Physical Chemistry Third Edition

12.2 Elementary Processes in Liquid Solutions 531

The reaction diameter for the reaction between hydrogen ions and hydroxide ions

in aqueous solution is larger than the sum of the radii of these ions. The explanation

for this fact is that water molecules can exchange hydrogen and hydroxide ions as

described in Chapter 10.

Exercise 12.5

Calculate the values of the electrostatic factorfforz 2 z 3 equal to 2, 1, 0,−1, and−2 at 298.15 K,

assuming a dielectric constant equal to 78.4 and a reaction diameter equal to 0.50 nm. Comment

on your results.

Equation (10.4-4) relates a diffusion coefficient to the effective radius of a diffusing

particle of substance 2 and the viscosity of the solvent:

D 2 

kBT

6 πηreff ,2

wherereff ,2is the effective radius of the particle of substance 2,ηis the viscosity of the

solvent,kBis Boltzmann’s constant, andTis the absolute temperature. If we use this

relation in Eq. (12.2-2) and assume thatd 23 reff ,2+reff ,3, we obtain the relation for

uncharged molecules

k

4 NAvkBT

6 η

(

1

reff ,2

+

1

reff ,3

)

(

reff ,2+reff ,3

)



2 RT

3 η

(

2 +

reff ,2

reff ,3

+

reff ,3

reff ,2

)

(12.2-6)

whereRis the ideal gas constant. Ifreff ,2andreff ,3are nearly equal, Eq. (12.2-6)

becomes

k≈

8 RT

3 η

(two reacting substances) (12.2-7)

so that the rate constantkdepends only on the viscosity of the solvent and the temper-

ature. If the two reacting molecules are of the same substance, we must divide by 2 in

order to avoid counting the same encounter twice:

k≈

4 RT

3 η

(single reacting substance) (12.2-8)

EXAMPLE12.5

Use the value of the viscosity of water at 25◦C to estimate the value of the rate constant

for any bimolecular diffusion-controlled reaction of uncharged molecules in water at that

temperature.

Solution

Use Eq. (12.2-7) as an approximation:

k

8(8.3145 J K−^1 mol−^1 )(298 K)

3

(

8. 904 × 10 −^4 kg m−^1 s−^1

)

 7. 4 × 106 m^3 mol−^1 s−^1  7. 4 × 109 L mol−^1 s−^1