Chemical Kinetics and Transition States Elementary Rate Laws k(T) Transition State Theory Catalysis

I. Rate Equation for Elementary Rate Laws Rate =d[A]/dt = -k [A] n assuming nth order –[A] is reactant in A  B (assume negligible reverse rxn) –k is rate constant, units of 1/[conc n-1 -time] –[A] = [A(0)] exp (- k f t) Now consider A ↔ B with k f rate and k r rate; i.e. there is a substantial back rxn – Then d[A]/dt = - k f [A] + k r [B] = - d[B]/dt Experiments  rate laws, k and rxn order

Review of Elementary Rate Laws Order Reactio n Differential Rate Law - d[A]/dt = Integrated Rate Law [A] = f(t) Units of kPlotHalf-life 0 A  P k[A] - [A] 0 = -ktM/s[A] vs t Slope = - k [A] 0 /2k 1 A  P k[A][A] = [A] 0 e -kt 1/sln[A] vs t Slope = -k (ln 2)/k 2 A  P k[A] 2 1/[A] 0 - 1/[A] = - kt(M-s) -1 1/[A] vs t Slope = k 1/k [A] 0 2 A+ B  P k[A][B]kt = ([B] 0 - [A] 0 ) -1 * ln {[A] 0 [B]/[B] 0 [A]} (M-s) -1 Assume [A] 0 < [B] 0 3 A  P k[A] 3 1/2 {[A] -2 - [A] 0 -2 } = kt 1/(M 2 -s)3/{2k A] 0 2 } n A  P k[A] n α 1/[A] 0 n-1

Equilibrium At equilibrium, d[A]/dt = 0  forward rate = k f [A] = k r [B] = reverse rate. –This is the Principle of Detailed Balancing and leads to –K =[B] eq /[A] eq = k f /k r (recall Eqn 13.23) –Principle of Microscopic Reversibility

II. Arrhenius Eqn: k(T) In Ch 13, we combined the Gibbs-Helmholtz Eqn (G(T)) and the Gibbs Eqn (  G = - RT ln K) –to get the van’t Hoff Eqn: d ln K/dT =  h o /kT 2 Arrhenius combined the van’t Hoff eqn with K = k f /k b to get –Differential eqn: d ln k f /dT = E a /kT 2 where E a = forward activation energy; assume constant to integrate –Integrated eqn: k f = A exp(-E a /kT); as T ↑, k f ↑ if E a > 0 (usual case shown in Fig 19.2 except see Prob 19.10) –Therefore a plot of ln k f vs 1/T  E a and A (Fig 19.5)

Activation Energy Diagram (Fig 19.3) E a = activation energy for forward rxn E a ‘= activation energy for reverse rxn ξ = rxn coordinate  h o = E a - E a ‘ Note that E a and E a ‘ > 0 (usual case) so k i ↑ with T for endo- and exothermic rxns Ex 19.1, Prob 8

III.Transition State Theory (TST) TS is at the top of the activation [‡] barrier between reactants and products. Energy landscape for chemical rxn –A + BC  [A--B--C] ‡ = TS  AB + C –Fig 19.7 for collinear rxn: D + H 2  HD + H See handouts for –H + H 2  H 2 + H –F + H 2  HF + H

Saddle Point (Fig 19.7) Rxn starts in LHS valley (Morse potential) of H 2 with D far away. D and H 2 approach, potential energy ↑ TS is at max energy along rxn coord; i.e. [H--H--D] ‡ exists. Then H moves away and valley (another Morse potential) is HD.

Potential Energy Contour Diagram (Fig 19.8) The information in Fig 19.7 can be shown as a contour diagram (Fig 19.8). Follow rxn A + BC  AB + C. Reactants are lower RH corner (energy min) and follow dotted line up to TS and then down to upper LH corner.

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A 360 degree view e_pictures.htmhttp://my.voyager.net/~desotosaddle/saddl e_pictures.htm

TST  Rate Constant, k 2 A + B --k 2  P overall rxn which proceeds via a TS: A + B  K ‡  (AB) ‡ --k ‡  P This 2-step mechanism involves an equilibrium between reactants and TS with eq. constant –K ‡ = [(AB) ‡ ]/[A][B] = [q ‡ /q A q B ] exp (  D ‡ /kT) and then the formation of products from the TS with rate constant k ‡. d[P]/dt = k ‡ [(AB) ‡ ] = k ‡ K ‡ [A][B] = k 2 [A][B]

TST: Reaction Coordinate d[P]/dt = k ‡ [(AB) ‡ ] = k ‡ K ‡ [A][B] = k 2 [A][B] k 2 = k ‡ K ‡ is the connection between kinetics and stat. thermo (partition functions) In TST, we hypothesize a TS structure and assume that the reaction coord ξ is associated with the vibrational degree of freedom of the A—B bond that forms.

