Chapter 7
Review of the Development of the Atomic and Quantum
Theories
The early history of Chemistry:
1)
First quantitative experiments that led to important chemical
laws by:

Albert Boyle (1627-1691)

Antoin Lavoisier (1743-1794)
Law of Conservation of Mass:
“Mass is neither created nor destroyed”
Law of Definite Proportion:
“A given compound always contains exactly the same
proportion of elements by mass”
Law of Multiple Proportions:
“When two elements form a series of compounds, the ratios
of the masses of the second element that combine with 1
gram of the first element can always be reduced to small
whole numbers”
Dalton’s Atomic Theory (1808):

Each element is made up of tiny particles called
atoms

The atoms of a given element are identical

A given compound always has the same relative
numbers and types of atoms

Chemical reaction is reorganization of atoms
However, Dalton provided no correct formulas of certain
compounds:
e.g: for water H : O ratio is 1: 8 and formula is HxOy (H2O)
Experimental work of measuring the volumes of gases by
Gay-Lussac (1778-1850) and the hypothesis of Avogadro
(1776-1856) provided the key to determining absolute
formulas of compounds
Avogadro’s hypothesis (1811):
Avogadro explained the results of Gay-Lussac’s experiments
and proposed that: “At the same temperature and pressure,
equal volumes of gases contain the same number of
particles”.
Early experiments to characterize the atom:
The electron:
Cathode-ray tubes experiments by J. J. Thomson:

Thomson postulated that the produced ray was a stream
of negatively charged particles called electrons:

By measuring the deflection of the beam in a magnetic
field:
e/m = -1.76 x 108 C/g

Since electrons could be produced from electrodes of
various types of metals, all atoms must contain
electrons.

Since atoms were known to be neutral, they must
contain some positive charge with the negative
electrons dispersed randomly in it (like an orange)
Robert Millikan (1909) with the e/m value was able to
Calculate the mass of electron as: m = 9.11 x 10-31 Kg
The nuclear atom:
Rutherford’s experiment on -particles bombardment of
metal foil:
The results could be explained only in terms of a nuclear atom
Atomic Masses:

The system of atomic masses was made based on 12C as the
standard of exactly 12 atomic mass units (amu).

The most accurate method for comparing the masses of
atoms is mass spectrometer:
____________________________________________
Particle
Mass
Charge
____________________________________________
Electron
9.11 x 10-31 Kg
-1
-27
Proton
1.67 x 10 Kg
+1
-27
Neutron
1.67 x 10 Kg
None
______________________________________________
*Charge of one electron is 1.60 x 10-19 C
Black body Problem:
Black body does not emit continuous radiation:
ΔE ≠
Energy quantization:
Max Planck suggested that energy is quantized it means
that matter cannot absorb or emit any quantity of energy; but
instead it can only gain or lose a specific quantity of λ.
ΔE α
ΔE = h
(change in energy of a system)
Where, h is called Planck’s constant = 6.626 x 10-34 Js
Then Albert Einstein suggested that electromagnetic
radiation is itself quantized and can be viewed as a stream of
particles called photons.
The energy of each photon is given by:
Ephoton = h = hc/
Also Einstein derived the following equation to explain
photoelectric effect:
E = mc2
That means light has some characteristic of matter
m = E/c2 = h/c2 = hc/c2
= h/c
Diffraction pattern of electrons:

Experimental work proves that electrons have wave
properties (diffraction experiments).

De Broglie (1923) proposed that for a particle of mass
m moving with a velocity v:
and
m = h/ v
(De Broglie equation)
 = h/m v
(for microscopic particles)
Atomic Spectra and The Bohr Model:

In 1913 Bohr developed a quantum model for the
hydrogen atom.

He proposed that the electron in a hydrogen atom
moves around the nucleus only in certain allowed
circular orbits:
E = -2.278 x 10-18 J (Z2/n2)
Where,
E is energy levels
n is an integer
Z is the nuclear charge
[The negative sign simply means that the energy of the
electron bound to the nucleus is lower than it would be if the
electron were at infinite distance (n  ) from the nucleus,
where there is no interaction and the energy is zero]
n + 1 __________ Excited state
ΔE= En+1 - En
n
h 
__________ Ground state
For hydrogen atom:
ΔE = -2.178 x 10-18 J x Z2 (1/nf2 – 1/ni2)
( Z = 1 for hydrogen atom)
The wavelength  of an emitted photon is:
ΔE = hc/
 = hc/ΔE
 1
1 

  R 2  2
n


n
f
i


or
1
 1

1
 R  2  2 
n


n
i
f


1
where,
R is Rydberg constant (1.09737316 x 107 m-1)
 is the wavelength of the transition between the
states ni and nf
(nf > ni)
The Quantum Mechanical Model of the Atom:
Heisenberg uncertainty principle:
We cannot determine the position and the momentum (i.e.
velocity) of a particle at the same time:
X (mv)  h/4
where, ΔX is uncertainty in position
Δ(mv) is uncertainty in momentum
Schrödinger equation:
ĤΨ = EΨ
where, Ψ is wave function
Ĥ is called an operator (represents a set of
mathematical instructions)
E represents energy values
The square of the wave function (Ψ2) indicates the
probability of finding an electron near a particular point in
space:
Ψ2 = probability or electron density
Chapters 21.4-7
Kinetics of Complex reactions
22.4 Relaxation Methods:

Flow techniques are limited by the speed with which it
is possible to mix solutions. For hydrodynamic reasons
it is impossible to mix two solutions in less than about
10-3 s.

