Chapter 18 Surface Chemistry and Colloid

18.1 Adsorption

~Physical adsorption (physisorption, van der Waals adsorption)

*van der Waals force

*The heat evolved is less than 20 kJ mol-1

*no activation energy rapidly process

*multimolecular layer

~Chemical adsorption (chemisorption, activated adsorption)

*covalent force

*The heat evolved is 100 to 500 kJ mol-1

*appreciable activation energy slow process

*unimolecular layer

Figure 18.1~Figure 18.1 presents the relationship between physisorption and chemisorption with the potential- energy diagram.

~It relates to the adsorption of hydrogen on a surface such as that of a metal.

~Physisorption occurs first, the intact molecule being held loosely to the surface by dispersion forces.

~The potential-energy curve for chemisorption, with dissociation of the hydrogen molecule, has a much deeper minimum, corresponding to stronger bonding.

~The crossing of the two curves shows that there is a small energy of activation for the chemisorption process.

~This diagram shows that initial physisorption is an important feature of the chemisorption process. If there were no physisorption, there would be a much higher activation energy for the chemisorption.

18.2 Adsorption Isotherms

An equation that relates the amount of a substance to a surface to its concentration in the gas phase or in solution, at a fixed temperature, is known as an adsorption isotherm.

The Langmuir Isotherm

~Assuming that all parts of the surface behave in exactly the same way.

~Suppose that, after equilibrium is established, a fraction of the surface is covered by adsorbed molecules; a fraction 1- will not be covered.

~The rate of adsorption will be proportional to the concentration [A] of the molecules in the gas or liquid phase and also proportional to the fraction of the surface that is bare, because adsorption can only occur when molecules strike the bare surface. The rate of adsorption va is thus

where ka is a rate constant relating to the adsorption process.

~The rate of desorption vd is proportional only to the number of molecules attached to the surface, which in turn is proportional to the fraction of surface covered:

where kd is a rate constant for the desorption process.

θ)1[A]( aa kv

θdd kv

~At equilibrium, the rates of adsorption and desorption are the same; thus

~The ratio ka/kd is an equilibrium constant and can be written as K; then

θθ)1[A]( da kk [A]θ1

θ

d

a

k

k

[A]θ1

θK

[A]1

[A]θ

K

K

Figure 18.2

~Figure 18.2a shows a graph of against [A].

~At sufficiently low concentrations we can neglect K[A] in comparison with unity, and then is proportional to [A].

~At very high concentrations

[A]1

[A]θ

K

K

[A]

1

[A]1

1θ1

KK

[A]1

[A]θ

K

K

[A]

11

θ

1

K

simple Langmuir isotherm

Example 18.1

Benzene adsorbed on graphite is found to obey the Langmuir isotherm to a good approximation. At a pressure of 1.00 Torr the volume of benzene adsorbed on a sample of graphite was found to be 4.2 mm3 at STP (0 C and 1 atm pressure); at 3.00 Torr it was 8.5 mm3. Assume a benzene molecule to occupy 30 Å2 and estimate the surface area of the graphite.

Solution

Suppose that the amount of adsorbed when the surface is saturated is x mm3; the fractions of surface covered at the two pressures are thus 4.2/x and 8.5/x. We can thus set up the two equations

K

K

x���

K

K

x 0.31

0.35.8

0.11

0.12.4

4.17and318.0 x����K

The maximum amount of benzene adsorbed is thus 17.4 mm3.

1 mol at STP occupies 22.4 L=2.24107 mm3

17.4 mm3 is thus 17.4/(2.24107) mol=7.7710-7 mol

=7.7710-76.0221023=4.681017 molecules.

The estimated area of the surface is thus

4.68101730 Å2=1.401019Å2=0.140 m2

Problem 18.1

A surface is half-covered by a gas when the pressure is 1 bar. If the simple Langmuir isotherm applies:a. What is K?b. What pressures give 75%, 90%, 99%, 99.9% coverage?c. What coverage is given by pressures of 0.1 bar, 0.5 bar, 1000 bar?

