
What can one do with the mass action
law? A lot  but it is not always very obvious. Lets ask a few "dumb"
questions and see how far we get. 

First we look at a really simple
reaction, e.g 





To keep it easy, we start with equal amounts of
H_{2} and Cl_{2} . 

How much HCl do we get? Notice
that we have the same number of mols
on both sides of the reaction equation. 


Well, in equilibrium (denoted by [..]) we
have 


[HCl]^{2}
[H_{2}] · [Cl_{2}] 
= 
K 




or, with [H_{2}] = [Cl_{2}] =
[equ] 




One equation with two unknowns; not
sufficient for calculating numbers. 


But then we also have the condition that the
number of the atoms involved stays constant, i.e.[H_{2}] + 2[HCl]
= constant = e.g. the number of H_{2} mols before the
reaction. 




Next, a little bit harder. Lets start
with arbitrary concentrations of something and see what we can say about the
yield of the reaction. For varieties sake
lets look at 


H_{2} + CO_{2} 
Û_{ } 
H_{2}O + CO 




Again a simple reaction with the same number of
mols on both sides, so we do not have to
worry about the precise form of the mass action law. 

We start with
n^{0}_{H2} and
n^{0}_{CO2} mols of the reacting gases and
define as the yield y the
number of mols of H_{2}O that the reaction will produce at
equilibrium. This leaves us with 


n_{H2O} 
= 
y 







n_{CO} 
= 
y 







n_{H2} 
= 
n^{0}_{H2} –
y 
= 
equilibrium
concentration of H_{2} 





n_{CO2} 
= 
n^{0}_{CO2} – y 
= 
equilibrium
concentration of CO 





S n 
= 
n^{0} =
n^{0}_{H2} +
n^{0}_{CO2} 




The last equation holds because the mol count
never changes in this example. 


The mass action law now gives 


y^{2}
(n^{0}_{H2} – y) ·
(n^{0}_{CO2} – y) 
= _{ } 
K_{ } 
y 
= 
æ
ç
è 
1
2 
æ
è 
1 – K 
ö
ø 
ö
÷
ø 
· 
æ
ç
è 
æ
è 
– n^{0} · K ± 
æ
è 
(n^{0} · K)^{2} + 4 · (1
– K) · n^{0}_{H2} · (n^{0}
– n^{0}_{H2}) · K 
ö
ø 
1/2 
ö
÷
ø 




The starting concentration of
CO_{2}, i.e n_{CO2}, is expressed as
n_{CO2} = n^{0} –
n^{0}_{H2}. 

Looks extremely messy, but this is
just the standard solution for a second order equation. Whatever this solution
means in detail, it tells us that the yield is a function of the starting
concentrations of the ingredients. 


What kind of starting concentrations will give us
maximum yield? To find out, we have to form
dy/dn^{0}_{H2} = 0. 


Well, go through the math yourself; this is
elementary stuff. The solution is 


n^{0}_{H2} 
= 
n^{0}
2 





n^{0}_{CO2} 
= 
n^{0}
2 
= n^{0}_{H2} 



In other words: maximum yield is
achieved if you mix just the right amounts of the starting stuff. This result
is always true, even for more complicated reactions. 


At this point we stop, again because otherwise we
might turn irreversibly into chemists. 