Chemical Kinetics and Transition States Elementary Rate Laws k(T) Transition State Theory Catalysis.
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Transcript of 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 LawsOrder Reaction Differential
Rate Law- d[A]/dt =
Integrated Rate Law[A] = f(t)
Units of k Plot Half-life
0 A P k [A] - [A]0 = -kt M/s [A] vs tSlope = - k
[A]0/2k
1 A P k[A] [A] = [A]0 e-kt 1/s ln[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 tSlope = 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/(M2-s) 3/{2k A]02}
n A P k[A]n α 1/[A]0 n-1
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.
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‡ - TS‡
– 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