Động hoá học và hoá lý hệ phân tán

ĐỘNG HÓA HỌC VÀ
HÓA LÝ HỆ PHÂN TÁN
- Tài liệu tham khảo chính:
1. Hóa lý và Hóa Keo, Nguyễn Hữu Phú, Nhà Xuất bản Khoa học
và Kỹ thuật, 2003
2. Bài tập Hóa lý cơ sở, Lâm Ngọc Thiềm, Trần Hiệp Hải, Nguyễn
Thị Thu, Nhà Xuất bản Khoa học và Kỹ thuật, 2003
3. Physical Chemistry, Third Edition, Robert G. Mortimer, Elsevier
Inc., 2008

Phần 1: Động hóa học

4. Physical Chemistry – Understanding our Chemical Word, Paul
Monk, John Wiley & Sons, Ltd, 2004
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Một số khái niệm cơ bản

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Một số khái niệm cơ bản
-Tốc độ phản ứng (Reaction rates)

- Động hóa học là môn học nghiên cứu về tốc độ và cơ
chế của quá trình hóa học.

- Định luật tác dụng khối lượng (Rate Law).
- Bậc phản ứng (Reaction Order).
- Phản ứng nguyên tố (Elementary step).
- Phân tử số (Molecularity).

- Nghiên cứu động học: Năng suất và Công nghệ của
một quá trình.

- Cơ chế phản ứng (Reaction Mechanism).
- Lý thuyết va chạm (The Collision Theory).
- Năng lượng hoạt hóa (Activation Energy).
-Xúc tác (Catalyst).

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Kinetics
At 298K, reaction: H2(g) + 1/2O2(g) =

H2O(l) ∆Go298(r) = -254,8 kJ.mol-1

Thermodynamics – does a reaction take place?
Kinetics – how fast does a reaction proceed?

• Studies the rate at which a chemical process
occurs.
• Besides information about the speed at
which reactions occur, kinetics also sheds
light on the reaction mechanism (exactly
how the reaction occurs).

Part 1: Chemical Kinetics

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Chemical Kinetics
(Simple Homogenous reaction)

Outline: Kinetics
Reaction Rates

How we measure rates.

Rate Laws

How the rate depends on amounts
of reactants.

Integrated Rate Laws

How to calc amount left or time to
reach a given amount.

Half-life

How long it takes to react 50% of
reactants.

Arrhenius Equation

How rate constant changes with T.

Mechanisms

Link between rate and molecular
scale processes.

Reaction rate is the change in the concentration of a
reactant or a product with time (M/s).
A

∆[A]
∆t

∆[B]
rate =
∆t
rate = -

B

∆[A] = change in concentration of A over
time period ∆t
∆[B] = change in concentration of B over
time period ∆t
Because [A] decreases with time, ∆[A] is negative.

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A

Exercise 1: Br2 (aq) + HCOOH (aq)

B

time

time

∆[B]
∆t

393 nm
light

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Exercise 1: Br2 (aq) + HCOOH (aq)

slope of
tangent

average rate = -

2Br- (aq) + 2H+ (aq) + CO2 (g)

slope of
tangent

slope of
tangent

∆[Br2] α ∆Absorption

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Exercise 1: Br2 (aq) + HCOOH (aq)

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2Br- (aq) + 2H+ (aq) + CO2 (g)

rate α [Br2]
rate = k [Br2]

[Br2]final – [Br2]initial
∆[Br2]
=∆t
tfinal - tinitial

rate
= rate constant
[Br2]
= 3.50 x 10-3 s-1

k=

instantaneous rate = rate for specific instance in time
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Detector

393 nm

Br2 (aq)

∆[A]
rate = ∆t
rate =

2Br- (aq) + 2H+ (aq) + CO2 (g)

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Exercise 2: 2H2O2 (aq)

Exercise 2: 2H2O2 (aq)

2H2O (l) + O2 (g)

