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For the reversible reaction $N H_{2} C O O N H_{4} (s) ⇌ 2 N H_{3} (g) + C O_{2} (g), K_{p} = 4 a t m^{3}$ Equilibrium pressure would be

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27 atm

3 atm

9 atm

6 atm

answer is C.

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Detailed Solution

$K_{p} = p_{N H_{3}}^{2} \cdot p_{{CO}_{2}}$

Since 2 moles of ${N H}_{3}$ and 1 mole of $C O_{2}$ are formed

$X_{N H_{3}} = \frac{2}{3}; X_{C O_{2}} = \frac{1}{3}$

$4 = {(\frac{2}{3} \times P_{tot})}^{2} (\frac{1}{3} \times P_{tot})$

$P_{tot} = 3 atm$

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