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

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