Adsorption, Partitioning and Interfaces
Accumulation of Solutes at
Interfaces
Air
KH = PoL/Csatw
Kow = Csato/Csatw
Koa = Csato/PoL
A gas is a gas is a gas
T, P
Koa
KH
Po L
Octanol
Water
Fresh, salt, ground, pore
T, salinity
Csatw
Kow
Pure Phase
(l) or (s)
Ideal behavior
NOM, biological lipids,
other solvents
Csato
Equilibrium partitioning between water and any organic
liquid:
Salinity will increase tendency to partition into the organic phase by
decreasing the solubility (increasing the activity coefficient) of the
solute in water.
Assuming that salts are largely insoluble in the organic phase.
Partitioning is driven by
Van der Waal forces
polarity/polarizability
H bonding
The better the match between chemical properties of solute and
solvent, the higher the equilibrium constant for better extraction
solvent
phase change costs cancel
LFERs for relating different organic
solvent-water systems
IF the two solvents are similar, then simple
linear FER can be used for a series of similar
compounds:
log Ki1w  a  log Ki 2 w  b
For example, hexadecane and octanol:
log Kihw  1.21  log Kiow  0.43
Works well for apolar and weakly polar solutes.
Does not work for very polar compounds, such as phenols
fig 7.1
Octanol-water partition coefficient:
K ow 
C o (mol / Lo )
C w (mol / Lw )
Importance
Huge database of Kow values is available
Method of quantifying the hydrophobic character of a compound.
Can be used to estimate aq. solubility
Can be used to predict partitioning of a compound into other
nonpolar organic phases:
other solvents
natural organic material (NOM)
biota (like fish, cells, lipids, etc.)
Why octanol?
Has both hydrophobic and small hydrophilic character due to OH bond
Therefore a broad range of compounds will have measurable Kow values
Methods of measuring Kow
Shake flask method (compounds with log Kow < 5)
Solute is equilibrated between the two phases by shaking:
octanol
water
Generator column method
octanolsaturated
water
glass beads coated with
compound of interest
solid
adsorbent
Chromatographic data
Relate Kow of known compounds to capacity factor,
k', on a reverse-phase C18 HPLC column
Measure k' for known compounds, develop
relationship between Kow and k'
Adsorption Equilibrium
• Adsorption vs. Absorption
– Adsorption is accumulation of molecules on a
surface (a surface layer of molecules) in
contact with an air or water phase
– Absorption is dissolution of molecules within a
phase, e.g., within an organic phase in contact
with an air or water phase
Adsorption
PHASE I
‘PHASE’ 2
Absorption (“partitioning”)
PHASE I
PHASE 2
Pgas  K H caq
Henry’s Law
Adsorption terminologies
Cu2+
Cu2+
Cu2+
Cu2+
Cu2+
Cu2+
Cu2+
Cu
2+
Cu2+
Adsorbed species present
at an overall concentration
of cCu(ads)
Dissolved
adsorbate, at
concentration cCu(aq)
Cu2+
Cu2+
Cu2+
Adsorbent, in
suspension at
concentration
csolid
Adsorbed species,
with adsorption
density q mg Cu per
g solid or per m2
Surface area per
gram of solid is the
specific surface area
 mg adsorbed 
 mg adsorbed 
 g solid per 
ci ,ads 
  qi 
 csolid 

per
L
of
solution
per
g
adsorbent
L
of
solution






Mechanisms of Adsorption
• Dislike of Water Phase – ‘Hydrophobicity’
• Attraction to the Sorbent Surface
– van der Waals forces: physical attraction
– electrostatic forces (surface charge interaction)
– chemical forces (e.g., - and hydrogen bonding)
Adsorbents in Natural & Engineered Systems
Natural Systems
Sediments
Soils
Engineered Systems
Activated carbon
Metal oxides (iron and aluminum as coagulants)
Ion exchange resins
Biosolids
Engineered Systems
• Activated carbon (chemical functional
groups)
– Adsorption of organics (esp. hydrophobic)
– Chemical reduction of oxidants
• Metal oxides (surface charge depends on
pH)
– Adsorption of natural organic matter (NOM)
– Adsorption of inorganics (both cations &
anions)
• Ion exchange resins
– Cations and anions
– Hardness removal (Ca2+, Mg2+)
– Arsenic (various negatively charged species),
NO3-, Ba2+ removal
Adsorptive Equilibration in a Porous
Adsorbent
Pore
Early
Later
Laminar
Boundary
Layer
GAC Particle
Adsorbed Molecule
Diffusing Molecule
Equilibrium
Adsorption Isotherms
Add Same Initial Target Chemical Concentration, Cinit, in each
Control
Different activated carbon dosage, Csolid, in each
 mg  cinit  c fin  mg/L 
q fin 

