Physical Chemistry Third Edition

10.4 Transport Processes in Liquids 471

In order for shearing flow to take place, layers of a liquid must flow past each other.

This requires disruption of cages and also raises the potential energy of the system.

The activation energy per molecule is similar to the activation energy of diffusion.

Note that the sign of the exponent in Eq. (10.4-6) is opposite from that in Eq. (10.4-5).

This is because the rate of shear is proportional to the Boltzmann probability factor in

Eq. (10.4-7), making the viscosity coefficient inversely proportional to it. The activation

energy for viscosity is often roughly equal to the activation energy for self-diffusion in

the same liquid, giving further plausibility to this argument.

EXAMPLE10.17

Following are data on the viscosity of liquid carbon tetrachloride:

T/◦C 0 15 20 30 40 50 60

η/cP 1.329 1.038 0.969 0.843 0.739 0.651 0.585

The viscosity coefficient values are given in centipoise (cP). The poise is an older unit of

viscosity, equal to 1 g cm−^1 s−^1 , so that 1 poise 0 .1kgm−^1 s−^1. Find the values ofη 0

andEaη.

Solution

We linearize Eq. (10.4-6) by taking logarithms:

ln(η/η 0 )

Eaη

RT

A linear least squares fit of ln(η/1cP)to1/Tgives an intercept of− 4 .279 and a slope of

1245 K, with a correlation coefficient of 0.9998 (a good fit).

Eaη(slope)×R(1245 K)

(

8 .3145 J K−^1 mol−^1

)

 1. 035 × 104 J mol−^1  10 .4 kJ mol−^1

η 0 (1 cP)e−^4.^279  1. 39 × 10 −^2 cP 1. 39 × 10 −^5 kgm−^1 s−^1

The value of this energy of activation is typical of liquids with small molecules. As might

be expected, it is somewhat smaller than the energy of vaporization of CCl 4 , 33.9 kJ mol−^1 ,

which corresponds to removing a molecule from all of its nearest neighbors.

EXAMPLE10.18

The value of the self-diffusion coefficient of carbon tetrachloride at 25◦C is equal to 1. 4 ×

10 −^9 m^2 s−^1. Estimate the value at 40◦C, assuming the same value of the energy of activation

as for the viscosity from the previous example.

Solution

D(313 K)

D(298 K)

exp

[

−

(

10350 J mol−^1

)(

8 .3145 J K−^1 mol−^1

)− 1 ( 1

313 K

−

1

298 K

)]

 1. 22

D(313 K)(1. 4 × 10 −^9 m^2 s−^1 )(1.22) 1. 7 × 10 −^9 m^2 s−^1