Question

Two moles of $PC{l}_5$ were heated in a closed vessel of 2 L. At equilibrium 40% of $PC{l}_5$ is dissociated into $PC{l}_{3}andC{l}_2$ . The value of equilibrium constant is

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

Answer 1

Answer: (B)

Degree of dissociation=0.4


$\text{a}=2,\text{x}=0.4,\text{V}=2$

$\left[{\mathrm{PCl}}_5\right]=\frac{2(1-0.4)}{2}=0.6\mathrm{mol}/\mathrm{L}$

$\left[\text{P}\text{C}{\text{l}}_3\right]=\frac{2\times 0.4}{2}=0.4$ mol/L

$\left[\text{C}{\text{l}}_2\right]=\frac{2\times 0.4}{2}=0.4$ mol/L

$\therefore$ ${\text{K}}_{\text{c}}=\frac{\left[\text{P}\text{C}{\text{l}}_3\right]\left[\text{C}{\text{l}}_2\right]}{\left[\text{P}\text{C}{\text{l}}_5\right]}$

$=\frac{0.4\times 0.4}{0.6}=0.267$