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For the reaction : (CH(3))(3)CBr(aq)+O...

For the reaction :
`(CH_(3))_(3)CBr(aq)+OH^(-)(aq)rarr(CH_(3))_(3)COH(aq)+Br^(-)(aq)`
it is found that halving the concentration of `(CH_(3))_(3)CBr` causes the reaction rate to be halved but halving the concentration of `OH^(-)` has no effect on the rate. What is the rate law?

A

Rate = `k[(CH_(3))_(3)CBr]^(1//2)[OH^(-)]`

B

Rate = `k[(CH_(3))_(3)CBr]^(2)[OH^(-)]`

C

`k[(CH_(3))_(3)CBr]^(1//2)`

D

`k[(CH_(3))_(3)CBr]`

Text Solution

Verified by Experts

The correct Answer is:
D
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