Answer:
0.720 M/s
Explanation:
Let's consider the following balanced equation.
5Br⁻(aq) + BrO⁻₃(aq) + 6H⁺(aq) → 3 Br₂(aq) + 3H₂O(l)
The molar ratio of Br⁻ to Br₂ is 5:3, that is, when 5 moles of Br⁻ are consumed, 3 moles of Br₂ are produced. If the average rate of consumption of Br⁻ is 1.20 × 10⁻⁴ M/s, the average rate of formation of Br₂ is:
[tex]\frac{1.20molBr^{-}}{L.s} .\frac{3molBr_{2}}{5molBr^{-}} =\frac{0.720molBr_{2}}{L.s}[/tex]
The average rate of formation of [tex]Br_2[/tex] during the same time interval is 0.720 M/s.
The average rate of reaction is calculated by the change in concentration divided by the change in time.
0.720 M/s
The given reaction is
[tex]5Br^- (aq)+BrO^-^3(aq)+6H+(aq) = 3Br_2(aq)+3H_2O(l)[/tex]
The molar ratio of reactant and product is 5:3. When 5 mol of Br⁻ are consumed then, 3 moles of Br₂ are produced.
The average rate of consumption of Br− is 1.20×10−4 M/s
To find the average rate of formation
[tex]\dfrac{1.20 \;mol\;Br^-}{L.s} \times \dfrac{3 \;mol\;Br^2}{5 \;mol\;Br^-} \\\\ =\dfrac{0.720}{L.s}[/tex]
Thus, the average rate of formation of Br₂ is 0.720 m/s.
Learn more about the average rate, here:
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