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

💡 Solution & Explanation

To form the HS radical: $\text{H}_2\text{S} \to \text{HS} + \text{H}$. $\Delta H = 376\text{ kJ/mol}$. Also, $\Delta H_f(\text{H}_2\text{S}) = -20\text{ kJ/mol}$ and $\Delta H_f(\text{H}) = 436 / 2 = 218\text{ kJ/mol}$. Thus, $\Delta H_f(\text{HS}) = \Delta H_{\text{reaction}} + \Delta H_f(\text{H}_2\text{S}) - \Delta H_f(\text{H}) = 376 + (-20) - 218 = 138\text{ kJ/mol}$.