Final answer:
The standard enthalpy of formation, ΔH°f , for NH3(g) at 298 K is 46 kJ/mol.
Step-by-step explanation:
The standard enthalpy of formation, ΔH°f, for NH3(g) at 298 K can be calculated using the given balanced equation and the enthalpy change of the reaction. The balanced equation is 2 NH3(g) → 3 H2(g) + N2(g). The given enthalpy change, ΔH°298, is 92 kJ/molrxn. To find the standard enthalpy of formation of NH3(g), we need to rearrange the equation so that NH3(g) is on the left side and then multiply the enthalpy change by the appropriate factor. Rearranging the equation gives us NH3(g) = (3/2) H2(g) + (1/2) N2(g). Since there are 2 moles of NH3(g) in the balanced equation, we need to divide the given enthalpy change by 2 to get the standard enthalpy of formation of NH3(g) at 298 K:
ΔH°f = (92 kJ/molrxn) / 2 = 46 kJ/mol