For the reaction,5Br^-{_{text{(aq)}}} + BrO_3 | vedclass

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Question

(D) For the given reaction: $5Br^-{_{	ext{(aq)}}} + BrO_3^-{_{	ext{(aq)}}} + 6H^+{_{	ext{(aq)}}} ightarrow 3Br_{2	ext{(aq)}} + 3H_2O_{	ext{(l)}

Vedclass · Accepted Answer

$0.01$

Vedclass · Answer

(D) For the given reaction: $5Br^-{_{	ext{(aq)}}} + BrO_3^-{_{	ext{(aq)}}} + 6H^+{_{	ext{(aq)}}} ightarrow 3Br_{2	ext{(aq)}} + 3H_2O_{	ext{(l)}}$
The rate of reaction is expressed as:
$Rate = -\frac{1}{5} \frac{\Delta[Br^{-}]}{\Delta t} = -\frac{\Delta[BrO_3^{-}]}{\Delta t} = -\frac{1}{6} \frac{\Delta[H^{+}]}{\Delta t} = \frac{1}{3} \frac{\Delta[Br_2]}{\Delta t}$
Given that $-\frac{\Delta[Br^{-}]}{\Delta t} = 0.05 \ mol \ L^{-1} \ min^{-1}$.
Equating the rates:
$-\frac{1}{5} \frac{\Delta[Br^{-}]}{\Delta t} = -\frac{\Delta[BrO_3^{-}]}{\Delta t}$
Substituting the given value:
$-\frac{\Delta[BrO_3^{-}]}{\Delta t} = \frac{1}{5} 	imes 0.05 = 0.01 \ mol \ L^{-1} \ min^{-1}$.

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