In a reaction , `A + B rarr` Product, rate is doubled when the concentration of `B` is doubled, and rate increases by a factor of `8` when the concentration of both the reactants (A and B) are doubled, rate law for the reaction can be written as

Options

A. `Rate = k[A][B]^(2)`
B. `Rate = k[A]^(2)[B]^(2)`
C. `Rate = k[A][B]`
D. `Rate = k[A]^(2)[B]`

Answer

D

Explanation

Let the rate law be
`Rate=k[A]^(x)[B]^(y)`
Let the initial concentrations `A` and `B` be `a` and `b` respectively. According to first observation
`R_(1)k[a]^(x)[b]^(y)`
`R_(2)=2R_(1)=k[a]^(x)[2b]^(y)`
Dividing the `2^(nd)` equation by `1^(st)` , we have
`2=2^(y)` or `2^(1)=2^(2)`
Hence, `y=1`
According to the second observation ltbr. `R_*(1)=k[a]^(x)[b]^(y)`
`R_(2)8R_(1)=k[2a]^(x)[2b]^(y)`
Dividing the `2^(nd)` equation by `1^(st)` , we get
`8=2^(x).2^(y)`
`2^(3)=2^(x+y)`
or `x+y=3`
Substituting the value of `y=1` , we have
`x+1=3`
or `x=3-1=2`
Thus, the correct rate law is
`Rate=K[A]^(2)[B]`