For 5Br⁻(aq) + BrO₃⁻(aq) + 6H⁺(aq) → 3Br₂(aq) + 3H₂O(l), the rate of reaction can be expressed as:
A−(1/5) Δ[Br⁻]/Δt = −Δ[BrO₃⁻]/Δt = −(1/6) Δ[H⁺]/Δt = (1/3) Δ[Br₂]/Δt
BSame as average rate of Br⁻ alone
C(1/5) Δ[Br⁻]/Δt only
D−Δ[Br⁻]/Δt = −Δ[BrO₃⁻]/Δt
Answer & Solution
Correct answer: A. −(1/5) Δ[Br⁻]/Δt = −Δ[BrO₃⁻]/Δt = −(1/6) Δ[H⁺]/Δt = (1/3) Δ[Br₂]/Δt
Divide each ΔC/Δt by the stoichiometric coefficient: rate = **−(1/5) Δ[Br⁻]/Δt = −Δ[BrO₃⁻]/Δt = −(1/6) Δ[H⁺]/Δt = (1/3) Δ[Br₂]/Δt = (1/3) Δ[H₂O]/Δt**.
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