For a first-order reaction, the integrated rate equation is given by: $t = \frac{2.303}{k} \log \frac{[R]_0}{[R]}$ Where: $[R]_0$ is the initial concentration. $[R]$ is the concentration at time $t$.

When the reaction is 99% complete: Amount reacted = $0.99 [R]_0$ Concentration remaining, $[R] = [R]_0 - 0.99 [R]_0 = 0.01 [R]_0$

Substituting these values into the rate equation: $t = \frac{2.303}{k} \log \frac{[R]_0}{0.01 [R]_0}$ $t = \frac{2.303}{k} \log (100)$ Since $\log(100) = 2$: $t = \frac{2.303}{k} \times 2$ $t = \frac{4.606}{k}$