CHAPTER 21: Reaction Dynamics

Transcription

1 CHAPTER 21: Reaction Dynamics I. Microscopic Theories of the Rate Constant. A. The Reaction Profile (Potential Energy diagram): Highly schematic and generalized. A---B-C B. Collision Theory of Bimolecular Gas-Phase Reactions. Consider the elementary reaction in gas phase: A + B P Rate Collision frequency Steric requirement = x x Energy requirement (collision freq. between A + B) = Z AB From kinetic theory of gases: Z AB = N A N B σ c rel f = collision density (# of collisions of A and B per volume per time) N A N B = number densities (# of A or B per unit volume) 1

2 σ=πd 2 = collision "cross section" or target area d = 1 d 2 ( + d A B) c rel = average relative velocity = (8k B T/πµ) 1/2 µ = m A m B /(m A + m B ) = "reduced mass" f = symmetry factor: = 1 if A B = 1/2 if A = B Special Case: A and B identical. µ = m A /2; f = 1/2 Z AA = 2 N A 2 πd A 2 (8k B T/πm A ) 1/2 In terms of molar concentrations [A] and [B]: Z AB = σ c rel f N avo 2 [A][B] The Energy requirement: Now, what about fraction of collisions with KE > E a : fraction... e -Ea/kT when Ea is on the "tail" of distribution. Turns out that the collision cross section can be viewed as a function of the relative kinetic energy of two colliding molecules, and that when you integrate over kinetic energy distribution, you get a simple exponential dependence on the activation energy. 2

3 Steric requirement: multiplicative factor θ, such that σ θ gives the effective crosssectional area. θ = fraction of collisions in a productive geometric arrangement. Total rate expression: Rate =- d[a]/dt = θ σ c rel fn avo e -Ea/kT [A][B] = k 2 [A][B] By comparison with Arrhenius theory. A = θ σ c rel fn avo In problems, substitutions of numerical values for parameters gives k 2 in units length 3 /mol-s, which should be converted to Liters/mol-s Consider rxn (Hydrogen exchange) at T = 300 K: H + H 2 H 2 + H Calculate Arrhenius A. (i) estimate r H, r H2 1.0 x m = (1Å) (ii) µ = (m H m H2 )/(m H + m H2 ) = {(1.0 * 2.0)/( )}N avo -1 (iii) k B = 1.38 x J/K J=kg m 2 /s 2 (iv) A = 4.1 x cm 3 /mol sec * 1 liter/1000 cm 3 = 4.1 x L/mol sec with steric P=1 (v) Experiment gives A = 5.4 x L/mol s, or A theory /A expt = 7.6 Discrepancy corrected by steric (or geometric) factor θ. 3

4 Example: angle of approach of reactants is important. Multiply k 2 by θ = steric factor = 1/7.6 C. Transition State Theory 1. A more general theory of rate constants, with no adjustable parameter like θ, the steric factor. 2. Potential energy surface of the rxn: H + HD H 2 + D One-dimensionalization along path: 4

5 3. TST theory says, H + HD H--H--D H 2 + D in quasi-equilibrium K A + B AB products Reactants are in pre-equilibrium with AB, the activated complex with equilibrium constant K. 4. Define: K = p AB p o p A p B p is the standard pressure to keep K unitless (=1 bar) 5. Now, we may convert to a concentration basis using p x = RT[X] [AB ] = RT p o K [A][B] The forward rate = k [AB ] with k = unimolecular decomposition rate constant of activated complex. Forward rate = k K [A][B](RT/p ) = k 2 [A][B] And so finally the TST second order rate constant is: k 2 = k K (RT/p ) 6. Our next task is to find K (= e = ΔG /RT according to Chapter 6) where ΔG = molar free energy of activation (standard) = G (AB ) - G (A + B) Remember that in general an equilibrium constant is a ratio of partition functions of products over reactants, with the product being in this case the activated complex. 5

6 o K = N q avo AB q o o A q B e ΔE o /RT where: ΔE o = E o (AB ) E o (A) E o (B) is the difference in zero-point electronic energies. 7. The A and B partition functions are given as usual (Chap 15), but the one for the activated complex is a bit tricky. The activated complex species has a vibration which leads to its decomposition. The partition function for this vibration is 1 q = 1 e 1 hν / kt hν / kt 1 (1 hν / kt +...) The activated complex also has other modes of vibration, possibly, o and these contribute to the remainder of its partition function q AB, so o q AB o (hν / kt)q AB 8. So finally: K = kt o N avo q AB hν q o o A q B e ΔE o /RT kt hν K 9. Now putting it all together we get the Eyring rate constant: # k 2 = k kt & % ( $ hν K (RT / p o ) ' 10. But here we make one more connection: the unimolecular decomp rate constant of the activated complex is the same as the vibration frequency ν assuming every vibration causes a decomposition, or if not, is related by a barrier-crossing transmission coefficient κ as: k = κν 6

