Question

Two moles of $PC{l}_5$ were heated in a closed vessel of $2L$ . At equilibrium $40\%$ of $PC{l}_5$ is dissociated into $PC{l}_3$ and equilibrium constant is :

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

Answer 1

Answer: (B)

Degree of dissociation $=0.4$
$\underset{\text{a(1-x)}}{\mathrm{PCl}{}_5}$ $\underset{\text{ax}}{\mathrm{PCl}{}_3}$ +$\underset{\text{ax}}{\mathrm{Cl}{}_2}$
$a=2,x=0.4,V=2$
$\left[PC{l}_5\right]=\frac{2\left(1-0.4\right)}{2}=0.6mol/L$
$\left[PC{l}_3\right]=\frac{2\times 0.4}{2}=0.4mol/L$
$\left[C{l}_2\right]=\frac{2\times 0.4}{2}=0.4mol/L$
$\because K_c=\frac{\left[PC{l}_3\right]\left[C{l}_2\right]}{\left[PC{l}_5\right]}$
$=\frac{0.4\times 0.4}{0.6}=0.267$