The bromination of acetone that occurs in acid solution is represented by this equation.
`CH_(3)COCH_(3) (aq) + Br_(2) (aq) rarr`
`CH_(3)COCH_(2) Br(aq) + H^(+) (aq) + Br(aq)`
These kinetic data were obtained for given reaction concentrations.
Initial concentration, `M`
`{:([CH_(2)COCH_(3)],[Br_(2)],[H^(+)],("Initail rate) (disappearance of "Br_(2)),),(0.30,0.05,0.05,5.7 xx 10^(-5),),(0.30,0.10,0.05,5.7xx10^(-5),),(0.30,0.10,0.10,1.2xx10^(-4),),(0.40,0.5,0.20,3.1xx10^(-4),):}`

Options

A. `Rate = k[CH_(3)COCH_(3)][Br_(2)][H^(+)]`
B. `Rate = k[CH_(3)COCH_(3)][H^(+)]`
C. `Rate = k[CH_(3)COCH_(3)][Br_(2)]`
D. `Rate = k[CH_(3)COCH_(3)][Br_(2)][H^(+)]^(2)`

Answer

B

Explanation

Writing the rate law, we have
Rate `=k[CH_(3)COCH_(3)]^(x)[Br_(2)]^(y)[H^(+)]^(z)`
According to the given data, we have
`Rate_(1)=k(0.30)^(x)0.05)^(y)(0.05)^(z)=5.7xx10^(-5)`
`Rate_(2)=k(0.30)^(x)0.10)^(y)(0.05)^(z)=5.7xx10^(-5)`
`Rate_(3)=k(0.30)^(x)0.10)^(y)(0.10)^(z)=1.2xx10^(-4)`
`Rate_(4)=k(0.40)^(x)0.05)^(y)(0.20)^(z)=3.1xx10^(-4)`
Dividing `Rate_(2)` by `Rate_(1)` , we get
`1=((0.10)/(0.05))`
Thus, `y=0`
Dividing `Rate_(3)` by `Rate_(2)` , we get
`(1.2xx10^(-4))/(5.7xx10^(-5))=((0.10)/(0.05))^(z)`
`2=2^(z)` implies `2^(1)=2^(z)`
Thus, `z=1`
Dividing `Rate_(4)` by `Rate_(1)` , we get `(y=0,z=1)`
`(3.1xx10^(-4))/(5.7xx10^(-5))=((0.40)/(0.30))^(x)((0.20)/(0.05))`
`5.44=(1.33)^(x)(4)`
`1.36=(1.33)^(x)`
or `x=1`
Thus, Rate low is Rate `=k[CH_(3)COOH_(3)]^(1)[Br_(2)]^(0)[H^(+)]^(1)`
or Rate `=k[CH_(3)COOH_(3)][H^(+)]`