Chapter 10, Problem 10.77QP

For the reaction

$2C_2{H}_6(\text{g})\text{ + }7O_2(\text{g})\to 4C{O}_2(\text{g})\text{ + }6H_{2}O(\text{g})$

(a) Predict the enthalpy of reaction from the average bond enthalpies in Table 10.4. (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:

The standard enthalpy of formation and bond enthalpies has to be used to calculate the enthalpy of the reaction.

Concept Introduction:

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,

$\text{ΔH°=}{\displaystyle \sum {\text{BE}}_{\text{(reactants)}}}\text{-}{\displaystyle \sum {\text{BE}}_{\left(\text{products}\right)}}$

Where,

BE= Bond enthalpy and ${\displaystyle \sum \text{=}}$ summation

To calculate: The enthalpy of the reaction from the bond enthalpy

Answer to Problem 10.77QP

The enthalpy of the reaction from the average bond enthalpy is found to be $\text{-2759}\text{kJ}{\text{mol}}^{\text{-1}}$

Explanation of Solution

Bonds Broken	No of broken bonds	Bond enthalpy ($\text{kJ}{\text{mol}}^{\text{-1}}$)	Enthalpy change ($\text{kJ}{\text{mol}}^{\text{-1}}$)
$\text{C-H}$	$\text{12}$	$414$	4968
$\text{C-C}$	$2$	$347$	$694$
$\text{O=O}$	$7$	$498.7$	$3491$

Bonds Formed	No of formed bonds	Bond enthalpy ($\text{kJ}{\text{mol}}^{\text{-1}}$)	Enthalpy change ($\text{kJ}{\text{mol}}^{\text{-1}}$)
$\text{C=O}$	$8$	$799$	$6392$
$\text{O-H}$	$12$	$460$	$5520$

$\text{ΔH°=}{\displaystyle \sum {\text{BE}}_{\text{(reactants)}}}\text{-}{\displaystyle \sum {\text{BE}}_{\left(\text{products}\right)}}$

= $\text{(4968+694+3491)-(6392+5520)}$

= $\text{-2759}\text{kJ}{\text{mol}}^{\text{-1}}$

The enthalpy of the reaction from the bond enthalpy = $\text{-2759}\text{kJ}{\text{mol}}^{\text{-1}}$

Interpretation Introduction

Interpretation:

The standard enthalpy of formation and bond enthalpies has to be used to calculate the enthalpy of the reaction.

Concept Introduction:

Standard enthalpy of formation:

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 enthalpy of the reaction from standard enthalpy of formation

Answer to Problem 10.77QP

The enthalpy of the reaction from the standard enthalpy of formation = $\text{-2855}\text{.4}\text{kJ}{\text{mol}}^{\text{-1}}$

Explanation of Solution

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

Standard enthalpy of ${\text{H}}_{\text{2}}\text{O=-241}\text{.8}\text{kJ}{\text{mol}}^{\text{-1}}$

Standard enthalpy of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{=-84}\text{.7}\text{kJ}{\text{mol}}^{\text{-1}}$

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

$\begin{array}{l}{\text{ΔH°}}_{\text{reaction}}{\text{=4ΔH°}}_{\text{f}}{\text{(CO}}_{\text{2}}{\text{)+6ΔH°}}_{\text{f}}{\text{(H}}_{\text{2}}\text{O)-}\left[{\text{2ΔH°}}_{\text{f}}{\text{(C}}_{\text{2}}{\text{H}}_{\text{6}}{\text{)+7ΔH°}}_{\text{f}}{\text{(O}}_{\text{2}}\text{)}\right]\\ \\ {\text{ΔH°}}_{\text{reaction}}\text{=(4)(-393}\text{.5}\text{kJ}{\text{mol}}^{\text{-1}}\text{)+(6)(-241}\text{.8}\text{kJ}{\text{mol}}^{\text{-1}}\text{)-}\left[\text{(2)(-84}\text{.7}\text{kJ}{\text{mol}}^{\text{-1}}\text{)+(7)(0)}\right]\\ \\ {\text{ΔH°}}_{\text{reaction}}\text{=-2855}\text{.4}\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

The enthalpy of the reaction from the standard enthalpy of formation = $\text{-2855}\text{.4}\text{kJ}{\text{mol}}^{\text{-1}}$

The enthalpy of the reaction from the standard enthalpy of formation is calculated by plugging in the standard enthalpies of formation of reactants and products.  The enthalpy of the reaction from the standard enthalpy of formation is found to be $\text{-2855}\text{.4}\text{kJ}{\text{mol}}^{\text{-1}}$.

To compare the enthalpy of the reaction

The enthalpy of the reaction is different because in one case average bond enthalpy was used and in other case standard enthalpy of formation was used.

The enthalpy of the reaction from the standard enthalpy of formation = $\text{-2855}\text{.4}\text{kJ}{\text{mol}}^{\text{-1}}$

The enthalpy of the reaction from the bond enthalpy= $\text{-2759}\text{kJ}{\text{mol}}^{\text{-1}}$

The enthalpy of the reaction is different because in one case average bond enthalpy was used and in other case standard enthalpy of formation was used.