Question

The rate lawfor a reaction between the substancesaand bis given by rate = k [a]n [b]m on doubling the concentration ofaand halving the concentration ofb, the ratio of the newrate to the earlier rate of the reactionwill be as

Answer 1

Ratio = 2^(n-m).

Original rate = k[a]^n.[b]^m. --1

New rate = k[2a]^n.[b/2]^m. --2

Dividing 2 by 1 we get the answer.