If the half-life is less than 10-3 s, the reaction will be
completed by the time it takes for mixing to be
achieved. For example the neutralization of an acid by a
bass H+ + OH-  H2O, under ordinary conditions has a
half-life of 10-6 s or less and the rate therefore cannot be
measured by any technique involving the mixing of
solutions.

In the relaxation methods we start with the system at
equilibrium under a given set of conditions. We then
change these conditions very rapidly, the system is no
longer at equilibrium and it relaxes to a new state of
equilibrium.

The speed with which it relaxes can be measured,
usually by spectrophotometry, and we can then
calculate the rate constant.

There are various ways in which the conditions are
disturbed. One is by changing the hydrostatic pressure.
Another, the most common technique, is to increase the
temperature suddenly, usually by rapid discharge of a
capacitor, this method is called the temperature-jump
or T-Jump method.

It is possible to raise the temperature of a tiny cell
containing a reaction mixture by a few degrees in less
than 10-3 s, which is sufficiently rapid to allow us to
study even the fastest chemical processes.

Suppose that the reaction is the simple type:
AZ
(See Relaxation Figure)

We measure the displacement (x) at time (t) during
relaxation. The deviation of (x) from final equilibrium
is:
Δx = x – xe
and
d (x) / dt  x
(in first-order way)
dx
 k1 ( xe  x)  k 1 ( xe  x)  k1x  k 1x
dt
or
dx
 (k1  k1 )x
dt
 dx / x  (k
1
 k1 )  dt
The quantity Δx thus varies with time in the same
manner as does the concentration of a reaction in a firstorder reaction:

ln (Δx) = -(k1 + k-1) t + I
By integration and applying the boundary conditions
that:
Δx = (Δx)o
when t = 0
I = lnΔxo

or
ln(Δx) – ln(Δx)o = -(k1 + k-1) t
 (x) 
ln 
  (k1  k1 )t
 (x)o 
leads to
x
and
or

 (x)o e( k1 k1 )t
 (x) o 
ln 
  (k1  k 1 )t
(

x
)


(2)
We define a relaxation time t* as the time
corresponding to:
 (xo ) 
ln 
 1
(

x
)


(3)
From equation (3) into equation (2):
1 = (k1 + k-1) t*
t* 
1
(k1  k1 )
K = (k1/k-1)
and
and
(4)
(5)

Therefore, if we determine t* experimentally for
such a system we can then calculate (k1 + k-1).

The ratio (k1/k-1) is the equilibrium constant and
can be determined directly experimentally too,
from which the individual constants k1 and k-1 can
be determined from equations 4 and 5.
22.5 The temperature dependence of Reaction Rates:
The Arrhenius Law:

It states the temperature dependence of the rate
constant of a reaction.

It was found empirically that the rate constant k is
related to the absolute temperature T by the equation:
k = Ae-B/T
where A and B are constants. This relationship was
expressed by Van’t Hoff and Arrhenins in the form:
k = Ae-Ea/RT
where
(Arrhenius law)
(1)
R is the gas constant (8.314 J mol-1 K-1)
Ea is known as activation energy.
A is pre-exponential factor
By taking logarithms:
ln k  ln A 
Ea
RT
(2)
A plot of (ln k) versus (1/T) gives straight line of slope
 Ea 
equals to   R 


or
 Ea

log K  log A  

 2.303RT 
Plot of (log k) versus (1/T) gives slope equals to
(-Ea/2.303 R)
Example:
A given reaction in solution has a rate constant of 5.6x10-5 M-1s-1 at
25.0oC and 1.64 x 10-4 M-1s-1 at 40.0oC. Calculate the activation
energy Ea.
Answer:
k1 = 5.6 x 10-5 M-1s-1
at
T1 = 273.15 + 25 = 298.15 K
k2 = 1.64x10-4 M-1s-1
at
T2 = 273.15 + 40 = 313.15 K
k2 e Ea / RT2
  Ea / RT2  e( Ea / RT1  Ea / RT2 )
k1 e
 k2  Ea
Ea
Ea  1 1  Ea  T2  T1 


