Chapter 8, Problem 79QP

For the reaction

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

(a) Predict the enthalpy of reaction from the average bond enthalpies in Table 8.6. (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 $\Delta {\text{H}}_{\text{rxn}}$ for the given reaction by using the average bond enthalpies and the standard enthalpy of formation is to be determined.

Concept Introduction:

The enthalpy of the reaction by using the bond enthalpy is calculated by the expression, which is as:

$\Delta H_{\text{rxn}}{}^{\circ }=\sum \text{BE}\left(\text{reactants}\right)-\sum \text{BE}\left(\text{products}\right).$

Here, $\Delta H_{\text{rxn}}{}^{\circ }$ is the enthalpy of the reaction, $\sum \text{BE}\left(\text{reactants}\right)$ is the sum of the average bond enthalpy of the reactant, and $\sum \text{BE}\left(\text{products}\right)$ is the sum of the average bond enthalpy of the product.

The enthalpy of the reaction by using the standard enthalpy of formation is calculated by the expression, which is as:

$\Delta {\text{H}}^{\circ }=\sum \Delta {\text{H}}_{\text{f}}{}^{\circ }\left(\text{products}\right)-\sum \Delta {\text{H}}_{\text{f}}{}^{\circ }\left(\text{reactants}\right).$

Here, $\Delta {\text{H}}^{\circ }$ is the standard change in the enthalpy of the reaction, $\sum \Delta {\text{H}}_{\text{f}}{}^{\circ }\left(\text{reactants}\right)$ is the sum of the standard change in the enthalpy of the reactant, and $\sum \Delta {\text{H}}_{\text{f}}{}^{\circ }\left(\text{products}\right)$ is the standard change in the enthalpy of the product.

Rules to write the Lewis structure are as:

The skeletal structure of the compound is drawn in which the less electronegative element is placed as the central atom, which is surrounded by substituent atoms.

The total number of valence electrons for the compound is determined.

Subtract two electrons for each bond in the skeletal structure from the total number of valence electrons to know the number of remaining electrons.

Complete the octet of each terminal atom by placing a pair of electrons from the remaining electrons.

If any of the electrons are remaining after completing the octet of terminal atoms, place the remaining electrons as a pair on the central atom.

Answer to Problem 79QP

Solution:

The $\Delta {\text{H}}_{\text{rxn}}$ value for the reaction from the average bond enthalpy is equal to $-2759\text{ kJ/mol}$.

The $\Delta {\text{H}}_{\text{rxn}}$ value for the reaction from the standard enthalpy of formation is equal to $-2855.4\text{ kJ/mol}$.

Explanation of Solution

a)The enthalpy of reaction from the average bond enthalpies.

The given reaction is ${\text{C}}_2{\text{H}}_6\left(g\right)+7{\text{O}}_2\left(g\right)\to 4{\text{CO}}_2\left(g\right)+6{\text{H}}_2\text{O}\left(g\right)$.

The bond enthalpy value of $\text{C-H}$ from the table $8.6$ is $\text{414 kJ/mol}$.

The bond enthalpy value of $\text{C-C}$ from the table $8.6$ is $347\text{kJ/mol}$.

The bond enthalpy value of $\text{O=O}$ from the table $8.6$ is $\text{498}\text{.7 kJ/mol}$.

The bond enthalpy value of $\text{C=O}$ from the table $8.6$ is $\text{799}\text{kJ/mol}$.

The bond enthalpy value of $\text{O-H}$ from the table $8.6$ is $\text{460 kJ/mol}$.

The Lewis structure of the given reaction is drawn in order to determine which bond is formed and which is broken.

In the given reaction, $\text{12 C-H}$, $\text{2 C-C}$, $\text{7O=O}$ bonds breaks while $8\text{ C=O}$ bond and $12\text{ O-H}$ bonds are formed.

The $\Delta {\text{H}}_{\text{rxn}}{}^{\circ }$ of the reaction is calculated by the expression, which is as:

$\Delta {\text{H}}_{\text{rxn}}{}^{\circ }=\sum \text{BE}\left(\text{reactants}\right)-\sum \text{BE}\left(\text{products}\right).$

Substitute the value $\text{414 kJ/mol}$ for $\text{C-H}$, $347\text{kJ/mol}$ for $\text{C-C}$, $\text{498}\text{.7 kJ/mol}$ for $\text{O=O}$, $\text{799}\text{kJ/mol}$ for $\text{C=O}$, and $\text{460 kJ/mol}$ for $\text{O-H}$ in the above expression.

$\begin{array}{c}\Delta H^{\circ }=\left[12\left(\text{414 kJ/mol}\right)+2\left(347\text{ kJ/mol}\right)+7\left(\text{ 498}\text{.7 kJ/mol}\right)\right]-\left[8\left(\text{799}\text{kJ/mol}\right)+12\left(460\text{kJ/mol}\right)\right]\\ =\left(\text{4968 kJ/mol}+\text{694 kJ/mol+3491 kJ/mol}\right)-\left(\text{6392 kJ/mol}+\text{5520 kJ/mol}\right)\\ =-2759\text{ kJ/mol}\text{.}\end{array}$

b) The enthalpy of reaction from the standard enthalpies of formation

The given reaction is ${\text{2C}}_2{\text{H}}_6\left(g\right)+7{\text{O}}_2\left(g\right)\to 4{\text{CO}}_2\left(g\right)+6{\text{H}}_2\text{O}\left(g\right)$.

