Download Notes on Debye-Hückel Theory

GEOS 20300/30300
Autumn 2004
Thermodynamics and Phase Change
M.S. Ghiorso
Notes on Debye-Hückel Theory
We seek:
µi = µio + RT ln Xi + µiex ,
where µ ex
i = RT ln λ i , or the deviation from ideal behavior due to long range interaction
of ions in solution. µ ex
i is electrical in nature.
From a theorem of electrostatics: “that the mutual energy of a system of charges is onehalf the sum of the products of the charges of each and the potentials due to the others,”
we have for the ith ion:
µiex =
zi eψ i
,
2
where zi e is the charge in coulombs and ψ i is the potential on the surface of the ith ion
due to all the other ions in solution. We want a simple expression for ψ i .
One of the fundamental laws of electrostatics (Maxwell’s first equation) is:
  4πϕ
,
∇•E =
ε

where E is the electric field vector, ϕ is the charge density per unit volume, and ε is the


dielectric constant of the medium. Now, E is just − ∇ ψ (just like the force is the gradient
of the potential energy), so:
 
4πϕ
4πϕ
2
or ∇ ψ = −
,
∇ • ∇ψ = −
ε
ε
which is the Poisson equation.
4πϕ i
, where we assume that ε is the same everywhere and is
ε
equal to the value for pure water. ϕ i is a function of radius about the ith ion.
For the ith ion: ∇ 2ψ i = −
Clearly, the charge on the ith ion added to the charge of every other ion in solution must
sum to zero, i.e. ∑ zi e = 0 , since the solution is neutral. Thus,
i
Notes on Debye-Hückel Theory, con’t
∞
∫
a
2
4π r 2ϕ i dr + zi e = 0 ,
2
where 4 π r is an element of surface area and a is the radius of the ith ion.
What does the charge density look like?
We assume the Boltzmann distribution,
(
)
C j ′ = C j exp − E j kT ,
where C j is the number of j ions per unit volume, and E j = z je ψ i , the energy of the ion,
is equal to its charge times the potential. Then,
(
)
ϕ i = ∑ z j eC j exp − z j eψ i kT .
j
Expanding in a Taylor series (and keeping only the first two terms):
(
)
ϕ i = ∑ z j eC j − ∑ z j eC j z j eψ i kT .
j
j
Note that the first of these sums is zero!
So, ∇ 2ψ i = −
4πϕ i
becomes
ε
∇ ψi = ∑
2
j
4π e2C j z 2j
ε kT
ψi .
4 πe2 C j z2j
Setting κ = ∑
, we have in Cartesian coordinates
ε kT
j
2
∇ 2ψ i = κ 2ψ i
or in Polar coordinates
2
ψ i ″ + ψ i ′ − κ 2ψ i = 0 .
r
A solution is: ψ i =
A −κ r B κ r
e + e
r
r
As r → ∞ , ψ i must be finite, so B is zero. Thus, ψ i =
A −κ r
e , so
r
Notes on Debye-Hückel Theory, con’t
3
C j z j 2 e 2 A −κ r
ϕi = −∑
e
kT r
j
How do we determine the constant A?
From
∫
∞
a
4 π r2 ρ idr + zie = 0 , we have
−∫
∞
a
C j z j 2 e 2 A −κ r
4π r ∑
e dr = −zi e ,
kT r
j
2
and
A=
zi e eκ a
ε 1+κa
At last:
zi e eκ a e−κ r
ψi =
ε 1+κa r
If we are only interested in the potential field of all ions except the ith ion, we must
subtract its potential, zi e ε r , from ψ i , yielding
ψ i* =
⎞
zi e ⎛ eκ a e−κ r
− 1⎟ .
⎜
εr ⎝ 1 + κ a
⎠
Therefore, the potential at the surface of the ith ion (r = a) due to the presence of all other
ions in solution is given by:
*
ψ i,a
=−
From µiex =
zi e ⎛ κ ⎞
⎜
⎟.
ε ⎝1+κa⎠
zi eψ i
, we get
2
zi2 e2 ⎛ κ ⎞
µ =−
⎜
⎟N,
2ε ⎝ 1 + κ a ⎠
ex
i
where N is Avogadro’s number, to put the potential on a molar basis. So:
Notes on Debye-Hückel Theory, con’t
4
ln λ i = −
Now we must modify κ 2 = ∑
4π e2C j z 2j
zi2 e 2 ⎛ κ ⎞
.
2ε kT ⎝ 1+ κa ⎠
to convert C j to M j units. We divide by N and
ε kT
convert cc to liters (1000 cc/liter):
j
κ2 = ∑
j
4π Ne2 M j z 2j
1000ε kT
.
To go from Molarity to molality we use the relation M j =
m j ρ*
1 + ∑ 0.001mkWk
k
, where Wk
is the molecular weight of the k component and ρ is the solution density. For
geological (natural) fluids this expression may be simplified to M j ≅ mj ρ , where ρ is the
density of pure water, since
*
th
⎛
⎞
ρ * ≅ ⎜ 1 + ∑ 0.001mkWk ⎟ ρ .
⎝
⎠
k
These manipulations give us
κ2 =
4π e2 N ρ ∑ m j z 2j
j
1000ε kT
.
If we define
I=
1
m j z 2j
∑
2 j
and call it the ionic strength, then
κ2 =
8π e2 N ρ I
.
1000ε kT
Hence:
8π e2 N ρ
I
−zi2 e2
1000
ε
kT
ln λi =
.
2ε kT
8π e2 N ρ
1+ a
I
1000ε kT
Notes on Debye-Hückel Theory, con’t
5
Define
8π e2 N ρ
1000ε kT
Bγ =
and
Aγ =
2π N e3 ρ
ln (10 ) 1000 ( ε kT )
3
,
2
and we get
log λi =
−Aγ zi2 I
1 + aBγ I
Now we must convert rational to molal activity coefficients.
First note that
Xi =
mi
,
55.51 + ∑ m j
j
and
µi = µio + RT ln ( λi Xi ) = µio,* + RT ln (γ i mi ) .
So:
⎛X ⎞
⎛λ ⎞
µio,* − µio
= ln ⎜ i ⎟ + ln ⎜ i ⎟
RT
⎝ mi ⎠
⎝γi⎠
As mi → 0 ,
µio,* − µio
⎛ 1 ⎞
→ ln ⎜
,
⎝ 55.51 ⎟⎠
RT
so
⎛
⎞
1
⎛ 1 ⎞
⎟ + ln ⎛ λi ⎞ .
ln ⎜
= ln ⎜
⎟
⎜⎝ γ ⎟⎠
⎜
⎟
⎝ 55.51 ⎠
i
⎜⎝ 55.51 + ∑ m j ⎟⎠
j
Thus
⎛
⎞
ln ( λi ) = ln (γ i ) + ln ⎜ 1 + ∑ m j 55.51⎟ .
⎝
⎠
j
Defining
⎛
⎞
Γ γ = − log ⎜ 1 + ∑ m j 55.51⎟ ,
⎝
⎠
j
Notes on Debye-Hückel Theory, con’t
6
we have finally:
log γ i =
−Aγ zi2 I
1 + aBγ I
+ Γγ
This is the Debye-Hückel law.
At low concentrations
log γ i → −Aγ zi2 I .
This is the Debye-Hückel Limiting Law. Note that a is an ion parameter for the solution.
Aγ and Bγ are f(T,P) and are tabulated by Helgeson and Kirkham (1974).