CH162 Kinetics

1
CH162Kinetics

2
5 lectures VGS

• Experimental methods in kinetics
• Reactions in liquid solution
• Photochemistry
• Femtochemistry

3
Gas-phase vs. Liquid phase
Solvent
Isolated molecules
Solute

• Intramolecular processes

• Intra/intermolecular processes
• Solute-solvent interactions

4
Experimental methods
A B ? Products
Change in conc.
dA dt
Rate -
Change in time
How do we do this?
5
Experimental methods
In real-time analysis, the composition of a
system is analysed while the reaction is in
progress through e.g. spectroscopic observation
of the reaction mixture.
Reagents are driven quickly into mixing chamber
and the time-dependence of the conc. is
monitored. Limitation Mixing times!!
6
Experimental methods
The transient species can be monitored by either

1. Absorption
2. Fluorescence
3. Conductivity
4. Pressure
5. NMR
6. Mass detection
7. Electron detection

7
Experimental methods
Absorption and fluorescence
A B ? Products
8
Experimental methods
Conductivity
C4H9Cl H2O ? C4H9OH H Cl-
9
Experimental methods
Pressure
2NOBr(g) ? 2NO(g) Br2(g)
2 moles ? 3 moles
10
Experimental methods
NMR
Different H chemical shifts in NMR
11
Experimental methods
Flash photolysis by-passes mixing times.
Reactants are premixed and flowed into a
photolysis cell. A pulse of light is then used to
produce a transient species whose concentration
is monitored as a function of time.
12
Experimental methods
A typical setup for transient absorption
Setup
Detector
Pulse 2
Pulse 1
13
Experimental methods
A typical setup for transient absorption
Setup
14
Absorption spectroscopy in a little detail..
Beer-Lambert law
if x ltlt 1
15
light sources for flash photolysis

• flash lamps

• pulse length ?s - ms

• lasers

• pulse length ? fs 10-15 s
• high power ? low precursor concentration
• high repetition rate ? 500Hz

16
Experimental methods
A typical setup for transient mass spectrometry
Step 1
HV ()
17
Experimental methods
A typical setup for transient mass spectrometry
Step 3
TOF tube
Time of flight (ion)
TOF (d)
Detector
HV ()
18
Experimental methods
-Ion time of flight mass spectrometry
Potential energy Ep of charged particle in
electric field is
When charged particle is accelerated into a TOF
tube by U, Ep is converted to kinetic energy Ek
19
Experimental methods
-Ion time of flight mass spectrometry
Potential energy is converted into kinetic energy
meaning
In field free region, velocity remains constant
20
Experimental methods
-Ion time of flight mass spectrometry
Rearranging so that the flight time is subject of
formula
As flight length and voltage (potential
difference) are constants
21
Experimental methods
-Ion time of flight mass spectrometry
Intensity (arb. units)
Time-of-flight (µs)
22
Experimental methods
Using a known mass to predict an unknown mass
? (4.24µs)
? (4.12µs)
4.12 µs OH 4.24 µs OH2
H (1µs)
Intensity (arb. units)
Time-of-flight (µs)
23
Experimental methods
-Ion time of flight mass spectrometry
What does one expect to observe with such
measurements?
Pulse 1 dissociates
Pulse 2 probes through ionization
pulse 1
OH intensity
Time delay between pulse 1 and 2
24
Experimental methods
NMR
The a and ß forms interconvert over a timescale
of hours in aqueous solution, to a final stable
ratio of aß 3664, in a process called
mutarotation
25
Experimental methods
NMR
We can monitor this interconversion using NMR.
How?
26
Experimental methods
NMR
Day 1 1053 am Only ?
27
Experimental methods
NMR
Day 1 127 pm
28
Experimental methods
NMR
Day 2 0913 am
29
Experimental methods
A typical setup for transient photoelectron
spectroscopy
Step 1
30
Experimental methods
-Photoelectron spectroscopy
Step 3

