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In the reaction BrO₃⁻(aq) + 5Br⁻(aq) + 6H⁺ → 3Br₂(l) + 3H₂O(l) The rate of appearance of bromine (Br₂) is related to rate of disappearance of bromide ions as following:

1. d[Br₂]/dt = 5/3 d[Br⁻]/dt
2. d[Br₂]/dt = 5/3 d[Br⁻]/dt
3. d[Br₂]/dt = 3/5 d[Br⁻]/dt
4. d[Br₂]/dt = -5/3 d[Br⁻]/dt

Correct answer: d[Br₂]/dt = 3/5 d[Br⁻]/dt

Solution

From the stoichiometry of the reaction, 5 moles of Br⁻ are consumed for every 3 moles of Br₂ formed. Thus, the rate of appearance of Br₂ is related to the rate of disappearance of Br⁻ by the ratio 3/5.

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