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

💡 Solution & Explanation

For acetonitrile CH3CN: 2C(graphite) + 1.5 $H_2$(g) + 0.5 $N_2$(g) → CH3CN(g); ΔfH° = +88.0 kJ/mol.<br>ΔfH° = 2 × 719.0 + 1.5 × 435.0 + 0.5 × 948.0 - [3 × BE(C-H) + BE(C-C) + BE(C≡N)].<br>88.0 = 1438.0 + 652.5 + 474.0 - [3 × 414.0 + 335.0 + BE(C≡N)].<br>88.0 = 2564.5 - [1577.0 + BE(C≡N)] → BE(C≡N) = 2564.5 - 1577.0 - 88.0 = 899.5 kJ/mol.