The reaction,

2BrO2(g) + O2(g) → 2BrO3(g)

obeys the rate law: rate = kobs[BrO2]2[O2] (where kobs is the experimentally observed rate constant). Which of the following mechanisms is consistent with the experimental rate law?

Question

The reaction,

2BrO2(g) + O2(g) → 2BrO3(g)

obeys the rate law:  rate = kobs[BrO2]2[O2] (where kobs is the experimentally observed rate constant). Which of the following mechanisms is consistent with the experimental rate law?

BlackTom AI · Accepted Answer

The question gives the experimental rate law for the reaction 2BrO2(g) + O2(g) → 2BrO3(g): rate = kobs[BrO2]^2[O2]. Now I will evaluate each proposed mechanism to see which one can produce that rate law.

Option 1: All of these.
- This would be correct only if every listed mechanism independently yields the rate law rate = kobs[BrO2]^2[O2]. We should test each mechanism; if any one mechanism fails to produce the observed rate expression, this option cannot be correct. Therefore we must assess the others before accepting this.

Option 2: Two-Step Mechanism: BrO2(g) + O2(g) ⇌ BrO3(g) + O    (fast equilibrium); BrO2(g) + O → BrO3(g)           (slow)
- In the proposed first step, BrO2 and O2 form BrO3 and O in a fast equilibrium. The second, slow step consumes BrO2 and O to form BrO3. If the slow step is BrO2 + O → BrO3, its rate law would be rate = kslow[BrO2][O]. The steady-state or pre-equilibrium assumption for the fast step would relate [O] to reactants, but there is no direct dependence on [O2] in the slow step or a clear path to generate a [BrO2]^2 term in the overall rate law. This setup more naturally leads to a rate law involving [O], not [BrO2]^2[O2]. Thus this mechanism does not align with the given rate law.

Option 3: Two-Step Mechanism: BrO2 + BrO2 ⇌ Br2O4       (fast equilibrium); Br2O4 + O2 → 2BrO3       (slow)
- Here, two BrO2 molecules rapidly form Br2O4 (pre-equilibrium): BrO2 + BrO2 ⇌ Br2O4. The slow step is Br2O4 + O2 → 2BrO3. The rate is governed by the concentration of the species in the slow step, i.e., rate = kslow[Br2O4][O2]. From the fast equilibrium, [Br2O4] is proportional to [BrO2]^2 (specifically [Br2O4] ≈ K_eq[BrO2]^2). Substituting gives rate ∝ [BrO2]^2[O2], which matches the experimental rate law. This mechanism is consistent with the observed rate law.

Option 4: Two-Step Mechanism: BrO2 + O2 → BrO4 (slow); BrO4 + BrO2 → 2BrO3 (fast)
- The slow first step would yield rate = kslow[BrO2][O2], not involving [BrO2]^2. Even with a subsequent fast step, the rate law is determined by the slow step, so the overall rate would not include a squared dependence on [BrO2]. Therefore this mechanism cannot reproduce the experimental rate law.

Option 5: Two-Step Mechanism: BrO2 + BrO2 ⇌ BrO3 + BrO    (fast equilibrium); BrO + O2 → BrO3 (slow)
- The fast equilibrium between BrO2, BrO3, and BrO would set relationships among their concentrations, but the slow step BrO + O2 → BrO3 would produce a rate law that depends on [BrO] and [O2], not [BrO2]^2[O2]. Without a clear pathway to convert [BrO2] into the needed [BrO] or to generate the exact squared dependence, this mechanism does not yield the observed rate expression.

In summary, only Option 3 provides a rate-determining step whose rate expression, combined with the fast pre-equilibrium relation [Br2O4] ∝ [BrO2]^2, yields rate ∝ [BrO2]^2[O2], matching the experimental rate law. The other options fail to reproduce the required dependence on [BrO2] and/or [O2].

The reaction, 2BrO2(g) + O2(g) → 2BrO3(g) obeys the rate law: rate = kobs[BrO2]2[O2] (where kobs is the experimentally observed rate constant). Which of the following mechanisms is consistent with the experimental rate law?

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