Foundations of Chemical Kinetics Lecture 30: Transition-state - PowerPoint PPT Presentation

Foundations of Chemical Kinetics Lecture 30: Transition-state theory in the solution phase Marc R. Roussel Department of Chemistry and Biochemistry

Transition-state theory in solution ◮ We revisit our original derivation of the thermodynamic formalism of transition-state theory. ◮ We decompose the reaction A + B k → product(s) into − K ‡ − TS k ‡ ⇀ A + B − ↽ − → product(s) − − ◮ For this representation, v = k ‡ [TS]. ◮ The equilibrium constant K ‡ is related to the activities of the transition state and reactants by K ‡ = a TS a A a B [TS] c ◦ = γ TS [A][B] γ A γ B

Transition-state theory in solution (continued) ◮ The derivation is completed as follows: [TS] = K ‡ γ A γ B [A][B] c ◦ γ TS ∴ v = k ‡ [TS] = k ‡ K ‡ γ A γ B [A][B] c ◦ γ TS ∴ k = k ‡ K ‡ γ A γ B c ◦ γ TS ◮ For an ideal solution, all the activity coefficients are equal to unity, and we get the limiting value k 0 = k ‡ K ‡ c ◦ ◮ Therefore γ A γ B k = k 0 γ TS

Transition-state theory in solution Constancy of first-order rate constants ◮ If we repeat the above derivation for a first-order reaction R → product(s), we get γ R k = k 0 γ TS ◮ In a unimolecular reaction, the reactant and transition state are typically very similar, the difference between the two involving small changes in charge distribution and geometry. ◮ Accordingly, γ R ≈ γ TS , so various factors that affect the activity coefficients (e.g. pressure, ionic strength) should have little effect on the rate constant.

Transition-state theory in solution Constancy of first-order rate constants (continued) ◮ Now imagine slowly decreasing the density of the solvent. ◮ In the limit, this operation converts a reaction in solution to one in the gas phase. ◮ By the argument above, the ratio of activity coefficients shouldn’t change much, so the rate constant should be more-or-less constant throughout this operation. ◮ This explains the experimental observation that, for first-order reactions, k is roughly the same whether the reaction occurs in the gas phase or in solution.

Variation of rate constant with pressure ◮ We return to the ideal solution case. In the thermodynamic formulation of transition-state theory, we have, for second-order reactions, − ∆ ‡ G ◦ � � k 0 = k B T c ◦ h exp RT ◮ Recall that dG = V dp − S dT or, for a reaction, d ∆ G = ∆ V dp − ∆ S dT so that � ∂ ∆ G � = ∆ V � ∂ p � T

Variation of rate constant with pressure (continued) ◮ For k 0 , we have − ∆ ‡ G ◦ � k B T � ln k 0 = ln c ◦ h RT ∂ ∆ ‡ G ◦ = − ∆ ‡ V ◦ � � ∴ ∂ = − 1 � � ∂ p ln k 0 � � RT ∂ p RT � � T T where ∆ ‡ V ◦ is the change in the molar volume (change in volume of solution per mole of reaction) on accessing the transition state. ◮ If ∆ ‡ V ◦ > 0, the rate constant decreases as p increases. The reverse is true if ∆ ‡ V ◦ < 0.

Gas-phase vs solution-phase rate constants ◮ We now consider the thermodynamic formulation of transition-state theory, which gives, for second-order reactions, − ∆ ‡ G ◦ � � k 0 = k B T c ◦ h exp RT ◮ The above equation should apply both in solution ( k 0 , s ↔ ∆ ‡ G ◦ s ) and in the gas phase ( k 0 , g ↔ ∆ ‡ G ◦ g ). ◮ We want to figure out how the two rate constants are related.

Gas-phase vs solution-phase rate constants (continued) ◮ Now consider the following thermodynamic cycle: ∆ ‡ G ◦ s A (sol) + B (sol) → TS (sol) − − − −  � − ∆ solv G ◦ (A) − ∆ solv G ◦ (B)  ∆ solv G ◦ (TS)   � ∆ ‡ G ◦ g A (g) + B (g) → TS (g) − − − − where ∆ solv G ◦ is the standard free energy of solvation. ◮ Reading off the cycle, we have ∆ ‡ G ◦ s = ∆ ‡ G ◦ g + ∆ solv G ◦ (TS) − [∆ solv G ◦ (A) + ∆ solv G ◦ (B)] = ∆ ‡ G ◦ g + ∆ ‡ ∆ solv G ◦

Gas-phase vs solution-phase rate constants (continued) ◮ From − ∆ ‡ G ◦ � � k 0 , s = k B T s c ◦ h exp RT and ∆ ‡ G ◦ s = ∆ ‡ G ◦ g + ∆ ‡ ∆ solv G ◦ we get � � − ∆ ‡ G ◦ − ∆ ‡ ∆ solv G ◦ � � k 0 , s = k B T g c ◦ h exp exp RT RT − ∆ ‡ ∆ solv G ◦ � � = k 0 , g exp RT

Gas-phase vs solution-phase rate constants (continued) − ∆ ‡ ∆ solv G ◦ � � k 0 , s = k 0 , g exp RT TS better solvated than reactants: ∆ ‡ ∆ solv G ◦ < 0 In this case, the argument of the exponential involving this solvation term is positive and k 0 , s > k 0 , g . Reactants better solvated than TS: ∆ ‡ ∆ solv G ◦ > 0 We then have the exponential of a negative value, so k 0 , s < k 0 , g .

Gas-phase vs solution-phase rate constants (continued) ◮ Under what conditions might we expect ∆ ‡ ∆ solv G ◦ < 0? ◮ One example might involve hydrophobic reactants and TS. If the TS is more compact (exposes less surface area to the solvent) than the two reactants, which will often be the case, then each of the individual ∆ solv G ◦ terms will be positive, but the solvation free energy of the transition state would be much smaller than that of either reactant. ◮ Differences in solvation free energies of ∼ − 10 kJ mol − 1 are not unreasonable under these conditions, which would give − ∆ ‡ ∆ solv G ◦ � � exp = 56 RT at 25 ◦ C. Rate enhancements of this order of magnitude have been seen in some reactions.