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For a second order reaction the half-life is 1.0 hour. After how many hours will the amount of reactant fall to 25% of its initial value?
- 2 hr
- 3 hr
- 4 hr
- 1.5 hr
Correct answer: 3 hr
Solution
For second order, t = (1/k)(1/[A] - 1/[A]0). Using t1/2 = 1/(k[A]0) = 1, the time to reach 25% of [A]0 works out to 3 hours.
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