Solutions to exercises
155
2H 2 O (g) + 2V
••
O m 2O
#
O + 4H
•
i
with K
V P
H
H O
O
H O
i
2 2
4
••
•
2
2
=
6
6 @
@
, where P H 2 O denotes the water partial pressure.
From this we deduce the concentration [H
•
i ]:
[H
•
i ] = K
¼
H 2 O [V
••
O ]
½
P
½
H 2 O
2. Let us write that the sum of the hydrogen and water vapor partial pressures
is constant near the sample:
P H 2 O + P H 2 = α
From this we obtain
P
1 P
P
H O
H O
H
1
2
2
2
α =
+
−
e
o
By inserting this result into the expression for [H
•
i ] from question B.1, we
obtain
H
P
P
K V
1
i
H O
H
O
2
•
••
2
2
1 2
=
+
e
o
6
6
@
@
with K = K
¼
H 2 O α
½
3. The equilibrium constant K eq for the formation of water vapor,
2H 2 + O 2 m 2H 2 O
is expressed as
K
e
eq
RT
G
r T
0
=
Δ
−
and
K
e .
.
eq
T
T
8 314 10
494 0 112
3
–
=
#
−
−
+
K
P P
P
eq
H O
H O
2
2
2
2
2
=
from which we obtain
P
P
K P
H O
H
eq
O
2
2
2
1 2
1 2
=
−
−
Equation for [H
•
i ] in terms of oxygen partial pressure (see fig. 47)
Starting from the expression
H
P
P
K V
1
i
H O
H
O
•
••
2
2
1 2
1 2
=
+
e
o
6
6
@
@