• # question_answer 2 moles of $PC{{l}_{5}}$ were heated in a closed vessel of 2 litre capacity. At equilibrium, 40% of $PC{{l}_{5}}$ is dissociated into $PC{{l}_{3}}$ and $C{{l}_{2}}$. The value of equilibrium constant is A) 0.266 B) 0.53 C) 2.66 D) 5.3

 [a] $\underset{2}{\mathop{PC{{l}_{5}}}}\,$$\rightleftharpoons$ $\underset{0}{\mathop{PC{{l}_{3}}}}\,+\underset{0}{\mathop{C{{l}_{2}}}}\,$ $\frac{2\times 60}{100}$ $\frac{2\times 40}{100}$ $\frac{2\times 40}{100}$ Volume of container = 2 litre. ${{K}_{c}}=\frac{\frac{2\times 40}{100\times 2}\times \frac{2\times 40}{100\times 2}}{\frac{2\times 60}{100\times 2}}=0.266$.