Question

For the reaction ${3BrO}^{-}\to {BrO}_3^{-}+{2Br}^{-}$ in an alkaline aquesous solution, the value of the second order $\left(in{BrO}^{-}\right)$ rate at $80$ in the rate law for $-\frac{\Delta \left[{BrO}^{-}\right]}{\Delta t}$ was found to be 0.056 mol $L^{-1}s-1$. What is the rate when the rate law is written for (a) $\frac{\Delta {BrO}_3^{-}}{\Delta t}$,(b)$\frac{\Delta {Br}^{-}}{\Delta t}$

Answer 1

$3Br{O}^{-}\to Br{P}_3^{-}+2B{r}^{-}$

$-\frac{1}{3}\frac{\Delta \left[Br{O}^{-}\right]}{\Delta t}=\frac{\Delta \left[Br{O}_3^{-}\right]}{\Delta t}=\frac{1}{2}\frac{\Delta \left[B{r}^{-}\right]}{\Delta t}$

$\frac{-\Delta \left[Br{O}^{-}\right]}{\Delta t}=3\frac{\Delta \left[Br{O}_3^{-}\right]}{dt}$

$\frac{\Delta \left[Br{O}_3^{-}\right]}{\Delta t}=\frac{1}{3}\times 0.056=0.0181$

$\frac{\Delta \left[B{r}^{-}\right]}{\Delta t}=\frac{2}{3}\frac{-\Delta \left[Br{O}^{-}\right]}{\Delta t}$

$=\frac{2}{3}\times 0.056=0.0382$