In the reaction ${BrO}_{3}^{-} (aq) {+ 5Br}^{-} (aq) {+ 6H}^{+} (aq) \to {3Br}_{2} (l) {+ 3H}_{2} O (l)$ The rate of appearance of bromine $({Br}_{2})$ is related to rate of disappearance of bromide ions as following :

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$\frac{d [{Br}_{2}]}{dt} = \frac{5}{3} \frac{d [{Br}^{-}]}{dt}$

$\frac{d [{Br}_{2}]}{dt} = - \frac{5}{3} \frac{d [{Br}^{-}]}{dt}$

$\frac{d [{Br}_{2}]}{dt} = \frac{3}{5} \frac{d [{Br}^{-}]}{dt}$

$\frac{d [{Br}_{2}]}{dt} = - \frac{3}{5} \frac{d [{Br}^{-}]}{dt}$

answer is B.

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Detailed Solution

: Rate of disappearance of ${Br}^{-}$ = rate of appearance of ${Br}_{2}$

$\Rightarrow - \frac{1}{5} \frac{d [{Br}^{-}]}{dt} = \frac{1}{3} \frac{d [{Br}_{2}]}{dt}$

$\Rightarrow \frac{1}{3} \frac{d [{Br}_{2}]}{dt} = - \frac{1}{5} \frac{d [{Br}^{-}]}{dt}$

$\Rightarrow \frac{d [{Br}_{2}]}{dt} = - \frac{3}{5} \frac{d [{Br}^{-}]}{dt}$

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