Q. The rate constant for the first order reaction is $6\ s^{- 1}.$ How much time will it take to reduce the concentration of the reactant to $\frac{1}{16}\ th$ value?
A) $4.6 \times 10^{- 2}\ s$
B) $4.6 \times 10^{4}\ s$
C) $4.6 \times 10^{2}\ s$
D) $4.6 \times 10^{- 4}\ s$
Explanation
(a)
$t = \frac{2.303}{k}\log\frac{\lbrack A\rbrack_{0}}{\lbrack A\rbrack}$
$= \frac{2.303}{60}\log{\frac{a}{\frac{a}{16}} = \frac{2.303}{60}\log 16}$
$= \frac{2.303}{60} \times 1.204$
$= 0.0462\ s\$
$= 4.6 \times 10^{- 2}s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.