Question

What is `DeltaH_f^@` for `NO(g)`?

Answer 1

For `"NO"`, `DeltaH_f""^@=+91.3*kJ*mol^-1`; and thus here `DeltaH_f""^@-=DeltaH_"rxn"^@`

Explanation:

To account for the positive value, consider what `DeltaH_f""^@` represents:

" the enthalpy associated with the formation of 1 mole of substance from its constituent elements in their standard states under given conditions....... "

So here, `DeltaH_f""^@` is the enthalpy associated with the following reaction:

`1/2N_2(g) + 1/2O_2(g) rarr NO(g)` `;DeltaH""^@""_"rxn"=91.3*kJ*mol^-1`

In order to make the `N=O` bond, we have to break exceptionally strong `N-=N` and `O=O` bonds. The energy shortfall is anticipated.