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  4. If 70% of a first order reaction was completed in 52 minutes, 50% of the same reaction would be completed in approximately (log 3 = 0.47)

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Q. If 70% of a first order reaction was completed in 52 minutes, 50% of the same reaction would be completed in approximately (log 3 = 0.47)

Solution:

$\frac{0.693}{t_{\frac{1}{2}}}=\frac{2.303}{52}log \frac{100}{\left(100-70\right)}$
$t_{1/2} \simeq 30 minute$