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 equilibrium constant is :

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

Answer 1

Answer: (B)

Degree of dissociation $ = 0.4$
$\underset{\text{a(1-x)}}{\ce{PCl5}}$ $\qquad$ $\underset{\text{ax}}{\ce{PCl3}}$ +$\underset{\text{ax}}{\ce{Cl2}}$
$a = 2, x = 0.4, V = 2$
$\left[PCl_{5}\right] = \frac{2\left(1-0.4\right)}{2} = 0.6\, mol/ L $
$ \left[PCl_{3}\right] = \frac{2\times 0.4}{2} = 0.4 \,mol/ L $
$\left[Cl_{2}\right] = \frac{2\times 0.4}{2} = 0.4\, mol/ L $
$ \because K_{c} = \frac{\left[PCl_{3}\right]\left[Cl_{2}\right]}{\left[PCl_{5}\right]} $
$ = \frac{0.4\times 0.4}{0.6} = 0.267 $