Chapter 9, Problem 9.70QP

(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 9.70QP

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.

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

(b)

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 9.70QP

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)

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