The chemical reaction `2O_(3) overset(k_(1))rarr 3O_(2)` proceeds as follows:
`O_(3) overset(k_(eq))hArr O_(2)+O` (fast)
`O + O_(3) overset(k)rarr 2O_(2)` (slow)
What should be the rate law expresison ?

Options

A. `r=k[O_(3)]^(2)[O_(2)]^(-1)`
B. `r=k[O_(3)]^(2)`
C. `r=k[O_(3)][O_(2)]`
D. None of these

Answer

A

Explanation

Rate `= k[O][O_(3)]`
`K=([O_(2)][O])/([O_(3)])`
`:. [O]=(K[O_(3)])/([O_(2)])`
`:.` Rate `= (k.K.[O_(3)])/([O_(2)])xx[O_(3)]`
`=(k[O_(3)]^(2))/([O_(2)])=K'[O_(3)]^(2)[O_(2)]^(-1)`.