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 of the earlier rate of the reaction will be as

Options

A. (m+n)
B. (n-m)
C. `2^(n-m)`
D. `(1)/(3^(m+n))`

Answer

C

Explanation

`Rate_(1) =k[A]^(n)[B]^(m),Rate_(2)=k[2A]^(n)[1//2B]^(m)`
`therefore (Rate_(2))/(Rate_(1))=[2]^(n)[1//2]^(m)=2^(n).2^(-m)=2^(n-m)`