Transcript Lecture 20 - University of Windsor

Quenching
• The presence of a quencher, Q, opens an additional channel for
deactivation of S*
S* + Q → S + Q
vQ = kQ[Q][S*]
Now the steady-state approximation for [S*] gives:
Iabs - kf[S*] - kIC[S*] - kISC[S*] - kQ[Q][S*] = 0
The fluorescence quantum yield in the presence of quencher
becomes

kf
k f  k IC  k ISC  kQ [Q ]
• The ratio of Φ/ Φf is then given by
f
 1   o k Q [Q ]

• Therefore a plot of the left-hand side of the above equation against
[Q] should produce a straight line with the slope τ0kQ. Such a plot is
called Stern-Volmer plot. (fluorescence intensity and life time)
• The fluorescence intensity and lifetime are both
proportional to the fluorescence quantum yield, plot of
If,0/I0 and t0/t against [Q] should also be linear with the
same slope and intercept as
f
 1   o k Q [Q ]

• Self-test 23.4 The quenching of tryptophan
fluorescence by dissolved O2 gas was monitored by
measuring emission lifetimes at 348 nm in aqueous
solutions. Determine the quenching rate constant for
this process
[O2]/(10-2 M) 0
2.3
5.5
8
10.8
Tau/(10-9 s) 2.6
1.5
0.92 0.71 0.57
Three common mechanisms for
bimolecular quenching
• Collisional deactivation:
S* + Q → S + Q
is particularly efficient when Q is a
heavy species such as iodide ion.
• Resonance energy transfer:
S* + Q → S + Q*
• Electron transfer: S* + Q → S+ + Q- or
S* + Q → S- + Q+
Energy Transfer Processes
• (Forster theory,1952) Energy transfer is more
efficient when
1. The energy donor and acceptor are separated
by a short distance, in the nanometer scale
2. Photons emitted by the excited state of the
donor can be absorbed directly by the acceptor
• The efficiency of energy transfer, ET, equals
f
ET 1 
 f ,0
R06
ET  6
R0  R6
Where R is the distance between the donor
and the acceptor. R0 is a parameter that is
characteristic of each donor-acceptor pair.
• Fluorescence resonance energy transfer (FRET)
Electron transfer reactions
(Marcus theory)
• The distance between the donor and acceptor, with
electron transfer becoming more efficient as the distance
between donor and acceptor decrease.
• The reaction Gibbs energy, ∆rG, with electron transfer
becoming more efficient as the reaction becomes more
exergonic.
• The reorganization energy, the energy cost incurred by
molecular rearrangements of donor, acceptor, and
medium during electron transfer.
The electron transfer rate is predicted to increase
as this reorganization energy is matched closely by the
reaction Gibbs energy.
23.8 Complex photochemical
processes
• Overall quantum yield: the number of reactant
molecules consumed per photon absorbed (can
be larger than 1):
For example:
HI + hv
→ H. + I.
HI + H.
→ H2 + I.
I. + I. + M → I2 + M*
Here the overall quantum yield is two, because the
absorption of one photon destroys two reactant
molecules HI. Therefore, in a chain reaction the overall
quantum yield can be very large.
Rate laws of complex
photochemical reactions.
• See example: 23.21 (8th edition)
Photosensitization
• Example: hydrogen gas containing trace
amount of mercury.
• The synthesis of formaldehyde
H. + CO -> HCO.
HCO. + H2 -> HCHO + H.
HCO. + HCO. -> HCHO + CO
• Photodynamic therapy
• Example: When a sample of 4-heptane was irradiated
for 100s with 313 nm radiation with a power output of
50W under conditions of total absorption, it was found
that 2.8 mmol C2H4 was formed. What is the quantum
yield for the formation of ethylene?
Solution: First calculate the number of photons generated in the
interval 100s.
Then divide the amount of ethylene molecules formed by
the amount of photons absorbed.
N(photons) = P∆t/(hc/λ)
Ф = n(C2H4)*NA/N
= 0.21
Chapter 24. Molecular Reaction
Dynamics
Purpose: Calculation of rate constants for simple
elementary reactions.
For reactions to take place:
1. Reactant molecules must meet.
2. Must hold a minimum energy.
Gas phase reactions:
Collision theory.
Solution phase reactions: Diffusion controlled.
Activation controlled.
24.1 Collision theory
• Consider a bimolecular elementary reaction
A + B → P
v = k2[A][B]
The rate of v is proportional to the rate of collision, and
therefore to the mean speed of the molecules,
v   (T / M )1/ 2 N A N B   (T / M )1/ 2[ A][B]
• Because a collision will be successful only if the kinetic energy
exceeds a minimum value. It thus suggests that the rate constant
should also be proportional to a Boltzmann factor of the form, e  E / RT .
a
k2   (T / M )1 / 2 e  Ea / RT
• Consider the steric factor, P,
k2  P (T / M )1 / 2 e  Ea / RT
• Therefore, k2 is proportional to the product of steric requirement x
encounter rate x minimum energy requirement
Collision rate in gases
•
Collision density, ZAB, is the number of (A, B) collisions in a region of the
sample in an interval of time divided by the volume of the region and the
duration of the interval.
1/ 2
Z AB
 8kT 


 u 
N A2 [ A][B]
where σ =  d2
d = ½(dA + dB)
and u is the reduced mass
u
•
m AmB
m A  mB
when A and B are the same, one gets
1/ 2
Z AA
•
 4kT 

