Chapter 5, Problem 83AP

Ethanol $\left({\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\right)$ and gasoline (assumed to be all octane. ${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}$) are both used as automobile fuel. If gasoline is selling for $\text{\$2}\text{.20, gal}$, what would the price of ethanol have to be in order to provide the same amount of heat per dollar? The density and $\Delta H_{\text{f}}^{\text{°}}$ of octane are $\text{0}\text{.7025 g/mL and -249}\text{.9 kJ/mol}\text{.}$ respectively, and of ethanol are $\text{0}\text{.7894 g/mL and -277}\text{.0 kJ/mol}\text{.}$ respectively $\left(\text{1 gal = 3}\text{.785 L}\right)$.

Interpretation Introduction

Interpretation:

The price of ethanol is to be calculated.

Concept introduction:

The standard enthalpy for a reaction is the amount of enthalpy that occurs under standard conditions.

The standard enthalpy of reaction is determined by using the equation given below:

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

Here, the stoichiometric coefficients are represented by m for the reactants and n for the products and the enthalpy of formation under standard conditions is represented by $\Delta H_{\text{f}}^{°}$.

The value of the enthalpy of formation of an element is zero at its most stable state.

Answer to Problem 83AP

Solution: $\$1.57/\text{gal}$

Explanation of Solution

Given information: Density $d_{{\text{C}}_2{\text{H}}_5\text{OH}}=0.7894\text{ g}/\text{mL}$

$\Delta H{°}_{\text{f}}\left({\text{C}}_2{\text{H}}_5\text{OH}\right)=-277.0\text{ kJ}/\text{mol}$

Density $d_{{\text{C}}_8{\text{H}}_{18}}=0.7025\text{ g}/\text{mL}$

$\Delta H{°}_{\text{f}}\left({\text{C}}_{\text{8}}{\text{H}}_{18}\right)=-249.9\text{ kJ}/\text{mol}$

The combustion of octane takes place as follows:

${\text{C}}_{\text{8}}{\text{H}}_{18}\left(l\right)+\frac{25}{2}{\text{O}}_2\left(g\right)\to 8{\text{ CO}}_2\left(g\right)\text{+}9{\text{ H}}_{\text{2}}\text{O}\left(l\right)$

From appendix 2,

$\begin{array}{c}\Delta H{°}_{\text{f}}\left({\text{C}}_{\text{8}}{\text{H}}_{18}\right)=-249.9\text{ kJ}/\text{mol}\\ \Delta H{°}_{\text{f}}\left({\text{H}}_{\text{2}}\text{O}\right)=-285.8\text{ kJ}/\text{mol}\\ \Delta H{°}_{\text{f}}\left({\text{CO}}_2\right)=-393.5\text{ kJ}/\text{mol}\end{array}$

The oxygen gas is in its most stable form, so its enthalpy of formation is zero.

Calculate the standard enthalpy of the reaction for the combustion of octane as follows:

$\Delta H{°}_{\text{rxn}}=\left[8\Delta H{°}_{\text{f}}\left({\text{CO}}_2\right)+9\Delta H{°}_{\text{f}}\left({\text{H}}_2\text{O}\right)\right]-\left[\Delta H{°}_{\text{f}}\left({\text{C}}_{\text{8}}{\text{H}}_{\text{18}}\right)+\frac{25}{2}\Delta H{°}_{\text{f}}\left({\text{O}}_2\right)\right]$

Substitute $-249.9\text{ kJ}/\text{mol}$ for $\Delta H{°}_{\text{f}}\left({\text{C}}_{\text{8}}{\text{H}}_{18}\right)$, $-285.8\text{ kJ}/\text{mol}$ for $\Delta H{°}_{\text{f}}\left({\text{H}}_{\text{2}}\text{O}\right)$, $-393.5\text{ kJ}/\text{mol}$ for $\Delta H{°}_{\text{f}}\left({\text{CO}}_2\right)$, and $\text{0 kJ}/\text{mol}$

for $\Delta H{°}_{\text{f}}\left({\text{O}}_{\text{2}}\right)$

in the above equation

$\begin{array}{c}\Delta H{°}_{\text{rxn}}=\left[8\left(-393.5\text{ kJ}/\text{mol}\right)+9\left(-285.8\text{ kJ}/\text{mol}\right)\right]-\left[\left(-249.9\text{ kJ}/\text{mol}\right)+\frac{25}{2}\left(0\right)\right]\\ =\left(-3148\text{ kJ}/\text{mol}-2572.2\text{ kJ}/\text{mol}\right)-\left(-249.9\text{ kJ}/\text{mol}\right)\\ =-5470.3\text{ kJ}/\text{mol}\end{array}$

