If 75% of a first order reaction was completed in 90 minutes, 60% of the same reaction would be completed in approximately (in minutes) _______________ .(Take : log 2 = 0.30; log 2.5 = 0.40)

According to the first order kinetics:

$k = \frac{2.303}{t} \log \frac{[A_{o}]}{[A_{t}]}$

When 75% reaction was completed.

$k = \frac{2.303}{90} \log \frac{100}{25} k = \frac{2.303}{90} \log 4 k = 0.0154 \min^{- 1}$

Time required to completed 60% of reaction.

$t = \frac{2.303}{0.0154} \log \frac{100}{40} t = \frac{2.303 \times 0.4}{0.0154 \min^{- 1}} = 60 \min$

If 75% of a first order reaction was completed in 90 minutes, 60% of the same reaction would be completed in approximately (in minutes) _______________ .
(Take : log 2 = 0.30; log 2.5 = 0.40)