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Question
In a first-order reaction, the concentration of the reactant decreases from 20 mmol dm−3 to 8 mmol dm−3 in 38 minutes. What is the half-life of reaction?
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Solution
Given:
[A]0 = 20 m mol dm−3,
[A]t = 8 m mol dm−3,
t = 38 min
To find:
Half-life of reaction t1/2 = ?
Formulae:
i. `k = 2.303/t log_10 [A]_0/[A]_t`
ii. t1/2 = `0.693/k`
Calculation:
Substituting given value in
`k = 2.303/t log_10 [A]_0/[A]_t`
`k = 2.303/(38 min) log_10 20/8`
= `2.303/(38 min) log_10 (2.5)`
= `2.303/(38 min) xx 0.3979`
= 0.0241 min−1
t1/2 = `0.693/k`
= `0.693/0.0241`
= 28.7 min
The half life of reaction is 28.7 min.
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