To solve this problem, we need to determine the half-life of a first-order reaction using the given rate data.$\newline$For a first-order reaction, the rate constant $k$ is related to the rate $R$ and concentration $[A]$ by:$\newline R = k[A] \newline$Given:$\newline$• Rate at $10 \,$s$= 0.04 \,$mol L$^{-1}$s$^{-1} \newline$• Rate at $20 \,$s$= 0.03 \,$mol L$^{-1}$s$^{-1} \newline$For a first-order reaction, the concentration $[A]$ at time $t$ is given by:$\newline [A] = [A]_0 e^{-kt} \newline$The rate at time $t$ is:$\newline R = k[A]_0 e^{-kt} \newline$Taking the ratio of rates at $10 \,$s$$ and $20 \,$s$: \newline \frac{R_{10}}{R_{20}} = \frac{k[A]_0 e^{-k \cdot 10}}{k[A]_0 e^{-k \cdot 20}} \newline \frac{0.04}{0.03} = \frac{e^{-k \cdot 10}}{e^{-k \cdot 20}} \newline \frac{4}{3} = e^{k \cdot 10} \newline$Taking the natural logarithm on both sides:$\newline \ln\left(\frac{4}{3}\right) = k \cdot 10 \newline k = \frac{\ln\left(\frac{4}{3}\right)}{10} \newline$The half-life $t_{1/2}$ for a first-order reaction is given by:$\newline t_{1/2} = \frac{\ln(2)}{k} \newline$Substitute the value of $k: \newline t_{1/2} = \frac{\ln(2)}{\frac{\ln\left(\frac{4}{3}\right)}{10}} \newline t_{1/2} = 10 \times \frac{\ln(2)}{\ln\left(\frac{4}{3}\right)} \newline$Calculate the values:$\newline \ln(2) \approx 0.693 \quad$and$\quad \ln\left(\frac{4}{3}\right) \approx 0.2877 \newline t_{1/2} = 10 \times \frac{0.693}{0.2877} \newline t_{1/2} \approx 24.1 \,$s$\newline$Therefore, the half-life period of the reaction is $24.1 \,$s$,$ which corresponds to Option 1.