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 : \(\log 2 = 0.30\); \(\log 2.5 = 0.40\))
(Take : \(\log 2 = 0.30\); \(\log 2.5 = 0.40\))
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Correct Answer: 60
Solution and Explanation
75% completion: \(90 = \frac{2.303}{k}\log\frac{100}{25} = \frac{2.303}{k}\log 4 = \frac{2.303}{k}(2 \times 0.30) = \frac{2.303}{k}(0.60)\).
60% completion: \(t = \frac{2.303}{k}\log\frac{100}{40} = \frac{2.303}{k}\log 2.5 = \frac{2.303}{k}(0.40)\).
Dividing: \(\frac{t}{90} = \frac{0.40}{0.60} = \frac{2}{3}\), so \(t = 60\) minutes.
Answer: 60
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