   


ln   


 k1  RT1 RT2 R  T1 T2  R  T1T2 
 k  E  T 

ln  2   a 
 k1  R  T1T2 
 1.64 x104 
Ea
15K



ln 



5 
1 1
5
.
7
x
10
8
.
314
Jmol
K
298
.
15
Kx
313
.
15
K




 1.64 x10 4 
1
1  298.15 Kx313.15 K 

Ea  ln 

8
.
314
Jmol
K


5 
15
K
5
.
7
x
10




Ea  54689.6 Jmol 1  54.7 KJmol 1
22.4 Reactions Approaching Equilibrium:

One complication is that a reaction may proceed to a state of
equilibrium that differs appreciably from completion. The
simplest case is when both forward and reverse reactions are
of the first order:
AZ
At time = 0
[A]o = ao
At time = t
[A] = ao – x and [Z] = x
Net rate = rate of forward – rate of reverse
Then the net rate is given as:
dx
 k1[ A]  k1[ Z ]  k1 (ao  x)  k1 x
dt
at equilibrium:
or
net rate = zero
and x = xe
rate of forward = rate of reverse
[A] = [A]e = (ao-xe)
and [Z] = [Z]e = xe
k1 (ao-x) = k-1xe
k-1 = k1[(ao-xe)/xe]
Form (2) into (1) for k-1
(1)
(2)
 ( a o  xe ) 
dx
 k1 (ao  x)  k1 
x
dt
 xe 
 k1 
  [( ao  x) xe  (ao  xe ) x]
 xe 
 k1 
  [( ao xe  xxe  ao x  xxe ]
 xe 
k
 1 ( a o xe  a o x )
xe
dx k1ao

( xe  x)
dt
xe
By integration and applying boundary condition:
at t = 0, x = 0 and we get:
xe  xe 

k1t  ln 
ao  xe  x 
(3)
 xe  xe 

ln 
A plot of  a
against t gives straight time of slope
 o  xe  x 
equals to k1. The constant k-1 for the reverse reaction can be
obtained from the values of the equilibrium constant K where,
k1
K
k 1
(4)
22.6 Molecularity and order:
Molecularity: is the number of molecules enter given
reaction:.
eg:
AP
A+BP
Order:
(monomolecular)
(bimolecular)
is an experimental quantity
Evidences for complex (composite) mechanism:
1)
When kinetic law does not correspond to stoichiometric
Equation: e.g: H2 + 2ICl  I2 + 2HCl

If the above reaction involved a single elementary step, it
would be third order; first order in H2 and second order in
ICl. However, the rate is:   [H2] [ICl]

This can be explained if there is initially a slow reaction
between one molecule of H2 and one of ICl:
H2 + ICl  HI + HCl
R.D.S
HI + ICl  I2 + HCl
__________________
H2 + 2ICl  I2 + 2HCl

The rate of the second step (fast) has no effect on the
overall rate.

In this scheme the first step is said to be the ratedetermining step (RDS).
Another example is the oxidation of bromide ion by
hydrogen peroxide in aqueous aid solution:
2Br- + H2O2 + 2H+  Br2 + 2H2O

The rate is given by the expression:
  [H2O2] [H+] [Br-]

This reaction occurs in two steps:
H+ + Br- + H2O2  HOBr + H2O
R.D.S
HOBr + H+ + Br-  Br2 + H2O

2)
The fact that the rate is proportional to [H2O] [H+] [Br]
suggests at once that reaction 1 (slow) is the rate
determining step
When the kinetic equation is complex:
e.g: for the reaction : H2 + Br2  2HBr
The rate is:
3)
k [ H 2 ] [ Br2 ]1 / 2
v
1  [ HBr ] / m[ Br2 ]
When intermediate can be detected experimentally.
22.7 The Steady-State Approximation :
(1)
Consider the consecutive reaction (first-order):
k1
k2
A 
X 
Z
d [ A]
  k1[ A]
dt
By integration:
[ A]
t
d [ A]
 k1  dt

[ A]
[ A]o
0
ln
[ A]
  k1t
[ A]o
[ A]  [ A]o e  k1t
(1)
Then
dX
 k1[ A]  k 2 [ X ]
dt
 k1t
= k1[ A]o e
 k2[X]
By integration we get:
k1
[ X ]  [ A]o
e k1t  e k2t
k 2  k1

(2)

From (1) and (3) into (2) for [A] and [X]:
(3)
 k2e( k1k2 )t  k1  k1t
d[ X ]
 k1[ A]o 
e
dt
k

k
2
1



A complex equation to be integrated.

By applying the conditions of the Steady State
Approximation (SSA) conditions that when:
a)
k2 >> k1 the term (k2 – k1)  k2


d[ X ]  k1 
  [ A]o k 2 e ( k1 k2 )t  k1 e k1t
dt
 k2 
b)
t >> 1/k
e
 k1t
then the term k1t >> 1 and
0
and
d[ X ]
0
dt
(SAA)

This means that once X is formed it converts to Z very
rapidly.

