Bond Enthalpy Question - 6

QuestionEstimate the enthalpy change for the combustion of gaseous ethanol:C₂H₅OH(g) + 3O₂(g) → 2CO₂(g) + 3H₂O(g)Mean bond enthalpies / kJ mol^-1: C-H 413, C-C 347, C-O 358, O-H 464, O=O 498, C=O 805Step 1: count the bonds in ethanol. Draw the displayed formula of CH₃CH₂OH. It has 5 C-H, 1 C-C, 1 C-O and 1 O-H bond. The O-H bond is easy to miss.Step 2: bonds broken.
5 C-H = 2065
1 C-C = 347
1 C-O = 358
1 O-H = 464
3 O=O = 1494
Total = 4728 kJStep 3: bonds made.
4 C=O (two CO₂) = 3220
6 O-H (three H₂O) = 2784
Total = 6004 kJStep 4: subtract.ΔH = 4728 - 6004 = -1276 kJ mol^-1CommentThe data book Δ_cH^θ of ethanol is -1367 kJ mol^-1. Most of the difference is due to state: the standard value uses liquid ethanol and forms liquid water. Condensing three moles of water releases extra energy, and this outweighs the energy needed to vaporise the ethanol. The rest of the difference comes from using mean bond enthalpies rather than exact values for these molecules.