ch162 kinetics

CH162Kinetics

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

5 lectures VGS

Gas-phase vs. Liquid phase

•Intra/intermolecular processes•Solute-solvent interactions

•Intramolecular processes

Solute

SolventIsolated

molecules

Experimental methods

Change in conc.

Rate = -d[A]

dt Change in time

How do we do this?

A + B Products

Experimental methodsIn 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 methodsThe transient species can be monitored by either:

1. Absorption

2. Fluorescence

3. Conductivity

4. Pressure

5. NMR

6. Mass detection

7. Electron detection

[conc.]


Absorption and fluorescence:

A + B Products

[A](t)

[Products](t)

0

2

4

6

8

10

12

14

0 10 20 30 40

time

conce

ntr

ati

on


Conductivity:

C4H9Cl + H2O C4H9OH + H+ + Cl-


2NOBr(g) 2NO(g) + Br2(g)

2 moles 3 moles

pressure increase (at constant volume)

P =nV

RT

Pressure:

NMR:

A I

B

C

FAST

Slow


CA

B

Time

[ ]

Different H chemical shifts in NMR

CB

A

I


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 methodsA typical setup for transient

absorption:

Pulse 2

Pulse 1

Setup

Dete

ctor

Pulse 1

Pulse 2

Time

Experimental methodsA typical setup for transient

absorption:

Time

I0

Inte

nsi

ty

I(t)

Data

Dete

ctor

Pulse 2

Pulse 1

Setup

Absorption spectroscopy in a little detail…..

Beer-Lambert law cleII 0

2462

1432 xxx

xex

x1 if x << 1

clIIII abs 00

clI

Iabs 0

clIIclII 000 )1( small c

light sources for flash photolysis

- pulse length fs = 10-15 s- high power low precursor

concentration- high repetition rate

500Hz

• flash lamps

• lasers

- pulse length s - ms


HV (+)

Step 1

+

+

HV (+)

Step 2

A typical setup for transient mass spectrometry:

Pulse 1Pulse 2 (Δt)

+

+

HV (+)

Step 3

Detector

Time of flight (ion)

++

Experimental methodsA typical setup for transient mass spectrometry:

TOF (d)

TOF tube

-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:

Ep =qU U=potential differenceq=charge of particle

12

Ek = mV2 m=mass of particleV=velocity of particle



Potential energy is converted into kinetic energy meaning:

qU= 12

mV2

In field free region, velocity remains constant:

qU= 12

m dt( )

2 d=length of flight tubet=time of flight



Rearranging so that the flight time is subject of formula:

As flight length and voltage (potential difference) are constants:

2Ut = d

q( )m

t = K q( )m


Time-of-flight (μs)

Inte

nsi

ty(a

rb.

un

its)


Using a known mass to predict an unknown mass:

Time-of-flight (μs)

Inte

nsi

ty(a

rb.

un

its)

H (

1μs)

? (4

.24μs)

? (4

.12μs)

4.12 μs = OH4.24 μs = OH2


t1 =K q( )m1

t2 =K q( )m2

m1 =m2

t1

t2( )

2q=+1


-Ion time of flight mass spectrometryWhat does one expect to observe with such measurements?

Time delay between pulse 1 and 2

OH

+ inte

nsi

ty

pulse 1

Pulse 1 dissociates

Pulse 2 probes through ionizationpulse 2

pulse 2

pulse 2

pulse 2

Experimental methodsNMR:

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

http://upload.wikimedia.org/wikipedia/commons/0/09/Glucoseab.png


We can monitor this interconversion using NMR. How?

H H

H

H

http://upload.wikimedia.org/wikipedia/commons/6/60/Beta-D-glucopyranose-2D-skeletal.png

http://upload.wikimedia.org/wikipedia/commons/4/48/Alpha-D-glucopyranose-2D-skeletal.png


Day 110:53 am

Only


Day 11:27 pm


Day 209:13 am


Step 1

A typical setup for transient photoelectron spectroscopy:

Pulse 1Pulse 2 (Δt)

Step 2

+

+ e1-

e2-

1

2

-Photoelectron spectroscopy

Step 3

Detector

e1-


1

2 e2-

+

+

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.

e-

X

X+

hv

hv-Ii

Ii

Orbital i


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 = mev2 + Ii12

Kinetic energy

By knowing the kinetic energy of the photoelectron and the frequency of the incoming radiation, Ii may be measured.


