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204 Chapter 5
Copyright © 2017 Pearson Education, Inc.
The calculation shows that for every 1000 molecules in the chair conformation, there is 0.13 molecule in a
twist-boat conformation.
This agrees with the statement on page 127 of the text that for every 10,000 chair conformers of cyclohexane,
there is no more than one twist-boat conformer.
65.
A step-by-step description of how to solve this problem is given in the box entitled “Calculating Kinetic
Parameters” on page 224 of the text.
E
a
can be determined from the Arrhenius equation
1
ln
k
=
-
E
a
>
RT
2
, because a plot of ln
k
versus 1
>
T
gives a slope
=
-
E
a
>
R
.
ln 2.11
*
10
-
5
=
-
10.77
T
=
304
1
>
T
=
3.29
*
10
-
3
ln 4.44
*
10
-
5
=
-
10.02
T
=
313
1
>
T
=
3.19
*
10
-
3
ln 1.16
*
10
-
4
=
-
9.06
T
=
324.5
1
>
T
=
3.08
*
10
-
3
ln 2.10
*
10
-
4
=
-
8.47
T
=
332.8
1
>
T
=
3.00
*
10
-
3
ln 4.34
*
10
-
4
=
-
7.74
T
=
342.2
1
>
T
=
2.92
*
10
-
3
slope
=
-
8290
E
a
=
-
1
slope
2
R
E
a
=
-
1
-
8290
2
*
1.98
*
10
-
3
kcal
>
mol
E
a
=
16.4 kcal
>
mol
To find
G
:
-
∆
G
°
=
RT
ln
kh
>
Tk
b
From the graph used to determine
E
a
, one can find the rate constant
1
k
2
at 30
°
.
1
It is 1.84
*
10
-
5
s
-
1
.
2
-
∆
G
°
=
1.98
*
10
-
3
*
303 ln
1
1.84
*
10
-
5
*
1.58
*
10
-
31
2>1
303
*
3.30
*
10
-
19
2
-
∆
G
°
=
1.98
*
10
-
3
*
303 ln
1
2.90
*
10
-
36
2>1
1.00
*
10
-
16
2
-
∆
G
°
=
1.98
*
10
-
3
*
303 ln 2.90
*
10
-
20
-
∆
G
°
=
1.98
*
10
-
3
*
303
*
1
-
3.50
2
∆
G
°
=
2.10 kcal
>
mol
To find
H
:
∆
H
°
=
E
a
-
RT
∆
H
°
=
16.4
-
1.98
*
10
-
3
*
303
∆
H
°
=
16.4
-
0.6
∆
H
°
=
15.8 kcal
>
mol
To find
S
:
∆
S
°
=
1
∆
H
°
-
∆
G
°
2>
T
∆
S
°
=
1
15.8
-
2.10
2>
303
∆
S
°
=
1
-
13.7
2>
303
∆
S
°
=
0.045 kcal
>1
mol deg
2
=
45 cal
>1
mol deg
2