Rate law for a reaction between A and B is given by r = k[A]n[B]m If concentration of A is doubled and concentration of B is halved from their initial value, the ratio of new rate of reaction to the initial rate of reaction ( r2r1 ) is
- (n-m)
- 2(n-m)
- 12m+n
- (m+n)
Answer
(2) 2(n-m)
Solution
r1 = k[A]n[B]m
r2 = k(2[A])n ( [B]2 )m
r2 = 2n 2-m k[A]n[B]m
r2r1 = 2(n-m)
Answer: (2)