Initially, 0.8 mol of $\mathrm{P}\mathrm{C}{\mathrm{l}}_5$ and 0.2 mol of $\mathrm{P}\mathrm{C}{\mathrm{l}}_3$ are mixed in a one litre vessel. At equilibrium, 0.4 mol of $\mathrm{P}\mathrm{C}{\mathrm{l}}_3$ is present. The value of ${\mathrm{K}}_{\mathrm{c}}$ for the reaction, $\mathrm{P}\mathrm{C}{\mathrm{l}}_5\left(\mathrm{g}\right)\rightleftharpoons \mathrm{P}\mathrm{C}{\mathrm{l}}_3\left(\mathrm{g}\right)+\mathrm{C}{\mathrm{l}}_2\left(\mathrm{g}\right),$ would be

• Correct answer
  A

  $0.13\text{ mol }{\text{L}}^{-1}$
• B

  $0.66\text{ mol }{\text{L}}^{-1}$
• C

  $0.013\text{ mol }{\text{L}}^{-1}$
• D

  $0.05\text{ mol }{\text{L}}^{-1}$

Solution

So $x=0.2$

${\mathrm{K}}_{\mathrm{c}}=\frac{\left[\mathrm{P}\mathrm{C}{\mathrm{l}}_3\right]\left[\mathrm{C}{\mathrm{l}}_2\right]}{\left[\mathrm{P}\mathrm{C}{\mathrm{l}}_5\right]}=\frac{0.4\times 0.2}{0.6}$

$=0.13\text{ mol }{\text{L}}^{-1}$

Hence, the correct answer is option: 'A'.