2H2O (l) + O2 (g)

PV = nRT
P=

n
RT = [O2]RT
V
1
[O2] =
P
RT

rate =

∆[O2]
1 ∆P
=
RT ∆t
∆t

measure ∆P over time
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Reaction Rates and Stoichiometry

The “classical” methods for determining the reaction rate
1. The absorbance of radiation at some wavelength at which a given

product or reactant absorbs.
2. The intensity of the emission spectrum of the system at a
wavelength at which a given product or reactant emits.
3. The volume of a solution required to titrate an aliquot removed from
the system.
4. The pressure of the system (for a reaction at constant volume).
5. The volume of the system (for a reaction at constant pressure).
6. The electrical conductance of the system.
7. The mass spectrum of the system.
8. The ESR or NMR spectrum of the system.
9. The dielectric constant or index of refraction of the system.
10. The mass loss if a gas is evolved.
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Factors That Affect Reaction Rates
• Concentration of Reactants
–

As the concentration of reactants increases, so does the probability
that reactant molecules will collide.

• Temperature
–

At higher temperatures, reactant molecules have more kinetic

energy, move faster, and collide more often and with greater energy.

• Catalysts
–

Speed reaction by changing
mechanism.

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Reaction Rates

Reaction Rates

C4H9Cl(aq) + H2O(l) → C4H9OH(aq) + HCl(aq)
Average Rate, M/s

C4H9Cl(aq) + H2O(l) → C4H9OH(aq) + HCl(aq)

The average rate of the
reaction over each
interval is the change in
concentration divided
by the change in time:

average rate = −

∆[C4 H 9Cl ]
∆t

average rate = −

∆[C4 H 9Cl ] 0.1000 − 0.0905 M
=
50.0 − 0.0 s
∆t

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• Note that the average rate
decreases as the reaction
proceeds.
• This is because as the
reaction goes forward,
there are fewer collisions
between reactant
molecules.

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Reaction Rates

Reaction Rates

C4H9Cl(aq) + H2O(l) → C4H9OH(aq) + HCl(aq)

C4H9Cl(aq) + H2O(l) → C4H9OH(aq) + HCl(aq)

• A plot of concentration vs.
time for this reaction
yields a curve like this.
• The slope of a line tangent
to the curve at any point is
the instantaneous rate at
that time.

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• The reaction slows down
with time because the
concentration of the
reactants decreases.

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Reaction Rates and Stoichiometry

Reaction Rates and Stoichiometry

C4H9Cl(aq) + H2O(l) → C4H9OH(aq) + HCl(aq)
• In this reaction, the ratio
of C4H9Cl to C4H9OH is
1:1.
• Thus, the rate of
disappearance of C4H9Cl
is the same as the rate of
appearance of C4H9OH.

Rate =

-∆[C4H9Cl]
=
∆t

• What if the ratio is not 1:1?
H2(g) + I2(g) → 2 HI(g)
• 2 mol of HI are made for each mol of H2 used.

∆[C4H9OH]
∆t

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Reaction Rates and Stoichiometry

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Write the rate expression for the following reaction:
CH4 (g) + 2O2 (g)

• To generalize, for the reaction
aA + bB

Reactants (decrease)

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rate = -

cC + dD

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CO2 (g) + 2H2O (g)

∆[CH4]
∆[CO2]
1 ∆[O2]

1 ∆[H2O]
=
==
∆t
∆t
∆t
2 ∆t
2

Products (increase)

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The Rate Law

The Rate Law
The rate law expresses the relationship of the rate of a reaction
to the rate constant and the concentrations of the reactants
raised to some powers.
aA + bB

cC + dD

Rate = k [A]x[B]y

Concentration and Rate
Each reaction has its own equation that gives its rate as
a function of reactant concentrations.
⇒
this is called its Rate Law
To determine the rate law we measure the rate at
different starting concentrations.

reaction is xth order in A
reaction is yth order in B
reaction is (x +y)th order overall
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Example:

F2 (g) + 2ClO2 (g)

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2FClO2 (g)

• A rate law shows the relationship between the reaction
rate and the concentrations of reactants.
– For gas-phase, reactants use PA instead of [A].

