rate of formation of bromine ∆[Br2 ]/∆t, is 6.0 x 10-6 M/s What is the rate of consumption of H+ , ∆[H^+ ]/∆t? 5Br−+BrO3−+6H+⟶3Br2 +3H2O?

Answer 1

6.0x10^-6 M/s Br2 * (6 H+ / 3 Br2) = 1.2x10^-5 M/s H+

Answer 2

5 Br{-} + BrO3{-} + 6 H{+} ⟶ 3 Br2 + 3 H2O

(6.0 x 10^-6 M/s) x (6 mol H{+} / 3 mol Br2) = - 1.2 x 10^-5 M/s H{+}