Question

Ammonium carbamate dissociates as:
$N{H}_{2}COON{H}_4\left(s\right)\rightleftharpoons 2N{H}_3\left(g\right)+C{O}_2\left(g\right)$
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 $C{O}_2$ :

• A. 4
• B. 9
• C. $\frac{4}{9}$
• D. $\frac{2}{9}$

Answer 1

The correct option is A 4

$\underset{Initial}{{N{H}_{2}COON{H}_4}_{\left(s\right)}}\rightleftharpoons \underset{2P}{{2N{H}_3}_{\left(g\right)}}+\underset{P}{{C{O}_2}_{\left(g\right)}}$

$K_P=P_{N{H}_3}^2{P}_{C{O}_2}$

$\Rightarrow K_P=2P.P⟶\left(1\right)$

Now, in second case $P_t=3P$

$\underset{Initial}{{N{H}_{2}COON{H}_4}_{\left(s\right)}}\rightleftharpoons \underset{3P}{{2N{H}_3}_{\left(g\right)}}+\underset{P^{\prime }}{{C{O}_2}_{\left(g\right)}}$

$K_P=\left(3P{)}^2\right(P^{\prime })⟶(2)$

From $\left(1\right)$ & $\left(2\right)$, $(2P{)}^{2}P=(3P{)}^2\left(P^{\prime }\right)$

$\Rightarrow P^{\prime }=\frac{4P^3}{9P^2}=\frac{4P}{9}$

$\therefore \frac{P^{\prime }}{P}=\frac{4}{9}$