1
e1-

e2-
2
Detector
31
Experimental methods
Photoelectron spectroscopy is essentially the
photoelectric effect in the gas phase. By
measuring the kinetic energy of the ejected
photoelectrons, we can infer the orbital
energies, not only of the molecular ion but also
of the neutral molecule.
32
Experimental methods
As the energy is conserved when a photon ionizes
a sample, the energy of the incident photon (hv)
must be equal to the sum of the Ii and the KE of
the photoelectron.
By knowing the kinetic energy of the
photoelectron and the frequency of the incoming
radiation, Ii may be measured.
33
Experimental methods
Photoelectron spectra are interpreted in terms of
an approximation called Koopmans theorem. This
states that the ionization energy (Ii) is equal
to the orbital energy of the ejected
photoelectron. The theory however ignores that
the remaining electrons adjust their
distribution when ionization occurs (i.e.
ionization is assumed to be instantaneous). The
ejection of the electron can leave the ion in a
vibration-ally excited state. As a result, not
all the excess energy of the photon appears as
kinetic energy of the photoelectron. As a result,
we write
34
Experimental methods
Example Photoelectrons ejected from N2 with
He(I) radiation had kinetic energies of 5.63 e.V.
Helium(I) radiation of wavelength 58.43 nm
corresponds to 1.711 x 105 cm-1 which
corresponds to an energy of 21.22 e.V. What is
the ionization energy of the molecular orbital?
35
Experimental methods
therefore
This corresponds to the removal of an electron
from the HOMO (3sg) orbital.
36
Further reading material

• In addition to the books already suggested,
  further information regarding these lectures can
  be found in
• Review of Scientific Instruments V26,
  PP1150-1157, yr1955 (lectures 12)
• Foundations of Spectroscopy, Duckett and Gilbert,
  Oxford Chemistry Primers (lectures 12)
• Elements of Physical Chemistry, Atkins and de
  Paula, 4th edition (lectures 23)

37
Reactions in liquid solution

• Activation control and diffusion control
• 2. Diffusion and Ficks laws
• 3. Activation/diffusion revisited
• 4. Thermodynamic formulation of rate coefficient
• 5. Ionic-strength effects

38
Reactions in liquid solution
With reactions in solution, the reactant
molecules do not fly freely through a gaseous
medium and collide with each other. Instead, the
molecules wriggle past their closely packed
neighbours as gaps open up in the structure.
For example, in nitrogen gas, at 298 K and 1
atm., the molecules occupy only 0.2 of the total
volume. In a liquid, this figure typically rises
to more than 50.
39
Activation diffusion control
The rate determining step plays a critical role
in solution-phase reactions, leading to a
distinction between diffusion control and
activation control. Suppose that a reaction
between two solute molecules A and B occurs
through the following mechanism. A and B move
into each others vicinity through diffusion
forming an encounter pair, AB at a rate
proportional to the concentration of A and B,
40
Activation diffusion control
The encounter pair persists as a result of the
cage effect caused by the surrounding solvent.
However the encounter pair can break up if A and
B have opportunity to diffuse apart
The competing process is the reaction between A
and B while an encounter pair, forming products.
41
Activation diffusion control
The rate of formation of products is given by
Since
Apply SSA to AB gives
kdAB-kdAB-kaAB0
42
Activation diffusion control
Or
Therefore
There are two limits for this expression, kagtgtkd
and kaltltkd
43
Activation diffusion control
Diffusion controlled limit kagtgtkd
Rate governed by rate of diffusion of reactants
Activation controlled limit kaltltkd
Rate governed by rate in which energy is
accumulated in encounter pair
44
Diffusion
Plays a critical role involving reactions in
solution. Molecular motion in liquids involves a
series of short steps, with constant changes of
molecular direction. The process of migration
by random jostling motion is called diffusion and
the molecular motion in random directions is
known as a random walk. Suppose that there
is an initial concentration gradient in a liquid,
the rate at which molecules spread out is
proportional to the concentration gradient, ?c/?x
or
Rate of diffusion ? concentration gradient
45
Diffusion
The rate of diffusion is measured by the flux, J.
This corresponds to the number of particles
passing through an imaginary window in a given
time interval, divided by the area of the window
and duration of the interval
Number of particles passing through window
J

