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

Answer

B

Explanation

`O_(3) underset(k_(-1))overset(k_(1))(hArr) O_(2) + [O]` (fast)
`[O] + O_(3) overset(K_(2))(to) 2O_(2)` (slow)
Rate of reaction is determined by slow step hence, Rate `= K_(2) [O][O_(3)]` ltbr? [O] is unstable intermediate so subsitute the value of
[O] in above equation.
Rate of forward reaction `= k_(1) [O_(3)]`
Rate of backward reaction `= k_(-1) [O_(2)][O]`
At equilibrium,
Rate of forward reaction = Rate of backward reaction
`k_(1) [O_(3)] = k_(-1) [O_(2)][O]`
`[O] = (k_1)[O_(3)])/(k_(-1)[O_(2)])`
Rate = `k_(2) ((k_(1)[O_(3)])/(k_(-1)[O_(2)])) [O_(3)] = (k[O_(3)]^(2))/([O_(2)])`