Time required for completion of $99.9 \%$ of a First order reaction is ________ times of half life $\left(t_{1 / 2}\right)$ of the reaction.
Answer (integer)
10
Solution
<p>$$\frac{\mathrm{t}_{99.9 \%}}{\mathrm{t}_{1 / 2}}=\frac{\frac{2.303}{\mathrm{k}}\left(\frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}\right)}{\frac{2.303}{\mathrm{k}} \log 2}=\frac{\log \left(\frac{100}{100-99.9}\right)}{\log 2}=\frac{\log 10^3}{\log 2}=\frac{3}{0.3}=10$$</p>
About this question
Subject: Chemistry · Chapter: Chemical Kinetics · Topic: Rate of Reaction
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