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Temperature Dependence of K
∆
G
o
=
∆
H
o
 T
∆
S
o
= RT ln K
Dividing both sides by RT and rearranging:
ln K = 
Τ
∆Η
ο
R
+
R
S
ο
∆
Compare:
y
=
mx
+
b
(straight line equation)
where y = ln K
and x =
Plot ln K vs and get a straight line.
The slope of the line = 
R
ο
∆Η
and the yintercept =
R
S
ο
∆
This slope will be positive or negative depending on the sign of
∆
H.
Suppose the plot of ln K vs gives a positive slope.
Since the slope

R
ο
∆Η
, a positive slope
implies
∆
H is negative (exothermic reaction).
A positive slope also means:
As ln K increases, increases
As ln K increases, T decreases
As K increases, T decreases
As K decreases, T increases
This is consistent with LeChatelier’s principle, which implies that for an exothermic reaction, a
lower
temperature will shift the reaction to the right leading to a higher value of K, or conversely, a
higher temperature will shift the reaction to the left, leading to a lower value of K.
To compare two temperatures and two equilibrium constants:
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View Full Document ln K
2
= 
2
R
Τ
∆Η
ο
+
R
S
ο
∆
ln K
1
= 
1
R
Τ
∆Η
ο
+
R
S
ο
∆
Subtracting the second equation from the first:
ln K
2
– ln K
1
= 
R
ο
∆Η
(
2
T
1

1
T
1
)
This is equivalent to:
ln
1
2
K
K
=
R
ο
∆Η
(
1
T
1

2
T
1
)
or
ln
1
2
K
K
=
R
ο
∆Η
(
2
1
1
2
T
T
T
T

)
N
2
(g) + 3H
2
(g)
⇄
2NH
3
(g)
K = 6.0 x 10
5
at 298.15 K.
What is value of K at 600 K?
For this reaction,
∆
H
o
= 92.22 kJ
∆
G
o
= 32.96 kJ
∆
S
o
= 198.7 J/K
ln
1
2
K
K
=
R
ο
∆Η
(
1
T
1

2
T
1
)
ln
5
10
x
0
.
6
x
=
314
.
8
220
,
92

(
)
15
.
298
)(
600
(
15
.
298
600

)
Solving:
x = 4.46 x 10
3
Alternate method:
∆
G
o
= RT ln K
∆
G
o
= 32.98 kJ at 298 K.
But this value will change at other temperatures.
∆
G
o
=
∆
H  T
∆
S
o
AT 600 K:
∆
G
o
= 92.220 600(
1000
7
.
198

) = +27.0 kJ
Note that reaction is spontaneous in the reverse direction at this high temperature.
+27.0 =  ln K
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This note was uploaded on 07/17/2010 for the course 160 162 taught by Professor Kimmel during the Spring '10 term at Rutgers.
 Spring '10
 Kimmel

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