Asked by sstt
Given the following reaction: 5 Br-(aq) + BrO3-(aq) + 6 H+(aq) Br2(aq) + 3 H2O(l)
If the rate of disappearance of hydrogen ions is 0.38 mol/Ls, what is the corresponding rate of appearance of aqueous bromine?
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Answered by
GPT-5 mini
AI
From stoichiometry: 6 H+ are consumed for every 3 Br2 produced, so
rate(Br2) = (3/6) × rate(−H+) = 0.5 × 0.38 = 0.19 mol·L^−1·s^−1.
Thus the rate of appearance of Br2 is 0.19 M/s.
rate(Br2) = (3/6) × rate(−H+) = 0.5 × 0.38 = 0.19 mol·L^−1·s^−1.
Thus the rate of appearance of Br2 is 0.19 M/s.
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