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Can you explain the answer of this question below:
In a reversible chemical reaction having two reactants in equilibrium, if the concentration of the reactants are doubled then the equilibrium constant will
- A:also be doubled
- B:be halved
- C:becomes one fourth
- D:remains the same
The answer is d.
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Can you explain the answer of this question below:In a reversible chem...
Change in concentration, pressure, catalyst, inert gas addition, etc. have NO effect on Equilibrium CONSTANT. Only temperature does. Temperature, concentration, pressure, catalyst, inert gas addition lead to a shift in Equilibrium position.
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Can you explain the answer of this question below:In a reversible chem...
Explanation:
The equilibrium constant is a ratio of the product of the concentrations of the products raised to their stoichiometric coefficients to the product of the concentrations of the reactants raised to their stoichiometric coefficients.
Kc = [C]^c [D]^d / [A]^a [B]^b
Where, A, B are reactants and C, D are products and a, b, c, d are their stoichiometric coefficients.
If the concentration of the reactants are doubled, then the new equilibrium concentrations will be:
[A] = 2[A]0
[B] = 2[B]0
Substituting these values in the equilibrium constant expression, we get:
K'c = [C]^c [D]^d / (2[A]0)^a (2[B]0)^b
K'c = [C]^c [D]^d / 2^a [A]0^a 2^b [B]0^b
K'c = 1/2^a [C]^c [D]^d / [A]0^a [B]0^b
As we can see, the equilibrium constant expression remains the same even after doubling the concentration of the reactants. Therefore, the equilibrium constant remains the same.
Hence, the correct answer is option D.
The equilibrium constant is a ratio of the product of the concentrations of the products raised to their stoichiometric coefficients to the product of the concentrations of the reactants raised to their stoichiometric coefficients.
Kc = [C]^c [D]^d / [A]^a [B]^b
Where, A, B are reactants and C, D are products and a, b, c, d are their stoichiometric coefficients.
If the concentration of the reactants are doubled, then the new equilibrium concentrations will be:
[A] = 2[A]0
[B] = 2[B]0
Substituting these values in the equilibrium constant expression, we get:
K'c = [C]^c [D]^d / (2[A]0)^a (2[B]0)^b
K'c = [C]^c [D]^d / 2^a [A]0^a 2^b [B]0^b
K'c = 1/2^a [C]^c [D]^d / [A]0^a [B]0^b
As we can see, the equilibrium constant expression remains the same even after doubling the concentration of the reactants. Therefore, the equilibrium constant remains the same.
Hence, the correct answer is option D.
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Can you explain the answer of this question below:In a reversible chemical reaction having two reactants in equilibrium, if the concentration of the reactants are doubled then the equilibrium constant willA:also be doubledB:be halvedC:becomes one fourthD:remains the sameThe answer is d.
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Can you explain the answer of this question below:In a reversible chemical reaction having two reactants in equilibrium, if the concentration of the reactants are doubled then the equilibrium constant willA:also be doubledB:be halvedC:becomes one fourthD:remains the sameThe answer is d. for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Can you explain the answer of this question below:In a reversible chemical reaction having two reactants in equilibrium, if the concentration of the reactants are doubled then the equilibrium constant willA:also be doubledB:be halvedC:becomes one fourthD:remains the sameThe answer is d. covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Can you explain the answer of this question below:In a reversible chemical reaction having two reactants in equilibrium, if the concentration of the reactants are doubled then the equilibrium constant willA:also be doubledB:be halvedC:becomes one fourthD:remains the sameThe answer is d..
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