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The rate constant for a first order reaction is 60 s^-1. Approximately how much time is required to reduce the initial concentration to 1/16 of its value? (ln 2 = 0.693)
A0.0462 s
B0.0231 s
C0.116 s
D0.693 s
Answer & Solution
Correct answer: A. 0.0462 s
1. Reducing to $1/16$ means four half-lives, since $(1/2)^4 = 1/16$.
2. $t_{1/2} = \dfrac{0.693}{k} = \dfrac{0.693}{60} = 0.01155\ \text{s}$.
3. $t = 4\times t_{1/2} = 4\times0.01155$.
4. $= 0.0462\ \text{s}$.
5. Distractor B takes only two half-lives; distractor C miscounts as ten half-lives.
_Source: NCERT Class 12 Chemistry Ch 3 "Chemical Kinetics", p.26_
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