Chapter 10, Problem 10.75P

(a)

Interpretation Introduction

Interpretation:

$\Delta H_{\text{rxn}}^{°}$ for the combustion of gaseous ethanol is to be calculated.

Concept introduction:

The heat of the reaction $\left(\Delta H_{\text{rxn}}^{°}\right)$ is defined as the heat released or absorbed during a chemical reaction as a result of the difference in the bond energies (BE) of reactant and product in the reaction. $\Delta H_{\text{rxn}}^{°}$ is negative for exothermic reaction and $\Delta H_{\text{rxn}}^{°}$ is positive for an endothermic reaction.

The formula to calculate $\Delta H_{\text{rxn}}^{°}$ of reaction is as follows:

  $\Delta H_{\text{rxn}}^{°}=\sum \Delta H_{\text{reactant bond broken}}^{°}+\sum \Delta H_{\text{product bond formed}}^{°}$

Or,

  $\Delta H_{\text{rxn}}^{°}=\sum {\text{BE}}_{\text{reactant bond broken}}-\sum {\text{BE}}_{\text{product bond formed}}$

The bond energy of reactants is positive and the bond energy of products is negative.

Answer to Problem 10.75P

$-1267\text{kJ}$ is the heat of combustion of ethanol.

Explanation of Solution

The given chemical equation for the combustion of gaseous ethanol is as follows:

The number of broken bonds is $\text{5 C}-\text{H}$ bond, $\text{1 C}-\text{C}$ bond, $\text{1 C}-\text{O}$ bond, $\text{1 O}-\text{H}$ bond, and $\text{3 O=O}$ bonds.

The number of bonds formed is $4\text{ C=O}$ bonds and $6\text{ O}-\text{H}$ bonds.

The formula to the enthalpy of the given reaction is as follows:

  $\Delta H_{\text{rxn}}^{°}=\left[\begin{array}{l}\left(5{\text{BE}}_{\text{C}-\text{H}}+1{\text{BE}}_{\text{C}-\text{C}}+1{\text{BE}}_{\text{C}-\text{O}}+1{\text{BE}}_{\text{O}-\text{H}}+3{\text{BE}}_{\text{O=O}}\right)\\ -\left({\text{4BE}}_{\text{C=O}}+6{\text{BE}}_{\text{O}-\text{H}}\right)\end{array}\right]$        (1)

Substitute $413\text{kJ/mol}$ for ${\text{BE}}_{\text{C}-\text{H}}$, $347\text{kJ/mol}$ for ${\text{BE}}_{\text{C}-\text{C}}$, $358\text{kJ/mol}$ for ${\text{BE}}_{\text{C}-\text{O}}$, $498\text{kJ/mol}$ for ${\text{BE}}_{\text{O=O}}$, $799\text{kJ/mol}$ for ${\text{BE}}_{\text{C=O}}$ and $467\text{kJ/mol}$ for ${\text{BE}}_{\text{O}-\text{H}}$ in the equation (1).

  $\begin{array}{c}\Delta H_{\text{rxn}}^{°}=\left[\begin{array}{l}\left(\begin{array}{l}\left(\text{5 mol}\right)\left(413\text{kJ/mol}\right)+\left(\text{1 mol}\right)\left(347\text{kJ/mol}\right)+\left(\text{1 mol}\right)\left(358\text{kJ/mol}\right)\\ +\left(\text{1 mol}\right)\left(467\text{kJ/mol}\right)+\left(\text{3 mol}\right)\left(498\text{kJ/mol}\right)\end{array}\right)-\\ \left(\text{4 mol}\right)\left(799\text{kJ/mol}\right)+\left(\text{6 mol}\right)\left(467\text{kJ/mol}\right)\end{array}\right]\\ =-1267\text{kJ}\text{.}\end{array}$

Conclusion

The number of broken bonds is $\text{5 C}-\text{H}$ bond, $\text{1 C}-\text{C}$ bond, $\text{1 C}-\text{O}$ bond, $\text{1 O}-\text{H}$ bond, and $\text{3 O=O}$ bond while the number of bonds formed is $4\text{ C=O}$ bonds and $6\text{ O}-\text{H}$ bonds.

(b)

Interpretation Introduction

Interpretation:

The enthalpy of reaction for the combustion of liquid ethanol is to be calculated.

Concept introduction:

The heat of the reaction $\left(\Delta H_{\text{rxn}}^{°}\right)$ is defined as the heat released or absorbed during a chemical reaction as a result of the difference in the bond energies (BE) of reactant and product in the reaction. $\Delta H_{\text{rxn}}^{°}$ is negative for exothermic reaction and $\Delta H_{\text{rxn}}^{°}$ is positive for an endothermic reaction.

