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  4. 75 % of a first order reaction was completed in 32 min, when was 50 % of the reaction completed?

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Q. $75 \%$ of a first order reaction was completed in $32\, min$, when was $50 \%$ of the reaction completed?

BHUBHU 2007

Solution:

First order reaction,
$k=\frac{2.303}{t} \log \frac{a}{a-x}$
Here, $ \therefore a=100$
$\therefore x=75$
$\because a-x=(100-75)$
$k=\frac{2.303}{32} \log \frac{100}{25} $
$=\frac{2.303}{32} \log 4$
$=\frac{2.303 \times \log 2^{2}}{32} $
$=\frac{2.303 \times 2 \log 2}{32} $
$=\frac{2.303 \times 2 \times 0.3010}{32}$
$=\frac{2 \times 0.63}{32}$
$\therefore t_{1 / 2}=\frac{0.693}{k} $
$\therefore t_{1 / 2}=\frac{0.693}{2 \times 0.693 / 32}4$
$=16\, \min $