For the reaction, \(BrO_3^-(aq)+5Br^-(aq)+6H^+ \to 3Br_2(l)+3H_2O(l)\), the rate equation can be expressed in two ways
\(\dfrac{d[Br_2]}{dt}= k [Br^-][BrO_3][H^+]^2\) and
\(-\dfrac{d[Br^-]}{dt}= k ' [Br^-][BrO_3][H^+]^2\)
Here, k and k' are related as

For a reaction:
\(aA \to bB\)
\(a,b \text{ are stoichiometric coefficients of reactant A and product B respectively}\)
\(\text{Rate}=-\dfrac{1}{a}\times\dfrac{d[A]}{dt}=+\dfrac{1}{b}\times \dfrac{d[B]}{dt}\)
So, for the reaction:
\(BrO_3^-(aq)+5Br^-(aq)+6H^+ \to 3Br_2(l)+3H_2O(l)\)

Writing rate in terms of \(Br^-\) and \(Br_2\)
\(\text{Rate}=-\dfrac{1}{5}\times\dfrac{d[Br^-]}{dt}=+\dfrac{1}{3}\times \dfrac{d[Br_2]}{dt}\)

\( \Rightarrow\dfrac{d[Br_2]}{dt}=-\dfrac{3}{5}\dfrac{d[Br^-]}{dt}.....eqn1\)
Given:
\(\dfrac{d[Br_2]}{dt}= k [Br^-][BrO_3][H^+]^2......eqn(2)\)
\(-\dfrac{d[Br^-]}{dt}= k ' [Br^-][BrO_3][H^+]^2.......eqn(3)\)
Substituting equation (1) in (3), we get,
\(\dfrac{5}{3}\dfrac{d[Br_2]}{dt}= k ' [Br^-][BrO_3][H^+]^2\)
\(\dfrac{d[Br_2]}{dt}= \dfrac{3}{5}k ' [Br^-][BrO_3][H^+]^2..... eqn (4)\)

From equation (2) and (4),
\(1 = \dfrac{5}{3} \dfrac{k}{k'}\)
\(3k' = 5k\)