Chemical Kinetics and Nuclear ChemistryHard
Question
The rate law for a reaction between the substances A and B is given by Rate = k[A]n [B]m On doubling the concentration of A and halving the concentration of B, the ratio of the new rate to the earlier rate of the reaction will be as
Options
A.(m + n)
B.(n - m)
C.2(n - m)
D.
Solution
Rate1 = k [A]n [B]m; Rate2 = k[2A]n [1/2B]m
∴
= [2]n [1/2m] = 2n.2-m = 2n-m
∴
= [2]n [1/2m] = 2n.2-m = 2n-mCreate a free account to view solution
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