Question

For the reaction given below,
$ 5Br^- _{(aq)} BrO^{-}_{3 (aq)} + 6H^{+}_{aq} \to 3Br_{2(l)} +3H_2O_{(l)} $
The rate of formation of $ Br_2 $ is related to the rate of consumption of $ Br^- $ by the following relation

• A. $ \frac{d\left[Br_{2}\right]}{dt} = -\frac {5}{3}\frac {d\left[Br^{-}\right]}{dt} $
• B. $ \frac{d\left[Br_{2}\right]}{dt} = -\frac{d\left[Br^{-}\right]}{dt} $
• C. $ \frac{d\left[Br_{2}\right]}{dt} = \frac{5}{3}\frac{d\left[Br^{-}\right]}{dt} $
• D. $ \frac{d\left[Br_{2}\right]}{dt} = -\frac{3}{5}\frac{d\left[Br^{-}\right]}{dt} $

Answer 1

Answer: (D)

$5Br^- + BrO_3^- + 6H^+ \to 3Br_2 + 3H_2O$
$\therefore \frac{1}{3} \cdot \frac{d[Br_2]}{dt} = - \frac{1}{5} \cdot \frac{d[Br^-]}{dt}$
$ = - \frac{d[BrO_3^-]}{dt}$
$\therefore \frac{d[Br_2]}{dt} = - \frac{3}{5} \cdot \frac{d[Br]}{dt}$