Chapter 10, Problem 10.60PAE

Interpretation Introduction

Interpretation: Calculate ${\text{ΔG}}^0$

for the complete combustion of 1 mole of the following fossil fuels: methane (CH₄), ethane (C₂H₆), propane (C₃H₈), and n-butane (C₄H₁₀). Identify any trends that are apparent from these calculations.

Concept introduction:

The change of Gibbs free energy goes with the formation of 1 mole of a substance in its standard state from its constituent elements in their standard state is known as the standard Gibbs free energy of formation.

Answer to Problem 10.60PAE

Solution: ${\text{ΔG}}^0$ of methane $\left({\text{CH}}_{\text{4}}\right)$ = -800.8 kJ

${\text{ΔG}}^0$ of ethane $\left({\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\right)$ = -1441.5 kJ

${\text{ΔG}}^0$ of propane $\left({\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\right)$ = -2073.1 kJ

${\text{ΔG}}^0$ of butane $\left({\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\right)$ = -2703.9 kJ

The value of ${\text{ΔG}}^0$ for each reaction shows trend that as the number of carbons in the hydrocarbon increases the ${\text{ΔG}}^0$ value of the combustion reaction increases.

Explanation of Solution

The change in Gibbs free energy of a reaction is calculated as follows:

${\text{ΔG}}^0$ = sum of ${\text{ΔG}}^0$ (product) - sum of ${\text{ΔG}}^0$ (reactant)

Balanced equation for combustion of methane is as follows:

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

Thus, change in Gibbs free energy of combustion of methane will be:

$\begin{array}{c}{\text{ΔG}}^0\text{=}\left[\begin{array}{l}1\times \Delta {\text{G}}^0\left({\text{CO}}_{\text{2}}\left(\text{g}\right)\right)\\ \text{+2}\times \Delta {\text{G}}^0\left({\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\right)\end{array}\right]-\left[\begin{array}{l}1\times \Delta {\text{G}}^0\left({\text{CH}}_{\text{4}}\left(\text{g}\right)\right)\\ \text{+1}\times \Delta {\text{G}}^0\left({\text{O}}_2\left(\text{g}\right)\right)\end{array}\right]\\ =\left[\begin{array}{l}\left(-\text{394}\text{.4 }\frac{\text{kJ}}{\text{mol}}\right)\left(\text{1 mol}\right)\\ \text{+}\left(\left(-\text{228}\text{.60}\frac{\text{kJ}}{\text{mol}}\right)\left(\text{2 mol}\right)\right)\end{array}\right]-\left[\left(-\text{50}\text{.81}\frac{\text{kJ}}{\text{mol}}\right)\left(\text{1 mol}\right)\right]\\ \text{= }-\text{800}\text{.8 kJ}\end{array}$

Balanced equation for the ethane

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

${\text{ΔG}}^0$ = sum of ${\text{ΔG}}^0$ (product) - sum of ${\text{ΔG}}^0$ (reactant)

Thus, change in Gibbs free energy of combustion of ethane will be:

$\begin{array}{c}{\text{ΔG}}^0\text{=}\left[\begin{array}{l}4\times \Delta {\text{G}}^0\left({\text{CO}}_{\text{2}}\left(\text{g}\right)\right)\\ \text{+6}\times \Delta {\text{G}}^0\left({\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\right)\end{array}\right]-\left[\begin{array}{l}2\times \Delta {\text{G}}^0\left({\text{C}}_2{\text{H}}_{\text{6}}\left(\text{g}\right)\right)\\ \text{+7}\times \Delta {\text{G}}^0\left({\text{O}}_2\left(\text{g}\right)\right)\end{array}\right]\\ =\left[\begin{array}{l}\left(-\text{394}\text{.4 }\frac{\text{kJ}}{\text{mol}}\right)\left(\text{4 mol}\right)\\ \text{+}\left(\left(-\text{228}\text{.60}\frac{\text{kJ}}{\text{mol}}\right)\left(\text{6 mol}\right)\right)\end{array}\right]-\left[\left(-\text{32}\text{.89}\frac{\text{kJ}}{\text{mol}}\right)\left(\text{2 mol}\right)\right]\\ \text{= }-\text{2883 kJ}\end{array}$

Now, the above value is calculated for 2 mol of ethane. Now calculating ${\text{ΔG}}^0$ for 1 mol: -

$-\text{2883 kJ/2 }\left({\text{mol of C}}_{\text{2}}{\text{H}}_{\text{6}}\right)\text{ = }-\text{1441}\text{.5 kJ}$

Therefore, the ${\text{ΔG}}^0$ for the combustion of 1 mole ethane is -1441.5 kJ

Balanced equation for the propane

${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\left(\text{g}\right){\text{+ 5O}}_{\text{2}}\left(\text{g}\right)\to {\text{ 3CO}}_{\text{2}}\left(\text{g}\right){\text{+ 4H}}_{\text{2}}\text{O}\left(\text{g}\right)$

