Final answer:
The enthalpy of combustion of 1 mole of ethanol is -1377 kJ/mol.
Step-by-step explanation:
The balanced equation for the combustion of ethanol to CO₂(g) and H₂O(g) is:
C₂H₅OH(g) + 3O₂(g) → 2CO₂(g) + 3H₂O(g)
To calculate the enthalpy of combustion of 1 mole of ethanol, we need to use the bond energies. Since we don't have the bond energies for ethanol, we can approximate it using the average bond energies.
The average bond energies are as follows:
C-C bond: 347 kJ/mol
C-H bond: 414 kJ/mol
O-H bond: 464 kJ/mol
Using these bond energies, we can calculate the enthalpy of combustion:
(2 x C-C bond) + (6 x C-H bond) + (3 x O=O bond) - (2 x O-H bond) - (6 x C=O bond)
Substituting the values:
(2 x 347) + (6 x 414) + (3 x 498) - (2 x 464) - (6 x 736) = -1377 kJ/mol
The enthalpy of combustion of 1 mole of ethanol is -1377 kJ/mol.