Q. $t_{1/4}$ can be taken as the time taken for the concentration of reactant to drop to $\frac{3}{4}$ of its initial value. If the rate constant for a first order reaction is k the$t_{1/4}$ can be written as
A) 0.75 /k
B) 0.69 /k
C) 0.29 /k
D) 0.10 /k
Explanation
(c)
$\ \ \ A\ \ \ \ \ \ \ \ \rightarrow \ \ \ \ product$
Initially a 0
After time t (a-x) x
After $t_{1/4}\ \ \ \ \ \ \ \ \ \ \left( a - \frac{a}{4} \right)\ \ \ \ \ \ \frac{a}{4}$
For the first order kinetics ,
$k = \frac{2.303}{t}\log\left( \frac{a}{a - x} \right)$
$k = \frac{2.303}{t_{1/4}}\log\frac{a}{\frac{3a}{4}}$
$t_{1/4} = \frac{2.303log\frac{4}{3}}{k}$
$= \frac{0.29}{k}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.