1 Chemical Kinetics Chapter 15 An automotive catalytic muffler.

1

Chemical KineticsChemical KineticsChapter 15Chapter 15


An automotive catalytic muffler.

2

• We can use thermodynamics to tell if a We can use thermodynamics to tell if a reaction is product or reactant favored.reaction is product or reactant favored.

• But this gives us no info on HOW FAST But this gives us no info on HOW FAST reaction goes from reactants to products.reaction goes from reactants to products.

• KINETICSKINETICS — the study of — the study of REACTION REACTION RATESRATES and their relation to the way the and their relation to the way the reaction proceeds, i.e., its reaction proceeds, i.e., its MECHANISMMECHANISM..

• We can use thermodynamics to tell if a We can use thermodynamics to tell if a reaction is product or reactant favored.reaction is product or reactant favored.

• But this gives us no info on HOW FAST But this gives us no info on HOW FAST reaction goes from reactants to products.reaction goes from reactants to products.

• KINETICSKINETICS — the study of — the study of REACTION REACTION RATESRATES and their relation to the way the and their relation to the way the reaction proceeds, i.e., its reaction proceeds, i.e., its MECHANISMMECHANISM..



3

Thermodynamics or Thermodynamics or KineticsKinetics

• Reactions can be one of the following:

1. Not thermodynamically favored (reactant favored)

2. Thermodynamically favored (product favored), but not kinetically favored (slow)

3. Thermodynamically favored (product favored) and kinetically favored (fast)

4


• Not thermodynamically favored (reactant favored)

• Sand (SiO2) will not decompose into Si and O2

5Thermodynamics or Thermodynamics or KineticsKinetics

• Thermodynamically favored (product favored), but not kinetically favored (slow)

• Diamonds will turn into graphite, but the reaction occurs very slowly

6


• Thermodynamically favored (product favored) and kinetically favored (fast)

• Burning of paper in air will turn to ash very quickly

7

Reaction MechanismsReaction MechanismsReaction MechanismsReaction MechanismsThe sequence of events at the molecular level that The sequence of events at the molecular level that

control the speed and outcome of a reaction.control the speed and outcome of a reaction.

Br from biomass burning destroys stratospheric ozone. Br from biomass burning destroys stratospheric ozone. (See R.J. Cicerone, Science, volume 263, page 1243, (See R.J. Cicerone, Science, volume 263, page 1243, 1994.)1994.)

Step 1:Step 1:Br + OBr + O33 ---> BrO + O ---> BrO + O22

Step 2:Step 2:Cl + OCl + O33 ---> ClO + O ---> ClO + O22

Step 3:Step 3:BrO + ClO + light ---> Br + Cl + OBrO + ClO + light ---> Br + Cl + O22

NET: NET: 2 O2 O33 ---> 3 O ---> 3 O22

The sequence of events at the molecular level that The sequence of events at the molecular level that control the speed and outcome of a reaction.control the speed and outcome of a reaction.

Br from biomass burning destroys stratospheric ozone. Br from biomass burning destroys stratospheric ozone. (See R.J. Cicerone, Science, volume 263, page 1243, (See R.J. Cicerone, Science, volume 263, page 1243, 1994.)1994.)

Step 1:Step 1:Br + OBr + O33 ---> BrO + O ---> BrO + O22

Step 2:Step 2:Cl + OCl + O33 ---> ClO + O ---> ClO + O22

Step 3:Step 3:BrO + ClO + light ---> Br + Cl + OBrO + ClO + light ---> Br + Cl + O22

NET: NET: 2 O2 O33 ---> 3 O ---> 3 O22

8

•Reaction rate = change in Reaction rate = change in concentration of a reactant or concentration of a reactant or product with time.product with time.

•Know about Know about initial rateinitial rate, , average average raterate, and , and instantaneous rateinstantaneous rate. See . See Screen 15.2. Screen 15.2.

Reaction Rates Reaction Rates Section 15.1Section 15.1

Reaction Rates Reaction Rates Section 15.1Section 15.1

10

Determining a Reaction RateDetermining a Reaction RateDetermining a Reaction RateDetermining a Reaction Rate

Blue dye is oxidized Blue dye is oxidized with bleach. with bleach.

Its concentration Its concentration decreases with time.decreases with time.

