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
Φ =
kf
k f + k IC + k ISC + k Q [Q ]
The ratio of Φ/ Φf is then given by
ϕf
= 1 + τ o k Q [Q ]
ϕ
If
= 1 + τ o k Q [Q ]
I
τf
= 1 + τ o kQ [Q ]
τ
(Note that here φ represents the case in the presence of quencher, so do τ and
I. In the class, we use φnf to represent the primary quantum yield in the
presence of quencher, another significant difference here is τ0 =
1
. This notation is consistent with decay time!!!)
k f + k IC + k IYC
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.
Three common mechanisms for bimolecular quenching:
(a) Collisional deactivation: S* + Q → S + Q
(b) Energy transfer:
S* + Q → S + Q*
(c) Electron transfer:
S* + Q → S+ + Q- or S- + Q+
Energy transfer processes (Forster theory)
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
R06
ET = 6
R0 + R 6
Where R is the distance between the donor and the acceptor. R0 is a
parameter that is characteristic of each donor-acceptor pair.
Electron transfer reactions (Marcus theory)
1. The distance between the donor and acceptor, with electron transfer
becoming more efficient as the distance between donor and acceptor
decrease.
2. The reaction Gibbs energy, ∆rG, with electron transfer becoming more
efficient as the reaction becomes more exergonic.
3. 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.
The overall quantum yield of a photochemical reaction
Overall quantum yield: the number of reactant molecules consumed per
photon absorbed:
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. In a chain reaction, the overall quantum
yield can be very large.
Example 26.11a: 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 ethene?
Solution: First calculate the number of photons generated in the interval
100s. Then divide the amount of ethene molecules formed by the amount of
photons absorbed.
N(photons) = P∆t/(hc/λ)
Ф = n(C2H4)*NA/N
= 0.21
27 Molecular reaction dynamics
27.1 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 ∝ σ (T / M )
1/ 2
N A N B ∝ σ (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 − Ea / RT .
k2 ∝ σ (T / M )1 / 2 e − Ea / RT
Consider the steric factor, P,
k2 ∝ Pσ (T / M )1 / 2 e − Ea / RT
k2 ∝ steric requirement x encounter rate x minimum energy requirement
(a) 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
 8kT 
2
ZAB = σ 
 N A [ A][ B ]
 πu 
2
where σ = π d
d = ½(dA + dB)
and u is the reduced mass
m AmB
u=
m A + mB
when A and B are the same, one gets
ZAA
 4kT 

= σ 
m
π

A 
1/ 2
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.
(b) The energy requirement
For a collision with a specific relative speed of approach vrel
d [ A]
= − σ (ε )v rel N A [ A][ B ]
dt
d [ A]
=−
dt
(∫ σ (ε )v
∞
0
rel
)
f (ε )dε N A [ A][ B ]
reorganize the rate constant as
k2 = N A
(∫ σ (ε )v
∞
0
rel
f (ε )dε
)
Assuming that the reactive collision cross-section is zero below εa
ε ·
§
σ (ε ) = ¨ 1 − a ¸σ
ε ¹
©
−
k2 = N Aσ c rel e − Ea / RT
(c) The steric requirement
Steric factor, P
Reactive cross-section, σ*
σ* = P σ
§ 8kT ·
k2 = P σ ¨
¸
© πu ¹
1/ 2
N A e − Ea / RT
Example 27.1a 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 AmB
= 3.12 x 10-27 kg
m A + mB
1/ 2
 8kT 


 πu 
= 2.66 x 103 ms-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 = σ

 πu 
N A = 7.37 x 1011 L mol-1s-1
P = 1.24 x 106 L mol-1 s-1/7.37 x 1011 L mol-1s-1
= 1.7 x 10-6
Harpoon mechanism: Electron transfer preceded the atom extraction. It
extends the cross-section for the reactive encounter.