Question

Ammonium carbamate dissociates as:
     \[N{{H}_{2}}COON{{H}_{4}}(s)\rightleftharpoons 2N{{H}_{3}}(g)+C{{O}_{2}}(g)\]
In a closed vessel containing ammonium carbamate in equilibrium, ammonia is added such that the partial pressure of $N{{H}_{3}}$ now equals the original total pressure. Calculate the ratio of final pressure to the original total pressure.
[A] $\dfrac{31}{27}$
[B] 9
[C] $\dfrac{4}{9}$
[D] $\dfrac{2}{9}$

Answer 1

Hint: To solve this question, firstly find out the ${{K}_{p}}$ for the given reaction assuming the original pressure to be P. Use this to find the value of partial pressure of carbon dioxide after addition of ammonia at equilibrium. Use the partial pressure of the two products after equilibrium to find the final pressure and then find the required ratio.

Complete step by step answer:
The reaction given to us is:
\[N{{H}_{2}}COON{{H}_{4}}(s)\rightleftharpoons 2N{{H}_{3}}(g)+C{{O}_{2}}(g)\]

The reactant given is ammonium carbamate and it gives us ammonia and carbon dioxide gas. We know that the partial pressure exhibited by a solid is negligible therefore we will consider only the partial pressure of ammonia gas and carbon dioxide gas and we can write the total pressure in terms of the partial pressure of these gases.
Let us assume that the total pressure is ‘P’. We have a total 3 components on the product side out of which 2 is ammonia and 1 is carbon dioxide. Therefore, we can write that-
Partial pressure of ammonia, $p\left( N{{H}_{3}} \right)=\dfrac{2}{3}P$
And similarly, partial pressure of carbon dioxide, $p\left( C{{O}_{2}} \right)=\dfrac{1}{3}P$

Therefore, we can write the equilibrium constant in terms of partial pressure, ${{K}_{p}}$ as -
     \[{{K}_{p}}=p{{\left( N{{H}_{3}} \right)}^{2}}\times p\left( C{{O}_{2}} \right)={{\left( \dfrac{2}{3}P \right)}^{2}}\times \left( \dfrac{1}{3}P \right)=\dfrac{4}{27}{{P}^{3}}\]
Now, at equilibrium we have added ammonia and the partial pressure of ammonia becomes equal to the original total pressure.

We have assumed the original total pressure to be P.
Therefore, now partial pressure of ammonia, $p\left( N{{H}_{3}} \right)=P$
Hence the equilibrium constant will become \[{{K}_{p}}={{P}^{2}}\times p\left( C{{O}_{2}} \right)\]
Or, we can write that- \[\dfrac{4}{27}{{P}^{3}}={{P}^{2}}\times p\left( C{{O}_{2}} \right)\]
Solving this we can find the partial pressure of carbon dioxide which will be \[\dfrac{4}{27}P\].
Thus, now the total pressure will become, ${{P}_{final}}={{p}_{N{{H}_{3}}}}+{{p}_{C{{O}_{2}}}}$
Putting the values we will get- ${{P}_{final}}=P+\dfrac{4}{27}P=\dfrac{31}{27}P$
Now we know the original total pressure and the change in pressure after addition of ammonia at equilibrium so we can find out the ratio.
     \[\dfrac{{{P}_{final}}}{{{P}_{initial}}}=\dfrac{\dfrac{31}{27}P}{P}=\dfrac{31}{27}\]

We can see from the above discussion that the ratio of the pressure after addition of ammonia i.e. the final pressure to that of the original pressure is $\dfrac{31}{27}$.
Therefore, the correct answer is $\dfrac{31}{27}$.

So, the correct answer is “Option A”.

Note: Partial pressure is the force exerted by a gas. It is basically the force exerted by individual gases in a mixture. We know that gases are most compressible followed by liquids and then solids. Therefore, changing the pressure does not affect a solid as much it affects a gas and hence we neglect it like we did in the question above.
Also the pure solids do not affect the equilibrium constant as their concentration stays the same throughout the reaction.