Q: Solid Ammonium carbamate dissociates as:

NH_{2}COONH_{4}(s) $\rightleftharpoons $ 2NH_{3}(g) +CO_{2}(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 NH_{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.

Solution: NH_{2}COONH_{4}(s) $\rightleftharpoons $ 2NH_{3}(g) + CO_{2}(g)

Let P = original equilibrium pressure, from the mole ratio of NH_{3} and CO_{2} 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 NH_{3} is added such that, P_{NH3} = P

Find the pressure of CO_{2}

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

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

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

Total new pressure is

P_{new} = P_{NH3} + P_{CO2}

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

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