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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. Calculate $K_{p}$ . If ${N H}_{4} C N (s)$ is allowed to decompose in presence of ${N H}_{3}$ at 0.25 atm, calculate partial pressure of HCN at equilibrium.

K_{p}

= 0.149

{a t m}^{2}

HCN partial pressure =0.0694 atm

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K_{p}

= 0.0222

{a t m}^{2}

HCN partial pressure =0.25 atm

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K_{p}

= 0.0222

{a t m}^{2}

HCN partial pressure =0.0694 atm

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Solution

The correct option is C $K_{p}$ = 0.0222 ${a t m}^{2}$
HCN partial pressure =0.0694 atm
${N H}_{4} C N (s) ⇌ {N H}_{3} (g) + H C N (g)$
Partial Pressure at equilibrium,
$∵$ Total pressure at equilibrium = 2P = 0.298 atm $∴$ P = 0.149 atm
Also, $K_{p}$ $=$ $P_{{N H}_{3}}^{'} \times P_{H C N}^{'}$
$= 0.149 \times 0.149 = 0.0222 {a t m}^{2}$
Now the dissociation is made when, $P_{{N H}_{3}}$ = 0.25 atm
${N H}_{4} C N (s) ⇌ {N H}_{3} + H C N$

Pressure at equilibrium
$∴$ $K_{p} = P \times (P + 0.25)$
$∴$ $0.0222 = P^{2} + 0.25 P$
$∴$ $P = 0.0694 a t m$

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