Chapter 18, Problem 18.98QP

When 1.000 g of ethylene glycol, C₂H₆O₂, is burned at 25°C and 1.00 atmosphere pressure, H₂O(l) and CO₂(g) are formed with the evolution of 19.18 kJ of heat.

1. a Calculate the molar enthalpy of formation of ethylene glycol. (It will be necessary to use data from Appendix C.)
2. b $\Delta G_f^{°}$ of ethylene glycol is −322.5 kJ/mol. What is ΔG° for the combustion of 1 mol ethylene glycol?
3. c What is ΔS° for the combustion of 1 mol ethylene glycol?

Interpretation Introduction

Interpretation:

The molar enthalpy of formation of ethylene glycol, standard free energy change for the combustion of ethylene glycol and entropy change for the combustion of ethylene glycol has to be calculated.

Concept introduction:

Standard free energy change:

Standard free energy change is measured by subtracting the product of temperature and standard entropy change from the standard enthalpy change of a system.

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{where,}\\ {\text{ΔG}}^{\text{o}}\text{-}\text{standard}\text{free}\text{energy change}\\ {\text{ΔH}}^{\text{o}}\text{-}\text{standard}\text{enthalpy}\text{change}\\ {\text{ΔS}}^{\text{o}}\text{-}\text{standard}\text{entropy change}\text{and T}\text{-}\text{temperature}\text{.}\end{array}$

Answer to Problem 18.98QP

The molar enthalpy of formation $\left({\text{ΔH}}^{\text{o}}{}_{\text{f}}\right)$ of ethylene glycol is $\text{-454}\text{kJ/mol}\text{.}$

Explanation of Solution

To calculate: The value of ${\text{ΔH}}^{\text{o}}{}_{\text{f}}$ for ethylene glycol

Given reaction and information,

$\begin{array}{l}{\text{C}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}{}_{\text{(l)}}\text{+ }{\text{5/2O}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{ 2CO}}_{\text{2}}{}_{\left(\text{g}\right)}\text{+}{\text{3H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\\ {\text{ΔH}}^{\text{o}}{}_{\text{f}}\text{:}\text{?}\text{2}\text{×}\text{-393}\text{.5}\text{3}\text{×}\text{-285}\text{.8}\text{kJ}\\ {\text{ΔH}}^{\text{o}}\text{is}\text{-19}\text{.18}\text{kJ}\text{.}\end{array}$

Calculate the value of ${\text{ΔH}}^{\text{o}}$ per mole

$\begin{array}{l}{\text{ΔH}}^{\text{o}}\text{=}\frac{\text{-19}\text{.18}\text{kJ}}{\text{1}\text{.000}\text{g}}\text{×}\frac{\text{62}\text{.07}\text{g}}{\text{1}\text{mol}}\\ \text{=}\text{-1190}\text{.4}\text{kJ/mol}\text{.}\end{array}$

Calculate the value of ${\text{ΔH}}^{\text{o}}{}_{\text{f}}$ of ethylene glycol

$\begin{array}{l}{\text{ΔH}}^{\text{o}}\text{=}{\displaystyle \sum {\text{nΔH}}^{\text{o}}{}_{\text{f}}}\left(\text{products}\right)\text{-}{\displaystyle \sum {\text{mΔH}}^{\text{o}}{}_{\text{f}}}\left(\text{reactants}\right)\\ \text{-1190}\text{.4}\text{kJ}\text{=}\left[\left(\text{2}\text{×}\text{-393}\text{.5}\right)\text{+}\left(\text{3}\text{×}\text{-285}\text{.8}\right)\text{-}\left(\text{1}\text{mol}\right)\text{×}{\text{ΔH}}^{\text{o}}{}_{\text{f}}\left({\text{C}}_{\text{2}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}\right)\right]\text{kJ}\\ \text{-1190}\text{.4}\text{kJ}\text{=}\left[\text{-787}\text{-}\text{857}\text{.4}\text{-}\left(\text{1}\text{mol}\right)\text{×}{\text{ΔH}}^{\text{o}}{}_{\text{f}}\left({\text{C}}_{\text{2}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}\right)\right]\text{kJ}\\ {\text{ΔH}}^{\text{o}}{}_{\text{f}}\left({\text{C}}_{\text{2}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}\right)\text{=}\text{-454}\text{.0}\text{=}\text{-}\text{454}\text{kJ/mol}\text{.}\end{array}$

The molar enthalpy of formation $\left({\text{ΔH}}^{\text{o}}{}_{\text{f}}\right)$ of ethylene glycol is $\text{-454}\text{kJ/mol}\text{.}$

Interpretation Introduction

Interpretation:

The molar enthalpy of formation of ethylene glycol, standard free energy change for the combustion of ethylene glycol and entropy change for the combustion of ethylene glycol has to be calculated.

