Chemical Kinetics Question 4
Question 4 - 2024 (27 Jan Shift 2)
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.
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Answer (10)
Solution
$\frac{t _{99.9 \%}}{t _{1 / 2}}=\frac{\frac{2.303}{k}\left(\frac{a}{a-x}\right)}{\frac{2.303}{k} \log 2}=\frac{\log \left(\frac{100}{100-99.9}\right)}{\log 2}=\frac{\log 10^{3}}{\log 2}=\frac{3}{0.3}=10$
