PCl₅ dissociates as ${PCl}_{5} (g) ⇌ {PCl}_{3} (g) + {Cl}_{2} (g)$
5 moles of ${PCl}_{5}$ are placed in a 200 litre vessel which contains 2 moles of $N_{2}$ and is maintained at $600 K$ . The equilibrium pressure is $2.46 atm$ . The equilibrium constant $K_{p}$ for the dissociation of ${PCl}_{5}$ is____________ $\times 10^{- 3}$ . (nearest integer)
(Given: $R = 0.082 L$ atm $K^{- 1} {mol}^{- 1}$ : Assume ideal gas behaviour)

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answer is 1107.

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Detailed Solution

Given : 2 mole of N₂ gas was present as inert gas. Equilibrium pressure = 2.46 atm

${PCl}_{5} (g) ⇌ {PCl}_{3} (g) + {Cℓ}_{2} (g) \begin{array}{llll} t = 0 & 5 & 0 & 0 \\ t = {Eq}^{m} & 5 - x & x & x \end{array}$

from ideal gas equation

$\begin{aligned} 2.46 \times 200 = (5 - x + x + x + 2) \times 0.082 \times 600 \\ x = 3 \\ K_{P} = \frac{n_{{PCl}_{3}} \times n_{{Cl}_{2}}}{n_{{PCl}_{5}}} \times [\frac{P_{total}}{n_{total}}] \\ \frac{3 \times 3}{2} \times \frac{2.46}{10} = 1.107 = 1107 \times 10^{- 3} \end{aligned}$