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Question
The rate constant for a first order reaction is 60 s−1. How much time will it take to reduce the initial concentration of the reactant to its `1/16`th value?
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Solution
Given: Rate constant (k) = 60 s−1
[R]0 = [R]0
[R] = `[R]_0/16`
t = ?
By using the formula:
t = `2.303/k log_10 ([R]_0)/([R])`
= `2.303/60 log_10 ([R]_0)/(1/16 xx [R]_0)`
= 0.3838 log10 (16)
= 0.3838 × 1.2041
= 4.62 × 10−2 s
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