Verified answer

Na(s), N2(g), Fe(s), and 02(g) all possess enthalpies of formation of 0 since they are formed from their component elements and lie at their given states of matter at standard conditions. To verify this one only needs to reference a period table and see if the states of matter of the elements matches up with that given to you in the question. N2(g) and O2(g) are unique cases since they share a covalent bond; in order to be stable they naturally form a diatomic molecule at STP, so even though there is a bond, it would take more energy to break the bond to produce, per say, O(g), than it would to simply leave O2(g) as it is and yield an enthalpy of formation of 0.

O2 is the hardship-loose state of oxygen. So the enthalpy of formation of O2 is 0. that's no longer, notwithstanding, an 'absolute' 0. we are waiting to on no account diploma truthfully enthalpies (in assessment to entropy), in reality enthalpy variations (consequently the 'delta'H, somewhat than purely H), and to offer a baseline for those variations the enthalpies of formation of things of their hardship-loose states have been set at 0.