Lecture Notes (PDF) - Aqueous and Environmental Geochemistry

Environmental Geochemistry
DM Sherman, University of Bristol
2014/2015
Kinetics and Geochemical
Dynamics
Environmental Geochemistry
D.M. Sherman, University of Bristol
Open Systems
A system (e.g., a geochemical reservior) is open if it
can exchange matter with the surroundings.
Atmosphere
Hydrothermal
processes
Ppt’n
Hydrosphere
Evap’n
Weathering
Chemical
Sedimentation
Nutrient
Uptake
Respiration
Photosynthesis
Biosphere
Biomineralization
Volcanic Emission
Lithosphere
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Environmental Geochemistry
DM Sherman, University of Bristol
2014/2015
Simple Box Model!
Input
Reservoir
Output
F in = QC in
M res = VC res
F out = QC out
F = flux of chemical species (mass/time)
Q = flow rate (volume/time)
V = volume of reservoir
C = concentration (mass/volume)
Simple Box Model!
Input
Reservoir
Output
F in = QC in
M res = VC res
F out = QC out
The system will be in steady state when F in = F out
The residence time (t R ) of a chemical species is
the average time an atom of that species is in the
reservoir.
t R = V resC res
F out
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Environmental Geochemistry
DM Sherman, University of Bristol
2014/2015
Residence Times of Chemical Species!
Water in the Oceans:
Volume of Oceans (V ocean ) = 1.37 x 10 21 L
Global Riverine Input (F Riv ) = 3.6 x 10 16 L/y
t H oceans = V oceans
F Riv
= 1.37 x1021 L
3.6 x10 16 L / yr = 3.81x 104 years
€
Residence Times of Chemical Species!
Sodium in the Oceans:
Concentration in Oceans (C ocean ) = 0.47 mol/L
Avg. Conc. in Rivers = 2.2 x 10 -4 mol/L
oceans = C Na
oceans
=
Na F River
t Na
C River
V oceans
(0.47mol / L)(1.37 x10 21 L)
(2.2x10 −4 mol / L)(3.6 x10 16 L / yr ) = 8.1 x 107 years
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Page 3

Environmental Geochemistry
DM Sherman, University of Bristol
2014/2015
Non-Steady State:
Conservative (Non-reactive) Species!
Input
Reservoir
Output
F in = QC in
M res = VC res
F out = QC res
Conservation of mass requires that:
dM res
dt
= F in − F out
V dC res
dt
= Q(C in − C res )
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Conservative (Non-reactive) Species!
Input
Reservoir
Output
F in = QC in
M res = VC res
F out = QC res
Rearranging gives:
V dC res
+ C
Q dt res = C in
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Page 4

Environmental Geochemistry
DM Sherman, University of Bristol
2014/2015
Conservative (Non-reactive) Species!
Input
Reservoir
Output
F in = QC in
M res = VC res
F out = QC res
Suppose we shut off the input (set C in = 0) and
let the system evolve:
V dC res
+ C res = 0
Q dt
C res = C 0
⎛
res exp −t ⎞
⎜ ⎟
⎝ V / Q⎠
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€
Conservative (Non-reactive) Species!
The quantity V/Q is simply the residence time of
water in the reservoir. This will be designated t H
⎛
C res = C 0 exp −t ⎞
⎜ ⎟
⎝ ⎠
t H
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Environmental Geochemistry
DM Sherman, University of Bristol
2014/2015
Box Models (Example: Hg Cycle)!
Atmosphere (40)
Ppt’n (167)
Volatilization (167)
Ppt’n (83)
Volatilization (83)
Oceans
(42.5 x10 5 )
Rivers (13)
Continental Crust
(1x10 5 )
Deposition (13) Uplift (13)
Sediments (3.3 x10 9 )
Fluxes in 10 8 Gt/y; Masses in 10 8 Gt
Box Models (Example: Hg Cycle)!
What would be the
effect on atmospheric
Hg if we perturb the
steady state system by
the anthropogenic flux
of 102 x 10 8 g/yr?
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Environmental Geochemistry
DM Sherman, University of Bristol
2014/2015
Box Models (Example: Hg Cycle)!
Atmosphere (40)
Ppt’n (F out )
Volatilization (F in )
Land+Oceans
Hg
dM Atm
dt
= F in
− F out
If we assume first-order kinetics for F out :
Hg
dM Atm
dt
€
Hg
= F in
− F out
= F in
− kM Atm
€
Box Models (Example: Hg Cycle)!
Atmosphere (40)
Ppt’n (F out )
Volatilization (F in )
Land+Oceans
Hg
dM Atm
dt
Hg
= F in
− kM Atm
Solving for M atm (t) gives:
€
Hg (t) = F in
k (1− e−kt ) + M atm
M atm
Hg (0)e −kt
€
Page 7

