In the reaction,BrO_{3(aq)}^{-} + 5Br_{(aq)}^ | vedclass

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(D) For the reaction $BrO_{3(aq)}^{-} + 5Br_{(aq)}^{-} + 6H^{+} 	o 3Br_{2(l)} + 3H_2O_{(l)}$,the rate of reaction is given by:Rate $= -\frac{d[BrO_3^

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$\frac{d[Br_2]}{dt} = -\frac{3}{5} \frac{d[Br^{-}]}{dt}$

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(D) For the reaction $BrO_{3(aq)}^{-} + 5Br_{(aq)}^{-} + 6H^{+} 	o 3Br_{2(l)} + 3H_2O_{(l)}$,the rate of reaction is given by:Rate $= -\frac{d[BrO_3^-]}{dt} = -\frac{1}{5} \frac{d[Br^-]}{dt} = -\frac{1}{6} \frac{d[H^+]}{dt} = \frac{1}{3} \frac{d[Br_2]}{dt} = \frac{1}{3} \frac{d[H_2O]}{dt}$.We need to relate the rate of appearance of $Br_2$ to the rate of disappearance of $Br^-$.From the expression: $\frac{1}{3} \frac{d[Br_2]}{dt} = -\frac{1}{5} \frac{d[Br^-]}{dt}$.Multiplying both sides by $3$,we get:$\frac{d[Br_2]}{dt} = -\frac{3}{5} \frac{d[Br^-]}{dt}$.

In the reaction,$BrO_{3(aq)}^{-} + 5Br_{(aq)}^{-} + 6H^{+} \to 3Br_{2(l)} + 3H_2O_{(l)}$. The rate of appearance of bromine $(Br_2)$ is related to the rate of disappearance of bromide ions as:

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