Physical Chemistry Third Edition

1190 28 The Structure of Solids, Liquids, and Polymers

On the average, there will be a slightly smaller concentration of component 2 above the

vacancy than below. The rate at which molecules move into the vacancy from below

is equal to

Ratebkc 2 (z′−a) (28.6-3)

whereais an averagezcomponent of the displacement into the vacancy, and where

c 2 (z′−a) is the concentration of component 2 evaluated atz′−a. The proportionality

constantkis analogous to a rate constant and we assume that it is given by Eq. (28.6-2).

The rate at which molecules move into the vacancy from above is equal to

Rateakc 2 (z′+a) (28.6-4)

We assume that the molecules coming from above on the average have the same rate

constant as those coming from below.

The net contribution to the diffusion flux is

J 2 zk[c 2 (z′−a)−c 2 (z′+a)]a (28.6-5)

Let us representc 2 (z′+a) as a Taylor series around the pointz′−a:

c 2 (z′+a)c 2 (z′−a)+

∂c 2

∂z

∣

∣

∣

∣

z′−a

( 2 a)+ ··· (28.6-6)

where the subscript indicates that the derivative is evaluated atzz′−a. If we neglect

the terms not shown in Eq. (28.6-6), Eq. (28.6-5) becomes

J 2 z−

kBT

h

(

2 a^2

)

e−∆G

‡◦/RT

(

∂c 2

∂z

)

(28.6-7)

Comparison with Fick’s law, Eq. (28.6-2), gives an expression for the diffusion coef-

ficient:

D 2  2 a^2

kBT

h

e−∆G

‡◦/RT

(28.6-8)

EXAMPLE28.13

Many liquids with molecules of ordinary size have diffusion coefficients approximately equal

to 10−^9 m^2 s−^1. Assume thatais equal to 1× 10 −^10 m and estimate the value of∆G‡◦for

T300 K.

Solution

∆G‡

◦

−RTln

(

D 2 h

2 a^2 kBT

)

−(8.3JK−^1 mol−^1 )(300 K)ln

(

(10−^9 m^2 s−^1 )(6. 6 × 10 −^34 Js)

2(10−^10 m)^2 (1. 4 × 10 −^23 JK−^1 )(300 K)

)

 1 × 104 J mol−^1 10 kJ mol−^1