csolid  g/L 
 g 
An adsorption ‘isotherm’ is a q vs. c
relationship at equilibrium
Theory of adsorption
Gibb’s free energy at the interface
dG (s )  S(s ) dT    i(s ) dn i(s )  dA
i
Integrating at constant temperature,
G (s )    i(s ) n i(s )  A
i
Taking the total derivative,
dG (s )    i(s ) dn i(s )   n i(s ) d i(s )  dA  Ad
i
i
(s )
(s )
n
d

 i i  Ad  0
i
d   d i( s )  2 d (2s )
Or
For
i
 (2s )  RT ln C(2s )
The Gibb’s adsorption isotherm can be written as
C (2l) d
d
2  

(l)
RT dC (2l )
RTd (ln C 2 )
Accumulation of solutes at the interface lowers the interface tension.
Adsorption at Solid-Liquid and Solid-Gas
Interfaces
Langmuir isotherm, gas/solid:
rate of evaporation = k1S1
rate of condensation = k2p(S – S1)
At steady-state
dS1
 k 2 p(S  S1 )  k 1S1
dt
S1 = occupied sites
k 2 pS  k 2 pS1  k 1S1  0
k2
p
S1
k1
bp
 

k2
S
1  bp
1
p
k1
Langmuir Isotherm, liquid/solid:
backward rate of desorption = k1S1
forward rate of adsorption = k2C(S – S1)
dS1
 k 2 C(S  S1 )  k 1S1
dt
At steady-state
bC e
q
 0 
1  bC e
Q
Q 0 bC e
q
1  bC e
k 2 C e S  k 2 C e S1  k 1S1  0
k2
Ce
bC e
S
k1
 1 

