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: $\boldsymbol{\operatorname{log}} 2 \boldsymbol{=} \mathbf{0.30} \boldsymbol{;} \boldsymbol{\operatorname{log}} 2.5 \boldsymbol{=} \mathbf{0.40}$)

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Chemical Kinetics

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

)

Answer: 60

For a first-order reaction

k = \frac{2.303}{t} lo g \frac{a _{0}}{a _{0} - x}

. From

75%

completion in

90

min,

k = \frac{2.303}{90} lo g 4 = \frac{2.303 \times 0.60}{90} = 0.01535

min

^{- 1}

. For

60%

completion,

t = \frac{2.303}{k} lo g \frac{a _{0}}{0.40 a _{0}} = \frac{2.303}{0.01535} lo g 2.5 = \frac{2.303 \times 0.40}{0.01535} \approx 60

min.

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Integer answer