Chapter 8, Problem 95SCQ

Bromine-containing species play a role in environmental chemistry. For example, they are evolved in volcanic eruptions.

1. (a) The following molecules are important in bromine environmental chemistry: HBr, BrO, and HOBr. Which is an odd-electron molecule?
2. (b) Use bond dissociation enthalpies to estimate ${\Delta }_{\text{r}}H$ for three reactions of bromine:

   Br₂(g) → 2 Br(g)

   2 Br(g) + O₂(g) → 2 BrO(g)

   BrO(g) + H₂O(g) → HOBr(g) + OH(g)

3. (c) Using bond dissociation enthalpies, estimate the standard enthalpy of formation of HOBr(g) from H₂(g), O₂(g), and Br₂(g).
4. (d) Are the reactions in parts (b) and (c) exothermic or endothermic?

Interpretation Introduction

Interpretation:

An odd electron molecule from the given list has to be determined.

Concept Introduction:

In a balanced equation the number of atoms of each element as a reactant is equal to the number of atoms of that element as a product.

Coefficient is a number placed before a formula in a chemical equation.

A balanced equation should be obeying the law of conservation of mass. Law of conservation of mass states that, the number of atoms remains constant throughout the reaction, simply it can be stated as follows, “during a chemical reaction atoms are neither be created nor be destroyed”.

Explanation of Solution

Given molecules are:

$\text{HBr,BrO}\text{and}\text{HOBr}$

$\text{HBr}$ have $8$ electrons.

$\text{BrO}$ have $13$ electrons.

$\text{HOBr}$ have $14$ electrons.

Thus the molecule with odd electron is $\text{BrO}$ with $13$ number of electrons.

Interpretation Introduction

Interpretation:

The enthalpy changes of the given reactions have to be determined.

Concept Introduction:

Bond energy or more correctly the bond dissociation enthalpy is the enthalpy change when breaking a bond in a molecule with the reactant and products in the gas phase.

${\Delta }_{r}H=\sum \Delta H\left(bondsbroken\right)-\sum \Delta H\left(bondsformed\right)$

Enthalpy of formation:

${\Delta }_{r}H=\sum {\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}\left(products\right)-\sum {\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}\left(\text{reactants}\right)$

Explanation of Solution

Given:

• ${\text{Br}}_{\text{2}}\left(\text{g}\right)\underset{}{\to }\text{2 Br}\left(\text{g}\right)$

Bond dissociation enthalpies of bonds are given below:

$\Delta H_{Br-Br}=193kJ/mol$

${\Delta }_{r}H=\sum \Delta H\left(bondsbroken\right)-\sum \Delta H\left(bondsformed\right)$

There is one $Br-Br$ bond broken and no bonds are formed.

$\begin{array}{c}{\Delta }_{r}H=\sum \Delta H\left(bondsbroken\right)-\sum \Delta H\left(bondsformed\right)\\ \\ =\left(1\times 193kJ/mol\right)-\left(0kJ/mol\right)\\ \\ {\Delta }_{r}H=193kJ/mol\end{array}$

• $\text{2}\text{Br}\left(\text{g}\right)\text{+}{\text{O}}_{\text{2}}\left(\text{g}\right)\underset{}{\to }\text{2 BrO}\left(\text{g}\right)$

Bond dissociation enthalpies of bonds are given below:

$\begin{array}{l}\Delta H_{Br-O}=201kJ/mol\\ \Delta H_{O=O}=498kJ/mol\end{array}$

${\Delta }_{r}H=\sum \Delta H\left(bondsbroken\right)-\sum \Delta H\left(bondsformed\right)$

There is one $O=O$ bond broken and two $Br-O$ bonds are formed.

$\begin{array}{c}{\Delta }_{r}H=\sum \Delta H\left(bondsbroken\right)-\sum \Delta H\left(bondsformed\right)\\ \\ =\left(1\times 498kJ/mol\right)-\left(2\times 201kJ/mol\right)\\ \\ {\Delta }_{r}H=96kJ/mol\end{array}$

• $\text{BrO}\left(\text{g}\right)\text{+}{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\underset{}{\to }\text{HOBr}\left(\text{g}\right)\text{+ OH}\left(\text{g}\right)$

Bond dissociation enthalpies of bonds are given below:

$\Delta H_{O-H}=463kJ/mol$

${\Delta }_{r}H=\sum \Delta H\left(bondsbroken\right)-\sum \Delta H\left(bondsformed\right)$

There are two $O-H$ bonds broken and two $O-H$ bonds are formed.

