Transcript CH162 Kinetics

CH162
Kinetics
5 lectures VGS
• Experimental methods in kinetics
• Reactions in liquid solution
• Photochemistry
• Femtochemistry
Gas-phase vs. Liquid phase
Isolated
molecules
•Intramolecular processes
Solvent
Solute
•Intra/intermolecular processes
•Solute-solvent interactions
Experimental methods
A + B  Products
Change in conc.
Rate = -
d[A]
dt
Change in time
How do we do this?
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!!
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
 [conc.]
Experimental methods
Absorption and fluorescence:
A + B  Products
concentration
14
12
[Products](t)
10
8
6
[A](t)
4
2
0
0
10
20
time
30
40
Experimental methods
Conductivity:
C4H9Cl + H2O  C4H9OH + H+ + Cl-
Experimental methods
Pressure:
2NOBr(g)  2NO(g) + Br2(g)
2 moles  3 moles
P = n RT
V
 pressure increase (at constant volume)
Experimental methods
NMR: Different H chemical
shifts in NMR
B
A
I
A
[]
B
C
A
B
Time
C
I
C
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.
Experimental methods
A typical setup for transient absorption:
Setup
Detector
Pulse 2
Pulse 2
Pulse 1
Pulse 1
Time
Experimental methods
A typical setup for transient absorption:
Setup
Data
Pulse 2
Pulse 1
Intensity
Detector
I0
I(t)
Time
Absorption spectroscopy in a little detail…..
Beer-Lambert law
I  I 0 e cl
2
3
4
x
x
x
ex  1 x   
 
2 6 24
1  x
if x << 1
I  I 0 (1  cl )  I 0  I 0cl
I 0  I  I abs  I 0  cl
I abs
 cl
I0
small c
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
Experimental methods
A typical setup for transient mass spectrometry:
Step 1
Pulse 1
Pulse 2 (Δt)
Step 2
+
+
HV (+)
HV (+)
Experimental methods
A typical setup for transient mass spectrometry:
Step 3
TOF tube
+
+
+
+
Time of flight (ion)
TOF (d)
HV (+)
Detector
Experimental methods
-Ion time of flight mass spectrometry
Potential energy Ep of charged particle in electric field is:
Ep = qU
U=potential difference
q=charge of particle
When charged particle is accelerated into a TOF tube by U, Ep is
converted to kinetic energy Ek:
Ek = 1 mV2
2
m=mass of particle
V=velocity of particle
Experimental methods
-Ion time of flight mass spectrometry
Potential energy is converted into kinetic energy meaning:
qU= 1 mV2
2
In field free region, velocity remains constant:
2
d
qU= 1 m ( )
t
2
d=length of flight tube
t=time of flight
Experimental methods
-Ion time of flight mass spectrometry
Rearranging so that the flight time is subject of formula:
d
t=
2U
m
(q)
As flight length and voltage (potential difference) are constants:
t=K
m
(q)
Experimental methods
Intensity
(arb. units)
-Ion time of flight mass spectrometry
Time-of-flight (μs)
Experimental methods
Using a known mass to predict an unknown mass:
q=+1
Time-of-flight (μs)
m1 = m2
? (4.24μs)
? (4.12μs)
H (1μs)
Intensity
(arb. units)
m1
t1 = K ( q )
m2
t2 =K ( q )
t1
t2
2
( )
4.12 μs = OH
4.24 μs = OH2
Experimental methods
-Ion time of flight mass spectrometry
What does one expect to observe with such measurements?
pulse 2
OH+ intensity
pulse 2
pulse 2
pulse 2
pulse 1
Time delay between pulse 1 and 2
Pulse 1 dissociates
Pulse 2 probes through
ionization
Experimental methods
NMR:
The α and β forms interconvert over a timescale of hours in aqueous
solution, to a final stable ratio of α:β 36:64, in a process called
mutarotation
Experimental methods
NMR:
H
H
H
H
We can monitor this interconversion using NMR. How?
Experimental methods
NMR:
Day 1
10:53 am
Only

Experimental methods
NMR:
Day 1
1:27 pm
Experimental methods
NMR:
Day 2
09:13 am
Experimental methods
A typical setup for transient photoelectron spectroscopy:
Step 1
Pulse 1
Pulse 2 (Δt)
Step 2
1
+
2 +
e1e2-
Experimental methods
-Photoelectron spectroscopy
Step 3
1
2
+
+
e1-
e2Detector
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.
hv-Ii
e
X+
hv
Orbital i
Ii
X
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.
hv = 1 mev2 + Ii
2
Kinetic energy
By knowing the kinetic energy of the photoelectron and
the frequency of the incoming radiation, Ii may be
measured.