Environmental Geochemistry
DM Sherman, University of Bristol
2014/2015
Box Models (Example: Hg Cycle)!
Atmosphere (40)
Ppt’n (250)
Volatilization (250)
Land+Oceans
We can evaluate the rate constant k: under steady state
conditions, F in - F out = 0 so that
k =
F in
Hg
M atmos
= 250 x108 g / y
40 x10 8 g
= 6.25 y −1
€
Box Models (Example: Hg Cycle)!
We can now evaluate the effect of the anthropogenic
perturbation on input flux:
Atmosphere (40)
Ppt’n Volatilization (250 + 102 Anthropogenic)
60
Land+Oceans
M Hg
/10 8 g
50
40
30
20
10
0
0 0.2 0.4 0.6 0.8 1
Time (years)
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Environmental Geochemistry
DM Sherman, University of Bristol
2014/2015
Reactive Species in Open Systems
Chemical Equilibrium vs. Steady State!
The link between steady state and
chemical equilibrium is kinetics
Transition State Theory: Energy Barriers
and Rate Constants
k = k B T
h e−ΔG± /RT
Page 9

Environmental Geochemistry
DM Sherman, University of Bristol
2014/2015
Transition State Theory: Reaction Rate
and Disequilibrium
Rate net
= Rate forward
− Rate reverse
€
For an elementary reaction
A → B
Rate forward
= k f
[A] = [A] k T B ±
h e−ΔG f
/RT
Rate reverse
= k r
[B] = [B] k BT
h e−ΔG r ± / RT
€
Transition State Theory: Reaction Rate
and Disequilibrium
Since
ΔG = ΔG f ± − ΔG r
±
Rate forward
± = e−ΔG f
/RT
Rate reverse e −ΔG r
± /RT = e−ΔG/RT
Rate net
= Rate forward
(1− e −ΔG / RT )
€
Rate net
= Rate forward
(1− Q K )
Note: don’t confuse this
Q with the discharge Q.
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Page 10

Environmental Geochemistry
DM Sherman, University of Bristol
2014/2015
d[A]
dt
Integrating Rate laws..!
Suppose we have a simple first-order rate law for
the transformation of A:
= −k f
[A]
Multiply both sides by dt and rearranging gives..
€
dA = −k f
[A]dt
€
d[A]
[A]
= −k f
dt
€
Integrating Rate laws (cont)..!
Now we integrate both sides. We have our initial
condition that at t = 0, [A] = [A] 0 .
A d[A] t
∫ = −kf ∫ dt
[A] 0
A 0
€
⎛
ln [A] ⎞
⎜ ⎟ = −k f
t
⎝ [A] 0 ⎠
€
[A] = [A] 0
e −k f t
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Page 11

Environmental Geochemistry
DM Sherman, University of Bristol
2014/2015
Half-Life!
The half-life of A is the time it takes for half of A
to disappear (no reverse reaction):
If
A = A o
e −k f t
€
Then when A = A 0 /2:
t 1/ 2 = 0.693/ k f
€
Half-Life!
Process
Ion complexation
Adsorption
Gas dissolution
Hydrolysis
Precipitation/Dissolution
Mineral Recrystallization
Half-life

Environmental Geochemistry
DM Sherman, University of Bristol
2014/2015
Reactive Species in Open Systems
Chemical Equilibrium vs. Steady State!
Input
C in = 0
Reservoir
rate = k(C sat
−C)
Output
F out = QC
A mineral dissolves in the reservoir with a rate = k(C sat -C)
€
Conservation of mass gives us:
V dC
dt
= Vk(C sat
−C) −QC
€
Reactive Species in Open Systems
Chemical Equilibrium vs. Steady State!
Input
C in = 0
Reservoir
rate = k(C sat
−C)
Output
F out = QC
At steady state:
€
V dC = Vk(C sat
−C) −QC = 0
dt
€
C =
kC sat
k + (Q /V)
Page 13

Environmental Geochemistry
DM Sherman, University of Bristol
2014/2015
Reactive Species in Open Systems
Chemical Equilibrium vs. Steady State!
Input
C in = 0
Reservoir
rate = k(C sat
−C)
Output
F out = QC
t 1/2
= ln(2) / k = time for reaction to go 1/2 way
€
t H
=V / Q = residence time of water
⎛ t
C = C sat
1+ 1/2
⎞
⎜
⎝ (ln2)t
⎟
H ⎠
−1
Hence, C will only
achieve chemical
equilibrium if t H >> t 1/2 .
Reaction Rates and H 2 O Residence
Times!
Page 14