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) [O_(2)]^(-1)`
B. `r = k[O_(3)]^(2)`
C. `r =k [O_(3)][O_(2)]`
D. unpridicatable

Answer

A

Explanation

Let k be the rate constant of given reaction
`2O_(3) rarr 3O_(2)`
From the slowest step r = k `[O] [O_(3)]`
To eliminate [O] (From fast step)
`k_(eq)= ([O_(2)][O])/([O_(3)])`
`[O]=(k_(eq)[O_(3)])/([O_(2)])`
Now, substituting value of [O] in Eq (i)
`r = (k.k_(eq)[O_(3)][O_(3)])/([O_(2)])`
Let k.`k_(eq)=k`
thus r= `k [O_(3)]^(2) [O_(2)]^(-1)`