Final answer:
The question pertains to calculating the enthalpy of combustion for ethanol using its chemical equation and provided enthalpies of formation. The heat of combustion for one mole of ethanol is calculated and then extended to find the enthalpy for 1 L of ethanol based on its density.
Step-by-step explanation:
The question involves calculating the heat of combustion of ethanol, C₂H₅OH(I), when water (H₂O(l)) and carbon dioxide (CO₂(g)) are formed. This calculation requires the use of enthalpies of formation. The balanced chemical equation for the combustion of ethanol is: C₂H₅OH(I) + 3O₂(g) → 2CO₂(g) + 3H₂O(l). According to the enthalpies of formation provided (−278 kJ/mol for ethanol, −286 kJ/mol for water, and −394 kJ/mol for carbon dioxide), the heat of combustion can be determined using the following formula: ∆H = [Σ∆Hf(products)] − [Σ∆Hf(reactants)]. For one mole of ethanol, this results in: ∆H = [(2 × −394) + (3 × −286)] − [− 278] = −1366 kJ/mol.
To find the enthalpy of combustion for 1 L of ethanol, the density of ethanol (0.7893 g/mL) is used to convert the volume to mass, and then the mass to mol using the molar mass of ethanol. Finally, the molar heat of combustion is multiplied by the number of moles to get the total enthalpy released when 1 L of ethanol is combusted.