The bond enthalpy of C-C bond (kJ/mol) is — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

For propane $C_3H_8$: 3C(graphite) + 4$H_2$(g) → $C_3H_8$(g); ΔfH° = -85.0 kJ/mol.<br>Using Born-Haber energy balance: ΔfH° = 3 × ΔsubH(C) + 4 × ΔdissH($H_2$) - [2 × BE(C-C) + 8 × BE(C-H)].<br>-85.0 = 3 × 719.0 + 4 × 435.0 - [2 × BE(C-C) + 8 × 414.0].<br>-85.0 = 2157.0 + 1740.0 - 2 × BE(C-C) - 3312.0.<br>-85.0 = 585.0 - 2 × BE(C-C) → 2 × BE(C-C) = 670.0 → BE(C-C) = 335.0 kJ/mol.