Solutions to exercises
149
( )
c
e
10
.
.
Br
10
8 314 900
2 96 480
2 5
2
=
−
−
#
#
#
#
Pl
^
h
( )
c
bar
10
Br
38
–
2
=
Pl
Referring to the Brouwer diagram at
( )
c
bar
10
Br
38
2
=
−
Pl
, we observe
that the concentration of color centers is practically equal to that of the
non-ionized bromine vacancies:
[V
•
Br(SE) ] = [V
#
Br ] = 10
18
cm
−3
5. a. The dissociation reaction of solid KBr in the conditions given in the
problem statement is
KBr (s) m K (g) + 2
1
Br 2(g)
The corresponding equilibrium constant is
K
e
eq
RT
G
r T
=
Δ
−
°
K
e .
eq
8 314 900
327 400
=
−
#
K
10
eq
19
=
−
Moreover, we have
K
P P
eq
K
Br 2
1 2
=
#
with pressure expressed in bars.
From this we deduce P
b ar
10
Br
30
2
=
−
The electrochemical chain to consider is
P K ,Me / KBr (s) / Me,P Br2
(−)
(+)
The crystal becomes colored at the (−) pole, but much less so than for
P Br2 = 10
−38
bar.
The concentrations of the majority species, as read from the diagram
(fig. 59), are
[V
#
Br ] ≈ 10
15
cm
−3
[e ′ ] ≈ 10
9
cm
−3
[V
•
Br ] = [V ′
K ] ≈ 10
18
cm
−3