Page 115 - Chemical equilibria Volume 4

Molecular Chemical Equilibria 91
In order to find the equation of a curve expressed in the system of
perpendicular axes A 2xy, we first need to be capable of transposing a point P
characterized by two molar fractions into coordinates A 2xy. Thus, let us
consider a point P in the equilateral triangle characterized by the coordinates
x 1 and x 3 (Figure 3.14). We can write:
x = PQ QN NP [3.70]
+
=
However, in the triangle NQA 2, we have a 30° angle at A 2, and therefore:

QA x
QN = 2 = 3 [3.71]
3 3

and in the triangle MPN, we have a 30° angle at P:

3PN
PB = [3.72]
2
Thus, in view of the value of PB:

2x
PN = 1 [3.73]
3

By feeding back expressions [3.71] and [3.73] into equation [3.70], we
find:
x + 2x
x = 3 1 [3.74]
3

Conversely, we can easily calculate:

y
x = [3.75a]
3
x = x 3 − y [3.75b]
1
2
and:

2 −− x 3
y
x = [3.75c]
2
2