che 553 lecture 9 statistical mechanics of adsorption 1
TRANSCRIPT
ChE 553 Lecture 9
Statistical Mechanics Of Adsorption
1
Objective
• Start to discuss thermodynamics of surface phase transitions
• Landau analysis – what phases do we see
• Solve one dimensional surface – what is phase behavior like
2
Background• So far we discussed simple adsorption
with no interactions between adsorbed species, complete mobility– Results in Langmuir behavior
• At saturation densities like liquid densities – Molecules interact
• Diffusion rates not always fast– Leads to defects in adsorbed layer
• Predicts no ordering of adsorbed layer
3
Real Surface systems Are Different
• Usually a series of phases that vary with coverage and temperature
4
P(2x2)
C(2x2)
(2x1)
(1x2)
i
1b
1a1a
2 2
3
33
4b
4a 4a
4b
5 5
1b 22
3
4b
4a4a
4b
55
i
1b
1a1a
2 2
3
33
4b
4a 4a
4b
5 5
1b 22
3
4b
4a4a
4b
55
i
1b
1a1a
2 2
3
33
4b
4a 4a
4b
5 5
1b 22
3
4b
4a4a
4b
55
i
1b
1a1a
2 2
3
33
4b
4a 4a
4b
5 5
1b 22
3
4b
4a4a
4b
55
Typical Surface phases
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Figure 4.22 The absorption of molecules in a P(2x2), C(2x2), (2x1) overlayer. The dark circles represent sites, the red circles represent adsorption on the sites.
Why Do These Phases Arise?
• Forces Between Adsorbates keep them apart– Direct forces (Van der waals repulsions)– Indirect forces (Adsorbate attracts local
electrons, depleting nearby regions for adsorption
6
Today
• Consider the effects of pair-wise interactions on the behavior of adsorbed layer– Qualitative features: Landau analysis– Quantifying results, analytically and with
monte carlo
• Try to predict surface phase behavior
7
Landau Analysis
• Attempt to predict what phases are produced as gas adsorbs on solid surfaces
8
P(2x2)
C(2x2)
(2x1)
(1x2)
i
1b
1a1a
2 2
3
33
4b
4a 4a
4b
5 5
1b 22
3
4b
4a4a
4b
55
i
1b
1a1a
2 2
3
33
4b
4a 4a
4b
5 5
1b 22
3
4b
4a4a
4b
55
i
1b
1a1a
2 2
3
33
4b
4a 4a
4b
5 5
1b 22
3
4b
4a4a
4b
55
i
1b
1a1a
2 2
3
33
4b
4a 4a
4b
5 5
1b 22
3
4b
4a4a
4b
55
Example:When will Each Of The Following Phases Form?
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Figure 4.22 The absorption of molecules in a P(2x2), C(2x2), (2x1) overlayer. The dark circles represent sites, the red circles represent adsorption on the sites.
Steps in Landau Analysis
• Make a list of possible arrangements of the adsorbate molecules to consider
• Calculate the energy of each arrangement at 0 oK– Usually use pairwise additive assumption to
calculate energy
• The structure with the lowest energy fills first, the one with the second lowest energy fills second, …, until the surface fills up
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Energy Expression
-Ei=H1+nnhnn+n2nnh2nn
Ei=Energy molecule on site
nn = Number of nearest
neighbors occupied
n2nn= = Number of second
newest neighbors occupied
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Solution
Random adsorption on a (1x1) layer: For random absorption the probability of any site being filled is .
the energy of a atom absorbed at site i, becomes
4 because 4 first and second nearest neighbors
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,iE
nn2nn1)1x1( h4h4HE(4.202)
Random Absorption On A C(2x2) Layer:
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With a C(2x2) layer there are no first nearest neighbors, but there are several second nearest neighbors. If it is assumed that the C(2x2) layer is partially filled, then
)22(21)22( 4 xcnnxc hHE 3
1)22( xc
(4.203)
Random Absorption On A P(2x2) Layer:
With a P(2x2) layer there are no first or second nearest neighbors
(4.204)
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1)2x2(p HE
Random Absorption On A (2x1) Layer:
With a (2x1) layer there are two first nearest neighbors and no second nearest neighbors
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)12(1)12( 2 xnnx hHE 2
1)12( x
Summary Of The Equations
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nn2nn1i h4h4HE
nn2h2HE 1i
(1x1)
C(2x2)
1i HE
nn1i hHE
(p2x2)
(2x1)
When Does The (1x1) Have The Lowest Energy?
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E(1x1)< E(2x1)
E(1x1)< EC(2x2)
E(1x1)< EP(2x2) 11
2121
2121
44
244
44
HhhH
hHhhH
hHhhH
nnnn
nnnnnn
nnnnnn
Solving hnn+h2nn>O hnn>O
Strong attractions and insufficiently weak
second nearest neighbors to prevent
1x1
When Does The P(2x2) Have The Lowest Energy?
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P(2x2)< E(2x1)
P(2x2)< EC(2x2)
P(2x2)< EP(2x2)
Solving h2nn<0 hnn<0
Occurs when First and second
neighbors are repulsive
What Happens If We Squeeze On More Molecules When P(2x2) Is Lowest Energy?
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Also Get Incommensurate Adsorption
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Figure 3.15 The domain wall structure of CO on Pt(100). (Proposed by Persson et al. [1990].)
Domain wallDomain wall
(2x2) Domain
Actual Phase Behavior More Complex
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Need Statistical Mechanics To Solve Real System
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Define occupancy number by:ξ=1 (occupied site) ξ =-1 (empty site)
Partition function definition:
Combining with expression
nEexpQS
11211
S
iiiiiSS
FhexpHexpQ
S 111 2411211
(4.78)
(4.67)
Analytical Solution Of Equations In One Dimension
• Assume a circular chain with So sites
• Calculate partition function analytically (see p 270 in Adsorption)
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Qso=exp(Hso){(1)So+(2)So}
h) (-0.5+F) ( h) (0.5+F) ( )4
h( = 2
1 expsinhexpcoshexp
h) (-0.5+F) ( h) (0.5-F) ( )4
h( = 2
2 expsinhexpcoshexp
(4.103)
(4.104)
Calculate Coverage Analytically
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)4h
exp(- + F)(sinh
F)sinh(
2
2S +
2S = > N < 00
(4.115)
Fig 4.17 A series of isotherms calculated from Equation 4.115 for βh = 3, 2, 1, 0.5, 0, -0.5, -1. The Langmuir
line (βh = 0) is shaded.
Replot vs. Pressure
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Figure 4.18 A replot of the data from Figure 4.17 versus dimensionless pressure.
Fig 4.17 A series of isotherms calculated from Equation 4.115 for βh = 3, 2, 1, 0.5, 0, -0.5, -1. The
Langmuir line (βh = 0) is shaded.
Key Features
• Characteristic S like behavior
• No first order phase transitions
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Fig 4.17 A series of isotherms calculated from Equation 4.115 for βh = 3, 2, 1, 0.5, 0, -0.5, -1. The Langmuir
line (βh = 0) is shaded.
Real Surfaces Show Phase Transitions
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Need Monte Carlo Calculation
Summary
• Can use statistical mechanics to calculate surface phase behavior
• Qualitatively – almost Langmuir behavior if only nearest neighbor interactions in 1 D
• More complex in 2 D.
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