Convert kilojoule per mole to kilojoule per gallon as follows:

$\begin{array}{c}\Delta H{°}_{\text{rxn}}=\left(-5470.3\text{ kJ}/\cancel{\text{mol}}\right)\left(\frac{\text{0}\text{.7025 g}/\cancel{\text{ml}}}{114.2\text{ g}/\cancel{mol}}\right)\left(3785\cancel{\text{ml}}/\text{gal}\right)\\ =-1.27\times {10}^5\text{ kJ}/\text{gal}\end{array}$

The combustion of ethanol takes place as follows:

${\text{C}}_2{\text{H}}_5\text{OH}\left(l\right)+3{\text{ O}}_2\left(g\right)\to 2{\text{ CO}}_2\left(g\right)\text{+}3{\text{ H}}_{\text{2}}\text{O}\left(l\right)$

From appendix 2,

$\begin{array}{c}\Delta H{°}_{\text{f}}\left({\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\right)=-277\text{ kJ}/\text{mol}\\ \Delta H{°}_{\text{f}}\left({\text{H}}_{\text{2}}\text{O}\right)=-285.8\text{ kJ}/\text{mol}\\ \Delta H{°}_{\text{f}}\left({\text{CO}}_2\right)=-393.5\text{ kJ}/\text{mol}\end{array}$

Calculate the standard enthalpy of the reaction for the combustion of ethanol as follows:

$\Delta H{°}_{\text{rxn}}=\left[2\Delta H{°}_{\text{f}}\left({\text{CO}}_{\text{2}}\right)+3\Delta H{°}_{\text{f}}\left({\text{H}}_2\text{O}\right)\right]-\left[\Delta H{°}_{\text{f}}\left({\text{C}}_2{\text{H}}_5\text{OH}\right)+3\Delta H{°}_{\text{f}}\left({\text{O}}_2\right)\right]$

Substitute $-393.5\text{ kJ}/\text{mol}$ for $\Delta H{°}_{\text{f}}\left({\text{CO}}_2\right)$, $-285.8\text{ kJ}/\text{mol}$ for $\Delta H{°}_{\text{f}}\left({\text{H}}_{\text{2}}\text{O}\right)$, $-277\text{ kJ}/\text{mol}$ for $\Delta H{°}_{\text{f}}\left({\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\right)$, and $\text{0 kJ}/\text{mol}$

for $\Delta H{°}_{\text{f}}\left({\text{O}}_{\text{2}}\right)$ in the above equation and solve as

$\begin{array}{c}\Delta H{°}_{\text{rxn}}=\left[2\left(-393.5\text{ kJ}/\text{mol}\right)+3\left(-285.8\text{ kJ/mol}\right)\right]-\left[\left(-277\text{ kJ}/\text{mol}\right)+3\left(0\right)\right]\\ =\left(-787\text{ kJ}/\text{mol}-857.4\text{ kJ}/\text{mol}\right)-\left(-277\text{ kJ}/\text{mol}\right)\\ =-1644.4\text{ kJ}/\text{mol}+277\text{ kJ}/\text{mol}\\ \text{=}-1367.4\text{ kJ}/\text{mol}\end{array}$

Convert kilojoule per mole to kilojoule per gallon as follows:

$\begin{array}{c}\Delta H{°}_{\text{rxn}}=\left(-1367.4\text{ kJ}/\cancel{\text{mol}}\right)\left(\frac{\text{0}\text{.7894 g}/\cancel{\text{ml}}}{46\text{ g}/\cancel{mol}}\right)\left(3785\cancel{\text{ml}}/\text{gal}\right)\\ =-8.8\times {10}^4\text{ kJ}/\text{gal}\end{array}$

Now, the cost of ethanol per gallon is

$\begin{array}{l}\frac{\$2.20}{1.27\times {10}^{\cancel{5}}\cancel{\text{kJ}/\text{gal}}}\times 8.8\times \cancel{{10}^4}\cancel{\text{kJ}/\text{gal}}\\ =\frac{8.8}{12.7}\times \$2.79\\ =\$1.57\end{array}$

Conclusion

The price of ethanol is $\$1.57/\text{gal}$.