# Why entropy is not zero while free energy of formation of element in its standard form is zero?

Jul 31, 2017

I'm assuming you're asking, why is ${S}^{\circ} \ne 0$ at ${25}^{\circ} \text{C}$ and $\text{1 atm}$ for an element in its elemental state, but $\Delta {G}_{f}^{\circ}$ at ${25}^{\circ} \text{C}$ and $\text{1 atm}$ for an element in its elemental state is $0$.

STANDARD MOLAR ENTROPY

${S}^{\circ}$ is defined to be the absolute molar entropy, setting the reference state to S("0 K") = "0 J/mol"cdot"K". It has nothing to do with formation reactions. Here is how one would calculate this for ${\text{N}}_{2} \left(g\right)$.

Any substance at room temperature will have some form of energy dispersal, so its absolute molar entropy cannot be zero. That is, anything at room temperature will have some form of motion, whether it is internal or external.

For example (Physical Chemistry, Levine, Appendix), ul(S_("Cl"_2(g))^@ = "223.066 J/mol"cdot"K").

CHANGE IN STANDARD MOLAR GIBBS' FREE ENERGY

On the other hand, $\Delta {G}_{f}^{\circ}$ for an element "X"("phase") in its elemental state & phase must be $\text{0 kJ/mol}$, because the formation reaction this involves is:

overbrace("X"("phase"))^("Elemental state") -> overbrace("X"("phase"))^"Product formed"

Thus, nothing happens, and the free energy of formation is trivially zero (there is exact "conversion"). Same with the enthalpy of formation, because there is zero energy input/output resulting from the "conversion".

SIDENOTE

As a sidenote, this is only true for elements in their ELEMENTAL state. So, if one had to form ${\text{Cl}}^{-} \left(a q\right)$, let's say... the formation reaction would be:

$\frac{1}{2} {\text{Cl"_2(g) + e^(-) -> "Cl}}^{-} \left(a q\right)$

This would have (Physical Chemistry, Levine, Appendix)

$\Delta {G}_{f}^{\circ} = - \text{131.228 kJ/mol}$,
${S}^{\circ} = \text{56.5 J/mol"cdot"K}$,

neither of which are zero.

Thus, ${\text{Cl}}^{-} \left(a q\right)$ is NOT the elemental state of ${\text{Cl}}_{2} \left(g\right)$ (which is obvious).