Answer to Question #160061 in Chemistry for Arvin Shopon

Question

Calculate the enthalpy of the methanol combustion reaction using the binding energies.

2CH₃OH (I) + 3O₂ (g) → 2CO₂ (g) + 4H₂O (I)

Answer 1

Solution:

2CH₃OH(I) + 3O₂(g) → 2CO₂(g) + 4H₂O(I)

To find enthalpy of the methanol combustion reaction, use the following formula:

ΔH_rxn = ∑(bonds broken) − ∑(bonds formed)

In this reaction, 6 C-H bonds, 2 C-O bonds, 2 O-H bonds and 3 O=O bonds must be broken.

Also, 4 C=O bonds and 8 O-H bonds are formed.

Bond energies of these bonds:

• C-H bond: 413 kJ/mol
• C-O bond: 358 kJ/mol
• O-H bond: 463 kJ/mol
• O=O bond: 495 kJ/mol
• C=O bond in CO₂: 799 kJ/mol

Bonds broken = 6(C-H) + 2(C-O) + 2(O-H) + 3(O=O)

Bonds broken = 6×(413 kJ/mol) + 2×(358 kJ/mol) + 2×(463 kJ/mol) + 3×(495 kJ/mol) = 5605 kJ/mol

Bonds broken = 5605 kJ/mol

Bonds formed = 4(C=O) + 8(O-H)

Bonds formed = 4×(799 kJ/mol) + 8×(463 kJ/mol) = 6900 kJ/mol

Bonds formed = 6900 kJ/mol

ΔH_rxn = Bonds broken - Bonds formed = 5605 kJ/mol - 6900 kJ/mol = -1295 kJ/mol

ΔH_rxn = -1295 kJ/mol

Answer: The enthalpy of the methanol combustion reaction (ΔH_rxn) is -1295 kJ/mol.