`NH_(2)COONH_(4)(s)hArr2NH_(3)(g)+CO_(2)(g)`
Let `P`= original equilibrium pressure.
From the mole ratio of `NH_(3)` and `CO_(2)` at equilibrium, we have
`p_(NH_(3))=2/3 P` and `p_(CO_(2))=P/3`
`rArr K_(p)=(p_(NH_(3)))^(2).p_(CO_(2))=(2/3P)^(2) (P/3)=4/27 P^(3)`
Now `NH_(3)` is added such that `p_(NH_(3))=P`
Find the pressure of `CO_(2)`
`rArr 4/27 P^(3)=P^(2) p_(CO_(2)) rArr p_(CO_(2))=4/27 P`
Total new pressure `=P_("new")=p_(NH_(3))+p_(CO_(2))`
`rArr P_("new")=P+4/27 P=31/27 P`
`rArr` Ratio =`P_("new")/P_("original")=(31/27P)/P=31/27`
Let `x` be the partial pressure of `NH_(3)` added at original equilibrium.
`NH_(2)COONH_(4)(s)hArr2NH_(3)(g)+CO_(2)(g)`
`{:("At equilibrium",,2/3 P,,1/3 P),("When" NH_(2) " is added",,2/3P+x,,1/3 P),("At new equilibrium",,2/3P+x-2y,,1/3P-y):}`
`rArr 2/3 P+x-2y=p_(NH_(3))` and `1/3 P-y=P_(CO_(2))=4/27P`
`rArr` Solve to get: `x=19/27 P`