previously in chem 104: how to determine rate law how to determine rate constant, k recognizing...
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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 itThis weekend!
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[rgt]o
Time, sec
[rgt] t½ = ½ [rgt]o
[rgt] t¼= ¼ [rgt]o
t ½ t¼
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Radioactive Decay and Half Lives
Technetium Radiopharmaceuticals, Tc
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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 formationCollision Theory Connects Macroscopic and Microscopic Perspectives of Kinetics
-The more molecules in a volume, the more collisions, or, the reaction occurrence depends on concentration
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Collision Theory and:The Rate Law: the Macroscopic View
[H2O2]o
Time, sec
Rate = k[H2O2][I-]
Why concentrations affect rate
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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)
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Energetics of a Reaction are Summarized in a Reaction Coordinate
Ex. 1: For a single step reaction: A + A B
Ea : the “sufficient energy” in collision
Hf : net reaction enthalpy
2Argts
Bprdt
en
erg
y
Reaction progress
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Collision Theory and:The Rate Law: the Macroscopic View
The Rate Law: the Microscopic View
[H2O2]o
Time, sec
Rate = k[H2O2][I-]Why concentrations affect rate
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The Importance of the Rate Law
The Rate Law specifies thethe molecularity of theRate-Determining Step,it specifies which collisionsmost affect rate.
The Rate Determining Stepis the process (collision) that has Ea, the energy of activation,the most energetic step of reaction.
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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
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The rate determining step at the Burgmayer’s:
Andrew…
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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?
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the Rate Law is: Raterxn = k’ [H2O2]o [KI]o
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
If this is the slow step, how do we get to products?
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the Rate Law is: Raterxn = k’ [H2O2]o [KI]o
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
These steps are called Elementary Reaction Steps.
Here, all are bi-molecular (involve 2 species)
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The Arrhenius Equation
k = Ae-Ea/RT
Activation energy:We now understand what that is
What is this?
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H2O2 I-
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Bad orientation: no productive reaction occurs
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I-O-HOH-
If collision orientation is favorable, a reaction occurs
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There may be several good collision orientations
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The Arrhenius Equation
k = Ae-Ea/RT
Activation energy:We now understand what that is
Orientation Factor
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Energetics of a Reaction Summarized in a Reaction Coordinate
Ex. 2: For a multi step reaction: 2 H2O2 O2 + 2 H2O
Hf
2 H2O2
en
erg
y
Reaction progress
2H2O + O2
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Energetics of a Reaction Summarized in a Reaction Coordinate
Ex. 2: For a multi step reaction: 2 H2O2 O2 + 2 H2O
Ea
Hf
en
erg
y
Reaction progress
H-O-I + OH-
intermediates
2 H2O2
2H2O + O2
Step 1: + I-
- I-
Transition state
Step 2
Step 3
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