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Three common mechanisms for bimolecular quenching

Three common mechanisms for bimolecular quenching. 1. Collisional deactivation: S* + Q → S + Q is particularly efficient when Q is a heavy species such as iodide ion. 2. Resonance energy transfer: S* + Q → S + Q *

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Three common mechanisms for bimolecular quenching

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  1. Three common mechanisms for bimolecular quenching 1. Collisional deactivation: S* + Q → S + Q is particularly efficient when Q is a heavy species such as iodide ion. 2. Resonance energy transfer: S* + Q → S + Q* 3. Electron transfer: S* + Q → S+ + Q- or S* + Q → S- + Q+

  2. (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) Energy Transfer Processes

  3. 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.

  4. 23.8 Complex photochemical processes • The overall quantum yield of a photochemical reaction. (can be larger than 1) • Rate laws of complex photochemical reactions. • Photosensitization (no direct absorption).

  5. Quantum yield of a complex 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. Therefore, in a chain reaction the overall quantum yield can be very large.

  6. Rate laws of complex photochemical reactions. • See example: 23.21 (8th edition)

  7. 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

  8. 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

  9. 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.

  10. 24.1 Collision theory • Consider a bimolecularelementary 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

  11. 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 = d2 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.

  12. 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

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

  14. 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 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

  15. 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

  16. 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]

  17. 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)

  18. 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*

  19. 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.

  20. (b) Solutions of the equation

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