1. ChE 452 Lecture 25 Non-linear Collisions 1

2. Background: Collision Theory Key equation Method Use molecular dynamics to simulate the collisions Integrate using statistical mechanics 2

3. Summary Of Collision Theory For Linear Collisions One can approximate the collision of two molecules as a collision between two classical particles following Newton’s equations of motion. The reactants have to have enough total energy to get over the saddle point in the potential energy surface. It is not good enough for the molecules to just have enough energy. Rather, the energy needs to be correctly distributed between vibration and transition. 3

4. Summary Continued Coordinated motions of the atoms are needed. In particular, it helps to have C moving away from B when A collides with BC. It is necessary to localize energy and momentum into the B-C bond for reaction to happen. The detailed shape of the potential energy surface has a large influence on the rate. Want banked curves, that funnel molecules to saddle point. 4

5. Summary Continued Leads to very complex behavior, but not big rate changes. Latter four effects only cause perhaps a factor of 10 or 100 in rate. There are always some trajectories which make it over the barrier, even though the molecules have barely enough energy to cross the barrier. If 1% of the trajectories make it those trajectories will have an important effect on the rate. 5

6. Today: Non-linear Case Non-linear case: Reaction probability varies with the impact parameter 6 Figure 8.27 The typical trajectory for the collision of an A atom with a BC molecule.

7. Variation Of Reaction Probability With Impact Parameter 7 Leads to finite cross section. Figure 8.26 The variation in P A  BC with changing impact parameter.

8. Why Does Reaction Probability Vary With Impact Parameter? Molecules still have enough energy to go over the barrier Energy is not coupled to the bond that breaks. Reactants miss. Angular momentum, Coriolis force carries molecules apart. 8

9. Angular Momentum Creates Extra Barrier To Non-Linear Collisions Component of velocity carries reactants apart. Must overcome that velocity component for reaction to occur. 9 Figure 8.27 The typical trajectory for the collision of an A atom with a BC molecule.

10. Derivation Of The Angular Momentum Barrier To Reaction A+B  Products Classical equations of motion of A and B. Pages of Algebra 10 Kinetic energy Extra energy barrier due to Coriolis force Potential energy 

11. Result: Atom Moves In Effective Potential: 11 (8.144) (8.56)

12. Angular Momentum Barrier To Reaction 12 The angular momentum barrier prevents reactions from occurring when molecules approach with large impact parameters. As a result, no reaction occurs unless the reactants get close to each other.

13. Angular Momentum Barrier Results In Drop Off Of Reaction Probability At Large b. 13 Figure 8.26 The variation in P A  BC with changing impact parameter.

14. Next: Quantifying The Effects: P reaction Varies with: v A  BC, the velocity that the A molecule approaches the BC molecule; E BC, the internal state (i.e., vibrational, rotational energy) of the BC molecule before collision occurs; The “impact parameter” b A  BC, which is a measure of how closely A collides with BC; “The angle of approach” where  is a measure of the angle of the collision The initial position R BC and velocity v BC of B relative to C when collision occurs. 14

15. Next Equation 15 (8.57) Structureless molecules (8.58)

16. Example 8.E Calculating The Cross Section A program called ReactMD is available from Dr. Masel’s website. Assume you used the program to calculate the reaction probability as a function of impact parameter, and the data in Table 8.E.1 were obtained. Calculate the cross section for the reaction. 16

17. 17

18. Solution: The Cross Section Is Given By: 18 We can integrate using the trapezoid rule.

19. Spreadsheet To Do The Calculations: 19

20. Numerical Results 20

21. Example 8.D: Calculating The Rate Constant Using Equation 8.60 Figure 8.17 shows some data for the cross section for the reaction H+H 2  H 2 +H as a function of E T, the translational energy of H approaching H 2. Assume that the cross section follows: 21 where E T is the translational energy. Calculate the rate constant for the reaction at 300K. (8.D.2)

22. Solution: Step 1: Derive Equation According to equation 8.60, if there is no E BC dependence 22 (8.D.3)

23. Derive Equation Continued According to results in Example 6.D: Where µ ABC is the reduced mass of ABC, k B is Boltzman’s constant and T is the temperature. 23 kBkB (8.D.3)

24. Derive Equation Continued Looking up the integral in the CRC yields: Combining equations (8.D.3) and (8.D.5) and substituting yields: 24 (8.D.5) (8.D.7) kBkB kBkB kBkB kBkB kBkB kBkB

25. Final Equation Note Therefore, 25 (8.D.6) (8.D.8) kBkB

26. Define Average Cross-Section I A  BC 26 (8.D.8) Equation 8.C.8 becomes: (8.D.9) kBkB kBkB kBkB

27. Step 2: Simplify Equation Let’s define a new variable W by: Substituting equation (8.C.6) into equation (8.C.5) yields: 27 (8.D.10) (8.D.11) kBkB kBkB kBkB kBkB kBkB

28. Step 3: Substitute In Numbers For future reference, it is useful to note: k B T = 0.6 kcal/mole = 0.026 eV/molecule 28 (8.D.12) (8.D.13)

29. Step 4: Calculate The Velocity According to equation (7.A.4): 29 (8.D.15)

30. Step 5: Further Algebra Substituting equation 8.C.10 into 8.C.8 and adding the appropriate conversion factors yields: 30 (8.D.16) kBkB kBkB kBkB

31. Step 6: Further Simplify Equation Note:  = 0 for E T < 0.35eV. Therefore, 31 (8.D.15) kBkB kBkB -0.35eV/k B T

32. More Simplification Let’s define I A  BC and F(W) by: 32 (8.D.17) (8.D.18) kBkB kBkB

33. Combining Equations (8.D.15) And (8.D.16) Yields: 33 (8.D.19)

34. One Can Conveniently Integrate Equation 8.D.17 Using The Laguere Integration Formula 34 (8.D.20) Where the B i and W i s are given in the spreadsheet on the next page.

35. Spreadsheet For The Calculations 35

36. Results: 36

37. Solution 37 (8.D.21) (8.D.24) (8.D.25) kBkB kBkB

38. Notes Notice that the activation barrier is about 0.35 eV (i.e., the minimum energy to get reaction) even though the reaction probability is small below 0.5eV. It is not exactly 0.35eV though. In the problem set, we ask the reader to calculate the rate constant at other temperatures. If you make an Arrhenius plot of the data, you find that the activation barrier is not exactly 0.35eV, but close to 0.35eV, even though the reaction probability is negligible at E = 0.35eV. 38

39. Summary Can use MD to calculate reaction probability as a function of b. Can integrate to get reaction rate. Results differ from TST due to dynamic corrections (turning and angular momentum barriers to reaction). 39

40. Question What did you learn new in this lecture? 40