Practice question
Question
Calculate the bond enthalpy of N≡N in N₂(g) given: Δ Hf° (NH₃,g) = -46.10 kJ/mol , N-H = 391.0 kJ/mol , Δ Ha (H,g) = 218.0 kJ/mol .
Explanation
For N₂(g) + 3H₂(g) → 2NH₃(g) , Δ H = 2 × (-46.10) = -92.20 kJ . Bonds broken: N≡N + 3 × H-H = N≡N + 3 × 436.0 = N≡N + 1308.0 kJ . Bonds formed: 6 × N-H = 6 × 391.0 = 2346.0 kJ . Using Δ H = (bonds broken) - (bonds formed) , -92.20 = (N≡N + 1308.0) - 2346.0 , N≡N = 945.8 kJ/mol .
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