1 / 23

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 )}

ilyssa
Télécharger la présentation

Ch 18 Chemical Kinetics and Ch 12 Thermodynamics

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related