Question

Ammonium carbamate dissociate 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 original total pressure. The ratio of total pressure now to the original pressure is $x$, then find the value of $\frac{3^4}{31}\times x$.

Answer 1

Consider the original equilibrium.
The decomposition of one mole of ammonium carbamate gives 2 moles of ammonia and 1 mole of $C{O}_2$. The mole fractions of ammonia and $C{O}_2$ will be $\frac{2}{3}$ and $\frac{1}{3}$ respectively. If P is the total pressure, then the partial pressures of ammonia and $C{O}_2$ will be $\frac{2}{3}P$ and $\frac{1}{3}P$ respectively.
The equilibrium constant will be $K_p=P_{N{H}_3}^2{P}_{C{O}_2}=(\frac{2}{3}P{)}^2\times (\frac{1}{3}P)=\frac{4}{27}{P}^3$
Now ammonia is added. The partial pressure of ammonia becomes P. The partial pressure of $C{O}_2$ is calculated below.
$\frac{4}{27}{P}^3=P^2\times P_{C{O}_2}$
$P_{C{O}_2}=\frac{4}{27}P$
The total pressure will be $P_T=P+\frac{4}{27}P=\frac{31}{27}P$
The ratio of total pressure now to the original pressure
$\frac{P_T}{P}=\frac{\frac{31}{27}P}{P}=\frac{31}{27}$

Hence, $\frac{31}{27}\times \frac{3^4}{31}=3$