
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 





If we denote with N_{i} the
quantity of substance A_{i} in mols, the free enthalpy G of the
mixture contains the sum of the free enthalpies g_{i} of
the constituents and the mixing entropy S_{m}, we have 


G = S_{i}
(N_{i} · g_{i}) – T
S_{m} 



S_{m} is calculated
in the usual way by
considering the number of possibilities for mixing the substances in question,
as a result one obtains 


S_{m}^{ } 
= – R · 
æ
ç
è 
S_{i} N_{i} ·
ln 
N_{i}
S_{i} N_{i} 
ö
÷
ø 




In this formulation the ln is negative
because (N_{i}/S_{i}N_{i}) << 1, and we
thus must assign a negative sign to the total entropy, cf. the
link. 


The total free enthalpy now is 


G^{ } = S_{i} · (N_{i} ·
g_{i}) + T · R 
æ
ç
è 
S_{i} N_{i} · ln

N_{i}
S_{i} N_{i} 
ö
÷
ø 



We are looking for the minimum of the
free enthalpy. For that we consider what happens if we change the
N_{i} by some DN_{i}. 


This changes G by some
DG expressible as a total differential
DG = S_{i}(¶G/¶N_{i}) · DN_{i}. It is not much fun to go
through the motions, but it is just a simple differentiation job without any
problems. We obtain 
DG_{ } 
= 
S_{i} (DN_{i} · g_{i}) +
T · R 
æ
ç
è 
S_{i} ln
(N_{i} · DN_{i})
+ S_{i} (DN_{i}) – S_{i} 
æ
è 
DN_{i} · ln
(S_{i} N_{i}) 
ö
ø 
– S_{i} 
æ
è 
N_{i} · 
S_{i}
DN_{i}
S_{i} N_{i} 
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 





We had
that before; we obtain 


DG 
= 
Dx
· 
æ
è 
S_{i}
g_{i} · n_{i}
+ RT · S_{i}
n_{i} · ln N_{i}
– RT · S_{i}
N_{i} · ln (S_{i}n_{i}) 
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 
S_{i}
g_{i} · n_{i}
+ RT · S_{i}
n_{i} · ln N_{i}
– RT · S_{i}
N_{i} · ln (S_{i}
n_{i}) 
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



P_{i} N
^{n} 
= 
æ
è 
S_{i}N 
ö
ø 
Sn 
· 
exp – 
S_{i}
g_{i} · n_{i}
RT 




which is the mass action law in its
most general form. 

No chemical potentials m, no standard chemical potential m_{0}, no fugacities or activities. Everything is clear. 


The catch, of course, is the entropy formula. It
is only valid for classical noninteracting
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 g_{i} 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. 