${\text{ΔG}}^0$ = sum of ${\text{ΔG}}^0$ (product) - sum of ${\text{ΔG}}^0$ (reactant)

Thus, change in Gibbs free energy of combustion of propane will be:

$\begin{array}{c}{\text{ΔG}}^0\text{=}\left[\begin{array}{l}3\times \Delta {\text{G}}^0\left({\text{CO}}_{\text{2}}\left(\text{g}\right)\right)\\ \text{+4}\times \Delta {\text{G}}^0\left({\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\right)\end{array}\right]-\left[\begin{array}{l}1\times \Delta {\text{G}}^0\left({\text{C}}_3{\text{H}}_{\text{8}}\left(\text{g}\right)\right)\\ \text{+5}\times \Delta {\text{G}}^0\left({\text{O}}_2\left(\text{g}\right)\right)\end{array}\right]\\ =\left[\begin{array}{l}\left(-\text{394}\text{.4 }\frac{\text{kJ}}{\text{mol}}\right)\left(\text{4 mol}\right)\\ \text{+}\left(\left(-\text{228}\text{.60}\frac{\text{kJ}}{\text{mol}}\right)\left(\text{6 mol}\right)\right)\end{array}\right]-\left[\left(-\text{24}\text{.5}\frac{\text{kJ}}{\text{mol}}\right)\left(\text{1 mol}\right)\right]\\ \text{= }-\text{2073}\text{.1 kJ}\end{array}$

Therefore the ${\text{ΔG}}^0$ for the combustion of 1 mole propane is -2073.1 kJ

Balanced equation for the butane

${\text{2 C}}_{\text{4}}{\text{H}}_{\text{10}}\left(\text{g}\right){\text{+ 13O}}_{\text{2}}\left(\text{g}\right)\to {\text{8CO}}_{\text{2}}\left(\text{g}\right){\text{+ 10H}}_{\text{2}}\text{O}\left(\text{g}\right)$

${\text{ΔG}}^0$ = sum of ${\text{ΔG}}^0$ (product) - sum of ${\text{ΔG}}^0$ (reactant)

Thus, change in Gibbs free energy of combustion of butane will be:

$\begin{array}{c}{\text{ΔG}}^0\text{=}\left[\begin{array}{l}8\times \Delta {\text{G}}^0\left({\text{CO}}_{\text{2}}\left(\text{g}\right)\right)\\ \text{+10}\times \Delta {\text{G}}^0\left({\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\right)\end{array}\right]-\left[\begin{array}{l}2\times \Delta {\text{G}}^0\left({\text{C}}_4{\text{H}}_{\text{10}}\left(\text{g}\right)\right)\\ \text{+13}\times \Delta {\text{G}}^0\left({\text{O}}_2\left(\text{g}\right)\right)\end{array}\right]\\ =\left[\begin{array}{l}\left(-\text{394}\text{.4 }\frac{\text{kJ}}{\text{mol}}\right)\left(\text{8 mol}\right)\\ \text{+}\left(\left(-\text{228}\text{.60}\frac{\text{kJ}}{\text{mol}}\right)\left(\text{10 mol}\right)\right)\end{array}\right]-\left[\left(-\text{16}\text{.7}\frac{\text{kJ}}{\text{mol}}\right)\left(\text{2 mol}\right)\right]\\ \text{= }-\text{5407}\text{.8 kJ}\end{array}$

Now, the above value is calculated for 2 mol of butane. Now calculating ${\text{ΔG}}^0$

for 1 mol: -

$-\text{5407}{\text{.8 kJ/2mol C}}_{\text{4}}{\text{H}}_{\text{10}}\text{ = }-\text{2703}\text{.9 kJ}$

Hence, the ${\text{ΔG}}^0$ for the combustion of 1 mole butane is -2703.9 kJ.

From the calculation of the standard free energy ${\text{ΔG}}^0$ for each reaction shows trend that as the number of carbons in the hydrocarbon increases the ${\text{ΔG}}^0$ value of the combustion reaction increases.

Conclusion

The change of Gibbs free energy goes with the formation of 1 mole of a substance in its standard state from its constituent elements in their standard state is known as the standard Gibbs free energy of formation. ${\text{ΔG}}^0$

for the complete combustion of 1 mole of the following fossil fuels: methane (CH₄), ethane (C₂H₆), propane (C₃H₈), and n-butane (C₄H₁₀) is:

${\text{ΔG}}^0$ of combustion of methane $\left({\text{CH}}_{\text{4}}\right)$

1. = -800.8 kJ
2. ${\text{ΔG}}^0$ of combustion of ethane $\left({\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\right)$
3. = -1441.5 kJ
4. ${\text{ΔG}}^0$ of combustion of propane $\left({\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\right)$ = -2073.1 kJ
5. ${\text{ΔG}}^0$ of combustion of butane $\left({\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\right)$
6. = -2703.9 kJ