Page 118 - Chemical equilibria Volume 4

94 Chemical Equilibria
and the total quantity of matter is:
∑ β
n =+ i z [3.84]
1
β
1
Thus, the molar fraction x 3 of A 3 is:
zβ / β
x = ∑ β 1 [3.85]
3
3
1+ i z
β 1
This function is an increasing function of z, so the way of finding the
maximum of x 3 is identical to that for finding the maximum of z.

Let us write the law of mass action, feeding back the molar fractions into
relation [3.78]:

⎛
z 3 β = Const. 1+ ∑ β i z ⎟ ⎞ ∑ i β [3.86]
⎜
1 β ⎛ β ⎞ 2 β ⎜ ⎝ β 1 ⎟ ⎠
−
−
(uz ) ⎜ 1 u − 2 z ⎟
⎝ β 1 ⎠
By taking the logarithmic derivative of both sides in relation to u, and by
cancelling all the terms dz/du, we obtain:
β β
− 1 + 2 = 0 [3.87]
u − z β
1 u−− 2 z
β 1

From this, we deduce that:
β
u = 1 [3.88a]
β + β 2
1
and:
β
1 u−= 2 [3.88b]
β + β 2
1

This demonstrates our theorem.