Chapter 9, Problem 9.72QP

For the reaction

$2{\text{C}}_2{\text{H}}_6(g)+7{\text{O}}_2(g)\stackrel{}{\to }4{\text{CO}}_2(g)+6{\text{H}}_2\text{O}(g)$

1. (a) Predict the enthalpy of reaction from the average bond enthalpies in Table 9.4.
2. (b) Calculate the enthalpy of reaction from the standard enthalpies of formation (see Appendix 2) of the reactant and product molecules, and compare the result with your answer for part (a).

Interpretation Introduction

Interpretation:

Enthalpy of the reaction using average bond enthalpies of reactants and products has to be calculated.

Concept Introduction:

$\Delta H°$ refers to change in enthalpy.  Change in enthalpy in a reaction and bond energy (BE) are related as,

$\text{ΔH°}\text{=}\text{ΣBE(reactants)-ΣBE(products)}$

Answer to Problem 9.72QP

Enthalpy of the reaction using bond enthalpies of reactants and products is $-2759kJ/mol.$

Explanation of Solution

Calculation of $\Delta H°$

The given reaction is,

${\text{2 C}}_{\text{2}}{\text{H}}_{\text{6(g)}}\text{+}\text{7}{\text{O}}_{2(g)}\stackrel{}{\to }\text{ 4}{\text{CO}}_{\text{2}}{}_{\text{(g)}}+6H_2{O}_{(g)}$

Reactants $O_2$ has one $O=O$ bond, and two moles of ethane has two $C-C$ and twelve $C-H$ bonds.

Products are four moles of $C{O}_2$ and it has totally eight $C=O$ bonds and six moles of water has twelve $O-H$ bonds.

$\Delta H°$ for the reaction is calculated using average bond enthalpy values given in text book.

$\begin{array}{l}\text{ΔH°}\text{=}\text{ΣBE(reactants)-ΣBE(products)}\\ \text{ΔH°}\text{=}\left[\text{2BE}\left(C-C\right)\text{+7BE}\left(O=O\right)+12BE\left(C-H\right)\right]\text{-}\left[\text{8BE}\left(C=O\right)+12BE\left(O-H\right)\right]\\ \Delta H°=\left[\left(2\times 347kJ/mol\right)+\left(7\times 498.7kJ/mol\right)+\left(12\times 414kJ/mol\right)\right]-\left[\left(8\times 799kJ/mol\right)+\left(12\times 460kJ/mol\right)\right]\\ =\left[4968kJ/mol+694kJ/mol+3491kJ/mol\right]-\left[6392kJ/mol+5520kJ/mol\right]\\ =-2759kJ/mol\end{array}$

Interpretation Introduction

Interpretation:

Enthalpy of the reaction using heat of formation of reactants and products has to be calculated.

Concept Introduction:

$\Delta H$ refers to change in enthalpy.  Change in enthalpy in a reaction and enthalpy of formation are related by the formula,

$\Delta H_{rxn}^{°}=\Sigma \Delta H_f^{°}(product)-\Sigma \Delta H_f^{°}(reac\tan t)$

Where,

$\begin{array}{l}\Delta H_{rxn}^{°}=\text{ enthalpy change in reaction}\\ \Delta H_f^{°}(product)=\text{ enthalpy or heat of formation of product}\\ \Delta H_f^{°}(reac\tan t)=\text{ enthalpy or heat of formation of }reac\tan t\end{array}$

Answer to Problem 9.72QP

Enthalpy of the reaction using enthalpy of formation of reactants and products is calculated as $-2855kJ/mol.$

Explanation of Solution

The given reaction is,

${\text{2 C}}_{\text{2}}{\text{H}}_{\text{6(g)}}\text{+}\text{7}{\text{O}}_{2(g)}\stackrel{}{\to }\text{ 4}{\text{CO}}_{\text{2}}{}_{\text{(g)}}+6H_2{O}_{(g)}$

Known data from appendix 2:

$\begin{array}{l}{\text{ΔH}}_{\text{f}}^{\text{°}}{\text{ for CO}}_{\text{2(g)}}\text{= -393}\text{.5 kJ/mol}\\ {\text{ΔH}}_{\text{f}}^{\text{°}}{\text{ for H}}_{\text{2}}{\text{O}}_{\text{(g)}}\text{= -241}\text{.8 kJ/mol}\\ {\text{ΔH}}_{\text{f}}^{\text{°}}{\text{ for C}}_{\text{2}}{\text{H}}_{\text{6}}{}_{\text{(g)}}\text{= -84}\text{.7 kJ/mol}\\ {\text{ΔH}}_{\text{f}}^{\text{°}}{\text{ for O}}_{\text{2(g)}}\text{= 0 kJ/mol}\end{array}$

$\Delta H°$ for the reaction is calculated using enthalpy of formation of products and reactants given in text book (chapter 6)

$\begin{array}{l}\text{ΔH°}\text{=}\left[{\text{4ΔH}}_{\text{f}}^{\text{°}}\left({\text{CO}}_{\text{2}}{}_{\text{(g)}}\right){\text{+6ΔH}}_{\text{f}}^{\text{°}}\left({\text{H}}_{\text{2}}{\text{O}}_{\text{(g)}}\right)\right]\text{-}\left[{\text{2ΔH}}_{\text{f}}^{\text{°}}\left({\text{C}}_{\text{2}}{\text{H}}_{\text{6}}{}_{\text{(g)}}\right){\text{+7ΔH}}_{\text{f}}^{\text{°}}\left({\text{O}}_{\text{2}}{}_{\text{(g)}}\right)\right]\\ \text{=}\left[\left(\text{4×-393}\text{.5 kJ/mol}\right)+\left(6\times -241.8kJ/mol\right)\right]\text{-}\left[\left(\text{2}\times \text{-84}\text{.7 kJ/mol}\right)\text{+}\left(\text{7}\times \text{0 kJ/mol}\right)\right]\\ \text{=}\text{-}\text{2855 kJ/mol}\end{array}$