The rate law for a reaction between the substances 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 will be as:

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)` …….(1)
`r_(1)=K[2A]^(n)[1/2B]^(m)`………(2)
`(2)/(1) rArr r_(1)/r_(2) = 2^(n-m) , r_(1) =2^(n-m).r`