Question

$2$ moles of $PC{l}_5$ were heated in a closed vessel of $2$ liter capacity. At equilibrium, $40\%$ of $PC{l}_5$ dissociated into $PC{l}_3$ and $C{l}_2$.

The value of equilibrium constant is:

• A. $0.267$
• B. $0.53$
• C. $2.63$
• D. $5.3$

Answer 1

The correct option is A $0.267$

${PCl}_5\rightleftharpoons {PCl}_3+{Cl}_2$ Volume $=2$ lit

$t=0$ $2$ $0$ $0$

$t=t_{eq}$ $2-\alpha$ $\alpha$ $\alpha$

Given $40$% dissociated

$PC{l}_5$ remains at eqm. $=2\times 60/100=1.2=2-\alpha$

So, $\alpha =0.8$

$\left[{Cl}_2\right]=\frac{0.8}{2}=0.4$

$\left[{PCl}_3\right]=\frac{0.8}{2}=0.4$

$\left[{PCl}_5\right]=\frac{1.2}{2}=0.6$

$K=\frac{\left[{Cl}_2\right]\left[{PCl}_3\right]}{\left[{PCl}_5\right]}=\frac{0.4\times 0.4}{0.6}=0.267$

So, the correct option is $A$