Calculate the enthalpy of the following homogeneous gaseous reaction:
CH₃COCH₃ + 2O₂ → CH₃COOH + CO₂ + H₂O
from the following data:
Bond energies (kJ/mol): C-H = 414; C-C = 348; C=O = 580; C-O = 354; O=O = 610; O-H = 462;
Magnitude of resonance energies (kJ/mol): COOH = 118; CO₂ = 140.

Question

Calculate the enthalpy of the following homogeneous gaseous reaction:

CH₃COCH₃ + 2O₂ → CH₃COOH + CO₂ + H₂O

from the following data:

Bond energies (kJ/mol): C-H = 414; C-C = 348; C=O = 580; C-O = 354; O=O = 610; O-H = 462;

Magnitude of resonance energies (kJ/mol): COOH = 118; CO₂ = 140.

TestHub · Accepted Answer

Bonds broken (acetone: 6 C–H, 2 C–C, 1 C=O; plus 2 O=O): 6(414) + 2(348) + 580 + 2(610) = 4980 kJ.

Bonds formed (CH_₃COOH: 3 C–H, 1 C–C, 1 C=O, 1 C–O, 1 O–H; CO_₂: 2 C=O; H_₂O: 2 O–H): 3(414) + 348 + 580 + 354 + 462 + 2(580) + 2(462) = 5070 kJ.

The enthalpy change is calculated as:

Δ H = Bonds broken - Bonds formed = 4980 - 5070 = -90 kJ

Resonance stabilizes the products by 118 + 140 = 258 kJ. Therefore, the final enthalpy change is:

Δ H = -90 - 258 = -348 kJ

This corresponds to option (d).

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