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27/01/2022 5:21 pm
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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 1/16th value?
1 Answer
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27/01/2022 5:24 pm
It is known that,
t = \(\frac{2.303}{k}\)log \(\frac{[R]_0}{[R]}\)
= \(\frac{2.303}{60\;s^{-1}}\)log \(\frac{1}{\frac{1}{16}}\)
= \(\frac{2.303}{60\;s^{-1}}\)log 16
= 4.6 x 10-2 (approximately)
Hence, the required time is 4.6 x 10-2s.