Transition Theory Abbreviated Derivation [ A - B - C] # E o. Reaction Coordinate. [ ] # æ Æ

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1 Transition Theory Abbreviated Derivation A + BC æ Æ AB + C [ A - B - C] # E A BC D E o AB, C Reaction Coordinate A + BC æ æ Æ æ A - B - C [ ] # æ Æ æ A - B + C The rate of reaction is the frequency of the complex crossing the barrier # -r A = r AB = u I C ABC (1) where the frequency u I can be thought of as a loose vibration. We are going to assume that the activated complex is in virtual equilibrium with the reactants Then K # C = C # ABC (2) C A C BC -r A = u I K C # C A C BC (3) From thermodynamics we know Then DG = -RT ln K (4) K = e -DG RT (5) where K is the dimensionless true equilibrium constant and related to the concentration equilibrium constant K C by K = K g K C V d m = K C V d m = e -DG RT (6) Then for the equilibrium between the reactants and the transition state K # C = V -d m e -DG # RT 1

2 Part I Relate K C # to Partition Functions We are now going use statistical mechanics to derive the equation for K C #. We start by recalling that the number of ways, W, of arranging N particles among m energy states is N! W = n 1!n 2!Kn m! where n i is the number of particles in the i th state. For a fixed total number of molecules N = Ân i and a fixed total energy E = Ân i e i the most probable distribution, the one that overwhelms all the others, is found by setting dw = 0. The result is where n i N = e -be i q q = Âe -be i, b = 1 k B T n i N = fraction of molecules that occupy energy state e i (9) q = molecular partition function The molecular partition function, q, gives a measure of how the molecules are distributed (partitioned) among the energy states. It gives an indication of the average number of states that are accessible at a particular temperature for non interacting molecules. Fundamental postulate relating t and W Sterling s approximation Using Sterling s approximation, Equation (10) becomes S = k ln W = k[ ln N!-Â ln n i!] (10) (7) (8) ln X!= X ln X - X (11) n i Substituting for using Equation (8) N S = -k Â ln n i N 2 (12)

3 U-U 67 8 o N} S = kbâ n i e i + kâ n i ln q The sum Â n i e i is the internal energy relative to the ground state, i.e. (U U o ) S = U - U o + kn ln q (13) T This result is for non interacting particles. For interacting particles, the result is S = U - U o T + k ln q N From thermodynamic we know the relationship between Combining Equations (14) and (15) N! (14) G = U - T S + PV = U - T S + nrt (15) G = U o - nrtln q N (16) We now define molar partition function, q m q N = q = q m n q m = q n (17) Dividing Equation (16) by the number of moles n when Typical units of G and U are (J/mol). G = U o - RTln q m (18) G G = n, U o = U n We now will apply Equation (18) to each species in the reaction The change in Gibbs free energy for reaction (19) is Using Equation (18) we obtain aa+ bb Æ cc + dd (19) G i = U oi - RTln q mi (20) DG = cg C + dg D - bg B - ag A (21) 3

4 where and using Equation (22) The molar partition function is DG = DE 0 - RTln q c mc a q ma d q md b q mb N -d Avo (22) DE 0 = cu C + du Do - bu Bo - au Ao (23) K = e -DE 0 RT d = d + c - b - a q c mc q mi = q i and the partition function per unit volume is then d q md b a q mb q ma n q i = q i V N -d Avo (24) V q mi = q i n = q i V m (25) K = e -DE 0 RT q C q B ( ) c ( q D ) d ( ) b ( ) a q A Recalling Equation (6) and equating it to Equation (26) K C K g V d m = K = e -DE 0 RT q C q B V m d -d ( ) c ( q D ) d ( ) b ( ) a q A V m d -d For ideal mixtures Kg = 1 and canceling the molar volumes we arrive at the main result we have been looking for K C = e -DE 0 RT q C q B ( ) c ( q D ) d ( ) b ( ) a q A N -d Avo We now apply Equation (28) to our transition state reaction (d = 1) A+ BC æ Æ ABC # æ Æ AB+ C K # C = e - ( DE 0 RT) q ABC -1 q A q BC # (26) (27) (28) (29) 4