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: hv = mev2 + Ii + Evib

12

Vibrationally excited ion


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 = mev2 + Ii12 Hence hv - mev2 = Ii

12


21.22 eV. - 5.63 eV. = Ii= 15.59 eV.

therefore…

This corresponds to the removal of an electron from theHOMO (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 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,

d

A + B AB d[AB]dt

=kd[A][B]

diffusional


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]dt

=k’d[AB]-

The competing process is the reaction between A and B while an encounter pair, forming products.

AB Products d[AB]dt

=ka[AB]-

activated


The rate of formation of products is given by:

AB Products d[P]dt

=ka[AB]

Since:d[AB]dt

=kd[A][B]-k’d[AB]-ka[AB]

Apply SSA to [AB] gives:

kd[A][B]-k’d[AB]-ka[AB]=0


Or:

Therefore

kd[A][B]k’d+ka

[AB]=

d[P]dt

=ka[AB]=kakd[A][B]k’d+ka

There are two limits for this expression, ka>>k’d and ka<<k’d


Diffusion controlled limit: ka>>k’d

d[P]dt

= kd[A][B]

d[P]dt

=kakd[A][B]

k’d

Activation controlled limit: ka<<k’d

Rate governed by rate of diffusion of reactants

Rate governed by rate in which energy is accumulated in encounter pair

DiffusionPlays 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

DiffusionThe 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:

J =Number of particles passing through window

Area of window x time interval

J =-D x concentration gradient

J = -Ddcdx D = diffusion coefficient

= -D

Diffusion-Fick’s first law

J dcdx

dcdx<0

Negative sign makes the flux positive. Greatest flux is found when gradient is steepest

c

x

Diffusion-Fick’s first law

Ar in tetrachloromethane3.63C12H22O11 (sucrose) in water 0.522CH3OH in water 1.58H2O in water 2.26NH2CH2COOH in water 0.673O2 in tetrachloromethane3.82

Diffusion coefficients at 25 oC, D/10-9m2s-1

Molecules that diffuse fast have a larger diffusion coefficient

DiffusionSuppose 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

dm3cmm2s-1mol

= 5.22 x 10-11

(10-3m3) x (10-2m)m2s-1mol

= 5.22 x 10-6molm-2s-1

DiffusionCalculate the amount of sucrose passing through a 1 cm square window in 10 minutes:

=(5.22 x 10-6 molm-2s-1) x (1x10-2m)2 x (10x60s)

n = J x A x Δt

=3.1 x 10-7 mol

Diffusion-Fick’s second lawFick’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 non-uniform solution

Rate of change of concentration in a region=D x curvature of concentration in region

orc = D

2cx2t

Diffusion-Fick’s second law

c

x

5

4

32 1

dcdx<0

J dcdx

c 2cx2t

x

5

4

3

2

1

dcdx

Positive curvature

Diffusion-Fick’s second law

Diffusion equation tells us that a concentration with unvarying slope through a region results in no net change in concentration.

c

x

Fills

Spreads

xdcdx

Positive curvature

xdcdx

Negative curvature

‘constant’

Diffusion-random walk

(a) Snapshot picture of the instantaneous configuration of a liquid. (b) Trajectories of particles in the 2ps following snapshot (1ps = 1x10-12s)

(a) (b)

Diffusion-random walkThe 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 walkThe 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?

t =d2

2D

(1x10-2m)2

2 x 2.26x10-9m2s-1t1cm=

= 22124s= 6.1h

(2x10-2m)2

2 x 2.26x10-9m2s-1

= 88496s= 25h

t2cm=

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).

D =

λ2

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?

τ =

λ2

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:

D =

kBT6πηa

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.

