To solve this problem, we need to determine the rate constant for a first-order reaction.$\newline$The reaction is given as $A \rightarrow$Products.$\newline$For a first-order reaction, the rate equation is:$\newline k = \frac{1}{t} \ln \left(\frac{[A]_0}{[A]}\right) \newline$where:$\newline$• $k$ is the rate constant.$\newline$• $t$ is the time.$\newline$• $[A]_0$ is the initial concentration of $A. \newline$• $[A]$ is the concentration of $A$ at time $t. \newline$Given:$\newline$• Initial concentration $[A]_0 = 0.1 \,$M$\newline$• Final concentration $[A] = 0.001 \,$M$\newline$• Time $t = 5 \,$minutes$\newline$Substitute these values into the rate equation:$\newline k = \frac{1}{5} \ln \left(\frac{0.1}{0.001}\right) \newline$Simplify the expression inside the logarithm:$\newline \frac{0.1}{0.001} = 100 \newline$Thus, the equation becomes:$\newline k = \frac{1}{5} \ln (100) \newline$Calculate $\ln (100): \newline \ln (100) = \ln (10^2) = 2 \ln (10) \newline$Using the approximation $\ln (10) \approx 2.302,$ we have:$\newline \ln (100) = 2 \times 2.302 = 4.604 \newline$Substitute back to find $k: \newline k = \frac{1}{5} \times 4.604 = 0.9208 \,$min$^{-1} \newline$Rounding to four significant figures, we get $k \approx 0.9212 \,$min$^{-1}. \newline$Therefore, the rate constant for the reaction is $0.9212 \,$min$^{-1},$ which corresponds to Option 2.