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

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

The Observed Rate Law for the Reaction 2 O3(g) →\to→ 3 O2(g)is Rate = K[O3]2/[O2]

The Observed Rate Law for the Reaction 2 O₃(g) $\to$ 3 O₂(g)is Rate = K[O₃]²/[O₂]