On the basis of the steady state approximation the rate of
change of the concentration of an active intermediate to a
good approximation, can be set equal to zero. Whenever the
intermediate is formed slowly it disappears very rapidly.
SSA:
dX
 0 where, X is an active intermediate
dt

Will apply steady state approximation to the following two
reactions:
1)
AXZ
2)
A+BX
XZ
1)
For the reaction:
AXZ
d [ A]
 k1 [ A]
dt
By integration:
ln
[ A]
  k1t
[ A]o
[ A]  [ A]o e  k1t
By applying SSA:
d[ X ]
 k1[ A]  k2 [ X ]  0
dt
(1)
or
k1[A] = k2[X]
[X ] 
k1
[ A]
k2
But
[A]o = [A] + [X] + [Z]

[Z] = [A]o – [A] – [X]
and
(2)
k 
[ Z ]  [ A]o  [ A]o e k1t   1 [ A]
 k2 
k 
[ Z ]  [ A]o  [ A]o e k1t   1 [ A]o e k1t
 k2 




k 
[ Z ]  [ A]o 1  e k1t   1 e k1t 
 k2 


(3)
The steady–state-treatment is of great importance in the
analysis of composite mechanism, since often there are
mathematical difficulties that make it impossible to obtain an
exact solution of the rate equation for the reaction.
(2)
Pre-equilibria:
For the reactions:
k1
1.
A+BX
k-1
k2
2.
XZ
Determine the rate of reaction () in terms of the rate of
formation of Z (z):
d[Z ]
 k2 [ X ]
dt
(1)
d[ X ]
 k1[ A][ B]  k 1[ X ]  k 2 [ X ]
dt
(2)
By applying steady-state approximation to equation (2):
d[ X ]
 0  k1 [ A][ B]  k 1 [ X ]  k 2 [ X ]
dt
or
k1[ A][ B]  k1[ X ]  k2[ X ]  [ X ](k1  k2 )
k1[ A][ B]
[
X
]

 k[ A][ B]
and
k 1  k 2
From (4) into (1) for [X] we get:
(4)
v  vZ 
or
d [ Z ] k 1k 2 [ A][ B]

dt
k 1  k 2
v  vZ  k[ A][ B]
(4)
(2nd order kinetic)
Similarly we can write:
d [ A]
[ A][ B]
 k1 [ A][ B]  k 1 k1 
k 1k 2 
dt
d [ A]
k
 (k1  1 )[ A][ B]
dt
k2
or
d [ A]
 k '[ A][ B]
dt
(2nd order kinetic)
Rate–Determining (Controlling) Steps:
For the reactions:
k1
1.
A+BX
k-1
k2
2.
XZ
To test the validity of SSA we consider two possibilities:
1.
When:
k2 >> k-1
(reaction 2 is fast)
This means that as soon X is formed in reaction (1) it
disappears very fast to Z in reaction (2). Therefore
reaction (1) will be the rate-determining step and
2.
When:
 = k1 [A] [B]
(2nd order kinetic)
k2 << k-1
(reaction 1 is fast)
Therefore, reaction (2) will be the rate-determining step
and:
 = k2 [X]
But
Kc = (k1/k-1) = [X] / [A][B]

[ X ]  (k1 / k1)[ A][ B]
or
  (k1k2 / k1 )[ A][ B]
or
  k "[ A][ B]
(2nd order kinetic)
(Therefore, SAA is valid approximation)
Relationship between the rate constants and the
equilibrium constant Kc:
Consider the reaction:
H2 + 2ICl  I2 + 2HCl
[ I 2 ][ HCl ]2
Kc 
[ H 2 ][ ICl ]2
The above reaction occurs in two steps:
1.
H2 + ICl  HI + HCl
2.
HI + ICl  HCl + I2
At equilibrium the rate of each elementary reaction and its
reserve must be the same:
Rate forward = rate reverse
k1 [H2] [ICl] = k-1 [HI] [HCl]
k2 [HI] [ICl] = k-2 [HCl] [I2]
The equilibrium constant for each reaction is:
K c1  (
 [ HI ][ HCl ] 
k1

)  
k 1  [ H 2 ][ ICl ]  eq
Kc2  (
 [ HCl ][ I 2 ] 
k2
)  

k 2
[
HI
][
ICl
]

eq
The product of these two equilibrium constants is:
 [ I 2 ][ HCl ]2 
k1k2
  Kc
K c1K c 2  (
)  
2 
k1k2  [ H 2 ][ ICl ] 
where, Kc is the equilibrium constant for the overall reaction.
In general:
Kc = Kc1 Kc2 Kc2 …… = k1 k 2 k 3…/ k -1 k -2 k -3 ……

The principle that at equilibrium each elementary
process is exactly balanced by its reverse reaction is
known as the principle of detailed balancing.
Applications of Steady-State-Approximation:
1)
Organic free radical chain reactions:
(Dehydrogentation of ethane to ethylene):
C2H6  C2H4 + H2
1.
C2H6  2CH3
(initiation)
2.
CH3 + C2H6  CH4 + C2H5
(initiation)
3.
C2H5  C2H4 + H
(chain propagation)
4.
H + C2H6  H2 + C2H5
(chain propagation)
5.
2C2H5  C4H10
(termination)

The initiation process involves the breaking of a C-C
bond, which is the weakest bond in the molecules.

