Why is N2 enthalpy zero, entropy 191.5, yet G is listed as zero and not [-T*0.1915]? In standard tables, $\ce{N2}$ at STP has an enthalpy formation ($\Delta H_\mathrm f$) listed as $0\ \mathrm{kJ/mol}$, entropy ($S$) as $0.1915\ \mathrm{kJ/K}$, and free energy of formation ($\Delta G_\mathrm f$) as $0\ \mathrm{kJ/mol}$.

If we use the std eq.: $\Delta G = \Delta H-T\cdot\Delta S$, I don't get zero with those numbers, obviously. (assuming STP and all that) But, $\Delta G$ is zero for nitrogen?

My initial thought is that $0.1915\ \mathrm{kJ/K}$ is not $\Delta S$ as required by the equation, but just $S$, so I need to figure out $\Delta S$ for $\ce{N2}$ and use that instead. Okay, that would make sense, but it doesn't follow for this example, which I see everywhere:

$$\ce{N2 + 3H2 -&gt; 2NH3}$$

$\Delta H$ is calculated for the reaction using the appropriate $\Delta H_\mathrm f$

$\Delta S$ is calculated using the appropriate $S$ values (So $0.1915\ \mathrm{kJ/K}$ for $\ce{N2}$, not $0$)

Then they use these reaction $H$ and $S$ to get $\Delta G$

However, if I just use the $\Delta G$ values found in databases like the CRC Handbook (so $0$ for $\ce{N2}$), I don't get the same answer using $\Delta G_\mathrm r=\Delta G_\text{products}-\Delta G_\text{reactants}$

This is driving me nuts, I know I must be missing something simple here.
Enthalpies of formation and Free energies of formation are expressed relative to the standard state of 25 C and 1 bar.  So, in the standard state, they are taken to be zero.  The entropy value you cited is the absolute entropy, relative to absolute zero.  If you wanted the entropy of formation of nitrogen relative to the standard state of 25C and 1 bar, that would be 0.