Question

Solid 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, solid ammonium carbonate is in equilibrium with its dissociation products. At equilibrium, ammonia is added such that the partial pressure of $N{H}_3$ at new equilibrium now equals the original total pressure. Calculate the ratio of total pressure at new equilibrium to that of original total pressure.

• A. $\frac{31}{27}$
• B. $\frac{17}{7}$
• C. $\frac{49}{29}$
• D. None of these

Answer 1

Answer: (A)

$\frac{31}{27}$

Let P be the total pressure at the first equilibrium.

For dissociation of 1 mole of solid ammonium carbonate, 2 moles of ammonia and 1 mole of carbon dioxide are obtained.

Hence, the mole fractions of ammonia and carbon dioxide are $\frac{2}{3}$ and $\frac{1}{3}$ respectively.

The partial pressure is the product of mole fraction and total pressure.
Hence, the partial pressures of ammonia and carbon dioxide are $\frac{2}{3}P$ and $\frac{1}{3}P$ respectively.

The equilibrium constant expression is $K_p=P_{N{H}_3}^2{P}_{C{O}_2}=(\frac{2}{3}P{)}^2\frac{1}{3}P=\frac{4}{27}{P}^3$

At second equilibrium, the partial pressure of ammonia is equal to the original total pressure P

The partial pressure of carbon dioxide is calculated from the equilibrium constant expression.

$K_p=P_{N{H}_3}^2{P}_{C{O}_2}$

$\frac{4}{27}{P}^3=P^2{P}_{C{O}_2}$

$P_{C{O}_2}=\frac{4}{27}P$

The total pressure at new equilibrium is $P_{N{H}_3}+P_{C{O}_2}=P+\frac{4}{27}P=\frac{31}{27}P$

The ratio of total pressure at new equilibrium to that of original total pressure is:
$\frac{\frac{31P}{27}}{P}=\frac{31}{27}$

Hence, the correct option is A.