Answer:
[tex]1.236\times 10^{-4} M/s[/tex] is the average rate of formation of bromine gas during the same time interval.
Explanation:
The given reaction :
[tex]5Br^-(aq)+BrO^{3-}(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)[/tex]
The average rate of consumption of [tex]Br^-=-\frac{d[Br^-]}{dt}= 2.06\times 10^{-4} M/s[/tex]
Rate of the reaction = R
[tex]R=\frac{-1}{5}\frac{d[Br^-]}{dt}[/tex]
[tex]R=\frac{1}{5}\times 2.06\times 10^{-4} M/s[/tex]
[tex]R=4.12\times 10^{-5}M/s[/tex]
The average rate of formation of [tex]Br_2=\frac{d[Br_2]}{dt}[/tex]
[tex]R=\frac{1}{3}\frac{d[Br_2]}{dt}[/tex]
[tex]\frac{d[Br_2]}{dt}=R\times 3=4.12\times 10^{-5}M/s\times 3=1.236\times 10^{-4} M/s[/tex]
[tex]1.236\times 10^{-4} M/s[/tex] is the average rate of formation of bromine gas during the same time interval.
