1. 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.

2. 24.1 Collision theory Consider a bimolecular elementary reaction A + B → P v = k 2 [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, k 2 is proportional to the product of steric requirement x encounter rate x minimum energy requirement

3. Collision rate in gases Collision density, Z AB, 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 2 d = ½(d A + d B ) 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 10 34 m -3 s -1.

4. The energy requirement For a collision with a specific relative speed of approach v rel reorganize the rate constant as Assuming that the reactive collision cross-section is zero below ε a

6. 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 Br 2 reaction

7. Example 24.1 Estimate the steric factor for the reaction H 2 + C 2 H 4 -> C 2 H 6 at 628K given that the pre-exponential factor is 1.24 x 10 6 L mol -1 s -1. Solution:Calculate the reduced mass of the colliding pair From Table 24.1 σ(H 2 ) = 0.27 nm 2 and σ(C 2 H 4 ) = 0.64 nm 2, given a mean collision cross-section of σ = 0.46 nm 2. P = 1.24 x 10 6 L mol -1 s -1 /7.37 x 10 11 L mol -1 s -1 = 1.7 x 10 -6

8. Example 24.2: Estimate the steric factor for the reaction: K + Br 2 → KBr + Br Solution: The above reaction involves electron flip K + Br 2 → K + + Br 2 - Three types of energies are involved in the above process: (1) Ionization energy of K, I (2) Electron affinity of Br 2, E ea (3) Coulombic interaction energy: Electron flip occurs when the sum of the above three energies changes sign from positive to negative

9. 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 = k d [A][B] The encounter pair, AB, has the following two fates: AB → A + B v = k d ’[AB] AB → P v = k a [AB] The net rate of change of [AB]: = k d [A][B] - k d ’[AB] - k a [AB]

10. Invoking steady-state approximation to [AB] The net rate of the production: When k d ’<< k a k 2 = k d (This is diffusion-controlled limit) When k d ’>> k a (This is activation-controlled reaction)

11. 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. where η is the viscosity of the medium. R A and R B are the hydrodynamic radius of A and B. If we assume R A = R B = 1/2R*

12. 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.

13. (b) Solutions of the equation