
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 


dG 
= 
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 


