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Chemistry Forums for Students => High School Chemistry Forum => Chemistry Olympiad and other competitions => Topic started by: Jai on August 21, 2014, 10:10:22 AM

Title: Chemical equilibrium
Post by: Jai on August 21, 2014, 10:10:22 AM

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 kp for equilibrium assuming total pressure to be p. So kp= (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.

Title: Re: Chemical equilibrium
Post by: Borek on August 21, 2014, 10:18:05 AM

Why do you go through Kc instead of Kp?

Title: Re: Chemical equilibrium
Post by: Jai on August 21, 2014, 10:42:05 AM

Ok even if I go through kp I get the same answer.

Title: Re: Chemical equilibrium
Post by: Jai on August 21, 2014, 10:45:16 AM

And by the way I did go through ko and not kc

Title: Re: Chemical equilibrium
Post by: Borek on August 21, 2014, 10:54:58 AM

What is Ko?

Title: Re: Chemical equilibrium
Post by: Jai on August 21, 2014, 11:02:15 AM

I am sorry wrote that by mistake.

Title: Re: Chemical equilibrium
Post by: mjc123 on August 21, 2014, 12:11:31 PM

Did you forget to square the ammonia pressure in your second expression for K_p? Then you cancel p² to give
p/3 + x = 4p/27
which gives you the right answer.

Title: Re: Chemical equilibrium
Post by: Jai on August 23, 2014, 03:47:41 PM

Thanks a lot....made a stupid mistake:)