Bond Enthalpy Question - 8

QuestionThe standard enthalpy change of formation of ammonia is -46 kJ mol^-1. Use this and the bond enthalpies below to calculate the mean N-H bond enthalpy.Bond enthalpies / kJ mol^-1: N≡N 945, H-H 436Step 1: write the formation equation for one mole of NH₃.½N₂(g) + 1½H₂(g) → NH₃(g)Step 2: bonds broken.
½ N≡N = ½ × 945 = 472.5 kJ
1½ H-H = 1.5 × 436 = 654 kJ
Total = 1126.5 kJStep 3: bonds made. One NH₃ has 3 N-H bonds. Call the N-H bond enthalpy x, so the bonds made total 3x.Step 4: set up and solve.Δ_fH = Σ(broken) - Σ(made)
-46 = 1126.5 - 3x
3x = 1126.5 + 46 = 1172.5
x = +391 kJ mol^-1 (to 3 s.f.)CommentThis matches the data book mean N-H bond enthalpy of 391 kJ mol^-1. It is a mean because the three N-H bonds in ammonia need different energies to break one after another, and the calculation gives their average.Point to watchUse the fractions: the formation equation is for one mole of ammonia, so only half a mole of N≡N bonds is broken. Using N₂ + 3H₂ → 2NH₃ also works, but then ΔH must be doubled to -92 kJ mol^-1 and 6 N-H bonds are made.