Question #119747
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?
1
Expert's answer
2020-06-04T10:28:32-0400

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=[Br]5tr = -\frac{∆[Br^-]}{5∆t} .

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

r=[Br2]3tr =\frac{∆[Br_2]}{3∆t} .

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

[Br2]t=3[Br]5t=1.061043/5=6.36105\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.


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