To solve this problem, we need to use the formula for a first-order reaction:$\newline t = \frac{2.303}{k} \log \left(\frac{[A]_0}{[A]}\right) \newline$where:$\newline$• $t$ is the time required for the concentration to change from $[A]_0$ to $[A]. \newline$• $k$ is the rate constant.$\newline$• $[A]_0$ is the initial concentration of the reactant.$\newline$• $[A]$ is the final concentration of the reactant.$\newline$Given:$\newline$• $k = 4.606 \times 10^{-3} \,$s$^{-1} \newline$• Initial mass of reactant $= 2.0 \,$g$\newline$• Final mass of reactant $= 0.2 \,$g$\newline$Since the reaction is first-order, the concentration is directly proportional to the mass.$\newline$Therefore, we can use the masses in place of concentrations:$\newline t = \frac{2.303}{4.606 \times 10^{-3}} \log \left(\frac{2.0}{0.2}\right) \newline$Calculate the logarithm:$\newline \log \left(\frac{2.0}{0.2}\right) = \log (10) = 1 \newline$Substitute the values into the equation:$\newline t = \frac{2.303}{4.606 \times 10^{-3}} \times 1 \newline t = \frac{2.303}{4.606 \times 10^{-3}} \newline t = 500 \,$s$\newline$Therefore, the time required to reduce the reactant from 2.0 g to 0.2 g is 500 s.$\newline$This corresponds to Option 2.