7.2 Reactions Involving Gases and Pure Solids or Liquids
311
and Eq. (7.1-20) for the equilibrium constant is
K
c
i1
P i,eq
P ◦
v i
(ideal gas reaction)
(7.2-2)
E X A M P L E 7.3
Consider the reaction:
0 2 NO 2 (g) − N 2 O 4 (g)
a. Calculate the value of ∆G ◦ at 298.15 K.
b. Calculate the value of K at 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∆ f G ◦ (NO 2 ) − ∆ f G ◦ (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
P eq (NO 2 )/P
2
P eq (N 2 O 4 )/P ◦
exp
⎛
⎝
−4729 J mol −1
8.3145 J K −1 mol −1
(298.15 K)
⎞
⎠
0.148
c. Let α be the degree of dissociation (the fraction of the initial N 2 O 4 that dissociates at
equilibrium).
P eq (NO 2 )
n(NO 2 ) RT
V
(1.000 mol)(2α) RT
V
P eq (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 )
4α 2 (1.0135)
1 − α