Solution

[A]1

[A]θ

K

K

a.

bar11

bar15.0

K

K 1bar1 �K

b.

1

1

bar11

bar1θ

P

P

θ1

θ

P

bar999999.0θ

bar9999.0θ

bar990.0θ

bar375.0θ

�P

�P

�P

�P

1

1

bar11

bar1θ

P

Pc.

999.0θbar1000

33.0θbar5.0

091.0θbar1.0

�P

�P

�P

Problem 18.3

The following results were reported by Langmuir for the adsorption of nitrogen on mica at 20 C:

Pressure/atm 2.8 4.0 6.0 9.4 17.1 33.5

Amount of gas adsorbed/mm3 at 20 C and 1 atm

12.0 15.1 19.0 23.9 28.2 33.0

a. Make a linear plot of these values in order to test the Langmuir isotherm. If it applies, evaluate the constant K.

b. Suppose that 1015 molecules cover 1 cm2 of the surface. Make an estimate of the effective area in Langmuir’s experiment.

Solution

The amount x adsorbed is proportional to and therefore,

[A]1

[A]θ

K

aKax

To convert atmospheres to concentrations:

3dmmol15.29308205.0

/atm

�P

RT

P

V

n

To convert amount of gas adsorbed to moles:

mol15.29308205.010

mm/

mol�atmdm�15.29308205.0

(atm)16

3

13

V

�V

RT

PVn

Concentration/mol dm-3 0.116 0.166 0.249 0.391 0.711 1.39

Amount adsorbed/10-7 mol 4.99 6.28 7.90 9.94 11.7 13.7

a. A linear plot may be obtained by plotting 1/x against 1/[A]:

[A]1

[A]

K

aKx

aaKx

1

[A]

11

[A]-1/mol dm-3 8.62 6.02 4.02 2.56 1.41 0.719

x-1/106 mol 2.00 1.59 1.27 1.01 0.85 0.73

6

6

10161.01

10615.01

aK

a

0 2 4 6 8 101/[A ]

0.0

0.5

1.0

1.5

2.0

2.5

1/x

Y = 0 .1 6 1 * X + 0 .6 1 5

13

6

moldm82.3

mol1063.1

��K

a

b.Complete coverage corresponds to

1.6310-6 mol=9.821017 molecules

The surface area was thus about

9.82 1017/1015=103 cm2=10-1 m2

[A]1

[A]θ

K

K

Adsorption with Dissociation

S represents a surface site and A the substance being adsorbed.

In certain cases there is evidence that the process of adsorption is accompanied by the dissociation of the molecule when it becomes attached to the surface. For example, when hydrogen gas is adsorbed on the surface of many metals, the molecules are dissociated into atoms each of which occupies a surface site. This type of adsorption may be represented as

~The process of adsorption is now a reaction between the gas molecule and two adjacent surface sites, and the rate of adsorption is therefore

~The desorption process involves reaction between two adsorbed atoms, and the rate is therefore proportional to the square of the fraction of surface covered,

2θ)1[A]( aa kv

2θdd kv

~At equilibrium the rates are equal, and therefore,

2/12/12/1

[A][A]θ1

θK

k

k

d

a

K is equal to ka/kd.

2/12/1

2/12/1

[A]1

[A]θ

K

K

Figure 18.2

~Figure 18.2c shows a graph of against [A]1/2.

~When the concentration is very small, K1/2[A]1/2 is much smaller than unity, and is then proportional to [A]1/2.

~At very high concentrations, K1/2[A]1/2 is >>1,

2/12/1 [A]1

1θ1

K

2/12/1 [A]

1θ1

K

2/12/1

2/12/1

[A]1

[A]θ

K

K

2/12/1 [A]

11

θ

1

K

Competitive Adsorption

The isotherm for two substances adsorbed on the same surface is of importance in connection with inhibition and with the kinetics of surface reactions involving two reactants.

