Hess's Law: The total enthalpy change of a reaction is independent of the pathway — it depends only on the initial and final states. This means we can calculate ΔH using standard enthalpies of formation.
Standard enthalpy of formation (ΔfH°): Enthalpy change when 1 mole of a compound is formed from its elements in their standard states at 298 K and 1 bar pressure.
Key formula: $\Delta_r H^\circ = \sum n_p\Delta_f H^\circ(\text{products}) - \sum n_r\Delta_f H^\circ(\text{reactants})$, where $n_p$ and $n_r$ are stoichiometric coefficients.
Elements in standard state: ΔfH° = 0 for elements in their standard state. Here O₂(g) is the standard state of oxygen, so ΔfH°(O₂) = 0. This is a very common point where students make errors — always check if a reactant is an element in standard state.
Exothermic reaction: ΔH < 0 means heat is released. Combustion reactions like this (burning H₂S) are always exothermic. The negative sign means the products have lower enthalpy than reactants.
Per mole convention: ΔrH° is per mole of reaction as written. Here the reaction involves 2 mol H₂S, so the coefficients multiply each formation enthalpy.
ΔrH° = Σ(n×ΔfH°products) − Σ(n×ΔfH°reactants). Products: 2H₂O(l) + 2SO₂(g). Reactants: 2H₂S(g) + 3O₂(g). ΔfH°(O₂)=0 always.
Products side: 2×(−286)+2×(−297)=−572+(−594)=−1166 kJ. Reactants side: 2×(−20.1)+3×0=−40.2 kJ. ΔrH°=−1166−(−40.2)=−1125.8 kJ. Magnitude=1126 kJ.
Standard enthalpy of formation is defined for compounds, not elements. O₂ is already in its standard elemental state. All elements in standard state have ΔfH°=0 by definition.
Burning H₂S releases 1126 kJ per 2 moles. This is a highly exothermic process. H₂S is found in natural gas and volcanic emissions; controlled combustion converts it to SO₂ and H₂O in the Claus process for sulfur recovery.