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 the partial pressure of $N{H}_3$ now equals the original total pressure. Calculate the ratio of the 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 C $\frac{4}{9}$

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

$K_p=\left(2p{)}^2\right(p)=4p^3$

If $N{H}_3$ is added and pressure of $N{H}_3$ after addition at equilibrium is 3p.

$K_p=4p^3=(p_{N{H}_3}^{{}^{\prime }}{)}^2\times p_{C{O}_2}^{{}^{\prime }}$
$K_p=4p^3=(3p{)}^2\times p^{{}^{\prime }}$
$\frac{p^{{}^{\prime }}}{p}$ = $\frac{4}{9}$