~Suppose that the fraction of surface covered by molecules of type A is A and that the fraction covered by B is B. The fraction bare is 1-A-B. If both substances are adsorbed without dissociation, the rate of adsorption of A and B are

~The rates of desorption are

)θθ1[A]( BAAA aa kv )θθ1[B]( BA

BB aa kv

AAA θdd kv

BBB θdd kv

[A][A]θθ1

θAA

A

BA

A Kk

k

d

a

[B][B]θθ1

θBB

B

BA

B Kk

k

d

a

[B][A]1

[A]θ

BA

AA KK

K

[B][A]1

[B]θ

BA

BB KK

K

~The fraction of the surface covered by one substance is reduced if the amount of the other substance is increased. This is because the molecules of A and B are competing with one another for a limited number of surface sites.

~There is evidence that sometimes two substances are adsorbed on two different sets of surface sites, in which case there is no competition between them.

Other Isotherms

~Systems that obey the Langmuir type are often referred to as showing ideal adsorption.

~Systems frequently deviate significantly from the Langmuir equation. This may be because the surface is not uniform, and also there may be interactions between adsorbed molecules; a molecule attached to a surface may make it more difficult, or less difficult, for another molecule to become attached to a neighboring site, and this will lead to a deviation from the ideal adsorption equation.

Freudlich Isotherm

nkcx

~x is the amount of a substance adsorbed; c is the concentration of a substance.

~k and n are empirical constants. The value of n is usually less than unity.

BET Isotherm

000

0 1

)( V

P

KVPPV

PP

~V is the volume of gas adsorbed at pressure P, and V0 the volume that be adsorbed as a monolayer; P0 is the saturation vapor pressure, and K is the equilibrium constant for the adsorption.

~The BET isotherm is particularly useful for determining surface area.

Example 18.2

The following data relate to the adsorption of nitrogen at 77 K on a 1.00-g sample of silica gel:

P/kPa 15.2 54.8

V/cm3 (STP) 135 247

At 77 K the saturation vapor pressure P0 of nitrogen is 101.3 kPa. Estimate the surface area of the gel, taking the molecular area of nitrogen to be 1.6210-19 m2.

Solution

Insertion of the data into the BET equation gives the following two simultaneous equations:

00

2.151132.0

VKV

00

8.541483.0

VKV )(kPa17.3)(cm113 13

0 �K����V

At STP, 22.4 L=22400 cm3 is the volume occupied by 1 mol.

A volume of 113 cm3 thus contains 113/22400=0.00504 mol; 0.005046.0221023=3.041021 molecules.

Since each molecule occupies 1.6210-19 m2, the estimated surface area is

3.0410211.6210-19=492 m2

18.4 Chemical Reactions on Surfaces

~An important concept in connection with surface reactions is the molecularity, which is the number of reactant molecules that come together during the course of reaction; we do not count the surface site.

~The molecularity of a surface reaction is deduced from the kinetics on the basis of the experimental results and of theoretical considerations.

~Reactions involving a single reacting substance are usually, but invariably, unimolecular.

*The mechanism of the surface-catalyzed ammonia decomposition is, for example, usually unimolecular.

~Reaction involving two reacting substance are usually bimolecular.

*The kinetics of the decomposition of acetaldehyde on various surfaces can only be interpreted on the hypothesis that two acetaldehyde molecules, adsorbed on neighboring surface sites, undergo a bimolecular.

*When reactant molecules are dissociated on the surface, the reaction may involve interaction between an atom or radical and a molecule; for example, the exchange reaction between ammonia and deuterium on iron is a bimolecular interaction between a deuterium atom and an ammonia molecule.

Unimolecular Reactions

Surface reactions involving one molecule may be treated in terms of the Langmuir adsorption isotherm. In the simplest case the rate of reaction is proportional to and is thus

[A]1

[A]θ

K

kKkv

It has been derived on assumption of a rapid adsorption

equilibrium followed by slow chemical reaction.

Figure 18.3

~The dependence of v on [A], shown in Figure 18.3, is exactly the same as that given by the Langmuir isotherm (Figure 18.2).

~At sufficiently high concentrations the rate is independent of the concentration, which means that the kinetics are zero order.

~At low concentrations, when K[A]<<1, the kinetics are first order.

~A good example of this type of behavior is the decomposition of ammonia into nitrogen and hydrogen on a tungsten surface.

