Bond Enthalpy Question - 1

QuestionUse mean bond enthalpies to estimate the enthalpy change of combustion of methane, with all substances as gases:CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)Mean bond enthalpies / kJ mol^-1: C-H 413, O=O 498, C=O 805, O-H 464Step 1: count the bonds. CH₄ has 4 C-H bonds. Each O₂ has one O=O double bond, so 2O₂ has 2. CO₂ is O=C=O, which has 2 C=O bonds. Each H₂O has 2 O-H bonds, so 2H₂O has 4.Step 2: bonds broken.
4 C-H = 4 × 413 = 1652
2 O=O = 2 × 498 = 996
Total = 2648 kJStep 3: bonds made.
2 C=O = 2 × 805 = 1610
4 O-H = 4 × 464 = 1856
Total = 3466 kJStep 4: subtract.ΔH = 2648 - 3466 = -818 kJ mol^-1CommentThe data book value of Δ_cH^θ for methane is -890 kJ mol^-1. The estimate is less exothermic mainly because the standard value forms liquid water: condensing water vapour releases extra energy that a gas-phase bond calculation does not include. Using mean rather than exact bond enthalpies adds a smaller error.The most common slip in this question is counting only one C=O bond in CO₂.