Answer to Question #119747 in Chemistry for A

Question

5Br−(aq)+BrO−3(aq)+6H+(aq)→3Br2(aq)+3H2O(l)

The average rate of consumption of Br− is 1.06×10−4 M/s over the first two minutes. What is the average rate of formation of Br2 during the same time interval?

Answer 1

In the reaction

5Br⁻(aq)+BrO₃^-(aq)+6H⁺(aq)→3Br₂(aq)+3H₂O(l)

5 bromide ions react and produce 3 bromine molecules. The rate of reaction equals the rate of disappearance of a reagent divided by its stoichiometric coefficient:

$r = -\frac{∆[Br^-]}{5∆t}$ .

The same relation is valid for the rate of formation of the bromine:

$r =\frac{∆[Br_2]}{3∆t}$ .

Using these two equations, we can easily calculate the rate of formation of Br₂:

$\frac{∆[Br_2]}{∆t} = -\frac{3∆[Br^-]}{5∆t} = 1.06·10^{-4}·3/5 = 6.36·10^{-5}$ M/s.

Answer: the average rate of formation of Br₂ during the same time interval is 6.36·10^-5 M/s.