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Dalton's Law of Partial Pressures

Ammonium carb...

Question

Ammonium carbamate dissociates as:
$N H_{2} C O O N H_{4} (s) ⇌ 2 N H_{3} (g) + C O_{2} (g)$
In a closed vessel containing ammonium carbamate in equilibrium, ammonia is added such that partial pressure of $N H_{3}$ now equals to the original total pressure. Calculate the ratio of partial pressure of $C O_{2}$ now to the original partial pressure of $C O_{2}$ :

4

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\frac{4}{9}

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\frac{2}{9}

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Solution

The correct option is C $\frac{4}{9}$
Given reaction is,
$N H_{2} C O O N H_{4} (s) ⇌ 2 N H_{3} (g) + C O_{2} (g)$
Let partial pressure of $C O_{2}$ at equlibrium be $P$ , then $P_{N H_{3}} = 2 P$ and total pressure at equlibrium = $3 P$
$∴ K_{p} = (2 P)^{2} \times P = 4 P^{3}$
If $N H_{3}$ is added and the pressure of $N H_{3}$ after addition at equlibrium is $3 P$ then,
$K_{p} = (P_{N H_{3}}^{/})^{2} \times P_{C O_{2}}^{/} = 4 P^{3}$
$\Rightarrow P_{C O_{2}}^{/} = \frac{4}{9} P$
So the ratio of $\frac{P_{C O_{2}}^{/}}{P_{C O_{2}}} = \frac{4}{9}$

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