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In the reaction, BrO3−(aq)+5Br−(aq)+6H+→3Br2(l)+3H2O(l)BrO_3^-(aq) + 5Br^-(aq) + 6H^+ \rightarrow 3Br_2(l) + 3H_2O(l). The rate of appearance of bromine (Br2Br_2) is related to the rate of disappearance of bromide ions (Br−Br^-) as:

A

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

B

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

C

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

D

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

Step-by-Step Solution

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: BrO3−+5Br−+6H+→3Br2+3H2OBrO_3^- + 5Br^- + 6H^+ \rightarrow 3Br_2 + 3H_2O

The rate expression is: Rate=−d[BrO3−]dt=−15d[Br−]dt=−16d[H+]dt=+13d[Br2]dt=+13d[H2O]dt\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 Br2Br_2 and the rate of disappearance of Br−Br^-, we equate their respective terms: +13d[Br2]dt=−15d[Br−]dt+\frac{1}{3}\frac{d[Br_2]}{dt} = -\frac{1}{5}\frac{d[Br^-]}{dt}

Multiplying both sides by 3: d[Br2]dt=−35d[Br−]dt\frac{d[Br_2]}{dt} = -\frac{3}{5}\frac{d[Br^-]}{dt}

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