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This module is registered in the
"advanced" part, because it uses the concept of the
chemical potential also developed in
the advanced part. |
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We use a rather general derivation,
but do not go too deep into the details. |
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The mass action law is usually taught
in high school chemistry, so we know what we want to find: We look at some
chemical reaction, e.g. |
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The mass action law,
as we know it, than asserts that the concentrations of the particles (=
molecules in this case) in equilibrium can be written as |
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[H2]2 · [O2]
[H2O]2 |
= |
K(T, p) |
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With K = reaction constant. |
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Or in words: The product of the concentration of
the reaction partners with all concentrations always taken to the power of their stoichiometric factors, equals
a constant K which has a numerical value that depends on the temperature
and pressure. The constant K is called
reaction constant. |
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This statement,
however, includes already a generalization
and a convention: |
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There can be any number of particles reacting or
resulting from the reaction, and we always bring the results of the reaction, (in the example the
H2O), to the right side
of the equation and assign a negative value to its
stoichiometric factors - the reaction products thus end up in the
denominator of the concentration products.
We mostly use integers for the
stoichiometric factors, but that is not de rigeur. |
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An alternative way of writing the reaction
equations that shows the "minus" sign more clearly, is |
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The mass action law is deceptively simple, it is
however not so trivial to derive it from thermodynamics including a value for the reaction constant, and it
is often quite tricky to use for real
cases! |
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We will now give a standard
derivation; an alternative way is given
in another module. |
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First we define arbitrary reactions
of any kind by the equation |
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ν1 ·
A1 + ν2
· A2 + .... + νf · Af |
= |
νg · Ag
+ νg+1 ·
Ag+1 + .... + νi · Ai |
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The Ax
denote the particles (or reactions partners involved) - atoms, ions, molecules,
vacancies, electrons, holes, .. - we want to be very general at this point. The
corresponding stoichiometric factors are the νx, and they are usually (but not always)
integers. Bringing the products of the
reaction to the left side of the equation which gives their stoichiometric factors a negative
sign, leads to the simple version |
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Chemical reactions as written down in
standard notation always inherently assume that we have exactly the right
amount of the chemicals (or, as we prefer to call it, particles) that are
needed. |
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The reaction above, for our example,
thus takes two mols of H2
(= A1) for every
mol of O2 (= A2); or in our lingo,
two H2 particles (=
molecules in this case) for one
O2 particle., yielding two H2O (=
A3) particles. |
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We have ν1 = 2, ν2 = 1, ν3 = 2. |
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Real life is
different. You mix some number of
H2 particles with some
number of O2 particles, and after the reaction you have
some number of all three particles involved
(with one number probably being very low, or ideally zero, if the most scarce
particle was completely used up in the reaction). |
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In deriving the mass action law, we have to allow
for this by allowing arbitrary starting concentrations
ci0 of the particles involved including, if we
wish, some concentration of the reaction products even before a reaction took
place - nobody keeps us from filling some water into the container with
H2 and O2 before we start the
reaction. |
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We want to get a statement about the
concentration of the particles in
equilibrium for an arbitrary mix of
concentrations at the start of the reaction in non-equilibrium; for
ease of writing we denote the equilibrium concentration of the component
i with ci; the concentration at the
start than is ci0, and an arbitrary
concentration is Ci. |
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The various
ci may be the number of mols, the absolute number of
particles, or the concentration relative to some fixed value - it doesn't
matter as long as the same definition is used throughout. |
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As pointed out above, it is
important to realize, that the ci0 can have
any initial values whatsoever - you always can throw into a closed container
whatever you want - but the dCi; the
changes in the concentrations, are tied to
each other via the reaction equation. |
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If you produce one mol of
H2O from any initial quantity of H2 and
O2; you will have reduced the H2
concentration by 1 mol and the O2 concentration by
0,5 mol - the dCi thus are not independent. |
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The whole mixture of stuff - at
whatever composition, i.e. for the whole range of the
Ci - will have some free enthalpy
G(Ci, p, T). |
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The important question is: For which
concentration values of the various particles, do we have equilibrium and thus
the minimum of G? |
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In other words: For what conditions is
dG = 0? |
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Lets write it down. With G =
G(Ci, p, T) we have for
dG |
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dG |
= |
∂G
∂C1 |
· dC1 + |
∂G
∂C2 |
· dC1 + ... + |
∂G
∂Ci |
· dCi + |
∂G
∂T |
· dT + |
∂G
∂p |
· dp |
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The (∂G/∂Ci) by definition are the
chemical potentials μi 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 |
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Now comes a
decisive step. We know that our dCi are tied somehow,
but how? |
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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"
dξ; a somewhat artificial, but useful
quantity (without a unit). |
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The changes in the concentrations of the various
particles of our system then must be proportional to dξ and the proportionality
constants are the stoichiometric indices νi. Think about it!
However you wiggle - if the concentration of O2 changes some,
the concentration of H2 will change twice as much. |
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In other words, or better yet, in math, we have
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Substituting that
into the equation for dG from above, we obtain |
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dG |
= |
i
S
1 |
μi · νi · dξ |
= |
dξ · |
i
S
1 |
μi · νi = 0 |
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Since dξ is some arbitrary number, the sum term must be zero
by itself and we have as equilibrium
condition |
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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. |
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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. |
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In order to stay within our particle scheme, we
use k instead of R and the activity Ai of the component
i instead of its concentration Ci. Feel free to
read "activity" as "somewhat corrected concentration" if you are
unfamiliar or uncomfortable with activities. We have |
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And, since we are treating equilibrium, the
activity Ai now is the equilibrium activity
ai (= concentration ci if
everything would be "ideal") instead of the arbitrary concentration
Ci because we are treating equilibrium now by
definition. |
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Inserting this formula in the equilibrium
condition from above (and omitting the index "i" at the sum
symbol for ease of writing) yields |
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Σ (νi
· μi0) +
kT · S (νi · ln ai) |
= |
0 |
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Going through the
mathematical motions now is easy. |
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Expressing the sum of
ln's as the ln of the products of the arguments, and
rearranging a bit gives |
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ln |
P |
(ai)ni |
= |
1
kT |
· Σ μi0 · νi = |
1
kT |
· ΔG0 |
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Because Σνi ·
μi0 is just the sum over
all standard reaction enthalpies
involved, which we call ΔG0. |
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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 2H2
+ O2 ↔ H2 + O
from above. Putting everything in the exponent
finally yields the mass action law: |
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P (ai)ni |
= exp |
G0
kT |
= K 1 = |
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(Reaction Constant) 1 |
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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. |
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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 ξ) - and none of this shows up in the final formula!
There are certainly some questions. |
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What's left are only equilibrium concentrations
(or activities) - what happened to the starting concentrations? |
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Can't we derive the mass action law then without
introducing quantities that seem not to be needed? |
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Some short answers: |
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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. |
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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. |
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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? |
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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. |
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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. |
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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. |
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Pitfalls and
extensions of the mass action law |
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Some
standard (chemical) examples of applying mass action law |
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Alternative
derivations of the mass action law |
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Some defects in ionic crystal related
applications of the mass action law |
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© H. Föll