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

Correct option is 1k1=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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k1=5k2=1.66k3

3k1=k2=5k3

5k1=3k2=k3

k1=k2=k3

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detailed solution

Correct option is A

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

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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