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The chemical reaction , 2O(3)rarr3O(2) p...

The chemical reaction , `2O_(3)rarr3O_(2)` proceeds as follows :
Step I : `O_(3)hArrO_(2)+O` (fast)
Step II : `O+O_(3)rarr2O_(2)` (slow)
The rate law expression should be-

A

`r=k'[O_(3)][O_(2)]`

B

`r=k'[O_(3)]^(2)[O_(2)]^(-1)`

C

`r=k'[O_(3)]^(2)`

D

unpredicatable

Text Solution

Verified by Experts

As, the slowest step is the rate determining step , hence from step II,
`r= k[O_(3)][O]" "....[1]`
From step I, `K_(eq)=([O_(2)][O])/([O_(3)]" "...[2]`
or, `[O]=(K_(eq)[O_(3)])/([O_(2)])`
From eq. [1] and [2],
`r=kK_(aq)([O_(3)]^(2))/([O_(2)])=k'[O_(3)]^(2)[O_(2)]^(-1)[becausek'=kK_(eq)]`
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