A combustion reaction is that reaction in which reactant is reacted with molecular oxygen to form the product. Heat is released and the energy is produced in the reaction. Molecular oxygen is employed as an oxidizing agent in these reactions.

Answer to Problem 10.75P

$-1226\text{kJ}$ is the enthalpy of reaction for the combustion of liquid ethanol.

Explanation of Solution

The heat of vaporization of ethanol $\left(\Delta H_{\text{vap}}^{°}\right)$ is $40.5\text{kJ/mol}$. The formula to calculate the enthalpy combustion of liquid ethanol is as follows:

  $\Delta H_{\text{com(liq ethanol)}}^{°}=\Delta H_{\text{vap}}^{°}+\Delta H_{\text{rxn}}^{°}$        (2)

Substitute $40.5\text{kJ/mol}$ for $\Delta H_{\text{vap}}^{°}$ and $-1267\text{kJ}$ for $\Delta H_{\text{rxn}}^{°}$ in the equation (2).

  $\begin{array}{c}\Delta H_{\text{com(liq ethanol)}}^{°}=\left(\text{1}\text{mol}\right)\left(40.5\text{kJ/mol}\right)+\left(-1267\text{kJ}\right)\\ =-1226.7\text{kJ}\\ \approx -1226\text{kJ}\text{.}\end{array}$

Conclusion

The vaporization of a liquid occurs when the intermolecular forces between the molecules break and the molecules are free to vaporize.

(c)

Interpretation Introduction

Interpretation:

The enthalpy of reaction for the combustion of liquid ethanol calculated from the bond energies and standard enthalpies of formation is to be compared.

Concept introduction:

The standard enthalpy of reaction is calculated by the summation of standard enthalpy of formation of the product minus the summation of standard enthalpy of formation of reactant at the standard conditions. The formula to calculate the standard enthalpy of reaction $\left(\Delta H_{\text{rxn}}^{°}\right)$ is as follows:

  $\Delta H_{\text{rxn}}^{°}=\sum m\Delta H_{\text{f (products)}}^{°}-\sum n\Delta H_{\text{f (reactants)}}^{°}$

Here, m and n are the stoichiometric coefficients of reactants and product in the balanced chemical equation.

Answer to Problem 10.75P

The enthalpy of reaction for the combustion of liquid ethanol calculated from the bond energies is close to the enthalpy of reaction for the combustion of liquid ethanol calculated from standard enthalpies of formation.

Explanation of Solution

The given chemical equation for the combustion of ethanol is as follows:

The standard state of oxygen is ${\text{O}}_2\left(g\right)$ so the standard enthalpy of formation for ${\text{O}}_2$ is zero.

The formula to calculate the standard enthalpy for the reaction $\left(\Delta H_{\text{rxn}}^{°}\right)$ is as follows:

  $\Delta H_{\text{rxn}}^{°}=\left[\begin{array}{l}\left\{2\Delta H_{\text{f}}^{°}\left[{\text{CO}}_2\left(g\right)\right]+3\Delta H_{\text{f}}^{°}\left[{\text{H}}_2\text{O}\left(g\right)\right]\right\}\\ -\left\{1\Delta H_{\text{f}}^{°}\left[{\text{C}}_{\text{2}}{\text{H}}_5\text{OH}\left(l\right)\right]+3\Delta H_{\text{f}}^{°}\left[{\text{O}}_2\left(g\right)\right]\right\}\end{array}\right]$        (3)

Substitute $-393.5\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{CO}}_2\left(g\right)\right]$, $-241.826\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{H}}_2\text{O}\left(g\right)\right]$, $-277.3\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{C}}_{\text{2}}{\text{H}}_5\text{OH}\left(l\right)\right]$ and 0 for $\Delta H_{\text{f}}^{°}\left[{\text{O}}_2\left(g\right)\right]$ in the equation (3).

  $\begin{array}{c}\Delta H_{\text{rxn}}^{°}=\left[\begin{array}{l}\left\{\left(2\text{mol}\right)\left(-393.5\text{kJ/mol}\right)+\left(3\text{mol}\right)\left(-241.826\text{kJ/mol}\right)\right\}\\ -\left\{\left(1\text{mol}\right)\left(-277.3\text{kJ/mol}\right)+\left(3\text{mol}\right)\left(0\right)\right\}\end{array}\right]\\ =–1234.848\text{kJ}\\ \approx –1234.8\text{kJ}\text{.}\end{array}$

The calculated value is close to the value calculated in part (b). Therefore the enthalpy of reaction for the combustion of liquid ethanol calculated from the bond energies and standard enthalpies of formation are in good agreement.

