# Solid Ammonium carbamate dissociates as: NH2COONH4(s) → 2NH3(g) +CO2(g). In a closed vessel solid ammonium carbamate is in equilibrium with its dissociation products. At equilibrium ammonia is added such that the partial pressure of NH3 at new equilibrium now equals the original total pressure. Calculate the ratio of total pressure at new equilibrium to that of original total pressure.

Q: Solid Ammonium carbamate dissociates as:

NH2COONH4(s) $\rightleftharpoons$ 2NH3(g) +CO2(g). In a closed vessel solid ammonium carbamate is in equilibrium with its dissociation products. At equilibrium ammonia is added such that the partial pressure of NH3 at new equilibrium now equals the original total pressure. Calculate the ratio of total pressure at new equilibrium to that of original total pressure.

Solution:    NH2COONH4(s) $\rightleftharpoons$  2NH3(g) + CO2(g)

Let P = original equilibrium pressure, from the mole ratio of NH3 and CO2 at equilibrium, we have

$\large P_{NH_3} = \frac{2P}{3} , P_{CO_2} = \frac{P}{3}$

$\large K_p = (P_{NH_3})^2 (P_{CO_2})$

$\large K_p = (\frac{2P}{3})^2 (\frac{P}{3}) = \frac{4 P^3}{27}$

Now NH3 is added such that, PNH3 = P

Find the pressure of CO2

$\large K_p = (P_{NH_3})^2 (P_{CO_2})$

$\large \frac{4 P^3}{27} = (P)^2 (P_{CO_2})$

$\large P_{CO_2} = \frac{4 P}{27}$

Total new pressure is

Pnew = PNH3 + PCO2

$\large P_{new} = P + \frac{4P}{27} = \frac{31P}{27}$

$\large Ratio = \frac{P_{new}}{P_{original}} = \frac{31}{27}$