For the equilibrium of the reaction:
${N H}_{4} C l (s) ⇌ {N H}_{3} (g) + H C l (g); K_{p} = 81 {a t m}^{2}$
Total pressure at equilibrium will be $x$ times the pressure of ${N H}_{3}$
The value of $x$ will be:

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Solution

For the equilibrium of the reaction:
${N H}_{4} C l (s) ⇌ {N H}_{3} (g) + H C l (g); K_{p} = 81 {a t m}^{2}$
Total pressure at equilibrium will be $x$ times the pressure of ${N H}_{3}$
The value of $x$ will be $2$ .
$K_{p} = P_{N H_{3}} P_{H C l} = 81$
But $P_{N H_{3}} = P_{H C l}$
Hence, $K_{p} = P_{N H_{3}}^{2} = 81$
$P_{N H_{3}} = 9 = P_{H C l}$
Total pressure $= P_{N H_{3}} + P_{H C l} = 9 + 9 = 18$
$\frac{Total pressure}{P_{N H_{3}}} = \frac{18}{9} = 2$

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