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Chem 125 Lecture 38 1/12/09 This material is for the exclusive use of Chem 125 students at Yale and may not be copied or distributed further. It is not.

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Presentation on theme: "Chem 125 Lecture 38 1/12/09 This material is for the exclusive use of Chem 125 students at Yale and may not be copied or distributed further. It is not."— Presentation transcript:

1 Chem 125 Lecture 38 1/12/09 This material is for the exclusive use of Chem 125 students at Yale and may not be copied or distributed further. It is not readily understood without reference to notes or the wiki from the lecture.

2 Semester 1 : Bonds and Molecular Structure (and some thermodynamics) Welcome Back Semester 2 : Functional Group Reactions (and some spectroscopy)

3 Classical Trajectories & The Potential Energy Surface Visualizing Reaction

4 Time-Lapse “Classical” (Molecular Mechanics) Trajectory for non-reactive collision of 13 atoms 6 molecules 40 Dimensions (3n + time) by E. Heller faster slower heavier lighter rotating slowly rotating rapidly & vibrating Too Complicated (for our purposes)

5 Using Energies to Predict Equilibria and Rates for One-Step Reactions: Free-Radical Halogenation H CH 3 Cl H Cl CH 3 Cl CH 3 Cl Cl "free-radical chain"

6 Unreactive Trajectory: (A bounces off vibrating B-C) Potential Energy Surface for Linear Triatomic A-B-C

7 C flies away from vibrating A-B Reactive Trajectory A approaches non-vibrating B-C Potential Energy Surface for Linear Triatomic A-B-C “classical” trajectory (not quantum)

8 “I wanted to catch a little one” John McBride (1973)

9 Studying Lots of Random Trajectories Provides Too Much Detail Summarize Statistically with Collective Enthalpy (H) & Entropy (S)

10 “Reaction Coordinate” Diagram (for a one-step atom transfer) Not a trajectory, but a sequence of three species Starting Materials Products Transition “State” G each with H and S, i.e. Free Energy (G)

11 Free Energy determines what can happen (equilibrium) K = e -  G/RT = 10 -(3/4)  G kcal/mole @ room Temp and how rapidly (kinetics) k (/sec) = 10 13 e -  G /RT ‡ ‡ = 10 13-(3/4)  G kcal/mole @ room Temp Amount of ts (universal) Velocity of ts theory

12 H 3 C H + X X  H 3 C X + H X F Cl Br I 37 58 46 36 105 ” 142 163 151 141 251 187 160 129 136 103 88 71 115 84 72 58 Possibility of Halogenation (Equilibrium) 109 19 9 12 CostReturnProfit

13 H H 2 H 2 H HHHHHH H H H Henry Eyring (1935) Dissociation followed by association requires high activation energy. SLOW Make-as-you-break “displacement” is much easier. FAST

14 Free-Radical Chain Substitution X-HR-H X-X R-X X R cyclic machinery

15 H 3 C-H + X 2  HX + H 3 CX F Cl Br I 37 58 46 36 105 ” 142 163 151 141 251 187 160 129 136 103 88 71 115 84 72 58 Possibility of Halogenation (Equilibrium) 109 24 9 12 CostReturnProfit H 3 C-H  HX X X2  H3CXX2  H3CX H 3 C 37 58 46 36 136 103 88 71 Step 1 31 2 17 34 Step 2 78 26 22 (Mechanism for Reasonable Rate) How can we predict activation energy & k?

16 Digression on Reaction Order The kinetic analogue of the Law of Mass Action i.e. dependance of rate on concentration(s).

17 Rate (amount per second) Doubled Rate Chemists can also change [Concentration]

18 Rate “Laws”: Kinetic Order Rate = d [Prod] / d t 0 th Order: Rate = k Simple One-Step Reactions = k  concentration(s) ? Dependent on Mechanism Discovered by Experiment

19 0 th Order Kinetics Would more sheep give a faster rate? NO! (saturation) Catalyst e.g. enzyme “Substrate” Rate  But the catalysis was not originally recognized. [Catalyst] [Substrate] 0 1  Photo: Antonio Vidigal by permission

20 Rate “Laws”: Kinetic Order 1 st Order: Rate = k [A] Rate = d [Prod] / d t 0 th Order: Rate = k Simple One-Step Reactions = k  concentration(s) ? Dependent on Mechanism Discovered by Experiment (Reasonable)

21 Product Time (sec) Concentration First-Order Kinetics k = 0.69/sec

22 Product Time (sec) Concentration First-Order Kinetics Starting Material 1/2 1/4 1/8 1/16 Exponential Decay  Constant “Half Life” = 0.69 / k k = 0.69/sec

23 Reversible First-Order Kinetics Starting MaterialProduct k -1 k1k1 at Equilibrium forward rate = reverse rate k 1  [Starting Material] = k -1  [Product] = k -1 k1k1 [Product] [Starting Material] K

24 Time (sec) Concentration Reversible First-Order Kinetics Starting Material Product k 1 = 0.69/sec k -1 = 0.23/sec Exponential Decay to Equilibrium Mixture Half Life = 0.69 / (k 1 + k -1 ) Starting MaterialProduct k -1 k1k1 ( K = 3 )

25 Rate Laws: Kinetic Order 2 nd Order: Rate = k [A] 2 “1 st Order in A” Rate = d [Prod] / d t 1 st Order: Rate = k [A] 0 th Order: Rate = k Simple One-Step Reactions = k  concentration(s) ? Dependent on Mechanism Discovered by Experiment or Rate = k [A] [B] “Pseudo” 1 st Order If [B] is (effectively) constant k or [B] >> [A] [B] a catalyst

26 Time (sec) Concentration Second- vs First-Order Kinetics First Order Second Order Slows Faster Not Exponential No Constant Half Life

27 Rate Laws: Kinetic Order Rate = d [Prod] / d t Complex Reactions = k  concentration(s) ? Dependent on Mechanism Discovered by Experiment The Rate-Limiting Step Who Cares? Rapid pre- equilibrium reactive intermediate(low concentration) “ ”

28 Starting MaterialIntermediate k -1 k1k1 Product k2k2 Actual 2 nd Step rate limiting 1 st Step rate limiting A B Click for Excel Program

29 End of Lecture 38 Jan. 12, 2009


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