Double [F2] with [ClO2] constant
x=1
Quadruple [ClO2] with [F2] constant

Rate quadruples

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Rate Laws

rate = k [F2]x[ClO2]y

Rate doubles

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rate = k [F2][ClO2]
This equation is called
the rate law, and k is
the rate constant.

• k is a constant that has a specific value for each reaction.
• The value of k is determined experimentally.

rate = k [F2][ClO2]
“Constant” is relative here - k is unique for each reaction,
k changes with T

y=1
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Rate Laws

Exercise 3: Determine the rate law and calculate the rate
constant for the following reaction from the following data:
2SO42- (aq) + I3- (aq)
S2O82- (aq) + 3I- (aq)

•

Rate laws are always determined experimentally.

•

Reaction order is always defined in terms of reactant
(not product) concentrations.

•

The order of a reactant is not related to the
stoichiometric coefficient of the reactant in the
balanced chemical equation.
F2 (g) + 2ClO2 (g)

[S2O82-]

[I-]

Initial Rate
(M/s)

1

0.08

0.034

2.2 x 10-4

2

0.08

0.017

1.1 x 10-4

3

0.16

0.017

2.2 x 10-4

Double [S2O82-], rate doubles (experiment 2 & 3)

rate = k [F2][ClO2] 1

k=
29

2.2 x 10-4 M/s
rate
=
= 0.08/M•s
[S2O82-][I-] (0.08 M)(0.034 M)

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Consider a simple first order reaction: A → B

Experiment

[S2O82-]

[I-]

Initial Rate
(M/s)

1

0.08

0.034

2.2 x 10-4

2

0.08

0.017

1.1 x 10-4

rate = k [S2O82-]x[I-]y
y=1
x=1

3

0.16

0.017

2.2 x 10-4

rate = k [S2O82-][I-]

•
•

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Integrated Rate Laws
First order reaction

Exercise 3: Determine the rate law and calculate the rate
constant for the following reaction from the following data:
S2O82- (aq) + 3I- (aq)
2SO42- (aq) + I3- (aq)

•
•

rate = k [S2O82-]x[I-]y
y=1
x=1
rate = k [S2O82-][I-]

Double [I-], rate doubles (experiment 1 & 2)

2FClO2 (g)

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Experiment

Differential form:
How much A is left after time t? Integrate:

Exponents tell the order of the reaction with respect to each reactant.
This reaction is
First-order in [S2O82-]
First-order in [I-]
The overall reaction order can be found by adding the exponents on the
reactants in the rate law.
This reaction is second-order overall.

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First-Order Reactions

Integrated Rate Laws
First order reaction

Manipulating this equation produces…

The integrated form of first order rate law:

Can be rearranged to give:

…which is in the form

= mx + b

y

[A]0 is the initial concentration of A (t=0).
[A]t is the concentration of A at some time, t, during the course of
the reaction.
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First-Order Reactions

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Exercise 4: The reaction 2A
B is first order in A
-2
-1
with a rate constant of 2.8 x 10 s at 800C. How long
will it take for A to decrease from 0.88 M to 0.14 M ?
[A]0 = 0.88 M

ln[A] = ln[A]0 - kt

[A] = 0.14 M

kt = ln[A]0 – ln[A]
t=

ln[A]0 – ln[A]
=
k

ln

[A]0
[A]
k

ln
=

0.88 M
0.14 M

2.8 x 10-2 s-1

= 66 s

If a reaction is first-order, a plot of ln [A]t vs. t will yield a
straight line with a slope of -k.
So, use graphs to determine reaction order.
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First-Order Reactions

Second-Order Reactions

The half-life, t½, is the time required for the concentration of a
reactant to decrease to half of its initial concentration.

