The bond dissociation enthalpy of the free radical HS is — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

Dissociation of the HS radical: $\text{HS} \to \text{H} + \text{S}$. The bond dissociation energy is $\Delta H_f(\text{H}) + \Delta H_f(\text{S}) - \Delta H_f(\text{HS}) = 218 + 277 - 138 = 495 - 138 = 357\text{ kJ/mol}$.