There are two possible oxidation routes for sulfur. $\text{S}(s) + \text{O}_2(g) \rightarrow \text{SO}_2(g) \quad \Delta H^\circ = -296.5\,\text{kJ mol}^{-1}$ $2\text{S}(s) + 3\text{O}_2(g) \rightarrow 2\text{SO}_3(g) \quad \Delta H^\circ = -791.4\,\text{kJ mol}^{-1}$ Sulfur trioxide is formed when sulfur dioxide reacts with oxygen. $2\text{SO}_2(g) + \text{O}_2(g) \rightarrow 2\text{SO}_3(g)$ Calculate the standard enthalpy change, $\Delta H^\circ$, for this reaction.
- A$-1384.4\,\text{kJ mol}^{-1}$
- B$-989.8\,\text{kJ mol}^{-1}$
- C$-494.9\,\text{kJ mol}^{-1}$
- D$-198.4\,\text{kJ mol}^{-1}$