$5 Br^{-}_{(aq)} + BrO^{-}_{3(aq)} + 6 H^{+}_{(aq)} \rightarrow 3 Br_{2(aq)} + 3 H_{2}O_{(l)}$
The rate of consumption of $H^{+}$ is $x \ mol \ L^{-1} \ s^{-1}$.
$(a)$ What is the rate of consumption of $Br^{-}$?
$(b)$ What is the rate of formation of $Br_{2}$?

For the reaction: $5 Br^{-} + BrO^{-}_{3} + 6 H^{+} \rightarrow 3 Br_{2} + 3 H_{2}O$
The rate of reaction is given by:
$Rate = -\frac{1}{5} \frac{d[Br^{-}]}{dt} = -\frac{d[BrO^{-}_{3}]}{dt} = -\frac{1}{6} \frac{d[H^{+}]}{dt} = \frac{1}{3} \frac{d[Br_{2}]}{dt}$
Given that the rate of consumption of $H^{+}$ is $-\frac{d[H^{+}]}{dt} = x \ mol \ L^{-1} \ s^{-1}$.
$(a)$ Rate of consumption of $Br^{-}$:
$-\frac{1}{5} \frac{d[Br^{-}]}{dt} = -\frac{1}{6} \frac{d[H^{+}]}{dt}$
$-\frac{d[Br^{-}]}{dt} = \frac{5}{6} \times (-\frac{d[H^{+}]}{dt}) = \frac{5}{6}x \ mol \ L^{-1} \ s^{-1}$.
$(b)$ Rate of formation of $Br_{2}$:
$\frac{1}{3} \frac{d[Br_{2}]}{dt} = -\frac{1}{6} \frac{d[H^{+}]}{dt}$
$\frac{d[Br_{2}]}{dt} = \frac{3}{6} \times (-\frac{d[H^{+}]}{dt}) = \frac{1}{2}x \ mol \ L^{-1} \ s^{-1}$.