# 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 n0 CO2 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 · (n0 – n0H2) · K | ö ø | 1/2 | ö ÷ ø
|  | The starting concentration of CO2, i.e n CO2, is expressed as nCO2 = n0 – n 0H2.
| 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 |
--- | --- | --- | ---
n 0CO2 | = | 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.

With frame

Mass Action Law

Pitfalls and Extensions of the Mass Action Law

Solution to Exercise 2.1-5

© H. Föll (Defects - Script)