Page 120 - Chemical equilibria Volume 4

96 Chemical Equilibria
A 3M (the ratio of the distances from a point on that line to the sides A 3A 1
and A 3A 2 is, indeed, constant and equal to MA 1/MA 2).
If β 2 is zero (with component A 2 being inert), the iso-Q curves pass
through A 2. In order to know whether they can have a maximum, we take the
logarithmic derivatives by making β 2 = 0 and dy = 0, which leads us to:

β
− 1 = 0 [3.90]
x 3 − y

This equation is impossible: the curves only exhibit a maximum if the
polycomponent phase contains only two components that are involved in the
reaction (Figure 3.15(a)).

We shall now examine a number of remarkable cases:
– First case: β = β = 1and β = (case of synthesis of IH)
2
1
3
2
Equation [3.77] at equilibrium becomes:
4y 2
= K
(I)
y
( x 3 − y )( 2 − − x ) 3
The maximum of x 3 is easy to obtain because, in all cases, x = 1/ 3 ,
m
which gives us:

y = K (I) /4
m
1+ K (I) / 4
In order to examine the properties of the curve, let us change the axis:

X =− 1
x
3
In this new coordinate system, the equation of the curve becomes:

K (I) (1 y− 2 ) 4y− 2
X =
2
(3K r )