5 Part II Partition Function We now focus on the partition function per unit volume Dividing by V q = Âe -be i = Âe -b ( e el +e T +e V +e Rot ) i = Âe -be el Âe -be T Âe -be V Âe -be Rot = q el q T q V q R (30) q T = q T V (31) q = q el q T q V q R (32) The Transitional Partition Function, q T The translational partition function is obtained by solving the wave equation for a particle in a box. q T = 1 L V 3 q T = 1 L = 2pmkT 3 h ( ) 1 2 Where L is the thermal wave length, h is Plank s constant, m = mass of the molecule and k is Boltzmann s constant. Substituting for k and h and simplifying m q T = AB m 3 1 amu T 300K 3 2 (33) The Vibrational Partition Function, q V For the harmonic oscillator For small arguments of Evaluating hu i kt q V = q V1 q V2 q V3 K (34) 1 q Vi = 1- exp - hu i kt (35) q vi = kt hu i (36) 5

6 hu k B T = hc u u = K k B T 1 cm T c = speed of light The Rotational Partition Function, q R For linear molecules we solve the wave equation for the rigid rotator model to find the rotational partition function to be where B = Rotational Constant where I = moment of inertia Evaluating k, h, c, and simplifying q R = kt 2hcB B = h 8pcI I = Â m i r i 2 c = speed of light T I q R =12.4 AB 300K 1 amu Ang 2 1 S y The overall or total partition function for the activated complex is # q ABC = q # e q # T q # # V q R We now consider the loose imaginary vibration u I. The total vibrational partition function is the product of the partition functions for each vibrational mode, i.e., 6 (37) (38) q # V = q # Vi q # # V1 q V2 = kt q # # V1 q V2 = kt q V# (39) hu I hu I where q V# is the partition function that includes the loose vibration and q V# is the vibrational partition function without the loose vibration. Similarly for the total partition function substituting into Equation (39) # q ABC = kt q ABC# (40) hu I K # C = kt e -DE o RT q ABC# (41) hu I q A q BC

7 Substituting Equation (41) into Equation (3) we note that the loose vibration frequency cancels and we obtain -r A = kt e -DE o RT q ABC# C A C BC (42) h q A q BC 43 K C# k B T T = h molecules s 300K 23 molecules = mol Note: If B and C are the same molecule, B 2, in the reaction A = kt h q ABC# N q A q Avo (43) BC A+ B 1 - B 2 Æ AB+ B There will be a factor of 2 ( 2kT h) in the rate constant because A can attach either B molecule, B 1 or B 2. It s sometimes easier to make TST calculations by taking ratios The Eyring Equation q A-B-C = q A q BC q A-B-C q A q BC Tran q A-B-C q A q BC Vib q A-B-C q A q BC Letting K C# represent the equilibrium constant with the loose vibration removed we have -r A = k B T h K C# C A C BC the true dimensionless equilibrium constant is then K = K g K C V m d Rot K C# = K # V m -d = K # V m = K # C T where C T is the total concentration 7

8 DG # = -RT ln K # -DG RT K # = e DG = DH - TDS K C# = K C# = -r A = k BT 1 h DS # # = S ABC = C T e -DG # RT C T e DS # R e DS# R e -DH # RT e -DH # RT C T C A C BC (44) -S A -S B = a negative number because we are going from a less ordered state to a more ordered state. DH # = H ABC-# - H A - H BC = a positive number e DS# R Configurations Leading to Reactions = Total Number of Configurations (45) Definitions q = overall partition function = Âe -be i, b = 1 kt q m = q n = molar partition function q = q e q Tr q Vib q Rot q T = q T V q V = q Vi q V1 q V2 q V # = vibrational partition for activated state including loose vibration q V# = vibrational partition function with loose vibration removed q # = overall partition function per unit volume with loose vibration removed 8

3. RATE LAW AND STOICHIOMETRY

Page 1 of 39 3. RATE LAW AND STOICHIOMETRY Professional Reference Shelf R3.2 Abbreviated Lecture Notes Full Lecture Notes I. Overview II. Introduction A. The Transition State B. Procedure to Calculate