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For the reaction, 5Br−+BrO3−+6H+→3Br2+3H2O5Br^- + BrO_3^- + 6H^+ \rightarrow 3Br_2 + 3H_2O, the correct representation of the consumption and formation of reactants and products is:

A

d[Br−]dt=−35d[Br2]dt\frac{d[Br^-]}{dt} = -\frac{3}{5}\frac{d[Br_2]}{dt}

B

d[Br−]dt=35d[Br2]dt\frac{d[Br^-]}{dt} = \frac{3}{5}\frac{d[Br_2]}{dt}

C

d[Br−]dt=−53d[Br2]dt\frac{d[Br^-]}{dt} = -\frac{5}{3}\frac{d[Br_2]}{dt}

D

d[Br−]dt=53d[Br2]dt\frac{d[Br^-]}{dt} = \frac{5}{3}\frac{d[Br_2]}{dt}

Step-by-Step Solution

For the reaction 5Br−+BrO3−+6H+→3Br2+3H2O5Br^- + 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:

Rate=−15d[Br−]dt=−d[BrO3−]dt=−16d[H+]dt=13d[Br2]dt=13d[H2O]dt\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−Br^- and Br2Br_2: −15d[Br−]dt=13d[Br2]dt-\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−Br^- in terms of the rate of formation of Br2Br_2, we get: d[Br−]dt=−53d[Br2]dt\frac{d[Br^-]}{dt} = -\frac{5}{3}\frac{d[Br_2]}{dt}

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