Page 101 - Chemical equilibria Volume 4

Molecular Chemical Equilibria 77
with the number of gaseous moles being the same on both sides of the
equation, the entropy and therefore the slope of the Ellingham line is
practically non-existent. In reaction [3R.11], there is an increase in the
number of gaseous moles because of the reaction, and therefore an increase
in entropy, resulting in a negative slope. For reaction [3R.12], there is a
decrease in the number of gaseous moles, so the standard entropy decreases
and the slope of the Ellingham line is positive.

3.3.2.4. Oxygen pressure at equilibrium at a given temperature in a
metal-oxide Ellingham diagram

Consider the Ellingham line relative to a reaction r 1 (Figure 3.7(a)). Let Ω
denote the origin of the axes – i.e. the point with the coordinates (-273°C, 0).
We place ourselves at a temperature T, the abscissa of a point H and let M 1
be the point of intersection of the vertical issuing from H and the Ellingham
line of the reaction r 1. At another given temperature T’ (the abscissa of
point H’), we again place a vertical axis y’y. The line ΩM 1 cuts that axis at
point M’ 1. We can show that all the points on the line ΩM’ 1 correspond to
the same oxygen pressure.

In the two homothetic triangles ΩHM 1 and ΩH’M’ 1, we can write:

HM 1 = ΩH [3.47a]
H'M' 1 ΩH'

so

T
Δ g 0 = RlogP 1 = T
1
T
Δ ' g 0 R 'log ' P 1 ' T [3.47b]
1
From this we deduce that the pressure at point M’ 1 is the same as at point
M 1, and therefore that, whatever the position of the y’y axis, we have:

log P = log ' [3.48]
P
1 1
Now consider a second reaction r 2 and let Δ g be the standard Gibbs
0
2
energy associated with the temperature T, which is k times higher than that
of reaction r 1 at the same temperature, and thus: