285
TABLE 17-2
Standard Free Energies of Formation at 25°C (AG° In kcal/mole)
Substance
Any element
Ag
+
AgCI
Br (atom)
Br
BrCI
C (atom)
C (diamond)
Ca
2+
Cd
z+
CH 3 OH
CH 3 OH
C 2 H 5 OH
C 2 H 5 OH
C 4 H 10
C 6 H 6
C 6 H 6
Cl (atom)
Cl
cio:
State
standard
aq
s
g
aq
g
g
s
aq
aq
1
9
1
9
9
1
g
g
aq
aq
AG?
000
1843
-2622
1969
-2457
-021
16084
068
-132 18
-1858
-3973
-3869
-41 77
-4030
-375
2976
3099
2519
-31 35
-257
When changes occur with all
would be used. For example
Substance
CO
C0 2
Cu
2+
F (atom)
FFeO
Fe 2 O 3
H (atom)
i-r
HBr
HCI
HI
H 2 0
H 2 0
H 2 S
I (atom)
IICI
K+
Ll+
materials
, for AS
0
\ v °
iiO —
State
AG?
g
g
aq
g
aq
s
s
9
aq
9
9
g
1
g
9
9
aq
9
aq
aq
-3281
-9426
1553
1420
-6608
-5840
-177 10
4858
000
-1272
-2277
031
-5669
-5464
-789
1677
-1235
-1 32
-6747
-7022
Substance
State
N (atom)
Na (atom)
Na
+
NH,
NH 3
NH 4
+
NO
NO 2
NOf
N 2 O
0 (atom)
OHP (atom)
PCI 3
PCI 5
S (atom)
S
2 ~
SO 2
SO3Zn
2+
in their standard states, the
we
>H°
would have
- AG°
T
9
9
aq
9
aq
aq
9
9
aq
9
9
aq
a
9
9
9
aq
9
aq
aq
AG?
81 47
1867
-6259
-398
-637
-1900
2072
1239
-2641
2476
5499
-3760
6671
-6842
-7759
4357
2000
-71 79
-17734
-3518
superscript °
(17-20)
PROBLEM:
Calculate the change in standard entropy for the dissociation of PC1 5 into PC1 3 and
C1 2 at 25.0°C.
SOLUTION:
Take the value of AG° = +9170 cal as obtained in the preceding problem, and
the value of A//° = +22,130 cal as obtained on p. 218 in Chapter 14. Substitute
these values into Equation 17-20 and get
(22. 130 cal) -(9170 cal)
2yo.2 K.
cal
K.