Question

For the reaction given below,
$5B{r}_{(aq)}^{-}Br{O}_{3(aq)}^{-}+6H_{aq}^{+}\to 3B{r}_{2(l)}+3H_2{O}_{(l)}$
The rate of formation of $B{r}_2$ is related to the rate of consumption of $B{r}^{-}$ by the following relation

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

Answer 1

Answer: (D)

$5B{r}^{-}+Br{O}_3^{-}+6H^{+}\to 3B{r}_2+3H_{2}O$
$\therefore \frac{1}{3}\cdot \frac{d[B{r}_2]}{dt}=-\frac{1}{5}\cdot \frac{d[B{r}^{-}]}{dt}$
$=-\frac{d[Br{O}_3^{-}]}{dt}$
$\therefore \frac{d[B{r}_2]}{dt}=-\frac{3}{5}\cdot \frac{d[Br]}{dt}$