Question

Question 66: The rate law for the reaction$Br{O_3}^{-}\left(aq\right)+3S{O}_3^{2-}\left(aq\right)\to B{r}^{-}\left(aq\right)+3S{O}_4^{2-}\left(aq\right)$ is$Rate=k\left[Br{O_3}^{-}\right]\left[S{O_3}^{2-}\right]\left[H^{+}\right]$

The first step in a proposed mechanism is $S{O_3}^{2-}\left(aq\right)+H^{+}\left(aq\right)\stackrel{k_1}{\to }HS{O_3}^{-}\left(aq\right)$

fast. The second step is rate determining. Write a possible second step for the mechanism.

Short answer

Answer

Possible second step for the mechanism can be$Br{O_3}^{-}+HS{O_3}^{-}\to HBr{O}_2+S{O_4}^{2-}$

Step-by-step solution

Step-by-step solutionStep 1: Analyzing the rate law

The rate law contains three species, two of them are included in first step. So, second step must contain$Br{O_3}^{-}$

Completing rate-determining step (second step) using rate law

Reacting the product in first step with $Br{O_3}^{-}$assecond step can’t contain reactants of first step because they are already included in rate law.

So, we get second step as,

$Br{O_3}^{-}+HS{O_3}^{-}\to HBr{O}_2+S{O_4}^{2-}$