Chemical Kinetics and Transition States

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/[concn-1-time] • [A] = [A(0)] exp (- kft) • Now consider A ↔ B with kf rate and kr rate; i.e. there is a substantial back rxn • Then d[A]/dt = - kf[A] + kr[B] = - d[B]/dt • Experiments  rate laws, k and rxn order

Review of Elementary Rate Laws

Equilibrium • At equilibrium, d[A]/dt = 0  forward rate = kf[A] = kr[B] = reverse rate. • This is the Principle of Detailed Balancing and leads to • K =[B]eq/[A]eq = kf/kr (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 = ho/kT2 • Arrhenius combined the van’t Hoff eqn with K = kf/kb to get • Differential eqn: d ln kf/dT = Ea/kT2 where Ea = forward activation energy; assume constant to integrate • Integrated eqn: kf = A exp(-Ea/kT); as T ↑, kf↑ if Ea > 0 (usual case shown in Fig 19.2 except see Prob 19.10) • Therefore a plot of ln kf vs 1/T  Eaand A (Fig 19.5)

Activation Energy Diagram (Fig 19.3) • Ea= activation energy for forward rxn • Ea‘= activation energy for reverse rxn • ξ = rxn coordinate • ho = Ea- Ea‘ • Note that Ea and Ea‘ > 0 (usual case) so ki↑ 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 + H2  HD + H • See handouts for • H + H2  H2 + H • F + H2  HF + H

Saddle Point (Fig 19.7) • Rxn starts in LHS valley (Morse potential) of H2 with D far away. • D and H2 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 • http://my.voyager.net/~desotosaddle/saddle_pictures.htm

TST  Rate Constant, k2 • A + B --k2 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‡/qAqB] 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] = k2[A][B]

TST: Reaction Coordinate • d[P]/dt = k‡[(AB)‡] = k‡K‡[A][B] = k2[A][B] • k2 = 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 qξ as the partition function of this weak vibrational deg of freedom and separate it from other degs of freedom (q‡*) • Then q‡ = q‡* qξ = product of TS q except rxn coord x q of rxn coord • The reaction coord ξ is associated with a weak bond (small kξ and small ν ξ).

TST • Then q‡ ≈ κkT/hνξ. • κ= transmission coefficient; 0 < κ≤ 1. • K‡= [q‡/qAqB] exp(D‡/kT) = q‡*qξ/[qAqB] exp(D‡/kT) = q‡* kT/hνξ /[qAqB] exp(D‡/kT) • k2 = k‡K‡=νξ(kT/hνξ){q‡*/[qAqB]} exp(D‡/kT) = (kT/h) q‡*/[qAqB] 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 k2 = (kT/h) q‡*/[qAqB] exp(D‡/kT)

Isotope Effect • Example in text is for breaking the CH (kH) or CD (kD) 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 νCX or reduced mass.

Isotope Effect • kH /kD = exp [(DCH‡ - DCD‡ )/kT] • = exp {-(h/2kT)[νCD - νCH ]} • = exp {-(hνCH/2kT)[2-1/2 - 1]} since • ν = (1/2π)√(ks/μ) ks= 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‡ • k2 = (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 Ea Prob 6 • k vs T expts  H‡, S‡, G‡

IV. Catalysis • k0 = (kT/h) [AB‡*]/[A][B]= (kT/h) K0‡* w/o catalyst • kc = (kT/h) [ABC‡*]/[A][B][C]= (kT/h) Kc‡* w/catalyst • Rate enhancement = kc/k0 = [ABC‡*]/[AB‡][C] = measure of binding of C to TS = binding constant = KB* • KB* 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 = ka [HA] [R] • H+ is produced in AH ↔ H+ + A- • Ka = [H+][ A-]/ [AH] • These two rxns are coupled

Brønsted Law • log ka = α log Ka + ca • or log ka = - α pKa + ca • α > 0 and ca = constant • As Ka ↑ (stronger acid), rxn rate constant ka ↑ • Plot log ka vs pKa (Fig 19.15) • This law proposes that presence of acid stabilizes the product. • Omit pp 361-365

Chemical Kinetics and Transition States