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Question
Ammonium carbamate dissociates as shown below. $$\ce{NH2COONH4 (s) -> 2NH3 (g) + CO2 (g)}$$ In a closed vessel containing ammonium carbamate in equilibrium with ammonia and carbon dioxide, ammonia is added such that partial pressure of $\ce{NH3}$ now equals to the original total pressure. What is the ratio of the new total pressure to the original pressure?

I tried solving the problem and I know I am quite close to the answer which is $\frac{31}{27}$. Firstly I took out $K_p$ for equilibrium assuming total pressure to be $p$. So,

$K_\mathrm{p} = \left(\frac{2p}{3}\right)^2 \times \frac{p}{3} = \frac{4p^3}{27}$

After this the par pressure of ammonia becomes $p$ so let's say that of $\ce{CO2}$ is $\frac{p}{3} + x$.

Now since $K_\mathrm{p}$ doesn't change so:

$\frac{p}{3} + x = \frac{4p^2}{27}$ (cancelling one of the $p$)

But solving further and finding new pressure and dividing by the old one doesn't give the answer

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As follows:

Just to simplify the math a little bit, assume pressure of $\ce{CO2_{(g)}}$ is p. Thus, pressure of $\ce{NH3_{(g)}}$ is 2p at equilibrium. Total pressure in this case is 3p.

Let 3p equal new pressure of $\ce{NH3_{(g)}}$.

Added pressure of $\ce{NH3_{(g)}}$ would shift the equilibrium to the left, so say we lose 2x pressure of $\ce{NH3_{(g)}}$ and x pressure of $\ce{CO2_{(g)}}$. New equilibrium pressures will be 3p-2x for $\ce{NH3_{(g)}}$ and p-x for $\ce{CO2_{(g)}}$. Thus, ratio of new to old is $\frac{4p-3x}{3p}$. You can solve for x in terms of p using $K_p$.

Solve for x and find the ratio, which should be $\frac{31}{27}$

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  • $\begingroup$ how do i solve further?please explain. multiplying 3p-2x and p-x and equating it to 4p^3/27 and solving for x is a crazy thing to do. please explain to me. $\endgroup$ – Jai Mahajan Aug 21 '14 at 17:43
  • $\begingroup$ what will be the initial pressure for co2 in the 2nd case wont it be p minus something? $\endgroup$ – Jai Mahajan Aug 21 '14 at 17:50
  • $\begingroup$ After equilibrium shifts to the left due to addition of NO2, the CO2 pressure will be p minus something or I denote as x $\endgroup$ – user2759975 Aug 21 '14 at 18:19

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