Physical Chemistry Third Edition

7.6 The Temperature Dependence of Chemical Equilibrium. The Principle of Le Châtelier 337

Solution

∆H◦ 2 ∆fH◦(NO 2 )−∆fH◦(N 2 O 4 )2(33.095 kJ mol−^1 )+(−1)(9.179 kJ mol−^1 )

 57 .011 kJ mol−^1

ln

(

K( 373 .15 K)

K( 298 .15 K)

)

−

57011 J mol−^1

8 .3145 J K−^1 mol−^1

[

1

373 .15 K

−

1

298 .15 K

]

 4. 622

Using the value ofK(298.15 K) from Example 7.3,

K(373.15 K)K(298.15 K)e^4.^622

(0.148)(101.7) 14. 92

∆G◦(373.15 K)−RTln(K(373.15 K))

−(8.3145 J K−^1 mol−^1 )(373.15 K) ln(14.92)

−8390 J mol−^1 − 8 .39 kJ mol−^1

Exercise 7.18

Find the value ofKfor the reaction of Example 7.4 at 1000 K.

If the assumption of constant∆H◦is not sufficiently accurate, the next simplest

assumption is that the heat capacities are constant, so that

∆H◦(T)∆H◦(T 1 )+∆CP(T−T 1 ) (7.6-10)

When Eq. (7.6-10) is substituted into Eq. (7.6-4) and an integration is carried out from

T 1 toT 2 , the result is

ln

(

K(T 2 )

K(T 1 )

)

−

∆H◦(T 1 )

R

[

1

T 2

−

1

T 1

]

+

∆CP

R

[

ln

(

T 2

T 1

)

+

T 1

T 2

− 1

]

(7.6-11)

Exercise 7.19

a.Verify Eq. (7.6-11).

b.Using heat-capacity data from Table A.8 of Appendix A and assuming the heat capacities to

be temperature-independent, evaluateKand∆G◦for the reaction of Example 7.18 at 100◦C.

Calculate the percent difference between your value forKand that in Example 7.18.

This principle isnamedfor Henri Louis
Le Châtelier,1850–1936,aFrench
chemist.

The Principle of Le Châtelier

The behavior of a system at chemical equilibrium when subjected to a change in

temperature illustrates theprinciple of Le Châtelier. This principle states:If possible,