Question

Solid ammonium carbamate dissociates as:
${\text{N}}{{\text{H}}_{\text{2}}}{\text{COON}}{{\text{H}}_{\text{4}}}\left( {\text{s}} \right) \rightleftharpoons {\text{2N}}{{\text{H}}_{\text{3}}}\left( {\text{g}} \right){\text{ + C}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right)$
In a closed vessel, solid ammonium carbamate is in equilibrium with its dissociation products. At equilibrium, ammonia is added such that the partial pressure of ${\text{N}}{{\text{H}}_{\text{3}}}$ at new equilibrium now equals the original total pressure. Calculate the ratio of total pressure at new equilibrium to that of original total pressure.

A.$\dfrac{{{\text{31}}}}{{{\text{27}}}}$
B.$\dfrac{{{\text{17}}}}{{\text{7}}}$
C.$\dfrac{{{\text{49}}}}{{{\text{29}}}}$
D.None of these

Answer 1

Hint: The ratio of the product of partial pressures of products to the product of partial pressures of reactants, each raised to the power of their stoichiometric coefficients in the balanced chemical equation is known as the equilibrium constant. The equilibrium constant is denoted by ${K_p}$.

Step by step answer: The dissociation of solid sodium ammonium carbamate is as follows:

${\text{N}}{{\text{H}}_{\text{2}}}{\text{COON}}{{\text{H}}_{\text{4}}}\left( {\text{s}} \right) \rightleftharpoons {\text{2N}}{{\text{H}}_{\text{3}}}\left( {\text{g}} \right){\text{ + C}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right)$
Let the initial partial pressure of ${\text{C}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right)$ be $P$.
From the reaction stoichiometry,
${\text{1 mol C}}{{\text{O}}_2} = 2{\text{ mol N}}{{\text{H}}_3}$
Thus, the initial partial pressure of ${\text{N}}{{\text{H}}_{\text{3}}}\left( {\text{g}} \right)$ will be ${\text{2}}P$.
The equilibrium constant in terms of partial pressures is the ratio of the product of partial pressures of products to the product of partial pressures of reactants, each raised to the power of their stoichiometric coefficients in the balanced chemical equation.
Thus,
${K_p} = \dfrac{{{{\left( {{P_{{\text{N}}{{\text{H}}_{\text{3}}}}}} \right)}^2}\left( {{P_{{\text{C}}{{\text{O}}_2}}}} \right)}}{{\left( {{P_{{\text{N}}{{\text{H}}_{\text{2}}}{\text{COON}}{{\text{H}}_{\text{4}}}}}} \right)}}$
The solid ammonium carbamate,

${\text{N}}{{\text{H}}_{\text{2}}}{\text{COON}}{{\text{H}}_{\text{4}}}$ is in solid state. Thus, it does not exert any pressure.
Thus,
${K_p} = {\left( {{P_{{\text{N}}{{\text{H}}_{\text{3}}}}}} \right)^2}\left( {{P_{{\text{C}}{{\text{O}}_2}}}} \right)$
Substitute ${\text{2}}P$ for the initial partial pressure of ${\text{N}}{{\text{H}}_{\text{3}}}$, $P$ for the initial partial pressure of ${\text{C}}{{\text{O}}_{\text{2}}}$. Thus,
${K_p} = {\left( {2P} \right)^2}\left( P \right)$ …… (1)
The original total pressure is the summation of initial partial pressure of ${\text{N}}{{\text{H}}_{\text{3}}}$ and initial partial pressure of ${\text{C}}{{\text{O}}_{\text{2}}}$.
Thus,
${\text{Original total pressure}} = 2P + P$
${\text{Original total pressure}} = 3P$
Thus, the original total pressure is ${\text{3}}P$.
Calculate the partial pressure of ${\text{C}}{{\text{O}}_{\text{2}}}$ at new equilibrium as follows:
At new equilibrium, the partial pressure of ${\text{N}}{{\text{H}}_{\text{3}}}$ is equal to the original total pressure. Thus, the partial pressure of ${\text{N}}{{\text{H}}_{\text{3}}}$ is ${\text{3}}P$. Thus,
${K_p} = {\left( {{P_{{\text{N}}{{\text{H}}_{\text{3}}}}}} \right)^2}\left( {{P_{{\text{C}}{{\text{O}}_2}}}} \right)$
Substitute ${\text{3}}P$ for the partial pressure of ${\text{N}}{{\text{H}}_{\text{3}}}$. Thus,
${K_p} = {\left( {3P} \right)^2}\left( {{P_{{\text{C}}{{\text{O}}_2}}}} \right)$
Equate equation (1) and equation (2). Thus,
${\left( {2P} \right)^2}\left( P \right) = {\left( {3P} \right)^2}\left( {{P_{{\text{C}}{{\text{O}}_2}}}} \right)$
$\Rightarrow 4{P^3} = 9{P^2} \times {P_{{\text{C}}{{\text{O}}_2}}}$
$\Rightarrow {P_{{\text{C}}{{\text{O}}_2}}} = \dfrac{{4{P^3}}}{{9{P^2}}}$
$\Rightarrow {P_{{\text{C}}{{\text{O}}_2}}} = \dfrac{{4P}}{9}$
Thus, the partial pressure of ${\text{C}}{{\text{O}}_{\text{2}}}$ at the new equilibrium is $\dfrac{{4P}}{9}$.
Calculate the ratio of total pressure at new equilibrium to that of original total pressure as follows:
$\dfrac{{{\text{Total pressure at new equilibrium}}}}{{{\text{Original total pressure}}}} = \dfrac{{3P + \dfrac{{4P}}{9}}}{{3P}}$
$\Rightarrow \dfrac{{{\text{Total pressure at new equilibrium}}}}{{{\text{Original total pressure}}}} = \dfrac{{27P + 4P}}{{27P}}$
$\Rightarrow \dfrac{{{\text{Total pressure at new equilibrium}}}}{{{\text{Original total pressure}}}} = \dfrac{{31P}}{{27P}}$
$\Rightarrow \dfrac{{{\text{Total pressure at new equilibrium}}}}{{{\text{Original total pressure}}}} = \dfrac{{31}}{{27}}$
Thus, the ratio of total pressure at new equilibrium to that of original total pressure is $\dfrac{{31}}{{27}}$.

Thus, the correct option is (A) $\dfrac{{31}}{{27}}$.

Note: The relationship between the products and reactants when equilibrium is attained is expressed by the equilibrium constant. While writing the equation for the equilibrium constant, reactants and products in solid and liquid state are not taken into consideration.