Ethanol, \({{\bf{C}}_{\bf{2}}}{{\bf{H}}_{\bf{5}}}{\bf{OH}}\), is used as a fuel for motor vehicles, particularly in Brazil.

(a) Write the balanced equation for the combustion of ethanol to CO₂(g) and H₂O(g), and, using the data in Appendix G, calculate the enthalpy of combustion of 1 mole of ethanol.

(b) The density of ethanol is 0.7893 g/ml. Calculate the enthalpy of combustion of exactly 1 L of ethanol.

(c) Assuming that an automobile’s mileage is directly proportional to the heat of combustion of the fuel, calculate how much farther an automobile could be expected to travel on 1 L of gasoline than on 1 L of ethanol. Assume that gasoline has the heat of combustion and the density of n–octane, \({{\bf{C}}_{\bf{8}}}{{\bf{H}}_{{\bf{18}}}}\) (ΔH_f=

- 208.4 kJ/mol; density = 0.7025 g/mL).

The amount of heat released when a combustible substance gets burned in the presence of oxygen is called the enthalpy of combustion.

We have to write the balanced equation for the combustion of ethanol and calculate the enthalpy of combustion of 1 mol of ethanol.

\(\begin{array}{l}{\rm{Balanced chemical reaction:}}\\{{\bf{C}}_{\bf{2}}}{{\bf{H}}_{\bf{5}}}{\bf{OH(l) + }}{\bf{3}}{{\bf{O}}_{\bf{2}}}{\bf{(g)}} \to {\bf{2C}}{{\bf{O}}_{\bf{2}}}{\bf{(g)}}{\bf{ + }}{\bf{3}}{{\bf{H}}_{\bf{2}}}{\bf{O(g)}}\\\\{\rm{Enthalpy chnage in the reaction, }}\\{\rm{\Delta }}{{\rm{H}}_{{\rm{reaction}}}}{\rm{ = }}\sum {{\rm{\Delta }}{{\rm{H}}_{{\rm{products}}}}} {\rm{ - }}\sum {{\rm{\Delta }}{{\rm{H}}_{{\rm{reactants}}}}} \\{\rm{\Delta }}{{\rm{H}}_{{\rm{reaction}}}}{\rm{ = }}\left( {{\rm{2 \times \Delta }}{{\rm{H}}_{{\rm{C}}{{\rm{O}}_{\rm{2}}}{\rm{(g)}}}}{\rm{ + 3 \times \Delta }}{{\rm{H}}_{{{\rm{H}}_{\rm{2}}}{\rm{O(g)}}}}} \right){\rm{ - }}\left( {{\rm{\Delta }}{{\rm{H}}_{{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{OH(l)}}}}{\rm{ + 3 \times \Delta }}{{\rm{H}}_{{{\rm{O}}_{\rm{2}}}{\rm{(g) }}}}} \right)\\{\rm{\Delta }}{{\rm{H}}_{{\rm{reaction}}}}{\rm{ = }}\left( {{\rm{2 \times - 393}}{\rm{.51 - 3 \times 241}}{\rm{.82}}} \right){\rm{ - ( - 277}}{\rm{.6 + 0)}}\\{\rm{\Delta }}{{\rm{H}}_{{\rm{reaction}}}}{\rm{ = }}\left( {{\rm{ - 1512}}{\rm{.48 + 277}}{\rm{.6}}} \right){\rm{ kJ}}\\{\rm{\Delta }}{{\rm{H}}_{{\rm{reaction}}}}{\rm{ = - 1234}}{\rm{.88 kJ}}\end{array}\)

Hence, the enthalpy of combustion of ethanol is-1234.88 kJ.

On combustion of one mole of ethanol, the amount of heat produced is -1234.88 kJ.

The density of ethanol is 0.7893 g/mL.

So, the mass of 1 L ethanol will be = \(0.7893{\rm{ g/ml }} \times {\rm{1000 ml = 789}}{\rm{.3 g}}\)

Hence, the enthalpy of combustion for exactly 1 L of ethanol will be -21157.6515 kJ.

\(\begin{array}{l}{\rm{ = 789}}{\rm{.3 g of }}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{OH(g) \times }}\frac{{{\rm{1 mol of }}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{OH(g)}}}}{{{\rm{46}}{\rm{.068 g of }}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{OH(g)}}}}{\rm{ \times }}\frac{{{\rm{ - 1234}}{\rm{.88 kJ}}}}{{{\rm{1 mol of }}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{OH(g)}}}}\\{\rm{ = }}{\rm{ - 21157}}{\rm{.6515}}\;{\rm{kJ}}\end{array}\)

The enthalpy of combustion of ethanol for 1 L is equal to -21157.6515 kJ.

We have to calculate the enthalpy of combustion for gasoline in terms of per liter.

\(\begin{array}{l}{\rm{Combustion reaction for gasoline, }}\\{{\rm{C}}_{\rm{8}}}{{\rm{H}}_{{\rm{18}}}}{\rm{(g) + }}{\rm{ }}\frac{{27}}{2}{{\rm{O}}_{\rm{2}}}{\rm{(g)}} \to {\rm{8C}}{{\rm{O}}_2}(g) + 9{{\rm{H}}_{\rm{2}}}{\rm{O(g)}}\\\Delta {{\rm H}_{reaction}} = \left( {9 \times \Delta {{\rm H}_{{{\rm{H}}_{\rm{2}}}{\rm{O(g)}}}} + 8 \times \Delta {{\rm H}_{{\rm{C}}{{\rm{O}}_2}(g)}}} \right) - \left( {\Delta {{\rm H}_{{{\rm{C}}_{\rm{8}}}{{\rm{H}}_{{\rm{18}}}}{\rm{(g)}}}} + \frac{{27}}{2} \times \Delta {{\rm H}_{{{\rm{O}}_{\rm{2}}}{\rm{(g)}}}}} \right)\\\Delta {{\rm H}_{reaction}} = (9 \times - 241.82 - 8 \times 393.51) - ( - 208.4 + 0){\rm{ kJ}}\\\Delta {{\rm H}_{reaction}} = - 5324.46 + 208.4{\rm{ kJ}}\\\Delta {{\rm H}_{reaction}} = - 5116.06{\rm{ kJ}}\end{array}\)

We have to calculate the amount of heat of combustion for 1 L of gasoline.

\(\begin{array}{l}{\rm{Density of gasoline = 0}}{\rm{.7025 g/ml}}\\{\rm{Mass of 1L of gasoline = 702}}{\rm{.5 g}}\\{\rm{702}}{\rm{.5 g of }}{{\rm{C}}_{\rm{8}}}{{\rm{H}}_{{\rm{18}}}}{\rm{(g) \times }}\frac{{{\rm{1 mol of }}{{\rm{C}}_{\rm{8}}}{{\rm{H}}_{{\rm{18}}}}{\rm{(g)}}}}{{{\rm{114}}{\rm{.224 g of }}{{\rm{C}}_{\rm{8}}}{{\rm{H}}_{{\rm{18}}}}{\rm{(g)}}}}{\rm{ \times }}\frac{{{\rm{ - 5116}}{\rm{.06 kJ}}}}{{{\rm{1 mol of }}{{\rm{C}}_{\rm{8}}}{{\rm{H}}_{{\rm{18}}}}{\rm{(g)}}}}\\{\rm{ = }}{\rm{ - 31464}}{\rm{.8}}\;{\rm{kJ}}\end{array}\)

Hence, the amount of heat released on burning 1 L of gasoline will be 31464.8 kJ.

An automobile can travel 1.48 times further on gasoline than on ethanol,and its calculation is as follows:

\(\frac{{31464.8}}{{21157.7}} = 1.48\)