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Liquid-Vapour Equillibrium

At 500 K, 0.1...

Question

At $500 K$ , $0.10$ mole of $P C l_{5}$ are heated in a $8.2 L$ flask. The pressure of equilibrium mixture is found to be $0.1 a t m$ . Calculate $K_{p} (i n a t m)$ for the reaction.
$P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)$
$G i v e n : R = 0.082 {L atm K}^{- 1} m o l^{- 1}$

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Solution

$\begin{matrix} P C l_{5} (g) ⇌ & P C l_{3} (g) & + & C l_{2} (g) At equilibrium (mole) & (0.1 - x) & x & x \end{matrix}$
$Total number of mole, n = (0.1 - x) + x + x = (0.1 + x) P V = n R T 0.1 \times 8.2 = (0.1 + x) \times 0.082 \times 500 o r x = 0.08 K_{c} = \frac{[P C l_{3}] [C l_{2}]}{[P C l_{5}]} = \frac{x^{2}}{(0.1 - x) \times 8.2} = \frac{0.08 \times 0.08}{(0.1 - 0.08) \times 8.2} = \frac{0.32}{8.2} {mol L}^{- 1} K_{p} = K_{c} (R T)^{Δ n} = K_{c} R T (∵ Δ n = + 1) \Rightarrow \frac{0.32}{8.2} \times 0.082 \times 500 = 1.6 a t m$

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