Question

Question 65: The reactionis expected to obey the mechanism .

Write the rate law for this reaction.

Short answer

Answer

$R=k_3\left(\frac{k_{-2}}{k_2}\right)\left(\frac{k_{-1}}{k_1}\right)\left[B{r}^{-}\right]{\left[H^{+}\right]}^2{{\left[Br{O_3}^{-}\right]}_{}}_{}$

To simplify the rate laws for the reactions, we have always assumed that the rate is being studied under conditions where only the forward reaction is important. This produces rate laws that contain only reactant concentration.

Step-by-step solution

Step-by-step solutionStep 1: Write the rate law equation from the slowest reaction that is rate determine step [RDS].

In the given question equation 3 is the rate determining step. Therefore, from eq 3

$R=k_3\left[B{r}^{-}\right]\left[H_{2}Br{O_3}^{+}\right]$

With the help of equation 2, Writing down the further reaction.

$r=\frac{K\left[HBrO\right]}{K\left[H\right]\left[HBrO\right]}$

solving the value of $H_{2}Br{O_3}^{+}=\frac{k_2}{k_{-2}}\left[H^{+}\right]\left[Br{O_3}^{-}\right]$

From equation 1, determine the rate of reaction, then solving the value of HBrO3.  

$$\begin{gathered} R=\frac{k_1}{k_{-1}}\left[HBr{O}_3\right]/\left[H^{+}\right]\left[Br{O_3}^{-}\right] \\ HBr{O}_3=\frac{k_1}{k_{-2}}\left[H^{+}\right]\left[Br{O_3}^{-}\right] \end{gathered}$$

Substituting the values