Foundations of Chemical Kinetics. Lecture 30: Transition-state theory in the solution phase

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

2 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 A + B K TS k product(s) For this representation, v = k [TS]. The equilibrium constant K is related to the activities of the transition state and reactants by a TS K = a A a B = γ TS [TS]c γ A γ B [A][B]

3 Transition-state theory in solution (continued) The derivation is completed as follows: [TS] = K c γ A γ B γ TS [A][B] v = k [TS] = k K c γ A γ B γ TS [A][B] 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 k = k 0 γ A γ B γ TS

4 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 k = k 0 γ R 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, γr γ TS, so various factors that affect the activity coefficients (e.g. pressure, ionic strength) should have little effect on the rate constant.

5 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.

6 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, k 0 = k BT c h exp ( G ) Recall that dg = V dp S dt or, for a reaction, d G = V dp S dt so that G p = V T

7 Variation of rate constant with pressure (continued) For k0, we have ( ) kb T ln k 0 = ln c G h p ln k 0 = 1 G T p T = V 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.

8 Gas-phase vs solution-phase rate constants We now consider the thermodynamic formulation of transition-state theory, which gives, for second-order reactions, k 0 = k BT ( c h exp G ) 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.

9 Gas-phase vs solution-phase rate constants (continued) Now consider the following thermodynamic cycle: solv G (A) solv G (B) A (sol) + B (sol) A (g) + B (g) G s TS (sol) G g TS (g) solv G (TS) 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

10 Gas-phase vs solution-phase rate constants (continued) From and k 0,s = k ( BT c h exp Gs ) G s = G g + solv G we get ( ) k 0,s = k BT c h exp Gg exp ( solv G ) = k 0,g exp ( solv G )

11 Gas-phase vs solution-phase rate constants (continued) k 0,s = k 0,g exp ( solv G ) 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.

12 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 exp ( solv G ) = 56 at 25 C. Rate enhancements of this order of magnitude have been seen in some reactions.

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