Conclusion

The enthalpy of reaction for the combustion of liquid ethanol calculated from standard enthalpies of formation is $–1234.8\text{kJ}$.

(d)

Interpretation Introduction

Interpretation:

$\Delta H_{\text{rxn}}^{°}$ for the formation of gaseous ethanol from ethylene gas with water vapor is to be calculated.

Concept introduction:

The heat of the reaction $\left(\Delta H_{\text{rxn}}^{°}\right)$ is defined as the heat released or absorbed during a chemical reaction as a result of the difference in the bond energies (BE) of reactant and product in the reaction. $\Delta H_{\text{rxn}}^{°}$ is negative for exothermic reaction and $\Delta H_{\text{rxn}}^{°}$ is positive for an endothermic reaction.

The formula to calculate $\Delta H_{\text{rxn}}^{°}$ of reaction is as follows:

  $\Delta H_{\text{rxn}}^{°}=\sum \Delta H_{\text{reactant bond broken}}^{°}+\sum \Delta H_{\text{product bond formed}}^{°}$

Or,

  $\Delta H_{\text{rxn}}^{°}=\sum {\text{BE}}_{\text{reactant bond broken}}-\sum {\text{BE}}_{\text{product bond formed}}$

The bond energy of reactants is positive and the bond energy of products is negative.

Lewis structure is generally considered as a simplified structure of any molecule or atom. Lewis structure for any atom or molecule depicts the valence electrons as dots around the element’s symbol present in the molecule along with the bonds that connect them. Every element tries to complete an octet except the hydrogen atom.

Every element in the Lewis structure tries to attain eight electrons in its valence shell by transfer or share of electrons. This rule is known as the octet rule.

Answer to Problem 10.75P

$-37\text{kJ}$ is the heat of formation of gaseous ethanol from ethylene gas.

Explanation of Solution

The Lewis structures for the reaction of the formation of gaseous ethanol are as follows:

The number of broken bonds is $\text{4 C}-\text{H}$ bonds, $\text{1 C=C}$ bond, and $\text{2 O}-\text{H}$ bond.

The number of bonds formed is $\text{1 C}-\text{C}$ bond, $\text{5 C}-\text{H}$ bonds, $\text{1 C}-\text{O}$ bond, and $\text{1 O}-\text{H}$ bond.

The formula to the enthalpy of the given reaction is as follows:

  $\Delta H_{\text{rxn}}^{°}=\left[\begin{array}{l}\left({\text{4BE}}_{\text{C}-\text{H}}+1{\text{BE}}_{\text{C=C}}+2{\text{BE}}_{\text{O}-\text{H}}\right)\\ -\left(5{\text{BE}}_{\text{C}-\text{H}}+1{\text{BE}}_{\text{C}-\text{C}}+1{\text{BE}}_{\text{C}-\text{O}}+1{\text{BE}}_{\text{O}-\text{H}}\right)\end{array}\right]$        (4)

Substitute $413\text{kJ/mol}$ for ${\text{BE}}_{\text{C}-\text{H}}$, $347\text{kJ/mol}$ for ${\text{BE}}_{\text{C}-\text{C}}$, $358\text{kJ/mol}$ for ${\text{BE}}_{\text{C}-\text{O}}$, $467\text{kJ/mol}$ for ${\text{BE}}_{\text{O}-\text{H}}$ and $614\text{kJ/mol}$ for ${\text{BE}}_{\text{C=C}}$ in the equation (4).

  $\begin{array}{c}\Delta H_{\text{rxn}}^{°}=\left[\begin{array}{l}\left(\left(\text{4 mol}\right)\left(413\text{kJ/mol}\right)+\left(\text{1 mol}\right)\left(614\text{kJ/mol}\right)+\left(\text{2 mol}\right)\left(467\text{kJ/mol}\right)\right)-\\ \left(\text{5 mol}\right)\left(413\text{kJ/mol}\right)+\left(\text{1 mol}\right)\left(347\text{kJ/mol}\right)+\left(\text{1 mol}\right)\left(358\text{kJ/mol}\right)\\ +\left(\text{1 mol}\right)\left(467\text{kJ/mol}\right)\end{array}\right]\\ =-37\text{kJ}\text{.}\end{array}$

Conclusion

The number of broken bonds is $\text{4 C}-\text{H}$ bonds, $\text{1 C=C}$ bond, and $\text{2 O}-\text{H}$ bond while the number of bonds formed is $\text{1 C}-\text{C}$ bond, $\text{5 C}-\text{H}$ bonds, $\text{1 C}-\text{O}$ bond and $\text{1 O}-\text{H}$ bond.