Respuesta :
Answer:
1) [tex]7.04\times 10^{-5} M/s[/tex] is the average rate of formation of bromine gas during the same time interval.
2) [tex]3.42\times 10^{-4} M/s[/tex] is the average rate of consumption of [tex]H^+[/tex] during the same time interval.
Explanation:
[tex]5Br^-(aq)+BrO_3^{-}(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)[/tex]
1 ) Rate of the reaction : R
[tex]R=-\frac{1}{5}\frac{d[Br^-]}{dt}=\frac{1}{2}\frac{d[Br_2]}{dt}[/tex]
The average rate of consumption of [tex]Br^-=1.76\times 10^{-4} M/s[/tex]
[tex]-\frac{d[Br^-]}{dt}=1.76\times 10^{-4} M/s[/tex]
The average rate of formation of [tex]Br_2=\frac{d[Br_2]}{dt}[/tex]
[tex]-\frac{1}{5}\frac{d[Br^-]}{dt}=\frac{1}{2}\frac{d[Br_2]}{dt}[/tex]
[tex]\frac{1}{5}\times 1.76\times 10^{-4} M/s=\frac{1}{2}\frac{d[Br_2]}{dt}[/tex]
[tex]\frac{d[Br_2]}{dt}=7.04\times 10^{-5} M/s[/tex]
[tex]7.04\times 10^{-5} M/s[/tex] is the average rate of formation of bromine gas during the same time interval.
2 ) Rate of the reaction : R
[tex]R=-\frac{1}{6}\frac{d[H^+]}{dt}=\frac{1}{2}\frac{d[Br_2]}{dt}[/tex]
The average rate of formation of [tex]Br_2=\frac{d[Br_2]}{dt}=1.14\times 10^{-4} M/s[/tex]
The average rate of consumption of [tex]H^+=-\frac{d[H^+]}{dt}[/tex]
[tex]-\frac{1}{6}\frac{d[H^+]}{dt}=\frac{1}{2}\frac{d[Br_2]}{dt}[/tex]
[tex]-\frac{1}{6}\frac{d[H^+]}{dt}=\frac{1}{2}\times 1.14\times 10^{-4} M/s[/tex]
[tex]-\frac{d[H^+]}{dt}=3.42\times 10^{-4} M/s[/tex]
[tex]3.42\times 10^{-4} M/s[/tex] is the average rate of consumption of [tex]H^+[/tex] during the same time interval.
