For the reaction: 3BrO^- → BrO3^- + 2Br^- in alkaline aqueous solution, the value of the second-orde — Chemical Kinetics Chemistry Question
Question
For the reaction: 3BrO^- → BrO3^- + 2Br^- in alkaline aqueous solution, the value of the second-order rate constant (in BrO^-) in rate law for -d[BrO^-]/dt was found to be 0.06 M^-1 s^-1. The rate constant (in M^-1 s^-1), when the rate law is written as +d[BrO3^-]/dt is a and as +d[Br^-]/dt is b, then the value of (a + b) × 100 is
Answer: 6
💡 Solution & Explanation
Rate = -1/3 d[BrO^-]/dt = +d[BrO3^-]/dt = +1/2 d[Br^-]/dt = k[BrO^-]^2. Given -d[BrO^-]/dt = 0.06[BrO^-]^2. Thus, +d[BrO3^-]/dt = 1/3 (0.06) [BrO^-]^2 = 0.02[BrO^-]^2 (so a = 0.02). +d[Br^-]/dt = 2/3 (0.06) [BrO^-]^2 = 0.04[BrO^-]^2 (so b = 0.04). (a + b) × 100 = (0.02 + 0.04) × 100 = 6. Therefore, correct answer is 6.
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