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 vibrationally 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:
hv = 1 mev2 + Ii + Evib
2
Vibrationally
excited ion
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?
hv = 1 mev2 + Ii
2
Hence
hv - 1 mev2 = Ii
2
Experimental methods
therefore…
21.22 eV. - 5.63 eV. = Ii
= 15.59 eV.
This corresponds to the removal of an electron from the
HOMO (3sg) orbital.
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 1&2)
•Foundations of Spectroscopy, Duckett and Gilbert, Oxford
Chemistry Primers (lectures 1&2)
•Elements of Physical Chemistry, Atkins and de Paula, 4th
edition (lectures 2&3)
Reactions in liquid solution
1. Activation control and diffusion control
2. Diffusion and Fick’s laws
3. Activation/diffusion revisited
4. Thermodynamic formulation of rate coefficient
5. Ionic-strength effects
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%.
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
d
A and B occurs through the
following mechanism. A
and B move into each other’s vicinity through
diffusion forming an encounter pair, AB at a rate
proportional to the concentration of A and B,
diffusional
A
+
B
AB
d[AB]
=kd[A][B]
dt
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
AB
A
+
B
-
d[AB]
=k’d[AB]
dt
The competing process is the reaction between A and
B while an encounter pair, forming products. activated
AB
Products
d[AB]
=ka[AB]
dt
Activation & diffusion control
The rate of formation of products is given by:
AB
Since:
Products
d[P]
=ka[AB]
dt
d[AB]
=kd[A][B]-k’d[AB]-ka[AB]
dt
Apply SSA to [AB] gives:
kd[A][B]-k’d[AB]-ka[AB]=0
Activation & diffusion control
Or:
kd[A][B]
[AB]=
k’d+ka
Therefore
kakd[A][B]
d[P]
=ka[AB]=
dt
k’d+ka
There are two limits for this expression, ka>>k’d and ka<<k’d
Activation & diffusion control
Diffusion controlled limit: ka>>k’d
d[P]
= kd[A][B]
dt
Rate governed by rate of
diffusion of reactants
Activation controlled limit: ka<<k’d
kakd[A][B]
d[P]
=
dt
k’d
Rate governed by rate in
which energy is accumulated
in encounter pair
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, Δc/Δx or:
Rate of diffusion  concentration gradient
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
dx
D = diffusion coefficient
Diffusion-Fick’s first law
dc
J = -D
dx
c
dc
dx
Negative sign makes the
flux positive. Greatest flux
is found when gradient is
steepest
<0
x
Diffusion-Fick’s first law
Diffusion coefficients at 25 oC, D/10-9m2s-1
Ar in tetrachloromethane
C12H22O11 (sucrose) in water
CH3OH in water
H2O in water
NH2CH2COOH in water
O2 in tetrachloromethane
3.63
0.522
1.58
2.26
0.673
3.82
Molecules that diffuse fast have a larger diffusion
coefficient
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:
J = -(0.522 x 10-9m2s-1) x (-0.1 moldm-3cm-1)
= 5.22 x
10-11
= 5.22 x 10-11
m2s-1mol
dm3cm
m2s-1mol
(10-3m3) x (10-2m)
= 5.22 x 10-6 molm-2s-1
Diffusion
Calculate the amount of sucrose passing through a 1 cm
square window in 10 minutes:
n = J x A x Δt
= (5.22 x 10-6 molm-2s-1) x (1x10-2m)2 x (10x60s)
= 3.1 x 10-7 mol
Diffusion-Fick’s second law
Fick’s 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 nonuniform solution
Rate of change of concentration in a region
= D x curvature of concentration in region
or
2
c = D  c
t
x2
Diffusion-Fick’s second law
2
c   c
t
x2
dc
J
dx
5
c
4
dc
dx
x
<0
1
2
3
3
2
x
1
4
dc
dx
5
Positive curvature
Diffusion-Fick’s second law
Spreads
x
‘constant’
dc
dx
Negative curvature
c
Fills
x
dc
dx
Positive curvature
x
Diffusion equation tells us that a concentration with
unvarying slope through a region results in no net change in
concentration.
