Notes on Transition State Theory, Lecture notes of Chemistry

Download Notes on Transition State Theory and more Lecture notes Chemistry in PDF only on Docsity! Transition state theory 1. The equilibrium constant Equilibrium constants can be calculated for any chemical system from the partition functions for the species involved. In terms of the partition function q, the chemical potential of a species (essentially the molar Gibb’s free energy) is given by  = RT ln N q Consider the reaction aA + bB pP + qQ. At equilibrium the free energy of the reactants is equal to the free energy of the products. Greact = Gprod a A + b B = p P + q Q RT ln(NA/qA)a + RT ln(NB/qB)b = RT ln(NP/qP)p + RT ln(NQ/qQ)q Rearranging gives NPpNQq NAaNBb = qPpqQq qAaqBb = Kn We recognise this as the equilibrium constant formulated in terms of concentrations. We are often interested in Kp, the equilibrium constant in terms of partial pressures. For the case in which all species involved are gases, we can convert from Kn to Kp using the ideal gas equation, p = nRT/V. Kp = pPppQq pAapBb = NPpNQq NAaNBb      kBT V (p+qab) = qPpqQq qAaqBb      kBT V n where n = p + q a b is the change in the number of moles during reaction. If we use vibrational partition functions defined with energies measured from the zero-point energy for each reactant and product species, then we also need to include the extra factor eD0/kBT (see Statistical Mechanics tutorial) in order to ensure that all of the molecular partition functions are referenced to the same zero of energy. The equilibrium constant then becomes Kp = qPpqQq qAaqBb     kBT V n exp      p0P+qD0QaD0AbD0B kBT = qPpqQq qAaqBb     kBT V n exp      D0 kBT The quantity D0 = (pD0P + qD0Q aD0A bD0B) is the difference between reactant and product zero-point energies. Usually, we can set D0 = H0, with H0 the enthalpy of reaction. 2. The rate constant We can also use statistical mechanics to obtain rate constants from partition functions. For a reaction A + B AB‡ + P, we assume that the reactions involved in forming the activated complex AB‡ are much faster than the formation of products from the complex. Kinetically, the reaction can then be treated as a pre-equilibrium (see first year Kinetics notes), with K‡ = [AB‡] [A][B] rate = k‡ [AB‡] = k‡K‡ [A][B] = kobs[A][B] where kobs=k‡K‡ is the observed rate constant, and k‡ is the rate constant for decomposition of the activated complex into products. The equilibrium constant K‡ can be rewritten in terms of the appropriate partition functions, giving kobs = k‡ qAB‡ qAqB eH ‡ /RT where the partition functions are evaluated with respect to the zero-point levels of the reactants and transition state. We can derive a complete expression for kobs by treating the motion of the activated complex over the top of the energy barrier (i.e. the reaction coordinate) as either a very loose vibration or as a translation. Derivation 1: Reaction coordinate as a vibration Reactants A and B have 3N6 vibrational degrees of freedom if non-linear, 3N5 if linear. The same is true of the activated complex, which has 3(NA+NB)6 vibrational modes if it is non-linear. One of these modes is of a different character from the rest, corresponding to a very loose vibration that allows the complex to dissociate into products. For this degree of freedom we can use a vibrational partition function q* in which the vibrational frequency  tends to zero. i.e. q* = 0 lim 1 1eh/kBT = 1 1(1h/kBT) = kBT h The rate constant is now given by kobs = k‡ kBT h qAB‡ qAqB eH ‡ /RT where qAB‡ is now less one degree of freedom corresponding to the reaction coordinate. The frequency  is the vibrational frequency of the activated complex in the degree of freedom corresponding to its decomposition. It is therefore the frequency of decomposition i.e. the rate constant k‡. This means that k‡q* = kBT/h, and the rate constant is given by kobs = kBT h qAB‡ qAqB eH ‡ /RT Derivation 2: Reaction coordinate as a translation Instead of using a vibrational partition function to describe the motion of the activated complex over the reaction barrier, we can also use a translational partition function. We consider all complexes lying within a distance x of the barrier (see diagram) to be activated complexes. The translational partition function for a particle of mass m in a box of length x is given by q* = (2m‡kBT)1/2x h