Question

2 mole each of $PC{l}_5$ and $C{l}_2$ and 1 mole of $PC{l}_3$ are taken in 5lt. flask at ${27}^{o}C$. After following equilibrium is attained in gaseous phase, $PC{l}_5\rightleftharpoons PC{l}_3+C{l}_2,PC{l}_5$ is found to be 40% dissociated. Calculate equilibrium constant $K_c$.

Answer 1

$P{Cl}_5\rightleftharpoons P{Cl}_3+{Cl}_2$

$2mol$ $1mol$ $2mol$

$2-2\alpha mol$ $(1+2\alpha )mol$ $(2+2\alpha )mol$

As degree of dissociation, $\alpha =40$%$=0.4$

Volume $=5l$ ------ (given)

$\therefore$ $\left[P{Cl}_5\right]=\frac{2(1-\alpha )}{V}=\frac{2(1-0.4)}{5}=0.24M$

$\left[P{Cl}_3\right]=\frac{1+2\alpha }{V}=\frac{1+2\times 0.4}{5}=0.36M$

$\left[{Cl}_2\right]=\frac{2+2\times 0.4}{5}=0.56M$

Now

$K_c=\frac{\left[P{Cl}_3\right]\left[{Cl}_2\right]}{\left[P{Cl}_5\right]}$

$K_c=\frac{0.36\times 0.56}{0.24}$

$K_c=0.84mol/l$