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)
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
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?
d2
t=
2D
(1x10-2m)2
t1cm=
2 x 2.26x10-9m2s-1
(2x10-2m)2
t2cm=
2 x 2.26x10-9m2s-1
= 22124s
= 88496s
= 6.1h
= 25h
Diffusion-Einstein-Smoluchowski
equation
The Einstein-Smoluchowski equation describes the relation
between D and the time a molecule takes its steps, τ (tau),
and the length of each step, λ (lambda).
λ2
D=
2τ
The larger and faster the step, the higher the diffusion
coefficient. τ represents the average lifetime of a molecule
near another molecule before it moves to the next position.
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?
λ2
τ=
2D
(200x10-12m)2
=
2 x 2.26x10-9m2s-1
= 9x10-12s = 9ps
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/τ) follows an
Arrhenius temperature dependence with an activation energy
Ea, then D will follow the relation:
D = D0 e-Ea/RT
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 η  1/D, where η is the
coefficient of viscosity. The Einstein relation states:
kBT
D=
6πηa
where a is the radius of the molecule and,
η = η0 eEa/RT
Diffusion-Temperature & viscosity
The viscosity of water at 40oC and 80oC corresponds to
η40=6.8x10-4 kgm-1s-1 and η80=3.5x10-4 kgm-1s-1 respectively.
Calculate the activation energy for the viscosity of water.
R = 8.314 JK-1mol-1 and 0oC ~ 273K
η40 = η0 eEa/RT
η80 = η0 eEa/RT
40
80
} ln(η
ln(η40) = ln(η0)+Ea/RT40
80)
= ln(η0)+Ea/RT80
ln(η40/η80) = Ea(1/RT40 -1/RT80)
ln(η40/η80)
(1/RT40 -1/RT80)
= Ea
= 15.2 kJmol-1
(1)
(2)
(1) – (2)
Recap. Activation & diffusion control
Diffusion controlled limit: ka>>k’d
d[P]
= kd[A][B]
dt
Rate governed by rate of
diffusion of reactants
Activation controlled limit: ka<<k’d
kakd[A][B]
d[P]
=
dt
k’d
Rate governed by rate in
which energy is accumulated
in encounter pair
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.
kd=4πR*DNA
where D=DA+DB
See Atkins 8th edition, page 877-878
kBT
DA =
6πηRA
kBT
DB =
6πηRB
Assuming RA=RB=(1/2)R*
kd =
Stokes-Einstein
8RT
3η
Diffusion and reaction
For a diffusion controlled reaction, given η=8.9x10-4 kgm-1s-1
at 25oC, calculate the diffusion controlled rate constant:
8 x 8.315JK-1mol-1 x 298K
kd =
3 x 8.9x10-4kgm-1s-1
8 x 8.315 x 298
kd =
3 x 8.9x10-4
JK-1mol-1K
kgm-1s-1
8 x 8.315 x 298
kd =
3 x 8.9x10-4
kgm2s-2mol-1
kgm-1s-1
kd = 7.4 x 106 m3s-1mol-1
J=kgm2s-2
Diffusion and reaction
Two neutral species A and B, with diameters 588pm and
1650pm respectively, undergo the diffusion controlled
reaction A+B->P in a solvent of viscosity η=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.
d[P]
=kd[A][B]
dt
kd=4πR*(DA+DB)NA
(
(
2kBTNx(R +R )x 1 + 1
=
A
B
RA RB
3η
Diffusion and reaction
(
(
2RT x(R +R )x 1 + 1
=
A
B
R A RB
3η
(2)x(8.314JK-1mol-1)x(313K) x(294+825)x 1 + 1
=
294 825
(3)x(2.37x10-3kgm-1s-1)
(
(
=3.8x106 mol-1m3s-1=3.8x109dm3mol-1s-1
d[P]
=(3.8x109dm3mol-1s-1)x(0.150 moldm-3)x(0.330 moldm-3)
dt
= 1.9x108 moldm-3s-1
Thermodynamic formulation of RC
In activation control reactions, the concentration of
encounter pairs {AB} is maintained at its equilibrium value,
determined by kd/k’d = KAB (equilibrium constant).
