SURFACE CHEMISTRY

CHAPTER 8

What is a surface?

Definition: A surface is the interface between two phases of matter

Sometimes the word interphase is used.

Generally, the word surface is used to refer to an interface of which one phase is gaseous and the other is condensed .. i.e, solid/gas and liquid/gas interfaces

The physicochemical nature of a surface is different from that either of the two associated bulk phases.

Surface Chemistry

Is a study of physical and chemical phenomena that occur at the interface of two phase:

i) gas/solid ii) liquid/solid

The boundary that separate the two phases is called “surface” or “interface”.

The adhesion of gas or liquid molecules to surface is known as adsorption.

Common Features of Adsorption

1. Highly selective (amount depends): i) nature of adsorbent surface ii) previous treatment of adsorbent surface iii) nature of adsorbed substance (adsorbate) 2. A rapid process rate increases with increase in temperature rate decreases with increase in amount

adsorbed 3. A spontaneous process It is a exothermic reaction.

Adsorbent

• Adsorbents are used usually in the form of spherical pellets, rods, moldings, or monoliths

• Hydrodynamic radius between 0.25 and 5 mm.

• Have high abrasion resistance, high thermal stability and small pore diameters

- higher exposed surface area and hence high capacity for adsorption.

• The adsorbents must also have a distinct pore structure that enables fast transport of the gaseous vapors.

Adsorbent

Characteristics of good absorbent:

• High surface area with proper pore structure and size distribution

• Good mechanical strength

• Possess thermal stability

Adsorbent Most industrial adsorbents fall into one of 3 classes:

• Oxygen-containing compounds

Hydrophilic and polar, including materials such as silica gel and zeolites.

• Carbon-based compounds Hydrophobic and non-polar, including materials such as activated carbon and graphite.

• Polymer-based compounds

Polar or non-polar functional groups in a porous polymer matrix. http://en.wikipedia.org/wiki/Adsorption

Adsorption of Gas and Vapours on Solids

When gas/vapour (adsorbate) is in contact with clean solid surface (adsorbent) – it get attached to the surface.

Surface (adsorbent)

Gas (adsorbate)

Adsorbed layer

Types of Adsorption

1. Physical adsorption (Physisorption)

- adsorbate and surface of adsorbent interact by van der Waals forces

- adsorbed molecules weakly bound to the surface

- heat of adsorption are low.

Types of Adsorption

2. Chemical adsorption (Chemisorption)

- adsorbate react chemically with surface (adsorbent)

- bonds are broken and formed – usually covalent or dative bonds

- heat of adsorption has same range of values as chemical reaction.

Characteristics of physisorption and chemisorption

Physisorption Chemisorption

The adsorbed layer vary in thickness from monolayer to multilayer. Van der waal forces can extend from one layer of molecules to another

The surface become saturated after is been covered with a single layer of adsorbed molecules. Only a monolayer can be formed

Amount of adsorption on surface is a function more of adsorbate that adsorbent

Amoung of adsorption characteristic of both adsorbate and adsorbent

The attainment of equilibrium is rapid. There is no activation involved

Very slow. It is a specific process which requires activation energy

Occur more readily at lower temperatures

Occurs at high temperatures

Characteristics of physisorption and chemisorption (cont)

Physisorption Chemisorption

Van der Waals interactions are weak. ΔHp is the same magnitude as enthalpy of condensation/vaporisation. (~ -

40kJmol-1 )

ΔHc > ΔHp (~ -200kJmol-1)

The amount adsorbed increases with increase in pressure of adsorbate. The process can be reversed

Amount adsorbed decreases with increase in pressure of adsorbate. Not readily reversible

More common Found only when the adsorbent and adsorbate tend to form a compound

Potential Energy Diagram

1. As adsorbate approaches adsorbent, physisorption takes place.

Dispersion

forces

Surface Held loosely

Physisorption

When temperature high enough and Ea is overcome, adsorbate approaches closely, stronger bond formed, chemisorption takes place.

Chemisorption

2. Chemisorption Curve

- characterize by deep minimum

∆Hc > ∆Hp

- chemisorption is not easily reversed due to

stronger bonding near surface.

3. Physisorption Curve

- first physically adsorbed

- involves approaching solid surface along a

low energy path.

- shallow ∆Hp - weak bond (dispersion forces)

- transition from physisorption takes place at

point where curve for C and P intersect.