7 So a cancelation takes place and, voila, this unknown frequency goes away: " k 2 = κ$ kt # h % ' K (RT / p o ) & The above equilibrium constant is a pressure-based one, which can be replaced by a concentration-based one like Eyring used: " k 2 = κ$ kt # h % ' K c & 11. Thermodynamic version of TST: The above rate constant is couched in terms of statistical mechanical partition functions, which is greatly useful when you have structural knowledge of the activated complex. Without this knowledge one can still make progress using thermodynamic description. Here we use the activation free energy: K = e ΔG /RT " k 2 = κ$ kt # h % ' (RT / po )e ΔG /RT & Activation free energy can be further decomposed into enthalpy and entropy of activation: ΔG = ΔH - TΔS 12. Relation of TST quantities to Arrhenius form: Need to find apparent E a in terms of ΔH and TST. E a = RT 2 ( lnk / T) = ΔH + 2RT for gas phase bimolecular reaction. 7

8 A = e 2 kt h RT p o eδs /R Note: entropy part is the only term unknown, and that part can be identified as the steric factor of collision theory: θ = e ΔS /R Determine A by expt - find ΔS, infer something about activated complex complexity. 13. Example bimolecular reaction H + H 2 (H-H-H)* H 2 + H From experiment: A = 5.4 x cm 3 /mole sec = 5.4 x L/mole sec E a = 8.22 kcal/mole Find ΔS at 300 K. ΔS = R (fairly large and negative) Indicating H 3 * state is much more orderly than H 2 (g) + H(g) state. 8

9 II. Reactions in Solution. A. Two limiting types of behavior. 1. Low activation energies: Rxn occurs every time there's a collision; rate then depends on how fast reactants can diffuse through solvent and encounter each other. Diffusion-Controlled Reaction: Smoluchowski expression: k = 4πdDN avo f relative diff coeff D = D A + D B (depends on solvent viscosity, size of A + B) d = sum of radii Example: H + (aq) + Ac-(aq) HAc f = electrostatic factor = 1 if non ionic 2. High activation energies: (Activation-controlled limit) Rxn rate depends on collision event itself, probability of success. Solvent plays a role by stabilizing effect on A + B independent species relative to AB complex. Will focus on case 2 for remainder of discussion. B. Reaction between non-polar molecules: solvent forces are negligible solvent acts simply as space filler treat same as gas-phase (use collision theory, e.g.) 9

10 C. Reaction between ions or polar molecules when solvent is polar: electrostatic effects important. Effect of dielectric constant (ε) of solvent: Find ΔG = G - G ( separation) Interparticle force f = Z A Z B e 2 /4πε o εx 2 f(x= ) = 0 (x = distance separ.) Calculate ΔG es (electrostatic portion of ΔG ) by computing work w done on ions bringing them togther. w = d d ( ) f x dx ( ) = Z A Z B e 2 / 4πε o εx 2 dx d ( ) ( ) = + Z A Z B e 2 / 4πε o εx = + Z A Z B e 2 / 4πε o εd = electrostatic contrib to ΔG per molecular pair molar ΔG = ΔG nes molar non-electrostatic contribution + ( Z A Z B e 2 / 4πε o εd) N avo to make per mole 10

11 Plug into TST: k 2 = constants e ΔG nes (( ) N avo) /RT k 2 = k 2 o e Z A Z B e2 /4πε o εd ( + ( Z A Z B e 2 /4πε o εd) N avo) /RT ln k = ln k o Z A Z B e 2 / 4πε o εdk B T Theory says: ln k vs. 1/ε gives straight line How do we change ε? Change solvents, change temperature. k as ε, why? D. Ionic strength (salt) effects on reactions between ions in solution Consider rxn in solution: A + B AB products In TST, we used K ~ [AB ]/[A][B]. With equilib constant of rxn in nonideal solution, should use activities. K = a /a A a B = ([AB ]/[A][B])( γ /γ A γ B ) act. coeff [AB ] = K [A][B]γ A γ B /γ TST: Rate = k [AB ] (However, rate still prop to [AB ], not activites) ~ K [A][B]γ A γ B /γ Rate = k o [A][B]γ A γ B /γ old rate const correction without activities factor considered 11

12 k = k o γ A γ B /γ log k = log k o + log γ A γ B /γ Use D-H limiting law log y = -BZ 2 I ionic strength log k = log k o Z A Z B I Origin of effect, non-independent behavior of ions at moderate conc: 12

13 Notes: 13