Alternative Derivations of the Mass Action Law

If despite all the efforts made in this Hyperscript you still don't like chemical potentials - here is the physicists way to deduce the mass action law without invoking chemical potentials and all that.
We start from the reaction equation with stoichiometric indices as before
Si ni · Ai   =  0
If we denote with Ni the quantity of substance Ai in mols, the free enthalpy G of the mixture contains the sum of the free enthalpies gi of the constituents and the mixing entropy Sm, we have
G  =  Si (Ni · gi)  –  T Sm
Sm is calculated in the usual way by considering the number of possibilities for mixing the substances in question, as a result one obtains
Sm   =  –  R ·  æ
ç
è
Si Ni · ln  Ni
Si Ni
ö
÷
ø
In this formulation the ln is negative because (Ni/SiNi) << 1, and we thus must assign a negative sign to the total entropy, cf. the link.
The total free enthalpy now is
G  = Si · (Ni · gi)  +  T · R æ
ç
è
Si Ni · ln   Ni
Si Ni
ö
÷
ø
We are looking for the minimum of the free enthalpy. For that we consider what happens if we change the Ni by some DNi.
This changes G by some DG expressible as a total differential DG = Si(G/Ni) · DNi. It is not much fun to go through the motions, but it is just a simple differentiation job without any problems. We obtain
DG   =  Si (DNi · gi)  +  T · R   æ
ç
è
Si ln (Ni · DNi)  +  Si (DNi)  –  Si æ
è
DNi · ln (Si Ni) ö
ø
 –  Si æ
è
Ni · Si DNi
Si Ni 
ö
ø
ö
÷
ø
Horrible, but it can be simplified (the third and fifth term actually cancel each other) and expressed via the reaction coordinate x using
DNi  =  ni · Dxi
We had that before; we obtain
DG  =  Dx ·  æ
è
Si gi · ni  +  RT · Si ni · ln Ni  –  RT · Si Ni · ln (Sini) ö
ø
For equilibrium we demand DG = 0 and this means that the expression in large brackets must be zero by itself:
0  =  æ
è
Si gi · ni  +  RT · Si ni · ln Ni  –  RT · Si Ni · ln (Si ni) ö
ø
Division with RT, putting both sides in the exponent, noticing that a sum in an exponent can be written as a product, and dropping the index i at n, g, and N because it is clear enough by now (and cannot be written properly in HTML anymore), yields
Pi N n  =  æ
è
SiN ö
ø
Sn  ·   exp – Si gi · ni
RT
which is the mass action law in its most general form.
No chemical potentials m, no standard chemical potential m0, no fugacities or activities. Everything is clear.
The catch, of course, is the entropy formula. It is only valid for classical non-interacting particles. However, if this is not the case, it is clear what need to be modified - it may not be so clear, however, how.
The other issue that takes perhaps a little thought, is the sum of the free enthalpies gi of the constituents. Since we look at the equilibrium situation, it is the free enthalpy of one mol of the substances present with respect to the prevailing condition.
 

With frame With frame as PDF

go to Vagaries in the Statistical Definition of the Entropy

go to Mass Action Law

go to Pitfalls and Extensions of the Mass Action Law

© H. Föll (Defects - Script)