Adsorption
Diagenesis
Lecture 19
Adsorption
• Physical
o attachment of atom or molecule to a surface through van der Waals or
intermolecular forces.
o Weak: ∆Had ≈ 4-12 kJ/mol
• Chemical
o formation of new chemical bond between adsorbed species and atoms
on surface
o Stronger: ∆Had >40 kJ/mol
Adsorption and Surface
Free Energy
• Adsorption changes the nature of the surface and
therefore the surface free energy.
• We can express the surface free energy as:
ds = -å
i
ni,s
d mi = -å G i d mi
A
i
• where we define Γi as the Gibbs Absorption Density:
moles of i absorbed on the surface per unit area.
Equilibrium Adsorption
• Consider adsorption of species M on surface S:
M + S = MS
• Let the fraction of surface sites occupied by M be θM.
• Fraction of free sites is then 1 - θM.
• Assuming elementary reactions, rate of adsorption is:
-
dM
= k+ [M ](1- Q M )
dt
dQ
= k- Q M
• Rate of desorption is
dt
• At equilibrium
k+ [M ](1- QM ) = k-QM
-
• and
QM =
k+ / k- [M ]
K ad [M ]
=
1+ k+ / k- [M ] 1+ K ad [M ]
Langmuir Isotherm
QM =
K ad [M ]
1+ K ad [M ]
• is known as the Langmuir Isotherm
o named for Irving Langmuir, who worked for GE.
• We can also express this in terms of the adsorption
density:
K ad [M ]
max
GM = GM
1+ K ad [M ]
• At large concentration then
G M = G max
M
Freundlich Isotherm
K ad [M ]
QM =
1+ K ad [M ]
• At low concentration,
QM @ K ad [M ]
• In other words, fraction of
sites occupied by M is
proportional to
concentration.
• This is known as the
Freundlich Isotherm.
• We’ll return to this in the
Red curve shows adsorption of Sr on
next chapter.
FeOOH calculated from Langmuir
Isotherm. Inset shows linear Freunlich
behavior at low concentration.
Diagenesis
Diagenesis is the transformation of a sediment into a
sedimentary rock. Physical processes involve compaction
and expulsion of water as the sediment is buried. We’ll briefly
consider some of the chemical processes that accompany
this.
• Consider a sedimentary layer ‘A’ being buried as
sediment accumulates above it.
• We can chose a reference frame fixed to the layer, or to
the sediment-water interface. In the latter case, the
layer will appear to move downward with time.
• Change in concentration at some fixed depth x is then:
dCi
æ ¶Ci ö
æ ¶Ci ö
-w ç
çè
÷ø =
÷ø
è
¶t x dt
¶x t
• First term on the right is any changes in concentration
within layer, second term is the concentration gradient
times the burial rate, ω.
• We can use this equation to change reference frames.
Steady-State Case
•
•
This is the case where:
dCi
æ ¶Ci ö
æ ¶Ci ö
=
w
çè
÷
çè
÷ =0
¶t ø x dt
¶x ø t
Solving, we have:
dCi
æ ¶C ö
=wç i÷
è ¶x ø t
dt
•
•
In this case, the
concentration in the
sediment at the surface is
constant, but diagenetic
changes result in decrease in
concentration with time. But
concentration at a fixed
depth is constant.
The diagenetic changes
impose a concentration
gradient.
No diagenesis
• This is the case where
• so that
dCi
=0
dt
æ ¶Ci ö
æ ¶Ci ö
=
w
çè
÷
çè
÷
¶t ø x
¶x ø t
• Change in
concentration at some
fixed depth is the burial
rate times the gradient.
Diagenetic Process
• Sediment consists of particles plus ‘pore water’. If
porosity is ϕ then fraction of particles is 1 - ϕ.
• As weight of sediment accumulating above
increases, water tends to be expelled - resulting in
upward velocity of pore fluid.
• In addition, dissolution and cementation
(precipitation) can change porosity.
• Due to compaction, rate of burial will not be equal
to sedimentation rate.
Fluxes and Bioturbation
•
•
Now imagine a layer of sediment
of thickness dx.
There are fluxes into and out of
this layer due to:
o
o
o
•
•
Water advection
Chemical diffusion
Bioturbation
Advective Flux is simply J = vCi
We can treat the biodiffusive flux
the same as chemical diffusion:
æ ¶C ö
J b = -DB ç
è ¶x ÷ø t
•
•
A difference is that the
biodiffusive flux is likely to be a
function of depth since most
animals live near the surface
(and O2 generally decreases with
depth).
For biodiffusion, Fick’s Second
Law becomes:
æ ¶C ö
ì ¶(DB ¶C / ¶x)) ü
ý
çè
÷ø = í
¶t x î
¶x
þt
Diffusion and Porosity
• Diffusion is so much faster through the pore water
than through solids, the latter can be effectively
ignored.
• Thus we can view the ‘molecular flux’ - the
chemical one we have already discussed, as
reduced to that occurring through pores:
æ ¶C ö
J M = -f DM ç
è ¶x ÷ø
• Fick’s Second Law becomes
æ ¶Ci ö 1 ¶(¶f D(¶Ci / ¶x)t
çè
÷ø =
¶t i f
¶x
Diagenetic Equation
• Thinking again about our
layer, conservation of
mass dictates that the
concentration of i will
change if the fluxes of i in
are different than the
fluxes out:
¶å J i
dCi
¶F
==- i
dt
¶x
¶x
o F is the sum of the three fluxes
• Concentration may also
occur as a result of
reactions involving i, Ri
occurring within the layer.
• Overall then
dCi
¶Fi
=+ å Ri
dt
¶x
Diagenetic Equation
•
•
•
dCi
¶F
= - i + å Ri
dt
¶x
This describes the changes in a
reference frame fixed to the layer.
We transform it to a reference frame
fixed to the sediment-water interface
by including the burial term:
dCi
¶C
æ ¶F ö
= - ç i ÷ - w i + å Ri
è ¶x ø t
dt
¶x
We can consider burial to also be a
flux. Combining the burial and flux
terms, we have
¶Fi
æ ¶Ci ö
=
+ å Ri
çè
÷
¶t ø x
¶x
•
Berner named this the Diagenetic
Equation.
Example: Reduction of
Sulfate
• Sulfate is abundant in seawater and will be reduced by
bacteria to sulfide once O2 is exhausted to metabolize
organic matter. We assume the (only) reaction is:
2CH2O + SO42- = H2S + 2HCO3• Assume the rate of reaction depends of availability of
organic matter:
1 d[CH 2O]
-
2
• Assuming steady-state
• Then
• Integrating:
dt
= k[CH 2O]
dCi
æ ¶C ö
=wç i÷
è ¶x ø t
dt
æ ¶[CH 2O] ö
-k[CH 2O] = w ç
è ¶x ÷ø t
[CH 2O](x) = [CH 2O]˚e- kx/w
Sulfate in Saanich Inlet
• Saanich Inlet is an anoxic
fjord in British Columbia.
• Since the consumption of
sulfate relates to the
consumption of organic
matter:
d[SO42- ] k[CH 2O] - kx/w
=
e
dt
2
• The sulfate is a dissolved
species, so we need to
consider diffusion.
Remembering the steady
state assumption, then:
æ ¶2 [SO42- ] ö
æ ¶[SO42- ] ö k[CH 2O]o - kx/w
fDç
-w ç
e
=0
2
÷
÷
2
è ¶x
ø
è ¶x ø t
• Integrating
w 2 [CH 2O]˚ kx/w
[SO ] =
(e -1)
2(w 2 + kD)f
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