The rate coefficient is given by:
kakd[A][B]
d[P]
=
= k[A][B]
dt
k’d
k= kaKAB
Thermodynamic formulation of RC
By applying transition state theory to ka, then the overall
rate constant can be expressed as:
kTS= κ(kBT/h) exp(-ΔG /RT)
where
K = exp(-ΔG /RT)
kTS = κ(kBT/h) K
and
Eyring equation
ΔG = Overall Gibbs free
energy of activation
K is the equilibrium constant between reactants and
activated complex
Thermodynamic formulation of RC
k= kaKAB
KAB = exp(-ΔGep/RT)
ka = κ(kBT/h) exp(-ΔGac/RT)
ep = encounter pair
ac = activated complex
Alternatively, we may write the Eyring equation in terms of
the enthalpy of activation ΔH and entropy of activation ΔS
kTS= κ(kBT/h) exp(ΔS /R) exp(-ΔH /RT)
Compare:
k= A exp(-Ea/RT)
Arrhenius
Ionic strength effects
Consider a solution of ionic strength I, where
I = 1 Σ cizi2
2
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?
kTS = κ(kBT/h) K
This equation is only valid for reactions of ions at infinite
dilution and needs modification at higher ionic strengths
Ionic strength effects
We start by considering the equilibrium between separate
reactants and the encounter complex. For non-ideal
solutions
KAB= ([AB]/[A][B]) γAB/γAγB
d[P]
= k[A][B]=ka[AB] (SSA)
dt
d[P] = kaKAB[A][B] (γAγB /γAB)
dt
and
k= kaKAB (γAγB /γAB)
Ionic strength effects
We can now re-write
kTS = (κkBT/h) K
to include activity coefficients:
kTS = (κkBT/h) K (γAγB /γAB)
The activity coefficient (γA 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.
Ionic strength effects
At low ionic strength, the activity coefficient of an ion may
be calculated from the limiting Debye-Hückel equation:
log(γA)=-A’zAI
2
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:
log(k/k0)= 1.018zAzBI
Kinetic salt effect
Note: The encounter pair has a charge zA+zB and k0 is rate
constant at zero ionic strength and equals kaKAB.
Ionic strength effects
A plot of log k vs. I is a straight line with slope~zAzB.
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.
Ionic strength effects
The slopes of the lines are
those given by the DebyeHückel limiting law.
Diffusion and reaction
The rate constant of the reaction:
H2O2(aq)+I-(aq)+H+(aq)->H2O(l) +HIO(aq)
is sensitive to the ionic strength of the aqueous solution in
which the reaction occurs. At 25 oC, k=12.2 dm6mol-2min-1 at
an ionic strength of 0.0525. Estimate the rate constant at
zero ionic strength.
log(k)=log(k0) + 1.018zAzBI
log(k0)= log(12.2) - (1.018)x(1)x(-1)x(0.0525)=1.32
k0= 20.9 dm6mol-2min-1
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
k/k0
0.005
0.995
I1/2
0.071
0.100
0.122
0.141
0.158
0.173
0.01
0.015
0.925 0.875
log(k/k0)
-0.002
-0.034
-0.058
-0.078
-0.097
-0.114
0.02
0.835
0.025
0.800
0.03
0.770
0
0.06
-0.02
-0.04
0.11
0.16
0.21
y = -1.0849x + 0.0747
2
R =1
-0.06
-0.08
-0.1
-0.12
Slope=1.018zAzB
-1.085=1.018zAzB
A=+1
B=-1
Ionic strength effects
At high ionic strength, log(γA)=-A’zAI breaks down and is
replaced by:
log(γA)=-A’zAI/(1+B’aI)
2
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.
Photochemistry
1. 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
Introduction to photochemistry
Vision-the ‘good’
hv
11-cis-Retinal
All-trans-Retinal
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)
Introduction to photochemistry
Multi-step organic synthesis- the ‘ugly’
Jablonski diagrams
Fluorescence vs. phosphorescence
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
GFP chromophore
p-hydroxybenzylidene-imidazolidone
Fluorescence vs. phosphorescence
kinetics
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.
Quantum yields in photophysics
Primary quantum yield
The primary quantum yield φ 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.
ν
number
of
events
rate
of
process
=
φ=
=
intensity of light
number of photons
Iabs
absorbed
absorbed
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, φi for all photophysical
and photochemical events i must equal 1 irrespective of the
number of reactions involving the excited state.
Σ φi = Σ
i
νi
=1
i Iabs
Quantum yields in photophysics
Primary quantum yield
For example, if the excited state only decays to the
ground state through photophysical processes, we write:
φf + φIC + φISC + φP = 1
where φf, φIC, φISC and φP are the quantum yields of
fluorescence, internal conversion, intersystem crossing
and phosphorescence respectively.
