Practice question
Question
Calculate the bond enthalpy of C-C in C₂H₆(g) given: Δ Hf° (C₂H₆,g) = -84.7 kJ/mol, Δ Ha (C,g) = 715 kJ/mol, Δ Ha (H,g) = 218 kJ/mol, C-H = 413 kJ/mol.
Explanation
For C₂H₆(g) → 2C(g) + 6H(g), Δ H = 2 × 715 + 6 × 218 - (-84.7) = 1430 + 1308 + 84.7 = 2822.7 kJ/mol. Bonds: 1(C-C) + 6(C-H), C-C = 2822.7 - 6 × 413 = 2822.7 - 2478 = 344.7 kJ/mol.
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