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)

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_