Given: S(s) + O2(g) → SO2(g), ΔH = -296.8 kJ, and SO2(g) + 1/2 O2(g) → SO3(g), ΔH = -98.9 kJ. What is ΔH for S(s) + 3/2 O2(g) → SO3(g)?

Core Concept

Adding the two equations cancels SO2 and leaves sulfur plus oxygen on the reactant side with SO3 as the sole product, exactly the target equation. Because enthalpy is a state function, the enthalpy changes of summed steps simply add. ΔH = (-296.8) + (-98.9) = -395.7 kJ. No reversal or scaling is needed here, only direct addition of the two given values.