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### Topic: Chemical equilibrium  (Read 6704 times)

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#### Jai

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##### Chemical equilibrium
« 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.
« Last Edit: August 21, 2014, 11:01:29 AM by Jai »

#### Borek

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##### Re: Chemical equilibrium
« Reply #1 on: August 21, 2014, 10:18:05 AM »
Why do you go through Kc instead of Kp?
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#### Jai

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##### Re: Chemical equilibrium
« Reply #2 on: August 21, 2014, 10:42:05 AM »
Ok even if I go through kp I get the same answer.

#### Jai

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##### Re: Chemical equilibrium
« Reply #3 on: August 21, 2014, 10:45:16 AM »
And by the way I did go through ko and not kc

#### Borek

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##### Re: Chemical equilibrium
« Reply #4 on: August 21, 2014, 10:54:58 AM »
What is Ko?
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#### Jai

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##### Re: Chemical equilibrium
« Reply #5 on: August 21, 2014, 11:02:15 AM »
I am sorry wrote that by mistake.

#### mjc123

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##### Re: Chemical equilibrium
« Reply #6 on: August 21, 2014, 12:11:31 PM »
Did you forget to square the ammonia pressure in your second expression for Kp? Then you cancel p2 to give
p/3 + x = 4p/27
which gives you the right answer.

#### Jai

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##### Re: Chemical equilibrium
« Reply #7 on: August 23, 2014, 03:47:41 PM »
Thanks a lot....made a stupid mistake:)