Question

The dissociation of ammonium carbamate occurs as follows -

$$ \mathrm{NH_2COONH_4}(s) \Leftrightarrow 2\mathrm{NH_3}(g) + \mathrm{CO_2}(g) $$

Ammonia is added to ammonium carbamate kept at equilibrium in a vessel so that the partial pressure of ammonia becomes equal to the total initial pressure. The ratio of the partial pressure of $\mathrm{CO}_2$ in the new equilibrium to the partial pressure of $\mathrm{CO}_2$ at the beginning is:

A 4
B 9

C $\frac{4}{9}$
D $\frac{2}{9}$

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Answer

The correct answer is:

C $$ \mathrm{NH}{2} \mathrm{COONH}{4}(s) \Leftrightarrow 2 \mathrm{NH}{3}(g) + \mathrm{CO}{2}(g) $$

Let's assume that the partial pressure of $\mathrm{CO}_{2}$ (carbon dioxide) at equilibrium is $P$. Thus, the partial pressure of $\mathrm{NH}_{3}$ (ammonia) would be $2P$.

At equilibrium: $$ P_{\mathrm{NH}3} = 2P $$ and the total pressure would be: $$ P{\mathrm{total}} = 3P $$

The equilibrium constant for the reaction in terms of pressure ($K_p$) can be expressed as: $$ K_p = (2P)^2 \times P = 4P^3 $$

Now, if additional $\mathrm{NH}_3$ is added such that the equilibrium partial pressure of $\mathrm{NH}_3$ becomes $3P$, we can set up the equation as follows:

$$ K_p = 4P^3 = (P_{\mathrm{NH}3})^2 \times P{\mathrm{CO}_2} $$

Given the new partial pressure of $\mathrm{NH}3$: $$ P{\mathrm{NH}_3} = 3P $$

So, substituting this back into the $K_p$ equation: $$ K_p = (3P)^2 \times P_{\mathrm{CO}2} = 9P^2 \times P{\mathrm{CO}_2} $$

Since $K_p = 4P^3$: $$ 9P^2 \times P_{\mathrm{CO}_2} = 4P^3 $$

Solving for $P_{\mathrm{CO}2}$: $$ P{\mathrm{CO}_2} = \frac{4P^3}{9P^2} = \frac{4P}{9} $$

Thus, the ratio of the initial partial pressure of $\mathrm{CO}_2$ to the new partial pressure of $\mathrm{CO}2$ is: $$ \frac{P{\mathrm{CO}2(new)}}{P{\mathrm{CO}_2(initial)}} = \frac{\frac{4P}{9}}{P} = \frac{4}{9} $$

Therefore, the final answer is:

C. $\frac{4}{9}$


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