A vessel of volume 0.5 L contains 2 moles of ${PCl}_{3}$ and 3 moles of ${Cl}_{2}$ at instants $t = t_{1} a n d t = 2 t_{1}$ Calculate number of moles of ${Cl}_{2}$ that should be added at $t = {3t}_{1}$ so that the weight of $P C l_{3}$ at equilibrium is found to decrease upto 75% .
Given: $P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g) K_{C} = 12 m o l e / l$ .

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answer is 3.5.

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Detailed Solution

Because concentration of $P C l_{3} (g)$ and $C l_{2} (g)$ is constant at $t = t_{1}$ and $t = 2 t_{1}$ , it indicates that this is the state of equilibrium.
$\underset{a}{P C l_{5} (g)} \underset{}{⇌} \underset{2}{P C l_{3}} + \underset{3}{C l_{2} (g)} \underset{- a t e q}{}$
$[P C l_{5}] = \frac{a}{V}$
$[P C l_{3}] = \frac{2}{V} K_{C} = \frac{\frac{3}{V} \times \frac{2}{V}}{\frac{a}{V}}$
$[C l_{2}] = \frac{3}{V} on solving a = 1$
$Let x {-moles of Cl}_{2} a r e a d d e d \begin{array}{l} \underset{1}{P C l_{5}} ⇌ \underset{2}{P C l_{3}} + \underset{3}{C l_{2}} {.........1}^{s t} e q \\ (1 + y) (2 - y) (3 + x - y) ..... n e w e q u i l i b r i u m \end{array} G i v e n : 2 - y = 1.5 y = 0.5 [P C l_{5}] = \frac{1.5}{V} [P C l_{3}] = \frac{1.5}{V} [C l_{2}] = \frac{(2.5 + x)}{V} K_{C} = \frac{2.5 + x}{V} x = 3.5$