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Q. The half life period of a first order chemical reaction is $6.93$ minutes. The time required for the completion of $99\%$ of the chemical reaction will be $(\log 2 = 0.301)$

VITEEEVITEEE 2017

Solution:

$K =\frac{0.693}{ t _{\frac{1}{2}}}$
$K =\frac{0.693}{6.93}=10^{-1} \min ^{-1}$
$10^{-1}=\frac{2.303}{ t } \log \frac{100}{1}$
$t =\frac{2.303 \times 2}{10^{-1}}=46.06\, \min$