For the reaction $5Br^- + BrO_3^- + 6H^+ \rightarrow 3Br_2 + 3H_2O$, the rate of reaction can be expressed in terms of the disappearance of reactants and appearance of products divided by their respective stoichiometric coefficients:

$\text{Rate} = -\frac{1}{5}\frac{d[Br^-]}{dt} = -\frac{d[BrO_3^-]}{dt} = -\frac{1}{6}\frac{d[H^+]}{dt} = \frac{1}{3}\frac{d[Br_2]}{dt} = \frac{1}{3}\frac{d[H_2O]}{dt}$

Equating the rate expressions for $Br^-$ and $Br_2$: $-\frac{1}{5}\frac{d[Br^-]}{dt} = \frac{1}{3}\frac{d[Br_2]}{dt}$

Rearranging this equation to express the rate of consumption of $Br^-$ in terms of the rate of formation of $Br_2$, we get: $\frac{d[Br^-]}{dt} = -\frac{5}{3}\frac{d[Br_2]}{dt}$