Ammonium carbonate dissociates as NH _ 2 COONH _ 4 (s) ⇌ 2NH _ 3...

{\text{NH}}_{\text{2}} {\text{COONH}}_{\text{4}} \text{(s)} ⇌ {\text{2NH}}_{\text{3}} \text{(g)}\text{+}{\text{CO}}_{\text{2}} \text{(g)}$<br />Suppose P is pressure of$C{O}_{2}$at equilibrium than<br />${P}_{N{H}_{3}} = 2P$<br />${P}_{Total} = 2P + P = 3P$<br />As${K}_{P} = {\left({P}_{N{H}_{3}}\right)}^{2}\left({P}_{C{O}_{2}}\right) = 4{P}^{3}$<br />On adding$N{H}_{3}$at equilibrium<br />${P}_{N{H}_{3}} = {P}_{Total} = 3P$<br />So${K}_{P} = 4{P}^{3} = {\left({P}_{N{H}_{3}}\right)}^{2} . \left(P{'}_{C{O}_{2}}\right)$<br />$4{P}^{3} = {\left(3P\right)}^{2}\left(P{'}_{C{O}_{2}}\right)$<br />$P{'}_{C{O}_{2}} = \dfrac{4}{9}P$<br />Now total$P' = 3P + \dfrac{4}{9}P = \dfrac{31}{9}P$<br />Hence pressure ratio \left(\dfrac{P'}{P}\right) = \dfrac{\dfrac{31}{9}P}{3P}$
$= \dfrac{31}{27}$