Answer
Pre-equilibrium Conditions in Reaction Mechanisms
In reactions involving multiple steps, some early steps are fast and reversible, establishing equilibrium before the slow, rate-determining step occurs.
This is the principle behind the pre-equilibrium approximation. The concentrations of intermediates are expressed in terms of original reactants using equilibrium constants, so the final rate law includes only measurable species.
Elementary Steps
- Fast equilibrium: BrO₃⁻ + H⁺ ⇌ HBrO₃
- Fast equilibrium: HBrO₃ + H⁺ ⇌ H₂BrO₃⁺
- Slow (rate-determining step): H₂BrO₃⁺ + Br⁻ → (Br–BrO₂) + H₂O
- Fast: (Br–BrO₂) + 4H⁺ + 4Br⁻ → Products
Rate Law Derivation
The slow step determines the rate:
Rate = k₃[H₂BrO₃⁺][Br⁻]
Using Equilibria:
K₁ = [HBrO₃] / ([BrO₃⁻][H⁺])⇒[HBrO₃] = K₁[BrO₃⁻][H⁺]K₂ = [H₂BrO₃⁺] / ([HBrO₃][H⁺])⇒[H₂BrO₃⁺] = K₂[HBrO₃][H⁺]-
Substituting:
[H₂BrO₃⁺] = K₂ ⋅ K₁ ⋅ [BrO₃⁻] ⋅ [H⁺]²
Substituting back into the rate expression:
Rate = k₃ ⋅ K₁ ⋅ K₂ ⋅ [BrO₃⁻] ⋅ [H⁺]² ⋅ [Br⁻]
Let k = k₃ ⋅ K₁ ⋅ K₂, then:
Final Rate Law: Rate = k [BrO₃⁻][Br⁻][H⁺]²
Correct Rate Expressions
- ✅
Rate = k [BrO₃⁻][Br⁻][H⁺]² - ✅
−d[Br⁻]/dt = k [BrO₃⁻][Br⁻][H⁺]² - ✅
d[(Br–BrO₂)]/dt = k [BrO₃⁻][Br⁻][H⁺]²
Incorrect Expressions & Why They’re Wrong
- ❌
k[H⁺][Br⁻][BrO₃⁻]²: BrO₃⁻ should be 1st order, H⁺ should be 2nd order. - ❌
−d[H⁺]/dt = k[Br⁻][BrO₃⁻][H⁺]: H⁺ is 1st order — it should be 2nd. - ❌
−d[Br⁻]/dt = k[Br⁻]²[BrO₃⁻][H⁺]: Br⁻ is 2nd order — should be 1st; H⁺ is only 1st order.
✅ The pre-equilibrium method allows the rate law to be written in terms of initial, measurable reactants only, eliminating intermediate species using known equilibrium relationships.