The rate law for 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 cocentration of B the ratio of the new rate to the earlier rate of reaction will be :

Options

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

Answer

C

Explanation

(C ) Earlier rate =`k a^(n) b^(m)`
New rate =k `(2a)^(n) ((1)/(2)b)^(m)`
`("New rate")/("Earlier rate") = (2^(n)a^(n)b^(m) 2^(-m))/(a^(n)b^(m))= 2n^(n) .2^(-m) =2^(n-m)`