To solve this problem, we need to use the integrated rate equation for a first-order reaction.$\newline$The integrated rate equation for a first-order reaction is:$\newline \ln \left( \frac{[R]_0}{[R]} \right) = k \cdot t \newline$where:$\newline$• $[R]_0$ is the initial concentration (or amount) of the reactant.$\newline$• $[R]$ is the concentration (or amount) of the reactant at time $t. \newline$• $k$ is the rate constant.$\newline$• $t$ is the time.$\newline$Given in the problem:$\newline$• Initial amount of reactant, $[R]_0 = 20 \,$g$\newline$• Final amount of reactant, $[R] = 5 \,$g$\newline$• Rate constant, $k = 10^{-2} \,$s$^{-1} \newline$Substitute these values into the integrated rate equation:$\newline \ln \left( \frac{20}{5} \right) = 10^{-2} \cdot t \newline$Simplify the fraction:$\newline \ln(4) = 10^{-2} \cdot t \newline$Calculate $\ln(4): \newline \ln(4) \approx 1.386 \newline$Substitute back to find $t: \newline 1.386 = 10^{-2} \cdot t \newline t = \frac{1.386}{10^{-2}} \newline t = 138.6 \,$s$\newline$Therefore, it will take 138.6 seconds for 20 g of the reactant to reduce to 5 g.$\newline$This corresponds to Option 1.