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Chapter 15 - Chemical Kinetics

- Objectives
- Determine rates of reactions from graphs of

concentration vs. time. - Recall the conditions which affect the rates.
- Recognize the order of reaction, give the rate

equation, calculate the rate constant. - Use integrated rate laws and half-life equations

in calculations. - Draw energy diagrams, find activation energy.
- Identify catalysts and their properties.
- Identify reaction intermediates.
- Recognize the rate equation given a mechanism,

and given the rate equation determine the

mechanism.

KINETICS the study of REACTION RATES

H2O2 decomposition in an insect

H2O2 decomposition catalyzed by MnO2

Chemical engineering, enzymology, environmental

engineering, etc.

Kinetics and Mechanisms

- KINETICS the study of REACTION RATES and their

relation to the way the reaction proceeds, i.e.,

its MECHANISM. - The reaction mechanism is our goal!
- The sequence of events at the molecular level

that control the speed and outcome of a reaction.

Br from biomass burning destroys stratospheric

ozone. (See R.J. Cicerone, Science, volume 263,

page 1243, 1994.)

- Step 1 Br O3 ---gt BrO O2
- Step 2 Cl O3 ---gt ClO O2
- Step 3 BrO ClO light ---gt Br Cl O2
- NET 2 O3 ---gt 3 O2
- Identify the intermediate
- Identify the catalyst(s)

Reaction Rates

- Reaction rate change in concentration of a

reactant or product with time.

- Three types of rates
- initial rate
- average rate
- instantaneous rate

Reaction Rates can be determined from a Plot

- Blue dye is oxidized with bleach.
- Its concentration decreases with time.
- The rate the change in dye conc with time

can be determined from the plot.

- The initial rate (in this case over the first

minute) is calculated from a tangent line

crossing the initial concentration. Then the

slope of the line is determined

Reaction Rates can be determined from a Plot

N2O5 ? NO2 O2

2

4

- The average rate is calculated from a time

interval. - The instantaneous rate is calculated at a single

point in time (or given concentration) by drawing

a tangent line crossing the point. Then the slope

of the line is determined.

- Compare average rates at the beginning and

end of reaction.

Factors affecting the Rates

- Concentrations
- Physical State of Reactants and Products
- Surface area
- Temperature
- Catalysts

Review Exp. 1 Factors affecting reactions rates

Concentration

Mg(s) 2 HCl(aq) ---gt

MgCl2(aq) H2(g)

0.3 M HCl

6 M HCl

- Increasing the concentration of reactants

_______________ the rate of the reaction.

Surface Area

- Increasing the surface area of reactants

_____________ the reaction rate.

Temperature

Bleach at 54 C

Bleach at 22 C

- Increasing the temperature ____________ the rate

of the reaction.

Catalysts

MnO2

2 H2O2 ---- gt 2 H2O O2

- A _________ is present at the beginning and at

the end of the reaction and it does not change

but it _______________ the rate of the reaction.

Factors affecting Reaction Rates

- Iodine clock reaction

1. Iodide is oxidized to iodine H2O2 2

I- 2 H -----gt 2 H2O I2 2. I2

reduced to I- with vitamin C I2 C6H8O6

----gt C6H6O6 2 H 2 I- When all vitamin

C is depleted, the I2 interacts with starch to

give a blue complex.

Factors affecting Reaction Rates

Concentration and Rate

- To postulate a mechanism we study
- - The reaction rate
- and its
- - Concentration dependence.
- Generate a Rate Law Equation.

Rate Laws

- In general for
- a A b B --gt x X
- Rate k AmBn
- The exponents m, n
- are the _______________
- can be 0, 1, 2 or fractions
- must be determined by ____________!

With a catalyst C

Cp

Interpreting Rate Laws

- Rate k AmBnCp
- If m 1, rxn. is 1st order in A
- Rate k A1
- If A doubles, then rate goes up by factor of

__________ - If m 2, rxn. is 2nd order in A.
- Rate k A2
- Doubling A increases rate by _____________
- If m 0, rxn. is zero order.
- Rate k A0
- If A doubles, rate ____________________

Deriving Rate Laws

- Derive rate law and k for
- CH3CHO(g) --gt CH4(g) CO(g)
- from experimental data for rate of disappearance

of CH3CHO

Expt. CH3CHO Disappear of CH3CHO

(mol/L) (mol/Lsec) 1 0.10 0.020 2 0.2

0 0.081 3 0.30 0.182 4 0.40 0.318

Rate k CH3CHOn

4 Rate k 2 CH3CHOn

n 2

Look at exp 1 and 2 concentration doubles rate

cuadruples

Deriving Rate Laws

- Rate of rxn
- Here the rate goes up by ________ when initial

concentration doubles. Therefore, we say this

reaction is __________ order. - Now determine the value of k. Use any exp.

Data - Using k you can calculate rate at other values

of CH3CHO at same T.

Concentration and Time

- What is the concentration of reactant as a

function of time? - Consider First Order reactions

Integrating we get

Integrated First Order Law

A / A0 fraction remaining after time t has

elapsed.

The decomposition of a certain insecticide in

water follows first-order kinetics with a rate

constant of 1.45 year-1 at 12oC. A quantity of

this insecticide is washed into a lake on June 1,

leading to a concentration of 5.0 x 10-7 g/cm3.

Assume that the average temperature of the lake

is 12oC.

- a) What is the concentration of the insecticide

on June 1 of the following year?b) How long will

it take for the concentration of the insecticide

to drop to 3.0 x 10-7 g/cm3?

Using Integrated Rate Laws

- All 1st order reactions have straight line plot

for ln A vs. time. - And 2nd order gives straight line for plot of

1/A vs. time.

