[Solution from source] Runs 1 and 2: doubling [Br₂] leaves the rate unchanged, so the reaction is zero order in Br₂. Runs 2 and 3: doubling [H⁺] doubles the rate, so it is first order in H⁺. Run 4 checks acetone: 5.7 × 10⁻⁵ × (0.40/0.30) × (0.20/0.05) ≈ 3.0 × 10⁻⁴, matching 3.1 × 10⁻⁴, so it is first order in acetone. Rate = k[CH₃COCH₃][H⁺]; any form containing [Br₂] contradicts runs 1 and 2. Answer (2).