JEE MAIN - Chemistry (2024 - 27th January Evening Shift - No. 24)
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
10
Explanation
$$\frac{\mathrm{t}_{99.9 \%}}{\mathrm{t}_{1 / 2}}=\frac{\frac{2.303}{\mathrm{k}}\left(\frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}\right)}{\frac{2.303}{\mathrm{k}} \log 2}=\frac{\log \left(\frac{100}{100-99.9}\right)}{\log 2}=\frac{\log 10^3}{\log 2}=\frac{3}{0.3}=10$$
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