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Chem 300 - Ch 29/#3 Today’s To Do List

Chem 300 - Ch 29/#3 Today’s To Do List. Unimolecular Reactions Chain Reactions Effect of a Catalyst Enzyme Catalysis. Unimolecular Reactions. CH 3 NC ==> CH 3 CN Rate = -k[CH 3 NC] Valid at high conc But at low conc Rate = -k[CH 3 NC] 2

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Chem 300 - Ch 29/#3 Today’s To Do List

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  1. Chem 300 - Ch 29/#3 Today’s To Do List • Unimolecular Reactions • Chain Reactions • Effect of a Catalyst • Enzyme Catalysis

  2. Unimolecular Reactions • CH3NC ==> CH3CN • Rate = -k[CH3NC] • Valid at high conc • But at low conc Rate = -k[CH3NC]2 • How come?? Is this really an elementary reaction? • Lindemann: Probably not.

  3. Lindemann-HinshelwoodUnimolecular Mechanism

  4. Lindemann Mechanism • A + M <==> A* + M • A* ==> B • Rate (B) = k2[A*] • SS condition: • d[A*]/dt = 0 = k1[A][M] – k-1[A*][M] – k2[A*] • [A*] = k1[M][A]/(k2 + k-1[M]) • Rate = k2 k1[M][A]/(k2 + k-1[M]) = k’[A]

  5. Rate = k2 k1[M][A]/(k2 + k-1[M]) = k’[A] • At high conc: k2 << k-1[M]) • Rate = k ‘ [A] • At low conc: k2 >> k-1[M]) • Rate = k1[M][A]

  6. CH3NC  CH3CN klc =k1[M] khc = k1k2/k-1

  7. Chain Reactions • Consider: H2 + Br2 2 HBr • Experim. Rate Law: • ½ d[HBr]/dt = k[H2][Br2]1/2/(1 +k’[HBr]/[Br2]) • How does it do that?? • It’s a chain reaction mechanism

  8. A Chain Reaction has Several Unique Steps • Initiation: Br2 + M ==> 2 Br + M k1 • (thermal or photochemical)(193 vs 436 kJ) • Propagation: Br + H2 ==> HBr + H k2 • H + Br2 ==> HBr + Br k3 • Inhibition: HBr + H ==> Br + H2 k-2 • HBr + Br ==> H + Br2 k-3 • Termination: 2 Br + M ==> Br2 + M k-1

  9. The Rate Laws • d[HBr]/dt = k2[Br][H2] – k-2[HBr][H] + k3[H][Br2] k-3  0 • d[H]/dt = k2[Br][H2] – k-2[HBr][H] - k3[H][Br2] • d[Br]/dt = 2k1[Br2][M] – 2k-1[Br]2[M] – k2[Br][H2] + k-2[HBr][H] +k3[H][Br] • Apply SS condition to: • d[H]/dt = d[Br]/dt = 0 • And solve the 2 simultaneously for [H] & [Br]

  10. Results • [Br] = (k1/k-1)1/2[Br2]1/2 • [H] = k2K1/2[H2][Br2]1/2/(k-2[HBr]+k3[Br2]) • Substitute into rate law for HBr: • ½ d[HBr]/dt = k2K1/2[H2][Br2]1/2/{1+(k-2/k3)[HBr]/[Br2]} • Same functional form as experimental law. • At the start of the reaction: [HBr] << [Br2] • ½ d[HBr]/dt = k2K1/2[H2][Br2]1/2

  11. Catalyst & Kinetics • Catalyst • Increases rate • Provides alternate pathway • Is not consumed • Lowers Ea • Homogeneous Catalyst: Same Phase • Heterogeneous Catalyst: Catalyst in different phase

  12. Effect of Catalyst on Ea

  13. Stratospheric Ozone • 2 O3 ==> 3 O2 • Mechanism (partial): • O3 O2 + O k1 • O2 + O  2 O3 k-1 • O + O3  2 O2 k2 • d[O3]/dt = -k1k2[O3]2/(k-1[O2] + k2[O3])

  14. 2 O3 ==> 3 O2 • d[O3]/dt = -k1k2[O3]2/(k-1[O2] + k2[O3]) • But O + O3  2 O2 Is slow • k2 small • d[O3]/dt -(k1k2 /k-1 )[O3]2/ [O2]

  15. Ozone Depletion • O3 + O ==> 2 O2 slow • Homogeneous Catalysis Proposed by Rowland & Molina (1974): • Chlorofluorocarbons (CFCl3 & CF2Cl2) • CFCl3 + h CFCl2 + Cl • O3 + Cl ==> ClO + O2 • ClO + O ==> O2 + Cl • ClO + O3 ==> 2O2 + Cl

  16. Antarctic Ozone Hole • Cl + CH4 CH3 + HCl • ClO + NO2  ClONO2 • Heterogeneous Catalysis: • HCl(g) + ClONO2(g)  Cl2(g) + HNO3(s) • Occurs on ice surface in polar strat clouds • In Spring: • Cl2(g) + h==> 2 Cl • Cl is regenerated & reacts with O3 & forms ClO & (ClO)2

  17. Enzyme CatalysisMichaelis-Menten Mechanism • -d[S]/dt =k[S]/(Km + [S]) • [S] = Substrate (molecule acted on) conc. • E + S <==> ES <==> E + P • -d[S]/dt = k1[E][S] – k-1[ES] • -d[ES]/dt = (k2 + k-1)[ES] –k1[E][S] –k-2[E][P] • d[P]/dt = k2[ES] – k-2[E][P] • [E]0 = [ES] + [E] = constant • Substitute & assume SS for [ES]

  18. SS Solution • [ES] = (k1[S] +k-2[P])[E]0/(k1[S] +k-2[P] + k-1 + k2) • Substitute into –d[S]/dt • Rate = (k1k2[S]– k-1k-2[P]) [E]0/(k1[S] +k-2[P] +k-1+ k2) • Initially: [S]  [S]0 & [P]  0 • Initial rate = k2[S]0[E]0/(Km + [S]0) • Km = (k-1 + k2)/k1 = Michaelis constant • Maximum rate = k2[E]0 • Turnover Rate = max rate/[E]0 = k2 • Catalase: 4.0 x 107/s

  19. Lineweaver-Burke Plot1/k = 1/k2 + Km/k2[S]

  20. Next Time • Start Chapter 27 • KMT

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