For a first-order reaction, the integrated rate equation is given by: $t = \frac{2.303}{k} \log \frac{[R]_0}{[R]}$ Given: Rate constant, $k = 2.303 \times 10^{-3} \text{ s}^{-1}$ Initial amount, $[R]_0 = 40 \text{ g}$ Final amount, $[R] = 10 \text{ g}$ Substituting the values into the formula: $t = \frac{2.303}{2.303 \times 10^{-3}} \log \left(\frac{40}{10}\right)$ $t = 10^3 \times \log 4$ $t = 1000 \times 2 \log 2$ Given $\log_{10} 2 = 0.3010$, $t = 1000 \times 2 \times 0.3010 = 1000 \times 0.6020 = 602 \text{ s}$.