Using Integrated Rate Laws

- In an experiment for
- 2 N2O5(g) ---gt 4 NO2(g) O2(g)
- Time (min) N2O50 (M)
- 0 1.00
- 1.0 0.705
- 2.0 0.497
- 5.0 0.173

If it were zero order

Data of conc. vs. time plot do not fit straight

line.

Using Integrated Rate Laws

- In an experiment for
- 2 N2O5(g) ---gt 4 NO2(g) O2(g)
- Time (min) N2O50 (M)
- 0 1.00
- 1.0 0.705
- 2.0 0.497
- 5.0 0.173

If it were first order

ln N2O50 0 -0.35 -0.70 -1.75

Plot of ln N2O5 vs. time is a straight line!

Calculate the ln

Using Integrated Rate Laws

Plot of ln N2O5 vs. time is a straight line!

Eqn. for straight line y mx b

The gas phase decomposition of hydrogen peroxide

at 400 oC is second order in H2O2. In one

experiment, when the initial concentration of

H2O2 was 0.246 M, the concentration of H2O2

dropped to 3.39 x 10-2 M after 25.9 seconds had

passed. What is the rate constant for the

reaction?

- 2 H2O2 ? 2 H2O O2

Half-Life

- HALF-LIFE is the time it takes for 1/2 a sample

is disappear. For 1st order reactions, the

concept of HALF-LIFE is especially useful.

Half-Life

- Reaction is 1st order decomposition of H2O2.
- Reaction after 1 half-life.
- 1/2 of the reactant has been consumed and 1/2

remains.

Half-Life

- After 2 half-lives _____ of the reactant remains.

- After 3 half-lives ____ of the reactant remains.

Half-Life

ln R / R0 k t

- A / A0 fraction remaining
- when t t1/2 then fraction remaining _________

- ln (____) - k t1/2

In an experiment, it is determined that 75 of a

sample of HCO2H (formic acid) has decomposed in

72 seconds following first-order kinetics.

Determine t1/2 for this reaction. HCO2H ? CO2

H2

ln R / R0 k t

t1/2 0.693 / k

Mechanisms

- Mechanism how reactants are converted to

products at the molecular level.

RATE LAW ----gt MECHANISM experiment

----gt theory

Activation Energy

- Molecules need a minimum amount of energy to

react. Visualized as an energy barrier -

activation energy, Ea.

Reaction coordinate diagram

Activation Energy

- Conversion of cis to trans-2-butene requires

twisting around the CC bond. - Rate k trans-2-butene

Transition State

Cis

Trans

Transition state

Mechanisms

- Reaction passes thru a TRANSITION STATE where

there is an activated complex that has sufficient

energy to become a product.

ACTIVATION ENERGY, Ea energy reqd to form

activated complex. Here Ea ___________

Mechanisms

- Also note that trans-butene is MORE STABLE than

cis-butene by about 4 kJ/mol. - Therefore, cis ---gt trans is _________________
- This is the connection between thermo-dynamics

and kinetics.

Effect of Temperature

In ice at 0 oC

- Reactions generally occur slower at lower T.

Room temperature

Iodine clock reaction, book page 705. H2O2 2

I- 2 H --gt 2 H2O I2

Activation Energy and Temperature

In general, differences in activation energy

cause reactions to vary from fast to slow.

- Reactions are __________ at a higher T because

a larger fraction of reactant molecules have

enough energy to convert to product molecules.

Collision Theory

- Molecules must collide with one another
- Molecules must collide with sufficient ________

to break bonds - Molecules must collide in an orientation that can

lead to rearrangement of the atoms.

Arrhenius Equation

Frequency factor related to frequency of

collisions with correct geometry.

Plot ln k vs. 1/T ---gt straight line. slope

-Ea / R

Collision Theory

- Reactions require
- (a) activation energy and
- (b) correct geometry.
- O3(g) NO(g) ---gt O2(g) NO2(g)

2. Activation energy and geometry

1. Activation energy

Mechanisms

- Most reactions involve a sequence of elementary

steps. - 2 I- H2O2 2 H ---gt I2 2 H2O
- Rate k I- H2O2
- NOTE
- 1. Rate law comes from experiment.
- 2. Order and stoichiometric coefficients not

necessarily the same! - 3. Rate law reflects all chemistry down to and

including the slowest step in multistep reaction.

Mechanisms

Most reactions involve a sequence of elementary

steps. 2 I- H2O2 2 H ---gt I2 2

H2O Rate k I- H2O2

Proposed Mechanism Step 1 slow HOOH I- --gt

HOI OH- Step 2 fast HOI I- --gt I2

OH- Step 3 fast 2 OH- 2 H --gt 2

H2O Rate of the reaction controlled by slow step

RATE DETERMINING STEP, Rate can be no faster

than RDS!

Mechanisms

Most reactions involve a sequence of elementary

steps. 2 I- H2O2 2 H ---gt I2 2

H2O Rate k I- H2O2

Proposed Mechanism Step 1 slow HOOH I- --gt

HOI OH- Step 2 fast HOI I- --gt I2

OH- Step 3 fast 2 OH- 2 H --gt 2 H2O

- Elementary Step 1 is bimolecular and involves I-

and HOOH. Therefore, this predicts the rate law

should be - Rate a I- H2O2 as observed!!
- The species HOI and OH- are ________________.

Catalysts and Activation Energy

MnO2 catalyzes decomposition of H2O2 2 H2O2 ---gt

2 H2O O2

Catalysts and Activation Energy

- Iodine-Catalyzed Isomerization of Cis-2-Butene

Remember

- Go over all the contents of your textbook.
- Practice with examples and with problems at the

end of the chapter. - Practice with OWL tutor.
- Practice with the quiz on CD of Chemistry Now.
- Work on your OWL assignment for Chapter 15.