Physical Chemistry Third Edition

7.2 Reactions Involving Gases and Pure Solids or Liquids 311

and Eq. (7.1-20) for the equilibrium constant is

K

∏c

i 1

(

Pi,eq

P◦

)vi

(ideal gas reaction) (7.2-2)

EXAMPLE 7.3

Consider the reaction:

0 2NO 2 (g)−N 2 O 4 (g)

a.Calculate the value of∆G◦at 298.15 K.

b.Calculate the value ofKat 298.15 K.

c.Calculate the equilibrium pressure of a system that initially consists of 1.000 mol of N 2 O 4

and that is confined in a fixed volume of 24.46 L at 298.15 K. Assume ideal gases.

Solution

a.From the Gibbs energy changes of formation,

∆G◦ 2 ∆fG◦(NO 2 )−∆fG◦(N 2 O 4 )

(2)(51.258 kJ mol−^1 )−(97.787 kJ mol−^1 )

 4 .729 kJ mol−^1

b.The equilibrium constant is

K

(

Peq(NO 2 )/P

) 2

Peq(N 2 O 4 )/P◦

exp

⎛

⎝ −4729 J mol

− 1

(

8 .3145 J K−^1 mol−^1

)

( 298 .15 K)

⎞

⎠

 0. 148

c.Letαbe the degree of dissociation (the fraction of the initial N 2 O 4 that dissociates at

equilibrium).

Peq(NO 2 )

n(NO 2 )RT

V



(1.000 mol)(2α)RT

V

Peq(N 2 O 4 )

n(N 2 O 4 )RT

V



(1.000 mol)(1−α)RT

V

We now can write

K 0. 148 

(2α)^2

1 −α

(1.000 mol)RT

P◦V



(2α)^2

1 −α

(1.000 mol)(8.3145 J K−^1 mol−^1 ) (298.15 K)

(100,000 Pa)(0.02446 m^3 )



4 α^2 (1.0135)

1 −α