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 factor f for z 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
k B T
6πηr eff ,2
where r eff ,2 is the effective radius of the particle of substance 2, η is the viscosity of the
solvent, k B is Boltzmann’s constant, and T is the absolute temperature. If we use this
relation in Eq. (12.2-2) and assume that d 23 r eff ,2 + r eff ,3 , we obtain the relation for
uncharged molecules
k
4N Av k B T
6η
1
r eff ,2
+
1
r eff ,3
r eff ,2 + r eff ,3
2RT
3η
2 +
r eff ,2
r eff ,3
+
r eff ,3
r eff ,2
(12.2-6)
where R is the ideal gas constant. If r eff ,2 and r eff ,3 are nearly equal, Eq. (12.2-6)
becomes
k ≈
8RT
3η
(two reacting substances)
(12.2-7)
so that the rate constant k depends only on the viscosity of the solvent and the temperature. 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 ≈
4RT
3η
(single reacting substance)
(12.2-8)
E X A M P L E 12.5
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 × 10 6 m 3 mol −1 s −1 7.4 × 10 9 L mol −1 s −1