The chemical reaction `2O_(3)rarr3O_(2)` proceeds as follows :
`O_(3)rarrO_(2)O" "` (fast)
`O+O_(3)rarr2O_(2)" "` (slow)
The rate law expression should be :

Options

A. `r=k'[O_3]^(2)`
B. `r=k'[O_(3)]^(2)[O_(2)]^(-1)`
C. `r=k'[O_(3)][O_(2)]^(-1)`
D. Unpredictable

Answer

B

Explanation

Rate depends upon the slowest step. Hence, from equation,
`O+O_(3)rarr2O_(2)`
`r=k[O_(3)][O]`
and from equation `O_(3)hArrO_(2)+O`
`K_(eq)=([O_(2)][O])/([O_(3)])`
`[O]=(K_(eq)[O_(3)])/([O_(2)])`
`:." " r=k[O_(3)](K_(eq[O_3]))/([O_2])`
`k'[O_3]^(2)[O_2]^(-1)`