Question

Ammonium carbamate dissociates as
$N{H}_{2}COON{H}_{4\left(s\right)}\leftrightharpoons 2N{H}_{3\left(g\right)}+C{O}_{2\left(g\right)}$
Ina closed vessel containing ammonium carbamate in equilibrium, ammonia is added such that the partial pressure of $N{H}_3$ now equals to the original total pressure. The ratio of total pressure now to the original pressure is:

• A. $\frac{27}{31}$
• B. $\frac{31}{27}$
• C. $\frac{4}{9}$
• D. $\frac{3.1}{27}$

Answer 1

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

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

Initial 2P P

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

$K_p=\left(2P{)}^2\right(P).........(i)$

Now, in second case $P_t\left(initial\right)=3P$

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

Initial 3P P'

$K_p=\left(3P{)}^2\right(P^{\prime }).........(ii)$

From equation (i) and (ii)

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

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

$\frac{P_T\left(new\right)}{P_T\left(old\right)}=\frac{3P+P^{\prime }}{3P}=\frac{3P+\frac{4P}{9}}{3P}=\frac{31}{27}$