The value of $\Delta {\text{H}}_{\text{f}}{}^{\circ }\left({\text{CO}}_2\right)$ from appendix $2$ is $\text{-393}\text{.5 kJ/mol}\text{.}$

The value of $\Delta {\text{H}}_{\text{f}}{}^{\circ }\left({\text{H}}_2\text{O}\right)$ from appendix $2$ is $\text{-241}\text{.8 kJ/mol}\text{.}$

The value of $\Delta {\text{H}}_{\text{f}}{}^{\circ }\left({\text{C}}_2{\text{H}}_6\right)$ from appendix $2$ is $\text{-84}\text{.7 kJ/mol}\text{.}$

The value of $\Delta {\text{H}}^{\circ }$ for the given reaction is calculated by using the relation given below:

$\Delta {\text{H}}^{\circ }=\left[4\Delta {\text{H}}_{\text{f}}{}^{\circ }\left({\text{CO}}_2\right)+6\Delta {\text{H}}_{\text{f}}{}^{\circ }\left({\text{H}}_2\text{O}\right)\right]-\left[2\Delta {\text{H}}_{\text{f}}{}^{\circ }\left({\text{C}}_2{\text{H}}_6\right)+7\Delta {\text{H}}_{\text{f}}{}^{\circ }\left({\text{O}}_2\right)\right].$

Here, $\Delta H_{\text{rxn}}{}^{\circ }$ is the enthalpy of the reaction, $\Delta {\text{H}}_{\text{f}}{}^{\circ }\left({\text{CO}}_2\right)$ is the change in the enthalpy of ${\text{CO}}_2$, $\Delta {\text{H}}_{\text{f}}{}^{\circ }\left({\text{H}}_2\text{O}\right)$ is the change in the enthalpy of ${\text{H}}_2\text{O,}$ $\Delta {\text{H}}_{\text{f}}{}^{\circ }\left({\text{C}}_2{\text{H}}_6\right)$ is the change in the enthalpy of ${\text{C}}_2{\text{H}}_6,$ and $\Delta {\text{H}}_{\text{f}}{}^{\circ }\left({\text{O}}_2\right)$ is the change in theenthalpy of ${\text{O}}_2$.

The value of change in the enthalpy for atoms in their standard state is zero. In the reaction, ${\text{O}}_2$ is present in thestandard state. Thus, the value of $\Delta {\text{H}}_{\text{f}}{}^{\circ }\left({\text{O}}_2\right)$ is zero.

Substitute $0$ for $\Delta {\text{H}}_{\text{f}}{}^{\circ }\left({\text{O}}_2\right)$, $\text{-393}\text{.5 kJ/mol}$ for $\Delta {\text{H}}_{\text{f}}{}^{\circ }\left({\text{CO}}_2\right)$, $\text{-241}\text{.8 kJ/mol}$ for $\Delta {\text{H}}_{\text{f}}{}^{\circ }\left({\text{H}}_2\text{O}\right)$ $\text{-84}\text{.7 kJ/mol}$ for $\Delta {\text{H}}_{\text{f}}{}^{\circ }\left({\text{C}}_2{\text{H}}_6\right)$ in the equation as:

$\begin{array}{c}\Delta {\text{H}}^{\circ }=\left[4\Delta {\text{H}}_{\text{f}}{}^{\circ }\left({\text{CO}}_2\right)+6\Delta {\text{H}}_{\text{f}}{}^{\circ }\left({\text{H}}_2\text{O}\right)\right]-\left[2\Delta {\text{H}}_{\text{f}}{}^{\circ }\left({\text{C}}_2{\text{H}}_6\right)+7\Delta {\text{H}}_{\text{f}}{}^{\circ }\left({\text{O}}_2\right)\right]\\ =\left[\left(4\right)\left(\text{-393}\text{.5 kJ/mol}\right)+\left(6\right)\left(\text{-241}\text{.8 kJ/mol}\right)\right]-\left[\left(2\right)\left(\text{-84}\text{.7 kJ/mol}\right)+\left(7\right)\left(0\right)\right]\\ =-2855.4\text{ kJ/mol}\text{.}\end{array}$

The values of $\Delta {\text{H}}_{\text{rxn}}{}^{\circ }$ calculated in part (a) and part (b) are not same; as in part (a), the value of $\Delta {\text{H}}_{\text{rxn}}{}^{\circ }$ is calculated by using the average bond enthalpies while in part (b), the value of $\Delta {\text{H}}_{\text{rxn}}{}^{\circ }$ is calculated by using the standard enthalpy of formation.