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 having the concentration of B, the ratio of the new rate to the earlier rate of the reactio will be:

Options

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

Answer

D

Explanation

`"Initial rate"=k[A]^(n)[B]^(n)`
when[A]=[2A] and `[B][(1)/(2)B]`, then
`"New rate"=k[2A]^(n)[1//28]^m`
On dividing Eq.(ii)by Eq.(i), we get
`"New rate"/"Initial rate"=(k[2A]^(n)[1//2B]^(m))/(k[A]^(n)[B}^(m))`
`"New rate"/"Initial rate"=(2)^(n)(1//2)^(m)=(2)^(n)(2)^(-m)=2(n-m)`
Hence, the ratio of new rate fo the reaction is `2^(n-m).`