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${NH}_{4} {COONH}_{2 (s)} ⇌ 2 {NH}_{3 (g)} + {CO}_{2 (g)}$ if equilibrium pressure is 3 atm for the above reaction $K_{P}$ for the reaction is

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Detailed Solution

$\begin{array}{l} P_{{NH}_{3}} : P_{{CO}_{2}} = 2 : 1 \\ ∴ P_{{NH}_{3}} = 2 atm, P_{{CO}_{2}} = 1 \\ K_{P} = {[P_{{NH}_{3}}]}^{2} \cdot [P_{{CO}_{2}}] = 2^{2} \times 1 = 4 \end{array}$

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