k=

t½ = t when [A] = [A]0/2
ln
t½ =

[A]0
[A]0/2
k

A

product

What is the half-life of N2O5 if it decomposes with a rate
constant of 5.7 x 10-4 s-1?
0.693
t½ = ln2 =
= 1200 s = 20 minutes
k
5.7 x 10-4 s-1

d[A]
dt

rate
M/s
=
= 1/M•s
M2
[A]2

1
1
=
+ kt
[A]
[A]0

ln2
0.693
=
=

k
k

rate = -

also in the form
y = mx + b

-

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Zero-Order Reactions
A
k=

product
rate
= M/s
[A]0

[A] = [A]0 - kt

also in the form
y = mx + b

rate = -

d[A]
dt

rate = k [A]0 = k
-

d[A]
=k
dt

[A] is the concentration of A at any time t

[A]0 is the concentration of A at time t=0

The half-life, t½
t½ = t when [A] = [A]0/2
1
k[A]0

The half-life, t½
t½ = t when [A] = [A]0/2

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Summary of the Kinetics of Zero-Order, First-Order
and Second-Order Reactions
Concentration-Time
Equation

Order

Rate Law

0

rate = k

[A] = [A]0 - kt

1

rate = k [A]

ln[A] = ln[A]0 - kt

2

rate = k [A]2

1
1
=
+ kt
[A]0
[A]

[A]0 is the concentration of A at time t=0

t½ =
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d[A]
= k [A]2
dt

[A] is the concentration of A at any time t

t½ =
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rate = k [A]2

Half-Life
t½ =

[A]0
2k

t½ = ln2
k
t½ =

1
k[A]0

[A]0
2k

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Nth Order Reactions

Forward Reactions with More Than One Reactant
Second-Order Reactions
CH3COOC2H5 + NaOH
H2 + I2

2HI

2NO2

rate =

41

A +
t=0 a
t
a-x

2NO + O2

B
b
b-x

product

dx
dx
. b − x) ⇒ ∫
= k .(a − x )(
= k .dt
(a − x )(. b − x ) ∫
dt

k=
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CH3COONa + C2H5OH

1
a−x b
ln
. 
t.(a − b )  b − x a 

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Complicated homogenous reaction

Complicated homogenous reaction

Reverse Reactions – Chemical Equilibrium

Reverse Reactions – Chemical Equilibrium
t=0
t
At Equilibrium t∞

The observable rate of the reaction is a net rate

a
a-x
a-x∞

rate =

At equilibrium,

0
x
x∞

dx
= k f [ A] − k r [B ] = k f (a − x ) − k r x
dt
k f .a

dx
with A =
= (k f + k r ).( A − x )
k f + kr
dt

− ln( A − x ) = (k f + k r ).t + const

A
1
+ k r ) = . ln
t
A− x
kfa
1
(k f + k r ) = . ln
t
k f a − x (k f + k r )

(k

K: equilibrium constant
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At t = 0, const = − ln A

f

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Complicated homogenous reaction

Complicated homogenous reaction

Reverse Reactions – Chemical Equilibrium

Reverse Reactions – Chemical Equilibrium

K=

Exercise 5: Transformation of γ-hydroxybutiric acid on γ-lacton

kf
kr

CH2-(CH2)2-CO

CH2OH-(CH2)2-COOH

k f .(a − x∞ ) = k r .x∞ or x∞ =

k f .a
k f + kr

=A

(k f +k r ) = 1 . ln x∞
t
x∞ − x

O

+ H2O

Time (min)

21

50

100

120

160

220

∞

Reacted concentration

2,41

4,96

8,11

8,9

10,35

11,15

13,28

Initial concentrations of the acid and the lacton are 18,23M and 0M
Determine the values of K, kf and kr .

