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

2. Unimolecular Reactions l 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 How come?? Is this really an elementary reaction? l Lindemann: Probably not.

3. Lindemann-Hinshelwood Unimolecular Mechanism

4. Lindemann Mechanism l A + M A * + M l A * ==> B l Rate (B) = k 2 [A * ] l SS condition: d[A * ]/dt = 0 = k 1 [A][M] – k -1 [A * ][M] – k 2 [A * ] [A * ] = k 1 [M][A]/(k 2 + k -1 [M]) Rate = k 2 k 1 [M][A]/(k 2 + k -1 [M]) = k’[A]

5. l At high conc: k 2 << k -1 [M]) Rate = k ‘ [A] l At low conc: k 2 >> k -1 [M]) Rate = k 1 [M][A]

6. CH 3 NC  CH 3 CN k lc =k 1 [M] k hc = k 1 k 2 /k -1

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

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

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

10. Results l [Br] = (k 1 /k -1 ) 1/2 [Br 2 ] 1/2 l [H] = k 2 K 1/2 [H 2 ][Br 2 ] 1/2 /(k -2 [HBr]+k 3 [Br 2 ]) l Substitute into rate law for HBr: l ½ d[HBr]/dt = k 2 K 1/2 [H 2 ][Br 2 ] 1/2 /{1+(k -2 /k 3 )[HBr]/[Br 2 ]} l Same functional form as experimental law. l At the start of the reaction: [HBr] << [Br 2 ] ½ d[HBr]/dt = k 2 K 1/2 [H 2 ][Br 2 ] 1/2

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

12. Effect of Catalyst on E a

13. Stratospheric Ozone l 2 O 3 ==> 3 O 2 l Mechanism (partial): O 3  O 2 + O k 1 O 2 + O  2 O 3 k -1 O + O 3  2 O 2 k 2 l d[O 3 ]/dt = -k 1 k 2 [O 3 ] 2 /(k -1 [O 2 ] + k 2 [O 3 ])

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

15. Ozone Depletion l O 3 + O ==> 2 O 2 slow l Homogeneous Catalysis Proposed by Rowland & Molina (1974): l Chlorofluorocarbons (CFCl 3 & CF 2 Cl 2 ) l CFCl 3 + h  CFCl 2 + Cl O 3 + Cl ==> ClO + O 2 ClO + O ==> O 2 + Cl ClO + O 3 ==> 2O 2 + Cl

16. Antarctic Ozone Hole l Cl + CH 4  CH 3 + HCl l ClO + NO 2  ClONO 2 l Heterogeneous Catalysis: HCl(g) + ClONO 2 (g)  Cl 2 (g) + HNO 3 (s) Occurs on ice surface in polar strat clouds In Spring: Cl 2 (g) + h ==> 2 Cl Cl is regenerated & reacts with O 3 & forms ClO & (ClO) 2

17. Enzyme Catalysis Michaelis-Menten Mechanism l -d[S]/dt =k[S]/(K m + [S]) [S] = Substrate (molecule acted on) conc. l E + S ES E + P -d[S]/dt = k 1 [E][S] – k -1 [ES] -d[ES]/dt = (k 2 + k -1 )[ES] –k 1 [E][S] –k -2 [E][P] d[P]/dt = k 2 [ES] – k -2 [E][P] l [E] 0 = [ES] + [E] = constant l Substitute & assume SS for [ES]

18. SS Solution l [ES] = (k 1 [S] +k -2 [P])[E] 0 /(k 1 [S] +k -2 [P] + k -1 + k 2 ) l Substitute into –d[S]/dt l Rate = (k 1 k 2 [S] – k -1 k -2 [P]) [E] 0 /(k 1 [S] +k -2 [P] +k -1 + k 2 ) l Initially: [S]  [S] 0 & [P]  0 Initial rate = k 2 [S] 0 [E] 0 /(K m + [S] 0 ) K m = (k -1 + k 2 )/k 1 = Michaelis constant l Maximum rate = k 2 [E] 0 l Turnover Rate = max rate/[E] 0 = k 2 Catalase: 4.0 x 10 7 /s

20. Lineweaver-Burke Plot 1/k = 1/k 2 + K m /k 2 [S]

21. Next Time Start Chapter 27 KMT