The given reaction represents the formation of 2 moles of water with an enthalpy change of -483.64 kJ. For the decomposition of water, the reaction is reversed, and accordingly, the sign of the enthalpy change is also reversed: 2H2O(g) → 2H2(g) + O2(g); \Delta rH° = +483.64 kJ. Since the question asks for the decomposition of exactly one mole of water, we must divide the equation and the enthalpy change by 2: H2O(g) → H2(g) + ½O2(g). Therefore, the enthalpy change is +483.64 / 2 = +241.82 kJ.