If we know K, then we can calculate kf and kr
Signification of A
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Complicated homogenous reaction

Complicated homogenous reaction

Parallel (Competing) Reactions

Parallel (Competing) Reactions

Simplest case: that two competing reactions are first
order with negligible reverse reaction.

Simplest case: that two competing reactions are first order
and the reverse reactions cannot be neglected.

If [Fo] and [Go] = 0, we have:

If [Fo] and [Go] = 0, we have

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Complicated homogenous reaction

Complicated homogenous reaction

Successive Reactions

Successive Reactions

(series or consecutive reaction)

(series or consecutive reaction)

Almost every chemical reaction takes place through a set of steps,
called the reaction mechanism.
The substance B is called
a reactive intermediate.

Case of k1 = 0.100 s−1 and k2 = 0.500 s−1
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Complicated homogenous reaction

Case of k1 = 0.50 s−1 and that k2 = 0.10 s−1.

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Complicated homogenous reaction

Successive Reactions

Successive Reactions

and non-negligible reverse reactions

and non-negligible reverse reactions

with

with
Both steps are at equilibrium when the entire reaction is at equilibrium:

the differential equations giving the rates are

The equilibrium constant K
for the overall reaction is equal to

and

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Determination of Reaction Order

Determination of Reaction Order

Using Integrated Rate Laws
Using the Half-Life

Using Integrated Rate Laws
Using the Half-Life
Method of Initial Rates
Method of Isolation

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Concentration-Time
Equation

Order

Rate Law

0

rate = k

[A] = [A]0 - kt

1

rate = k [A]

ln[A] = ln[A]0 - kt

2

rate = k [A]2

1
1
=
+ kt
[A]
[A]0

n

rate = k [A]n

Half-Life
t½ =

[A]0
2k

t½ = ln2

k
1
t½ =
k[A]0

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Temperature Dependence of the Rate Constant

Determination of Reaction Order

The Arrhenius Relation (1889)
(The Simplest and Empirical Equation)

Method of Initial Rates

Svante Arrhenius, 1859–1927,
was a Swedish chemist who
won the 1905 Nobel Prize in
chemistry for his theory of
dissociation and ionization of
electrolytes in solution.

Reaction rates depend strongly on temperature γ =

See the anterior exercises and examples

kT +10
kT

(γ

Arrhenius postulates:

Method of Isolation

Only “activated” molecules can react

For calculating the reaction order of each reactant

Numbers of activated molecules would be governed by the
Boltzmann probability distribution
This assumption leads to the Arrhenius relation:

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≈ 3)

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Temperature Dependence of the Rate Constant

Temperature Dependence of the Rate Constant
Activation Energy
Endothermic Reaction
Exothermic Reaction

The Arrhenius Relation
Arrhenius relation:
εa is the energy that the molecules must have in order to react and is
called the activation energy
A is called the pre-exponential factor
We can express Arrhenius relation in the form

lnk = -

Ea 1
+ lnA
R T

where Ea = NAvεa is the molar activation energy
R = kB.NAv is the ideal gas constant
Experimental molar activation energy values are usually in the range from 50 to
200kJ.mol−1, somewhat smaller than energies required to break chemical bonds.
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Temperature Dependence of the Rate Constant

The activation energy (Ea) is the minimum amount of
energy required to initiate a chemical reaction.
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Temperature Dependence of the Rate Constant
Exercise 6:

Activation Energy Determination
(Arrhenius plot)

lnk = -

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Ea 1
+ lnA
R T

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Exercise 7:

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Temperature Dependence of the Rate Constant
Exercise 5:

Reaction Mechanisms
The overall progress of a chemical reaction can be represented
at the molecular level by a series of simple elementary steps
or elementary reactions.
The sequence of elementary steps that leads to product
formation is the reaction mechanism.
2NO (g) + O2 (g)

2NO2 (g)

N2O2 is detected during the reaction!

