The rate of reaction is expressed by dividing the rate of change of concentration of any reactant or product by its stoichiometric coefficient. For reactants, the rate of change is negative (disappearance), and for products, it is positive (appearance).

For the reaction: $BrO_3^- + 5Br^- + 6H^+ \rightarrow 3Br_2 + 3H_2O$

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

To find the relationship between the rate of appearance of $Br_2$ and the rate of disappearance of $Br^-$, we equate their respective terms: $+\frac{1}{3}\frac{d[Br_2]}{dt} = -\frac{1}{5}\frac{d[Br^-]}{dt}$

Multiplying both sides by 3: $\frac{d[Br_2]}{dt} = -\frac{3}{5}\frac{d[Br^-]}{dt}$