JEE PYQ: Chemical Kinetics Question 16
Question 16 - 2020 (04 Sep Shift 1)
If 75% of a first order reaction was completed in 90 minutes, 60% of the same reaction would be completed in approximately ____.
(Take: $\log 2 = 0.30$; $\log 2.5 = 0.40$)
Show Answer
Answer: 60 min
Solution
$t = \frac{2.303}{k}\log\left[\frac{100}{100 - x%}\right]$. $t_{75%} = \frac{2.303}{k}\log\frac{100}{25} = 90$. $t_{60%} = \frac{2.303}{k}\log\frac{100}{40}$. $\frac{t_{75%}}{t_{60%}} = \frac{2\log 2}{\log 2.5} = \frac{90}{t_{60%}} = \frac{2 \times 0.3}{0.4}$. $t_{60%} = \frac{90 \times 4}{6} = 60$ min.
BETA