η40 = η0 eEa/RT40

η80 = η0 eEa/RT80

ln(η40) = ln(η0)+Ea/RT40

ln(η80) = ln(η0)+Ea/RT80}ln(η40/η80) = Ea(1/RT40 -1/RT80)

(1)

(2)

(1) – (2)

ln(η40/η80)

(1/RT40 -1/RT80)= Ea

= 15.2 kJmol-1

R = 8.314 JK-1mol-1and 0oC ~ 273K

Recap. Activation & diffusion control


d[P]dt

= kd[A][B]

d[P]dt

=kakd[A][B]

k’d




Diffusion and reactionDiffusion 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

kd=4πR*DNA where D=DA+DB

DA =kBT

6πηRA

DB =kBT

6πηRB

Assuming RA=RB=(1/2)R*

Stokes-Einstein

kd =8RT3η

Diffusion and reactionFor a diffusion controlled reaction, given η=8.9x10-4 kgm-1s-1 at 25oC, calculate the diffusion controlled rate constant:

kd =8 x 8.315JK-1mol-1 x 298K

3 x 8.9x10-4kgm-1s-1

kd =8 x 8.315 x 298 JK-1mol-1K

3 x 8.9x10-4 kgm-1s-1

kd =8 x 8.315 x 298 kgm2s-2mol-1

3 x 8.9x10-4 kgm-1s-1

kd =7.4 x 106 m3s-1mol-1

J=kgm2s-2

Diffusion and reactionTwo 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]dt

=kd[A][B]

kd=4πR*(DA+DB)NA

2kBTN3η

(RA+RB)x( RA RB

1 + 1 (

x=

Diffusion and reaction

2RT3η

(RA+RB)x( RA RB

1 + 1 (

x=

(2)x(8.314JK-1mol-1)x(313K)(3)x(2.37x10-3kgm-1s-1)

(294 8251 + 1 (

x(294+825)x=

=3.8x106 mol-1m3s-1=3.8x109dm3mol-1s-1

d[P]dt

=(3.8x109dm3mol-1s-1)x(0.150 moldm-3)x(0.330 moldm-3)

= 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:

d[P]dt

=kakd[A][B]

k’d

=k[A][B]

k= kaKAB

By applying transition state theory to ka, then the overall rate constant can be expressed as:

K = exp(-ΔG /RT)

kTS= κ(kT/h) exp(-ΔG /RT) Eyring equation

where

and K is the equilibrium constant between reactants and activated complex

ΔG = Overall Gibbs free energy of activation


kTS = κ(kT/h) K

Alternatively, we may write the Eyring equation in terms of the enthalpy of activation ΔH and entropy of activation ΔS

Compare:

kTS= κ(kT/h) exp(ΔS /R) exp(-ΔH /RT)


k= A exp(-Ea/RT) Arrhenius

KAB = exp(-ΔGep/RT)k= kaKAB ka = κ(kT/h) exp(-ΔGac/RT)

ep = encounter pair

ac = activated complex

Ionic strength effectsConsider a solution of ionic strength I, where

I = cizi2Σ1

2ci = Conc. in moldm-3 of ith ionzi

2= Charge of ith ionIn 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(kT/h) K

This equation is only valid for reactions of ions at infinite dilution and needs modification at higher ionic strengths

Ionic strength effectsWe start by considering the equilibrium between separate reactants and the encounter complex. For non-ideal solutions

KAB=([AB]/[A][B]) γAB/γAγB

d[P]dt

=k[A][B]=ka[AB] (SSA)

k= kaKAB (γAγB /γAB)

d[P]dt

kaKAB[A][B] (γAγB /γAB)=

and

Ionic strength effectsWe can now re-write

kTS = (κkT/h) K

to include activity coefficients:

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.

kTS = (κkT/h) K (γAγB /γAB)

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’zAI2

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 zA+zB and k0 is rate constant at zero ionic strength and equals kaKAB.

log(k/k0)= 1.018zAzBI Kinetic salt effect


A plot of log k vs. I is a straight line with slope~zAzB.

The slope gives information about the charge types involved in the activated complex of the rate determining step.

log(k)=log(k0) + 1.018zAzBI

y = c + mx


The slopes of the lines are those given by the Debye-Hückel limiting law.