Quantum yields in photophysics
Mechanism of decay of excited states
Absorption:
S + hvi
S*
νabs = Iabs
Fluorescence:
S*
S + hvf
νf = kf[S*]
IC:
S*
S
νIC = kIC[S*]
ISC:
S*
T*
νISC = kISC[S*]
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*]:
d[S*]
= Iabs – [S*](kf + kIC + kISC) = 0
dt
and
Iabs = [S*](kf + kIC + kISC)
Quantum yields in photophysics
and substituting into:
νf
kf[S*]
φf =
=
Iabs [S*](kf + kIC + kISC)
The final expression for the quantum yield for fluorescence
becomes:
φf =
kf
(kf + kIC + kISC)
Quantum yields in photophysics
We can express the fluorescence lifetime as:
φf
1
=
τ0 =
(kf + kIC + kISC) kf
No quencher
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.)
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:
Quenching:
S* + Q
S+Q
νQ = kQ[Q][S*]
Quantum yields in photophysics
Quenching
Three common mechanisms for bimolecular quenching of
an excited state are:
Collisional
deactivation
S* + Q
S + Q efficient when Q
is large e.g. I-
Energy transfer:
S* + Q
S + Q*
Electron transfer: S* + Q
S+ + Q-
or
S- + Q+
Quantum yields in photophysics
Invoking the SSA now gives:
d[S*]
= Iabs – [S*](kf + kIC + kISC + kQ[Q]) = 0
dt
The fluorescence quantum yield now becomes:
φ =
(kf + kIC
kf
+ kISC + kQ[Q])
Quantum yields in photophysics
Stern-Volmer plot
φf
= 1 + τ0kQ[Q]
φ
By plotting the LHS vs. [Q], we obtain a straight line with
slope τ0kQ. This is known as a Stern-Volmer plot.
φf
φ
1
0
τ0kQ
[Q]
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 τ0/τ vs.
[Q] should also be linear with same slope and intercept.
φf τ0
= 1 + τ0kQ[Q]
=
φ τ
therefore
1 + k [Q]
1
Q
τ = τ0
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 metalligand 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)
τ/(10-7 s)
0
1
2
3
4
5
5.69 4.91 4.33 3.86 3.49 3.18
We use the expression:
1 + k [Q]
1
Q
τ = τ0
6
7
8
2.92 2.70 2.52
9
2.35
10
2.21
Quantum yields in photophysics
(106 s)/τ
5
4
3
2
kQ = 2.7 x 109 L mol-1 s-1
τ0 = 5.77 x 10-7 s
1
0
0
0.5
[Fe3+]/(10-3 mol L-1)
1.0
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 τ0 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.
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.
A
+
hν
A*
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.
The laser pulses
ps
ns
10-9 s
10-12 s
fs
10-15
s
time
Depending on the dynamics being measured, different
laser-pulse durations are required
Probing products
Consider the following:
hν2 ‘Probe’
B
A
+
hν1
‘Pump’
+
C
A*
A
To probe the fate of A*, we can monitor the products of
dissociation B (shown) or C as a function of time.
Probing intermediates
Alternatively..
A
+
B
hν1
‘Pump’
+
C
A*
hν2 ‘Probe’
A+
we can monitor A* directly by ionizing it and looking at the
parent ion (A+) as a function of time.
Some typical data
In any case, one might see the following dynamics
[A+]
[B]
time
[B] rises while [A+] and hence [A*] falls.
time
The ‘pump’ and the ‘probe’
Probe (discrete step sizes)
[A+]
time
Pump (initiates excitation->A*)
After the pump excites A to A*, the probe arrives at
discrete delays (typically 10’s 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.
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
Pump & probe
temporally overlapped
Accurately
known
time
time zero
time
Changing time-delay
Once the temporal overlap is found, we can start to scan
the probe relative the pump, i.e.
pump
probe
time
time zero
pump
time
time zero
probe
pump
probe
time
time zero
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.
