# 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:

$$\ce{N2 + 3H2 -> 2NH3}$$

$\Delta H$ 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.

• Got it, thanks! Using $\ce{\Delta H}$ and $\ce{\Delta S}$ values to get $\ce{\Delta G}$, and then G values calculated from $\ce{\Delta H}$ and S to get $\ce{\Delta G}$ I was able to get -32.7 kJ for both methods. So, when determining $\ce{\Delta G}$, it is okay to use G values obtained from S and not $\ce{\Delta S}$, because the change is the same? This is what confused me...$\ce{G}$ for N2 was not 0, but rather -116.8. I see now that it is because S was used and not $\ce{\Delta S}$, but seems to be okay because the change is the same as if we were using calculated $\ce{\Delta S}$ values. Jan 14, 2017 at 15:23