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
Verified
122.4k+ 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.
Recently Updated Pages
The hybridization and shape of NH2 ion are a sp2 and class 11 chemistry JEE_Main
Total number of orbitals associated with the 3rd shell class 11 chemistry JEE_Main
Which of the following has the lowest boiling point class 11 chemistry JEE_Main
Which of the following compounds has zero dipole moment class 11 chemistry JEE_Main
Number of g of oxygen in 322 g Na2SO410H2O is Molwt class 11 chemistry JEE_Main
In the neutralization process of H3PO4 and NaOH the class 11 chemistry JEE_Main
Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility & More
JEE Main Login 2045: Step-by-Step Instructions and Details
JEE Main Chemistry Question Paper with Answer Keys and Solutions
JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking
JEE Main 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions
JEE Main Chemistry Exam Pattern 2025
Other Pages
NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reaction
NCERT Solutions for Class 11 Chemistry Chapter 5 Thermodynamics
NCERT Solutions for Class 11 Chemistry Chapter 8 Organic Chemistry
NCERT Solutions for Class 11 Chemistry Chapter 6 Equilibrium
NCERT Solutions for Class 11 Chemistry Chapter 9 Hydrocarbons
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs