# Chemical equilibrium

Watch
#1
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.
0
7 years ago
#2
Moved to Chemistry forum for you . They're better placed to help you
0
7 years ago
#3
(Original post by Jai1997)
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.
let the initial pressure of ammonia = p, so as the moles of carbon dioxide = ammonia/2 then carbon dioxide = p/2 = 0.5p

The equilibrium is given by p2(NH3) x p/2(CO2)

and initially the total pressure = the sum of the partial pressures = 1.5p

and the equilibrium constant can be expressed in terms of p as p3/2

If ammonia is increased to 2p and the equilibrium constant remains the same then:

p2(NH3) x p/2(CO2) becomes

2p.2p x new partial pressure of carbon dioxide = p3/2

p3/8p2 = new partial pressure of carbon dioxide = 0.125p

hence the final total pressure = 2.125p

and the original total pressure was 1.5p

So the ratio of final total to original total = 17 to 12 (using integers)

... better check the working
0
X

new posts
Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### Feeling behind at school/college? What is the best thing your teachers could to help you catch up?

Extra compulsory independent learning activities (eg, homework tasks) (7)
5%
Run extra compulsory lessons or workshops (24)
17.14%
Focus on making the normal lesson time with them as high quality as possible (24)
17.14%
Focus on making the normal learning resources as high quality/accessible as possible (20)
14.29%
Provide extra optional activities, lessons and/or workshops (41)
29.29%
Assess students, decide who needs extra support and focus on these students (24)
17.14%