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. The ratio of total pressure now to the original pressure is :
a.) $\dfrac{{27}}{{31}}$
b.) $\dfrac{{31}}{{27}}$
c.) $\dfrac{4}{9}$
d.) $\dfrac{{3.1}}{{27}}$

Answer 1

Hint: The total pressure of a system is the sum of partial pressure of all components. By assuming the total pressure to be P, we can find the partial pressure of ammonia, carbon dioxide. And from here, we can find the value of ${K_p}$. Then using, ${K_p}$; we will find the total new pressure and then divide the new with old to get the ratio.

Complete step by step answer :
The reaction given is as -
$N{H_2}COON{H_{4(s)}} \rightleftharpoons 2N{H_{3(g)}} + C{O_{2(g)}}$
The ammonium carbamate is in solid state. So, it will have zero partial pressure.
Thus, if total pressure is ‘P’, then
Partial pressure of ammonia = $\dfrac{2}{3}P$
Partial pressure of carbon dioxide = $\dfrac{1}{3}P$
So, the ${K_p}$=${({P_{N{H_3}}})^2} \times ({P_{C{O_2}}})$
Filling the values, we get -
${K_p}$=${\left( {\dfrac{2}{3}P} \right)^2} \times \left( {\dfrac{1}{2}P} \right)$
${K_p}$=$\dfrac{4}{{27}}{P^3}$
This was the initial state.
After this, it is said in question that at equilibrium; some ammonia is added such that the partial pressure of ammonia = P
So, at this state; we will find partial pressure of carbon dioxide. This can be found as -
${K_p}$=${({P_{N{H_3}}})^2} \times ({P_{C{O_2}}})$
Filling the values, we get -
$\dfrac{4}{{27}}{P^3}$=${\left( P \right)^2} \times \left( {{P_{C{O_2}}}} \right)$
$\left( {{P_{C{O_2}}}} \right)$=$\dfrac{4}{{27}}P$
So, the new total pressure can be found by adding partial pressures of two gases as -
Total pressure (now) = Partial pressure of ammonia + Partial pressure old carbon dioxide
Total pressure (now) = P + $\dfrac{4}{{27}}P$
Total pressure (now) = $\dfrac{{31}}{{27}}P$
So, now the ratio of two total pressures can be given as -
\[\dfrac{{Total{\text{ }}pressure{\text{ }}\left( {new} \right)}}{{Total{\text{ }}pressure{\text{ }}\left( {old} \right)}} = \dfrac{{31 \times P}}{{27 \times P}}\]
\[\dfrac{{Total{\text{ }}pressure{\text{ }}\left( {new} \right)}}{{Total{\text{ }}pressure{\text{ }}\left( {old} \right)}} = \dfrac{{31}}{{27}}\]

Thus, the option b.) is the correct answer.

Note: It must be noted that the equilibrium is a state where the rate of forward reaction is equal to rate of backward reaction. In this state, the amount of products formed from reactants is equal to the amount of reactants formed back from products. The overall concentration of reactants and products is constant.