Physical Chemistry Third Edition

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.

EXAMPLE 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,Pfor 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.

Ke−∆G

◦/RT

exp

⎛

⎝ 548560 J mol

− 1

(

8 .3145 J K−^1 mol−^1

)

(298.15 K)

⎞

⎠

 1. 27 × 1096

b.

(∂G/∂ξ)T,P∆G◦+RTln(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.Letmeq(CO)/m◦ 2 x, so thatmeq(O 2 )/m◦ 0. 00020 +xandmeq(CO 2 )/m◦

 0. 00015 − 2 x.

K 1. 27 × 1096 

[

meq(CO 2 )/m◦

] 2

[

meq(CO)/m◦

] 2

meq(O 2 )/m◦



(

(0. 00015 − 2 x)^2

(2x)^2 (0. 00020 +x)

)

We neglect 2xcompared with 0.00015 and neglectxcompared with 0.00020, obtaining

x^2 ≈

(

(0.00015)^2

4

(

1. 27 × 1096

)

(0.00020)

)

 2. 21 × 10 −^101

x≈ 4. 7 × 10 −^51 ;m(CO)≈ 9. 4 × 10 −^51 mol kg−^1

This value ofxindicates 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.