The rate — the The rate — the change in dye conc change in dye conc with time — can be with time — can be determined from the determined from the plot.plot.

Blue dye is oxidized Blue dye is oxidized with bleach. with bleach.

Its concentration Its concentration decreases with time.decreases with time.

The rate — the The rate — the change in dye conc change in dye conc with time — can be with time — can be determined from the determined from the plot.plot.

Dye

Co

nc

Dye

Co

nc

Dye

Co

nc

Dye

Co

nc

TimeTime

11

• ConcentrationsConcentrations and and physical physical statestate of reactants and products of reactants and products (Screens 15.3 and 15.4)(Screens 15.3 and 15.4)

• TemperatureTemperature (Screen 15.11) (Screen 15.11)

• CatalystsCatalysts (Screen 15.14) (Screen 15.14)

Factors Affecting Rates Factors Affecting Rates Section 15.2Section 15.2


12

• Concentrations Concentrations



Rate with 0.3 M HClRate with 0.3 M HCl

Rate with 6.0 M HClRate with 6.0 M HCl

13

• Physical state of reactantsPhysical state of reactants



14

Catalysts: catalyzed decomp of HCatalysts: catalyzed decomp of H22OO22

2 H2 H22OO22 --> 2 H --> 2 H22O + OO + O22



15

• Temperature Temperature



16

Concentrations and Rates Concentrations and Rates Concentrations and Rates Concentrations and Rates

To postulate a reaction To postulate a reaction mechanism, we studymechanism, we study

•• reaction ratereaction rate and and

•• its its concentration dependenceconcentration dependence

To postulate a reaction To postulate a reaction mechanism, we studymechanism, we study

•• reaction ratereaction rate and and

•• its its concentration dependenceconcentration dependence

17


Take reaction Take reaction where Clwhere Cl-- in in cisplatin cisplatin [Pt(NH[Pt(NH33))22ClCl33] ]

is replaced is replaced by Hby H22OO

Take reaction Take reaction where Clwhere Cl-- in in cisplatin cisplatin [Pt(NH[Pt(NH33))22ClCl33] ]

is replaced is replaced by Hby H22OO

CisplatinCisplatin

Rate of change of conc of Pt compd

= Am' t of cisplatin reacting (mol/L)

elapsed time (t)



elapsed time (t)

18


Rate of reaction is proportional to [Pt(NHRate of reaction is proportional to [Pt(NH33))22ClCl22]]

We express this as a We express this as a RATE LAWRATE LAW

Rate of reaction = k [Pt(NHRate of reaction = k [Pt(NH33))22ClCl22]]

where where k = rate constantk = rate constant

k is independent of conc. but increases with Tk is independent of conc. but increases with T

Rate of reaction is proportional to [Pt(NHRate of reaction is proportional to [Pt(NH33))22ClCl22]]

We express this as a We express this as a RATE LAWRATE LAW

Rate of reaction = k [Pt(NHRate of reaction = k [Pt(NH33))22ClCl22]]

where where k = rate constantk = rate constant

k is independent of conc. but increases with Tk is independent of conc. but increases with T

CisplatinCisplatin



elapsed time (t)



elapsed time (t)

19

Concentrations, Rates, Concentrations, Rates, and Rate Laws and Rate Laws Concentrations, Rates, Concentrations, Rates, and Rate Laws and Rate Laws

In general, forIn general, for

a A + b B ---> x Xa A + b B ---> x X with a catalyst Cwith a catalyst C

Rate = k [A]Rate = k [A]mm[B][B]nn[C][C]pp

The exponents m, n, and p The exponents m, n, and p

•• are the reaction orderare the reaction order

•• can be 0, 1, 2 or fractionscan be 0, 1, 2 or fractions

•• must be determined by experiment!!!must be determined by experiment!!!

In general, forIn general, for

a A + b B ---> x Xa A + b B ---> x X with a catalyst Cwith a catalyst C


The exponents m, n, and p The exponents m, n, and p

•• are the reaction orderare the reaction order

•• can be 0, 1, 2 or fractionscan be 0, 1, 2 or fractions

•• must be determined by experiment!!!must be determined by experiment!!!

