TITLE: Excess Gibbs Free Energy
QUESTION [0 upvotes]: I am having some confusion with something my professor has written and would like to have this issue cleared up.
I'll start from the beginning, I understand all this, I'll mention where it stops making sense. So, to determine the excess Gibbs free energy we first look at the definition of Gibbs free energy:
$$
G = \sum_{\alpha} \mu_{\alpha}N_{\alpha}
$$
All good...expanding this out:
$$
G = \sum_{\alpha} N_{\alpha}[\mu^o_{\alpha} + RT\ln(x_{\alpha} \gamma_{\alpha})]
$$
All good, let's split this up into  ideal and 'excess' non ideal bits:
$$
G = \underbrace{\sum_{\alpha} N_{\alpha}[\mu^o_{\alpha} + RT\ln(x_{\alpha})]}_{\text{Ideal}} + \underbrace{RT\sum_{\alpha} N_{\alpha}\ln( \gamma_{\alpha})}_{\text{Excess}}
$$
Ok so now there's a function for excess Gibbs free energy, writing this in partial molar form (dividing both sides by total number of mols):
$$
g^E= RT\sum_{\alpha} x_{\alpha}\ln( \gamma_{\alpha})
$$
Writing this for two components:
$$
\frac{g^E}{RT}= x_1\ln(\gamma_{1})  + x_2\ln(\gamma_{2})
$$
Ok, so next move. We're given an excess function as $Ax_1x_2$..fine, I know where that's from but then he goes and writes:
\begin{equation}
g^E = \frac{An_1n_2}{n_1 + n_2}
\end{equation}
I am convinced he has just made this up, if we look at the correct function $Ax_1x_2$, surely the excess energy will be given by:
$$
g^E = \frac{An_1n_2}{(n_1+n_2)^2}
$$
Because $x_1 = \frac{n_1}{n_1+n_2}$ and similar for $n_2$.
Could someone please advice where he got this from. Thank you.

REPLY [1 votes]: I think I figured out my issue, he doesn't know what he's doing.
The notation $g^E$ is actually used for total Gibbs excess energy, for example:
$$
g^E = RTn_1\ln(\gamma_1) + RTn_2\ln(\gamma_2)
$$
Therefore it makes sense to write something like:
$$
\frac{\partial g^E}{\partial n_i} = RT\ln(\gamma_i)
$$
Furthermore, if the PARTIAL MOLAR Gibbs free energy is given as $Ax_1x_2$ then the TOTAL MOLAR Gibbs free energy is simply:
$$
g^E = G^E\times N = \frac{n_1n_2}{(n_1+n_2)^2} \times (n_1 + n_2) = \frac{n_1n_2}{(n_1+n_2)}
$$
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