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The rate of first order reaction is 0 . 04 mol L - 1 s - 1 at 10 minutes and 0 . 03 mol L - 1 s - 1 at 20 minutes after initiation. Half life of the reaction is ______ minutes. (Given log 2 = 0 . 3010 , log 3 = 0 . 4771 ) Round off your answer to the nearest integer.

Correct answer

0

Step-by-step solution

Rate r = k A 0 . 04 = k A 0 10 0 . 03 = k A 20 ⇒ A 0 10 A 20 = 0 . 04 0 . 03 = 4 3 From t = 2 . 303 k log A 10 A 20 10 = 2 . 303 k log 4 3 10 = 2 . 303 k ( 0 . 6020 - 0 . 4771 ) k = 2 . 303 10 × 0 . 1249 = 2 . 876 × 10 - 2 min - 1 t 1 / 2 = 0 . 693 k = 0 . 693 2 . 876 × 10 - 2 = 0 . 24 × 10 2 min = 24 min .

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