Answer to Question #224173 in Chemistry for AK-74

Question

calculate enthalpy for ethanonl combustion using bond enthalpy calculations

Answer 1

Solution:

Complete combustion of ethanol (C₂H₅OH):

C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(g)

To find enthalpy for ethanol combustion, use the following formula:

ΔH_rxn = ∑(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

ΔH_rxn = Bonds broken - Bonds formed = 4741 kJ/mol - 6000 kJ/mol = -1259 kJ/mol

ΔH_rxn = -1259 kJ/mol

Answer: Enthalpy for ethanol combustion (ΔH_rxn) is -1259 kJ/mol.