Question

Two moles of $PCl_{5}$ were heated in a closed vessel of 2 L. At equilibrium 40% of $PCl_{5}$ is dissociated into $PCl_{3} \, and \, Cl_{2}$ . 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 $ $\text{K}_{\text{c}}=\frac{\left[\text{P} \text{C} \text{l}_{3}\right] \left[\text{C} \text{l}_{2}\right]}{\left[\text{P} \text{C} \text{l}_{5}\right]}$

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