For the reaction 6H^+ + BrO_3^- + 5Br^- to 3B | vedclass

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(C) The rate of reaction is defined as the rate of change of concentration of any reactant or product divided by its stoichiometric coefficient.For th

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

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(C) The rate of reaction is defined as the rate of change of concentration of any reactant or product divided by its stoichiometric coefficient.For the given reaction: $6H^+ + BrO_3^- + 5Br^- 	o 3Br_2 + 3H_2O$The rate of reaction is given by:Rate $= -\frac{1}{6} \frac{d[H^+]}{dt} = -\frac{d[BrO_3^-]}{dt} = -\frac{1}{5} \frac{d[Br^-]}{dt} = \frac{1}{3} \frac{d[Br_2]}{dt} = \frac{1}{3} \frac{d[H_2O]}{dt}$Comparing the terms for $Br^-$ and $Br_2$:$-\frac{1}{5} \frac{d[Br^-]}{dt} = \frac{1}{3} \frac{d[Br_2]}{dt}$Multiplying both sides by $5$,we get:$-\frac{d[Br^-]}{dt} = \frac{5}{3} \frac{d[Br_2]}{dt}$

For the reaction $6H^+ + BrO_3^- + 5Br^- \to 3Br_2 + 3H_2O$,which of the following relations is correct for the rate of consumption and formation of products?

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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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