Diffusion and reactionThe 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(k0)= log(12.2) - (1.018)x(1)x(-1)x(0.0525)=1.32

k0= 20.9 dm6mol-2min-1

Diffusion and reactionAn 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 0.03k/k0 0.995 0.925 0.875 0.835 0.800 0.770

I1/2 0.071 0.100 0.122 0.141 0.158 0.173

y = -1.0849x + 0.0747

R2 = 1

-0.12

-0.1

-0.08

-0.06

-0.04

-0.02

0

0.06 0.11 0.16 0.21log(k/k0) -0.002 -0.034 -0.058 -0.078 -0.097 -0.114

Slope=1.018zAzB

-1.085=1.018zAzB

A=+1B=-1


At high ionic strength, log(γA)=-A’zAI breaks down and is replaced by:

log(γA)=-A’zAI/(1+B’aI)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.

Photochemistry1. 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

hv

11-cis-Retinal

All-trans-Retinal

Vision-the ‘good’

http://upload.wikimedia.org/wikipedia/commons/1/11/RetinalCisandTrans.png

Introduction to photochemistryDNA 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 photochemistryMulti-step organic synthesis- the ‘ugly’

Jablonski diagrams

Fluorescence vs. phosphorescence

Green fluorescent protein

GFP

Osamu Shimomura Martin Chalfie Roger Y. Tsien

Nobel Prize in Chemistry 2008"for the discovery and development of the green fluorescent protein, GFP"

GFP chromophore

p-hydroxybenzylidene-imidazolidone

Fluorescence vs. phosphorescencekinetics

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.

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.

Iabs

number of eventsnumber of photons

absorbed

rate of processintensity of light

absorbed φ = = =

ν



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.

Iabs

φi Σi

=Σi

νi = 1



For example, if the excited state only decays to the ground state through photophysical processes, we write:

where φf, φIC, φISC and φP are the quantum yields of fluorescence, internal conversion, intersystem crossing and phosphorescence respectively.

φf + φIC + φISC + φP = 1


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*]


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*]dt

= Iabs – [S*](kf + kIC + kISC) = 0

and Iabs = [S*](kf + kIC + kISC)


and substituting into:

The final expression for the quantum yield for fluorescence becomes:

φf = Iabs

νf =[S*](kf + kIC + kISC)

kf[S*]

φf = (kf + kIC + kISC)

kf


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.)

τ0 = (kf + kIC + kISC)

1=

φf

kf

No quencher


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*]

Quenching

Three common mechanisms for bimolecular quenching of an excited state are:


Collisional S* + Q S + Q efficient when Qdeactivation is large e.g. I-

Energy transfer: S* + Q S + Q*

Electron transfer: S* + Q S+ + Q- or S- + Q+

dt

d[S*] =Iabs – [S*](kf + kIC + kISC + kQ[Q]) = 0

Invoking the SSA now gives:

The fluorescence quantum yield now becomes:

φ = (kf + kIC + kISC + kQ[Q])

kf


Stern-Volmer plot

φ φf =1 + τ0kQ[Q]

By plotting the LHS vs. [Q], we obtain a straight line with slope τ0kQ. This is known as a Stern-Volmer plot.

[Q]

φ φf τ0kQ

01


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.

φ φf =1 + τ0kQ[Q]τ

τ0 =

therefore

= + kQ[Q]τ 1

τ0 1


Stern-Volmer plot

The molecule 2,2’-bipyridine forms a complex with the Ru2+ ion. Ruthenium(II) tris-(2,2’-bipyridyl), Ru(bpy)3

2+, has a strong metal-ligand change transfer (MLCT) transition at 450 nm. The quenching of the *Ru(bpy)3

2+ 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

τ/(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:


= + kQ[Q]τ 1

τ0 1

0

1

2

3

4

5

0 0.5 1.0

[Fe3+]/(10-3 mol L-1)

(10

6 s

)/τ

kQ = 2.7 x 109 L mol-1 s-1

τ0 = 5.77 x 10-7 s


•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.


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.

A + hν A*

Femtochemistry

The laser pulses

time

ns10-9 s

ps10-12 s

fs10-15 s

Depending on the dynamics being measured, different laser-pulse durations are required

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.

A + hν1 A*

B + C

hν2

‘Pump’

A

‘Probe’

we can monitor A* directly by ionizing it and looking at the parent ion (A+) as a function of time.

Probing intermediates

Alternatively..

