Solid ammonium carbamate dissociates to give ammonia and carbon dioxide as follows:
${N H}_{2} C O O {N H}_{4} (s) ⇌ 2 {N H}_{3} (g) + C O_{2} (g)$
At equilibrium, ammonia is added such that partial pressure of ${N H}_{3}$ now equals to the original total pressure. Calculate the ratio of total pressure now to the original total pressure.

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Solution

$E x p l a n a t i o n :$

$N H_{2} C O O N H_{4_{(s)}} ⇌ 2 N H_{3 (g)} + C O_{2 (g)}$
Intial $2 P$ $P$
Final $3 P$ $P$

At equilibrium $P_{N H_{3}} = P_{T}$

$K_{P} = (P_{N H_{3}})^{2} (P_{C O_{2}})$

$K_{P} = (2 P)^{2} (P) \to (1)$

$P_{i n i t i a l} = 3 P$

$K_{P} = (P_{N H_{3}}) (P_{C O_{2}})$

$K_{P} = (3 P)^{2} (P^{'}) \to (2)$

$K_{P} = (3 P) (P^{'})$

From equation $(1)$ and $(2)$

$(2 P)^{2} (P) = (3 P)^{2} (P^{'})$

$P^{'} = \frac{4 P}{9}$

Now,

$\frac{P (n e w)}{P (o l d)} = \frac{3 P + P^{'}}{3 P} = \frac{3 P + \frac{4 P}{9}}{3 P} = \frac{31}{27}$

Hence the correct answer is $\frac{31}{27}$ .

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