CisplatinCisplatin

20

Interpreting Rate Laws Interpreting Rate Laws Interpreting Rate Laws Interpreting Rate Laws Rate = k [A]Rate = k [A]mm[B][B]nn[C][C]pp

• If m = 1, rxn. is 1st order in AIf m = 1, rxn. is 1st order in A

Rate = k [A]Rate = k [A]11

If [A] doubles, then rate goes up by factor of ? If [A] doubles, then rate goes up by factor of ?

• If m = 2, rxn. is 2nd order in A.If m = 2, rxn. is 2nd order in A.


Doubling [A] increases rate by ?Doubling [A] increases rate by ?

• If m = 0, rxn. is zero order.If m = 0, rxn. is zero order.


If [A] doubles, rate ?If [A] doubles, rate ?


• If m = 1, rxn. is 1st order in AIf m = 1, rxn. is 1st order in A


If [A] doubles, then rate goes up by factor of ? If [A] doubles, then rate goes up by factor of ?

• If m = 2, rxn. is 2nd order in A.If m = 2, rxn. is 2nd order in A.


Doubling [A] increases rate by ?Doubling [A] increases rate by ?

• If m = 0, rxn. is zero order.If m = 0, rxn. is zero order.


If [A] doubles, rate ?If [A] doubles, rate ?

CisplatinCisplatin

21

Deriving Rate Laws Deriving Rate Laws Deriving Rate Laws Deriving Rate Laws Derive rate law and k for Derive rate law and k for

CHCH33CHO(g) ---> CHCHO(g) ---> CH44(g) + CO(g)(g) + CO(g)

from experimental data for rate of disappearance of CHfrom experimental data for rate of disappearance of CH33CHOCHO

Expt. Expt. [CH[CH33CHO]CHO] Disappear of CHDisappear of CH33CHOCHO

(mol/L)(mol/L) (mol/L•sec)(mol/L•sec)

11 0.100.10 0.0200.020

22 0.200.20 0.0810.081

33 0.300.30 0.1820.182

44 0.400.40 0.3180.318

Derive rate law and k for Derive rate law and k for

CHCH33CHO(g) ---> CHCHO(g) ---> CH44(g) + CO(g)(g) + CO(g)

from experimental data for rate of disappearance of CHfrom experimental data for rate of disappearance of CH33CHOCHO

Expt. Expt. [CH[CH33CHO]CHO] Disappear of CHDisappear of CH33CHOCHO

(mol/L)(mol/L) (mol/L•sec)(mol/L•sec)

11 0.100.10 0.0200.020

22 0.200.20 0.0810.081

33 0.300.30 0.1820.182

44 0.400.40 0.3180.318

CisplatinCisplatin

22

Deriving Rate Laws Deriving Rate Laws Deriving Rate Laws Deriving Rate Laws

Rate of rxn = k [CHRate of rxn = k [CH33CHO]CHO]22

Here the rate goes up by ______ when initial conc. doubles. Here the rate goes up by ______ when initial conc. doubles. Therefore, we say this reaction is _________________ Therefore, we say this reaction is _________________ order.order.

Now determine the value of k. Use expt. #3 data—Now determine the value of k. Use expt. #3 data—

0.182 mol/L•s = k (0.30 mol/L)0.182 mol/L•s = k (0.30 mol/L)22

k = 2.0 (L / mol•s)k = 2.0 (L / mol•s)

Using k you can calc. rate at other values of [CHUsing k you can calc. rate at other values of [CH33CHO] at CHO] at

same T.same T.

Rate of rxn = k [CHRate of rxn = k [CH33CHO]CHO]22

Here the rate goes up by ______ when initial conc. doubles. Here the rate goes up by ______ when initial conc. doubles. Therefore, we say this reaction is _________________ Therefore, we say this reaction is _________________ order.order.

Now determine the value of k. Use expt. #3 data—Now determine the value of k. Use expt. #3 data—

0.182 mol/L•s = k (0.30 mol/L)0.182 mol/L•s = k (0.30 mol/L)22

k = 2.0 (L / mol•s)k = 2.0 (L / mol•s)

Using k you can calc. rate at other values of [CHUsing k you can calc. rate at other values of [CH33CHO] at CHO] at

same T.same T.