A + hν1 A*

B + C

hν2

‘Pump’

A+‘Probe’

In any case, one might see the following dynamics

[B] rises while [A+] and hence [A*] falls.

Some typical data

[B]

time

[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.

The ‘pump’ and the ‘probe’

[A+]

time

Probe (discrete step sizes)

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.

time

time zero

pump probe

Accuratelyknown

Pump & probetemporally overlapped

time

Accurate time-delay

Once the temporal overlap is found, we can start to scan the probe relative the pump, i.e.

Changing time-delay

time

time zero

pumpprobe

time

time zero

pump probe

time

time zero

pump probe

For example

ProbeDelay 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.

Delaying the fs pulses

Real examples

1. Photodissociation of Br2

X 1g

+

C 1u

1.5 2.5 3.5 4.5R (Å)

En

erg

y/c

m-1 (

10

4)

0

1

2

3

~~

400 nm

Br2 dissociates

Br Br

Photoabsorption followed by dissociation

5p 4P3/2

266.8 nm

3P2 (Br+)

Br


Detecting fragment Br atoms through resonance enhanced multiphoton ionisation (REMPI)

En

erg

y

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

Photodissociation of Br2 at 2.136 eV.

47 fs

Time (ps)-0.2 -0.1 0 0.1 0.2

3000

1500

Ion

coun

t

Br Br

Br Br

+/- 5 fs


2. Photodissociation of CH3I

X-state

A-state

C-I

En

erg

y

304 nmI HC

H

H

REMPI (277 nm)

+

The reaction of CH3I I + CH3

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

= Aniline

= Iodine

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.

JCP 105 (1996) 7864

The reaction of CH3I I + CH3

In summary….

Experimental methodsThe transient species can be monitored by either:

1. Absorption

2. Fluorescence

3. Conductivity

4. Pressure

5. NMR

6. Mass detection

7. Electron detection

[conc.]



d[P]dt

= kd[A][B]

d[P]dt

=kakd[A][B]

k’d




Diffusion control

Rate determining step of the reaction is the approach of reactants. Once in the solvent cage, reaction occurs

Solvent cage

kd=4πR*DNA

where D=DA+DB

R*=minimum distance for reaction

Activation control

Reactants (ep)

Products

Ea

Progress of reaction

En

erg

y

Activated complex

KAB = exp(-ΔGep/RT)k= kaKAB ka = κ(kT/h) exp(-ΔGac/RT)

ep = encounter pair

ac = activated complex

Reactions between ionsFrom the thermodynamic formulation of the rate constant, we can obtain a rate constant that depends on ionic strength.

k= kaKAB (γAγB /γAB)

d[P]dt

kaKAB[A][B] (γAγB /γAB)=


log(γA)=-A’zAI2

log(γB)=-A’zBI2

log(γAB)=-A’(zA+zB)I2 }

}Kinetic salt effect



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*]


d[S*]dt

= Iabs – [S*](kf + kIC + kISC) = 0

Iabs = [S*](kf + kIC + kISC)


φf = Iabs

νf =[S*](kf + kIC + kISC)

kf[S*]

φf = (kf + kIC + kISC)

kf

No quencher

φ = (kf + kIC + kISC +kQ[Q])

kf

Quencher

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.

φ φf =1 + τ0kQ[Q]τ

τ0 =

therefore

= + kQ[Q]τ 1

τ0 1


Revision question CH162

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?

Tryptophan

1.75E+095.70E-101.08E-01

1.41E+097.10E-108.00E-02

1.09E+099.20E-105.50E-02

6.67E+081.50E-092.30E-02

3.85E+082.60E-090.00E+00

[O2] t 1/t1.

0

4.0E+08

8.0E+08

1.2E+09

1.6E+09

2.0E+09

Gradient = 1.28x1010

0 0.04 0.08 0.12

[O2]

1/t

kQ = 1.28x1010 dm3mol-1s-1

2.τ0 = (kf + kIC + kISC)

1=

φf

kf

τ0

= kf = φf

2.6x10-9s

0.2= 7.7x107s-1

hv = PE + Iihv - PE = Ii

10.3 eV – 1.87 eV = 8.43 eV

3. i

3. ii-answer via email!

ch162 kinetics

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