~Sometimes a substance other than the reactant is adsorbed on the surface, with the result that the effective surface area, and therefore the rate, are reduced.

~Suppose that a substance A is undergoing a unimolecular reaction on a surface and that a nonreacting substance I, known as an inhibitor or a poison, is also adsorbed.

~If the fraction of the surface covered by A is and that covered by I is i, we have from the Langmuir isotherm with competition adsorption that

~The rate of reaction, equal to k, is thus

~A case of special interest is when the surface is only sparsely covered by the reactant but is fairly fully covered by the inhibitor. In other words

Ki[I]>>1+K[A] and the rate is then

[I][A]1

[A]θ

iKK

K

K and Ki are the adsorption constants for A and I.

[I][A]1

[A]

iKK

kKv

[I]

[A]

iK

kKv

A good example of this type reaction is provided by the decomposition of ammonia on platinum, the rate law for which is

Since hydrogen is a product of reaction, there is progressive inhibition as reaction proceeds. There is no appreciable inhibition by the other reaction product, nitrogen.

][H

][NH

2

3

iK

kv

Bimolecular Reactions

Langmuir-Hinshelwood Mechanism

~In the first step the two molecules A and B become adsorbed on neighboring sites on the surface. Reaction then takes place, by way of an activated complex, to give the reaction products.

~When this mechanism applies, the rate is proportional to the probability that A and B are adsorbed on neighboring sites, and this is proportional to the product of fractions of the surface, A and B, covered by A and B. The rate of reaction is therefore

[B][A]1

[A]θ

BA

AA KK

K

[B][A]1

[B]θ

BA

BB KK

K

2BA

BABA

[B])[A]1(

[B][A]θθ

KK

KkKkv

Figure 18.3The rate first rises, pass through a maximum, and then decreases toward zero.

~The physical explanation of the falling off of the rate at high concentrations is that one reactant displaces the other as its concentration is increased.

~The maximum rate corresponds to the presence of the maximum number of neighboring A-B pairs on the surface.

~At sufficiently low concentrations of A and B, the second-order reaction kinetics is predicted, and this has been observed in a number of systems.

~A case of special interest is when one reactant (e.g., A) is weakly adsorbed and the other strongly adsorbed. This means that KB[B]>>1+KA[A], and it follows that

2BA

BABA

[B])[A]1(

[A][B]θθ

KK

KkKkv

[A][B]BAKkKv

2BA

BABA

[B])[A]1(

[A][B]θθ

KK

KkKkv

[B]

[A]

B

A

K

kKv

The order is 1 for A and -1 for B. An example is the reaction between carbon monoxide and oxygen on quartz, where the rate is proportional to the pressure of oxygen and inversely proportional to the pressure of carbon monoxide.

Langmuir-Rideal Mechanism

The rate for this mechanism is proportional to the concentration of A and to the fraction of the surface that is covered by B:

[B][A]1

[A][B]θ[A]

BA

BB KK

kKkv

~The variation v with [A] is shown in Figure 18.3c; the same type of curve is found if v is plotted against [B].

~There is no maximum in the rate, and this provides a possible way of distinguishing between this mechanism an the Langmuir-Hinshelwood mechanism.

Figure 18.3

An alternative mechanism for a bimolecular surface process is for the reaction to occur between a molecule that is not absorbed (e.g, A) and an adsorbed molecule (B). The adsorption of A may occur; it is simply postulated, in this mechanism, that an adsorbed A does not react.

The fraction of the surface covered by hydrogen atoms is

and the rate of combination is thus

[H]1

[H]θ

K

K

[H]1

[H]θ[H]

2

K

kKkv

~At lower temperatures the surface may be fully covered, so that K[H]>>1 andv=k[H] and the kinetics are first order.

~At higher temperatures the coverage decreases and if 1>>K[H], the rate is v=kK[H]2

~An increase in order from 1 to 2 has in fact been observed as the temperature is raised.

~Not many ordinary chemical reactions occur by a Langmuir-Rideal mechanism. There is evidence, however, that radical combinations on surface sometimes occur in this way.