Area of window x time interval
J

-D x concentration gradient
dc
J
-D
D diffusion coefficient
dx
46
Diffusion-Ficks first law
dc
J
-D
dx
c
x
47
Diffusion-Ficks first law
Diffusion coefficients at 25 oC, D/10-9m2s-1
Ar in tetrachloromethane 3.63 C12H22O11
(sucrose) in water 0.522 CH3OH in
water 1.58 H2O in water 2.26 NH2CH2COOH in
water 0.673 O2 in tetrachloromethane 3.82
Molecules that diffuse fast have a larger
diffusion coefficient
48
Diffusion
Suppose in a region of unstirred aqueous solution
of sucrose the molar concentration gradient is
-0.1 moldm-3cm-1, calculate the flux, J
49
Diffusion
Calculate the amount of sucrose passing through a
1 cm square window in 10 minutes
50
Diffusion-Ficks second law
Ficks second law of diffusion, also known as the
diffusion equation, enables us to predict the
rate at which the concentration of the solute
changes in a non-uniform solution
Rate of change of concentration in a region

D x curvature of concentration in region
or
?2c
?c
D
?t
?x2
51
Diffusion-Ficks second law
?2c
?c
J
?
?
?t
?x2
52
Diffusion-Ficks second law
constant
Diffusion equation tells us that a concentration
with unvarying slope through a region results in
no net change in concentration.
53
Diffusion-random walk
(a)
(b)
(a) Snapshot picture of the instantaneous
configuration of a liquid. (b) Trajectories of
particles in the 2ps following snapshot (1ps
1x10-12s)
54
Diffusion-random walk
The nature of diffusion can be considered as the
outcome of a series of steps in random directions
and random distances molecules take-the random
walk. Although a molecule may take many steps,
it may only result in it being localized at a
point close to its initial starting point-some
steps take it away but others bring it back.
The net distance traveled in a time t from the
molecules initial starting point is measured by
the root mean square distance, d
d (2Dt)1/2
55
Diffusion-random walk
The diffusion coefficient of H2O in water is 2.26
x 10-9m2s-1 at 25 oC. How long does it take for
an H2O molecule to travel (a) 1 cm and (b) 2 cm
from its starting point in a sample of unstirred
water?
56
Diffusion-Einstein-Smoluchowski equation
The Einstein-Smoluchowski equation describes the
relation between D and the time a molecule takes
its steps, t (tau), and the length of each step,
? (lambda).
?2
D
2t
The larger and faster the step, the higher the
diffusion coefficient. t represents the average
lifetime of a molecule near another molecule
before it moves to the next position.
57
Diffusion-Einstein-Smoluchowski equation
Suppose an H2O molecule moves through one
molecular diameter (200 pm) each time it takes a
step in a random walk. What is the time for each
step at 25 oC?
58
Diffusion-Temperature
The diffusion coefficient, D, increases with
temperature as the increase in temperature
enables a molecule to escape more easily from the
attractive forces exerted by its
neighbours. Assuming that the rate of the random
walk (1/t) follows an Arrhenius temperature
dependence with an activation energy Ea, then D
will follow the relation
D D0 e-Ea/RT
59
Diffusion-Temperature viscosity
As the viscosity of the fluid increases, we
expect that the diffusion of particles through
the liquid should decrease. That is, we
anticipate that ? ? 1/D, where ? is the
coefficient of viscosity. The Einstein relation
states
kBT
D
6p?a
where a is the radius of the molecule and,
? ?0 eEa/RT
60
Diffusion-Temperature viscosity
The viscosity of water at 40oC and 80oC
corresponds to ?406.8x10-4 kgm-1s-1 and
?803.5x10-4 kgm-1s-1 respectively. Calculate the
activation energy for the viscosity of water.
61
Recap. Activation diffusion control
Diffusion controlled limit kagtgtkd
Rate governed by rate of diffusion of reactants
Activation controlled limit kaltltkd
Rate governed by rate in which energy is
accumulated in encounter pair
62
Diffusion and reaction
Diffusion controlled reactions are characterized
by a rate constant (kd) 109 dm3mol-1s-1. Their
rate depends on the rate at which reactants
diffuse together.
See Atkins 8th edition, page 877-878
Stokes-Einstein
Assuming RARB(1/2)R
63
Diffusion and reaction
For a diffusion controlled reaction, given
?8.9x10-4 kgm-1s-1 at 25oC, calculate the
diffusion controlled rate constant
64
Diffusion and reaction
Two neutral species A and B, with diameters 588pm
and 1650pm respectively, undergo the diffusion
controlled reaction AB-gtP in a solvent of
viscosity ?2.37x10-3 kgm-1s-1 at 40oC. Calculate
the initial rate if the initial concentration of
A and B are 0.150 moldm-3 and 0.330 moldm-3
respectively.
dP dt
kdAB
kd4pR(DADB)NA
(
1
1
2kBTN
(