Methane CH4 and butane C4H10 are observed as minor
products of the reaction.

The rate in terms of the rate of formation of product
C2H4 is:
v = vC2H4 = k3 [C2H5]

(1)
Apply SSA to the three active intermediates H, CH3
and C2H5 radicals:
d[H ]
 o  k 3 [C 2 H 5 ]  k 4 [ H ][C 2 H 6 ]
dt
(2)
d[CH 3 ]
 o  k1[C2 H 6 ]  k 2 [CH 3 ][C2 H 6 ]
dt
(3)
d [C2 H 5 ]
0
dt
 k 2 [CH 3 ][C2 H 6 ]  k3 [C2 H 5 ]  k 4 [ H ][C2 H 6 ]  k5 [C2 H 5 ]2
(4)
By adding equations (2), (3) and (4) we get:
0  k1[C2 H 6 ]  k5[C2 H 5 ]2
1/ 2
k 
[C2 H 5 ]   1  [C2 H 6 ]1 / 2
 k5 
From (5) into (1) for [C2H5] we get:
1/ 2
k 
v  k3  1  [C2 H 6 ]1/ 2
 k5 
v  k[C2 H 6 ]1/ 2
where, k = k3(k1/k5)1/2
(5)
(Thermal decomposition of acetaldehyde):
CH3CHO  CH4 + CO
1.
k1 CH  CHO
CH 3CHO 
3
2.
k2 CH  CH CO
CH 3  CH 3CHO 
4
3
3.
k3 CH  CO
CH 3CO 
3
4.
k4 C H
CH 3  CH 3 
2
6

The rate in terms of the rate of formation of product methane
CH4 is:
v = vCH4 = k2[CH3][CH3CHO]

(1)
By applying the SSA to the two radicals CH3 and CH3CO:
(CHO are side products and will not beused in the SSA)
d [CH 3 ]
0
dt
 k1[CH 3CHO]  k2 [CH 3 ][CH 3 CHO]  k3[CH 3CO]  k4 [CH 3 ]2
(2)
d [CH 3CO]
 0  k2 [CH 3 ][CH 3CHO ]  k3[CH 3CO]
dt
(3)
By adding equation (2) to equation (3) we get:
k1[CH 3CHO ]  k4 [CH 3 ]2  0
k
[CH 3 ]   1
 k4



and
1/ 2
[CH 3 CHO ]1 / 2
(4)
From (4) into (1) we get::
1/ 2
 k1 
v  k2   [CH 3CHO ]3 / 2
 k4 
This agrees with the experimental fact that the order is threehalves with respect to acetaldehyde.
3)
Explosions:

Explosions are chemical reactions that occur extremely
rapidly with the release of a considerable amount of energy.

The release of this energy causes every rapid rise in pressure
in the gaseous products of the reaction.

Gas-phase reactions that occur explosively occur by
mechanism that has a special feature. Some of the free
radicals involved in the mechanism are capable of
undergoing branching reactions, in which one radical
produces more than one radical:
e.g.: the catalytic reaction of H2 and O2 in the gas phase.
H + O2  OH + O
O + H2  OH + H

When reactions of this type occur to a significant extent, the
total number of free radical in the system may increase
rapidly, and the reaction occurs with very high velocity, and
since energy is released, the result is explosion.
4)
Catalysis:

The energy difference (ΔE) is the same for the catalyzed and
the un-catalyzed, but the activation energies in both
directions are lower for the catalyzed reaction.

The present section is concerned mainly with homo-gaseous
catalysis.

The rate of a catalyzed reaction is often proportional to the
concentration of the catalyst [C]:
v  [C ]
v = k[C]
I)
Acid-Base Catalysis:

It is highly dependent on the pH of the reaction.

There are regions of catalysis by hydrogen and
hydroxide ions, separated by a region in which the
amount of catalyst is unimportant in comparison with
the spontaneous reaction. When the catalyst is largely
by hydrogen ions:
S + H+  products
v = kH+ [H+][S]
but v/ [S] = k
(reaction rate constant of the uncatalyzed reaction)
k = kH+[H+]
and by taking logarithms we get:
log k = log kH+ + log [H+]
log k = log kH+ - pH
where, k is rate constant of un-catalyzed reaction.
kH+ is rate constant of acid catalyzed reaction.
when:
slope = 0
(un-catalyzed)
slope = ± 1
(catalyzed)
(II) Enzyme Catalysis:

The enzymes, which are proteins, are the biological catalysts.
Their action shows some resemblance to the catalytic action
of acids and bases but is considerably more complicated. The
details of the mechanisms of action of enzymes are still being
worked out, and much research remains to be done. Only
brief account can be given here.

