QAIPMT-2009-Ch-2122: NEET 2009 Chemistry - Class 12 AIPMT-2009-Ch-2122 Solution: \(HOCH_2CH_2OH\) on heating with periodic acid gives

1. Write the general rate expression for the reaction: For any generic reaction: $aA+bB\longrightarrow cC+dD$ The rate of the reaction can be written in terms of the disappearance of reactants (negative sign) and appearance of products (positive sign), each divided by their respective stoichiometric coefficient: $\text{Rate}=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=\frac{1}{c}\frac{d[C]}{dt}=\frac{1}{d}\frac{d[D]}{dt}$ 2. Apply this to the given reaction: $\mathrm{BrO}_{\mathrm{3}}^{\mathrm{-}}\mathrm{(aq)+5Br}^{\mathrm{-}}\mathrm{(aq)+6H}^{\mathrm{+}}\mathrm{(aq)\longrightarrow 3Br}_{\mathrm{2}}\mathrm{(l)+3H}_{\mathrm{2}}\mathrm{O(l)}$ Focusing only on the bromide ion ($\mathrm{Br}^{\mathrm{-}}$) and bromine ($\mathrm{Br}_{\mathrm{2}}$): $\text{Rate}=-\frac{1}{5}\frac{d[\mathrm{Br}^{\mathrm{-}}]}{dt}=\frac{1}{3}\frac{d[\mathrm{Br}_{\mathrm{2}}]}{dt}$ 3. Isolate the rate of appearance of bromine ($\frac{d[\mathrm{Br}_{\mathrm{2}}]}{dt}$): Multiply both sides of the equation by $3$: $\frac{d[\mathrm{Br}_{\mathrm{2}}]}{dt}=-\frac{3}{5}\frac{d[\mathrm{Br}^{\mathrm{-}}]}{dt}$