12.09.2011 - Chem 260 Today in Chemistry 260 Lecture...

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Unformatted text preview: Chem 260 Today in Chemistry 260 Lecture 39 •  Mechanisms and catalysis 12/9/2010 Next Week in Chemistry 260 •  Monday: Course Connections Lecture (clickers!) •  problem set 10 due Monday, December 12, 2011 by 3:00 pm •  Final exam: Fri. December 16, 8 ­10 am •  1202 School of Education (Section 100) •  1360 East Hall (Section 200) •  CONFLICTS: will announce time today 1 Collision Theory Dudley R. Herschbach 1932- Yuan T. Lee 1936- John C. Polanyi 1929- Rudolph A. Marcus (1923- ) H + H2 → H2 + H k = A∞ e ­Ea/RT 1.  Isotopes How fast? 2.  Catalysis 2 A = zAB = d AB 3.  Enzymes The value of A can be calculated using the kinetic theory of gases: 8! kBT µ µ= k = Pze! Ea / RT Reaction Coordinate But What About Entropy? k ΔH ~ΔU mA mA m =A mA + mA 2 P is a steric factor We plot Gibbs energy along reaction coordinate to account for entropy. Ea Reaction Coordinate Entropy accounts for the number of different states that contribute all along the reaction path. Gibbs Energy ΔH ~ΔU An endothermic reaction will have a large activation energy, Ea. Potential Energy Potential Energy Ea If A=B then µ = But, for complex molecules, rate constants tend to be smaller than predicted, sometimes by many orders of magnitude. Not every collision is productive – they must be properly oriented: 4.  Reactions in Liquids An exothermic reaction will occur rapidly if Ea < RT and only very slowly if Ea >> RT. mA mB mA + mB ΔGŧ ΔGr Reaction Coordinate ΔG‡ is the free energy or Gibbs energy of activation. ‡/RT k = kBT e ­ΔG h ‡ ­TΔS‡)/RT = kBT e ­(ΔH h = kBTeΔS h h = Planck s Constant ‡/R  ­ΔH‡/RT e Pre ­exponential factor A∞ ΔS‡ = Entropy of Activation ΔH‡ = Enthalpy of Activation ⇒ ~microscopic Ea 1 We plot Gibbs energy along reaction coordinate to account for entropy. Determining overall reaction order ΔGr =  ­RTln(K) ‡/RT k1 Gibbs Energy Consider the reaction: k1 = kBT e ­ΔG h ‡)/RT ŧ k ­1 = kBT e ­( ­ΔGr + ΔG h ΔG 2A + B -->> 3C + D Some data: 1 ΔGr Ex. # = Rate (M/hr) 0.120 2.00 0.120 0.120 0.500 3 0.240 0.0600 1.00 4 k1 = e( ­ΔGr)/RT k ­1 k1 =K k −1 [B] (M) 0.240 2 ‡)/RT k ­1 kBTh e ­( ­ΔGr + ΔG Reaction Coordinate [A] (M) 1 ‡/RT k1 kBTh e ­ΔG 0.0140 135 ? RTln(k1/k ­1) =  ­ΔGr 10 2A + B -->> 3C + D Determining overall reaction order 2A + B -->> 3C + D Determining overall reaction order Determining overall reaction order of species Ex. # [A] (M) [B] (M) Rate (M/hr) Ex. # [A] (M) [B] (M) Rate (M/hr) 1 0.240 0.120 2.00 1 0.240 0.120 2.00 2 0.120 0.120 0.500 2 0.120 0.120 0.500 3 0.240 0.0600 1.00 3 0.240 0.0600 1.00 4 0.0140 135 ? 