Final answer:
Using Hess's Law and the given enthalpy change for the overall reaction, it is determined that the standard molar enthalpy of formation for NO(g) is +90 kJ/mol, since the entire enthalpy of the reaction can be attributed to the formation of NO(g) in the absence of any contribution from N2(g), which is zero for elements in their standard state.
Step-by-step explanation:
To determine the standard molar enthalpy of formation (ΔH°f) of NO(g), we can use Hess's Law, which states that the total enthalpy change for a chemical reaction is the same, regardless of the pathway taken. The given reaction is 2H2(g) + 2NO(g) → 2H2O(g) + N2(g) with ΔH = -664 kJ/mol. The enthalpy of formation for H2O(g) is given as -242 kJ/mol, so the formation of 2 moles of H2O(g) would contribute -2 * 242 kJ/mol = -484 kJ/mol to the reaction enthalpy.
Therefore, the formation of 2 moles of NO(g) and 1 mole of N2(g) must account for the difference between the total reaction enthalpy and the contribution from water. The difference is: -664 kJ/mol (total) - (-484 kJ/mol for water) = -180 kJ/mol. Because we are dealing with 2 moles of NO(g), the enthalpy of formation of 1 mole of NO(g) is half of the difference, which is -180 kJ / 2 = -90 kJ/mol. However, the standard enthalpy of formation of elements in their standard state (like N2(g)) is zero, so all of the -180 kJ/mol is due to NO(g). Since the question asks for the heat of formation and does not specify the sign, the correct answer is 90 kJ/mol.