For the reaction 5Br^-(aq) + BrO3^-(aq) + 6H^+(aq) -> 3Br2(aq) + 3H2O…

The correct answer is Option (4) → A-III, B-I, C-IV, D-II

When the stoichiometric coefficients are not equal to one, the rate of disappearance of each reactant and the rate of appearance of each product is divided by its coefficient, so that every term gives the same rate of reaction r:

    \( \displaystyle r = -\frac{1}{5}\frac{\Delta[\ce{Br^-}]}{\Delta t} = -\frac{\Delta[\ce{BrO3^-}]}{\Delta t} = -\frac{1}{6}\frac{\Delta[\ce{H+}]}{\Delta t} = \frac{1}{3}\frac{\Delta[\ce{Br2}]}{\Delta t} \)

So the rate for each species is its coefficient times r.

(A) Rate of disappearance of Br⁻ — This matches (III) 5r; five Br⁻ ions are used up for every one reaction event.

(B) Rate of disappearance of BrO₃⁻ — This matches (I) r; its coefficient is one, so its rate equals the rate of reaction.

(C) Rate of disappearance of H⁺ — This matches (IV) 6r; the coefficient of H⁺ is six.

(D) Rate of appearance of Br₂ — This matches (II) 3r; three Br₂ molecules form for every reaction event.

NCERT (2026-27 edition):
3.1 Rate of a Chemical Reaction – “For expressing the rate of such a reaction where stoichiometric coefficients of reactants or products are not equal to one, rate of disappearance of any of the reactants or the rate of appearance of products is divided by their respective stoichiometric coefficients.” … “Similarly, for the reaction”

    \( \displaystyle \ce{5Br^-(aq) + BrO3^-(aq) + 6H^+(aq) -> 3Br2(aq) + 3H2O(l)} \)

    \( \displaystyle \text{Rate} = -\frac{1}{5}\frac{\Delta[\ce{Br^-}]}{\Delta t} = -\frac{\Delta[\ce{BrO3^-}]}{\Delta t} = -\frac{1}{6}\frac{\Delta[\ce{H+}]}{\Delta t} = \frac{1}{3}\frac{\Delta[\ce{Br2}]}{\Delta t} = \frac{1}{3}\frac{\Delta[\ce{H2O}]}{\Delta t} \)