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Industrial Preparation of Ammonia

The equilibri...

Question

The equilibrium pressure of $N H_{4} C N (s) ⇌ N H_{3} (g) + H C N (g)$ , is 0.298 atm. Value of $K_{p}$ if $N H_{4} C N (s)$ is allowed to decompose in pressure of $N H_{3}$ at 0.25 atm is x and partial pressure of $H C N$ at equilibrium is y.
10000(x+y) is

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Solution

$N H_{4} C N (s) ⇌ N H_{3} (g) + H C N (g)$
Pressure at equilibrium $P P$
$∴$ Total pressure at equilibrium, $2 P = 0.298$
$∴ P = 0.149 a t m$
$∴ K_{p} = P_{N H_{3}}^{'} \times P_{H C N}^{'}$
$= 0.149 \times 0.149 = 0.0222 {a t m}^{2}$
If dissociation is made in presence of $N H_{3}$ at 0.25 atm
$N H_{4} C N (s) ⇌ N H_{3} (g) + H C N (g)$
Initial pressure $0 0.25$
Pressure at equilibrium $(0.25 + P^{'}) P^{'}$
Also, $K_{p} = P^{'} (0.25 + P^{'})$
or $0.0222 = P^{'} (0.25 + P^{'})$
$∴ P^{'} = 0.0694 a t m$

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