For the reaction $5 Br_{(aq)}^{-} + BrO_{3_{(aq)}}^{-} + 6 H_{(aq)}^{+} \rightarrow 3 Br_{2_{(aq)}} + 3 H_{2}O_{(l)}$,write the rate of reaction expression.
The rate of reaction is expressed by dividing the rate of change of concentration of each reactant and product by its respective stoichiometric coefficient,with a negative sign for reactants and a positive sign for products:
Rate $= -\frac{1}{5} \frac{\Delta [Br^{-}]}{\Delta t} = -\frac{\Delta [BrO_{3}^{-}]}{\Delta t} = -\frac{1}{6} \frac{\Delta [H^{+}]}{\Delta t} = \frac{1}{3} \frac{\Delta [Br_{2}]}{\Delta t} = \frac{1}{3} \frac{\Delta [H_{2}O]}{\Delta t}$
Rate $= -\frac{1}{5} \frac{\Delta [Br^{-}]}{\Delta t} = -\frac{\Delta [BrO_{3}^{-}]}{\Delta t} = -\frac{1}{6} \frac{\Delta [H^{+}]}{\Delta t} = \frac{1}{3} \frac{\Delta [Br_{2}]}{\Delta t} = \frac{1}{3} \frac{\Delta [H_{2}O]}{\Delta t}$
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$t / s$ $0$ $30$ $60$ $90$ $[A] / mol \ L^{-1}$ $0.55$ $0.31$ $0.17$ $0.085$
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