Solid ammonium carbamate dissociates as: \[N{{H}_{2}}COON{{H}_{4}}\leftrightarrows2N{{H}_{3}}+C{{O}_{2}}\]. 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 \[N{{H}_{3}}\] at new equilibrium now equals the original total pressure. Calculate the ratio of total pressure at the new equilibrium to that of original total pressure.
(A) 13/27
(B) 12/34
(C) 14/37
(D)15/27
(E) 31/27
Answer
270.6k+ views
Hint: Equilibrium constant $Kp$ is used to express partial pressures of reactant and products. It is the ratio of the product of partial pressure of products to the product of partial pressure of reactants, each raised to their stoichiometric coefficients in the balanced equation.
Complete step by step solution:
-Sodium Ammonium Carbamate dissociates as \[N{{H}_{2}}COON{{H}_{4}}2N{{H}_{3}}+C{{O}_{2}}\]
Let initial partial pressure of Carbon dioxide be P and partial pressure of ammonia becomes 2P.
The equilibrium constant $Kp$ is used to express partial pressures of reactant and products. It is the ratio of the product of partial pressure of products to the product of partial pressure of reactants power raised to their stoichiometric coefficients in the balanced equation.
-pressure is exerted by gases so as ammonium carbamate exists in solid-state, it is not taken into account in the calculation of equilibrium constant.
\[Kp={{({{P}_{N{{H}_{3}}}})}^{2}}({{p}_{C{{O}_{2}}}})\]
\[Kp={{(2P)}^{2}}(P)\] (i)
In the second case, the initial total pressure is equal to 3P. Let us consider the partial pressure of carbon dioxide as ${{P}_{1}}$.
So equilibrium constant is equal to $Kp={{(3P)}^{2}}({{P}_{1}})$ (ii)
From both equations,
$\begin{align}
& {{(2P)}^{2}}(P)={{(3P)}^{2}}({{P}_{1}}) \\
& {{P}_{1}}=\dfrac{4P}{9} \\
\end{align}$
$\dfrac{{{P}_{T}}(new)}{{{P}_{T}}(old)}=\dfrac{3P+{{P}_{1}}}{3P}=\dfrac{3P+\dfrac{4P}{9}}{3P}=\dfrac{31}{27}$
Hence, the correct answer is (E).
Note: Equilibrium constant expresses the relationship between products and reactant when equilibrium is attained. It is the ratio of the product of partial pressure of products to the product of partial pressure of reactants, each raised to their stoichiometric coefficients in balanced equations. Solid and liquids are not taken into consideration as they do not exert pressure.
Complete step by step solution:
-Sodium Ammonium Carbamate dissociates as \[N{{H}_{2}}COON{{H}_{4}}2N{{H}_{3}}+C{{O}_{2}}\]
Let initial partial pressure of Carbon dioxide be P and partial pressure of ammonia becomes 2P.
The equilibrium constant $Kp$ is used to express partial pressures of reactant and products. It is the ratio of the product of partial pressure of products to the product of partial pressure of reactants power raised to their stoichiometric coefficients in the balanced equation.
-pressure is exerted by gases so as ammonium carbamate exists in solid-state, it is not taken into account in the calculation of equilibrium constant.
\[Kp={{({{P}_{N{{H}_{3}}}})}^{2}}({{p}_{C{{O}_{2}}}})\]
\[Kp={{(2P)}^{2}}(P)\] (i)
In the second case, the initial total pressure is equal to 3P. Let us consider the partial pressure of carbon dioxide as ${{P}_{1}}$.
So equilibrium constant is equal to $Kp={{(3P)}^{2}}({{P}_{1}})$ (ii)
From both equations,
$\begin{align}
& {{(2P)}^{2}}(P)={{(3P)}^{2}}({{P}_{1}}) \\
& {{P}_{1}}=\dfrac{4P}{9} \\
\end{align}$
$\dfrac{{{P}_{T}}(new)}{{{P}_{T}}(old)}=\dfrac{3P+{{P}_{1}}}{3P}=\dfrac{3P+\dfrac{4P}{9}}{3P}=\dfrac{31}{27}$
Hence, the correct answer is (E).
Note: Equilibrium constant expresses the relationship between products and reactant when equilibrium is attained. It is the ratio of the product of partial pressure of products to the product of partial pressure of reactants, each raised to their stoichiometric coefficients in balanced equations. Solid and liquids are not taken into consideration as they do not exert pressure.
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