Question
In the reaction, $$BrO_3^ - \left( {aq} \right) + 5B{r^ - }\left( {aq} \right) + 6{H^ + }$$ $$ \to 3B{r_2}\left( l \right) + 3{H_2}O\left( l \right)$$ the rate of appearance of bromine $$\left( {B{r_2}} \right)$$ is related to rate of disappearance of bromide ions as following.
A.
$$\frac{{d\left[ {B{r_2}} \right]}}{{dt}} = - \frac{3}{5}\frac{{d\left[ {B{r^ - }} \right]}}{{dt}}$$
B.
$$\frac{{d\left[ {B{r_2}} \right]}}{{dt}} = - \frac{5}{3}\frac{{d\left[ {B{r^ - }} \right]}}{{dt}}$$
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 :
$$\frac{{d\left[ {B{r_2}} \right]}}{{dt}} = - \frac{3}{5}\frac{{d\left[ {B{r^ - }} \right]}}{{dt}}$$
Solution :
$${\text{Rate of appearance/disappearance}}$$
\[=\pm \frac{1}{\begin{align}
& \text{stoichiometric} \\
& \text{coefficient} \\
\end{align}}\times \] $$\frac{{{\text{[reactant or product]}}}}{{{\text{time taken}}}}$$
$${\text{For the reaction,}}$$
$$BrO_3^ - \left( {aq} \right) + 5B{r^ - }\left( {aq} \right) + 6{H^ + }$$ $$ \to 3B{r_2}\left( l \right) + 3{H_2}O\left( l \right)$$
$${\text{Rate of appearance of bromine}}$$ $$\left( {B{r_2}} \right)$$
$$ = + \frac{1}{3}\frac{{d\left[ {B{r_2}} \right]}}{{dt}}$$
$${\text{Rate of disappearance of bromide}}$$ $$ion\,\left( {B{r^ - }} \right)$$
$$\eqalign{
& = - \frac{1}{5}\frac{{d\left[ {B{r^ - }} \right]}}{{dt}} \cr
& {\text{or}}\,\,\,\frac{{d\left[ {B{r_2}} \right]}}{{dt}} = - \frac{3}{5}\frac{{d\left[ {B{r^ - }} \right]}}{{dt}} \cr} $$