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Ch 18 Chemical Kinetics and Ch 12 Thermodynamics

Ch 18 Chemical Kinetics and Ch 12 Thermodynamics. Catalysis: Enzymatic Catalysis (solution): E + S   P Rate = -d[S]/dt=d[P]/dt= k 2 [E] 0 [S]/{[S] + K M } K M =(k -1 + k 2 )/k 1 Equilibrium Constant: K(T)=k f /k r =(A f /A r )exp{-(E af -E ar )}

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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 Catalysis: Enzymatic Catalysis(solution): E + S P Rate = -d[S]/dt=d[P]/dt=k2[E]0[S]/{[S] + KM} KM=(k-1 + k2)/k1 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 U is the Internal Energy= <E>=<PE> + <KE> The first Law of Thermodynamics U=q + w q: energy transferred as heat; w: transferred as work Detailed balance Leads to the Law of Mass Action: aA + bB  eE + fF K=[E]e[F]f/[A]a[B]b all concentrations are at Equilibrium

  2. Max Rate =k2 (k1/k-1)[NO]2 High PM Max Rate 2NO  N2 + O2 K=[N2] [O2]/[NO]2 K the equilibrium Constant is only a function temperature Rate Slope= k1[NO]2 Low PM [M]

  3. K=kf/kr=[Beq]b/[Aeq]a • 2A  3B • 2A  C • C  3B • Detailed Balance • k1[A]2=k-1 [C] • k2[C]=k-2[B]3 • k1/k-1 [A]2=(k-2/k2)[B]3 • (k1/k-1)(k2/k-2)=[Beq]3/[Aeq]2 • K=kf/kr=[Beq]3/[Aeq]2 • K(T)=kf/kr=(k1/k-1)(k2/k-2) • At Equilibrium kf aAbB kr (slow) (fast) Kinetics Equilibrium [B] [A] [Beq] [Aeq] time

  4. Catalysis Rxn path Diagram. Note the Catalyst does not affect the equilibrium Kcat=K=[Beq]b/[Aeq]a Internal Energy Change @ V=const Difference in the Stored between Reactants/Products

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

  6. 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

  7. 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

  8. 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

  9. 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

  10. Surface diffusion

  11. Diffusion in a gas, liquid, solid or on a Surface <Dr2>=6Dt D is the diffusion constant Drrms=√6Dt root-mean-square displacement Solid/liquid And surfaces Gas D=(3π/16)l<u>, <u>=Z1l, l mean free-path between collisions (the average distance traveled between collisions) Fig. 9-21, p. 427

  12. 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

  13. Enzyme Substrate Catalytic Reaction Steady State Approximation Enzyme(E), Substrate(S), the Complex(ES) and the Reaction Product(B) (1) E + S ES (2) ES  P 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.

  14. 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]

  15. Catalysis Rxn path Diagram Internal Energy Change @ V=const Difference in the Stored between Reactants/Products DU = q + w , q = heat: transfer of random Energy w = - PDV = - Force(F/A) x distance (ADL)

  16. aAbB kr kf (slow) (fast) Kinetics Equilibrium [B] [A] [Beq] [Aeq] time

  17. This Applies to Chemical as well as Physical Changes Fast V= VB - VA P= PB - PA P, V and T are Thermodynamic State Variables and defines a Thermodynamic State (A and B) In this case

  18. Fixed amount of a pure substance A mole of H 2O For example

  19. The First Law of Thermodynamics U= q(Heat transferred) + w(Work performed) U= q + w

  20. Heat flows from hot to cold? Hot q Cold (T) T1  T2 for T2 < T1 For the hot system q < 0 And for the cold system q > 0 qin < 0 and qou > 0 At V=const U=q for

  21. Work = - (force) x (distance moved) Pext Pext A A h1 h2 w = -F(h2-h1)= PextA (h2-h1)=Pext(V2 - V1) w = - PextV V =hA and P=F/A w < 0: system (gas in cylinder) does work: reduces U; V >0 w > 0: work done on the system: increases U; V <0

  22. U= q + w = q - PextV The Flame: CH4 + 2O2 CO2 + 2H2O(l) combustion gives off energy that is transferred as heat(q) to the gas in the piston which does work (-PextV) against the Pext

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