I am having a problem in the following question.
Ammonium carbamate dissociates as : In a closed vessel containing ammonium carbamate in equilibrium with ammonia and carbon dioxide, ammonia is added such that partial pressure of NH3 now equals to the original total pressure. The ratio of total pressure to the original pressure now is ?
Nh2coonh4(s)----2nh3(g)+co2(g)
I tried solving the problem and I know I am quite close to the answer which is 31/27. Firstly I took out ko for equilibrium assuming total pressure to be p. So kc= (2p/3)^2*p/3=4p^3/27
After this the par pressure of ammonia becomes p so let's say that of co2 is p/3+x. Now since kc doesn't change so p/3+x=4p^2/27(cancelling one of the p) But solving further and finding new pressure and dividing by the old one doesn't give the ans. please help me.
I am having a problem in the following question.
Ammonium carbamate dissociates as : In a closed vessel containing ammonium carbamate in equilibrium with ammonia and carbon dioxide, ammonia is added such that partial pressure of NH3 now equals to the original total pressure. The ratio of total pressure to the original pressure now is ?
Nh2coonh4(s)----2nh3(g)+co2(g)
I tried solving the problem and I know I am quite close to the answer which is 31/27. Firstly I took out ko for equilibrium assuming total pressure to be p. So kc= (2p/3)^2*p/3=4p^3/27
After this the par pressure of ammonia becomes p so let's say that of co2 is p/3+x. Now since kc doesn't change so p/3+x=4p^2/27(cancelling one of the p) But solving further and finding new pressure and dividing by the old one doesn't give the ans. please help me.
