For the second-order reaction: 2A - B, time taken for the [A] to fall to one-fourth value is how many times the time it takes for [A] to fall to half of its initial value?

Question

For the second-order reaction: 2A -> B, time taken for the [A] to fall to one-fourth value is how many times the time it takes for [A] to fall to half of its initial value?

Prashant Kotian · Accepted Answer

Correct answer: B. For a second-order reaction: 1/[A]_t - 1/[A]_0 = Kt. (1) Time to fall to half ([A]_t = 0.5 [A]_0): 2/[A]_0 - 1/[A]_0 = Kt_1/2 => t_1/2 = 1/(K[A]_0). (2) Time to fall to one-fourth ([A]_t = 0.25 [A]_0): 4/[A]_0 - 1/[A]_0 = Kt_3/4 => t_3/4 = 3/(K[A]_0). Therefore, t_3/4 = 3 × t_1/2.

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