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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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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 "↔" 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] |
= K3 = |
(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 μi0 that are contained
in the reaction constant (via Σi μi0) 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] |
= |
K11 |
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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 he
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 |
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[H2] |
= |
pH2 |
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[NH3] |
= |
pNH3 |
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And the total pressure is p =
Σpi |
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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 ν count the number of mols involved, we only have
identical mol numbers before and after the reaction if Σνi = 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 · V0 |
= |
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 p0,
respectively. |
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This equation is
also valid for the partial pressure pi of component
i (with Σi
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, Σi 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;
Σi 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/Σi 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/Σi 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.
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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). |
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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
Σi N = 4 |
[H2] =
0,3
[N2] = 0,1
[NH3] = 1,8
p = 0,55p0
Σi Ni = 2,2 |
KN = 1200 |
[O2] = 2
[O3] = 0
[O] = 0
p = p0
Σi N = 2 |
[O2] =
0,2
[O3] = 0,8
[O] = 0,8
p = p0
Σi 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
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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 |
[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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Π (ai)i |
= exp |
G0
kT |
= K = Kact =
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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 = Σpi = (ΣNi) ·
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 |
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pi
p |
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νi |
= ln P |
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Ni
ΣNi |
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νi |
= ln |
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( |
(ΣNi) |
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Σνi |
· |
( |
P Ni |
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νi |
) |
= 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 Ni, 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 |
= |
( |
ΣNi |
) |
Σνi
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· Kact =
KN |
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KN |
= |
( |
ΣNi |
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Σνi |
· Kact |
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Lets try it. For our ammonia example
we have |
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ΣNi
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= |
2,2 |
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Σνi |
= |
1 + 3 2 = 2 |
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( |
ΣNi |
) |
Σνi |
= |
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
K1 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 ΣNi by ΣNi = p ·
(N0/p0) which, inserted into the
expression between mass action constants from above, gives |
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KN |
= |
( |
p · |
N0
p0 |
) |
Σνi |
= 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