k2
S
1  bC e
1
Ce
k1
BET Isotherm – does not assume monolayer coverage
V
cX


Vm (1  X )(1  (c  1) X )
 H ads  H vap 

c  exp
RT


Freundlich Isotherm – empirical
q e  K f C en
1
Commonly Reported Adsorption Isotherms
q  klin c
n
Freundlich: q  k f c
Linear:
Langmuir:
K Lc
q  qmax
1  K Lc
Steps in Preparation of Activated Carbon
• Pyrolysis – heat in absence of oxygen to form graphitic char
• Activation – expose to air or steam; partial oxidation forms
oxygen-containing surface groups and lots of tiny pores
• Starting materials (e.g., coal vs. wood based) and
activation
• Pores and pore size distributions
• Internal surface area
• Surface chemistry (esp. polarity)
• Apparent density
• Particle Size: Granular vs. Powdered (GAC vs. PAC)
Characteristics of Some Granular Activated
Carbons
Characteristics of Activated Carbons (Zimmer, 1988)
F 300
H 71
C25
Bituminous Coal
Lignite
Coconut Shell
Bed Density, ρF (kg/m3)
500
380
500
Particle Density, ρP (kg/m3)
868
685
778
Particle Radius (mm)
0.81
0.90
0.79
Surface Area BET (m2/g)
875
670
930
0.33
0.21
0.35
----
0.38
0.14
----
0.58
0.16
----
1.17
0.65
Activated Carbon
Raw Material
Pore Volume (cm3/g)
Micro-
( radius < 1nm)
Meso-
(1nm < r < 25nm)
MacroTotal
(radius > 25nm)
Oxygen-Containing Surface Groups
on Activated Carbon
Mattson and Mark, Activated Carbon, Dekker, 1971
Kinetics of Atrazine Sorption onto
GAC
167 mg GAC/L
333 mg GAC/L
Metal Oxide Surfaces
Coagulants form precipitates of Fe(OH)3 and Al(OH)3
which have –OH surface groups that can adsorb humics
and many metals
Humic substances where R is organic
Sorption of NOM on Metal Oxide
Example. Adsorption of benzene onto activated carbon has been
reported to obey the following Freundlich isotherm equation, where c is
in mg/L and q is in mg/g:
0.533
qbenz  50.1 cbenz
A solution at 25oC containing 0.50 mg/L benzene is to be treated in a batch
process to reduce the concentration to less than 0.01 mg/L. The adsorbent is
activated carbon with a specific surface area of 650 m2/g. Compute the required
activated carbon dose.
Solution. The adsorption density of benzene in equilibrium with ceq of 0.010 mg/L
can be determined from the isotherm expression:
0.533
qbenz  50.1 cbenz
 4.30 mg/g
A mass balance on the contaminant can then be written and solved for the
activated carbon dose:
ctot ,benz  cbenz  qbenz cAC
0.50  0.010   4.30 mg/g  c AC
cAC  0.114 g/L  114 mg/L
Example If the same adsorbent dose is used to treat a solution containing 0.500
mg/L toluene, what will the equilibrium concentration and adsorption density be?
The adsorption isotherm for toluene is:
0.365
qtol  76.6 ctol
Solution. The mass balance on toluene is:
ctot ,tol  ctol  qtol cAC
0.50  ctol   76.6 ctol 0.365   0.114 g/L 
ctol  3.93x104 mg/L
Example
The following data were gathered in a laboratory study of the
adsorption of trichlorophenol (MW= 197.5 g/mol) on a commercial
powdered activated carbon for the following conditions:
Initial concentration of solute: 470 mg/L
Volume of solvent: 500 mL
Carbon Dose (mg)
Ce (mg/L)
13.4
4.1
9.3
6.2
5.6
27.9
3.3
91.1
2.0
168.0
Fit these data to the Langmuir and Freundlich isotherms, and
determine the constants of both the models. Which model fits
better?
Example
• To a 1 L container holding 0.01mg/L carbontetrachloride, 25 mg/L
of activated carbon was added. Determine the equilibrium
concentration of tetrachloride in the water if the Fruendlich isotherm
parameters for this carbon are given as: Kf = 11 (mg/g) (L/mg) 1/n
and 1/n =0.83.
Process Design Parameters
• Contactors provide large surface area
• Types of contactors
– Continuous flow, slurry reactors
– Batch slurry reactors (infrequently)
– Continuous flow, packed bed reactors
• Product water concentration may be
– Steady state or
– Unsteady state
Powdered Activated Carbon (PAC)
PAC +
Coagulants
Settled
Water
Sludge Withdrawal
PAC particles may or
may not be
equilibrated
PAC +
Coagulants
Flocculated
Water
Process Operates at Steady-State, cout = constant in time
Packed Bed Adsorption
v, cIN
Natural Packed Bed – subsurface
with groundwater flow
Engineered Packed Bed- granular
activated carbon
EBCT = empty-bed contact time (Vbed/Q)
Adsorptive capacity is finite (fixed
amount of adsorbent in bed)
v, cOUT
Process operates at unsteady state, cOUT
must increase over time
Linear Equilibrium Partitioning Between Two
Phases
p  pA  pb  
n A RT
pA 
V
pA
yA 
 mole fraction of component A in the gas phase
P
p A  x A p 0,A
cA
xA 
c
Henry’s Law: pA=HxA
Air–water partitioning coefficient:
K AW
c air
 A
cA
Octanol–water partitioning coefficient:
;
K OW
oct
A
c

cA
Example
• A tube containing 50 mL of octanol, 50 mL of
water, and 0.5 g of DDT is shaken until the DDT
reaches its equilibrium concentration in the
octanol and water phases. What is the
concentration of DDT (mg/L) in the water phase
when the Kow of DDT is 6.19?
Partitioning and Separation in Flow
Systems
(1)
2)
dc sys
dc (out
2)
V2
 V1
 Qc in( 2 )  Qc (out
dt
dt
(1)
2)
dc sys
 K 12dc (out
2)
2)
dc (out
dc (out
2)
V2
 V1K12
 Qc in( 2)  Qc (out
dt
dt
2)
dc (out
Q
Q
2)

(c in( 2)  c (out
)
dt
V2  V1K12
V2
2)
dc (out
1 ( 2) ( 2)

(c in  c out )
dt
tR
2)
c (out
 t 
 1  exp  
co
 tR 
1
1  K12
V1
V2
2)
(c in( 2)  c (out
)

V1

R  1  K12

V2




