Determine the standard enthalpy of formation of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) from its standard enthalpy of combustion \((-1367.4 \mathrm{~kJ} / \mathrm{mol})\)

We start by writing the chemical equation for the formation of ethanol from its elements: \[ 2\, C (s) + 6\, H (g) + O (g) \to C_2H_5OH (g) \] Here carbon is in solid form, hydrogen is in gaseous form, and oxygen is in gaseous form as they are in their standard states.

Next, we write the equation for the complete combustion of ethanol: \[ C_2H_5OH (g) + 3\, O_2 (g) \to 2\, CO_2 (g) + 3\, H_2O (g) \] This equation describes how a mole of ethanol (C2H5OH) reacts with oxygen to produce carbon dioxide and water.

We now use Hess's Law, which states that the total enthalpy change for a chemical reaction does not depend on the route by which the reaction is carried out. The combustion reaction can be viewed as the sum of the reverse formation reaction for ethanol and the formation reactions for the combustion products, carbon dioxide and water. So, we rearrange and add the equations as follows: \[ -[ 2\, C (s) + 6\, H (g) + O (g) \to C_2H_5OH (g) ] \] (We've reversed the first equation, indicating that we're considering the decomposition of ethanol into its elements rather than its formation.) \[ 2\, [ C (s) + O_2 (g) \to CO_2 (g) ] \] (The formation of two moles of carbon dioxide from carbon and oxygen.) \[ 3\, [ 2\, H_2 (g) + O_2 (g) \to 2\, H_2O (g) ] \] (The formation of three moles of water from hydrogen and oxygen.) Adding these equations results in the combustion reaction: \[ C_2H_5OH (g) + 3\, O_2 (g) \to 2\, CO_2 (g) + 3\, H_2O (g) \]

Lastly, we use the standard enthalpies of formation for carbon dioxide (\(-393.5 \, kJ/mol\)) and water (\(-285.8 \, kJ/mol\)) to calculate the standard enthalpy of formation for ethanol. The standard enthalpy change for the combustion reaction equals the sum of the standard enthalpies of formation of the products minus the sum of the standard enthalpies of formation of the reactants. Setting this equal to the given standard enthalpy of combustion for ethanol, we have: \[ -1367.4 \, kJ = [(2)(-393.5\, kJ) + (3)(-285.8\, kJ)] - [1\, \Delta H_f (C_2H_5OH)] \] Solving for \(\Delta H_f (C_2H_5OH)\), we find: \[ \Delta H_f (C_2H_5OH) = (2)(-393.5\, kJ) + (3)(-285.8\, kJ) + 1367.4 \, kJ = -277.6 \, kJ/mol \]