Previously in Chem 104: How to determine Rate Law TODAY How to determine rate constant, k...

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Previously in Chem 104:

• How to determine Rate Law

TODAY

• 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

Data on how the rate of H2O2 decomposition is affected by varying the Initial [I-] values.

2X4.1 X

4.1 X

2X

So Rate depends on [H2O2]o :

Raterxn = k [H2O2]o

AND Rate depends on [KI] o :

Raterxn = k* [KI]o Overall, Rate depends on two parameters:

Raterxn = k’ [H2O2]o [KI]o where k’= k k*

And we say the overall reaction is Second Order, 2o, First order, 1o, in H2O2 andFirst order, 1o, in KI

This expression where both dependences are written:

Raterxn = k’ [H2O2]o [KI]o

is the Rate law.

The Rate Law is the reason Kinetics studies are done:

It shows us the slowest step in reaction sequence:

the Rate Determining Step, r.d.s.

Obtaining Rate Constants from Kinetic Data

Examples of Plots of Different Reaction Orders

Integrated Rate Laws

[rgt]o

Time, sec

[rgt] t½ = ½ [rgt]o

[rgt] t¼= ¼ [rgt]o

t ½ t¼

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 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

Collision Theory and:The Rate Law: the Macroscopic View

[H2O2]o

Time, sec

Rate = k[H2O2][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

Ea : the “sufficient energy” in collision

Hf : net reaction enthalpy

2Argts

Bprdt

en

erg

y

Reaction progress

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

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.

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 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?

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?

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)

The Arrhenius Equation

k = Ae-Ea/RT

Activation energy:We now understand what that is

What is this?

H2O2 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 H2O2 O2 + 2 H2O

Hf

2 H2O2

en

erg

y

Reaction progress

2H2O + O2

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