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

`r=k [A]^(n) [B]^(n)`
`r'=k (2[A])^(n) (([B])/(2))^(m) =k[A]^(m) xx2^(n-m) rArr r'=2^(n-m).r`