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Answer: Ethanol (C2H5OH) and gasoline (assumed to be all

Chapter 6, Problem 93P

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QUESTION:

Ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and gasoline (assumed to be all octane, \(\mathrm{C}_{8} \mathrm{H}_{18}\) ) are both used as automobile fuel. If gasoline is selling for \(\$ 4.50\) /gal, what would the price of ethanol have to be in order to provide the same amount of heat per dollar? The density and \(\Delta H_{\mathrm{f}}^{\circ}\) of octane are \(0.7025 \mathrm{~g} / \mathrm{mL}\) and \(-249.9 \mathrm{~kJ} / \mathrm{mol}\) and of ethanol are \(0.7894 \mathrm{~g} / \mathrm{mL}\) and \(-277.0 \mathrm{~kJ} / \mathrm{mol}\), respectively. \(1 \mathrm{gal}=3.785 \mathrm{~L}\).

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QUESTION:

Ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and gasoline (assumed to be all octane, \(\mathrm{C}_{8} \mathrm{H}_{18}\) ) are both used as automobile fuel. If gasoline is selling for \(\$ 4.50\) /gal, what would the price of ethanol have to be in order to provide the same amount of heat per dollar? The density and \(\Delta H_{\mathrm{f}}^{\circ}\) of octane are \(0.7025 \mathrm{~g} / \mathrm{mL}\) and \(-249.9 \mathrm{~kJ} / \mathrm{mol}\) and of ethanol are \(0.7894 \mathrm{~g} / \mathrm{mL}\) and \(-277.0 \mathrm{~kJ} / \mathrm{mol}\), respectively. \(1 \mathrm{gal}=3.785 \mathrm{~L}\).

ANSWER:

Step 1 of 4

Here, we are going to calculate the price of ethanol which has to be in order to provide the same amount of heat per dollar.

The combustion reaction of octane is

                          \(\text{C}_8\text{H}_{18} + 25/2 \text{O}_2(g) \rightarrow 8\text{CO}_2(g)+9\text{H}_2\text{O}(l)\)

                         

Given that

                          \(\Delta H_{rxn}^{0}=\sum n\Delta H_{f}^{0}\ (product)-\sum \Delta H_{f}^{0}\ (reactant)\)

                                         \(=\left [8\times\Delta H_{f}^{0}(CO_2,g)+9\times\Delta H_{f}^{0}(H_2O,l)  \right ]-\left [\Delta H_{f}^{0}(C_8H_{18},g)+25/2\Delta H_{f}^{0}(O_2,g)  \right ]\)

                                         \(=\left [8\times\left ( -393.5\ kJ/mol \right )+9\times\left ( -285.8\ kJ/mol \right )  \right ]-\left ( -249.9\ kJ/mol \right )\)

                                         \(=-3148\ kJ/mol-2572.2\ kJ/mol + 249.73\ kJ/mol\)

                                         \(=-5470.5\ kJ/mol\)

 

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