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

Answer

C

Explanation

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