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Elementary step:

NO + NO

N2O2

+ Elementary step:

N2O2 + O2

2NO2

Overall reaction:

2NO + O2

2NO2

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Reaction Mechanisms

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Rate Laws and Elementary Steps

Intermediates are species that appear in a reaction
mechanism but not in the overall balanced equation.

Unimolecular reaction

An intermediate is always formed in an early elementary step
and consumed in a later elementary step.

Bimolecular reaction

A+B

products

rate = k [A][B]

Bimolecular reaction

A+A

products

rate = k [A]2

A

products

Elementary step:

NO + NO

N2O2

+ Elementary step:

N2O2 + O2

2NO2

Writing reasonable (plausible) reaction mechanisms:

Overall reaction:

2NO + O2

2NO2

•

The molecularity of a reaction is the number of molecules reacting in an
elementary step.
•

Unimolecular reaction – elementary step with 1 molecule

•

Bimolecular reaction – elementary step with 2 molecules

•

Termolecular reaction – elementary step with 3 molecules
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rate = k [A]

The sum of the elementary steps must give the overall balanced
equation for the reaction.

•

The rate-determining step should predict the same rate law that is
determined experimentally.
The rate-determining step is the slowest step in the sequence of
steps leading to product formation.

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Rate Laws and Elementary Steps

Chemical Reaction Mechanisms

The experimental rate law for the reaction between NO2
and CO to produce NO and CO2 is rate = k[NO2]2. The
reaction is believed to occur via two steps:
Step 1:

NO2 + NO2

NO + NO3

Step 2:

NO3 + CO

NO2 + CO2

- Sequential mechanism, with one step being
completed before the next step occurs.
Example:

What is the equation for the overall reaction?
NO2+ CO
What is the intermediate?

- Most chemical reactions occur through mechanisms
that involve at least two steps.

NO + CO2
NO3

What can you say about the relative rates of steps 1 and 2?
rate = k[NO2]2 is the rate law for step 1 so
step 1 must be slower than step 2
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In an elementary process, the order of any substance is
equal to the molecularity of that substance.
Unimolecular

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Bimolecular

Termolecular

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Chemical Reaction Mechanisms

Chemical Reaction Mechanisms

The Collision Theory of Bimolecular Elementary - Processes in Gases

The Collision Theory of Bimolecular Elementary - Processes in Gases

12

 8.k .T 
Z = π .d122  B  .N Av n A .nB
 π .µ12 
1
m .m
with µ12 = A B ; d12 = (d A + d B )
2
mB + mB

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Collisions leading to probable reaction:
2AB
A2(g) + B2(g)
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Chemical Reaction Mechanisms

Chemical Reaction Mechanisms

The Collision Theory of Bimolecular Elementary - Processes in Gases

The Collision Theory of Bimolecular Elementary - Processes in Gases

The steric factor, p, is the fraction of collisions in which the
molecules have a favorable relative orientation for the
reaction to occur.
p is sometimes called the probability factor, the orientation,
factor or the fudge factor.

In any group of reactant molecules, only a fraction of

molecules have energies at least equal to Eact, the activation
energy of the reaction.
This fraction is given by the expression:
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Chemical Reaction Mechanisms

Chemical Reaction Mechanisms

The Collision Theory of Bimolecular Elementary - Processes in Gases

Liquid-State Reactions

The temperature dependence of rate constants for both
gaseous and liquid-state reactions is usually well described by
the Arrhenius formula

Activation-limited reactions

Diffusion-limited reactions

Activation energies equal

Rate constant governed by the
to those for gas-phase reactions
diffusion coefficients
The activation energies of the diffusionlimited reactions (~10kJ.mol-1) are smaller
than for activation-limited reactions
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Diffusion coefficients in liquids given by Eq. which is also of
the same form as the Arrhenius formula:

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Chemical Reaction Mechanisms

Chemical Reaction Mechanisms

Change of mechanism with temperature (T in K)

Liquid-State Reactions

Example: For the reaction 2I → I2 in carbon tetrachloride, the value
of the rate constant at 23°C is 7.0 × 106 m3mol−1s−1. At 30°C, the
value is 7.7×106 m3mol−1s−1. Find the activation energy and compare

it with the activation energy for the viscosity of carbon tetrachloride
10400J.mol−1.

