For the heterogenous equilibrium with pressure P (At equilibrium)
$N H_{2} C O O N H_{4} (s) \to 2 N H_{3} + C O_{2}$
The value of $K_{P}$ is:

\frac{2}{27} p^{3}

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\frac{4}{27} p^{3}

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\frac{4}{27} p^{2}

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\frac{27}{4} p^{3}

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Solution

The correct option is B $\frac{4}{27} p^{3}$
$N H_{2} C O O N H_{4} \to 2 N H_{3} + C O_{2}$
Suppose total pressure of container Let pressure to $N H_{3} = 2 x$
pressure dir to $C o_{2} = x$
$2 x + x = P 3 x = P x = \frac{P}{3}$
$K_{p} = {[P_{N H_{3}}]}_{N H_{3}}^{2} [P_{C O_{2}}]$
$= {(\frac{2 P}{3})}^{2} \frac{P}{3}$
$= \frac{4 P^{3}}{27}$
$B$

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For the heterogenous equilibrium with pressure P (At equilibrium)NH2COONH4(s)→2NH3+CO2The value of KP is:

The correct option is B 427p3NH2COONH4→2NH3+CO2Suppose total pressure of container Let pressure to NH3=2xpressure dir to Co2=x2x+x=P3x=Px=P3Kp=[PNH3]2[PCO2]=(2P3)2P3=4P327B

For the heterogenous equilibrium with pressure P (At equilibrium)
$N H_{2} C O O N H_{4} (s) \to 2 N H_{3} + C O_{2}$
The value of $K_{P}$ is: