Calculate the heat of combustion of 1 mole of ethanol, \({{\bf{C}}_{\bf{2}}}{{\bf{H}}_{\bf{5}}}{\bf{OH}}\)(l), when \({{\bf{H}}_{\bf{2}}}{\bf{O}}\)(l) and \({\bf{C}}{{\bf{O}}_{\bf{2}}}\)(g) are formed.

Use the following enthalpies of formation: \({{\bf{C}}_{\bf{2}}}{{\bf{H}}_{\bf{5}}}{\bf{OH}}\)(l), −278 kJ/mol; \({{\bf{H}}_{\bf{2}}}{\bf{O}}\)(l), −286 kJ/mol; and \({\bf{C}}{{\bf{O}}_{\bf{2}}}\)(g), −394 kJ/mol.

By using the following special form of the Hess’ law, we can calculate the heat of combustion of 1 mole of ethanol.

\({\bf{\Delta H}}_{{\bf{reaction}}}^{\bf{o}}{\bf{ = \Sigma n \times \Delta Hf}}_{{\bf{products}}}^{\bf{o}}{\bf{ - \Sigma n \times \Delta Hf}}_{{\bf{reactants}}}^{\bf{o}}.\)

Given: Enthalpies of formation:

• \({{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{OH}}\)(l), −278kJ/mol
• \({{\rm{H}}_{\rm{2}}}{\rm{O}}\)(l), −286kJ/mol
• \({\rm{C}}{{\rm{O}}_{\rm{2}}}\)(g), −394 kJ/mol

The reaction of the combustion of ethanol is:

\({{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{OH(l) + 3}}{{\rm{O}}_{\rm{2}}}\left( {\rm{g}} \right){\rm{\;}} \to {\rm{3}}{{\rm{H}}_{\rm{2}}}{\rm{O(l)\; + \;2C}}{{\rm{O}}_{\rm{2}}}{\rm{(g)}}{\rm{.}}\)

From the combustion of ethanol, the products formed are water and carbon dioxide gas. The known values are substituted as shown below.

\(\begin{array}{c}{\rm{\Delta H}}_{{\rm{reaction}}}^{\rm{o}}{\rm{ = \Sigma n \times \Delta Hf}}_{{\rm{products}}}^{\rm{o}}{\rm{ - \Sigma n \times \Delta Hf}}_{{\rm{reactants}}}^{\rm{o}}\\{\rm{ = }}\left( {{\rm{3 \times \Delta Hf}}_{{{\rm{H}}_{\rm{2}}}{\rm{O(l)}}}^{\rm{o}}{\rm{\; + 2 \times \;\Delta Hf}}_{{\rm{C}}{{\rm{O}}_{\rm{2}}}{\rm{(g)}}}^{\rm{o}}} \right){\rm{ - \Delta Hf}}_{{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{OH(l)}}}^{\rm{o}}\\{\rm{ = 3}} \times \left( {{\rm{ - 286}}} \right){\rm{ + 2}} \times \left( {{\rm{ - 394}}} \right){\rm{ - }}\left( {{\rm{ - 278}}} \right)\\{\rm{ = - 1646 - 278 }}\\{\rm{ = \; - 1924 kJ/mol}}{\rm{.}}\end{array}\)