POTENTIAL ENERGY DIAGRAM: PHYSISORPTION AND CHEMISORPTION CURVE

Po

ten

tial E

nerg

y

Distance between adsorbate and adsorbent

Enthalpy of

chemisorption

Activation

energy

Enthalpy of

physisorption

Enthalpy of Adsorption ∆G = ∆H - T∆S ∆G = Gibbs free energy

∆H= enthalpy of adsorption

∆S = entropy

For adsorption to occur at an appreciable extent, ∆G = - x

At adsorption, molecules moving from chaotic to orderly state, ∆S = -y

For ∆G = - x, ∆H has to be negative (exothermic reaction)

Heat is released when adsorption takes place.

Adsorption Isotherms

Graphical presentations of relationship between the amount of gas adsorped on surface or the fractional coverage of surface (σ) and the pressure of gas at constant T.

Types of Isotherms

1. Chemisorption eg. H2 on Cu powder at 25oC.

Vo

lum

e a

dsorb

ed/c

m3(S

TP

)g-1

Pressure, bar

Strong tendency of surface to bind to

gas molecules

Saturation of adsorbing surface

Types of Isotherms

2. Physisorption eg. N2 on silica V

olu

me a

dsorb

ed/c

m3(S

TP

)g-1

Relative Pressure, P/Po

Pressure=vapour pressure

Condensation occurs

Increase in gas pressure,

Large increase in amount of gas

adsorbed

0 1

Fractional coverage (σ)

Is the extent of adsorption or extent of surface coverage

σ = number of adsorption sites occupied

number of adsorption sites available

Rate of adsorption

Rate of change of surface coverage is determined by change of fractional coverage with time.

Extent of adsorption depends:

1.Temperature

2.Pressure of adsorbate

3.Effective surface area of adsorbent

Before a reaction can take place on the surface of adsorbent in a heterogenous system (g/s, l/s) 1. Adsorbate molecules must be adsorbed

2. Dynamic equilibrium exists

Molecules in ⇄ molecules

bulk medium adsorbed on surface of

adsorbent

Examples of Well Established Isotherms

1. Langmuir Isotherm

2. Freundlich Isotherm

3. BET Isotherm

Langmuir Model of Adsorption

Assumptions:

1. Homogeneous surface – every site has same energy

2. Only adsorbate-adsorbent interactions considered

3. Adsorption limited to a single monolayer.

Langmuir Model of Adsorption

Elastic collision

Adsorption

Desorption

Langmuir Isotherm Kinetic Derivation of Langmuir Adsorption Eq.

Derivation 1:

σ = fraction of surface covered with molecules

(1- σ) = remaining fraction still available for further adsorption

ρ = given pressure of gas

Rate of molecules adsorbed on surface= kads ρ(1- σ) kads = proportionality constant

Derivation 2:

Rate molecule leaving the surface increase the larger the no. of molecules on the surface – proportional to σ (proportion of surface covered with molecules)

Rate of desorption= kdes σ

kdesis a second proportionality constant

Derivation 3:

At dynamic equilibrium:

Rate of adsorption = Rate of desorption

kads ρ(1- σ) = kdes σ

Divide above by kdes and let kads/kdes = K

Kρ(1- σ) = σ

Rearrange:

K ρ- K ρ σ = σ

Kρ = σ +Kρ σ

= σ (1 + Kρ )

σ = K ρ

1 + Kρ

Derivation 4:

i) Low pressure, kρ is very small

(1 + Kρ) ~ 1

σ = K ρ (σ ρ, corresponds initial steep

rise of the isotherm curve)

i) High pressure, Kρ >> 1

(1 + Kρ) = Kρ

σ = K ρ = 1 (σ is independent of change in

K ρ pressure, and approaching unity)

Chemisorption type – surface becomes saturated with molecules at high pressure.

At adsorption equilibrium

Where σ = V/Vm is the coverage (≤ 1)

Rate ads = Rate des

1.0

σ

0 Pressure

Note limits:

High ρ σ = 1

Low ρ σ = k ρ

Langmuir Isotherm without dissociation

1. As σ increases with increasing ρ, approaches σ=1 at very high pressure, gas molecules squashed to all available site of surface.

