Ordinary white phosphorus $P_{4}$ forms a vapour and dissociates into diatomic molecules at high temperature as : $P_{4} (g) ⇌ 2 P_{9} (g)$
A sample of white phosphorous, when heated to $1000 K$ , formed a mixture at equilibrium having a total pressure of $16 atm$ and a density of $18.6$ $g / litre$ . Use this information to evaluate $K_{P}$ for the above reaction [Use : $R = 0.08$ atm-litre/K-mole]

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answer is 8.

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Detailed Solution

$∵ P M = d R T$

$M = \frac{18.6 \times 0.08 \times 1000}{16} = 93$

Initially: 1 mole $P_{4} (g) ⇌ 2 P_{2} (g)$

At equilibrium $1 - x 2 x$

$∵ M_{avg} = 93 = \frac{(1 - x) \times 124 + (2 x \times 62)}{1 - x + 2 x}$

$x = \frac{1}{2}$

$∴ mole of ratio P_{4} : P_{2} :: \frac{2}{3} : \frac{2}{3} = 1 : 1$

If total pressure is 16 atm

$\begin{array}{l} ∴ P_{P_{4}} = 16 \times \frac{1}{2} = 8 atm \\ P_{P_{2}} = 16 \times \frac{1}{2} = 8 atm \\ K_{p} = \frac{(8)^{2}}{8} = 8 \end{array}$

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