How is the rate of formation of Br₂ is related to the rate of disappearance of Br, BrO₃^– and H⁺ in the reaction given below:
$5 Br - (aq) + {BrO}_{3}^{-} (aq) + 6 H^{+} (aq) \to 3 {Br}_{2} (aq) + 3 H_{2} O (l)$

Solution

This can be explain by the rate law

Rate = \frac{1}{3} \frac{d [{Br}_{2}]}{dt} = - \frac{1}{5} \frac{d [{Br}^{-}]}{dt} = \frac{- d [{BrO}_{3}^{-}]}{dt} = \frac{- 1}{6} \frac{d [H^{+}]}{dt}

thus form the above rate law explain the formation of Br₂ is realated to the rate of disapearance of Br.

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