x
(RARB)x

RA RB
3?
65
Diffusion and reaction
66
Thermodynamic formulation of RC
In activation control reactions, the
concentration of encounter pairs AB is
maintained at its equilibrium value, determined
by kd/kd KAB (equilibrium constant). The rate
coefficient is given by
67
Thermodynamic formulation of RC
By applying transition state theory to ka, then
the overall rate constant can be expressed as
where
and
K is the equilibrium constant between reactants
and activated complex
68
Thermodynamic formulation of RC
Alternatively, we may write the Eyring equation
in terms of the enthalpy of activation ?H and
entropy of activation ?S
Compare
69
Ionic strength effects
Consider a solution of ionic strength I, where
ci Conc. in moldm-3 of ith ion zi2 Charge of
ith ion
In solutions with non-zero ionic strength, there
are interactions between many ions whilst we have
only considered those between reactant pairs.
What happens in solutions with high concentration
of non-reactive ions?
This equation is only valid for reactions of ions
at infinite dilution and needs modification at
higher ionic strengths
70
Ionic strength effects
We start by considering the equilibrium between
separate reactants and the encounter complex. For
non-ideal solutions
and
71
Ionic strength effects
We can now re-write
to include activity coefficients
The activity coefficient (?A for species A etc.)
accounts for the deviation from ideal behaviour
of a mixture. In an ideal mixture, interaction
between each pair of chemical species is
identical.
72
Ionic strength effects
At low ionic strength, the activity coefficient
of an ion may be calculated from the limiting
Debye-Hückel equation
Where A is a constant (0.509 dm3/2mol-1/2 at
298K), zA is the charge of species A and I is the
ionic strength. Substituting this into the
overall rate constant, we obtain
Note The encounter pair has a charge zAzB and
k0 is rate constant at zero ionic strength and
equals kaKAB.
73
Ionic strength effects
A plot of log k vs. ?I is a straight line with
slopezAzB.
y c mx
log(k)log(k0) 1.018zAzB?I
The slope gives information about the charge
types involved in the activated complex of the
rate determining step.
74
Ionic strength effects
The slopes of the lines are those given by the
Debye-Hückel limiting law.
75
Diffusion and reaction
The rate constant of the reaction
H2O2(aq)I-(aq)H(aq)-gtH2O(l) HIO(aq)
is sensitive to the ionic strength of the aqueous
solution in which the reaction occurs. At 25 oC,
k12.2 dm6mol-2min-1 at an ionic strength of
0.0525. Estimate the rate constant at zero ionic
strength.
76
Diffusion and reaction
An ion A of charge number 1 is known to be
involved in the activated complex of a reaction.
Deduce the charge number of the other ion (B)
from the following data
I 0.005 0.01 0.015 0.02 0.025
0.03 k/k0 0.995 0.925 0.875
0.835 0.800 0.770
77
Ionic strength effects
At high ionic strength, log(?A)-AzA?I breaks
down and is replaced by
Where B is a constant and a is the radius of the
ion.
It is important to note that the above treatment
is only strictly applicable to activation
controlled reactions as we have assumed an
equilibrium concentration of the encounter
complex.
78
Photochemistry

