Question

For the reaction, $Br{O}_3^{-}\left(aq\right)+5B{r}^{-}\left(aq\right)+6H^{+}\to 3B{r}_2\left(l\right)+3H_{2}O\left(l\right)$, the rate equation can be expressed in two ways
$\frac{d\left[B{r}_2\right]}{dt}=k\left[B{r}^{-}\right]\left[Br{O}_3\right][H^{+}{]}^2$ and
$-\frac{d\left[B{r}^{-}\right]}{dt}=k^{\prime }\left[B{r}^{-}\right]\left[Br{O}_3\right][H^{+}{]}^2$
Here, k and k' are related as

Answer 1

For a reaction:
$aA\to bB$
$a,b\text{are stoichiometric coefficients of reactant A and product B respectively}$
$\text{Rate}=-\frac{1}{a}\times \frac{d\left[A\right]}{dt}=+\frac{1}{b}\times \frac{d\left[B\right]}{dt}$
So, for the reaction:
$Br{O}_3^{-}\left(aq\right)+5B{r}^{-}\left(aq\right)+6H^{+}\to 3B{r}_2\left(l\right)+3H_{2}O\left(l\right)$

Writing rate in terms of $B{r}^{-}$ and $B{r}_2$
$\text{Rate}=-\frac{1}{5}\times \frac{d\left[B{r}^{-}\right]}{dt}=+\frac{1}{3}\times \frac{d\left[B{r}_2\right]}{dt}$

$\Rightarrow \frac{d\left[B{r}_2\right]}{dt}=-\frac{3}{5}\frac{d\left[B{r}^{-}\right]}{dt}.....eqn1$
Given:
$\frac{d\left[B{r}_2\right]}{dt}=k\left[B{r}^{-}\right]\left[Br{O}_3\right][H^{+}{]}^2......eqn(2)$
$-\frac{d\left[B{r}^{-}\right]}{dt}=k^{\prime }\left[B{r}^{-}\right]\left[Br{O}_3\right][H^{+}{]}^2.......eqn(3)$
Substituting equation (1) in (3), we get,
$\frac{5}{3}\frac{d\left[B{r}_2\right]}{dt}=k^{\prime }\left[B{r}^{-}\right]\left[Br{O}_3\right][H^{+}{]}^2$
$\frac{d\left[B{r}_2\right]}{dt}=\frac{3}{5}{k}^{\prime }\left[B{r}^{-}\right]\left[Br{O}_3\right][H^{+}{]}^2.....eqn(4)$

From equation (2) and (4),
$1=\frac{5}{3}\frac{k}{k^{\prime }}$
$3k^{\prime }=5k$