 (1 / 2) 
 m A 
N A2 [ A]2
The collision density for nitrogen at room temperature and pressure, with d
= 280 pm, Z = 5 x 1034 m-3s-1.
The energy requirement
•
For a collision with a specific relative speed of approach vrel
d [ A]
  ( )v rel N A[ A][ B ]
dt

d [ A]
    ( )v rel f ( )d  N A[ A][B]
dt
 0

•
reorganize the rate constant as


k2  N A    ( )v rel f ( )d 
 0

•
Assuming that the reactive collision cross-section is zero below εa
 a 
 ( )  1  
 

for    a

k2  N A c rel e  Ea / RT
The steric effect
• Steric factor, P,
• Reactive cross-section, σ*,
•
σ* = P σ
•
1/ 2
 8kT 
k 2  P 


u


N Ae  Ea / RT
• Harpoon mechanism: Electron transfer preceded
the atom extraction. It extends the cross-section for the
reactive encounter.
• K and Br2 reaction
Example 24.1 Estimate the steric factor for the reaction
H2 + C2H4 -> C2H6 at 628K
given that the pre-exponential factor is 1.24 x 106 L mol-1 s-1.
Solution:
Calculate the reduced mass of the colliding pair
u
m Am B
 3.12 1027 kg
m A  mB
1/ 2
 8kT 


 u 
 2.66 103 m s 1
From Table 24.1 σ(H2) = 0.27 nm2 and σ(C2H4) = 0.64 nm2, given a
mean collision cross-section of σ = 0.46 nm2.
1/ 2
 8kT 
A 

 u 
N A  7.37 1011 Lmol1 s 1
P = 1.24 x 106 L mol-1 s-1/7.37 x 1011 L mol-1s-1
= 1.7 x 10-6
• Example 24.2: Estimate the steric factor for the
reaction: K + Br2 → KBr + Br
Solution: The above reaction involves electron flip
K + Br2 → K+ + Br2Three types of energies are involved in the above process:
(1) Ionization energy of K, I
(2) Electron affinity of Br2, Eea
(3) Coulombic interaction energy:
 e2
4 0 R
Electron flip occurs when the sum of the above three energies changes sign
from positive to negative
24.2 Diffusion-controlled reactions
• Cage effect: The lingering of one molecule near another on account
of the hindering presence of solvent molecules.
• Classes of reaction
Suppose that the rate of formation of an encounter pair AB is
first-order in each of the reactants A and B:
A + B →AB
v = kd[A][B]
The encounter pair, AB, has the following two fates:
AB → A + B
v = kd’[AB]
AB → P
v = ka[AB]
• The net rate of change of [AB]:
d [ AB ]
dt
= kd[A][B] - kd’[AB] - ka[AB]
• Invoking steady-state approximation to [AB]
[ AB ] 
kd [ A][B]
k d '  ka
• The net rate of the production:
k k [ A][B]
d[ P ]
 ka [ AB ]  a d
 k2 [ A][B]
dt
k d '  ka
k2 
k a kd
k d '  ka
• When kd’<< ka
k2 = kd
• When kd’>> ka
reaction)
k2 
ka kd
kd'
(This is diffusion-controlled limit)
(This is activation-controlled
Reaction and Diffusion
kd  4R* DN A
•
where R* is the distance between the reactant
molecules and D is the sum of the diffusion coefficients of the two
reactant species (DA + DB).
DA 
kT
6RA
DB 
kT
6RB
where η is the viscosity of the medium. RA and RB are
the hydrodynamic radius of A and B.
•
If we assume RA = RB = 1/2R*
kd 
8RT
3
24.3 The material balance equation
(a) The formulation of the equation
[ J ]
 2 [J ]
[ J ]
D
v
2
t
x
x
the net rate of change due to chemical reactions
[ J ]
  k[ J ]
t
the over rate of change
[ J ]
 2 [J ]
[ J ]
D
v
 k[ J ]
2
t
x
x
the above equation is called the material balance equation.
(b) Solutions of the equation
[ J ]
 2 [J ]
D
 k[ J ]
2
t
x
t
[J ]  k  [J ]e kt dt  [J ]e kt
*
0
[J ] 
n0e
 x 2 / 4 Dt
A(Dt )1 / 2