Question

$2molePC{l}_5$ were introduced in a $2L$, flask and heated at $625K$, to establish equilibrium when $60$% of $PC{l}_5$ was dissociated into $PC{l}_3$ and $C{l}_2$. Find the value of equilibrium constant.

• A. $0.85$
• B. $0.9$
• C. $0.72$
• D. None of these

Answer 1

Answer: (B)

$0.9$

The given reaction is :-

${PCl}_5\rightleftharpoons {PCl}_3+{Cl}_2$

Initial $1$ $0$ $0$

conc.

At eqn $1-0.6\times 1$ $0.6$ $0.6$ [$\because$ Initial concentration of ${PCl}_5=\frac{2mole}{2L}=1M/L$]

$=0.4$

(since, $60$% of ${PCl}_5$ is dissociated at equilibrium, so, $40$% of ${PCl}_5$ would be remaining at equilibrium)

Now, $K=\frac{\left[{PCl}_3\right]\left[{Cl}_2\right]}{\left[{PCl}_5\right]}$

$=\frac{0.6\times 0.6}{0.4}=0.09\times 10=0.9$

Option B is correct.