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

💡 Solution & Explanation

Reaction: $\text{CH}_3\text{CN} + 2\text{H}_2 \to \text{CH}_3\text{CH}_2\text{NH}_2$. Utilizing bond enthalpies, the process breaks 1 $\text{C}\equiv\text{N}$ ($899.5$) and 2 H-H ($2 \times 435 = 870$) to inherently form 1 C-N ($378$), 2 C-H ($2 \times 414 = 828$), and 2 N-H ($2 \times 426 = 852$). The unchanged $\text{CH}_3$-C bonds cancel. $\Delta H = \text{Bonds Broken} - \text{Bonds Formed} = (899.5 + 870) - (378 + 828 + 852) = 1769.5 - 2058 = -288.5\text{ kJ/mol}$.