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Lets look at ammonia synthesis, a major chemical
breakthrough at the beginning of the 20 th century, as a pretty simple chemical reaction between
gases (Remember : In the chemical formalism invoking the mass action
law, point defects behave like (ideal) gases). |
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The reaction equation that naturally comes to mind is |
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and the mass action law tells us that
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[N 2] · [H2]3
[NH3]2 | = |
K1 |
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With K1 = reaction constant for this process. We used the square
brackets [..] as the notation for concentrations, but lets keep in mind that the mass action law in full generality
is formulated for
activities or fugacities! |
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However, we also could look at the dissociation of ammonia
- equilibrium entails that some ammonia is formed, some decays; the "Û"
sign symbolizes that the reaction can go both ways. So lets write |
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The mass action law than gives |
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[NH3 ]2
[N2] · [H2]3
| = | K2 = |
1
K1 |
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To make things worse, we could write the two equations also like |
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[N2]1/2 · [H2]3/2
[NH3] | = K 3 = |
(K1)1/2 |
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and nobody keeps us from using the reaction as a source for
hydrogen via |
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[NH3]2/3 · [N2]1/3
[H2] | = |
K4 = ? |
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And so on. Now what does it mean ? What exactly
does the mass action law tell us? There are two distinct points in the examples which are important to realize: |
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1. Only the mass action law together
with the reaction equation and the convention of what we have in the nominator
and denominator of the sum of products makes any sense. A reaction constant given as some number (or function of p
and T) by itself is meaningless. |
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2. The standard chemical potentials m
i0 that are contained in the reaction constant (via
Si m
i0) where defined for reacting one standard unit, usually
1 mol. The reaction constant in the mass action law thus is the reaction constant for producing 1 unit, i.e.
one mol and thus applies, loosely speaking, to the component with the stoichiometry index 1. |
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That was N2 in the first example. Try it. Rearranging the reaction equation
to produce one mol of N2 gives |
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[NH3]2 · [H2]-3
[N2] | = |
K 1–1 |
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Which is just what we had for the inverse reaction before. |
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So the right equation for figuring out what it takes to make one
mol NH3 is actually the one with the fractional stoichiometry
indexes! |
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This looks worse than it is. All it takes is to remember the various conventions
underlying the mass action law, something you will get used to very quickly in actual work. The next point is the tricky
one! |
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Lets stick with the ammonia synthesis and give the concentrations symbolized by
[..] a closer look. What we have is a homogeneous reaction, i.e. only gases are
involved (a heterogeneous reaction thus involves that materials in more one kind of state are participating).
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We may then express the concentrations as partial pressures, (or, if we want to be totally precise, as fugacities). We thus have
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| [N2] | = |
pN2 | | | |
| | [H2] |
= | pH2 |
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| [NH 3] | = |
pNH3 |
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And the total pressure is p = S
pi |
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But what is the actual total pressure? |
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If we stick 1 mol N2 and 3 mols H2 in a
vessel keeping the pressure at the beginning (before the reaction takes place) at its standard value, i.e. at atmospheric
pressure, the pressure must have changed after the reaction, because we now might have
only 2 mols of a gas in a volume that originally contained 4! |
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If you think about it, that happens whenever the number of mols on both sides of a reaction
equation is not identical. Since the stoichiometry coefficients
n count the number of mols involved, we only have identical mol numbers before and after
the reaction if Sni = 0. |
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This is a tricky point and it is useful to illustrate it. Lets
construct some examples. We take one reaction where the mol count changes, and one example where it does not. For the first
example we take our familiar |
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We put 1 mol N2 and 3 mols H2, i.e.
N0 = mols into a vessel keeping the pressure at its standard value (i.e. atmospheric pressure p0).
This means we need 4 "standard" volumes which we call V0. |
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Now let the reaction take place until equilibrium is reached. Lets assume that 90 %
of the starting gases react, this leaves us with 0,1 mol N2, 0,3 mol of H2,
and 1,8 mols of NH3 . We now have N = 2,2 mols in our container |
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The pressure p must have gone down; as long as the gases are ideal, we have
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| p0 · V 0 | =
| N0 · RT | | |
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| p · V0 | = |
N · RT |
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N or N0 is the total number of mols contained in the reaction vessel at the pressure
p or p 0, respectively. |
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This equation is also valid for the partial pressure pi
of component i (with Si
pi = p) and gives for the partial pressure and the number of mols Ni
of component i, respectively |
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| pi | = |
p0 · |
Ni
N0 |
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| Ni | = |
N0 · |
pi
p0 |
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Now for the second example. Its actually not so easy to find a reaction between
gases where the mol count does not change (think about it!), Lets take the formal reaction
producing ozone, albeit a chemist might shudder: |
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Lets take comparable starting values: 2 mols O2, 90 % of which
react, leaving 0,2 mol of O2 and forming 0,8 mols of O3 and 0,8
mols of O (think about it!) - we always have two mols in the system. |
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The mass action law followed from the chemical potentials and the decisive factor
was ln ci with ci being a measure of the concentration of the component
i. We had several ways of measuring concentrations, and it is
quite illuminating to look closely at how they compare for our specific examples. |
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In real life, for measuring concentrations, we could use for example:
- The absolute number of mols
Ni,mol for component i. In general, the total number of mols in the reaction vessel, Si
Ni,mol, does not have to be constant as outlined above.
