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Q

The average rate of consumption of Br− is 1.06×10−4 M/s over the first two minutes

Consider the reaction: 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. The average rate of formation of Br2 during the same time interval is 6.36·10-5 M/s. What is the average rate of consumption of H+ during the same time interval?
A

Here’s the solution:

The rate of the reaction is R

R = – 1/5 * (d*[Br2])/dt

The average rate of consumption of bromine is 1.76 * 10-4 M/s, therefore,

– d*[Br]/dt = 1.76 * 10-4 M/s

The average rate of formation of bromine equals:

1/5 * 1.76 * 10-4 M/s = 7.04*10-5 M/s

Answer: The rate of formation of Br2 (bromine gas) is 7.04*10-5 M/s

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