The bromination of acetone that occurs in acid solution is represented by this equation
`CH_(3)COCH_(3)(aq.)rarr CH_(3)COCH_(2)Br(aq.)+H^(+)(aq.)+Br^(-)(aq.)`
These kinetic data were obtained from given reaction concentrations.
Initial concentrations, (M)
`{:("["CH_(3)COCH_(3)"]",[Br_(2)],[H^(+)]),(" 0.30",0.05,0.05),(" 0.30",0.10,0.05),(" 0.30",0.10,0.10),(" 0.40",0.05,0.20):}`
Initial rate, disappearance of `Br_(2),Ms^(-1)`
`5.7xx10^(-5)`
`5.7xx10^(-5)`
`1.2xx10^(-4)`
`3.1xx10^(-4)`
Based on these data, the rate equation is :

Options

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

Answer

A

Explanation

Let the order of reaction wrt `CH_(3)COOCH_(3) .Br_(2) and H^(+) are x,y,and z ` respectively thus .
`Rate (I) =[CH_(3)COCH_(3)]^(x)[Br_(2)]^(y)[H^(+)]^(z) `
`4.7xx10^(-5)=(0.30)^(x)(0.05)^(y)(0.05)^(z)`
`5.7xx10^(-5)=(0.30^(x)(0.10)^(y)(0.05)^(z)`
`1.2xx10^(-4)=(0.30)^(x)(0.10)^(y)(0.10)^(z)`
`3.1xx10^(-4)=(0.40)^(x)(0.05)^(y)(0.20)^(z)`
from Eqs (i) and(ii)
`1=((1)/(2))^(y)or 1^(@)=((1)/(2))^(y)`
`y=0`
from Eqs(ii) and (iii)
`z=1`
from Eqs.(i) and (iv)
`x=1`
thus rate law`prop |CH_(2)COCH_(3)|^(1)|Br_(2)|^(0)[H^(+)]^(1)`
`=K[CH_(3)COCH_(3)][H^(+)]`