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

Rate, `r = k[A]^(n)[B]^(m)`
In the first case, let `[A]=aM`
and `[B] = b M`
`:. r_(1)=k a^(n)b^(m)`
In the second case, `[A]=2aM`
and `[B]=(b//2)M`
`:. r_(2)=k(2a)^(n)(b//a)^(n) =ka^(n)b^(m)2^(n-m)`
`(r_(2))/(r_(1))=2^(n-m)`.