For complete combustion of ethanol, the amount of heat produced as measured in a bomb calorimeter is 1364.47 kJ/mol at 25°C. C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l) Assuming ideality, the enthalpy of combustion, ΔcH, for the reaction will be: [R = 8.314 J/mol]

C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l), ΔE = -1364.47 kJ/mol
ΔH = ?
T = 298 K
Δng = -1 (change in the number of gaseous moles)
R = 8.314 J/mol = 8.314 × 10⁻³ kJ/mol
ΔH = ΔE + PΔV
but, PΔV = ΔngRT
∴ ΔH = ΔE + ΔngRT
ΔH = -1364.47 + ((-1) × 8.314 × 10⁻³ × 298)
ΔH = -1366.95 kJ/mol
Hence, the correct option is -0.95 kJ/mol