Concept introduction:

Standard free energy change:

Standard free energy change is measured by subtracting the product of temperature and standard entropy change from the standard enthalpy change of a system.

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{where,}\\ {\text{ΔG}}^{\text{o}}\text{-}\text{standard}\text{free}\text{energy change}\\ {\text{ΔH}}^{\text{o}}\text{-}\text{standard}\text{enthalpy}\text{change}\\ {\text{ΔS}}^{\text{o}}\text{-}\text{standard}\text{entropy change}\text{and T}\text{-}\text{temperature}\text{.}\end{array}$

Answer to Problem 18.98QP

The value of free energy change $\left({\text{ΔG}}^{\text{o}}\right)$ for the combustion of $\text{1}\text{mol}\text{of}$ ethylene glycol is $\text{-1177}\text{.6}\text{kJ}\text{.}$

Explanation of Solution

To calculate: The value of ${\text{ΔG}}^{\text{o}}$ for the combustion of $\text{1}\text{mol}\text{of}$ ethylene glycol

Given information

${\text{ΔG}}^{\text{o}}{}_{\text{f}}\text{of}\text{ethylene glycol}\text{is}\text{-322}\text{.5}\text{kJ/mol}\text{.}$

Calculate the value of ${\text{ΔG}}^{\text{o}}$

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\displaystyle \sum {\text{nΔG}}^{\text{o}}{}_{\text{f}}}\left(\text{products}\right)\text{-}{\displaystyle \sum {\text{mΔG}}^{\text{o}}{}_{\text{f}}}\left(\text{reactants}\right)\\ \text{=}\left[\left(\text{2}\text{×}\text{-394}\text{.4}\right)\text{+}\left(\text{3}\text{×}\text{-237}\text{.1}\right)\text{-}\left(\text{-322}\text{.5}\right)\right]\text{kJ}\\ \text{=}\left[\text{-788}\text{.8}\text{-}\text{711}\text{.3}\text{+}\text{322}\text{.5}\right]\text{kJ}\\ \text{=}\text{-1177}\text{.6}\text{kJ}\text{.}\end{array}$

The value of free energy change $\left({\text{ΔG}}^{\text{o}}\right)$ for the combustion of $\text{1}\text{mol}\text{of}$ ethylene glycol is $\text{-1177}\text{.6}\text{kJ}\text{.}$

Interpretation Introduction

Interpretation:

The molar enthalpy of formation of ethylene glycol, standard free energy change for the combustion of ethylene glycol and entropy change for the combustion of ethylene glycol has to be calculated.

Concept introduction:

Standard free energy change:

Standard free energy change is measured by subtracting the product of temperature and standard entropy change from the standard enthalpy change of a system.

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{where,}\\ {\text{ΔG}}^{\text{o}}\text{-}\text{standard}\text{free}\text{energy change}\\ {\text{ΔH}}^{\text{o}}\text{-}\text{standard}\text{enthalpy}\text{change}\\ {\text{ΔS}}^{\text{o}}\text{-}\text{standard}\text{entropy change}\text{and T}\text{-}\text{temperature}\text{.}\end{array}$

Answer to Problem 18.98QP

The value of entropy change $\left({\text{ΔS}}^{\text{o}}\right)$ for the combustion of $\text{1}\text{mol}\text{of}$ ethylene glycol is $\text{-43}\text{.0}\text{J/K}\text{.}$

Explanation of Solution

To calculate: The value of entropy change $\left({\text{ΔS}}^{\text{o}}\right)$ for the combustion of $\text{1}\text{mol}\text{of}$ ethylene glycol

The relationship between ${\text{ΔG}}^{\text{o}}\text{and}{\text{ΔS}}^{\text{o}}$ is given by

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{Substitute}\text{the}\text{values}\text{of}{\text{ΔH}}^{\text{o}}\text{,}{\text{ΔG}}^{\text{o}}\text{and}\text{T}\left(\text{298}\text{K}\right)\\ \text{-1177}\text{.6}\text{kJ}\text{=}\text{-1190}\text{.4}\text{kJ}\text{-}\left(\text{298}\text{K}\right){\text{ΔS}}^{\text{o}}\\ {\text{ΔS}}^{\text{o}}\text{=}\frac{\left(\text{-1177}\text{.6}\text{kJ}\right)\text{+}\text{1190}\text{.4}\text{kJ}}{\text{298}\text{K}}\text{=}\text{-}\text{0}\text{.04295}\text{kJ/K}\\ \text{=}\text{-43}\text{.0}\text{J/K}\left(\text{for}\text{1}\text{mole}\right)\text{.}\end{array}$

The value of entropy change $\left({\text{ΔS}}^{\text{o}}\right)$ for the combustion of $\text{1}\text{mol}\text{of}$ ethylene glycol is $\text{-43}\text{.0}\text{J/K}\text{.}$