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? E a ? QUIZ! Watch for it This weekend!

[rgt] o Time, sec [rgt] t½ = ½ [rgt] o [rgt] t¼ = ¼ [rgt] o t ½ t¼t¼

Radioactive Decay and Half Lives Technetium Radiopharmaceuticals, Tc

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

Collision Theory and: The Rate Law: the Macroscopic View [H 2 O 2 ] o Time, sec Rate = k[H 2 O 2 ][I-] Why concentrations affect rate

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)

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

Collision Theory and: The Rate Law: the Macroscopic View The Rate Law: the Microscopic View [H 2 O 2 ] o Time, sec Rate = k[H 2 O 2 ][I-] Why concentrations affect rate

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 E a, the energy of activation, the most energetic step of reaction.

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

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

In the reaction: 2 H 2 O 2  O H 2 O the Rate Law is: Rate rxn = k’ [H 2 O 2 ] o [I - ] o And so the r.d.s. involves one H 2 O 2 and one I- Maybe like this?

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

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

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

H2O2H2O2 I-I-

Bad orientation: no productive reaction occurs

I-O-HOH - If collision orientation is favorable, a reaction occurs

There may be several good collision orientations

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

Energetics of a Reaction Summarized in a Reaction Coordinate Ex. 2: For a multi step reaction: 2 H 2 O 2  O2 + 2 H 2 O HfHf 2 H 2 O 2 energy Reaction progress 2H 2 O + O 2

Energetics of a Reaction Summarized in a Reaction Coordinate Ex. 2: For a multi step reaction: 2 H 2 O 2  O H 2 O EaEa HfHf energy Reaction progress H-O-I + OH - intermediates 2 H 2 O 2 2H 2 O + O 2 Step 1: + I - - I - Transition state Step 2 Step 3