0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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: $$ g^E = G^E\times N = \frac{n_1n_2}{(n_1+n_2)^2} \times (n_1 + n_2) = \frac{n_1n_2}{(n_1+n_2)} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.