Physical Chemistry , 1st ed.

If (^) rxnH° is assumed to not vary over the temperature range, it can be removed
from the integral along with R, and the expression becomes
ln 

K

K

2

1



(^) rx
R
nH°
T

1

1



T

1

2

 (5.20)

Using this expression, we can estimate the values of equilibrium constants at

different temperatures, knowing the standard enthalpy change. Or, we can es-

timate the standard enthalpy change knowing the equilibrium constant at two

different temperatures, rather than plotting data as suggested by equation 5.19.

Example 5.10

The dimerization of a protein has the following equilibrium constants at the

given temperatures:K(4°C) 1.3  107 ,K(15°C) 1.5  107. Estimate

the standard enthalpy of reaction for this process.

Solution

Using equation 5.20 and remembering to convert our temperatures into

kelvins:

ln 

1

1

.

.

3

5





1

1

0

0

7

 7   28

1

8K



27

1

7K



Solving for the enthalpy of reaction:

(^) rxnH°8630 J/mol 8.63 kJ/mol
How do we rationalize the effect of pressure on an equilibrium? Let us con-
sider a simple gas-phase reaction between NO 2 and N 2 O 4 :
2NO 2 N 2 O 4
The equilibrium constant expression for this reaction is
K
If the volume is decreased isothermally, the pressures of both NO 2 and N 2 O 4
increase. But the value of the equilibrium constant doesn’t change! Because the
partial pressure in the denominator is squared as a result of the stoichiometry
of the expression, the denominator increases faster relative to the numerator
ofKas the volume is decreased. In order to compensate, the denominator has
to decrease its relative value, and the numerator has to increase its relative
value, in order for Kto remain constant. In terms of the reaction, this means
that the partial pressure of N 2 O 4 (the product) goes up and the partial pres-
sure of NO 2 (the reactant) goes down. Generally speaking, the equilibrium
shifts toward the side of the reaction that has the lower number of gas mole-
cules; this is the simple expression of the Le Chatelier principle for pressure
effects. Inversely, lowering the pressure (for example, by increasing the volume
isothermally) will shift the reaction to the side with more gas molecules.
Example 5.11
In Example 5.6, the equilibrium partial pressures of I 2 and I in the gas phase
were 0.235 and 0.515 atm, with an equilibrium constant value of 1.13. Suppose
pN 2 O 4 /p°

(pNO 2 /p°)^2

JQPJ

(^) rxnH°

8.314 moJl K
134 CHAPTER 5 Introduction to Chemical Equilibrium