial pressure of $C{l}_2$ is found to be $0.324atm$. Calculate the equilibrium constant $\left(K_p\right)$ for the decomposition reaction.

• A. 0.35
• B. 0.426
• C. 0.298
• D. None of these

Answer 1

The correct option is C 0.298

Given reaction is :-

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

Initial $P$ $O$ $O$

pressure

At eqn $P-p$ $p$ $p$

Total pressure at equilibrium $=1$ atm

$\Rightarrow P-p+p+p=1$

$\Rightarrow P+p=1\to \left(1\right)$

& partial pressure of ${Cl}_2=0.324$

$\Rightarrow p=0.324\to \left(2\right)$

$\left(1\right)$ & $\left(2\right)$ $\Rightarrow P=0.676atm$

Now, $K_P=\frac{p{Cl}_2.{PCl}_3}{P{PCl}_5}$

where, ${pCl}_2,{PCl}_3,{Ppcl}_5$ are partial pressure of ${Cl}_2,{PCl}_3,{PCl}_5$ at equilibrium.

$\Rightarrow K_P=\frac{0.324\times 0.324}{0.676-0.324}=\frac{0.324\times 0.324}{0.352}=0.298$