For the reaction $5Br_{(aq)}^{-} + BrO_{3_{(aq)}}^{-} + 6H_{(aq)}^{+} \rightarrow 3Br_{2_{(aq)}} + 3H_{2}O_{(l)}$,if $\frac{-\Delta[Br^{-}]}{\Delta t} = 4.2 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$,calculate the rate of formation of $Br_{2}$,i.e.,$\frac{\Delta[Br_{2}]}{\Delta t}$.
$(2.52 \times 10^{-3} \ MOL \ L^{-1} \ S^{-1})$ From the stoichiometry of the balanced chemical equation,the rate of reaction is given by:
$\frac{-1}{5} \frac{\Delta[Br^{-}]}{\Delta t} = \frac{1}{3} \frac{\Delta[Br_{2}]}{\Delta t}$
Given that $\frac{-\Delta[Br^{-}]}{\Delta t} = 4.2 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$,we substitute this into the expression:
$\frac{1}{5} (4.2 \times 10^{-3}) = \frac{1}{3} \frac{\Delta[Br_{2}]}{\Delta t}$
$\frac{\Delta[Br_{2}]}{\Delta t} = \frac{3}{5} \times 4.2 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$
$\frac{\Delta[Br_{2}]}{\Delta t} = 2.52 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$
$\frac{-1}{5} \frac{\Delta[Br^{-}]}{\Delta t} = \frac{1}{3} \frac{\Delta[Br_{2}]}{\Delta t}$
Given that $\frac{-\Delta[Br^{-}]}{\Delta t} = 4.2 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$,we substitute this into the expression:
$\frac{1}{5} (4.2 \times 10^{-3}) = \frac{1}{3} \frac{\Delta[Br_{2}]}{\Delta t}$
$\frac{\Delta[Br_{2}]}{\Delta t} = \frac{3}{5} \times 4.2 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$
$\frac{\Delta[Br_{2}]}{\Delta t} = 2.52 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$
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