Write the elementary steps of the reaction `2" O"_(3)iff3" O"_(2)` and hence derive the rate law expression for this reaction. Comment on the order of reaction.

Explanation

Step 1. `O_(3)iffO_(2)+O ("fast")" "…(i)`
Step 2. `O+O_(3)to2" O"_(2)("slow")" "…(ii)`
From slow step, Rate `=k[O_(3)][O]" "...(iii)`
From eqn. (i), `K_("eqm.")=([O_(2)][O])/([O_(3)])" or "[O]=K_("eqm.")([O_(3)])/([O_(2)])`
Substituting this value in eqn. (iii), we get Rate `=k[O_(3)]xxK_("eqm.")[O_(3)]//[O_(2)]=k'[O_(3)]^(2)[O_(2)]^(-1)`
Thus, order w.r.t. `O_(3)=2" and order w.r.t. "O_(2)=-1`