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In an experiment the equilibrium constant for the reaction A+B\leftrightharpoons C+D is K when the initial concentration of A+B each is 0.1 mol L−1 Under the similar conditions in an another experiement if the initial concentration of A and B are taken 2 and 3 mol L−1 respectively then the value of equilibrium constant will be:?
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In an experiment the equilibrium constant for the reaction A+B\leftrig...
Equilibrium Constant and Reaction Stoichiometry
The equilibrium constant (K) is a fundamental concept in chemical equilibrium that quantifies the extent of a chemical reaction at equilibrium. It is defined as the ratio of the concentrations of the products to the concentrations of the reactants, each raised to the power of their stoichiometric coefficients.
The Reaction:
A + B ⇌ C + D
Initial Concentrations:
In the first experiment, the initial concentrations of A and B are both 0.1 mol L−1. Let's assume that at equilibrium, the concentrations of A, B, C, and D are represented by [A], [B], [C], and [D] respectively.
Expression for Equilibrium Constant:
Based on the stoichiometry of the reaction, the expression for the equilibrium constant is given by:
K = ([C] * [D]) / ([A] * [B])
Experiment 1:
Initial concentrations:
[A] = 0.1 mol L−1
[B] = 0.1 mol L−1
Experiment 2:
Initial concentrations:
[A] = 2 mol L−1
[B] = 3 mol L−1
Determining the Equilibrium Constant:
To find the equilibrium constant for the second experiment, we need to determine the concentrations of A, B, C, and D at equilibrium.
Using the Concentration Ratios:
The equilibrium concentrations can be determined by considering the ratio of the initial concentrations. In experiment 2, the initial concentration ratio of A to B is 2:3, which means that for every 2 moles of A, there are 3 moles of B.
Assuming Equilibrium Concentrations:
Let's assume that at equilibrium, the concentrations of A, B, C, and D are:
[A] = 2x
[B] = 3x
[C] = y
[D] = y
Applying the Stoichiometry:
Based on the stoichiometry of the reaction, the concentration of C and D can be related to A and B. Since the stoichiometric coefficients of C and D are equal to 1, the concentrations are directly proportional to the reactants' concentrations.
[C] = [A]
[D] = [B]
Substituting into the Equilibrium Expression:
Substituting the equilibrium concentrations into the equilibrium expression, we get:
K = ([C] * [D]) / ([A] * [B])
K = (y * y) / (2x * 3x)
K = y^2 / (6x^2)
Equating Initial and Equilibrium Concentrations:
To determine the values of x and y, we can equate the initial and equilibrium concentrations of A and B.
For A:
[A]initial = [A]equilibrium
0.1 mol L−1 = 2x
For B:
[B]initial = [B]equilibrium
0.1 mol L−1 = 3x
Solving these equations, we find:
x = 0.05 mol L−1
Calculating the Equilibrium Constant:
Substituting the values of x and y
The equilibrium constant (K) is a fundamental concept in chemical equilibrium that quantifies the extent of a chemical reaction at equilibrium. It is defined as the ratio of the concentrations of the products to the concentrations of the reactants, each raised to the power of their stoichiometric coefficients.
The Reaction:
A + B ⇌ C + D
Initial Concentrations:
In the first experiment, the initial concentrations of A and B are both 0.1 mol L−1. Let's assume that at equilibrium, the concentrations of A, B, C, and D are represented by [A], [B], [C], and [D] respectively.
Expression for Equilibrium Constant:
Based on the stoichiometry of the reaction, the expression for the equilibrium constant is given by:
K = ([C] * [D]) / ([A] * [B])
Experiment 1:
Initial concentrations:
[A] = 0.1 mol L−1
[B] = 0.1 mol L−1
Experiment 2:
Initial concentrations:
[A] = 2 mol L−1
[B] = 3 mol L−1
Determining the Equilibrium Constant:
To find the equilibrium constant for the second experiment, we need to determine the concentrations of A, B, C, and D at equilibrium.
Using the Concentration Ratios:
The equilibrium concentrations can be determined by considering the ratio of the initial concentrations. In experiment 2, the initial concentration ratio of A to B is 2:3, which means that for every 2 moles of A, there are 3 moles of B.
