In the reaction:
${N H}_{4} C O O {N H}_{4} (s) ⇌ 2 {N H}_{3} (g) + {C O}_{2} (g)$
Equilibrium pressure for this reaction is $3$ atm. The value of $K_{p}$ will be:

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Solution

In the reaction:
${N H}_{4} C O O {N H}_{4} (s) ⇌ 2 {N H}_{3} (g) + {C O}_{2} (g)$
Equilibrium pressure for this reaction is $3$ atm. The value of $K_{p}$ will be $4$ .

From the reaction stoichiometry,
$P_{N H_{3}} = 2 P_{C O_{2}}$
$P_{t o t a l} = P_{N H_{3}} + P_{C O_{2}}$

Substitute $P_{N H_{3}} = 2 P_{C O_{2}}$ in above equation.
$P_{t o t a l} = 2 P_{C O_{2}} + P_{C O_{2}}$
$P_{t o t a l} = 3 P_{C O_{2}} = 3$ atm
$P_{C O_{2}} = 1$ atm
$P_{N H_{3}} = 2 P_{C O_{2}} = 2 (1) = 2$ atm

$K_{P} = P_{N H_{3}}^{2} \times P_{C O_{2}}$
$K_{P} = 2^{2} \times 1 = 4$

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