Question

Solid ammonium carbamate dissociates to give ammonia and carbon dioxide as follows:
$N{H}_{2}COON{H}_4\left(s\right)\rightleftharpoons 2N{H}_3\left(g\right)+C{O}_2\left(g\right)$
At equilibrium, ammonia is added such that partial pressures of $N{H}_3$ now equals the original total pressure. Calculate the ratio of the total pressures now to the original total pressure.

• A. $\frac{31}{27}$
• B. $\frac{60}{40}$
• C. $\frac{31}{9}$
• D. $\frac{62}{27}$

Answer 1

The correct option is A $\frac{31}{27}$

$N{H}_{2}COON{H}_4\left(s\right)\rightleftharpoons 2N{H}_3\left(g\right)+C{O}_2\left(g\right)$
Initial $2P$ $P$
$K_P={\left(P_{N{H}_3}\right)}^2\left(P_{C{O}_2}\right)$
$K_P=\left(2P{)}^2\right(P)......(i)$
$P_T\left(initial\right)=3P$
$N{H}_{2}COON{H}_4\left(s\right)\rightleftharpoons 2N{H}_3\left(g\right)+C{O}_2\left(g\right)$
Final $3P$ $P^{{}^{\prime }}$
$K_p=\left(3P{)}^2\right(P^{\prime })......(ii)$
From eq. (i) and (ii)
$\left(2P{)}^2\right(P)=(3P{)}^2\left(P^{\prime }\right)$
$P^{\prime }=\frac{4P}{9}$
$\frac{P_T\left(New\right)}{P_T\left(Old\right)}=\frac{3P+P^{\prime }}{3P}=\frac{3P+\frac{4P}{9}}{3P}=\frac{31}{27}$.