Consider a certain reaction $A \rightarrow \text{Products}$ with $k = 2.0 \times 10^{-2} \ s^{-1}$. Calculate the concentration of $A$ remaining after $100 \ s$ if the initial concentration of $A$ is $1.0 \ mol \ L^{-1}$.

Given: $k = 2.0 \times 10^{-2} \ s^{-1}$,$t = 100 \ s$,$[A]_{0} = 1.0 \ mol \ L^{-1}$.
Since the unit of $k$ is $s^{-1}$,the reaction is a first-order reaction.
For a first-order reaction,the integrated rate equation is:
$k = \frac{2.303}{t} \log \frac{[A]_{0}}{[A]}$
Substituting the values:
$2.0 \times 10^{-2} = \frac{2.303}{100} \log \frac{1.0}{[A]}$
$\log \frac{1.0}{[A]} = \frac{2.0 \times 10^{-2} \times 100}{2.303} = \frac{2}{2.303} \approx 0.8684$
$\frac{1.0}{[A]} = \text{antilog}(0.8684) \approx 7.385$
$[A] = \frac{1.0}{7.385} \approx 0.135 \ mol \ L^{-1}$.
Alternatively,using $[A] = [A]_{0} e^{-kt}$:
$[A] = 1.0 \times e^{-(2.0 \times 10^{-2} \times 100)} = 1.0 \times e^{-2} \approx 1.0 \times 0.1353 = 0.1353 \ mol \ L^{-1}$.