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Catalysis - Catalyst

Catalysis - Catalyst
A catalyst is a substance that increases the rate of a chemical
reaction without itself being consumed.
k = A • exp( -Ea/RT )

Ea

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k

In heterogeneous catalysis, the reactants and the catalysts
are in different phases.
•

Haber synthesis of ammonia

•

Ostwald process for the production of nitric acid

•

Catalytic converters

In homogeneous catalysis, the reactants and the catalysts
are dispersed in a single phase, usually liquid.
uncatalyzed

catalyzed

•

Acid catalysis

•

Base catalysis

ratecatalyzed > rateuncatalyzed
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Ea‘ < Ea

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Ostwald Process

Haber Process
4NH3 (g) + 5O2 (g)

Pt catalyst

2NO (g) + O2 (g)
2NO2 (g) + H2O (l)

N2 (g) + 3H2 (g)

Fe/Al2O3/K2O
catalyst

4NO (g) + 6H2O (g)

2NO2 (g)
HNO2 (aq) + HNO3 (aq)

2NH3 (g)
Pt-Rh catalysts used
in Ostwald process

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Hot Pt wire
over NH3 solution

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Enzyme Catalysis

Catalytic Converters

Enzymes (which are large protein molecules) are nature's catalysts

CO + Unburned Hydrocarbons + O2
2NO + 2NO2

catalytic
converter

catalytic
converter

CO2 + H2O

Michaelis-Menten mechanism for the catalysis of biological chemical reactions

E is the enzyme, S is the "substrate" and ES is an enzyme-substrate complex

2N2 + 3O2

Solve for [ES],
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Enzyme Catalysis

Enzyme Catalysis

KM: the Michaelis-Menten constant

[E]o: the total enzyme concentration
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Enzyme Catalysis

uncatalyzed

enzyme
catalyzed

Part 2: Disperse Systems - Colloids
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Disperse Systems - Colloids

Disperse Systems - Colloids

Definition
A system in which one substance (particulate matter),

the

disperse

phase,

is

dispersed

as

particles

throughout another, the dispersion medium or the
continuous phase.

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Types of Disperse Systems - On the basis of particle size
Class

Molecular
dispersion

Size

Examples

Properties

< 1.0 nm

Oxygen gas,
ordinary ions,
glucose

- Invisible in electron microscope
-Pass through ultra filter and semipermeable membrane
- Rapid diffusion

Colloidal
dispersion

1.0 nm to
0.5 µm

Coarse
dispersion

> 0.5 µm

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Types of Disperse Systems
On the basis of physical states

- Detected by electron microscope but
Silver sols, natural not by ordinary
and synthetic
- Pass through ultra filter but not semipolymer lattices
permeable membrane
-- Very slow diffusion
Sand,
pharmaceutical
emulsions &
dispersions, red
blood cells

- Visible under ordinary microscope
- Do not pass through ultra filter and
semi-permeable membrane
-- Very slow diffusion
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Types of Disperse Systems
On the basis of the interaction of the dispersed
particles with the dispersion medium.

Preparation of Disperse Systems (Colloids)
Lyophilic colloids
Spontaneous, e.g.: gelatin soaked in water
Lyophobic colloids
A- Dispersion methods: breakdown of coarse particle
- Colloid mill
- Electric dispersion
- Ultrasonic irradiation

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surfactant micelles and phospholipid
vesicles, also known as association
colloids.
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Preparation of Disperse Systems (Colloids)
Lyophobic colloids
B- Condensation method: aggregation off subcolloidal
particles
1- Chemical reaction
- Reduction: e.g.: colloidal Ag
- Oxidation: e.g.: H2S
S
- Hydrolysis: e.g.: Fe(OH)3

