When the concentration of a reactant in reaction `A rarr B` is increased by `8` times but rate increases only `2` times, the order of the reaction would be

Options

A. 2
B. `1/3`
C. 3
D. `1/2`

Explanation

`R_(1) = k [A]^(x) …. i) R_(2) = [8A]^(x) therefore (R_(2))/(R_(1)) = (k [8A]^(x))/(k[A]^(x)) = 2`
`therefore 2 = 8^(x) or 2^(3x) = 2 or 3x = 1 or x = (1)/(3)` .