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

Answer

B

Explanation

`r_(0) = K[A]^(n)[B]^(m)`
`r_(1) = K[2A]^(n)[B//2]^(m)`
`r_(1)= K2^(n-m)[A]^(n)[B]^(m)`
or `r_(1) = r xx 2^(n - m)`