Question

$2$ mole of $P{Cl}_5$ is heated in a one litre vessel. If $P{Cl}_5$ dissociates to the extent of $80$%, the equilibrium constant for the dissociation of $P{Cl}_5$ is:

• A. $2\times {10}^{-2}$
• B. $6.4$
• C. $0.67$
• D. $0.32$

Answer 1

The correct option is B $6.4$

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

$2mole$ $0moles$ $0moles$ $80\%$ dissociation.

$2\alpha$ $0$ $0$

$2(1-\alpha )$ $2\alpha$ $2\alpha$ $\alpha =0.8mol$.

Volume of equilibrium mixture $V=1L$

$\left[PC{l}_5\right(g\left)\right]=\frac{2(1-\alpha )}{V}=\frac{2(1-0.8)}{1}=0.4mol{L}^{-1}$

$\left[PC{l}_3\right(g\left)\right]=\frac{2\alpha )}{V}=\frac{2\left(0.8\right)}{1}=1.6mol{L}^{-1}$

$\left[C{l}_2\right(g\left)\right]=\frac{2\alpha )}{V}=\frac{2\left(0.8\right)}{1}=1.6mol{L}^{-1}$

Equilibrium constant,

$K_c=\frac{\left[PC{l}_3\right(g\left)\right]\left[C{l}_2\right(g\left)\right]}{\left[PC{l}_5\right(g\left)\right]}=\frac{1.6\times 1.6}{0.4}=6.4$