• Introduction to photochemistry (day-to-day
  examples)
• 2. Review of Jablonski diagrams
• (a) Fluorescence
• (b) Phosphorescence
• (c) Intersystem crossing (ISC), internal
  conversion (IC)
• 3. Quantum yields for photophysical events
  (Stern-Volmer plots)
• 4. Complex photochemical processes

79
Introduction to photochemistry
Vision-the good
hv
11-cis-Retinal
All-trans-Retinal
80
Introduction to photochemistry
DNA damage by UV radiation-the bad
Photodimerization of adjacent Thymine bases upon
UV absorption
Cyclobutane thymine dimer (linked cell death)
The so-called 6-4 photoproduct (linked to DNA
mutations and tumours)
81
Introduction to photochemistry
Multi-step organic synthesis- the ugly
82
Jablonski diagrams
Fluorescence vs. phosphorescence
83
Green fluorescent protein
Osamu Shimomura Martin Chalfie
Roger Y. Tsien
Nobel Prize in Chemistry 2008 "for the discovery
and development of the green fluorescent protein,
GFP"
GFP
84
GFP chromophore
p-hydroxybenzylidene-imidazolidone
85
Fluorescence vs. phosphorescence kinetics
86
Quantum yields in photophysics
Primary process Products formed directly from
the excited state of a reactant. Examples are
fluorescence, phosphorescence etc. (we shall
focus on this mostly) Secondary
process Products formed from intermediates that
are produced directly from the excited state of a
reactant. An example of a secondary process is a
chain reaction.
87
Quantum yields in photophysics
Primary quantum yield The primary quantum yield
f is the number of photophysical or photochemical
events that lead to primary products divided by
the number of photons absorbed by the molecule in
the same interval.
88
Quantum yields in photophysics
Primary quantum yield An excited molecule must
either decay to the ground state or form
photochemical products. Therefore, the total
molecules deactivated by radiative processes,
nonradiative processes and photochemical
reactions must equal the number of excited
species produced by light absorption. The sum
of all primary quantum yields, fi for all
photophysical and photochemical events i must
equal 1 irrespective of the number of reactions
involving the excited state.
89
Quantum yields in photophysics
Primary quantum yield For example, if the
excited state only decays to the ground state
through photophysical processes, we write
where ff, fIC, fISC and fP are the quantum yields
of fluorescence, internal conversion, intersystem
crossing and phosphorescence respectively.
90
Quantum yields in photophysics
Mechanism of decay of excited states
91
Quantum yields in photophysics
Mechanism of decay of excited states If the
absorbance of the sample is low and the incident
light intensity is relatively high, we may assume
S is small and constant and therefore we can
invoke the steady state approx. for S
92
Quantum yields in photophysics
and substituting into
The final expression for the quantum yield for
fluorescence becomes
93
Quantum yields in photophysics
We can express the fluorescence lifetime as
The fluorescence lifetime can very easily be
measured with pulsed lasers. For example, the
sample is irradiated with a nanosecond laser and
the decay in the fluorescence intensity is
measured with a fast detector (photodiode,
photomultiplier etc.)
94
Quantum yields in photophysics
Quenching The shortening of the lifetime of an
excited state is called quenching. It may either
be a desired process (e.g. energy transfer) or
undesired side reaction that may decrease the
quantum yield of desired photochemical process.
The addition of a quencher Q opens up an
addition channel for deactivation
95
Quantum yields in photophysics
Quenching Three common mechanisms for
bimolecular quenching of an excited state are
96
Quantum yields in photophysics
Invoking the SSA now gives
The fluorescence quantum yield now becomes
97
Quantum yields in photophysics
Stern-Volmer plot
ff

