For the following first-order competing reaction: A + Reagent → Product; B + Reagent → Product. The ratio of K1/K2, if only 50% of 'B' will have been reacted when 94% of 'A' has been reacted is (log 2

Question

For the following first-order competing reaction: A + Reagent → Product; B + Reagent → Product. The ratio of K1/K2, if only 50% of 'B' will have been reacted when 94% of 'A' has been reacted is (log 2 = 0.3, log 3 = 0.48)

Prashant Kotian · Accepted Answer

Correct answer: A. ln(A0/A) / ln(B0/B) = K1/K2. K1/K2 = ln(100/6) / ln(100/50) = (log 100/6) / log 2 = (2 - log 2 - log 3) / 0.3 = (2 - 0.3 - 0.48) / 0.3 = 1.22 / 0.3 = 4.06. Therefore, correct answer is A.

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