For the reaction \(5Br^- (aq) + BrO_3^- (aq) + 6H^+ (aq) \longrightarrow 3Br_2 (aq) + 3H_2O(l)\)
The rate of the reaction is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → one-fifth of rate of disappearance of \(Br^-\) ions
When the stoichiometric coefficients are not all one, the rate of reaction is the rate of change of each species divided by its coefficient (minus sign for reactants):
\( \ce{5Br^- + BrO3^- + 6H^+ -> 3Br2 + 3H2O} \)
Rate = \( -\frac{1}{5}\frac{\Delta[Br^-]}{\Delta t} = -\frac{\Delta[BrO_3^-]}{\Delta t} = -\frac{1}{6}\frac{\Delta[H^+]}{\Delta t} = \frac{1}{3}\frac{\Delta[Br_2]}{\Delta t} = \frac{1}{3}\frac{\Delta[H_2O]}{\Delta t} \)
So the rate of reaction is one-fifth of the rate of disappearance of Br−.
Why the other options are incorrect
Option (2) one-fifth of rate of formation of \(Br_2\) — Incorrect. The coefficient of \( \ce{Br2} \) is 3, so the rate is one-third of its rate of formation.
Option (3) one-third of rate of disappearance of \(BrO_3^-\) ions — Incorrect. The coefficient of \( \ce{BrO3^{-}} \) is 1, so the rate equals its rate of disappearance.
Option (4) one-fourth of rate of formation of \(H_2O\) — Incorrect. The coefficient of \( \ce{H2O} \) is 3, so the rate is one-third of its rate of formation.
NCERT (2026-27 edition):
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.”