 | 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). |
|  | The reaction
equation that naturally comes to mind is |
| |
|
| | and the mass action law tells us that |
| |
|
| | 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! |
| 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 |
|
|
|
| | The mass action law than
gives |
| |
[NH3]2
[N2] ·
[H2]3 | = | K2 = | 1
K1 |
|
|
| To make things worse, we could write
the two equations also like |
| |
[N2]1/2 · [H2]3/2
[NH3] | = K3 = |
(K1)1/2 |
|
|
| | and nobody
keeps us from using the reaction as a source for hydrogen via |
| |
[NH3]2/3 · [N2]1/3
[H2] | = | K4 =
? |
|
|
| 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: |
| | 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. |
| | 2. The standard chemical potentials mi0 that are contained in the reaction constant (via Si mi0) 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. |
| | That was N2 in the first
example. Try it. Rearranging the reaction equation to produce one mol of
N2 gives |
| |
[NH3]2 · [H2]-3
[N2] | = | K1–1 |
|
|
| | Which is just what we had for the inverse
reaction before. |
| So the right equation for figuring out
what it takes to make one mol NH3 is actually the one with the fractional stoichiometry indexes! |
| 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! |
| |
| 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). |
| | We may then express the concentrations as partial
pressures, (or, if we want to be totally precise, as fugacities).
We thus have |
| |
| [N2] | = | pN2 | | | |
| | [H2] | =
| pH2 | | | | | | [NH3] | = | pNH3 |
|
|
| | And the total
pressure is p = Spi |
| But what is the actual total pressure? |
| | 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! |
| | 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. |
|
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 |
| |
|
| | 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. |
| | 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 |
| The
pressure p must have gone down; as long as the gases are ideal, we have |
| |
| p0 · V0 | = |
N0 · RT | | | | | | p ·
V0 | = | N · RT |
|
|
| | N or
N0 is the total number of mols contained in the reaction vessel at the pressure
p or p0, respectively. |
| | 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 |
| |
| pi | = | p0 · | Ni
N0 | | | | | | Ni | = | N0 · | pi
p0 |
|
|
| 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: |
| |
|
| | 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. |
| 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. |
| | 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 pi0 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·Ni,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.
|
| | For
ease of writing (especially in HTML), the various measures of concentrations will always be given by the square
bracket "[i]" for component i . |
| 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 | [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 | [O2] =
0,2/2 = 0,1
[O3] = 0,8/2 = 0,4
[O] = 0,82 = 0,4 | K = 16 | "Standard"
partial pressure pi0
| [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 | [O2] = 0,1
[O3] = 0,4
[O] = 0,4 | K = 16 |
|
| 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). |
| | 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. |
| | 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. |
| | 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. |
| Lets look at the the conversion from
activities to particle numbers; this essentially covers all important cases. |
| |
| Well, lets go back to the final stage
in the derivation of the mass action law and see what can be done. We had |
| |
| P (ai)i | = exp – | G0
kT | = K = Kact =
| Reaction constant
for
activities |
|
|
| | 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. |
| | 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 |
| |
|
| 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. |
| | We
can express pi by |
| |
|
| | With p0 =
standard pressure, and N0 = starting number of mols, and p = Spi = (SNi) ·
p0/N0. |
| | With this we can reformulate the mass action law by substituting |
| |
|
| | This gives (afer some fiddling around with
the products and sums) |
| |
| lnP (ai)ni | = ln P | æ
ç
è |
pi
p | ö
÷
ø | ni | = ln P | æ
ç
è | Ni
SNi |
ö
÷
ø | ni | = ln | æ
ç
è | æ
è | (SNi) | ö
ø | – Sni | · | æ
è | P Ni | ö
ø |
ni | ö
÷
ø | = ln Kact |
|
|
| | 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. |
| | We want the mass
action law for the particle numbers Ni, i.e. we want an expression of the form |
| |
|
| So if we write down
the mass action law now for particle number Ni we have |
| |
| P(Ni)ni | = | æ
è | SNi | ö
ø
| Sni |
· Kact = KN | | | | | | | | | | KN | = | æ
è | SNi |
ö
ø | – Sni | · Kact |
|
|
| Lets try it. For our ammonia example we
have |
| |
| SNi | = | 2,2 | | | | |
| Sni | = | 1 + 3 – 2 = 2 | | | | | æ
è | SNi | ö
ø
| Sni
| = | 2,22 =
4,84 |
|
|
| | 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? |
| | 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! |
| OK; this is unfair - but look at the
title of this subchapter! |
| One last word before we turn
irreversibly into chemists: |
| | With the equations that couple pressure and mol-numbers, we can express SNi by SNi = p ·
(N0/p0) which, inserted into the expression between mass action constants from
above, gives |
| |
| KN | = | æ
è | p · | N0
p0 |
ö
ø | Sni | = p · K' |
|
|
| | 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. |
| |
© H. Föll (Defects - Script)
2.1.1 Simple Vacancies and Interstitials 
With frame