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

Options

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

Answer

B

Explanation

We can write the rate law as
Rate `=k[A]^(x)[B]^(y)`
Let's take `a` and `b` as us intial concentrations of `A` and `B` respectively, then
`Rate_(1)=k(a)^(x)(b)^(y)`
When the concentration of `b` is doubled, then
`Rate_(2)=K(a)^(x)(2b)^(y)`
Dividing `Rate_(2)` by `Rate_(1)` , we get
`(Rate_(2))/(Rate_(1))=(1//4Rate_(1))/(Rate_(1))=2^(y)`
or `(1)/(4)=2^(y)`
`(1)/(2^(2))=2^(y)`
`2^(-2)=2^(y)`
Thus, `y=-2`
Hence, order with respectant `y` is `-2`