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Q. Ammonium carbonate dissociates as
$NH _{2} COONH _{4}( s ) \rightleftharpoons 2 NH _{3}( g )+ CO _{2}( g )$
In a closed vessel having ammonium carbonate in equilibrium, $NH_{3}$ is added such that partial pressure of $NH_{3}$ now, is equals to the original total pressure. The ratio of total pressure now and the original pressure is?

NTA AbhyasNTA Abhyas 2022

Solution:

$ NH _{2} COONH _{4}( s ) \rightleftharpoons 2 NH _{3}( g )+ CO _{2}( g ) $
Suppose $P$ is pressure of $CO _{2}$ at equilibrium than $ \begin{array}{l} P _{ NH _{3}}=2 P \\ P _{\text {Total }}=2 P + P =3 P \\ \text { As } K _{ P }=\left( P _{ NH _{3}}\right)^{2}\left( P _{ CO _{2}}\right)=4 P ^{3} \end{array} $
On adding $NH _{3}$ at equilibrium
$ P _{ NH _{3}}= P _{\text {Total }}=3 P $
So $K _{ P }=4 P ^{3}=\left( P _{ NH _{3}}\right)^{2} \cdot\left( P ^{\prime} CO _{2}\right)$
$ 4 P ^{3}=(3 P )^{2}\left( P ^{\prime} CO _{2}\right) $
$P ^{\prime} CO _{2}=\frac{4}{9} P$
Now total $P ^{\prime}=3 P +\frac{4}{9} P =\frac{31}{9} P$
Hence pressure ratio $\left(\frac{P^{\prime}}{ P }\right)=\frac{\frac{31}{9} P }{3 P }$ $=\frac{31}{27}$