# Solution to Exercise 2.1-5 "Do the Math for Mixed Point Defects"

| For obvious reasons some of the symbols deviate a little from the symbols used in the text; e.g. we have hFP instead of HFP.
| We start with the system of equations that came from the mass action law
| We start with the calculation of cv(C):
|  | Inserting the first and the second equation into the third equation yields:
| That was the first equation for cV(C). Next we calculate ci(C).
|  | Start with the third equation and eliminate cv(A) using the second. We have the final result after a series of mathematical manipulations:
| That was the third equation. Next we calculate cV(A).
|  | Start with the third equation and eliminate ci(C) using the first, we obtain
| That's it. Nothing to it. ;-)
|  | Well, not exactly. I myself certainly cannot solve problems like this without making some dumb mistakes in breaking down the math. Almost everybody does.
|  | However, I usually notice that I made a stupid mistake because the result just can't be true. And I can, if I really employ myself, get the right result eventually - because I did some exercises like this before. And that is why you should do it, too!.
| As a last comment we may note that solving equations coming from the mass action law can become rather tedious very quickly - compare the example in the link, which is about as simple as it could be.

| Now we look at the limiting cases of pure Schottky or pure Frenkel disorder.
|  | For pure Frenkel disorder we must have hFP << hS, and cV(A) = 0.
|  | For pure Schottky disorder we must have hFP >> hS, and ci(C) = 0.
| For the first case - pure Frenkel disorder - just look at the expression
æ ç è | 1 + | N N' | · exp | hS – hFP kT | ö ÷ ø | 1/2
|  | For hS >> hFP, the exponential in this case is positive which means
N N' | · exp | hS – hFP kT | >> 1
--- | --- | --- | ---
|  | So you may neglect the 1 in the above expression and replace the whole square root by
N N' | · exp | hS – hFP 2kT
--- | --- | ---
|  | This gives for ci(C)
ci(C) | = | N N' | · |
æ ç è | exp | hS – 2hFP kT | ö ÷ ø | 1/2
æ ç è | exp | hS – hFP kT | ö ÷ ø | 1/2
| = | N N' | · exp – | hFP kT
|  | This is the result as as it should be.
| With this we immediately obtain
cV(C) | = | N N' | · exp – | hFP 2kT
--- | --- | --- | --- | ---
cV(A) | = | 0
|  | This is so because
N N' | · exp | hS – hFP kT | >> | 1
--- | --- | --- | --- | ---
| Contrariwise, if hS << hFP, 1 + N/N' · exp[(hS – hFP)/kT] » 1 obtains.
|  | Because hS – 2hFP is a large negative number we get
ci(C) | = | N N' | · exp | hS – 2hFP 2kT | » | 0
--- | --- | --- | --- | --- | --- | ---
|  | The expressions for cV(C) and cV(A) immediately reduce to the proper equation
cV(C) | = cV(A) = exp – | hS 2kT
--- | --- | ---

With frame

Chemical Examples for Mass Action Law Applications

Exercise 2.1-5: Do the Math

© H. Föll (Defects - Script)