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 initial pressure to the original total pressure.

• A. $\frac{31}{27}$
• B. $9$
• C. $\frac{4}{9}$
• D. $\frac{2}{9}$

Answer 1

Answer: (A)

$\frac{31}{27}$

$N{H}_{2}COON{H}_4\left(s\right)\rightleftharpoons 2N{H}_3\left(g\right)+C{O}_2\left(g\right)initial2pp$

Hence,

$K_p=\left(p_{N{H}_3}{)}^2\right(p_{C{O}_2})$

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

Case second : $p_t\left(initial\right)=3p$

$N{H}_{2}COON{H}_4\left(s\right)\rightleftharpoons 2N{H}_3\left(g\right)+C{O}_2\left(g\right)initial3p{p}^{\prime }$

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

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

$p^{\prime }=\frac{4p}{9}$

$\frac{p_T\left(new\right)}{p_T\left(old\right)}=\frac{3p+p^{\prime }}{3p}=\frac{3p+\frac{4p}{9}}{3p}=\frac{31}{27}$
Option A is correct.