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

B

Explanation

Rate, `r_(1)=k[A]^(n),[B]^(m)`....(i)
Rate, `r_(2)=k[2A]^(n)[B//2]^(m)`....(ii)
Dividing (ii) by (i),
`(r_(2))/(r_(1))=(2^(n))/(2^(m))=2^(n-m)`
or `r_(2)=2^(n-m)r_(1)`.