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For a given first order reaction the reactant reduced to 1/4th its initial value in 10 minutes. The rate constant of the reaction is?
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For a given first order reaction the reactant reduced to 1/4th its ini...
First Order Reaction:
A first-order reaction is a chemical reaction in which the rate of conversion of reactants into products is proportional to the concentration of only one reactant.
Given:
The reactant reduced to 1/4th its initial value in 10 minutes.
Formula:
The rate of a first-order reaction can be expressed as:
rate = k[A]
where,
k = rate constant
[A] = concentration of reactant
Calculation:
Let the initial concentration of the reactant be [A]₀.
After 10 minutes, the concentration of the reactant becomes [A]/4.
The rate equation can be written as:
rate = k[A]₀
At t = 10 min, the concentration of the reactant is [A]/4.
rate = k[A]/4
Substituting the values in the equation above, we get:
k[A]₀ = k[A]/4
Simplifying the equation, we get:
k = 2.303/t log ([A]₀/[A])
where,
t = time taken for the concentration of the reactant to reduce to 1/4th of its initial value.
Substituting the values, we get:
k = 2.303/10 log ([A]₀/[A]/4)
k = 2.303/10 log ([A]₀/[A]) + 2.303/10 log 4
k = 0.2303 log ([A]₀/[A]) + 0.09212
Answer:
The rate constant of the reaction is 0.2303 log ([A]₀/[A]) + 0.09212.
A first-order reaction is a chemical reaction in which the rate of conversion of reactants into products is proportional to the concentration of only one reactant.
Given:
The reactant reduced to 1/4th its initial value in 10 minutes.
Formula:
The rate of a first-order reaction can be expressed as:
rate = k[A]
where,
k = rate constant
[A] = concentration of reactant
Calculation:
Let the initial concentration of the reactant be [A]₀.
After 10 minutes, the concentration of the reactant becomes [A]/4.
The rate equation can be written as:
rate = k[A]₀
At t = 10 min, the concentration of the reactant is [A]/4.
rate = k[A]/4
Substituting the values in the equation above, we get:
k[A]₀ = k[A]/4
Simplifying the equation, we get:
k = 2.303/t log ([A]₀/[A])
where,
t = time taken for the concentration of the reactant to reduce to 1/4th of its initial value.
Substituting the values, we get:
k = 2.303/10 log ([A]₀/[A]/4)
k = 2.303/10 log ([A]₀/[A]) + 2.303/10 log 4
k = 0.2303 log ([A]₀/[A]) + 0.09212
Answer:
The rate constant of the reaction is 0.2303 log ([A]₀/[A]) + 0.09212.
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For a given first order reaction the reactant reduced to 1/4th its initial value in 10 minutes. The rate constant of the reaction is?
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For a given first order reaction the reactant reduced to 1/4th its initial value in 10 minutes. The rate constant of the reaction is? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about For a given first order reaction the reactant reduced to 1/4th its initial value in 10 minutes. The rate constant of the reaction is? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For a given first order reaction the reactant reduced to 1/4th its initial value in 10 minutes. The rate constant of the reaction is?.
For a given first order reaction the reactant reduced to 1/4th its initial value in 10 minutes. The rate constant of the reaction is? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about For a given first order reaction the reactant reduced to 1/4th its initial value in 10 minutes. The rate constant of the reaction is? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For a given first order reaction the reactant reduced to 1/4th its initial value in 10 minutes. The rate constant of the reaction is?.
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