1 t0kQQ
f
By plotting the LHS vs. Q, we obtain a straight
line with slope t0kQ. This is known as a
Stern-Volmer plot.
ff
t0kQ
f
1
0
Q
98
Quantum yields in photophysics
Stern-Volmer plot As the fluorescence intensity
and lifetime are proportional to the fluorescence
quantum yield, plots of I0/I and t0/t vs. Q
should also be linear with same slope and
intercept.
99
Quantum yields in photophysics
Stern-Volmer plot The molecule 2,2-bipyridine
forms a complex with the Ru2 ion. Ruthenium(II)
tris-(2,2-bipyridyl), Ru(bpy)32, has a strong
metal-ligand change transfer (MLCT) transition at
450 nm. The quenching of the Ru(bpy)32 excited
state by Fe(H2O)63 in acidic solution was
monitored by measuring emission lifetimes at 600
nm. Determine the quenching rate constant for
this reaction from the following data
Fe(H2O)63/(10-4 mol L-1) 0 1 2
3 4 5 6 7
8 9 10 t/(10-7 s)
5.69 4.91 4.33 3.86 3.49 3.18 2.92
2.70 2.52 2.35 2.21
We use the expression
1
1
kQQ

t0
t
100
Quantum yields in photophysics
5
4
3
(106 s)/t
2
kQ 2.7 x 109 L mol-1 s-1
1
t0 5.77 x 10-7 s
0
0
0.5
1.0
Fe3/(10-3 mol L-1)
101
Quantum yields in photophysics

• The measurements of emission lifetimes are
  preferred as they yield values of kQ directly. To
  determine value of kQ from intensity or quantum
  yield measurements, an independent measurement of
  t0 must be made.
• Measuring the emission lifetimes at 600 nm is
  very easily achieved using a photomultiplier with
  a detection peak around this wavelength. Usually,
  fluorescence measurements are hampered by
  scattered radiation (450 nm in this case). These
  can be removed by a filter if the absorption and
  emission wavelengths are appreciably different.
• The fluorescence lifetimes are slow (10-7 s)
  which are easy to carry out with a pulsed
  nanosecond (10-9 s) laser.

102
Femtochemistry
Lasers are extremely powerful tools for measuring
processes that occur on a very short timescale.
Nobel prize for chemistry was awarded to Ahmed
Zewail for the studies of transition states of
chemical reactions using femtosecond spectroscopy.
The ability to follow the fate of A in real time
tells us about the mechanism by which A relaxes
either back to its ground state or dissociates
into various products.
103
The laser pulses
fs 10-15 s
ps 10-12 s
ns 10-9 s
time
Depending on the dynamics being measured,
different laser-pulse durations are required
104
Probing products
Consider the following
To probe the fate of A, we can monitor the
products of dissociation B (shown) or C as a
function of time.
105
Probing intermediates
Alternatively..
we can monitor A directly by ionizing it and
looking at the parent ion (A) as a function of
time.
106
Some typical data
In any case, one might see the following dynamics
A
B
time
time
B rises while A and hence A falls.
107
The pump and the probe
A
time
Pump (initiates excitation-gtA)
After the pump excites A to A, the probe arrives
at discrete delays (typically 10s of
femtoseconds (10-15 second)) to ionize A. The
arrival of the pump is referred to as the time
zero, i.e. the start of the chemical reaction.
108
Accurate time-delay
It is critical that the relative delay in time
between pump and probe be known accurately. This
is achieved by finding the temporal overlap
between the pump and probe laser beam.
Pump probe temporally overlapped
pump
probe
Accurately known
time
time
time zero
109
Changing time-delay
Once the temporal overlap is found, we can start
to scan the probe relative the pump, i.e.
pump
pump
probe
probe
time
time
time zero
time zero
110
Delaying the fs pulses
For example
Probe
Delay of 1 ?m
1 ?m
A delay of 1 ?m corresponds to 3 femtoseconds
(3.34 fs). We must however multiply by 2 as the
delay is doubled.
111
Real examples

112
1. Photodissociation of Br2
Photoabsorption followed by dissociation
3
C 1Pu
Br2 dissociates
2
Energy/cm-1 (104)
Br
Br
400 nm
1
X 1Sg
0
1.5 2.5 3.5 4.5
R (Å)
113
1. Photodissociation of Br2
Detecting fragment Br atoms through resonance
enhanced multiphoton ionisation (REMPI)

• Photoelectrons with 2.136 eV of kinetic energy
  are ejected
• Measure these photoelectrons in a TOF fashion
  using a magnetic bottle TOF mass spectrometer.

