NEET 1989

Question

$2	ext{ moles}$ of $	ext{PCl}_5$ were heated in a closed vessel of $2	ext{ litre}$ capacity. At equilibrium, $40\%$ of $	ext{PCl}_5$ is dissociated into $	ext{PCl}_3$ and $	ext{Cl}_2$. The value of equilibrium constant is: A). $0.266$ B). $0.53$ C). $2.66$ D). $5.3$

Edvaya · Accepted Answer

The correct answer is $0.266$. Reaction: $	ext{PCl}_5 ightleftharpoons 	ext{PCl}_3 + 	ext{Cl}_2$.
Initial moles: $	ext{PCl}_5 = 2$, $	ext{PCl}_3 = 0$, $	ext{Cl}_2 = 0$.
Dissociation is $40\%$, so moles of $	ext{PCl}_5$ dissociated = $2 	imes 0.40 = 0.8	ext{ mol}$.
Moles at equilibrium: $	ext{PCl}_5 = 2 - 0.8 = 1.2	ext{ mol}$; $	ext{PCl}_3 = 0.8	ext{ mol}$; $	ext{Cl}_2 = 0.8	ext{ mol}$.
Volume of vessel $V = 2	ext{ L}$.
Equilibrium concentrations: $[	ext{PCl}_5] = \frac{1.2}{2} = 0.6	ext{ M}$; $[	ext{PCl}_3] = \frac{0.8}{2} = 0.4	ext{ M}$; $[	ext{Cl}_2] = \frac{0.8}{2} = 0.4	ext{ M}$.
Equilibrium constant $K_c = \frac{[	ext{PCl}_3][	ext{Cl}_2]}{[	ext{PCl}_5]} = \frac{0.4 	imes 0.4}{0.6} = \frac{0.16}{0.6} = 0.266$.

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