4 0.0140 135 ? Double A, rate increases 4 fold Double A, rate increases 4 fold: second order in A Double B, rate increases 2 fold: first order in B 11 2A + B -->> 3C + D Determining overall reaction order 12 Rate-Determining Step 2NO2 (g) + F2 (g) Determining overall reaction order of species 2NO2F (g) v = kobs [ NO 2 ][ F2 ] Ex. # [A] (M) [B] (M) Rate (M/hr) NO2 (g) + F2 (g) k1 NO2F (g) + F (g) slow 1 0.240 0.120 2.00 NO2 (g) + F (g) k2 NO2F (g) fast 2 0.120 0.120 0.500 3 0.240 0.0600 1.00 4 0.0140 135 ? v = k1 [ NO 2 ][ F2 ] Mechanisms in which the rate-determining step occurs after one or more fast steps are often signaled by •  a reaction order greater than 2, or •  by a non-integral reaction order, or •  by an inverse concentration dependence on one of the species taking part in the reaction. Double A, rate increases 4 fold: second order in A Double B, rate increases 2 fold: first order in B Overall rate: rate = k[A]2[B] Determine k by putting in any of the values: 2.00 = k(0.240)2(0.120): k = 289 M-2hr-1 First step is rate-determining 13 15 2 Catalysis: Kinetic Isotope Effects Consider: 2 H2O2(l) → 2 H2O (l) + O2(g) The plants will grow – but not thrive. Mice fed D2O will die. k 1 = A∞e 1 ν= 2π E0 = ΔrG° = 2( ­237.13) + 0  ­ 2( ­120.35) =  ­233.56 kJ ΔrH° Why? = 2( ­285.83) + 0  ­ 2( ­187.78) =  ­196.10 kJ This reaction is exothermic and spontaneous! −Ea kBT In fact the equilibrium constant is huge. Ea(X ­D) > Ea(X ­H) ȹ k ȹ ȹ ȹ ȹ µ Ⱥ k1(X ­ ­D) < k1(X ­ ­H) 1 (hν ) 2 K = e ­ΔrG°/RT K = exp 233.56 × 103 J 8.31451 J K-1 298 = 1.15 × 1041 D A ­ ­H ­B D A ­H ­ ­B With deuterium substitution the reaction may slow by a factor of 5 or more. Catalysis: With this K, how is it possible to buy a bottle of H2O2? Catalysis: Consider: 2 H2O2(l) → 2 H2O (l) + O2(g) In aqueous solution H2O2 is also unstable with respect to disproportionation to O2 and H2O. What is the equilibrium constant for this reaction? O2(g) + 2H+ (aq) + 2 e– → H2O2(aq) H2O2(aq) + 2H+ (aq) + 2 e– → 2H2O (l) Eo=1.763 V ΔrG° = 2( ­237.13) + 0  ­ 2( ­120.35) =  ­233.56 kJ ΔrH° Eo=0.695 V = 2( ­285.83) + 0  ­ 2( ­187.78) =  ­196.10 kJ This reaction is exothermic and spontaneous! In fact the equilibrium constant is huge. O2 is a good oxidant – but H2O2 is a better oxidant Eo=1.763 V H2O2(aq) → O2(g) + 2H+ (aq) + 2 e– Eo=1.068 V K = exp H 233.56 × 103 J 8.31451 J K-1 298 H = 1.15 × 1041 Eo= ­0.695 V 2H2O2(aq) → 2H2O (l)+ O2(g) K = e ­ΔrG°/RT K = eνFE 0 RT = e2⋅96500⋅1.068 8.314⋅298 = 1.3 ×1036 O–O → H O–H http://www.youtube.com/watch?v=tnkKznd_Rc0& MnO2 2 H2O2(aq) 2 H2O (l) + O2(g) Add MnO2 Very high activation energy ΔrH° Reac. With this K, how is it p – but ttill h a bottle Somewhat smaller ossible so buy uge. of H2O2? Catalysis: + O• Energy H2O2(aq) + 2H+ (aq) + 2 e– → 2H2O (l) Q Prod. Catalysis: Some DeIinitions Heterogeneous Catalysis: Catalyst is in a different phase than reactants. MnO2 acts as a catalyst – that is, it provides a mechanism with a much smaller activation energy. Old Path H2O2 Q H2O + O2 Homogeneous Catalysis: Energy Energy Energy Along the Catalyzed Path Usually a solid surface catalyzing gas or liquid phase reactions. Examples: Rhodium, Platinum, Zeolites, Metal Oxides, … T Q CA Q Catalyst is in the same phase as the reactants. Usually in liquid solution. Examples: Complexes of Rhodium and Platinum, Enzymes, Strong Acids, … Initial and pinal states are unaffected by catalyst. The nature of the reaction coordinate is changed. 3 ...
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