Chapter 6, Problem 6.54QP

a)

Interpretation Introduction

Interpretation: The heat of combustion from standard enthalpies of formation has to be calculated.

Concept Introduction:

The change in enthalpy that is associated with the formation of one mole of a substance from its related elements being in standard state is called standard enthalpy of formation

(${\text{ΔH}}_{\text{f}}\text{°}$).  The standard enthalpy of formation is used to determine the standard enthalpies of compound and element.

The standard enthalpy of reaction is the enthalpy of reaction that takes place under standard conditions.

The equation for determining the standard enthalpies of compound and element can be given by,

${\text{ΔH°}}_{\text{reaction}}\text{=}{\displaystyle \sum {\text{nΔH°}}_{\text{f}}\text{(products)}}\text{-}{\displaystyle \sum {\text{mΔH°}}_{\text{f}}\text{(reactants)}}$

To calculate the heat of combustion

Answer to Problem 6.54QP

The heat of combustion is calculated to be $\text{ΔH°=-1410}\text{.9}\text{kJ}{\text{mol}}^{\text{-1}}$

Explanation of Solution

Standard enthalpy of formation of ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\text{=52}\text{.3}\text{kJ}{\text{mol}}^{\text{-1}}$

Standard enthalpy of formation of ${\text{O}}_{\text{2}}\text{=0}\text{kJ}{\text{mol}}^{\text{-1}}$

Standard enthalpy of formation of ${\text{CO}}_{\text{2}}\text{=-393}\text{.5}\text{kJ}{\text{mol}}^{\text{-1}}$

Standard enthalpy of formation of ${\text{H}}_{\text{2}}\text{O=-285}\text{.8}\text{kJ}{\text{mol}}^{\text{-1}}$

Heat of combustion, ${\text{ΔH}}^{\text{°}}$ =

$\begin{array}{l}{\text{ΔH°=[2ΔH}}^{\text{°}}{}_{\text{f}}{\text{(CO}}_{\text{2}}{\text{)+2ΔH}}^{\text{°}}{}_{\text{f}}{\text{(H}}_{\text{2}}{\text{O)]-[ΔH}}^{\text{°}}{}_{\text{f}}{\text{(C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{)+3ΔH}}^{\text{°}}{}_{\text{f}}{\text{(O}}_{\text{2}}\text{)]}\\ \\ \text{ΔH°=}\left[\left(\text{2}\right)\text{-(-393}\text{.5}\text{kJ}{\text{mol}}^{\text{-1}}\text{+}\left(\text{2}\right)\text{(-285}\text{.8}\text{kJ}{\text{mol}}^{\text{-1}}\text{)}\right]\text{-}\left[\text{(1)(52}\text{.3}\text{kJ}{\text{mol}}^{\text{-1}}\text{)+(3)(0)}\right]\\ \\ \text{ΔH°=-1410}\text{.9}\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

The heat of combustion for the given reaction = $\text{-1410}\text{.9}\text{kJ}{\text{mol}}^{\text{-1}}$

b)

Interpretation Introduction

Interpretation: The heat of combustion from standard enthalpies of formation has to be calculated.

Concept Introduction:

The change in enthalpy that is associated with the formation of one mole of a substance from its related elements being in standard state is called standard enthalpy of formation

(${\text{ΔH}}_{\text{f}}\text{°}$).  The standard enthalpy of formation is used to determine the standard enthalpies of compound and element.

The standard enthalpy of reaction is the enthalpy of reaction that takes place under standard conditions.

The equation for determining the standard enthalpies of compound and element can be given by,

${\text{ΔH°}}_{\text{reaction}}\text{=}{\displaystyle \sum {\text{nΔH°}}_{\text{f}}\text{(products)}}\text{-}{\displaystyle \sum {\text{mΔH°}}_{\text{f}}\text{(reactants)}}$

To calculate the heat of combustion

Answer to Problem 6.54QP

The heat of combustion is calculated to be $\text{ΔH°=-1124}\text{.1}\text{kJ}{\text{mol}}^{\text{-1}}$

Explanation of Solution

Standard enthalpy of formation of ${\text{H}}_2\text{S=20}\text{.15kJ}{\text{mol}}^{\text{-1}}$

Standard enthalpy of formation of ${\text{O}}_{\text{2}}\text{=0}\text{kJ}{\text{mol}}^{\text{-1}}$

Standard enthalpy of formation of ${\text{SO}}_{\text{2}}\text{=-296}\text{.4}\text{kJ}{\text{mol}}^{\text{-1}}$

Standard enthalpy of formation of ${\text{H}}_{\text{2}}\text{O=-285}\text{.8}\text{kJ}{\text{mol}}^{\text{-1}}$

Heat of combustion, ${\text{ΔH}}^{\text{°}}$ =

$\begin{array}{l}{\text{ΔH°=[2ΔH}}^{\text{°}}{}_{\text{f}}{\text{(H}}_{\text{2}}{\text{O)+2ΔH}}^{\text{°}}{}_{\text{f}}{\text{(SO}}_{\text{2}}{\text{)]-[2ΔH}}^{\text{°}}{}_{\text{f}}{\text{(H}}_{\text{2}}{\text{S)+3ΔH}}^{\text{°}}{}_{\text{f}}{\text{(O}}_{\text{2}}\text{)]}\\ \\ \text{ΔH°=}\left[\left(\text{2}\right)\text{-(-285}\text{.8 kJ}{\text{mol}}^{\text{-1}}\text{+}\left(\text{2}\right)\text{(-296}\text{.4}\text{kJ}{\text{mol}}^{\text{-1}}\text{)}\right]\text{-}\left[\text{(1)(20}\text{.15 kJ}{\text{mol}}^{\text{-1}}\text{)+(3)(0)}\right]\\ \\ \text{ΔH°=-1124}\text{.1}\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

The heat of combustion for the given reaction = $\text{-1124}\text{.1}\text{kJ}{\text{mol}}^{\text{-1}}$