Half-Life And Pseudo First Order Reactions

For a first order reaction, we can compare the time required for 99% completion to the time for 90% completion. The integrated rate equation is $t = \frac{2.303}{k} \log \frac{[R]_0}{[R]_t}$. For 99% completion, the remaining concentration is 1% of the initial, so we evaluate $t_{99\%} = \frac{2.303}{k} \log($ $)$. For 90% completion, the remaining concentration is 10% of the initial, so we evaluate $t_{90\%} = \frac{2.303}{k} \log($ $)$. Since $\log(100) = 2$ and $\log(10) = 1$, we can substitute these values. Dividing the two times gives $\frac{t_{99\%}}{t_{90\%}} = \frac{2}{1}$, proving that $t_{99\%}$ is exactly the time required for 90% completion.