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In the reaction, BrO3−(aq)+5Br−(aq)+6H+(aq)→3Br2(l)+2H2O(l)\text{BrO}_3^-(aq) + 5\text{Br}^-(aq) + 6\text{H}^+(aq) \rightarrow 3\text{Br}_2(l) + 2\text{H}_2\text{O}(l) The rate of appearance of bromine (Br2\text{Br}_2) is related to rate of disappearance of bromide ions as following:

A

d[Br2]dt=−35d[Br−]dt\frac{d[\text{Br}_2]}{dt} = -\frac{3}{5}\frac{d[\text{Br}^-]}{dt}

B

d[Br2]dt=53d[Br−]dt\frac{d[\text{Br}_2]}{dt} = \frac{5}{3}\frac{d[\text{Br}^-]}{dt}

C

d[Br2]dt=−53d[Br−]dt\frac{d[\text{Br}_2]}{dt} = -\frac{5}{3}\frac{d[\text{Br}^-]}{dt}

D

d[Br2]dt=35d[Br−]dt\frac{d[\text{Br}_2]}{dt} = \frac{3}{5}\frac{d[\text{Br}^-]}{dt}

Step-by-Step Solution

For the given reaction: BrO3−(aq)+5Br−(aq)+6H+(aq)→3Br2(l)+2H2O(l)\text{BrO}_3^-(aq) + 5\text{Br}^-(aq) + 6\text{H}^+(aq) \rightarrow 3\text{Br}_2(l) + 2\text{H}_2\text{O}(l) The overall rate of reaction can be expressed in terms of the rate of disappearance of reactants and the rate of appearance of products, divided by their respective stoichiometric coefficients: Rate=−d[BrO3−]dt=−15d[Br−]dt=−16d[H+]dt=+13d[Br2]dt=+12d[H2O]dt\text{Rate} = -\frac{d[\text{BrO}_3^-]}{dt} = -\frac{1}{5}\frac{d[\text{Br}^-]}{dt} = -\frac{1}{6}\frac{d[\text{H}^+]}{dt} = +\frac{1}{3}\frac{d[\text{Br}_2]}{dt} = +\frac{1}{2}\frac{d[\text{H}_2\text{O}]}{dt} Equating the terms for Br−\text{Br}^- and Br2\text{Br}_2: 13d[Br2]dt=−15d[Br−]dt\frac{1}{3}\frac{d[\text{Br}_2]}{dt} = -\frac{1}{5}\frac{d[\text{Br}^-]}{dt} Rearranging the equation to solve for the rate of appearance of bromine: d[Br2]dt=−35d[Br−]dt\frac{d[\text{Br}_2]}{dt} = -\frac{3}{5}\frac{d[\text{Br}^-]}{dt}

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