~The combination of hydrogen atoms, for example, is sometimes a first-order reaction, and it appears to occur by the mechanism

Problem 18.10

A first-order surface reaction is proceeding at a rate of 1.510-4 mol dm-3 s-1 and has a rate constant 2.010-3 s-1. What will be the rate and the rate constant if

a. the surface area is increased by a factor of 10?

b. the amount of gas is increased tenfold at constant pressure and temperature?

If these values of v and k apply to a reaction occurring on the surface of a spherical vessel of radius 10 cm:

c. What will be the rate and rate constant in a spherical vessel, of the same material, of radius 100 cm, at the same pressure and temperature?

d. Define a new rate constant k’ that is independent of the gas volume V and the area S of the catalyst surface.

e. What would be its SI unit?

Solution

a. The rate and the rate constant are both increased by a factor of 10:

v=1.5 10-3 mol dm-3 s-1; k= 2.010-2 s-1

b. The rate of conversion (mol s-1) remains the same, but since the volume is increased by a factor of 10 the rate is reduced by a factor of 10, as is the rate constant:

v=1.5 10-5 mol dm-3 s-1; k= 2.010-4 s-1

c. Increasing the radius by a factor of 10 increases the surface area by a factor of 100 and the volume by a factor of 1000. The rate and the rate constant are thus reduced by a factor of 10:

v=1.5 10-5 mol dm-3 s-1; k= 2.010-4 s-1

d. Since k is proportional to S and inversely proportional to V, the constant k’=kV/S is independent of V and S.

e. Its SI unit is m s-1.

Problem 18.16

Suggest explanations for the following observations, in each case writing an appropriate rate equation based on a Langmuir isotherm:

a. The decomposition of phosphine (PH3) on tungsten is first order at low pressures and zero order at higher pressures.

b. The decomposition of ammonia on molybdenum is retarded by the product nitrogen, but the rate does not approach zero as the nitrogen pressure is increased.

Solution

][PH1

][PH

3

3

K

kKv

a. low pressure][PH3kKv

kv high pressure

][N][NH1

][NH

23

3

iKK

kKv

b.

c. On certain surfaces (e.g., Au) the hydrogen-oxygen reaction is first order in hydrogen and zero order in oxygen, with no decrease in rate as oxygen pressure is greatly increased.

d.The conversion of para-hydrogen into ortho-hydrogen is zero order on several transition metals.

][O][H1

]][O[H

2O2H

22O

22

2

KK

kKv

c. smallis][H2H2

���K

islarge�][O2O2��K

][H2kv

d.

2/12

2/1 ][H

1θ1

K

22 θ)1]([H kv

K

kv

Solution

18.7 Surface Tension and Capillarity

Figure 18.8

~A molecule in the interior of a liquid is, on the average, attracted equally in all directions by its neighbors, and there is therefore no resultant force tending to move it in any direction. On the other hand, at the surface of a liquid that is in contact with vapor there is practically no force attracting the surface molecules away from the liquid, and there is therefore a net inward attraction on the surface molecules. This is represented in Figure 18.8a.

~If the surface area of a liquid is increased, more molecules are at the surface, and work must be done for this to occur. A surface therefore has an excess Gibbs energy, relative to the interior of the liquid; the SI unit of this energy per unit surface are is J m-2

=kg s-2=N m-1. This relationship show that the excess surface energy per unit area is force per unit length.

~Surfaces behave as if a membrane were stretched over them. An analogy is provided by a soap bubble, which assumes a spherical form because of the tension of the membrane. In the same way a liquid, particularly if it is suspended in liquid with which it is immiscible so as to eliminate the effects of gravity, tends to become spherical. Very small drops of any liquid are almost exactly spherical; larger ones are flattened by their weight.

