Bond Enthalpy Question - 7

QuestionUse bond enthalpies to estimate the standard enthalpy change of formation of water vapour, and compare it with the data book value of -241.8 kJ mol^-1.Bond enthalpies / kJ mol^-1: H-H 436, O=O 498, O-H 464Step 1: write the formation equation. An enthalpy change of formation is for one mole of product, so:H₂(g) + ½O₂(g) → H₂O(g)Half a mole of O₂ means half a mole of O=O bonds is broken.Step 2: bonds broken.
1 H-H + ½ O=O = 436 + 249 = 685 kJStep 3: bonds made.
2 O-H = 2 × 464 = 928 kJStep 4: subtract.Δ_fH^θ ≈ 685 - 928 = -243 kJ mol^-1CommentThe estimate is only 1.2 kJ mol^-1 more exothermic than the data book value. Agreement is very good because everything is gaseous and the H-H and O=O values are exact for these molecules. The only mean value is O-H, which is averaged over many compounds.If the question asked for liquid water, H₂O(l), a bond enthalpy estimate would be much less accurate, because condensing water vapour releases energy that bond enthalpies do not include. The data book value for liquid water is -285.8 kJ mol^-1.A common error is to use the whole equation 2H₂ + O₂ → 2H₂O, which gives -486 kJ, the value for two moles.