Question

For the reaction
$5B{r}^{-}(aq)+6H^{+}(aq)+Br{O}_3^{-}(aq)⟶3B{r}_2(aq)+3H_{2}O(I)$
if, $-\frac{\Delta \left[Br{O}_3^{-}\right]}{\Delta t}=0.01mol{L}^{-1}mi{n}^{-1}$
$\frac{\Delta \left[B{r}_2\right]}{\Delta t}$ in $mol{L}^{-1}{\min }^{-1}$ is

• A. 0.01
• B. 0.3
• C. 0.03
• D. 0.005

Answer 1

Answer: (C)

$5B{r}^{-}(aq)+6H^{+}(aq)+Br{O}_3^{-}(aq)⟶3B{r}_2(aq)+3H_{2}O(I)$

Since, we' ve $-\frac{1}{5}\frac{\Delta \left[B{r}^{-}\right]}{\Delta t}=-\frac{\Delta \left[Br{o}_3^{-}\right]}{\Delta t}=+\frac{1}{3}\frac{\Delta \left[B{r}_2\right]}{\Delta t}$

$\therefore \frac{\Delta Br{O}^{-}3}{\Delta t}=\frac{1}{3}\frac{\Delta \left[B{r}_2\right]}{\Delta t}$

$\Rightarrow \frac{\Delta \left[B{r}_2\right]}{\Delta t}=3\frac{\Delta \left[Br{O}_3^{-}\right]}{\Delta t}=3\times 0.01=0.03$