
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 = S
p_{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 
 
N_{ 2} + 3H_{2} 
Û  2NH_{3} 




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. 