2. Different curves with different k values at different temperature (k is temp dependent)

Fractional

Coverage σ

k = 0.1 atm-1

k = 1 atm-1

k = 10 atm-1

Pressure

Practice Exercise

The adsorption of nitrogen on Alumina is described by the Langmuir Isotherm at 25oC with a constant k = 0.68 kPa-1. Calculate the pressure at which the fractional surface coverage is 0.17.

Solution

σ = kρ

1 + kρ

0.17 = 0.68 ρ

1 + 0.68ρ

0.17(1 + 0.68ρ) = 0.68ρ

0.17 + 0.12 ρ = 0.68ρ

ρ = 0.17/0.56 = 0.30 kPa

Derivation 5: Experimental Isotherm data.(Just for Monolayer adsorption)

σ = V V=amount of gas adsorbed at pressure ρ

Vm Vm = amount of gas to form monolayer

(complete coverage)

V = 1 + kρ ∴ V = Vm (1 + kρ)

Vm

Rearrange: (linearised Langmuir Equation)

ρ = 1 ρ + 1 since y = mx + c

V Vm kVm plot graph, y = ρ/V, x = ρ

m = 1/Vm intercept = 1/kVm

y m x c

Plot of ρ/V Vs ρ

If data fit Langmuir theory, a straight line is obtained.

ρ/V

1

kVm

y

x

y = 1

x Vm

ρ

Deviation of Langmuir Plots

Plots may deviate from linearity at high concentration or high pressure due to:

1. Isotherm derived for adsorption of only monolayer coverage of adsorbate by chemisorption

2. High conc. or high pressure implies that layers of physisorbed adsorbates have formed on top of chemisorbed monolayer, forming multilayer adsorption pattern.

Practice Exercise

Refer to Pg 10, example 7.1

1. To confirm the data fit Langmuir isotherm

2. To determine the constants k, Vm

3. To determine the number of molecules involved for complete coverage

Solution Since x = ρ, y = ρ/V

Treat the data first.

ρ

(x-

axis)

100 200 300 400 500 600 700

V 10.2 18.6 25.5 31.5 36.9 41.6 46.1

ρ/V

y-axis

9.8 10.8 11.8 12.7 13.6 14.4 15.2

A straigth line confirms data fits Langmuir Isotherm

Langmuir Isotherm

0

2

4

6

8

10

12

14

16

100 200 300 400 500 600 700

p

p/V p/V

m (slope)= 0.009

1/Vm = 0.009

Vm = 111cm3

c(y-intercept) = 7.8

c = 1/kVm

k= 1/cVm

k = 1/7.8 x 111

= 1.16 x 10-3 Torr-1

To determine number of molecules for complete coverage

Molar Volume (at 273K) = 22.4 dm3/mol

= 22.4 x 103 cm3/mol

Since Vm = 111 cm3

∴111 cm3 gas required for complete coverage

nCO = 111 cm3 /22.4 x 103 cm3/mol

= 4.96 x 103 mol

No of CO molecule = nCO x NA

= 4.96 x 103 mol x 6.02 x 1023 molecules/mol

= 2.98 x 1021 molecules

Freundlich model of Adsorption

- Is a non ideal adsorption and deviate from Langmuir Isotherm:

1. Surface is not uniform

2. Interaction between adsorbed molecules

3. When a molecule is attached to a surface – it is difficult for another molecules to be attached to neighboring site

4. Energetically most favorable sites occupied first

5. Isotherm fails if concentration or pressure of adsorbate is too high.

Freundlich Equation X = mass of gas adsorbed by solid

m = mass of adsorbent

k,n = constant for given gas & solid

1/n unity

Since σ = V/Vm

Rearrange equation:

Log the equation

ln V = lnk + 1 lnp lnV = lnVmk + 1 lnp

Vm n n

y c m x

X = σ =kp1/n

m

V = kp1/n

Vm

X =kp1/n

m

Graph for Freundlich Isotherm

ln V

ln p

Slope = 1/n

lnVmK

BET Model of Adsorption

Assumptions:

1. Homogeneous surface – every site has same energy

2. Only adsorbate-adsorbent interactions considered

3. Multilayers are possible

4. Adsorbates adsorb/desorb one at a time

5. First layer different from the (liquid-like) others.

BET Model of Adsorption

Elastic collision

Adsorption

Desorption

BET Isotherm

Deals with multilayer adsorption on non-porous solid surfaces. The entire adsorption process include:

