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Effect on Kc

NH2COONH4s ⇌ ...

Question

$N H_{2} C O O N H_{4} (s) ⇌ 2 N H_{3} (g) + C O_{2} (g) .$
If equilibrium pressure is 6 atm for the above reaction; $K_{p}$ will be:

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Solution

The correct option is A 32
$N H_{2} C O O N H_{4} (s) - ⇌ 2 N H_{3} (g) 2 p + C O_{2} (g) p .$
As total equilibrium pressure is 6 atm
$\Rightarrow 2 p + p = 6 atm$
$p = 2$ atm
$∴ P_{N H_{3}} = 4 a t m; P_{C O_{2}} = 2 a t m$
$K_{p} = {[P_{N H_{3}}]}_{N H_{3}}^{2} [P_{C O_{2}}] = 4^{2} \times 2 = 32$

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Effect on Kc

Standard XII Chemistry

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NH2COONH4(s)⇌2NH3(g)+CO2(g). If equilibrium pressure is 6 atm for the above reaction; Kp will be:

The correct option is A 32NH2COONH4(s)−⇌2NH3(g)2p+CO2(g)p. As total equilibrium pressure is 6 atm ⇒2p+p=6 atm p=2 atm ∴ PNH3=4 atm; PCO2=2 atm Kp=[PNH3]2[PCO2]=42×2=32

$N H_{2} C O O N H_{4} (s) ⇌ 2 N H_{3} (g) + C O_{2} (g) .$
If equilibrium pressure is 6 atm for the above reaction; $K_{p}$ will be: