Time required for 99.9% completion of a first order reaction is _______ times the time required for completion of 90% reaction. (nearest integer)

We need to find the ratio of the time for 99.9% completion to the time for 90% completion of a first-order reaction.

For a first-order reaction, the time for a given fraction of completion is given by $$t = \frac{2.303}{k} \log_{10}\left(\frac{C_0}{C}\right) = \frac{2.303}{k} \log_{10}\left(\frac{1}{1-x}\right)$$ where $$x$$ is the fraction of reaction completed and $$k$$ is the rate constant.

In the case of 99.9% completion, $$x = 0.999$$ so that $$1 - x = 0.001$$, hence $$t_{99.9\%} = \frac{2.303}{k} \log_{10}\left(\frac{1}{0.001}\right) = \frac{2.303}{k} \log_{10}(1000) = \frac{2.303}{k} \times 3$$

For 90% completion, $$x = 0.9$$ so that $$1 - x = 0.1$$, giving $$t_{90\%} = \frac{2.303}{k} \log_{10}\left(\frac{1}{0.1}\right) = \frac{2.303}{k} \log_{10}(10) = \frac{2.303}{k} \times 1$$

Thus, the ratio of these times is $$\frac{t_{99.9\%}}{t_{90\%}} = \frac{\frac{2.303 \times 3}{k}}{\frac{2.303 \times 1}{k}} = \frac{3}{1} = 3$$ and the answer is 3.