Question

The rate constant for a first order reaction is 60 sec$^{-1}$. How much time will it take to reduce the concentration of the reaction to 1/10$^{th}$ of its initial value ?

Hint:

Memorize $\ln(10) \approx 2.303$. This conversion factor is ubiquitous in kinetics problems involving base-10 logs.

Answer 1

For a first-order reaction, the integrated rate equation is:
$t = \frac{2.303}{k} \log \frac{[A]_0}{[A]}$
Given:
$k = 60$ s$^{-1}$.
We want the final concentration $[A]$ to be $\frac{1}{10}$ of initial $[A]_0$.
So, $\frac{[A]_0}{[A]} = 10$.
Substitute values:
$t = \frac{2.303}{60} \log(10)$
Since $\log(10) = 1$:
$t = \frac{2.303}{60}$
$t \approx 0.03838$ s.
Rounding to significant figures, $t \approx 3.84 \times 10^{-2}$ s.