Reflecting on the Enthalpy of Formation and Combustion of C₂H₅OH

What is the calculation for the enthalpy of combustion of ethanol, C₂H₅OH?

Based on the enthalpies of formation provided, what is the enthalpy of combustion for C₂H₅OH?

Answer:

The heat of combustion of 1 mole of ethanol, C₂H₅OH(l), is calculated by subtracting the sum of the enthalpies of formation of the reactants from the products, resulting in -1368 kJ/mol.

Enthalpy of formation for ethanol, C₂H₅OH(l), is -278 kJ/mol. The enthalpy of formation for water, H₂O(l), is -286 kJ/mol, and for carbon dioxide, CO₂(g), is -394 kJ/mol. Oxygen, O₂(g), in its standard state, has an enthalpy of formation of 0 kJ/mol.

To calculate the enthalpy of combustion using Hess's law, we subtract the sum of the enthalpies of formation of the reactants from the products. Plugging in the values, we get ΔHₓₐᵃ₎ = [(2 × -394) + (3 × -286)] - [-278 + (3 × 0)] = -1368 kJ/mol.

Therefore, the heat of combustion of 1 mole of ethanol is -1368 kJ/mol, reflecting the energy released in the combustion reaction.