3P2 (Br)
5p 4P3/2
Energy
266.8 nm
Br
114
1. Photodissociation of Br2
3000
Br
Br
Ion count
1500
47 fs
/- 5 fs
Br
Br
-0.2 -0.1 0 0.1 0.2
Time (ps)
Photodissociation of Br2 at 2.136 eV.
115
2. Photodissociation of CH3I
The reaction of CH3I I CH3
REMPI (277 nm)
A-state

I
Energy
304 nm
X-state
C-I
n, s transition in which a non-bonding electron
on I is excited to an antibonding orbital on the
C-I framework
116
2. Photodissociation of CH3I
The reaction of CH3I I CH3
Aniline
Iodine
JCP 105 (1996) 7864
The reference aniline 1 1 REMPI transition (O)
defines the zero of time. The delay between
aniline and iodine indicates the dissociation
time of C-I which is 150 fs.
117
In summary.

118
Experimental methods
The transient species can be monitored by either

1. Absorption
2. Fluorescence
3. Conductivity
4. Pressure
5. NMR
6. Mass detection
7. Electron detection

119
Activation diffusion control
Diffusion controlled limit kagtgtkd
Rate governed by rate of diffusion of reactants
Activation controlled limit kaltltkd
Rate governed by rate in which energy is
accumulated in encounter pair
120
Diffusion control
Solvent cage
Rate determining step of the reaction is the
approach of reactants. Once in the solvent cage,
reaction occurs
where DDADB Rminimum
distance for reaction
121
Activation control
122
Reactions between ions
From the thermodynamic formulation of the rate
constant, we can obtain a rate constant that
depends on ionic strength.
dP dt
kaKABAB (?A?B /?AB)

k kaKAB (?A?B /?AB)


log(k)log(k0) 1.018zAzB?I
Kinetic salt effect
123
Quantum yields in photophysics
Mechanism of decay of excited states
124
Quantum yields in photophysics
Mechanism of decay of excited states
dS
Iabs S(kf kIC kISC) 0

dt
Iabs S(kf kIC kISC)
?f
kfS
ff

S(kf kIC kISC)
Iabs
No quencher
125
Quantum yields in photophysics
Stern-Volmer plot As the fluorescence intensity
and lifetime are proportional to the fluorescence
quantum yield, plots of I0/I and t0/t vs. Q
should also be linear with same slope and
intercept.
ff
t0
1 t0kQQ


t
f
therefore
1
1
kQQ

t0
t
126
Revision question CH162

• The quenching of Tryptophan fluorescence by
  dissolved O2 gas was monitored by measuring
  emission lifetimes at 348nm in aqueous solutions.
  Determine the quenching rate constant (kQ) for
  this process from the following data
• O2/(10-2 mol dm-3) 0 2.3 5.5 8 10.8
• t/(10-9 s) 2.6 1.5 0.92 0.71 0.57
• In water, the fluorescence quantum yield and
  observed fluorescence lifetime of Tryptophan are
  ff0.20 and t02.6ns, respectively. Calculate the
  fluorescence rate constant kf
• Single photon ionization from the ground state of
  Tryptophan to the ground state of the parent ion
  with VUV radiation of 10.3 eV. gave a peak in the
  photoelectron spectrum of 1.87 eV.
• i. Deduce the ionization potential of
  Tryptophan (eV.)
• ii. What might you expect with higher energy
  photons?

127
1.
2.0E09
O2
t
1/t
1.6E09
3.85E08
2.60E-09
0.00E00
1.2E09
6.67E08
1.50E-09
2.30E-02
1/t
8.0E08
1.09E09
9.20E-10
5.50E-02
Gradient 1.28x1010
1.41E09
7.10E-10
8.00E-02
4.0E08
kQ 1.28x1010 dm3mol-1s-1
1.75E09
5.70E-10
1.08E-01
0
0 0.04 0.08 0.12
O2
ff
1
3. i

t0
hv PE Ii
2.
(kf kIC kISC)
kf
hv - PE Ii
ff
10.3 eV 1.87 eV 8.43 eV
kf
7.7x107s-1
t0
3. ii-answer via email!