
The whole mixture of stuff  at whatever composition, i.e. for the whole range
of the C_{i}  will have some free enthalpy G(C_{i}, p, T).



The important question is: For which concentration values of the various particles, do we
have equilibrium and thus the minimum of G? 


In other words: For what conditions is dG = 0? 


Lets write it down. With G = G(C_{i}, p, T)
we have for dG 


dG  = 
¶ G
¶C_{1} 
· dC_{1} + 
¶ G
¶C_{ 2} 
· dC_{1} + ... + 
¶G ¶C
_{i}  · dC_{i} + 
¶G
¶T  · dT + 
¶G
¶p  · dp 




The (¶G/¶C
_{i}) by definition are the chemical potentials
m_{i} of the particle sort x in the mixture, and the two last terms
are simply = 0 if we look at it at constant pressure and temperature. For equilibrium.this leaves us with 
 
dG  = 
i S
1 
m_{i} · dC_{i} = 0 



Now comes a decisive step. We know that our dC_{i}
are tied somehow, but how? 


To see this, we "wiggle" the system a little and react some particles,
changing the concentrations a little bit. As a measure of this change we introduce a "reaction
coordinate" dx; a somewhat artificial, but useful quantity (without a unit). 


The changes in the concentrations of the various particles of our system then must be proportional
to dx
and the proportionality constants are the stoichiometric indices
n_{i}. Think about it! However you wiggle  if the concentration
of O_{2} changes some, the concentration of H_{2} will change twice as much. 


In other words, or better yet, in math, we have 
 


Substituting that into the equation for dG from
above, we obtain 


d G  = 
i S
1 
m_{i} · n
_{i} · dx  = 
dx · 
i S
1 
m_{i} · n_{i} =
0 




Since
dx is some arbitrary number, the sum term must be zero by itself and we have as equilibrium condition 
 



This looks (hopefully) familiar. It is the equilibrium
condition we had before for particles not reacting with each other when we looked at the meaning of the chemical potential. 

Now all we have to do is to take the "master
equation" for the chemical potential so beloved by the more chemically minded, and plug it into the equilibrium
condition for our reactions. 


In order to stay within our particle scheme, we use k instead of R and the activity A_{i}
of the component i instead of its concentration C_{i}. Feel free to read "activity" as " somewhat corrected concentration"
if you are unfamiliar or uncomfortable with activities. We have 
 
m_{i} 
= m
_{i}^{0} + kT · ln A_{i} 




And, since we are treating equilibrium, the activity A_{i}
now is the equilibrium activity a_{i} (= concentration c_{i}
if everything would be "ideal") instead of the arbitrary concentration C_{ i} because
we are treating equilibrium now by definition. 


Inserting this formula in the equilibrium condition from above (and omitting the index "i"
at the sum symbol for ease of writing) yields 
 
S (n_{ i} · m_{
i}^{0}) + kT · S (n
_{i} · ln a_{i} )  =  0




Going through the mathematical motions now is easy. 


Expressing the sum of ln's as the ln of the products
of the arguments, and rearranging a bit gives 
 
ln  P 
(a_{i})^{ni} 
= –  1 kT 
· S
m_{i}^{0} · n_{i} = – 
1^{ } kT 
· DG ^{0} 




Because Sn
_{i} · m_{ i}^{0} is just the sum over all standard reaction enthalpies involved, which we call DG^{0}
. 

The product on the right hand side is just a fancy way to write
down one part of the mass action law, it would give exactly what we formulated for the case of 2H_{2} + O_{2}
Û H_{2} + O
from above. Putting everything in the exponent finally yields the mass action law: 
 



P (a_{i})^{
ni}  = exp – 
G^{ 0} kT 
= K ^{–1} =  
(Reaction Constant) ^{–1} 




It doesn't matter much, but it is standard to write K ^{–1}. In other
words, put the products of the reaction in the nominator to get K. 

There seems to be a bit of magic involved: We started with arbitrary
amounts of components, let them react an arbitrary amount (we even defined a new quantity, the
reaction coordinate
x)  and none of this shows up in the final formula! There are certainly some questions. 


What's left are only equilibrium concentrations (or activities)  what happened to the starting
concentrations? 


Can't we derive the mass action law then without introducing quantities that seem not to be
needed? 

Some short answers: 


At some point, we essentially switched to changes (= derivatives)
of prime quantities  and everything not changing is now gone. It is still there, however, if we do real
calculations because then we need more information  the mass action law, after all, is just one
equation for several unknown concentrations. 


There probably is a more direct way to get the mass action law that does not involve the somehow
superfluous reaction coordinate. However  I do not know it and I'm in good company. Several text books I consulted do not
know a better way either. Still, try the link for some alternatives. 

Lets go back to our original question and mix
arbitrary amounts of whatever and than let the buggers react. What will we get, throwing in the reaction equation and possibly
some reaction enthalpies? 


The mass action law now gives us one relation between
the equilibrium concentration, but not the absolute amounts. There are, after all, just as many unknowns for the equilibrium
concentrations as you have components, and you need more than one equation to nail everything down. 


Additionally, the way we have spelled out the mass action law here also has a number of pitfalls; if you want to really use it, you must know a bit more, in particular about
conventions that must be strictly adhered to. 

All that is essentially beyond the scope of this "Defect" lecture, but
for the hell of it, a few more modules intertwining mass action law and chemical potentials were made; they are accessible
via the following links. 


Pitfalls and extensions of the mass action law 


Some standard (chemical) examples of applying mass action
law 


Alternative derivations of the mass action law 


Some defects in ionic crystal related applications
of the mass action law 
 