1. Attachment of molecules to sites on the solid surfaces – mostly by chemisorption

2. Attachment of molecules to site already occupied by adsorbed molecules – by physisorption

BET Isotherm At high presures the isotherm rise indefinitely:

V a

ds

ρ ρo

Vm

BET equation P Po = 1 + P

V(Po- P) VmK Vm

V = vol. of gas adsorped at pressure P

Vm = monolayer capacity

(vol. of gas required to cover unit mass of

adsorbent with a complete monolayer)

Po = saturation vapour pressure of adsorbate

K = equilibrium constant of adsorption

P = pressure of adsorbate

Application of BET equation

Rearrange BET equation: P

P = 1 + 1 P V(Po- P)

V(Po- P) VmKPo Vm Po

y b m x

A linear graph is obtain. P/Po

Vm calculate from the slope, and enable specific area for solid to be calculate.

K calculated from the intercept.

Exercise

P/kPa 0.160 1.87 6.11 11.67 17.02 21.92 27.29

V/mm3 601 720 822 935 1046 1146 1254

1. The following data were obtained for the adsorption of N2 on

1 g of TiO2 at 75 K.

At 75K, Po= 76 kPa and the volume V, in the table has been corrected

to 1 atm and 273 K (at STP). Confirm that the data given in the table fit

a BET isotherm in the range of pressures reported and evaluate Vm

which is the volume of gas required to cover unit mass of the adsorbent

with a complete monolayer.

Solution

P/kPa 0.160 1.87 6.11 11.67 17.02 21.92 27.29

V/mm3 601 720 822 935 1046 1146 1254

P/V(Po-P)

x 10-5

(y-axis)

0.35 3.50 10.6 19.4 27.6 35.4 44.7

P/Po x 10-3

(x-axis)

2,10 24.6 80.4 153.6 223.9 288.4 359

Graph: plot {P/V(Po-P)} vs (P/Po) and obtain a straight line, confirm the data fit BET

isotherm.

Slope = 1/Vm = 1.23 x 10-3 mm-3 Vm = 1/1.23 x 10-3 = 813 mm3

P = 1 + 1 P

V(Po- P) VmKPo Vm Po

Using the BET isotherm equation:

2. The data obtained for the adsorption of nitrogen, N2 on 1 g sample of silica gel at 77 K are found to fit a BET isotherm in the range of pressures reported. At this temperature, the saturation vapour pressure, Po of Nitrogen is 0.99 atm. The slope and the Y-intercept of the linear plot of P/V(Po-P) versus P/Po are recorded as:

slope = 1.25 x 10-3 cm-3

y-intercept = 3.3 x 10-6 cm-3

a) Determine the value of the constants, K and Vm

(volume of Nitrogen to form monomolecular layer)

b) Estimate the surface area of 1 g of silica gel if 1 molecule occupies an area of 1.62 x 10-19 m2. (Given the molar volume at 25 oC and 1 atm = 24.0 dm3mol-1, Avogadro’s No = 6.02 x 1023 mol-1)

a) The equation for BET isotherm:

P = 1 + 1 P

V(Po- P) VmKPo Vm Po

Slope = 1/Vm

Vm = 1/slope

= 1/1.25 x 10-3 cm-3

= 800 cm3

Intercept = 1/ VmKPO

K = 1/800cm3 x 0.99atm x 3.3 x 10-6 cm-3

= 382.6 atm-1

b) Estimate the surface area of 1 g of silica gel if 1

molecule occupies an area of 1.62 x 10-19 m2. (Given the

molar volume at 25 oC and 1 atm = 24.0 dm3mol-1,

Avogadro’s No = 6.02 x 1023 mol-1)

At 25 oC, molar volume = 24 dm3mol-1 = 24 x 103 cm3

no of mole of N2 = Vm/molar volume

= 800/24 x 103

= 0.033 mol

Surface area of 1 molecule = 1.62 x 10-19 m2

Surface area of adsorbent = 1.987 x 1022 molecules x

1.62 x 10-19 m2 molecule-1

= 3.22 x 103 m2

No of molecules on surface = 0.033 mol x 6.02 x 1023

molecules mol-1

= 1.987 x 1022 molecules

Figure 16.24 The metal-catalyzed hydrogenation of ethylene

H2C CH2 (g) + H2 (g) H3C CH3 (g)

Figure 16.24