- Double decomposition: colloidal AgI

- Peptisation: addition of preferentially adsorbed ions.
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Purification of Disperse Systems (Colloids)
1. Dialysis

Applications:
- Membrane filters
- Membrane diffusion

2- Change of solvent:
- Precipitation of colloidal S from alcoholic solution by
addition of water
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- Study of drug/protein binding
- Heamodyalysis
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Purification of Disperse Systems (Colloids)

Structure of a Colloids Micelle

2. Electrodialysis:
Application of an electric potential
across the semi-permeable membrane

3. Electrodecantation:
The x value depends on the concentration

Concentration of charged colloidal
particles at one side and at the base of
the membrane

At dilute concentration, the x value is
large, the colloid particles having the same
charges; the stability of the system is
good.

4. Ultrafiltration:

At higher concentration, the x value
decreases, the charge of colloid micelles
decreases and the system is less stable.

Application of a pressure or suction
across a filter
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Nature of Adsorption
Definition of adsorption

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Physical adsorption - Physisorption
General

Gas

- Physisorption is not adsorbent or adsorptive specific. This is a process
similar to condensation of gas on a surface

• Solid surfaces show strong affinity towards
gas molecules that it comes in contact with
and some of them are trapped on the surface

- Quantity of physisorbed molecules depends on the accessible surface area
and not on its chemical nature

• Process of trapping or binding of
molecules to the surface is called adsorption
• Desorption is removal of these gas
molecules from the surface

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- Physisorbed molecules progressively form successive layers when the gas
pressure P increases. When P reaches the vapor pressure P0, there is
condensation on the surface.

Solid

Two types of adsorption
– Physical adsorption :
* Van der Waals forces
* bond energy is less than 50 kJ/mole
– Chemical adsorption :
* bond energy is more than 50 kJ /mole
* direct chemical bond
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- Capillary condensation in the pores for P < P0 depending on the pore size.
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Langmuir theory (monolayer adsorption)

Theoretical approach

Thermodynamic Derivation

Langmuir theory (monolayer adsorption)
a. Thermodynamic Derivation

Langmuir theory based on three assumptions:

For molecules in contact with a solid
surface at a fixed temperature, the
Langmuir Isotherm, developed by Irving
Langmuir in 1916, describes the
partitioning between gas phase and
adsorbed species as a function of applied
pressure.

1. Adsorption cannot proceed beyond monolayer coverage.
2. All surface sites are equivalent and can accommodate, at most,
one adsorbed molecule.
3. The ability of a molecule to adsorb at a given site is independent
of the occupation of neighboring sites.

The adsorption process between gas phase molecules, A, vacant surface sites, S,
and occupied surface sites, SA, can be represented by the equation,

θ = Fraction of surface sites occupied (0 < θ < 1)
θ = V/Vm

assuming that there are a fixed number of surface sites present on the surface.
An equilibrium constant, K, can be written:

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Where: V = adsorbed volume of gas ;
Vm = adsorbed volume at saturation (θ = 1)
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Langmuir theory (monolayer adsorption)

Langmuir theory (monolayer adsorption)

Thermodynamic Derivation

Kinetic Derivation

[SA]

is proportional to the surface coverage of adsorbed molecules,
or proportional to θ

[S]

is proportional to the number of vacant sites, (1 - θ)

[A]

is proportional to the pressure of gas, P
Thus it is possible to define the equilibrium constant, b:

The rate of adsorption will be proportional to the pressure of the gas and the
number of vacant sites for adsorption. If the total number of sites on the surface
is N, then the rate of change of the surface coverage due to adsorption is:

The rate of change of the coverage due to the adsorbed species leaving the
surface (desorption) is proportional to the number of adsorbed species:

Rearranging gives the expression for surface coverage:
In these equations, ka and kd are the rate constants for adsorption and desorption
respectively, and p is the pressure of the adsorbed gas.
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