Chemical Examples for Mass Action Law Applications

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
H2  +  Cl2  Û  2HCl
To keep it easy, we start with equal amounts of H2 and Cl2 .
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
[H2] · [Cl2]
 =  K
or, with [H2] = [Cl2] = [equ]
[HCl] = [equ] · K1/2
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.[H2] + 2[HCl] = constant = e.g. the number of H2 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
H2  +  CO2  Û   H2O  +  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 n0H2  and n0CO2  mols of the reacting gases and define as the yield y the number of mols of H2O that the reaction will produce at equilibrium. This leaves us with
nH2O  =  y    
         
nCO  =  y    
         
nH2  =  n0H2  –  y  =   equilibrium
concentration of H2
         
nCO2  =  n0CO2  –  y  =   equilibrium
concentration of CO
         
S n  =  n0  =  n0H2  +  n0CO2
The last equation holds because the mol count never changes in this example.
The mass action law now gives
y2
(n0H2  –  y) · (n0CO2  –  y)
 =   K 

y  =  æ
ç
è
1
2
æ
è
1  –  K ö
ø
ö
÷
ø
 ·  æ
ç
è
æ
è
n0 · K  ±  æ
è
(n0 · K)2  +  4 · (1 – K) · n0H2 · (n0n0H2) · K ö
ø
1/2 ö
÷
ø
The starting concentration of CO2, i.e nCO2, is expressed as nCO2 = n0n0H2.
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/dn0H2 = 0.
Well, go through the math yourself; this is elementary stuff. The solution is
n0H2  =    n0
2  
       
n0CO2  =    n0
2
 =  n0H2
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.

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© H. Föll (Defects - Script)