
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
226.5k+ 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
JEE Main Admit Card 2026: Releasing Soon – Download Hall Ticket at jeemain.nta.nic.in

JEE Main Marks Vs Percentile 2026: Calculate Percentile Based on Marks

AP EAPCET 2026 Exam Dates (OUT), Eligibility, Syllabus, Result, and Counselling

JEE Main 2026 Exam Pattern Released: Total Questions, and Marks, and Marking Scheme

300 Marks in JEE Main Percentile Rank IIT CSE Chances

290 Marks in JEE Main 2026 Percentile Rank and IIT Chances

Trending doubts
JEE Main 2026: City Intimation Slip and Exam Dates Released, Application Form Closed, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Thermodynamics Class 11 Chemistry Chapter 5 CBSE Notes - 2025-26

Hydrocarbons Class 11 Chemistry Chapter 9 CBSE Notes - 2025-26

Equilibrium Class 11 Chemistry Chapter 6 CBSE Notes - 2025-26

Organic Chemistry Some Basic Principles And Techniques Class 11 Chemistry Chapter 8 CBSE Notes - 2025-26

Redox Reaction Class 11 Chemistry Chapter 7 CBSE Notes - 2025-26

