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Ch 18 Chemical Kinetics and Ch 12 Thermodynamics

Ch 18 Chemical Kinetics and Ch 12 Thermodynamics. Reaction Mechanism: When the first step is Rate-Limiting ( i.e., slow); the Rate Law is Integer Order: 1st, 2nd or 3rd Order. When the second step is Rate-Limiting; the Rate Law may or may not be Integer.

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Ch 18 Chemical Kinetics and Ch 12 Thermodynamics

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  1. Ch 18 Chemical Kinetics and Ch 12 Thermodynamics Reaction Mechanism: When the first step is Rate-Limiting ( i.e., slow); the Rate Law is Integer Order: 1st, 2nd or 3rd Order. When the second step is Rate-Limiting; the Rate Law may or may not be Integer. When neither step is Rate-Limiting, the Steady State Apprx: can be used to find the Rate Law! Catalysis: Increasing the Reaction Rate via a substance not Generated or Consumed in the Rxn, and that Provides a Lower Activation Barrier Rxn Path Equilibrium Constant: K(T)=kf/kr=(Af/Ar)exp{-(Eaf-Ear)} aA+bB  cC + dD Forward Rate= Reverse Rate U= Uprod - Ureact =(Eaf – Ear) Internal Energy Change @ V=const

  2. Reaction Mechanism • Given the overall reaction: • 2NO(g)  N2(g) + O2(g) • Propose a plausible 2 step mechanism • 2NO(g)  N2O(g) + O(g) (2) N2O(g) + O(g)  N2(g) + O2(g) • Assume Step (1) is rate limiting • Rate=k1[NO]22nd Order in NO Then is a plausible Mechanism

  3. Given the overall reaction: 2NO(g)  N2(g) + O2(g) • Propose a plausible 2 step mechanism • 2NO(g)  N2O(g) + O(g) (2) N2O(g) + O(g)  N2(g) + O2(g) Assume Step (2) is rate limiting • Forward Rate=k1[NO]2 and Reverse Rate =k-1[N2O][O] • Forward Rate =k2[N2O][O] Fast Equilibrium Forward Rate = Reverse Rate: k1[NO]2 =k-1[N2O][O] Overall Rate=k2[N2O][O] = k2(k1/k-1) [NO]2 = = k2K1[NO]2 Then this is also a plausible Mechanism where K=k1/k-1

  4. Same overall reaction: assume Steady-State Apprx Neither Step is rate limiting 2NO(g)+ M  N2O2(g) +M Propose a plausible 2 step mechanism (1) 2NO(g)+ M  N2O2(g) + M (2) N2O2(g)  N2(g) + O2(g) Assume No Step is rate limiting and [N2O2]~constant d/dt [N2O2]= k1[NO]2[M] - k-1[N2O2][M] - k2[N2O2]=0 Overall Rate =k2[N2O2 ] [N2O2]= k1[NO]2[M]/{ k-1[M] +k2} Overall Rate =k2k1[NO]2[M]/{ k-1[M] +k2} Then is a plausible Mechanism

  5. Overall Rate =k2k1[NO]2[M]/{ k-1[M] +k2} Rate [M](Buffer concentration) Fig. 18-19, p. 780

  6. Max Rate Rate Max Rate= k2 (k1/k-1)[NO]2 High PM k-1[M]>>k2 [M]

  7. Steady State Approximation Rate=d[N2]/dt=d[O2]/dt =k2k1[NO]2[M]/{ k-1[M] +k2} [M]=(nM/V) and PV=nRT Ideal Gas Law Therefore PM=[M]RT when PM is high, High Pressures then [M] is large and k-1[M]>>k2 Rate~ k2 (k1/k-1)[NO]2 Step (2) is rate-limiting, Step (1) is in equilibrium! The Rateis independent [M] for high PM

  8. Steady State Approximation Rate=d[N2]/dt=d[O2]/dt =k2k1[NO]2[M]/{ k-1[M] + k2} [M]=(nM/V) and PV=nMRT Ideal Gas Law Therefore PM=[M]RT when PM0, Low Pressures then [M]0 and k-1[M]<<k2 Rate~k1[NO]2[M] 3rd Order

  9. Max Rate =k2 (k1/k-1)[NO]2 High PM Max Rate Rate Slope= k1[NO]2 Low PM [M]

  10. Catalysis Rxn path Diagram Internal Energy Change @ V=const Difference in the Stored between Reactants/Products

  11. Catalytic Converters: e.g., Pt, Rh, ZrO2 etc Fig. 18-19, p. 780

  12. The hydrogenation reaction has a large activation barrier Ea and is Sterically hindered and must go throw a tight transition state, So k is small and the rate is slow

  13. Consider the gas phase(rxn path A) hydrogenation of ethylene to form ethane. This process requires H-H insertion into a C=C bond: a very tight transition state; high barrier and a low steric factor which mean the reaction is slow! C2H4(g) + H2(g)  C2H6(g) (1) H-H H-H

  14. Consider another path B with a lower activation barrier: heterogeneous hydrogenation on a Pt Catalyst C2H4(g) + H2(g)Pt(surface) C2H6(g) C2H4(g)+ H2(g) + Pt(surface)  C2H4(ad) + H2(ad) (1) C2H4(ad) + H2(ad)  C2H6(ad) (2) C2H6(ad)  Pt(surface) C2H6(g) (3) Pt-Pt-Pt Pt-Pt-Pt

  15. Consider another path B with a lower activation barrier: heterogeneous hydrogenation on a Pt Catalyst C2H4(g) + H2(g)Pt(surface) C2H6(g) C2H4(g)+ H2(g) + Pt(surface)  C2H4(ad) + 2H(ad) (1) C2H4(ad) + 2H(ad) C2H6(ad) (2) C2H6(ad)  Pt(surface) C2H6(g) (3) Pt-Pt-Pt Pt-Pt-Pt

  16. Surface diffusion

  17. Diffusion in a gas, liquid, solid or on a Surface Gas Solid/liquid surface Fig. 9-21, p. 427

  18. Enzyme Catalyzed Reaction The Enzyme plays the role of the surface and the kinetics can be modeled as a steady-state approximation Since the Enzyme/Substrate complex concentration ids nearly constant throughout the reaction

  19. Enzyme Substrate Catalytic Reaction Steady State Approximation Enzyme(E), Substrate(S), the Complex(ES) and the Reaction Product(B) E + S ES (1) ES  P (2) The Overall Rate = d[P]/dt = k2 [ES] Since [ES]~const d[ES]/dt = k1[E][S] - k-1[ES] - k2[ES]~0 Now since [E]0= [E] + [ES] and [E]= [E]0 - [ES] [ES]=k1[E]0[S]/{k1[S] + (k-1 + k2)} d[P]/dt=k2k1[E]0[S]/{k1[S] + (k-1 + k2)} let KM=(k-1 + k2)/k1 d[P]/dt=k2[E]0[S]/{[S] + KM} Michaelis-Menten Eq.

  20. Max Rate =k2 [E]0 High [S] Max Rate d[P]/dt=k2[E]0[S]/{[S] + KM} Slope= k2[E]0/KM Low [S]

  21. Catalysis Rxn path Diagram Internal Energy Change @ V=const Difference in the Stored between Reactants/Products

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