-d[BrO 3 - ]/dt -d[Br - ]/dt -d[H + ]/dt 5Br-(aq) + BrO3-(aq) + 6H+(aq) --> 3Br2

Question

-d[BrO3-]/dt

-d[Br-]/dt

-d[H+]/dt

5Br-(aq) + BrO3-(aq) + 6H+(aq) --> 3Br2(l) + 3H2O(l) The above reaction is expected to obey the mechanism: BrO3-(aq) + H+(aq) = HBrO3(aq) Fast equilibrium HBrO3(aq) + H+(aq) = H2BrO3+(aq) Fast equilibrium H2BrO3+(aq) + Br-(aq) --> (Br-BrO2)(aq) + H2O(l) Slow (Br-BrO2)(aq) + 4H+(aq) + 4Br-(aq) -->products Fast Here^'s the hint I^'m given (which I also don^'t understand): Think about it this way... Up to and including the rate determiningstep, 1 equivalent of BrO3- is consumed. And the subsequentreaction requires no more BrO3-. Therefore you can use the rateconstants simply to get the rate at which BrO3- disappears. Sincethe last step is rapid and consumes additional Br- and H+, thesespecies must disappear at faster rates than does BrO3- by theappropriate stoichiometric factors. I have determined that the rate equation (with respect to BrO3-)is: I know at least this is right. Other than that, I don^'t know what to do. Please help! Thanks.

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anyone?

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