The simplest case is that of an enzyme-catalyzed reaction
where there is a single substrate (S) an example is the
hydrolysis of an ester.

The rate varies linearly with the substrate concentration (i.e.
concentration dependent). At low concentration it is firstorder kinetic and at high concentration it is zero-order
kinetic.

Consider the single substrate enzyme-catalyzed reaction;
k1
E + S  ES
k-1
k2
ES  E + Z

The rate of reaction in terms of the rate of formation
of product Z is given as:
v = vZ = k2 [ES]
(1)

By applying the SSA to the active intermediate ES we get:
d [ ES ]
 0  k1[ E ][ S ]  k 1[ ES ]  k 2 [ ES ]
dt
0  k1[ E][ S ]  (k1  k2 )[ ES ]
But
[E]o = [E] + [ES]

[E] = [E]o – [ES]
(2)
(3)
Form equation (3) into equation (2) for [E] we get:
k1 ([E]o – [ES])[S] – (k-1 + k2)[ES] = 0
and
and
k1 [S][E]o – k1[S][ES] – (k-1 + k2) [ES] = 0
[ ES ] 
k1[ E ]o [ S ]
( k1  k2 )  k1[ S ]
From (4) into (1) for [ES] we get:
v
k1k 2 [ E ]o [ S ]
(k 1  k 2 )  k1[ S ]
Dividing by k1 we get:
v
k 2 [ E ]o [ S ]
[( k 1  k 2 ) / k1 ]  [ S ]
v
V [S ]
km  [ S ]
(4)
where,
V = k2[E]o
(Limiting rate)
km = (k-1 + k2)/k1
(Michaelis constant)
V [S ]
v
km  [ S ]
vkm  [ S ]  V [ S ]
or
Dividing by the term (v km [S]) we get:
1
1 V 1


[ S ] km km 
Plotting 1/v versus 1/[S] we get a slope of V/km and an
intercept of 1/km
Limiting cases:
v
1)
V [S ]
km  [ S ]
At very low [S]: [S] << km
and
v
v[ S ]
 k [ S ]`
km
(first-order kinetic)
2)
At high [S]:
and
3)
[S] = km
v = V[S] / 2[S] = V/2
(zero-order kinetic)
At very high [S]: [S] >> km
and
v = V[S] / [S] = V
v
V [S ]
km  [ S ]
(zero-order)
Chapter 25.3-5
Rates of Surface Processes
25.3 Adsorption processes:

A plane that separates two phases is known as a surface
or an interface.

Two main types of adsorption:
1.
Physical adsorption (Physisorption):
- The adsorbed molecules are held to the surface by
van der wals forces.
- The heat evolved is usually small of the order
20 kJ/mol.
2.
Chemical adsorption (Chemisorption):
- The adsorbed molecules are held to the surface by
covalent forces.
- The heat evolved is comparable to that evolved in
chemical bonding of the order 100-500 kJ/mol.
25.4 Adsorption Isotherms:

An equation that relates the concentration of a
substance in the gas phase or in solution to the fraction
of the surface where it is adsorbed at a fixed
temperature is known as an adsorption isotherm:
Langmuir Isotherm:
A
|
ka
|
A + S  S 
|
Product + S
kd


Two assumptions:
a.
All sites are identical
b.
Only uni-molecular layer is formed.
Suppose that after equilibrium is established, a fraction
 of the surface is covered by adsorbed molecules, then
a fraction (1- ) will not be covered. The rate of
adsorption va is given as:
va  (1   )
(Empty fraction)
va  [A]

va = ka [A] (1-)
(1)
The rate of desorption vd is:
vd  

(Occupied fraction)
vd = kd ()
At equilibrium: rate of adsorption = rate of desorption
ka [A] (1- ) = kd ()
or
Dividing by kd we get:
(ka/kd)[A](1- ) = 
K[A] (1- ) = 
 = K[A] (1- )
(2)
where, K is the equilibrium constant and equals (ka/kd).
 = K [A] – K [A]
or
(3)
 + K [A] = K [A]
 (1+ K [A]) = K [A]

K [ A]
1  K [ A]
(Langmuir Isotherm) (4)
Limiting cases:

1.
At very low [A]:
K [ A]
1  K [ A]
1 >> K [A]
 = K [A]
(5)
and  α [A]
2.
3.
At high [A]:
Equation (3) is:
/K [A] = (1- )
Equation (4) is:
/K [A] = 1/(1+ K [A])
Then
(1- ) = 1/(1+ K [A])
At very high [A]:
1 << K [A]
(1- ) = 1/ K [A]
(6)
(1- ) α 1/[A]

Langmuir emphasized that adsorption involves the formation
of a uni-molecular layer. The additional adsorption on the
layer already present is generally weak adsorption.
Applications of Langmuir Isotherm:
1) Adsorption with dissociation:
A A
|
|
ka
| |
A2 + S S  S S 
|
|
Product + S S
kd

When A2 (e.g. H2) is dissociated on the surface, where 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.

