reaction kineticsreaction kinetics. an introduction. 2 • a condition of equilibrium is reached in...

1

Reaction Kinetics

An Introduction

2

• A condition of equilibrium is reached in a system when 2 opposing changes occur simultaneously at the same rate.

• The rate of a chemical reaction may be defined as the # of mols of a substance which disappear or are formed by the reaction per unit volume in a unit of time.

3

Example

tI

ttIIRate

CI

forward ΔΔ

=−−

=

→

][][][

12

12

4

• The previous rate is for the DISAPPEARANCE of I, therefore:

""

][

oflossorngdisappearimeanssignnegativethewheretIrate

ΔΔ

−=

5

Backwards Example

IstudyratesreactionbothwhengformeanssignpositivethewheretC

tC

ttCC

Ratebkward

min

][][][][

12

12

ΔΔ

+=ΔΔ

=−−

=

6

More Complex Reactions

tZ

tXRate

ngdisappeariisXasfastasXappearingZtodueisthis

tZ

tXRate

ZX

ΔΔ

=ΔΔ

−=

∴

ΔΔ

≠ΔΔ

−=

→

31

3

3

7

In GeneralFor the Reaction:

pP + qQ → rR + sS

tS

stR

rtQ

qtP

pRate

ΔΔ

=ΔΔ

=ΔΔ

−=ΔΔ

−=1111

8

Reaction Order

9

In GeneralFor the Reaction:

pP + qQ → rR + sS

tconsrateortconsalityproportionkwhere

QPkRateor

QPtoalproportionisand

tS

stR

rtQ

qtP

pRate

mn

mn

tantan

][][

][][

1111

=

=

ΔΔ

=ΔΔ

=ΔΔ

−=ΔΔ

−=

10

k

A reaction with an incredibly large rate constant is faster than a reaction with an

incredibly small rate constant.

11

For the Reaction:pP + qQ → rR + sS

The reaction is “n” order in [P] and “m” order in [Q]

OR

Is OVERALL “(n + m)” order

12

KEY!!!!!!

“n” and “m” DO NOT necessarily equal “p”, “q”,

“r” or “s”

13

Example

][

][][41][

21

)()(4)(2

52

2252

2252

ONkor

tO

tNO

tON

Rate

gOgNOgON

=

ΔΔ

=Δ

Δ=

ΔΔ

−=

+→

14

• This reaction is FIRST order in N2O5, NOT SECOND order as one might intuit from the stoichiometry.

15

The Order of The Reaction• = the specification of the empirical (experimentally-

determined) dependence of the rate of the reaction on CONCENTRATIONS

• The order may = 0, a whole number or a non-whole number, e.g.,– 0– 1– 1½– 2

• We’ll focus on whole numbers and 0 (zero) for reactions of the type:

fF + gG → Products• AND! The order of the reaction is defined in terms of

REACTANTS not the products, therefore, products do not need to be specified

16

Zero-Order Reactions

formbmxylineStraighttatFttimeatF

ChemPherenotIntegrate

tkFarrange

ktF

and

kkftFtcoefficienaonlyisf

GFktF

fRate

o

ktoFF

+====

−−

Δ−=Δ

=ΔΔ

−

==ΔΔ

−∴

=ΔΔ

−=

−=

:0][;][

:)(

][:Re

][

'][

][]['][1

][][

00

17

For Zero-Order Reactions

• These reactions are relatively rare

• Occur on metal surfaces

• Reaction rate is INDEPENDENT of concentration of reactants

18

First Order Reactions• Assume reaction is 1st order in F and zero order

in G:

tko

o

eFFOR

tkFF

egrateandtkFFarrange

FktFkkf

GFktF

fRate

−=

−=

Δ=Δ

=ΔΔ

∴=

=ΔΔ

−=

][][

][][ln

:int:Re

]['

][]['1 01

19

• Many radioactive decays fit 1st order reactions:• 226Ra88 → 222Rn86 + 4He2

• 238U92 → 234Th90 + 4He2

• The rate is proportional to [F]

20

The Rate is Proportional to [F]

Double [F] k [F] k [F]2 k [F]3

Rate Change ⇑ X 2 ⇑ X 4 ⇑ X 8

][FktF

=ΔΔ

21

Second Order Reactions: Type 1

tkFF

egrateandarrangeGtorespectwithorderZero

FtorespectwithorderSecond

FktF

andkkf

GFktF

fRate

o

=−

=ΔΔ

−

=

=ΔΔ

−=

][1

][1

:intRe

][

'

][]['1

2

02

22

Second Order Reactions: Type 2

overallorderSECONDGinANDFinorderFIRSTisaction

tkGFFG

FG

partsbyegrateandarrange

GFktF

henceandkkf

GFktG

gtF

fRate

o

o

oo

∴

=−

=ΔΔ

−

=

=ΔΔ

−=ΔΔ

−=

,Re][][][][

ln][][

1)(intRe

][][

,,'

][]['11

Second order reactions are the most common reactions

The rate is proportional to [G], [G]o, [F], [F]o

23

Pseudo-First Order Reactions

• A special kind of 2d order reaction:• Example• 1 M acetyl chloride (AcCl) reacted with 56

M water (XSSV amount) to for HOAC and HCl

AcCl + H2O → HOAc + HCl

24

OrderFirstPSEUDOHenceorderfirstbetoAPPEARSaction

AcClkRatekOHk

soecttosmalltooisOHinchangeThe

OHAcClktd

AcCldRate

−

=∴=

=−=

:,""Re

][','][

,det][

][][][

2

2

2

25

Empirical Method in Determining Reaction Orders

studiedbeingareAinONLYdifferingreactionsdifferentWhere

AA

tt

X

StepSecondAreactionfororderunknowntheX

AA

tt

raterate

StepFirstX

][2

][][log

log

:""

][][

:

2

1

1

2

2

1

1

2

2

1

ΔΔ

=

=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

ΔