The observed rate law for the reaction 2 O₃ (g) $\rightarrow$ 3O₂ (g) is Rate =k[O₃]²/[O₂].Why is it improbable that the reaction occurs in one step: O₃ (g) + O₃ (g) $\rightarrow$ O₂ (g)+O₂ (g) + O₂ (g)?

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The Answer of The observed rate law for the reaction 2 O₃ (g) $\rightarrow$ 3O₂ (g) is Rate =k[O₃]²/[O₂].Why is it improbable that the reaction occurs in one step: O₃ (g) + O₃ (g) $\rightarrow$ O₂ (g)+O₂ (g) + O₂ (g)?

The Observed Rate Law for the Reaction 2 O3 (G) →\rightarrow→

The Observed Rate Law for the Reaction 2 O₃ (G) $\rightarrow$