For the given reaction: $\text{BrO}_3^-(aq) + 5\text{Br}^-(aq) + 6\text{H}^+(aq) \rightarrow 3\text{Br}_2(l) + 2\text{H}_2\text{O}(l)$ 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: $\text{Rate} = -\frac{d[\text{BrO}_3^-]}{dt} = -\frac{1}{5}\frac{d[\text{Br}^-]}{dt} = -\frac{1}{6}\frac{d[\text{H}^+]}{dt} = +\frac{1}{3}\frac{d[\text{Br}_2]}{dt} = +\frac{1}{2}\frac{d[\text{H}_2\text{O}]}{dt}$ Equating the terms for $\text{Br}^-$ and $\text{Br}_2$: $\frac{1}{3}\frac{d[\text{Br}_2]}{dt} = -\frac{1}{5}\frac{d[\text{Br}^-]}{dt}$ Rearranging the equation to solve for the rate of appearance of bromine: $\frac{d[\text{Br}_2]}{dt} = -\frac{3}{5}\frac{d[\text{Br}^-]}{dt}$