- The absolute particle number
Ni, p, which is the same as the absolute number of mols Ni, mol if you
multiply Ni, mol with Avogadros constant (or Lohschmidt's number) A = 6,02214 mol-1;
i.e. Ni, p = A·Ni, mol . Note that the absolute number of particles (= molecules)
does not have to stay constant, while the absolute number of
atoms , of course, never changes.
- The partial pressure
pi of component i, which is the pressure that we actually would find inside the reaction
vessel if only the the component i would be present. The sum of all partial pressures pi
thus gives the actual pressure
p inside the vessel; Si
pi = p and p does not have to be constant
in a reaction. This looks like a violation of our basic principle that we look at the minimum of the free enthalpy at constant pressure and temperature to find the mass action law. However, the mass
action law is valid for the equilibrium and the pressure at equilibrium - not for how
you reach equilibrium!
- The activity
ai (or the fugacity
fi) which for ideal gases is identical to ai = pi/p
= pi/Si
pi. This is more or less also what we called the concentration ci
of component i.
- The mol fraction Xi
, which is the number of mols divided by the total number of mols present in the system: Xi = Ni,
mol/Si
Ni, mol. This is the same thing as the concentration defined above because the partial pressure pi
of component i is proportional (for an ideal gas) to the number of mols in the vessel. We thus have Xi
= ci (= ai = fi as long as the gases are ideal).
- The "standard" partial pressure
pi 0 defined relative to the standard pressure p0
. This is the pressure that we would find in our reaction vessel if we multiply all absolute partial pressure with a
factor so that p = p0. We thus have
pi0 = (pi·N i,mol0)/Ni,
mol with Ni, mol0
= number of mols of component i at the beginning of the reaction (and p = standard pressure) as outlined above.
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For ease of writing (especially in HTML), the various measures of concentrations
will always be given by the square bracket "[i]" for component i . |
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We now construct a little table writing down the starting
concentrations and the equilibrium concentrations in the same system of measuring
concentrations. We then compute the reaction constant K for the respective concentrations, always by having the reaction
products in the denominator (i.e taking K = [NH3]2/[H2]3 · [N2]
or K = {[O3] · [O]}/[O2]2 , respectively). |
| Measure for c |
Starting values |
Equilibrium values |
Reaction constant |
NMol
absolute number of Mols
equivalent via
Ni,p = A·Ni, mol
to
Ni, p the absolute number of particles |
[H2] = 3
[N2] = 1
[NH3]
= 0
p = p0
Si
N = 4 |
[H2 ] = 0,3
[N2] = 0,1
[NH3]
= 1,8
p = 0,55p0
Si
Ni = 2,2 | KN = 1200 |
[O2] = 2
[O3] = 0
[O] = 0
p = p0
Si
N = 2 |
[O2] = 0,2
[O3] = 0,8
[O] =
0,8
p = p0
Si
Ni = 2 | KN = 16 |
Partial pressure pi
in units of p0 |
[H2] = 3/4
[N2] = 1/4
[H3] = 0 |
[H2] = 0,3/4 = 0,075
[N2] = 0,1/4
= 0,025
[NH3] = 1,8/4 = 0,450 |
K = 19 200 (p0)–2 |
[O2] = 2/2 = 1
[O3] = 0
[O] = 0
| [O2 ] = 0,2/2 = 0,1
[O3] = 0,4
[O] = 0,4 |
Kp = 16 |
Activity ai
identical to the concentration
ci
identical to the
Mol fraction
Xi |
[H2] = 3/4
[N2] = 1/4
[H3] = 0
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[H2] = 0,3/2,2 = 0,136
[N2] =
0,1/2,2 = 0,0454
[N3] = 1,8/2,2 = 0,818 |
Kact = 5 808 |
[O2 ] = 2/2 = 1
[O3] = 0
[O] = 0
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[O2] = 0,2/2 = 0,1
[O3] = 0,8/2
= 0,4
[O] = 0,82 = 0,4 | K = 16 |
"Standard" partial pressure pi0
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[H2] = 3/4
[N2] = 1/4
[NH3]
= 0 |
[H2] = 0,136
[N2] = 0,045
[NH3] = 0,818
pi0 = 1.1818pi |
K = 5 914 |
[O2] = 2/2 = 1
[O3] = 0
[O] = 0
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[O2] = 0,1
[O3] = 0,4
[O] = 0,4
| K = 16 |
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Well, you get the point. The reaction constant may be wildly
different for different ways of measuring the concentration of the components involved if the mol
count changes in the reaction (which it mostly does). |
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Well, at least it appears that we do not have any trouble calculating K if the concentrations
are given in whatever system. But this is not how it works! We do not want to compute
K from measured concentrations, we want to use known reactions constants assembled
from the standard reaction enthalpies or standard chemical potentials to calculate what we get. |
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So we must have rules telling us how to change the reaction constant if we go from from one
system of measuring concentrations to another one. |
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Essentially, we need a translation from absolute quantities like particle numbers (or partial
