
For the equilibrium $N_2 + 3H_2\rightleftharpoons 2NH_3$ at 1000K is $2.37\times {{10}^{-3}}$ . If at equilibrium $\left[ {{N}_{2}} \right]=2M$, $\left[ {{H}_{2}} \right]=3M$, the concentration of $N{{H}_{3}}$ is:
(A) 0.00358 M
(B) 0.0358 M
(C) 0.358 M
(D) 3.58 M
Answer
164.7k+ views
Hint: An equilibrium constant (K) is the relationship between the concentration of reactants and products present at equilibrium in a reversible chemical process at a given temperature. It is represented as the ratio of the concentration of products and reactants.
Formula Used: For the reaction: $A + 3B \rightleftharpoons 2C$
The equilibrium constant is given by $K=\frac{{{[C]}^{2}}}{[A]{{[B]}^{3}}}$
In this question, the value of the equilibrium constant and the concentration of the reactants are given. We have to find the concentration of the product.
Complete Step by Step Answer:
The given equation is:
$N_2 + 3H_2\rightleftharpoons 2NH_3$
The given equilibrium constant for this reaction is $K=2.37\times {{10}^{-3}}$
The product and reactant concentrations are,$\left[ {{N}_{2}} \right]=2M$, $\left[ {{H}_{2}} \right]=3M$
The equilibrium constant is given by $K=\frac{{{[N{{H}_{3}}]}^{2}}}{[{{N}_{2}}]{{[{{H}_{2}}]}^{3}}}$
$2.37\times {{10}^{-3}}=\frac{{{[N{{H}_{3}}]}^{2}}}{[2]{{[3]}^{3}}}$
${{\left[ N{{H}_{3}} \right]}^{2}}=2.37\times {{10}^{-3}}\times 2\times 27$
${{[N{{H}_{3}}]}^{2}}=12.79\times {{10}^{-2}}$
$[N{{H}_{3}}]=\sqrt{12.79\times {{10}^{-2}}}$
$[N{{H}_{3}}]=0.358M$
Thus, the concentration of $N{{H}_{3}}$ is 0.358 M.
Correct Option: (C) 0.358 M.
Additional Information: Chemical equilibrium is a state in which the rates of both the forward and backward reactions are equal and the concentration of both the reactants and products is constant. At equilibrium, there is no net change in the number of moles, although conversion from reactants to products or products to reactants is still occurring. According to the law of mass action, the value of the equilibrium constant, Kc, is constant at a constant temperature. The concentrations of reactants and products may vary, but the value for Kc remains the same.
Note: The equilibrium constant should always be calculated after balancing the equation. The stoichiometric coefficients of each of the reactants or products should be carefully written as the powers of concentration of each substance.
Formula Used: For the reaction: $A + 3B \rightleftharpoons 2C$
The equilibrium constant is given by $K=\frac{{{[C]}^{2}}}{[A]{{[B]}^{3}}}$
In this question, the value of the equilibrium constant and the concentration of the reactants are given. We have to find the concentration of the product.
Complete Step by Step Answer:
The given equation is:
$N_2 + 3H_2\rightleftharpoons 2NH_3$
The given equilibrium constant for this reaction is $K=2.37\times {{10}^{-3}}$
The product and reactant concentrations are,$\left[ {{N}_{2}} \right]=2M$, $\left[ {{H}_{2}} \right]=3M$
The equilibrium constant is given by $K=\frac{{{[N{{H}_{3}}]}^{2}}}{[{{N}_{2}}]{{[{{H}_{2}}]}^{3}}}$
$2.37\times {{10}^{-3}}=\frac{{{[N{{H}_{3}}]}^{2}}}{[2]{{[3]}^{3}}}$
${{\left[ N{{H}_{3}} \right]}^{2}}=2.37\times {{10}^{-3}}\times 2\times 27$
${{[N{{H}_{3}}]}^{2}}=12.79\times {{10}^{-2}}$
$[N{{H}_{3}}]=\sqrt{12.79\times {{10}^{-2}}}$
$[N{{H}_{3}}]=0.358M$
Thus, the concentration of $N{{H}_{3}}$ is 0.358 M.
Correct Option: (C) 0.358 M.
Additional Information: Chemical equilibrium is a state in which the rates of both the forward and backward reactions are equal and the concentration of both the reactants and products is constant. At equilibrium, there is no net change in the number of moles, although conversion from reactants to products or products to reactants is still occurring. According to the law of mass action, the value of the equilibrium constant, Kc, is constant at a constant temperature. The concentrations of reactants and products may vary, but the value for Kc remains the same.
Note: The equilibrium constant should always be calculated after balancing the equation. The stoichiometric coefficients of each of the reactants or products should be carefully written as the powers of concentration of each substance.
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

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

NCERT Solutions for Class 11 Chemistry In Hindi Chapter 1 Some Basic Concepts of Chemistry

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reaction

Instantaneous Velocity - Formula based Examples for JEE

Thermodynamics Class 11 Notes: CBSE Chapter 5