CisplatinCisplatin

23Concentration/Time Concentration/Time Relations Relations

Concentration/Time Concentration/Time Relations Relations

Need to know what conc. of reactant is as Need to know what conc. of reactant is as function of time. Consider FIRST ORDER function of time. Consider FIRST ORDER REACTIONSREACTIONS

For 1st order reactions, the rate law isFor 1st order reactions, the rate law is- (- ( [A] / [A] / time) = k [A] time) = k [A]

Need to know what conc. of reactant is as Need to know what conc. of reactant is as function of time. Consider FIRST ORDER function of time. Consider FIRST ORDER REACTIONSREACTIONS

For 1st order reactions, the rate law isFor 1st order reactions, the rate law is- (- ( [A] / [A] / time) = k [A] time) = k [A]

CisplatinCisplatin



Integrating - (Integrating - ( [A] / [A] / time) = k [A], we get time) = k [A], we getIntegrating - (Integrating - ( [A] / [A] / time) = k [A], we get time) = k [A], we get

CisplatinCisplatin




CisplatinCisplatin

[A] = - k tln[A]o

naturallogarithm [A] at time = 0

[A] = - k tln[A]o


[A] / [A][A] / [A]00 =fraction remaining after time =fraction remaining after time

t has elapsed.t has elapsed.






CisplatinCisplatin

[A] = - k tln[A]o


[A] = - k tln[A]o






Called the Called the integrated first-order rate lawintegrated first-order rate law..

27

Concentration/Time Relations Concentration/Time Relations Concentration/Time Relations Concentration/Time Relations

Sucrose decomposes Sucrose decomposes to simpler sugarsto simpler sugars

Rate of Rate of disappearance of disappearance of sucrosesucrose= k [sucrose]= k [sucrose]

k = 0.21 hrk = 0.21 hr-1-1

Initial [sucrose] = Initial [sucrose] = 0.010 M0.010 M

How long to drop How long to drop 90% (to 0.0010 M)?90% (to 0.0010 M)?

Sucrose decomposes Sucrose decomposes to simpler sugarsto simpler sugars

Rate of Rate of disappearance of disappearance of sucrosesucrose= k [sucrose]= k [sucrose]

k = 0.21 hrk = 0.21 hr-1-1

Initial [sucrose] = Initial [sucrose] = 0.010 M0.010 M

How long to drop How long to drop 90% (to 0.0010 M)?90% (to 0.0010 M)?

SucroseSucrose

28Concentration/Time RelationshipsConcentration/Time RelationshipsRate of disappear of sucrose = k [sucrose], k = 0.21 hrRate of disappear of sucrose = k [sucrose], k = 0.21 hr -1-1.. If If initial [sucrose] = 0.010 M, how long to drop 90% or to initial [sucrose] = 0.010 M, how long to drop 90% or to 0.0010 M?0.0010 M?

Use the first order integrated rate lawUse the first order integrated rate law



ln (0.100) = - 2.3 = - (0.21 hrln (0.100) = - 2.3 = - (0.21 hr -1-1) • time) • time



ln (0.100) = - 2.3 = - (0.21 hrln (0.100) = - 2.3 = - (0.21 hr -1-1) • time) • time

time = 11 hourstime = 11 hours

32

Using the Integrated Rate LawUsing the Integrated Rate LawUsing the Integrated Rate LawUsing the Integrated Rate LawThe integrated rate law suggests a way to tell if a The integrated rate law suggests a way to tell if a

reaction is first order based on experiment. reaction is first order based on experiment.

2 N2 N22OO55(g) ---> 4 NO(g) ---> 4 NO22(g) + O(g) + O22(g)(g)

Rate = k [NRate = k [N22OO55]]

Time (min)Time (min) [N[N22OO55]]00 (M) (M) ln [Nln [N22OO55]]00

00 1.001.00 00

1.01.0 0.7050.705 -0.35-0.35

2.02.0 0.4970.497 -0.70-0.70

5.05.0 0.1730.173 -1.75-1.75

33

Using the Integrated Rate LawUsing the Integrated Rate LawUsing the Integrated Rate LawUsing the Integrated Rate Law

2 N2 N22OO55(g) ---> 4 NO(g) ---> 4 NO22(g) + O(g) + O22(g) Rate = k [N(g) Rate = k [N22OO55]]

[N2O5] vs. time

time

1

0

0 5

[N2O5] vs. time

time

1

0

0 5

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

34


2 N2 N22OO55(g) ---> 4 NO(g) ---> 4 NO22(g) + O(g) + O22(g) Rate = k [N(g) Rate = k [N22OO55]]

[N2O5] vs. time

time

1

0

0 5

[N2O5] vs. time

time

1

0

0 5

l n [N2O5] vs. time

time

0

-2

0 5

l n [N2O5] vs. time

time

0

-2

0 5

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

Plot of ln [NPlot of ln [N22OO55] vs. ] vs.

time is a straight time is a straight line!line!

