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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