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Chemical KineticsChapter 15

An automotive catalytic muffler.

Chemical KineticsChapter 15

- We can use thermodynamics to tell if a reaction

is product or reactant favored. - But this gives us no info on HOW FAST reaction

goes from reactants to products. - KINETICS the study of REACTION RATES and their

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

its MECHANISM.

Thermodynamics or Kinetics

- Reactions can be one of the following
- Not thermodynamically favored (reactant favored)
- Thermodynamically favored (product favored), but

not kinetically favored (slow) - Thermodynamically favored (product favored) and

kinetically favored (fast)

Thermodynamics or Kinetics

- Not thermodynamically favored (reactant favored)
- Sand (SiO2) will not decompose into Si and O2

Thermodynamics or Kinetics

- Thermodynamically favored (product favored), but

not kinetically favored (slow) - Diamonds will turn into graphite, but the

reaction occurs very slowly

Thermodynamics or Kinetics

- Thermodynamically favored (product favored) and

kinetically favored (fast) - Burning of paper in air will turn to ash very

quickly

Reaction Mechanisms

- 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

Reaction Rates Section 15.1

- Reaction rate change in concentration of a

reactant or product with time. - Know about initial rate, average rate, and

instantaneous rate. See Screen 15.2.

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Determining a Reaction Rate

- 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.

Dye Conc

Time

Factors Affecting Rates Section 15.2

- Concentrations and physical state of reactants

and products (Screens 15.3 and 15.4) - Temperature (Screen 15.11)
- Catalysts (Screen 15.14)

Factors Affecting Rates Section 15.2

- Concentrations

Rate with 0.3 M HCl

Rate with 6.0 M HCl

Factors Affecting Rates Section 15.2

- Physical state of reactants

Factors Affecting Rates Section 15.2

- Catalysts catalyzed decomp of H2O2
- 2 H2O2 --gt 2 H2O O2

Factors Affecting Rates Section 15.2

- Temperature

Concentrations and Rates

- To postulate a reaction mechanism, we study
- reaction rate and
- its concentration dependence

Concentrations and Rates

- Take reaction where Cl- in cisplatin

Pt(NH3)2Cl3 is replaced by H2O

Concentrations and Rates

- Rate of reaction is proportional to Pt(NH3)2Cl2
- We express this as a RATE LAW
- Rate of reaction k Pt(NH3)2Cl2
- where k rate constant
- k is independent of conc. but increases with T

Concentrations, Rates, and Rate Laws

- In general, for
- a A b B ---gt x X with a catalyst C
- Rate k AmBnCp
- The exponents m, n, and p
- are the reaction order
- can be 0, 1, 2 or fractions
- must be determined by experiment!!!

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 0.20 0.081
- 3 0.30 0.182
- 4 0.40 0.318

Deriving Rate Laws

- Rate of rxn k CH3CHO2
- Here the rate goes up by ______ when initial

conc. doubles. Therefore, we say this reaction

is _________________ order. - Now determine the value of k. Use expt. 3 data
- 0.182 mol/Ls k (0.30 mol/L)2
- k 2.0 (L / mols)
- Using k you can calc. rate at other values of

CH3CHO at same T.

Concentration/Time Relations

- Need to know what conc. of reactant is as

function of time. Consider FIRST ORDER REACTIONS - For 1st order reactions, the rate law is - (D

A / D time) k A

Concentration/Time Relations

- Integrating - (D A / D time) k A, we get

Concentration/Time Relations

- Integrating - (D A / D time) k A, we get

A / A0 fraction remaining after time t has

elapsed.

Concentration/Time Relations

- Integrating - (D A / D time) k A, we get

A / A0 fraction remaining after time t has

elapsed.

Called the integrated first-order rate law.

Concentration/Time Relations

- Sucrose decomposes to simpler sugars
- Rate of disappearance of sucrose k sucrose
- k 0.21 hr-1
- Initial sucrose 0.010 M
- How long to drop 90 (to 0.0010 M)?

Sucrose

Concentration/Time RelationshipsRate of

disappear of sucrose k sucrose, k 0.21

hr-1. If initial sucrose 0.010 M, how long

to drop 90 or to 0.0010 M?

- Use the first order integrated rate law

Concentration/Time RelationshipsRate of

disappear of sucrose k sucrose, k 0.21

hr-1. If initial sucrose 0.010 M, how long

to drop 90 or to 0.0010 M?

- Use the first order integrated rate law

Concentration/Time RelationshipsRate of

disappear of sucrose k sucrose, k 0.21

hr-1. If initial sucrose 0.010 M, how long

to drop 90 or to 0.0010 M?

- Use the first order integrated rate law
- ln (0.100) - 2.3 - (0.21 hr-1) time

Concentration/Time RelationshipsRate of

disappear of sucrose k sucrose, k 0.21

hr-1. If initial sucrose 0.010 M, how long

to drop 90 or to 0.0010 M?

