Chapter 21, Problem 21.198QP

Interpretation Introduction

Interpretation:

The standard enthalpy change for the decomposition of water vapor to atoms in the gas phase, the $\text{O}-\text{H}$ bond enthalpy, the standard enthalpy change for the decomposition of hydrogen peroxide vapor to atoms in the gas phase and the $\text{O}-\text{O}$ bond enthalpy has to be calculated

Concept Introduction:

Enthalpy of reaction:

The enthalpy of a reaction is calculated by subtracting the heat of formation of reactatns from heat of formation of products.

$\Delta \text{H}°\text{ =}\sum \Delta \text{H}{°}_f{}_{\mathrm{Pr}\text{odcuts}}\text{ - }\sum \Delta \text{H}{°}_{f\text{ Reactants}}$

Bond enthalpy:

The measure of stability of molecule is bond enthalpy.  The change in enthalpy is related in breaking a specific bond of 1 mole of gaseous molecule.  In solids and liquids bond enthalpies are affected by neighboring molecules.  There is possibility to predict the enthalpy of reaction using the average bond enthalpies.  Energy is always needed for the breaking of chemical bonds and release of energy always takes place with the formation of chemical bonds.   The enthalpy of reaction is estimated by counting the total number of broken bonds and bonds formed in a reaction.

The enthalpy of reaction in gas phase is given by,

${\text{ΔH}}^{\text{o}}\text{=}{\displaystyle \sum {\text{BE}}_{\text{(reactants)}}\text{-}{\displaystyle \sum {\text{BE}}_{\text{(products)}}}}$

Answer to Problem 21.198QP

The standard enthalpy change for the decomposition of water vapor to atoms in the gas phase is $927.0\text{ kJ}$

The $\text{O}-\text{H}$ bond enthalpy is $\text{463}\text{.5 kJ/mol}$

The standard enthalpy change for the decomposition of hydrogen peroxide vapor to atoms in the gas phase is $1070.5\text{ kJ}$

The $\text{O}-\text{O}$ bond enthalpy is $\text{143}\text{.5 kJ/mol}$

Explanation of Solution

Given data:

Water vapor decomposes to atoms in gas phase.

Hydrogen peroxide vapor decomposes to atoms in gas phase.

To Calculate: The standard enthalpy change for the decomposition of water vapor to atoms in the gas phase

The decomposition reaction of water vapor to atoms is written in chemical equation and the enthalpies of formation are written beneath the equation as follows,

$\begin{array}{l}{\text{H}}_{\text{2}}\text{O(g) }\stackrel{}{\to }2\text{H}(g)+\text{ O(g)}\\ -241.8\text{ 218}\text{.0 249}\text{.2 kJ/mol}\end{array}$

The standard enthalpy change for the reaction is calculated as follows,

$\begin{array}{l}\Delta \text{H}°\text{=}\sum \Delta \text{H}{°}_f{}_{\mathrm{Pr}\text{odcuts}}\text{ - }\sum \Delta \text{H}{°}_{f\text{ Reactants}}\\ =\text{[2(218}\text{.0) + (249}\text{.2) - (-241}\text{.8)]kJ}\\ =\text{927}\text{.0 kJ}\end{array}$

The standard enthalpy change for the decomposition reaction of water vapor to atoms is $927.0\text{ kJ}$

To Calculate: The bond enthalpy of $\text{O}-\text{H}$ bond

The enthalpy for the above reaction in terms of bond energies is:

$\begin{array}{l}{\text{ ΔH}}^{\text{o}}\text{=}{\displaystyle \sum {\text{BE}}_{\text{(reactants)}}\text{-}{\displaystyle \sum {\text{BE}}_{\text{(products)}}}}\\ \text{927}\text{.0 kJ}=\text{[2 BE(O}-\text{H})-0]\\ \text{BE(O}-\text{H})=\frac{927.0\text{ kJ}}{2\text{ mol}}\\ =\text{463}\text{.5 kJ/mol}\end{array}$

The bond enthalpy of $\text{O}-\text{H}$ bond is $\text{463}\text{.5 kJ/mol}$

To Calculate: The standard enthalpy change for the decomposition of hydrogen peroxide vapor to atoms in the gas phase

The decomposition reaction of hydrogen peroxide vapor to atoms is written in chemical equation and the enthalpies of formation are written beneath the equation as follows,

$\begin{array}{l}{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\text{(g) }\stackrel{}{\to }2\text{H}(g)+\text{ 2O(g)}\\ -136.1\text{ 218}\text{.0 249}\text{.2 kJ/mol}\end{array}$

The standard enthalpy change for the reaction is calculated as follows,

$\begin{array}{l}\Delta \text{H}°\text{=}\sum \Delta \text{H}{°}_f{}_{\mathrm{Pr}\text{odcuts}}\text{ - }\sum \Delta \text{H}{°}_{f\text{ Reactants}}\\ =\text{[2(218}\text{.0) + 2(249}\text{.2) - (-136}\text{.1)]kJ}\\ =\text{1070}\text{.5 kJ}\end{array}$

The standard enthalpy change for the decomposition reaction of water vapor to atoms is $\text{1070}\text{.5 kJ}$

To Calculate: The bond enthalpy of $\text{O}-\text{O}$ bond

The enthalpy for the above reaction in terms of bond energies is:

$\begin{array}{l}{\text{ ΔH}}^{\text{o}}\text{=}{\displaystyle \sum {\text{BE}}_{\text{(reactants)}}\text{-}{\displaystyle \sum {\text{BE}}_{\text{(products)}}}}\\ \text{1070}\text{.5 kJ}=\text{[2 BE(O}-\text{H})+\text{BE(O}-\text{O})-0]\\ \text{Consider BE(O}-\text{H})\text{ from previous step }\\ \text{BE(O}-\text{O})=\text{1070}\text{.5 kJ/mol -2(463}\text{.5 kJ/mol)}\\ =\text{143}\text{.5 kJ/mol}\end{array}$

The bond enthalpy of $\text{O}-\text{O}$ bond is $\text{143}\text{.5 kJ/mol}$

From the table 9.5, the bond energies are:

$\begin{array}{l}\text{BE(O-H) = 459 kJ/mol}\\ \text{BE(O-O) = 142 kJ/mol}\end{array}$

Conclusion

The standard enthalpy change for the decomposition of water vapor to atoms in the gas phase is calculated as $927.0\text{ kJ}$

The $\text{O}-\text{H}$ bond enthalpy is calculated as $\text{463}\text{.5 kJ/mol}$

The standard enthalpy change for the decomposition of hydrogen peroxide vapor to atoms in the gas phase is calculated as $1070.5\text{ kJ}$

The $\text{O}-\text{O}$ bond enthalpy is calculated as $\text{143}\text{.5 kJ/mol}$