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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.

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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.

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  • $\begingroup$ Okay, but why then do all textbook examples following the above reaction use 191.5 for dinitrogen when the reaction is at stp? $\endgroup$ – prof.kvothe Jan 14 '17 at 2:47
  • $\begingroup$ Can you give an example of how you don't get the same answer for the standard free energy change of the reaction; because it shouldn't matter if the formation bases differ, since the values for the change associated with the reaction are independent of the formation bases. $\endgroup$ – Chet Miller Jan 14 '17 at 12:26
  • $\begingroup$ Chester, I still do not understand your temperature reference. The value, 191.5 J/mol-K is listed at 298 K rather than 0 K link. From..."Table 18.1. Standard Molar Entropies of Selected Substances at 298 K." Am I interpreting this incorrectly? An example with numbers to follow. $\endgroup$ – prof.kvothe Jan 14 '17 at 14:19
  • $\begingroup$ No. I meant that the 191.5 is at 298, but it doesn't have to be consistent with the enthalpy and the free energy. Only the changes as a result of the reaction need to be consistent. Please provide the example of the inconsistency that you found for the changes. $\endgroup$ – Chet Miller Jan 14 '17 at 14:53
  • $\begingroup$ 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. $\endgroup$ – prof.kvothe Jan 14 '17 at 15:23

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