Calculate the standard enthalpy of combustion for Ethanol (C₂H₅OH) using the following standard enthalpies of formation (ΔH°f):
C₂H₅OH(l) = -277 kJ/mol; CO₂(g) = -394 kJ/mol; H₂O(l) = -286 kJ/mol.
- -683 kJ/mol
- -1235 kJ/mol
- -1367 kJ/mol
- -1921 kJ/mol
📝 Detailed explanation
Reaction: C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O.ΔH° = [2(-394) + 3(-286)] - [-277 + 0] = [-788 - 858] + 277 = -1646 + 277 = -1369 kJ/mol (closest to -1367).