The rate of reaction between two reactants A and B increases by a factor of 4 if the concentration of reactant B is doubled. The order of this reaction with respect to reactant B is

Options

A. `-1`
B. `-2`
C. 1
D. 2

Answer

B

Explanation

Let rate of reaction
`r = k [A]^(n) [B]^(n)`
`r_(1) = k[A]^(n) [B]^(3)`
`r_2 = k [A]^(n) [2B]^(3) = r_(1) //4`
Thus `2^(y) = 1//4`
i.e. `y=-2`
Hence rate w.r.t. [B] becomes `-2`