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

💡 Solution & Explanation

Reaction: $H_2S$(g) → HS(g) + H(g); ΔH° = +376.0 kJ/mol. Since ΔH° = ΔfH°(HS) + ΔfH°(H) - ΔfH°($H_2S$), and ΔfH°(H) = 1/2 × BE(H-H) = 218 kJ/mol. Thus, 376.0 = ΔfH°(HS) + 218 - (-20) → ΔfH°(HS) = 376.0 - 238 = 138 kJ/mol.