An example is when hydrogen gas is adsorbed on the surface
of many metals.

The process of adsorption is now a reaction between the gas
molecule and two surface active sites,
and the rate of adsorption is:
va = ka [A] (1-)2
(1)
and the rate of desorption is:
vd = kd ()2
(2)
2 = K [A] (1- )2
(3)
at equilibrium:
 = K1/2 [A]1/2(1- )
= K1/2 [A]1/2 -  K1/2 [A]1/2
 +  K1/2 [A]1/2 = K1/2 [A]1/2
 (1 + K1/2 [A]1/2) = K1/2 [A]1/2
K 1 / 2[ A]1 / 2

1  K 1 / 2[ A]1 / 2
At very low [A]:
and
(4)
1 >> K [A]1/2
 = K1/2 [A]1/2
and
 α [A] 1/2
At very high [A]:
1 << K [A]1/2
(1- ) = 1/ K 1/2[A]1/2
(1- ) α 1/[A]1/2
(6)
2) Competitive adsorption:

When two substances A and B are adsorbed on the some
surface:
The rates of adsorption are:
 aA  kaA[ A](1   A   B )
(1)
 aB  kaB [ B](1   A   B )
(2)
The rates of desorption are:
 dA  k dA A
(3)
 dB  k dB B
(4)
From equation (1) and (3) at equilibrium:
A 
where,
K A[ A]
1  K A[ A]  K B [ B]
(5)
K A  (k aA / k dA )
Similarly, from equations (2) and (4) at equilibrium:
B 
K B [ B]
1  K A[ A]  K B [ B]
(6)
K B  (k aB / k dB )
where,
BET Isotherm:
(Brunaur, Emmett and Teller) in 1938 propose BET isotherm for
multilayer adsorption.
It is a multi layer adsorption.
The convenient simple form is:
PPo
1
P


V ( Po  P ) Vo K Vo
where, V is the volume of the gas adsorbed at pressure P and
Vo the volume that can be adsorbed as a monolayer.
Po is the saturation vapor pressure
K is equilibrium constant for the adsorption
Other Isotherms:

The various isotherms of the Langmuir type are based on the
simplest of assumptions, all sites on the surface are assumed
to be the same, and there are no interactions between
adsorbed molecules. Systems that obey these equations and
often referred to as showing ideal adsorption.

Systems frequently deviate significantly from the Langmuir
equations.
Freundlich-Isotherm:
The amount (x) adsorbed on the surface is related to the
concentration (C) as follows:
x α Cn
x = k Cn
where, k and n are constants
By taking logarithms:
log x = log k + nlog C
A plot of log x versus log C gives a straight line
Temkin-Isotherm:
where  is related to P as:
 α ln (aC)
 = (1/f) ln aC
where, f and a are constants.
25.5 Kinetics of catalyzed chemical reactions on surfaces:

An important concept in connection with surface
reactions is the molecularity. Reactions involving a
single reacting substance are usually unimolecular, and
these involving two reacting substances are usually bimolecular.
I)
Kinetic of Unimolecular Catalyzed Reactions:

In the simplest case the rate of reaction is proportional
to Θ and is thus:
α
v
1.
At low [A]:
kK [ A]
1  k[ A]
1 >> K [A]
 α [A]
(first order kinetic)
 = k K [A]
2.
At high [A]:
1 << K [A]
v=K
(zero-order kinetics)
The dependence of (v) on [A] is shown in the given
figure for unimolecular reaction.
2)
Kinetics of unimolecular catalyzed reactions with
inhibition:

Sometimes a substance (I) other than the reactant A is
adsorbed on the surface, with the result that the
effective surface area and, there fore the rate are
reduced (i.e. inhibition and I is said to be inhibitor or
poison):

K [ A]
1  K [ A]  Ki [ I ]
The rate of reaction (v) is proportional to () and thus given as:
v

kK[ A]
(1  K [ A])  K i [ I ]
A special case when the surface is fairly fully covered
by the inhibitor: i.e. I is strongly
Ki[I] >> (1+K[A])
v

kK[ A]
[ A]
 K'
Ki [ I ]
[I ]
A good example is provided by the decomposition of
ammonia on platinum, the rate low is:
v
K '[ NH 3 ]
[H 2 ]
(There is no inhibition by N2 but by H2)
II)
Bimolecular Catalyze Reaction:

