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

4

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20

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25

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15

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Solution

The correct option is A $4$
$N H_{2} C O O N H_{4 (s)} ⇌ 2 N H_{3 (g)} + C O_{2 (g)}$
At equilibrium, when volume and temperature are constant, the number of moles of a gas is proportional to its partial pressure.
Therefore:
2 moles of $N H_{3}$ have partial pressure = 2p
1 mole of $C O_{2}$ has partial pressure = p
Total partial pressure = 2p + p = 3p = 3 atm (given)
3p = 3
p = 1 atm

$K_{p} = \frac{p_{N H_{3}}^{2} \times p_{C O_{2}}}{p_{N H_{2} C O O N H_{4}}}$

$K_{p} = \frac{(2 p)^{2} \times (p)}{p}$

$K_{p} = \frac{4 p^{3}}{p}$

$K_{p} = 4 p^{2} = 4 (1)^{2} = 4$

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