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. (a) `n-m`
B. (b) `2^(n-m)`
C. (c ) `(1)/(2^(m+n))`
D. (d) `m+n`

Answer

b

Explanation

`r_(0)=K[A]^(n)[B]^(m)`
`r_(1)=K[2A]^(n)[B//2]^(m)`
`r_(1)=K 2^(n-m) [A]^(n)[B]^(m)` or `r_(1)=rxx2^(n-m)`