Chapter 9, Problem 9.91P

(a)

Interpretation Introduction

Interpretation:

$\Delta H_{\text{rxn}}^{°}$ for gas-phase hydration of ethylene to ethanol by the use of bond energies and enthalpies of formation is to be determined.

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.

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

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

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

Answer to Problem 9.91P

$\Delta H_{\text{rxn}}^{°}$ for gas-phase hydration of ethylene to ethanol by the use of bond energies and enthalpies of formation is $-37\text{kJ}$ and $-45.7\text{kJ}$ respectively.

Explanation of Solution

The given chemical equation for the gas-phase hydration of ethylene to ethanol is as follows:

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

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

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

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

Substitute $413\text{kJ/mol}$ for ${\text{BE}}_{\text{C}-\text{H}}$, $614\text{kJ/mol}$ for ${\text{BE}}_{\text{C=C}}$, $358\text{kJ/mol}$ for ${\text{BE}}_{\text{C}-\text{O}}$, $347\text{kJ/mol}$ for ${\text{BE}}_{\text{C}-\text{C}}$ and $467\text{kJ/mol}$ for ${\text{BE}}_{\text{O}-\text{H}}$ in the equation (1).

$\begin{array}{c}\Delta H_{\text{rxn}}^{°}=\left[\begin{array}{l}\left(\left(\text{4 mol}\right)\left(413\text{kJ/mol}\right)+\left(\text{1 mol}\right)\left(617\text{kJ/mol}\right)+\left(\text{2 mol}\right)\left(467\text{kJ/mol}\right)\right)-\\ \left(\text{5 mol}\right)\left(413\text{kJ/mol}\right)+\left(\text{1 mol}\right)\left(358\text{kJ/mol}\right)+\left(\text{1 mol}\right)\left(347\text{kJ/mol}\right)+\\ \left(\text{1 mol}\right)\left(467\text{kJ/mol}\right)\end{array}\right]\\ =-37\text{kJ}\end{array}$

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

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

Substitute $-235.1\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}\left(g\right)\right]$, $52.47\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{CH}}_{\text{2}}{\text{=CH}}_{\text{2}}\left(g\right)\right]$ and $-241.826\text{kJ}$ for $\Delta H_{\text{f}}^{°}\left[{\text{H}}_2\text{O}\left(g\right)\right]$ in the equation (2).

$\begin{array}{c}\Delta H_{\text{rxn}}^{°}=\left[\begin{array}{l}\left\{\left(1\text{mol}\right)\left(-235.1\text{kJ/mol}\right)\right\}\\ -\left\{\left(1\text{mol}\right)\left(52.47\text{kJ/mol}\right)+\left(1\text{mol}\right)\left(-241.826\text{kJ}\right)\right\}\end{array}\right]\\ =-45.744\text{kJ}\\ \approx -45.7\text{kJ}\end{array}$

Conclusion

$\Delta H_{\text{rxn}}^{°}$ for gas-phase hydration of ethylene to ethanol by the use of bond energies and enthalpies of formation is $-37\text{kJ}$ and $-45.7\text{kJ}$ respectively.

(b)

Interpretation Introduction

Interpretation:

$\Delta H_{\text{rxn}}^{°}$ for the hydrolysis of ethylene glycol 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 9.91P

$\Delta H_{\text{rxn}}^{°}$ for the hydrolysis of ethylene glycol is 0.

Explanation of Solution

The given chemical equation for the hydrolysis of ethylene glycol is as follows:

${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\text{O}\left(l\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\text{C}}_{\text{2}}{\text{H}}_6{\text{O}}_2\left(l\right)$

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

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

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

$\Delta H_{\text{rxn}}^{°}=\left[\begin{array}{l}\left({\text{4BE}}_{\text{C}-\text{H}}+1{\text{BE}}_{\text{C}-\text{C}}+{\text{1BE}}_{\text{C}-\text{O}}+2{\text{BE}}_{\text{O}-\text{H}}\right)\\ -\left({\text{4BE}}_{\text{C}-\text{H}}+1{\text{BE}}_{\text{C}-\text{C}}+{\text{1BE}}_{\text{C}-\text{O}}+2{\text{BE}}_{\text{O}-\text{H}}\right)\end{array}\right]$ (3)

Substitute $413\text{kJ/mol}$ for ${\text{BE}}_{\text{C}-\text{H}}$, $358\text{kJ/mol}$ for ${\text{BE}}_{\text{C}-\text{O}}$, $347\text{kJ/mol}$ for ${\text{BE}}_{\text{C}-\text{C}}$ and $467\text{kJ/mol}$ for ${\text{BE}}_{\text{O}-\text{H}}$ in the equation (3).

$\begin{array}{c}\Delta H_{\text{rxn}}^{°}=\left[\begin{array}{l}\left(\begin{array}{l}\left(\text{4 mol}\right)\left(413\text{kJ/mol}\right)+\left(\text{2 mol}\right)\left(358\text{kJ/mol}\right)+\left(\text{1 mol}\right)\left(347\text{kJ/mol}\right)+\\ \left(\text{2 mol}\right)\left(467\text{kJ/mol}\right)\end{array}\right)-\\ \left(\begin{array}{l}\left(\text{4 mol}\right)\left(413\text{kJ/mol}\right)+\left(\text{2 mol}\right)\left(358\text{kJ/mol}\right)+\left(\text{1 mol}\right)\left(347\text{kJ/mol}\right)+\\ \left(\text{2 mol}\right)\left(467\text{kJ/mol}\right)\end{array}\right)\end{array}\right]\\ =0\end{array}$

Conclusion

$\Delta H_{\text{rxn}}^{°}$ for the hydrolysis of ethylene glycol is 0.

(c)

Interpretation Introduction

Interpretation:

The reason for the relatively closer value for the hydration in part (a) but not that close for the hydrolysis in part (b) is to be determined.

Concept introduction:

Bond energy is the amount of energy needed to break a bond between two gaseous atoms. Bond energy is measured in $\text{kJ/mol}$. 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 9.91P

In part (b), the same number of similar bonds are broken in the reactant side and formed in the product side so the value of the enthalpy change is zero. Bond energy is generally specific for a bond and remains the same in different molecules.

Explanation of Solution

Bond energy is generally specific for a bond and depends upon the strength of the bond. The bond energy value is the average value so remain same in different molecules.

In part (b), the same number of similar bonds are broken in the reactant side and formed in the product side so the value of the enthalpy change is zero.

For example, the value of bond energy of $\text{O}-\text{H}$ is similar in water and ethyl glycol. Therefore, the bond energy depends upon the strength of the bond but not the molecule.

Conclusion

The bond energy value is the average value so the bond energy of the same bond in different molecules remains same.