1. 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
The ratio of Φ/ Φf is then given by
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)

2. 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
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)
Tau/(10-9 s)

3. 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+

4. 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
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)

6. 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.

7. 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 → H I.
I I M → I 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.

8. Rate laws of complex photochemical reactions.
See example: (8th edition)

9. 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

10. 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

11. 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.

12. 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,
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,
Consider the steric factor, P,
Therefore, k2 is proportional to the product of steric requirement x encounter rate x minimum energy requirement

13. 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.
where σ = d d = ½(dA + dB)
and u is the reduced mass
when A and B are the same, one gets
The collision density for nitrogen at room temperature and pressure, with d = 280 pm, Z = 5 x 1034 m-3s-1.

14. The energy requirement
For a collision with a specific relative speed of approach vrel
reorganize the rate constant as
Assuming that the reactive collision cross-section is zero below εa

16. The steric effect Steric factor, P, Reactive cross-section, σ*,
Harpoon mechanism: Electron transfer preceded the atom extraction. It extends the cross-section for the reactive encounter.
K and Br2 reaction

17. Example 24.1 Estimate the steric factor for the reaction
H 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
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.
P = 1.24 x 106 L mol-1 s-1/7.37 x 1011 L mol-1s-1
= 1.7 x 10-6

18. Solution: The above reaction involves electron flip
Example 24.2: Estimate the steric factor for the reaction: K + Br2 → KBr + Br
Solution: The above reaction involves electron flip
K + Br2 → K+ + Br2-
Three 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:
Electron flip occurs when the sum of the above three energies changes sign from positive to negative

19. 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]:
= kd[A][B] - kd’[AB] - ka[AB]

20. Invoking steady-state approximation to [AB]
The net rate of the production:
When kd’<< ka k2 = kd (This is diffusion-controlled limit)
When kd’>> ka (This is activation-controlled reaction)

21. Reaction and Diffusion
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).
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*

22. 24.3 The material balance equation
(a) The formulation of the equation
the net rate of change due to chemical reactions
the over rate of change
the above equation is called the material balance equation.

23. (b) Solutions of the equation