Solutions to exercises
33
K e = np
2 and the electroneutrality relation
p + 2[V
••
O ] = n + 4[V
4
′
Zr ]
At high oxygen partial pressure, the species V
4
′
Zr and h
•
dominate, which
gives the following approximation:
p = 4[V
4
′
Zr ]
from which
4
4
[ ]
[ ]
[ ]
[ ]
K
V
n P
V
p
K P
V
K
V
K
P
4
g
o
O
o
e
O
Zr
S
Zr
e
O
2
2
4
2
4
4
4
4
4
••
••
2
2
2
=
=
=
#
#
#
#
#
#
l
l
4
[ ]
.
V
K
K K
P
0 33
Zr
g
S e
O
2
4
2
1 5
1 5
#
=
l
e
o
3. The variation in the concentration n of electrons as a function of oxygen
partial pressure n = f (P O 2 ) is
4
[ ]
.
n
p
K
V
K
K
K K
K
P
4
4 0 33
1
e
Zr
e
e
S e
g
O
4
2
2
1 5
1 5
#
#
=
=
=
−
l
e
o
.
n
K
K K
P
0 76
S
g e
O
2
2
1 5
1 5
#
=
−
e
o
Solution 1.7 – The non-stoichiometry of iron monoxide
1. The problem statement gives a molar fraction of iron in the oxide of 0.4767.
The molar fraction of oxygen is thus
x O = 1 – x Fe
x O = 1 − 0.4767 = 0.5233
We deduce the two following formulas for the monoxide:
Fe 0.91 O
and
FeO 1.098
With respect to a perfect FeO crystal, the first formula indicates an iron
deficiency and the second indicates an excess of oxygen.