Chapter 11, Problem 11.70PAE

11.64 HBr is oxidized in the following reaction:

4 HBr(g) + O₂(g) —• 2 H₂O(g) + 2 Br,(g)

A proposed mechanism is

HBr + O₂ -* HOOBr

(slow)

HOOBr + HBr — 2 HOBr

(fast)

HOBr + HBr — H₂O + Bn

(fast)

1. Show that this mechanism can account for the correct stoichiometry.

Identify all intermediates in this mechanism.

What is the molecularity of each elementary’ step?

Write the rate expression for each elementary' step.

Identify the rate-determining step.

Interpretation Introduction

To determine:

Show that the given mechanism has correct stoichiometry.

Explanation of Solution

The molar ratio of the reactant molecules, products in a balanced chemical equation is called stoichiometry. Some reactions have so many steps to result in the final products. There can be intermediates which are not involved in the balanced equation. If a reaction has many steps to give their final products, by adding them together we can find the stoichiometric equation.

$\begin{array}{l}HBr+O_2\to HOOBr\text{(slow)}\\ HOOBr+HBr\to 2HOBr\text{ (fast)}\\ HOBr+HBr\to H_{2}O+B{r}_2\text{ (fast)}\end{array}$

$HBr+O_2+HOOBr+HBr+2HBr+2HOBr\to HOOBr+2HOBr+2H_{2}O+2B{r}_2$

Here, we have shown only one $\text{HOBr}$ molecule decomposition. But we should consider the decomposition of two $\text{HOBr}$ molecules.

Then we can cancel out same molecule type present in the both side.

$4HBr+O_2\to 2H_{2}O+2B{r}_2$

This reaction shows above mechanism can account for the correct stoichiometry.

But given reaction is

$4HBr(g)+O_2(g)\to 2H_{2}O(g)+2{\text{ Br}}_2(g)$

Therefore, mechanism matches the given stoichiometry.

Interpretation Introduction

To identify:

All the intermediates present in the mechanism.

Explanation of Solution

Some reactions should follow many steps to give their final product. In between these steps they will form intermediates. Intermediates don’t involve in the overall reaction. And final we can’t separate them out. Here, $\text{HOOBr}$ and $\text{HOBr}$ act as an intermediate.

Interpretation Introduction

To determine:

The molecularity of every elementary step.

Explanation of Solution

Molecularity of each step:

Molecularity is a theoretical concept and should not be negative, zero, fractional, infinite and imaginary. Molecularity is the total number of reactant molecules or atoms taking part in the chemical reaction.

Step 1 − molecularity is 2

Step 2 − molecularity is 2

Step 3 − molecularity is 2

Interpretation Introduction

To determine:

The rate expression for every elementary step.

Explanation of Solution

Rate equation:

Rate equation can be written using either product or reactants. If reactants are considered, we should take the reduction rate of reactants. If products are considered, we should take the growth rate of products.

$\begin{array}{l}\begin{array}{l}\begin{array}{l}Step1\to Rate=d\left[HBr\right]/dt=-k_1\left[HBr\right]\left[O_2\right]\\ \end{array}\hfill \\ Step2\to Rate=d\left[HBr\right]/dt=-k_2\left[HOOBr\right]\left[HBr\right]\hfill \\ \hfill \\ Step3\to Rate=d\left[HBr\right]/dt=-k_3\left[HOBr\right]\left[HBr\right]\hfill \end{array}\\ \begin{array}{l}\begin{array}{l}Here,\\ k_1=rateconstantforstep1\end{array}\hfill \\ k_2=rateconstantforstep2\hfill \\ k_3=rateconstantforstep3\hfill \end{array}\end{array}$

Interpretation Introduction

To identify:

The rate determining step.

Explanation of Solution

Rate determining step:

Rate determining step is the slowest step in a chemical reaction. This step determines the rate of the whole reaction. So here rate determining step is step 1.

$Step1HBr+O_2\to HOOBr\text{(slow)}$

Conclusion

A reaction mechanism can contain more than one step and among them slowest step is considered as rate determining step. Rate expressions can be written to all the steps involved. It can be recognized whether a reaction mechanism matches with the correct stoichiometry to determine if the mechanism is correct.