$\begin{array}{c}{\Delta }_{r}H=\sum \Delta H\left(bondsbroken\right)-\sum \Delta H\left(bondsformed\right)\\ \\ {\Delta }_{r}H=0kJ/mol\end{array}$

Interpretation Introduction

Interpretation:

The enthalpy change of the formation of $HOBr$ has to be determined.

Concept Introduction:

Bond energy or more correctly the bond dissociation enthalpy is the enthalpy change when breaking a bond in a molecule with the reactant and products in the gas phase.

${\Delta }_{r}H=\sum \Delta H\left(bondsbroken\right)-\sum \Delta H\left(bondsformed\right)$

Enthalpy of formation:

${\Delta }_{r}H=\sum {\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}\left(products\right)-\sum {\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}\left(\text{reactants}\right)$

Explanation of Solution

Given:

${\text{Br}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{O}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{H}}_{\text{2}}\text{O}\underset{}{\to }\text{HOBr}\text{+}\text{Br-O}\text{+}\text{OH}$

Bond dissociation enthalpies of bonds are given below:

$\begin{array}{l}\Delta H_{Br-O}=201kJ/mol\\ \Delta H_{O=O}=498kJ/mol\\ \Delta H_{Br-Br}=216kJ/mol\\ \Delta H_{O-H}=463kJ/mol\end{array}$

${\Delta }_{r}H=\sum \Delta H\left(bondsbroken\right)-\sum \Delta H\left(bondsformed\right)$

There is one $O=O$ bond, one $Br-Br$ bond broken and one $O-H$ bond broken and two $O-HandO-Br$ are formed.

$\begin{array}{c}{\Delta }_{r}H=\sum \Delta H\left(bondsbroken\right)-\sum \Delta H\left(bondsformed\right)\\ \\ =\left(1\times 216kJ/mol+1\times 498kJ/mol+1\times 463kJ/mol\right)-\left(2\times 463kJ/mol+2\times 201kJ/mol\right)\\ \\ {\Delta }_{r}H=-151kJ/mol\end{array}$

Interpretation Introduction

Interpretation:

The reactions which are endothermic or exothermic have to be identified.

Concept Introduction:

Bond energy or more correctly the bond dissociation enthalpy is the enthalpy change when breaking a bond in a molecule with the reactant and products in the gas phase.

${\Delta }_{r}H=\sum \Delta H\left(bondsbroken\right)-\sum \Delta H\left(bondsformed\right)$

Enthalpy of formation:

${\Delta }_{r}H=\sum {\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}\left(products\right)-\sum {\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}\left(\text{reactants}\right)$

Explanation of Solution

The reactions and its enthalpy changes are given below:

${\text{Br}}_{\text{2}}\left(\text{g}\right)\underset{}{\to }\text{2 Br}\left(\text{g}\right)$

$\begin{array}{c}{\Delta }_{r}H=\sum \Delta H\left(bondsbroken\right)-\sum \Delta H\left(bondsformed\right)\\ \\ =\left(1\times 193kJ/mol\right)-\left(0kJ/mol\right)\\ \\ {\Delta }_{r}H=193kJ/mol\end{array}$

$\text{2}\text{Br}\left(\text{g}\right)\text{+}{\text{O}}_{\text{2}}\left(\text{g}\right)\underset{}{\to }\text{2 BrO}\left(\text{g}\right)$

$\begin{array}{c}{\Delta }_{r}H=\sum \Delta H\left(bondsbroken\right)-\sum \Delta H\left(bondsformed\right)\\ \\ =\left(1\times 498kJ/mol\right)-\left(2\times 201kJ/mol\right)\\ \\ {\Delta }_{r}H=96kJ/mol\end{array}$

$\text{BrO}\left(\text{g}\right)\text{+}{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\underset{}{\to }\text{HOBr}\left(\text{g}\right)\text{+ OH}\left(\text{g}\right)$

$\begin{array}{c}{\Delta }_{r}H=\sum \Delta H\left(bondsbroken\right)-\sum \Delta H\left(bondsformed\right)\\ \\ {\Delta }_{r}H=0kJ/mol\end{array}$

The three equations given above are endothermic reactions since all of them have positive enthalpy change.

The reaction given below is exothermic since the enthalpy change is negative.

${\text{Br}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{O}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{H}}_{\text{2}}\text{O}\underset{}{\to }\text{HOBr}\text{+}\text{Br-O}\text{+}\text{OH}$

$\begin{array}{c}{\Delta }_{r}H=\sum \Delta H\left(bondsbroken\right)-\sum \Delta H\left(bondsformed\right)\\ \\ =\left(1\times 216kJ/mol+1\times 498kJ/mol+1\times 463kJ/mol\right)-\left(2\times 463kJ/mol+2\times 201kJ/mol\right)\\ \\ {\Delta }_{r}H=-151kJ/mol\end{array}$