
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). 


We may then express the
concentrations as partial
pressures, (or, if we want to be totally precise, as
fugacities). We thus have



[N_{2}] 
= 
p_{N2} 



[H_{2}] 
= 
p_{H2} 



[NH_{3}] 
= 
p_{NH3} 




And the total pressure is p =
Sp_{i} 

But what is
the actual total pressure? 


If we stick 1 mol N_{2}
and 3 mols H_{2} 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 Sn_{i} = 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 N_{2}
and 3 mols H_{2}, i.e. N_{0} = mols into a
vessel keeping the pressure at its standard value (i.e. atmospheric pressure
p_{0}). This means we need 4 "standard"
volumes which we call V_{0}. 


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 N_{2}, 0,3 mol of
H_{2}, and 1,8 mols of NH_{3}. 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



p_{0} · V_{0} 
= 
N_{0} · RT 



p · V_{0} 
= 
N · RT 
Þ 
p_{ } 
· 
N
N_{0} 
= 0,55 p_{0} 




N or N_{0}
is the total number of mols contained in the reaction vessel at the pressure
p or p_{0}, respectively. 


This equation is also valid for
the partial pressure p_{i} of component i
(with S_{i} p_{i} =
p) and gives for the partial pressure and the number of mols
N_{i} of component i, respectively 


p_{i} 
= 
p_{0} · 
N_{i}
N_{0} 




N_{i} 
= 
N_{0} · 
p_{i}
p_{0} 



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 O_{2}, 90 % of which react, leaving 0,2 mol
of O_{2} and forming 0,8 mols of O_{3} 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 c_{i}
with c_{i} 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
N_{i,mol} for component i. In general, the total
number of mols in the reaction vessel, S_{i} N_{i,mol}, does not have to be constant as outlined above.
 The absolute particle number
N_{i, p}, which is the same as the absolute number of
mols N_{i, mol} if you multiply N_{i,
mol} with Avogadros constant (or Lohschmidt's number) A =
6,02214 mol^{1}; i.e. N_{i, p} =
A·N_{i, 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
p_{i} 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
p_{i} thus gives the actual pressure p inside the vessel;
S_{i} p_{i} = 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 a_{i} (or
the fugacity
f_{i}) which for ideal gases is identical to
a_{i} = p_{i}/p =
p_{i}/S_{i}
p_{i}. This is more or less also what we called the concentration c_{i} of component
i.
 The mol fraction X_{i},
which is the number of mols divided by the total number of mols present in the
system: X_{i} = N_{i, mol}/S_{i} N_{i, mol}. This is the
same thing as the concentration defined above because the partial pressure
p_{i} of component i is proportional (for
an ideal gas) to the number of mols in the vessel. We thus have
X_{i} = c_{i} (= a_{i} =
f_{i} as long as the gases are ideal).
 The "standard" partial
pressure p_{i}^{0} defined relative to
the standard pressure p^{0}. 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 = p^{0}. We thus have p_{i}^{0} =
(p_{i}·N_{i,mol}^{0})/N_{i,
mol} with N_{i, mol}^{0} = 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 =
[NH_{3}]^{2}/[H_{2}]^{3} · [N_{2}]
or K = {[O_{3}] · [O]}/[O_{2}]^{2} ,
respectively). 
Measure for
c 
Starting
values 
Equilibrium
values 
Reaction constant 
N_{Mol}
absolute number of Mols
equivalent via
N_{i,p} = A·N_{i, mol} to
N_{i, p} the absolute number of particles 
[H_{2}]
= 3
[N_{2}] = 1
[NH_{3}] = 0
p = p^{0}
S_{i} N_{} = 4 
[H_{2}]
= 0,3
[N_{2}] = 0,1
[NH_{3}] = 1,8
p = 0,55p^{0}
S_{i} N_{i} = 2,2 
K_{N} = 1200 
[O_{2}]
= 2
[O_{3}] = 0
[O] = 0
p = p^{0}
S_{i} N_{} = 2 
[O_{2}]
= 0,2
[O_{3}] = 0,8
[O] = 0,8
p = p^{0}
S_{i} N_{i} = 2 
K_{N} = 16 
Partial pressure
p_{i}
in units of p^{0} 
[H_{2}] = 3/4
[N_{2}] = 1/4
[H_{3}] = 0 
[H_{2}]
= 0,3/4 = 0,075
[N_{2}] = 0,1/4 = 0,025
[NH_{3}] = 1,8/4 = 0,450 
K = 19 200
(p^{0})^{–2} 
[O_{2}]
= 2/2 = 1
[O_{3}] = 0
[O] = 0 
[O_{2}] = 0,2/2 = 0,1
[O_{3}] = 0,4
[O] = 0,4 
K_{p} =
16 
Activity
a_{i}
identical to the concentration c_{i}
identical to the
Mol fraction
X_{i} 
[H_{2}]
= 3/4
[N_{2}] = 1/4
[H_{3}] = 0 
[H_{2}]
= 0,3/2,2 = 0,136
[N_{2}] = 0,1/2,2 = 0,0454
[N_{3}] = 1,8/2,2 = 0,818 
K_{act} = 5 808 
[O_{2}]
= 2/2 = 1
[O_{3}] = 0
[O] = 0 
[O_{2}]
= 0,2/2 = 0,1
[O_{3}] = 0,8/2 = 0,4
[O] = 0,82 = 0,4 
K = 16 
"Standard" partial
pressure p_{i}^{0}

[H_{2}]
= 3/4
[N_{2}] = 1/4
[NH_{3}] = 0 
[H_{2}]
= 0,136
[N_{2}] = 0,045
[NH_{3}] = 0,818
p_{i}^{0} = 1.1818p_{i} 
K = 5 914 
[O_{2}]
= 2/2 = 1
[O_{3}] = 0
[O] = 0 
[O_{2}]
= 0,1
[O_{3}] = 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. 