35


All 1st order reactions have straight line plot All 1st order reactions have straight line plot for ln [A] vs. time. for ln [A] vs. time.

(2nd order gives straight line for plot of 1/[A] (2nd order gives straight line for plot of 1/[A] vs. time)vs. time)

ln [N2O5] vs. time

time

0

-2

0 5

l n [N2O5] vs. time

time

0

-2

0 5

Plot of ln [NPlot of ln [N22OO55] vs. time ] vs. time

is a straight line! is a straight line! Eqn. for straight line: Eqn. for straight line: y = ax + b y = ax + b

Plot of ln [NPlot of ln [N22OO55] vs. time ] vs. time

is a straight line! is a straight line! Eqn. for straight line: Eqn. for straight line: y = ax + b y = ax + b

ln [N2O5] = - kt + ln [N 2O5]o

conc at time t

rate const = slope

conc at time = 0

ln [N2O5] = - kt + ln [N 2O5]o

conc at time t

rate const = slope

conc at time = 0

36

Graphical Methods for Graphical Methods for Determining Reaction Order Determining Reaction Order

and Rate Constantand Rate Constant• Zero-order

• Straight line plot of [A]t vs. t• Slope is –k (mol/L*s)

mxby

ktoAtA

][][

37


and Rate Constantand Rate Constant• First-order

• Straight line plot of ln[A]t vs. t• Slope is – k (s-1)

mxby

ktoAtA

]ln[]ln[

38


and Rate Constantand Rate Constant• Second-order

• Straight line plot of 1/[A]t vs t• Slope is k (L/mol*s)

mxby

ktoAtA

][

1

][

1

39

Half-LifeHalf-LifeSection 15.4 and Screen 15.8Section 15.4 and Screen 15.8


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

For 1st order reactions, For 1st order reactions, the concept of HALF-the concept of HALF-LIFE is especially LIFE is especially useful.useful.

40

• Reaction is 1st order Reaction is 1st order decomposition of decomposition of HH22OO22..

Half-LifeHalf-LifeHalf-LifeHalf-Life

41


• Reaction after 654 Reaction after 654 min, 1 half-life.min, 1 half-life.

• 1/2 of the reactant 1/2 of the reactant remains.remains.

42


• Reaction after Reaction after 1306 min, or 2 1306 min, or 2 half-lives.half-lives.


43


• Reaction after 3 Reaction after 3 half-lives, or 1962 half-lives, or 1962 min.min.


44


• Reaction after 4 Reaction after 4 half-lives, or 2616 half-lives, or 2616 min.min.

• 1/16 of the 1/16 of the reactant remains.reactant remains.

45



Sugar is fermented in a 1st order process (using Sugar is fermented in a 1st order process (using an enzyme as a catalyst).an enzyme as a catalyst).

sugar + enzyme --> productssugar + enzyme --> products

Rate of disappear of sugar = k[sugar]Rate of disappear of sugar = k[sugar]

k = 3.3 x 10k = 3.3 x 10-4-4 sec sec-1-1

What is the half-life of this reaction?What is the half-life of this reaction?

46



SolutionSolution

[A] / [A][A] / [A]00 = 1/2 when t = t = 1/2 when t = t1/21/2

Therefore, ln (1/2) = - k • tTherefore, ln (1/2) = - k • t1/21/2

- 0.693 = - k • t- 0.693 = - k • t1/21/2

tt1/21/2 = 0.693 / k = 0.693 / kSo, for sugar, So, for sugar,

tt1/21/2 = 0.693 / k = 2100 sec = = 0.693 / k = 2100 sec = 35 min35 min

Rate = k[sugar] and k = 3.3 x 10Rate = k[sugar] and k = 3.3 x 10-4-4 sec sec-1-1. What is the half-. What is the half-life of this reaction?life of this reaction?

