The rate law for a reaction between `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 becomes

Options

A. `(1)/(2)^(m+n)`
B. `(m+n)`
C. `(n-m)`
D. `2^(n-m)`

Answer

D

Explanation

Rate `= k[A]^(n)[B]^(m)`
When `A' = 2A` and `B' = 0.5 B`
The new rate becomes Rate' `= [2A]^(n)[0.5]^(m)`
`(Rate')/(Rate) = (k2^(n)[A]^(n)(0.5)^(m)[B]^(m))/(k[A]^(n)[B]^(m))`
`= 2^(n)((1)/(2))^(m) = 2^((n-m))`