Question

Ammonium carbonate 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 equal to the original total pressure. The ratio of total pressure now to the original pressure is:
a.) $\dfrac{27}{31}$
b.) $\dfrac{31}{27}$
c.) $\dfrac{4}{9}$
d.) $\dfrac{5}{9}$

Answer 1

To solve this, question the most important step is to write the balanced chemical reaction involved. In the question the equation given is already balanced but if this equation was not balanced then first we should balance the chemical reaction.

Complete Solution :
Given in the question:
Ammonium carbonate dissociates as:
$N{{H}_{2}}COON{{H}_{4}}(s)\rightleftharpoons 2N{{H}_{3}}(g)+C{{O}_{2}}(g)$

And, ammonia is added in a closed vessel containing ammonium carbonate in equilibrium, such that the partial pressure of $N{{H}_{3}}$ now equal to the original total pressure
According to the given equation: the initial pressure of ammonia and carbon dioxide will be:
$N{{H}_{2}}COON{{H}_{4}}(s)\rightleftharpoons 2N{{H}_{3}}(g)+C{{O}_{2}}(g)$

| 2P| P| Initial pressure
---|---|---|---

The value of equilibrium constant in terms of pressure will be:
$\begin{aligned} & {{K}_{p}}={{({{p}_{N{{H}_{3}}}})}^{2}}({{p}_{C{{O}_{2}}}}) \\\ & {{K}_{p}}={{(2P)}^{2}}(P) \\\ \end{aligned}$
Now in the second case, the total pressure will be:
$N{{H}_{2}}COON{{H}_{4}}(s)\rightleftharpoons 2N{{H}_{3}}(g)+C{{O}_{2}}(g)$

| 3P| ${{P}^{i}}$ | Initial pressure
---|---|---|---

The value of equilibrium constant in terms of pressure will be:
$\begin{aligned} & {{K}_{p}}={{({{p}_{N{{H}_{3}}}})}^{2}}({{p}_{C{{O}_{2}}}}) \\\ & {{K}_{p}}={{(3P)}^{2}}({{P}^{i}}) \\\ \end{aligned}$
According to the question we have to find the ratio of total pressure now to the original pressure is:

& {{(2P)}^{2}}(P)={{(3P)}^{2}}({{P}^{i}}) \\\ & {{P}^{i}}=\dfrac{4P}{9} \\\ & \dfrac{{{P}_{T}}(new)}{{{P}_{T}}(old)}=\dfrac{3P+{{P}^{i}}}{3P}=\dfrac{3P+\dfrac{4P}{9}}{3P}=\dfrac{31}{27} \\\ \end{aligned}$$ Hence, the correct answer is the ratio of total pressure now to the original pressure is $\dfrac{31}{27}$. **So, the correct answer is “Option B”.** **Note:** ${{K}_{p}}$ is defined as the equilibrium constant in terms of partial pressure. It is used to express the relationship between the product pressure and the reactant pressure. An equilibrium constant in terms of concentration is defined as the product of concentration of product to the product of concentration of reactant each raise to the power of their respective stoichiometric coefficient.