 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 
 
[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. 
 