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Questions
In a first-order reaction A → B, 60% of a given sample of a compound decomposes in 45 mins. What is the half-life of reaction? Also, write the rate law equation for the above first-order reaction.
60% of the reactant decomposes in 45 minutes in a first order reaction. Calculate the half life period of the reaction. Write the relation between half life period and initial concentration for zero order reaction.
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Solution
Given:
[A]0 = 100%
[A]t = 100 − 60 = 40%
t = 45 min
To find: Half life of reaction (t1/2)
Formula: k = `2.303/t log_10 [A]_0/[A]_t`
Calculation: Substitution of these in above
k = `2.303/"t" log_10 100/40`
= `2.303/(45 "min") log_10 (2.5)`
= `2.303/(45 "min") xx 0.3979`
= 0.0203 min-1
For a first-order reaction, the half-life formula is:
`t_(1/2) = 0.693/k`
= `0.693/(0.0203 "min"^-1)`
= 34 min
∴ The half life of reaction is 34 min.
Relation between half life and initial concentration for zero-order reaction:
`t_(1/2) = [A]_0/(2k)`
In first-order reactions, the half-life of a zero-order reaction is directly proportional to the initial concentration.
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