Complete combustion of ethanol (C2H5OH):
C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g)
To find enthalpy for ethanol combustion, use the following formula:
ΔHrxn = ∑(bonds broken) − ∑(bonds formed)
In this reaction, 5 C-H bonds, 1 C-C bond, 1 C-O bond, 1 O-H bond and 3 O=O bonds must be broken.
Also, 4 C=O bonds and 6 O-H bonds are formed.
Bond energies of these bonds:
C-H bond: 415 kJ/mol
C-C bond: 350 kJ/mol
C-O bond: 358 kJ/mol
O-H bond: 464 kJ/mol
O=O bond: 498 kJ/mol
C=O bond in CO2: 804 kJ/mol
Bonds broken = 5(C-H) + 1(C-C) + 1(C-O) + 1(O-H) + 3(O=O)
Bonds broken = 5×(415) + (350) + (358) + (464) + 3×(498) = 4741 (kJ/mol)
Bonds broken = 4741 kJ/mol
Bonds formed = 4(C=O) + 6(O-H)
Bonds formed = 4×(804) + 6×(464) = 6000 (kJ/mol)
Bonds formed = 6000 kJ/mol
ΔHrxn = Bonds broken - Bonds formed = 4741 kJ/mol - 6000 kJ/mol = -1259 kJ/mol
ΔHrxn = -1259 kJ/mol
Answer: Enthalpy for ethanol combustion (ΔHrxn) is -1259 kJ/mol.
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