Foundations of Chemical Kinetics Lecture 29: Diffusion-influenced - PowerPoint PPT Presentation

Foundations of Chemical Kinetics Lecture 29: Diffusion-influenced reactions, Part II Marc R. Roussel Department of Chemistry and Biochemistry

Diffusion-influenced reactions ◮ Consider an elementary reaction A + B k − → product(s). ◮ If { AB } is the encounter pair formed from the reactants A and B, we can break down a reaction in solution into two steps: k D k 2 − − ⇀ A + B ↽ − − { AB } − → product(s) k − D ◮ An encounter pair is not ◮ a chemical intermediate or ◮ the transition state of a reaction. ◮ The encounter pair has a very short lifetime (to be estimated later). We should therefore be able to apply the steady-state approximation to the encounter pair.

Diffusion-influenced reactions (continued) k D k 2 − − ⇀ A + B ↽ − − { AB } − → product(s) k − D d [ { AB } ] = k D [A][B] − ( k − D + k 2 )[ { AB } ] ≈ 0 dt k D ∴ [ { AB } ] ≈ [A][B] k − D + k 2 k D k 2 ∴ v = k 2 [ { AB } ] ≈ [A][B] k − D + k 2 ◮ Since v = k [A][B], k D k 2 k = k − D + k 2

Comparison of the two approaches ◮ In the last lecture, we derived the equation k D k R k = � � U ( R AB ) k R + k D exp k B T for the case of weak intermolecular forces (hard-sphere potential), where k R is the second-order rate constant for A reacting with B molecules that have already reached distance R AB . ◮ We now have k D k 2 k = k − D + k 2 where k 2 is the first-order rate constant for the formation of products from the encounter pair.

Comparison of the two approaches (continued) ◮ These are two different ways of describing the same thing, so the two k expressions must be equal. k D k R k D k 2 � = ∴ � k − D + k 2 U ( R AB ) k R + k D exp k B T 1 1 � = ∴ � 1 + k − D / k 2 U ( R AB ) 1 + k D k R exp k B T � U ( R AB ) � ∴ k R = k D exp k 2 k − D k B T or � � 1 − U ( R AB ) k 2 = k R exp K D k B T where K D = k D / k − D is the equilibrium constant for formation of the encounter pair.

Comparison of the two approaches (continued) 1 � − U ( R AB ) � k 2 = k R exp K D k B T ◮ K − 1 has units of concentration. D It is a characteristic concentration scale for the equilibrium of the formation of the encounter pair. (If [A] , [B] ∼ K − 1 D , then [ { AB } ] ∼ K − 1 D .) ◮ Thus, k R is the second-order rate constant we would get if the encounter pair were in equilibrium with A and B at the characteristic concentration. (Compare the last equation in lecture 27: � � − U ( r ) c i ( r ) = c ◦ i exp .) k B T

Comparison of the two approaches (continued) ◮ k 2 ought in principle be evaluable from variational transition-state theory or other similar approaches. ◮ In either theory, the same equation for k D holds.

Encounter pair formation and breakup ◮ For the special case of molecules whose intermolecular forces with the solvent are about the same as those between each other, we can estimate K D by a statistical argument. This will also lead to an estimate of k − D since we know how to calculate k D . ◮ Focus again on an A molecule. ◮ Suppose that the coordination number of the reactive site of A is N , i.e. that the reactive site of A makes contacts with N neighboring molecules in solution. ◮ Let [S] be the mole density of the solvent. The probability that any given molecule of S has been replaced by a B in the first solvation sphere of A is [B]/[S].

Encounter pair formation and breakup ◮ If [B] ≪ [S], the probability that one of the N solvent molecules around A has been replaced by a B is N [B] / [S]. ◮ Another way to think about it is that N [B] / [S] is the fraction of A molecules that have a B molecule in their first solvation sphere. ◮ Therefore, [ { AB } ] ≈ N [B] [S] [A] ◮ Since K D = [ { AB } ] [A][B] we get K D = N [S]

Example: Reaction of bromphenol blue with hydroxide ion +OH − k − → quinoid form (blue) carbinol form (colorless) k = 9 . 30 × 10 − 4 L mol − 1 s − 1 in water at 25 ◦ C. To do: Estimate k 2 .

Example: Reaction of bromphenol blue with hydroxide ion Estimate of k D ◮ Given: D OH − = 5 . 30 × 10 − 9 m 2 s − 1 , D BPB = 4 . 4 × 10 − 10 m 2 s − 1 , C-O bond length = 143 pm, κ H 2 O = 78 . 37 (all at 25 ◦ C) ◮ Take R AB ≈ C-O bond length. ◮ The two reactants are both anions with a single negative charge. ǫ = κǫ 0 = 6 . 939 × 10 − 10 C 2 J − 1 m − 1 z A z B e 2 β = � � � � z A z B e 2 4 πǫ k B T exp − 1 4 πǫ k B TR AB = 4 . 846 × 10 − 12 m .

Example: Reaction of bromphenol blue with hydroxide ion Estimate of k D (continued) ∴ k D = 4 π LD AB β = 4 π (6 . 022 142 × 10 23 mol − 1 )[(5 . 30 + 0 . 44) × 10 − 9 m 2 s − 1 ] × (4 . 846 × 10 − 12 m) = 2 . 11 × 10 5 m 3 mol − 1 s − 1 ≡ 2 . 11 × 10 8 L mol − 1 s − 1

Example: Reaction of bromphenol blue with hydroxide ion Estimate of K D ◮ For this structure, N = 2 seems reasonable. ◮ The mole density of water at 25 ◦ C is 55 . 33 mol L − 1 . ◮ Therefore N 2 55 . 33 mol L − 1 = 4 × 10 − 2 L mol − 1 K D ≈ [H 2 O] =

Example: Reaction of bromphenol blue with hydroxide ion Estimate of k − D ◮ Since K D = k D / k − D , we have k − D = k D / K D = 6 × 10 9 s − 1 Aside: This value of k − D implies a half-life of the encounter pair of t 1 / 2 = ln 2 / k − D = 1 × 10 − 10 s ≡ 100 ps

Example: Reaction of bromphenol blue with hydroxide ion Estimate of k 2 ◮ By rearranging k D k 2 k = k − D + k 2 we get k 2 = kk − D k D − k (9 . 30 × 10 − 4 L mol − 1 s − 1 )(6 × 10 9 s − 1 ) = 2 . 11 × 10 8 L mol − 1 s − 1 − 9 . 30 × 10 − 4 L mol − 1 s − 1 = 3 × 10 − 2 s − 1 ◮ The very small value of this rate constant compared to k − D means that many, many encounter pairs are formed and broken up for each one that reacts.