- Use the first order integrated rate law
- ln (0.100) - 2.3 - (0.21 hr-1) time
- time 11 hours

Using the Integrated Rate Law

- The integrated rate law suggests a way to tell if

a reaction is first order based on experiment. - 2 N2O5(g) ---gt 4 NO2(g) O2(g)
- Rate k N2O5
- Time (min) N2O50 (M) ln N2O50
- 0 1.00 0
- 1.0 0.705 -0.35
- 2.0 0.497 -0.70
- 5.0 0.173 -1.75

Using the Integrated Rate Law

- 2 N2O5(g) ---gt 4 NO2(g) O2(g) Rate k

N2O5

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

line.

Using the Integrated Rate Law

- 2 N2O5(g) ---gt 4 NO2(g) O2(g) Rate k

N2O5

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

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

line.

Using the Integrated Rate Law

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

Eqn. for straight line y ax b

- All 1st order reactions have straight line plot

for ln A vs. time. - (2nd order gives straight line for plot of 1/A

vs. time)

Graphical Methods for Determining Reaction Order

and Rate Constant

- Zero-order
- Straight line plot of At vs. t
- Slope is k (mol/Ls)

Graphical Methods for Determining Reaction Order

and Rate Constant

- First-order
- Straight line plot of lnAt vs. t
- Slope is k (s-1)

Graphical Methods for Determining Reaction Order

and Rate Constant

- Second-order
- Straight line plot of 1/At vs t
- Slope is k (L/mols)

Half-LifeSection 15.4 and Screen 15.8

- 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.

Half-Life

- Reaction after 654 min, 1 half-life.
- 1/2 of the reactant remains.

Half-Life

- Reaction after 1306 min, or 2 half-lives.
- 1/4 of the reactant remains.

Half-Life

- Reaction after 3 half-lives, or 1962 min.
- 1/8 of the reactant remains.

Half-Life

- Reaction after 4 half-lives, or 2616 min.
- 1/16 of the reactant remains.

Half-LifeSection 15.4 and Screen 15.8

- Sugar is fermented in a 1st order process (using

an enzyme as a catalyst). - sugar enzyme --gt products
- Rate of disappear of sugar ksugar
- k 3.3 x 10-4 sec-1
- What is the half-life of this reaction?

Half-LifeSection 15.4 and Screen 15.8

Rate ksugar and k 3.3 x 10-4 sec-1. What

is the half-life of this reaction?

- Solution
- A / A0 1/2 when t t1/2
- Therefore, ln (1/2) - k t1/2
- - 0.693 - k t1/2
- t1/2 0.693 / k
- So, for sugar,
- t1/2 0.693 / k 2100 sec 35 min

Half-LifeSection 15.4 and Screen 15.8

Rate ksugar and k 3.3 x 10-4 sec-1.

Half-life is 35 min. Start with 5.00 g sugar. How

much is left after 2 hr and 20 min?

- Solution
- 2 hr and 20 min 4 half-lives
- Half-life Time Elapsed Mass Left
- 1st 35 min 5.00 g
- 2nd 70 2.50 g
- 3rd 105 1.25 g
- 4th 140 0.625 g

Half-LifeSection 15.4 and Screen 15.8

Radioactive decay is a first order process.

Tritium ---gt electron helium

3H 0-1e 3He If you have 1.50 mg of tritium,

how much is left after 49.2 years? t1/2 12.3

years

- Solution

Half-LifeSection 15.4 and Screen 15.8

Start with 1.50 mg of tritium, how much is left

after 49.2 years? t1/2 12.3 years

- Solution
- ln A / A0 -kt
- A ? A0 1.50 mg t 49.2 years
- Need k, so we calc k from k 0.693 /

t1/2 - Obtain k 0.0564 y-1
- Now ln A / A0 -kt - (0.0564 y-1)

(49.2 y) - - 2.77
- Take antilog A / A0 e-2.77 0.0627
- 0.0627 is the fraction remaining!

Half-LifeSection 15.4 and Screen 15.8

Start with 1.50 mg of tritium, how much is left

after 49.2 years? t1/2 12.3 years

- Solution
- A / A0 0.0627
- 0.0627 is the fraction remaining!
- Because A0 1.50 mg, A 0.094 mg
- But notice that 49.2 y 4.00 half-lives
- 1.50 mg ---gt 0.750 mg after 1
- ---gt 0.375 mg after 2
- ---gt 0.188 mg after 3
- ---gt 0.094 mg after 4

Half-Lives of Radioactive Elements

- Rate of decay of radioactive isotopes given in

terms of 1/2-life. - 238U --gt 234Th He 4.5 x 109 y
- 14C --gt 14N beta 5730 y
- 131I --gt 131Xe beta 8.05 d
- Element 106 - seaborgium263Sg 0.9 s

- Darleane Hoffman of UC-Berkeley studies the

newest elements. - Hoffman is Director of Seaborg Institute. Recd

ACS Award in Nuclear Chem and the Garvan Medal.