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Question:

The chemical reaction 2O3 → 3O2 proceeds as follows:
Step 1: O3 ⇌ O2 + O (fast)
Step 2: O + O3 → 2O2 (slow)
The rate law expression should be:

r=k′[O3][O2]

r=k′[O3]2[O2]

r=k′[O3]2

unpredictable

Solution:

The slow step (step-2) is the rate-determining step. The rate law is determined by the stoichiometry of the slow step. Therefore, the rate law is:
r = k[O][O3]
However, O is an intermediate. We need to express the rate law in terms of reactants only. From the equilibrium in step 1:
K = [O2][O]/[O3]
Solving for [O]:
[O] = K[O3]/[O2]
Substituting this into the rate law:
r = k(K[O3]/[O2])[O3] = kK[O3]2/[O2]
Since k and K are constants, we can combine them into a new rate constant k':
r = k′[O3]2/[O2]
Therefore, the rate law expression is r = k′[O3]2/[O2]