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In a chemical equilibrium the rate constant of the backward reaction is 7.5 x 10^-4 and the equilibrium constant is 1.5. So, the rate of the forward reaction is A) 2 x 10^-3 B) 15 x 10^-4 C) 1.125 x 10^-3 D) 9.0 x 10^-4?
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In a chemical equilibrium the rate constant of the backward reaction i...
Answer:
Understanding the problem:
The given information is as follows:
- Rate constant of the backward reaction, kb = 7.5 x 10^-4
- Equilibrium constant, Kc = 1.5
We need to find the rate of the forward reaction, kf.
Using the equilibrium constant:
The expression for the equilibrium constant is given as:
Kc = [products] / [reactants]
For the given reaction, let's assume the balanced equation as:
aA + bB ⇌ cC + dD
Then, the equilibrium constant expression becomes:
Kc = ([C]^c [D]^d) / ([A]^a [B]^b)
Substituting the given values:
1.5 = ([C]^c [D]^d) / ([A]^a [B]^b)
We can assume any arbitrary value for the stoichiometric coefficients a, b, c, and d, as long as they are in the proper ratio.
Using the rate constants:
The rate law for the forward reaction can be written as:
ratef = kf [A]^a [B]^b
Similarly, the rate law for the backward reaction can be written as:
rateb = kb [C]^c [D]^d
At equilibrium, the rates of the forward and backward reactions are equal, i.e., ratef = rateb. Therefore, we can write:
kf [A]^a [B]^b = kb [C]^c [D]^d
Calculating the rate of the forward reaction:
Using the equilibrium constant expression, we can write:
1.5 = ([C]^c [D]^d) / ([A]^a [B]^b)
Simplifying and rearranging the terms, we get:
[A]^a [B]^b = ([C]^c [D]^d) / 1.5
Substituting this value in the rate law for the forward reaction, we get:
ratef = kf ([C]^c [D]^d) / 1.5
Using the rate law for the backward reaction, we can write:
[C]^c [D]^d = kf [A]^a [B]^b / kb
Substituting this value in the expression for the rate of the forward reaction, we get:
ratef = kf kf [A]^a [B]^b / (1.5 kb)
Simplifying the expression, we get:
ratef = (kf)^2 / (1.5 kb)
Substituting the given values, we get:
ratef = (kf)^2 / (1.
Understanding the problem:
The given information is as follows:
- Rate constant of the backward reaction, kb = 7.5 x 10^-4
- Equilibrium constant, Kc = 1.5
We need to find the rate of the forward reaction, kf.
Using the equilibrium constant:
The expression for the equilibrium constant is given as:
Kc = [products] / [reactants]
For the given reaction, let's assume the balanced equation as:
aA + bB ⇌ cC + dD
Then, the equilibrium constant expression becomes:
Kc = ([C]^c [D]^d) / ([A]^a [B]^b)
Substituting the given values:
1.5 = ([C]^c [D]^d) / ([A]^a [B]^b)
We can assume any arbitrary value for the stoichiometric coefficients a, b, c, and d, as long as they are in the proper ratio.
Using the rate constants:
The rate law for the forward reaction can be written as:
ratef = kf [A]^a [B]^b
Similarly, the rate law for the backward reaction can be written as:
rateb = kb [C]^c [D]^d
At equilibrium, the rates of the forward and backward reactions are equal, i.e., ratef = rateb. Therefore, we can write:
kf [A]^a [B]^b = kb [C]^c [D]^d
Calculating the rate of the forward reaction:
Using the equilibrium constant expression, we can write:
1.5 = ([C]^c [D]^d) / ([A]^a [B]^b)
Simplifying and rearranging the terms, we get:
[A]^a [B]^b = ([C]^c [D]^d) / 1.5
Substituting this value in the rate law for the forward reaction, we get:
ratef = kf ([C]^c [D]^d) / 1.5
Using the rate law for the backward reaction, we can write:
[C]^c [D]^d = kf [A]^a [B]^b / kb
Substituting this value in the expression for the rate of the forward reaction, we get:
ratef = kf kf [A]^a [B]^b / (1.5 kb)
Simplifying the expression, we get:
ratef = (kf)^2 / (1.5 kb)
Substituting the given values, we get:
ratef = (kf)^2 / (1.
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In a chemical equilibrium the rate constant of the backward reaction i...
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In a chemical equilibrium the rate constant of the backward reaction is 7.5 x 10^-4 and the equilibrium constant is 1.5. So, the rate of the forward reaction is A) 2 x 10^-3 B) 15 x 10^-4 C) 1.125 x 10^-3 D) 9.0 x 10^-4?
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In a chemical equilibrium the rate constant of the backward reaction is 7.5 x 10^-4 and the equilibrium constant is 1.5. So, the rate of the forward reaction is A) 2 x 10^-3 B) 15 x 10^-4 C) 1.125 x 10^-3 D) 9.0 x 10^-4? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about In a chemical equilibrium the rate constant of the backward reaction is 7.5 x 10^-4 and the equilibrium constant is 1.5. So, the rate of the forward reaction is A) 2 x 10^-3 B) 15 x 10^-4 C) 1.125 x 10^-3 D) 9.0 x 10^-4? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a chemical equilibrium the rate constant of the backward reaction is 7.5 x 10^-4 and the equilibrium constant is 1.5. So, the rate of the forward reaction is A) 2 x 10^-3 B) 15 x 10^-4 C) 1.125 x 10^-3 D) 9.0 x 10^-4?.
In a chemical equilibrium the rate constant of the backward reaction is 7.5 x 10^-4 and the equilibrium constant is 1.5. So, the rate of the forward reaction is A) 2 x 10^-3 B) 15 x 10^-4 C) 1.125 x 10^-3 D) 9.0 x 10^-4? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about In a chemical equilibrium the rate constant of the backward reaction is 7.5 x 10^-4 and the equilibrium constant is 1.5. So, the rate of the forward reaction is A) 2 x 10^-3 B) 15 x 10^-4 C) 1.125 x 10^-3 D) 9.0 x 10^-4? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a chemical equilibrium the rate constant of the backward reaction is 7.5 x 10^-4 and the equilibrium constant is 1.5. So, the rate of the forward reaction is A) 2 x 10^-3 B) 15 x 10^-4 C) 1.125 x 10^-3 D) 9.0 x 10^-4?.
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In a chemical equilibrium the rate constant of the backward reaction is 7.5 x 10^-4 and the equilibrium constant is 1.5. So, the rate of the forward reaction is A) 2 x 10^-3 B) 15 x 10^-4 C) 1.125 x 10^-3 D) 9.0 x 10^-4?, a detailed solution for In a chemical equilibrium the rate constant of the backward reaction is 7.5 x 10^-4 and the equilibrium constant is 1.5. So, the rate of the forward reaction is A) 2 x 10^-3 B) 15 x 10^-4 C) 1.125 x 10^-3 D) 9.0 x 10^-4? has been provided alongside types of In a chemical equilibrium the rate constant of the backward reaction is 7.5 x 10^-4 and the equilibrium constant is 1.5. So, the rate of the forward reaction is A) 2 x 10^-3 B) 15 x 10^-4 C) 1.125 x 10^-3 D) 9.0 x 10^-4? theory, EduRev gives you an
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