For a first order reaction, show that the time required for 99.9% completion is how many times the half-life t1/2?

1. At 99.9% completion, $[R] = [R]_0 - 0.999[R]_0 = 0.001[R]_0$. 2. $t = \dfrac{2.303}{k}\log\dfrac{[R]_0}{0.001[R]_0} = \dfrac{2.303}{k}\log 10^3 = \dfrac{6.909}{k}$. 3. Half-life $t_{1/2} = \dfrac{0.693}{k}$. 4. $\dfrac{t}{t_{1/2}} = \dfrac{6.909}{0.693} = 10$. 5. Distractor A would correspond to 99% type fractions; this is exactly ten times. _Source: NCERT Class 12 Chemistry Ch 3 "Chemical Kinetics", p.16_