47



SolutionSolution

2 hr and 20 min = 4 half-lives2 hr and 20 min = 4 half-lives

Half-life Half-life Time ElapsedTime Elapsed Mass LeftMass Left

1st1st 35 min35 min 5.00 g5.00 g

2nd2nd 7070 2.50 g2.50 g

3rd3rd 105105 1.25 g1.25 g

4th4th 140140 0.625 g0.625 g

Rate = k[sugar] and k = 3.3 x 10Rate = k[sugar] and k = 3.3 x 10-4-4 sec sec-1-1. Half-life . Half-life is 35 min. Start with 5.00 g sugar. How much is is 35 min. Start with 5.00 g sugar. How much is left after 2 hr and 20 min?left after 2 hr and 20 min?

48



SolutionSolution

Radioactive decay is a first order process. Radioactive decay is a first order process.

Tritium ---> electron + heliumTritium ---> electron + helium

33HH 00-1-1ee 33HeHe

If you have 1.50 mg of tritium, how much is left If you have 1.50 mg of tritium, how much is left after 49.2 years? tafter 49.2 years? t1/21/2 = 12.3 years = 12.3 years

49Half-LifeHalf-LifeSection 15.4 and Screen 15.8Section 15.4 and Screen 15.8


SolutionSolution

ln [A] / [A]ln [A] / [A]00 = -kt = -kt

[A] = ?[A] = ? [A][A]00 = 1.50 mg = 1.50 mg t = 49.2 yearst = 49.2 years

Need k, so we calc k from: k = 0.693 / tNeed k, so we calc k from: k = 0.693 / t1/21/2

Obtain k = 0.0564 yObtain k = 0.0564 y-1-1

Now ln [A] / [A]Now ln [A] / [A]00 = -kt = - (0.0564 y = -kt = - (0.0564 y-1-1) • (49.2 y) ) • (49.2 y)

= - 2.77= - 2.77

Take antilog: [A] / [A]Take antilog: [A] / [A]00 = e = e-2.77-2.77 = 0.0627 = 0.0627

0.0627 is the 0.0627 is the fraction remainingfraction remaining!!

Start with 1.50 mg of tritium, how much is left after 49.2 Start with 1.50 mg of tritium, how much is left after 49.2 years? tyears? t1/21/2 = 12.3 years = 12.3 years

50Half-LifeHalf-LifeSection 15.4 and Screen 15.8Section 15.4 and Screen 15.8


SolutionSolution

[A] / [A][A] / [A]00 = 0.0627 = 0.0627

0.0627 is the 0.0627 is the fraction remainingfraction remaining!!

Because [A]Because [A]00 = 1.50 mg, [A] = 0.094 mg = 1.50 mg, [A] = 0.094 mg

But notice that 49.2 y = 4.00 half-livesBut notice that 49.2 y = 4.00 half-lives

1.50 mg ---> 0.750 mg after 11.50 mg ---> 0.750 mg after 1

---> 0.375 mg after 2---> 0.375 mg after 2

---> 0.188 mg after 3---> 0.188 mg after 3

---> 0.094 mg after 4---> 0.094 mg after 4

Start with 1.50 mg of tritium, how much is left after 49.2 Start with 1.50 mg of tritium, how much is left after 49.2 years? tyears? t1/21/2 = 12.3 years = 12.3 years

51

Half-Lives of Radioactive ElementsHalf-Lives of Radioactive ElementsHalf-Lives of Radioactive ElementsHalf-Lives of Radioactive Elements

Rate of decay of radioactive isotopes given in Rate of decay of radioactive isotopes given in terms of 1/2-life. terms of 1/2-life.

238238U --> U --> 234234Th + HeTh + He 4.5 x 104.5 x 1099 y y1414C --> C --> 1414N + betaN + beta 5730 y5730 y131131I --> I --> 131131Xe + betaXe + beta 8.05 d8.05 d

Element 106 - seaborgiumElement 106 - seaborgium263263Sg Sg 0.9 s0.9 s

1 Chemical Kinetics Chapter 15 An automotive catalytic muffler.

Documents

Transcript of 1 Chemical Kinetics Chapter 15 An automotive catalytic muffler.