Consider the following reactions, O3⇌O2+O (fast .....(i)O+O3⟶2O2 (slow) .......(ii) The rate law expression should be

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r=kO32O2−1

r=kO3O2

r=kO32

r=kO2−1

answer is A.

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Detailed Solution

From Eq. (i) Keq =O2[O]O3or [O]=KeqO3O2From Eq.(ii) rate law =kO3[O]or r=kO3O3O2Keqor r=k′O32O2−1k′=k×Keq

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