# Is the stoichiometric coefficient equivalent to the mole fraction?

I have a small problem to understand (or distinguish between) two concepts of free mixing enthalpy.

In several textbooks, I found the following formula for the molar free mixing enthalpy (binary system):

$\Delta_{mix}G_m = RT (x_A \ln{a_A} + x_B \ln{a_B}) \tag{1}$

When it comes to free reaction enthalpies, the given formula is:

$\Delta_{r}G_m = \Delta_{r}G{^\circ}_m + RT \sum \nu_i \ln{a_i} \tag{2}$

Where $RT \sum \nu_i \ln{a_i}$ represents the influence of mixing of the products and reagents. As far as I understand this, the second term of Eqn. (2) , e.i. $RT \sum \nu_i \ln{a_i}$, should be equal to $\Delta_{mix}G_m$ (Eqn. (1)).

But this would only be true, if $x_i = \nu_i$. My problem is to verify this.

If one considers Eqn. (3):

$d\xi = \frac{dn_i}{\nu_i} \tag{3}$

With $x_i = \frac{n_i}{\sum n_i}$ and, therefore $dn_i = dx_i\sum n_i$, follows that

$\nu_i d\xi = n dx_i \tag{4}$

with $n = \sum n_i$. To solve Eqn. (4) an integration is necessary, but I am not sure, how to set the integration boundaries. My guess is:

Since $\xi$ goes from 0 to 1, what means at 0 no substances were converted and at 1 all (= n) substances have transformed, the boundaries would be as follows:

$\nu_i \int_{0}^{n} d\xi = n \int_{0}^{x_i}dx_i \tag{5}$

In fact, this would lead to $\nu_i = x_i$. Unfortunately, I think this might be a fallacy because I might have chosen the boundaries to fulfill my expectation ^^

What is the correct way to interpret this?

$\nu_i$ and $x_i$ are not equal (in general). This can simply be verified by understanding that in general the $\nu_i$ can be negative (for educts) and greater than one which are both properties $x_i$ does not have since it's values are by definition confined to the range from zero to one.
• The difference comes from the very definition of these two enthalpies. While $\Delta_{mix}G_m = G_{after} - G_{before}$, the reaction enthalpy is defined as $\Delta_rG_m = \left(\frac{\partial G}{\partial \xi}\right)_{p,T}$. As you can see one is a difference whilst the other is a differentiation. Therefore it makes sense that the corresponding expressions differ. If you happen to understand German, I could give you a link to a script dealing with this in more detail. – Raven Aug 4 '18 at 9:24
• Based on your commend, I came over another possibility which might solve the issue: Since $\Delta_rG_m = \left(\frac{\partial G}{\partial \xi}\right)_{p,T}$ the following should be valid: $\int_{before}^{after} dG = RT\int_{before}^{after}\ln{a_i}dn_i$, with $d\xi=\frac{dn_i}{\nu_i}$. The problem is now, that I don't know $a_i(n_i)$ and therefore I cannot integrate the right side. If one could consider $\ln{a_i}$ to not depend on $n_i$, it would lead to the equality of the both mixing terms, but this is not true, as far as I know. – user65662 Aug 4 '18 at 10:32