For the reaction, 5Br−+BrO3−+6H+⟶3Br2+3H2OIf −dBr−dt=k1Br−,−dBrO3−dt=k2Br− dBr2dt=k3Br− then k1,k2 and k3are related as

For the reaction, 5Br−+BrO3−+6H+⟶3Br2+3H2O −15dBr−dt=−dBrO3−dt=13dBr2dt 15k1Br−=k2Br−=13k3Br−or 3k1=15k2=5k3 k1=5k2=1.66k3

For the reaction, 5Br−+BrO3−+6H+⟶3Br2+3H2OIf −dBr−dt=k1Br−,−dBrO3−dt=k2Br− dBr2dt=k3Br− then k1,k2 and k3are related as

For the reaction, 5Br−+BrO3−+6H+⟶3Br2+3H2O −15dBr−dt=−dBrO3−dt=13dBr2dt 15k1Br−=k2Br−=13k3Br−or 3k1=15k2=5k3 k1=5k2=1.66k3

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For the reaction, $5 {Br}^{-} + {BrO}_{3}^{-} + 6 H^{+} ⟶ 3 {Br}_{2} + 3 H_{2} O$
If $- \frac{d [{Br}^{-}]}{dt} = k_{1} [{Br}^{-}], - \frac{d [{BrO}_{3}^{-}]}{dt} = k_{2} [{Br}^{-}]$ $\frac{d [{Br}_{2}]}{dt} = k_{3} [{Br}^{-}]$ then $k_{1}, k_{2} and k_{3}$ are related as

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$5 k_{1} = 3 k_{2} = k_{3}$

$k_{1} = k_{2} = k_{3}$

$k_{1} = 5 k_{2} = 1 .66 k_{3}$

$3 k_{1} = k_{2} = 5 k_{3}$

answer is A.

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Detailed Solution

For the reaction, $5 {Br}^{-} + {BrO}_{3}^{-} + 6 H^{+} ⟶ 3 {Br}_{2} + 3 H_{2} O$
$\frac{- 1}{5} \frac{d [{Br}^{-}]}{dt} = - \frac{d [{BrO}_{3}^{-}]}{dt} = \frac{1}{3} \frac{d [{Br}_{2}]}{dt}$
$\frac{1}{5} k_{1} [{Br}^{-}] = k_{2} [{Br}^{-}] = \frac{1}{3} k_{3} [{Br}^{-}]$
or $\begin{array}{l} 3 k_{1} = 15 k_{2} = 5 k_{3} \\ k_{1} = 5 k_{2} = 1 .66 k_{3} \end{array}$