Question

Two moles of $PC{l}_5$ were heated in closed vessel of 2 litre capacity. At equilibrium, 40% of $PC{l}_5$ was dissociated into $PC{l}_3$ and $C{l}_2$.

The value of the equilibrium constant is :

• A. $0.267$
• B. $0.53$
• C. $2.67$
• D. $5.3$

Answer 1

The correct option is A $0.267$

The given reaction is , $PC{l}_5\rightleftharpoons PC{l}_3+C{l}_2$

At equilibrium 40% of $PC{l}_5$ was dissociated into $PC{l}_3$ and $C{l}_2$.

At equilibrium,

$\left[PC{l}_5\right]=\frac{2-0.8}{2}=0.6$

$\left[PC{l}_3\right]=\frac{0.8}{2}=0.4$

$\left[C{l}_2\right]=\frac{0.8}{2}=0.4$

$K_c=\frac{\left[PC{l}_3\right]\left[C{l}_2\right]}{\left[PC{l}_5\right]}$

Therefore, $K_C=\frac{0.4\times 0.4}{0.6}=0.267$

So, the correct option is $A$.