The enthalpy of hydrogenation of CH3CN (kJ/mol) is — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

Hydrogenation: CH3CN(g) + 2$H_2$(g) → CH3CH2NH2(g).<br>Bonds broken: 1 × C≡N, 2 × H-H (Total energy = 899.5 + 2 × 435.0 = 1769.5 kJ).<br>Bonds formed: 1 × C-N, 2 × N-H, 2 × C-H (Total energy = 378.0 + 2 × 426.0 + 2 × 414.0 = 2058.0 kJ).<br>ΔH = ∑BE(broken) - ∑BE(formed) = 1769.5 - 2058.0 = -288.5 kJ/mol.