Assuming Equilibrium Concentrations:
Let's assume that at equilibrium, the concentrations of A, B, C, and D are:
[A] = 2x
[B] = 3x
[C] = y
[D] = y
Applying the Stoichiometry:
Based on the stoichiometry of the reaction, the concentration of C and D can be related to A and B. Since the stoichiometric coefficients of C and D are equal to 1, the concentrations are directly proportional to the reactants' concentrations.
[C] = [A]
[D] = [B]
Substituting into the Equilibrium Expression:
Substituting the equilibrium concentrations into the equilibrium expression, we get:
K = ([C] * [D]) / ([A] * [B])
K = (y * y) / (2x * 3x)
K = y^2 / (6x^2)
Equating Initial and Equilibrium Concentrations:
To determine the values of x and y, we can equate the initial and equilibrium concentrations of A and B.
For A:
[A]initial = [A]equilibrium
0.1 mol L−1 = 2x
For B:
[B]initial = [B]equilibrium
0.1 mol L−1 = 3x
Solving these equations, we find:
x = 0.05 mol L−1
Calculating the Equilibrium Constant:
Substituting the values of x and y
Community Answer
In an experiment the equilibrium constant for the reaction A+B\leftrig...
Equilibrium constant doesn't depend upon concentration it only depend upon temperature so the equilibrium constant remain same so the ans is K
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In an experiment the equilibrium constant for the reaction A+B\leftrightharpoons C+D is K when the initial concentration of A+B each is 0.1 mol L−1 Under the similar conditions in an another experiement if the initial concentration of A and B are taken 2 and 3 mol L−1 respectively then the value of equilibrium constant will be:?
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In an experiment the equilibrium constant for the reaction A+B\leftrightharpoons C+D is K when the initial concentration of A+B each is 0.1 mol L−1 Under the similar conditions in an another experiement if the initial concentration of A and B are taken 2 and 3 mol L−1 respectively then the value of equilibrium constant will be:? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about In an experiment the equilibrium constant for the reaction A+B\leftrightharpoons C+D is K when the initial concentration of A+B each is 0.1 mol L−1 Under the similar conditions in an another experiement if the initial concentration of A and B are taken 2 and 3 mol L−1 respectively then the value of equilibrium constant will be:? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In an experiment the equilibrium constant for the reaction A+B\leftrightharpoons C+D is K when the initial concentration of A+B each is 0.1 mol L−1 Under the similar conditions in an another experiement if the initial concentration of A and B are taken 2 and 3 mol L−1 respectively then the value of equilibrium constant will be:?.
In an experiment the equilibrium constant for the reaction A+B\leftrightharpoons C+D is K when the initial concentration of A+B each is 0.1 mol L−1 Under the similar conditions in an another experiement if the initial concentration of A and B are taken 2 and 3 mol L−1 respectively then the value of equilibrium constant will be:? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about In an experiment the equilibrium constant for the reaction A+B\leftrightharpoons C+D is K when the initial concentration of A+B each is 0.1 mol L−1 Under the similar conditions in an another experiement if the initial concentration of A and B are taken 2 and 3 mol L−1 respectively then the value of equilibrium constant will be:? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In an experiment the equilibrium constant for the reaction A+B\leftrightharpoons C+D is K when the initial concentration of A+B each is 0.1 mol L−1 Under the similar conditions in an another experiement if the initial concentration of A and B are taken 2 and 3 mol L−1 respectively then the value of equilibrium constant will be:?.
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In an experiment the equilibrium constant for the reaction A+B\leftrightharpoons C+D is K when the initial concentration of A+B each is 0.1 mol L−1 Under the similar conditions in an another experiement if the initial concentration of A and B are taken 2 and 3 mol L−1 respectively then the value of equilibrium constant will be:?, a detailed solution for In an experiment the equilibrium constant for the reaction A+B\leftrightharpoons C+D is K when the initial concentration of A+B each is 0.1 mol L−1 Under the similar conditions in an another experiement if the initial concentration of A and B are taken 2 and 3 mol L−1 respectively then the value of equilibrium constant will be:? has been provided alongside types of In an experiment the equilibrium constant for the reaction A+B\leftrightharpoons C+D is K when the initial concentration of A+B each is 0.1 mol L−1 Under the similar conditions in an another experiement if the initial concentration of A and B are taken 2 and 3 mol L−1 respectively then the value of equilibrium constant will be:? theory, EduRev gives you an
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