Bond Enthalpy Question - 3

QuestionEstimate the enthalpy change for the Haber process reaction and compare it with the data book value.N₂(g) + 3H₂(g) → 2NH₃(g)Bond enthalpies / kJ mol^-1: N≡N 945, H-H 436, N-H 391Step 1: bonds broken.
N₂ has one N≡N triple bond. 3H₂ has 3 H-H bonds.
945 + 3(436) = 945 + 1308 = 2253 kJStep 2: bonds made.
Each NH₃ has 3 N-H bonds, so 2NH₃ has 6.
6(391) = 2346 kJStep 3: subtract.ΔH = 2253 - 2346 = -93 kJ mol^-1This value is for the equation as written, which forms 2 mol of NH₃. Per mole of ammonia it is -93 ÷ 2 = -46.5 kJ mol^-1.CommentThe data book value for the equation is -92 kJ mol^-1, very close to the estimate. Agreement is good because every substance is a gas, and the only mean value used is N-H. The N≡N and H-H values are exact, since those bonds occur only in N₂ and H₂.The very large N≡N bond enthalpy also explains why nitrogen is so unreactive: a lot of energy is needed to break the triple bond, which is one reason the Haber process uses an iron catalyst.