Real examples
1. Photodissociation of Br2
Energy/cm-1 (104)
Photoabsorption followed by dissociation
~
~
3
C 1P u
Br2 dissociates
2
400 nm
1
0
Br
X 1Sg+
1.5
2.5
3.5
R (Å)
4.5
Br
1. Photodissociation of Br2
Energy
Detecting fragment Br atoms through resonance enhanced
multiphoton ionisation (REMPI)
•Photoelectrons with 2.136 eV of
3P (Br+)
kinetic energy are ejected
2
•Measure these photoelectrons in
5p 4P3/2
a TOF fashion using a magnetic
bottle TOF mass spectrometer.
266.8 nm
Br
1. Photodissociation of Br2
3000
Ion count
Br
1500
Br
-0.2
47 fs
+/- 5 fs
Br
-0.1
0
0.1
Time (ps)
0.2
Photodissociation of Br2 at 2.136 eV.
Br
2. Photodissociation of CH3I
The reaction of CH3I
REMPI (277 nm)
A-state
Energy
I + CH3
I + C
304 nm
H
H
H
X-state
C-I
n, σ* transition in which a non-bonding electron on I is
excited to an antibonding orbital on the C-I framework
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.
In summary….
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
 [conc.]
Activation & diffusion control
Diffusion controlled limit: ka>>k’d
d[P]
= kd[A][B]
dt
Rate governed by rate of
diffusion of reactants
Activation controlled limit: ka<<k’d
kakd[A][B]
d[P]
=
dt
k’d
Rate governed by rate in
which energy is accumulated
in encounter pair
Diffusion control
Solvent cage
Rate determining step of the reaction is the approach of
reactants. Once in the solvent cage, reaction occurs
kd=4πR*DNA
where D=DA+DB
R*=minimum distance
for reaction
Activation control
Energy
Activated complex
Ea
Reactants (ep)
Products
Progress of reaction
k= kaKAB
KAB = exp(-ΔGep/RT)
ka = κ(kBT/h) exp(-ΔGac/RT)
ep = encounter pair
ac = activated complex
Reactions between ions
From the thermodynamic formulation of the rate constant,
we can obtain a rate constant that depends on ionic
strength.
d[P] = kaKAB[A][B] (γAγB /γAB)
dt
k= kaKAB (γAγB /γAB)
}
}
log(γA)=-A’zAI
2
log(γB)=-A’zBI
2
log(γAB)=-A’(zA+zB)I
2
log(k)=log(k0) + 1.018zAzBI
Kinetic salt effect
Quantum yields in photophysics
Mechanism of decay of excited states
Absorption:
S + hvi
S*
νabs = Iabs
Fluorescence:
S*
S + hvf
νf = kf[S*]
IC:
S*
S
νIC = kIC[S*]
ISC:
S*
T*
νISC = kISC[S*]
Quantum yields in photophysics
Mechanism of decay of excited states
d[S*]
= Iabs – [S*](kf + kIC + kISC) = 0
dt
Iabs = [S*](kf + kIC + kISC)
φf =
φf =
νf
Iabs
kf
(kf + kIC + kISC)
No quencher
=
kf[S*]
[S*](kf + kIC + kISC)
φ=
kf
(kf + kIC + kISC +kQ[Q])
Quencher
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 τ0/τ vs.
[Q] should also be linear with same slope and intercept.
φf τ0
= 1 + τ0kQ[Q]
=
φ τ
therefore
1 + k [Q]
1
Q
τ = τ0
Revision question CH162
Tryptophan
1.
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
τ/(10-9 s)
2.6
1.5
0.92
0.71
0.57
2.
In water, the fluorescence quantum yield and observed fluorescence lifetime of Tryptophan
are φf=0.20 and τ0=2.6ns, respectively. Calculate the fluorescence rate constant kf
3.
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?
1.
[O2]
t
2.0E+09
1/t
1.6E+09
0.00E+00
2.60E-09
3.85E+08
2.30E-02
1.50E-09
6.67E+08
5.50E-02
9.20E-10
1.09E+09
8.0E+08
8.00E-02
7.10E-10
1.41E+09
4.0E+08
1.08E-01
5.70E-10
1.75E+09
φf
1
=
2. τ0 =
(kf + kIC + kISC) kf
φf
0.2
= 7.7x107s-1
= kf =
2.6x10-9s
τ0
1/t
1.2E+09
0
3. i
Gradient = 1.28x1010
kQ = 1.28x1010 dm3mol-1s-1
0
0.04
[O2]
0.08
0.12
hv = PE + Ii
hv - PE = Ii
10.3 eV – 1.87 eV = 8.43 eV
3. ii-answer via email!