Q.Calculate the internal energy at 298K for the formation of one mole of ammonia, if the enthalpy change at constant pressure is $-42.0$ kJ $mol^{-1}$. (Given: R = 8.314 J $K^{-1} mol^{-1}$)

Enthalpy (H) is a thermodynamic property defined as H = U + PV, the internal energy of a system plus the energy equivalent of the PV work it could in principle do; being built entirely from state functions (U, P and V), H is itself a state function, so ΔH = H2 - H1 depends only on the initial and final states of the system. Because most real chemical reactions are carried out in open vessels at constant atmospheric pressure rather than at constant volume, it can be shown (starting from Qp = ΔU + Pext ΔV, and setting Pext = P for a system at mechanical equilibrium with its surroundings) that ΔH = ΔU + PΔV = Qp -- in other words, the heat exchanged by a system at constant pressure is exactly equal to its enthalpy change, which is why ΔH is the quantity actually measured and tabulated for most chemical reactions. For reactions involving only solids and liquids, the volume change ΔV is negligible, so ΔH is essentially equal to ΔU; but for reactions involving gases, ΔV cannot be ignored, and applying the ideal gas law to both the gaseous reactants (n1 mol) and gaseous products (n2 …