Two mechanisms to be studied:
1)
Langmuir–Hinshelwood mechanism:
(when two molecules are adsorbed on the surface)
2)
Langmuir–Rideal mechanism:
(when only one molecule is adsorbed on the
surface)
1)
Langmuir–Hinshelwood mechanism:
A….B
|
|
|
|
|
|
A + B + S S  S S  Product + S S
The rate is given as:
v   A B
v  k A B
v
kK A K B [ A][ B]
(1  K A [ A]  K B [ B]) 2
v

kK A K B [ A][ B]
(1  K A [ A]  K B [ B]) 2
At very low concentrations of A and B:
1 >> (KA[A] + KB[B])
v = K [A][B]

(second-order kinetics)
A special case is when one reactant (e.g: A) is
weakly adsorbed and the other is strongly
adsorbed (e.g: B):
KB[B] >> (1 + KA[A])
v

kK A K B [ A][ B]
[ A][ B]
[ A]

K
'

K
'
( K B [ B]) 2
[ B]2
[ B]
An example is the reaction between carbon
monoxide and oxygen on quartz, where the rate is
directly proportional to the pressure of oxygen
and inversely proportional to the pressure of CO:
v = K (PO2/PCO)
2)
Langmuir–Rideal mechanism:

In which one molecule is not adsorbed (e.g: A)
reacts with or adsorbed molecule (e.g: B):
B
B…. A
|
|
|
A + S  S  Product + S

v  [ A] B
v  k[ A] B
v
kKB [ A][ B]
(1  K B [ B])
(See figure 17.2c)

Not many ordinary chemical reactions occur by a
Longmuir-Rideal mechanism.
e.g: The combination of hydrogen atoms, for
example, is sometime a first-order reaction and it
appear to occur by the mechanism.
kK[ H ]2
v  k[ H ] 
(1  K [ H ])
kK[ H ]2
v
(1  K [ H ])

At low temperature:
(Rate of adsorption >> Rate of desorption)
i.e. high [H] and K [H] >> 1
v = kK[H]

(first-order)
At high temperature:
(Rate of ads. << Rate of des.)
i.e. low [H] and K [H] << 1
v = kK[H]2
(second-order)
(An increase in the order from 1 to 2 has in fact
been observed experimentally as the temperature
is raised)
Electronegativity:
The calculations for the hydrogen chloride led to the calculations that there
is a piling up of electron density between the nuclei. The H2 molecule is
symmetrical, SO that the electron cloud lies symmetrically between the
nuclei.
The quantum-mechanical calculations for hydrogen chloride, on the other
hand: show that the electron could lies more tow and the chlorine atom.
The consequence of this asymmetry is that the molecule has a dipole
moment. The dipole moment of a diatomic molecule is equal to the effective
charge q at the +ve and –ve ends multiplied by the distance between them:
+q
d
-q
+
=qd
Where,
q = 1.602 X 10-19C and d = internuclear distance
The SI unit of dipole moment is meter, mm: 1D = 3.336 X 10-30 cm
The percentage ionic character of the bond is :
 exp
x100
 ionic
% ionic character =
If we know the distance between two atoms in a diatomic molecule, we can
calculate a dipole moment ionic. The term exp is the experimental dipole
moment.
The theoretical percent ionic character is :
%
ionic
character
=
(2 / 1  2 ) x100
Liuus Pauling :
Ecovalent
= [D(AA)D(BB)]12
Eionic
= D(AB) – [D(AA)D(BB)]12
(Eionic)1/2 = Kl A   Bl
Pauling eq~
D(AA) is dissociation energy, K is a coefficient ~ and ,  A   B  M Dipole
moment
Eig: MHCl = 2.8 – 2.1 = 0.7 D
Mullikan:
M 
( I  A) / ev
5.6
Mullikan eq~
See table for electronegativities :
Table Atomic Electronegativities of the Pauling (Mulliken) Scale
H
2.20 (3.06)
Li
Be
B
C
N
O
F
0.98 (1.28)
1.57 (1.99)
2.04 (1.83)
2.55 (2.67)
3.04 (3.08)
3.44 (3.22)
3.98 (4.44)
Na
Mg
Al
Si
P
S
Cl
0.93 (1.21)
1.31 (1.63)
1.61 (1.37)
1.90 (2.03)
2.19 (2.39)
2.58 (2.65)
3.16 (3.54)
K
Ca
Ga
Ge
As
Se
Br
0.82 (1.03)
1.00 (1.30)
1.81 (1.34)
2.01 (1.95)
2.18 (2.26)
2.55 (2.51)
2.96 (3.24)
Rb
Sr
Im
Sn
Sb
Te
I
0.82 (0.99)
0.95 (1.21)
1.78 (1.30)
1.96 (1.83)
2.05 (2.06)
2.10 (2.34)
2.66 (2.88)