211 / 912
Chapter 5 203
Copyright © 2017 Pearson Education, Inc.
61.
∆
G
°
=
∆
H
°
-
T
∆
S
°
a.
∆
G
°
=
20
-
1
298
21
0.05
2
∆
G
°
=
20
-
14.9
=
5.1
∆
G
°
=
-
RT
ln
K
eq
5.1
=
-
1.986
*
10
-
3
*
298
*
ln
K
eq
5.1
=
-
0.59 ln
K
eq
-
8.6
=
ln
K
eq
K
eq
=
1.8
*
10
-
4
b.
∆
G
°
=
20
-
1
398
21
0.05
2
∆
G
°
=
20
-
20
=
0
K
eq
=
1.0
62.
C
O
R R
+
C
OH
R R
+
C R
OH
OH
H
R
H O H
H
H
2
O
+
C R
OH
OH
R
H
3
O
+
H
2
O
63.
a.
∆
G
°
=
-
RT
ln
K
eq
=
-
1.986
*
10
-
3
*
298
*
ln10
-
3
=
-
2.72 kcal
>
mol
∆
G
°
=
-
1.986
*
10
-
2
*
298
*
ln10
-
2
=
-
1.36 kcal
>
mol
∆∆
G
°
=
-
2.72
-
(
-
1.36)
=
-
1.36 kcal
>
mol
Thus,
∆
G
°
must change by 1.36 kcal/mol.
b.
∆
G
°
=
∆
H
°
-
0
-
1.36
=
∆
H
°
-
0
∆
H
°
=
-
1.36 kcal
>
mol
c.
∆
G
°
=
0
-
T
∆
S
°
-
1.36
=
0
-
298
∆
S
°
∆
S
°
=
1.36
>
298
=
4.56
*
10
-
3
kcal
>1
mol deg
2
64.
∆
G
°
=
-
RT
ln
K
eq
ln
K
eq
=
-
∆
G
°
>
RT
ln
K
eq
=
-
∆
G
°
>
0.59 kcal
>
mol
ln
K
eq
=
-
5.3
>
0.59 kcal
>
mol
ln
K
eq
=
-
9.0
K
eq
=
3
B
4 > 3
A
4
=
0.00013
3
B
4 > 3
A
4
=
0.00013
>
1
=
0.13
>
1000