At 298 K the value of -frac{Delta[Br^{-}]}{D | vedclass

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Question

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

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$\frac{x}{5}$

Vedclass · Answer

(A) 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 defined as the rate of disappearance of reactants divided by their stoichiometric coefficients.
Rate $= -\frac{1}{5} \frac{\Delta[Br^{-}]}{\Delta t} = -\frac{\Delta[BrO_3^{-}]}{\Delta t} = -\frac{1}{6} \frac{\Delta[H^{+}]}{\Delta t} = \frac{1}{3} \frac{\Delta[Br_2]}{\Delta t} = \frac{1}{3} \frac{\Delta[H_2O]}{\Delta t}$
Given that $-\frac{\Delta[Br^{-}]}{\Delta t} = x \ mol \ L^{-1} \ min^{-1}$.
Substituting this into the rate expression:
Rate $= \frac{1}{5} 	imes (-\frac{\Delta[Br^{-}]}{\Delta t}) = \frac{1}{5} 	imes x = \frac{x}{5} \ mol \ L^{-1} \ min^{-1}$.

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