Question

In the following reaction, how is the rate of appearance of $B{r}_2$ related to the rate of disappearance of the ${Br}^{-}$?
$Br{O}^{3-}\left(aq\right)+5B{r}^{-}\left(aq\right)+6H^{+}\to 3B{r}_2\left(l\right)+3H_{2}O\left(l\right)$

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

Answer 1

Answer: (C)

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

For a given equation, rate law is written as: $Rate=-\frac{d\left[reactant\right]}{dt}=\frac{d\left[product\right]}{dt}$
where, $\frac{d\left[reactant\right]}{dt}=$ change in concentration of reactant with time

$\frac{d\left[product\right]}{dt}=$ change in concentration of product with time

Therefore, for the rate law of the given equation,
$Rate=5\frac{d\left[B{r}_2\right]}{dt}=-3\frac{d\left[B{r}^{-}\right]}{dt}$

$Rate=\frac{d\left[B{r}_2\right]}{dt}=-\frac{3}{5}\frac{d\left[B{r}^{-}\right]}{dt}$