TST Define q ξ as the partition function of this weak vibrational deg of freedom and separate it from other degs of freedom (q ‡* ) Then q ‡ = q ‡* q ξ = product of TS q except rxn coord x q of rxn coord The reaction coord ξ is associated with a weak bond (small k ξ and small ν ξ ).

TST Then q ‡ ≈ κ kT/h ν ξ. κ = transmission coefficient; 0 < κ ≤ 1. K ‡ = [q ‡ /q A q B ] exp(  D ‡ /kT) = q ‡* q ξ /[q A q B ] exp(  D ‡ /kT) = q ‡* kT/h ν ξ /[q A q B ] exp(  D ‡ /kT) k 2 = k ‡ K ‡ = ν ξ (kT/h ν ξ ){q ‡* /[q A q B ]} exp(  D ‡ /kT) = (kT/h) q ‡* /[q A q B ] exp(  D ‡ /kT) = (kT/h) K ‡ * Ex 19.2

Primary Kinetic Isotope Effect When an isotopic substitution is made for an atom at a reacting position (i.e. in the bond that breaks or forms in the TS), the reaction rate constant changes. These changes are largest for H/D/T substitutions. And can be calculated using eqn for k 2 = (kT/h) q ‡* /[q A q B ] exp(  D ‡ /kT)

Isotope Effect Example in text is for breaking the CH (k H ) or CD (k D ) bond. Assume that –q(CH ‡ )≈q(CD ‡ ) and q(CH) ≈ q(CD) –C-H and C-D have the same force constants. Then C-H and C-D bond breakage depends on differences in vibration of reaction coord (ξ) or ν CX or reduced mass.

Isotope Effect k H /k D = exp [(  D CH ‡ -  D CD ‡ ) /kT] = exp {-(h/2kT)[ ν CD - ν CH ]} = exp {-(h ν CH /2kT)[ 2 -1/2 - 1] } since ν = (1/2π)√(k s /μ) k s = force const, μ = reduced mass Ex 19.3; see Fig 19.9 Prob 3

Thermodynamic Properties of TS or Activated State (Arrhenius) K ‡ * = equilibrium constant from reactants to TS without the rxn coordinate ξ. Define a set of thermody properties for the TS: –  G ‡ = - kT ln K ‡ * =  H ‡ - T  S ‡ – k 2 = (kT/h) K ‡ * = (kT/h) exp(-  G ‡ /kT) = [(kT/h) exp(-  S ‡ /k)] exp(-  H ‡ /kT) [term] is related to Arrhenius A and  H ‡ is related to E a Prob 6 k vs T expts   H ‡,  S ‡,  G ‡

IV. Catalysis k 0 = (kT/h) [AB ‡* ]/[A][B] = (kT/h) K 0 ‡ * w/o catalyst k c = (kT/h) [ABC ‡* ]/[A][B][C] = (kT/h) K c ‡ * w/catalyst Rate enhancement = k c /k 0 = [ABC ‡* ]/[AB ‡ ][C] = measure of binding of C to TS = binding constant = K B * K B * increases as C-TS binding increases Fig 19.12, 19.13

Catalysis Mechanisms Catalysts stabilize ‡ relative to reactants; this lowers activation barrier. T 19.1: create favorable reactant orientation T 19.2 and Fig 19.14: reduce effect of polar solvents on dipolar transition state

Acid and Base Catalysis Consider a rxn R  P catalyzed by H + Then rate might be = k a [HA] [R] H + is produced in AH ↔ H + + A - –K a = [H + ][ A - ]/ [AH] These two rxns are coupled

Brønsted Law log k a = α log K a + c a –or log k a = - α pK a + c a –α > 0 and c a = constant –As K a ↑ (stronger acid), rxn rate constant k a ↑ –Plot log k a vs pK a (Fig 19.15) This law proposes that presence of acid stabilizes the product. Omit pp