Physical Chemistry Third Edition

28.6 Approximate Theories of Transport Processes in Liquids 1191

This value is reasonable, being about one-fourth of a typical energy change of vapor-

ization.

The temperature dependence of a diffusion coefficient is sometimes represented by

a formula due to Arrhenius that is used for chemical reaction rate constants:

DAde−∆Ead/RT (28.6-9)

whereEadis the activation energy andAdis the preexponential factor for the diffusion

process. To compare Eq. (28.6-8) with this equation, we assume that the enthalpy

change of activation and the energy change of activation are nearly equal, since there

is little change inPVin a liquid-state process. Therefore,

D 2  2

kBT

h

a^2 e−∆S

‡◦/R

e−∆U

‡◦/RT

(28.6-10)

and if we identify∆U‡◦withEa, the preexponential factor is

Ad 2

kBT

h

a^2 e−∆S

‡◦/R

(28.6-11)

This preexponential factor is temperature-dependent, but its temperature dependence

is much weaker than that of the exponential factor.

EXAMPLE28.14

The diffusion coefficient for 1,1,1-trichloroethane in a mixed solvent of 2,2-dichloropropane

and carbon tetrachloride was measured to be 1.41× 10 −^9 m^2 s−^1 at 25◦C and 2. 02 × 10 −^9

m^2 s−^1 at 45◦C. Find the value of the apparent energy of activation and the value of the

preexponential factor.

Solution

Ea

Rln

(

k 2

k 1

)

1

T 1

−

1

T 2



(

8 .3145 J K−^1 mol−^1

)

ln

(

2. 02 × 10 −^9 m^2 s−^1

1. 41 × 10 −^9 m^2 s−^1

)

1

298 .15 K

−

1

318 .15 K

14200 J mol−^1

AdkeEa/RT

(1. 4110 −^9 m^2 s−^1 )exp

⎛

⎝

(

14200 J mol−^1

)

(

8 .3145 J K−^1 mol−^1

)

( 298 .15 K)

⎞

⎠

 4. 29 × 10 −^7 m^2 s−^1

Exercise 28.11

Estimate the value of∆S‡◦for the data of the preceding example, assuming thata 4 ×

10 −^10 m. Comment on the magnitude and sign of your answer.