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

Answer

B

Explanation

`Rate_(1)=kk[A]^(n)[B]^(m)`
On doubling the concentration of `A` and having the concentration of `B` , we get
`Rate_(2)=k[2A]^(n)[B//2]^(m)`
Dividing `Rate_(2)` by `Rate_(1)` , we get
`(Rate_(2))/(Rate_(1))=(k[2A]^(n)[B//2]^(m))/(k[A]^(n)[B]^(m))=(2)^(n)((1)/(2))^(m)`
`=2^(n-m)`