JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics
Time required for completion of \(99.9 \%\) of a First order reaction is_________ times of half life \(\left(\mathrm{t}_{1 / 2}\right)\) of the reaction.
- A \(15\)
- B \(17\)
- C \(10\)
- D \(5\)
Answer & Solution
Correct Answer
(C) \(10\)
Step-by-step Solution
Detailed explanation
\(\frac{t_{99.9 \%}}{t_{1 / 2}}=\frac{\frac{2.303}{k}\left(\frac{a}{a-x}\right)}{\frac{2.303}{k} \log 2}=\frac{\log \left(\frac{100}{100-99.9}\right)}{\log 2}=\frac{\log 10^3}{\log 2}=\frac{3}{0.3}=10\)
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