Previously in Chem 104: How to determine Rate Law How to determine rate constant, k Recognizing Plots

1. Previously in • Chem 104: • How to determine Rate Law • How to determine rate constant, k • Recognizing Plots • Using Integrated Rate Laws to determine concentrations vs time • Using the Arrhenius Eqn to find k at new Temp • TODAY • Half lives • the collision theory of kinetics • Reaction Coordinates • Importance of Rate Law: Mechanism • What is A? Ea? • QUIZ! Watch for it • This weekend!

2. [rgt]o [rgt] t½ = ½ [rgt]o [rgt] t¼= ¼ [rgt]o t ½ t¼ Time, sec

3. Radioactive Decay and Half Lives Technetium Radiopharmaceuticals, Tc

4. The Collision Theory of Reactions • Reactions result when atoms/molecules collide with • sufficient energy to break bonds • - Molecules must collide in an orientation that leads to • productive bond cleavage and/or formation Collision Theory Connects Macroscopic and Microscopic Perspectives of Kinetics • The more molecules in a volume, the more collisions, or, the reaction occurrence depends on concentration

5. Collision Theory and: The Rate Law: the Macroscopic View Rate = k[H2O2][I-] [H2O2]o Why concentrations affect rate Time, sec

6. The Collision Theory: why higher temperatures help • Reactions result when atoms/molecules collide with • sufficient energy to break bonds • Molecules at a higher temperature move faster— • have a greater energy (energy distribution increases)

7. Energetics of a Reaction are Summarized in a Reaction Coordinate Ex. 1: For a single step reaction: A + A  B energy Ea : the “sufficient energy” in collision 2A rgts DHf : net reaction enthalpy B prdt Reaction progress

8. Collision Theory and: The Rate Law: the Macroscopic View Rate = k[H2O2][I-] Why concentrations affect rate [H2O2]o Time, sec The Rate Law: the Microscopic View

9. The Importance of the Rate Law The Rate Law specifies the the molecularity of the Rate-Determining Step, it specifies which collisions most affect rate. The Rate Determining Step is the process (collision) that has Ea, the energy of activation, the most energetic step of reaction.

10. Connecting Hoses to Water the Garden ½ inch, 4 gal/min 3/4 inch, 8 gal/min 1 inch, 16 gal/min How do you connect these 3 hoses to deliver water at the fastest rate? All will have same rate— Limited by the 4 gal/min, ½ inch hose

11. The rate determining step at the Burgmayer’s: Andrew…

12. In the reaction: 2 H2O2 O2 + 2 H2O the Rate Law is: Raterxn = k’ [H2O2]o [I-]o And so the r.d.s. involves one H2O2 and one I- Maybe like this?

13. If this is the slow step, how do we get to products? Step 1: H2O2 + I- H-O-I + OH-slow Step 2: H-O-I + H2O2 O2 + H2O + I- + H+ fast Step 3: H+ + OH- H2O v. fast Net reaction: 2 H2O2 O2 + 2 H2O the Rate Law is: Raterxn = k’ [H2O2]o [KI]o

14. These steps are called Elementary Reaction Steps. Here, all are bi-molecular (involve 2 species) Step 1: H2O2 + I- H-O-I + OH-k = 10-3 sec-1 Step 2: H-O-I + H2O2 O2 + H2O + I- + H+ k = ?sec-1 Step 3: H+ + OH- H2O k = 1013 sec-1 Net reaction: 2 H2O2 O2 + 2 H2O k = 10-3 sec-1 the Rate Law is: Raterxn = k’ [H2O2]o [KI]o

15. The Arrhenius Equation What is this? k = Ae-Ea/RT Activation energy: We now understand what that is

17. H2O2 I-

18. Bad orientation: no productive reaction occurs

19. If collision orientation is favorable, a reaction occurs OH- I-O-H

20. There may be several good collision orientations

21. The Arrhenius Equation Orientation Factor k = Ae-Ea/RT Activation energy: We now understand what that is

22. Energetics of a Reaction Summarized in a Reaction Coordinate Ex. 2: For a multi step reaction: 2 H2O2 O2 + 2 H2O energy 2 H2O2 DHf 2H2O+ O2 Reaction progress

23. Energetics of a Reaction Summarized in a Reaction Coordinate Ex. 2: For a multi step reaction: 2 H2O2 O2 + 2 H2O Transition state Step 2 energy Step 1: + I- Ea H-O-I + OH- intermediates - I- Step 3 2 H2O2 DHf 2H2O+ O2 Reaction progress