7.3 Chemical Equilibrium in Solutions
317
and the reactant molecules are uncharged, we can often assume that activity coefficients
are nearly equal to unity.
E X A M P L E 7.7
a. Find the value of the equilibrium constant for the reaction of Example 7.6 at 298.15 K.
b. Find the value of (∂G/∂ξ) T ,P for the case that the molality of carbon monoxide is equal
to 0.00010 mol kg −1 , that of carbon dioxide is equal to 0.00015 mol kg −1 , and that of
oxygen is equal to 0.00020 mol kg −1 . Assume activity coefficients equal to unity.
c. Find the equilibrium composition for an initial molality of carbon dioxide equal to
0.00015 mol kg −1 , an initial molality of oxygen equal to 0.00020 mol kg −1 , and an initial
molality of carbon monoxide equal to zero. Assume activity coefficients equal to unity.
Solution
a. From Example 7.6, ∆G ◦ −548.56 kJ mol −1 at 298.15 K.
K e −∆G ◦ /RT exp
⎛
⎝
548560 J mol −1
8.3145 J K −1 mol −1
(298.15 K)
⎞
⎠
1.27 × 10 96
b.
(∂G/∂ξ) T ,P ∆G ◦ + RT ln(Q)
−548,560 J mol −1
+
8.3145 J K −1 mol −1
(298.15 K) ln
(0.00015) 2
(0.00010) 2 (0.00020)
−548560 J mol −1 + 23000 J mol −1
−525000 J mol −1 −525 kJ mol −1
c. Let m eq (CO)/m ◦ 2x, so that m eq (O 2 )/m ◦ 0.00020 + x and m eq (CO 2 )/m ◦
0.00015 − 2x.
K 1.27 × 10 96
m eq (CO 2 )/m ◦ 2
m eq (CO)/m ◦
2 m eq (O 2 )/m ◦
(0.00015 − 2x) 2
(2x) 2 (0.00020 + x)
We neglect 2x compared with 0.00015 and neglect x compared with 0.00020, obtaining
x 2 ≈
(0.00015) 2
4
1.27 × 10 96
(0.00020)
2.21 × 10 −101
x ≈ 4.7 × 10 −51 ; m(CO) ≈ 9.4 × 10 −51 mol kg −1
This value of x indicates that the reaction goes essentially to completion. A very large
volume of solution would be required to contain a single CO molecule at equilibrium.
Exercise 7.6
Calculate the volume of solution that contains one CO molecule at equilibrium. Comment on
the likelihood that equilibrium will actually be attained.