pressures) to relative quantities (= concentrations), which are always absolute quantities divided by some reference state
like total number of particles or total pressure. The problem clearly comes from the changing reference state if the mol
count changes in a reaction. |
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Lets look at the the conversion from activities to particle numbers; this essentially
covers all important cases. |
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Well, lets go back to the final stage
in the derivation of the mass action law and see what can be done. We had |
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| P (ai)i |
= exp – |
G0
kT |
= K = Kact = |
Reaction constant
for activities |
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The ai are the activities, which we defined when discussing
the chemical potential analogous to the fugacities for gases. Fugacities,
in turn, were introduced to take care of non-ideal behavior of gases. |
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However, as long as we look at gases and as long as they are ideal, the fugacity (or activity),
the prime quantity in the chemical potential for gases was the concentration of gas
i given by its partial pressure
pi divided by the actual pressure p, a relative quantity. For the purpose of this
paragraph it is sufficient to consider |
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Lets now switch to an absolute quantity. We take the number of mols of gas i.
Ni, mol; now lets see how the mass action law changes. |
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We can express
pi by |
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With p0 = standard pressure, and N0 = starting
number of mols, and p = S
pi = (SNi) · p0 /N0. |
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With this we can reformulate the mass action law by substituting |
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This gives (afer some fiddling around with the products and sums) |
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| lnP (ai)ni
| = ln P |
æ
ç
è |
pi
p |
ö
÷
ø |
n i |
= ln P |
æ
ç
è |
Ni
SNi |
ö
÷
ø |
ni |
= ln |
æ
ç
è |
æ
è | (S
Ni) | ö
ø |
– Sn i |
· | æ
è
| P Ni |
ö
ø |
ni |
ö
÷
ø |
= ln Kact |
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If this looks a bit like magic, you are encouraged to go through the motions in fiddling around
the products and the sums yourself. If you don't want to - after all we are supposed to be dealing with defects, not with
elementary albeit tricky math - look it up. |
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We want the mass action law for the particle numbers N
i, i.e. we want an expression of the form |
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So if we write down the mass action law now for particle number
Ni we have |
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| P(Ni)ni
| = | æ
è
| SNi |
ö
ø |
Sni |
· Kact = KN |
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| | K N |
= | æ
è |
SNi |
ö
ø |
– Sn
i | · Kact |
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Lets try it. For our ammonia example we have |
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| SNi |
= | 2,2 |
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| Sn
i | = |
1 + 3 – 2 = 2 |
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æ
è |
SNi | ö
ø
| Sni
| = |
2,22 = 4,84 |
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Well, the two constants from the table above are KN = 1200 and Kact
= 5 808; Kact/KN = 4,84 as it should be? Great - but shouldn't it be the other way around? |
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Indeed, we should have KN/Kact = 4,84 according to the formula
above - just the other way around. However, the way we formulated the mass action law above, we
should have written K–1 to compare with the values in the table! |
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OK; this is unfair - but look at the title of this subchapter! |
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One last word before we turn irreversibly into chemists: |
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With the equations that couple pressure and mol-numbers,
we can express SNi by S Ni
= p · (N0/p0) which, inserted into the expression between mass action
constants from above, gives |
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| K N | = |
æ
è |
p · |
N0
p0 |
ö
ø |
Sni |
= p · K' |
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In words: The reaction constant is proprotional to the pressure. If you do not just accept
whatever pressure you will get after a reaction, but keep the system at a certain pressure, you can influence how much (or
little) of the reaction products you will get. |
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© H. Föll (Defects - Script)
2.1.1 Simple Vacancies and Interstitials 
With frame