Final answer:
The enthalpy of combustion of 1 mole of ethanol, based on the given enthalpies of formation, is calculated to be -1368 kJ.
Step-by-step explanation:
Calculating the Enthalpy of Combustion of Ethanol
The enthalpy of combustion of a compound is the heat change when one mole of the substance burns completely in oxygen. To determine the enthalpy of combustion of ethanol (C₂H₅OH), we can use the enthalpies of formation of the reactants and products involved in the reaction.
The balanced chemical equation for the combustion of ethanol is:
- C₂H₅OH(l) + 3 O₂(g) → 2 CO₂(g) + 3 H₂O(l)
Using the provided enthalpies of formation:
- For ethanol (C₂H₅OH(l)): -278 kJ/mol
- For water (H₂O(l)): -286 kJ/mol
- For carbon dioxide (CO₂(g)): -394 kJ/mol
We calculate the enthalpy of combustion using the equation:
- ΔH°₁combustion = [∑ΔH°₁f(products)] - [∑ΔH°₁f(reactants)]
- ΔH°₁combustion = [(2 mol x -394 kJ/mol) + (3 mol x -286 kJ/mol)] - [1 mol x -278 kJ/mol]
- ΔH°₁combustion = [-788 kJ + (-858 kJ)] - [-278 kJ]
- ΔH°₁combustion = -1646 kJ + 278 kJ = -1368 kJ/mol
So, the heat of combustion of 1 mole of ethanol is -1368 kJ.