Figure 18.8

Figure 18.8b shows a thin film, such as a soap film, stretched on a wire frame having a movable side of length l. The film has two sides, and the total length of the side of the film is 2l. The force F required to stretch the film is proportional to 2l,

)2γ( lF

The proportionality constant is known as the surface tension. Its SI unit is N m-1=J m-2, and it is therefore the surface energy per unit area. Thus if the piston in Figure 18.8b moves a distance dx. the area (on the two sides) increase 2ldx, and the work done is 2ldx. The ratio of the work done to the increase in the surface is therefore

γ2

γ2

areansurface� �increasei�workdone�

ldx

dxl

Note that this work done is the increase in Gibbs energy; the increase in Gibbs energy is thus the increase in surface area multiplied by the surface tension.

Figure 18.9

Figure 18.9 illustrates the simplest and most commonly employed is the capillary-rise method.

γ21 rF

ghrF ρ22 2

ργ

grh

If the angle between the meniscus and the surface is zero

If the angle between the meniscus and the surface is

cosθ2

ργ

grh

Problem 18.17

The surface tension of water at 20 C is 7.2710-2 N m-1 and its density is 0.998 g cm-3. Assuming a contact angle of zero, calculate the rise of water at 20 C in a capillary tube of radius (a) 1mm and (b) 10-3 cm.

Solution

2

ργ

grh

(a)

grh

ρ

γ2

m1049.1)s(m81.9)m(kg998(m)10

)m(N1027.72 2233

12

���

�h

r=10-3 m

(b)

m49.1)s(m81.9)m(kg998(m)10

)m(N1027.72235

12

���

�h

r=10-5 m

Problem 18.20

The two arms of a U-tube have radii of 0.05 cm and 0.10 cm. A liquid of density 0.80 g cm-3 is placed in the tube, and the height in the narrow arm is found to be 2.20 cm higher than that in the wider arm. Calculate the surface tension of the liquid, assuming =0.

Solution

grh

ρ

γ2

21

11

ρ

γ2Δ

rrgh

001.0

1

0005.0

1

)s(m81.9)m(kg108.0

γ2m022.0

233 ��

12 mN086.0skg086.0γ ��

Suppose that the ordinary vapor pressure of a liquid is P0 and that when it is present in droplets of radius r, the vapor pressure is P. The Gibbs energy change when dn mol of liquid is transferred from a plane surface to a droplet is then

0

lnP

PdnRTdG

This change can also be calculated from the surface energy change that results from the increase in surface area. The volume of dn mol is Mdn/, where M is the molar mass and is the density. The droplet has a surface area of 4r2, and if the radius increase by dr, the increase in volume is 4 r2dr; thus

drrMdn 2π4

ρ dn

r

Mdr

ρπ4 2

The increase in surface area of the droplet is

rdrrdrrdA π8π4)π(4 22

The increase in surface Gibbs energy is

drrdG γπ8

drrdG γπ8

dnr

Mdr

ρπ4 2

dnr

MdG

ρ

γ2

0

lnP

PdnRTdG

dnr

M

P

PdnRT

ρ

γ2ln

0

rRT

M

P

P

ρ

γ2ln

0

Kelvin equation

The fact that the vapor pressure of a tiny droplet can be so much higher than that of the bulk liquid poses an interesting question regarding the process of condensation of a vapor.

Suppose that air saturated with water is chilled. The vapor then becomes supersaturated and is in a metastable state. However, if the condensation first produces tiny droplets, these will have vapor pressures many times that of the bulk liquid, and they should immediately evaporate again. How, then, can condensation ever get started?

~a larger number of molecules may come together to form a droplet large enough that it does not at once reevaporate.

~dust particles act as nuclei for supersaturated vapors or the condensation may occur at the surface of the vessel.

Problem 18.19

The density of water at 20 C is 0.998 g cm-3 and the surface tension is 7.2710-2 N m-1. Calculate the ratio the vapor pressure of a mist droplet having a mass of 10-12 g and the vapor pressure of water at a plane surface.

Solution

331831212

π3

4m10002.1cm10002.1

998.0

10dropletVolumeof� � r

m1021.6 7r

0017.0

(K)15.298)K�(Jmol�3145.8(m)1021.6)m(kg10998.0

)mol(kg1002.18)m(N1027.72

ρ

γ2ln

11733

1312

0

������

�������

rRT

M

P

P

0017.10

P

P