Question

For the reaction $H^{+}+Br{O}_3^{-}+3B{r}^{-}\to 5B{r}_2+H_{2}O$ which of the following relation correctly represents the consumption & formation of reactants and product:

• A. $\frac{d\left[B{r}^{-}\right]}{dt}=-\frac{3}{5}\frac{d\left[B{r}_2\right]}{dt}$
• B. $\frac{d\left[B{r}^{-}\right]}{dt}=\frac{3}{5}\frac{d\left[B{r}_2\right]}{dt}$
• C. $\frac{d\left[B{r}^{-}\right]}{dt}=-\frac{5}{3}\frac{d\left[B{r}_2\right]}{dt}$
• D. $\frac{d\left[B{r}^{-}\right]}{dt}=\frac{5}{3}\frac{d\left[B{r}_2\right]}{dt}$

Answer 1

The correct option is A $\frac{d\left[B{r}^{-}\right]}{dt}=-\frac{3}{5}\frac{d\left[B{r}_2\right]}{dt}$

$H^{+}+{BrO}_3^{-}+3{Br}^{-}⟶{5Br}_2+H_{2}O.$

Applying rate law.

$\frac{-1}{3}\frac{d\left[{Br}^{-}\right]}{dt}=\frac{1}{5}\frac{d\left[{Br}_2\right]}{dt}$

$\Rightarrow \frac{d\left[{Br}^{-}\right]}{dt}=\frac{-3}{5}\frac{d\left[{Br}_2\right]}{dt}$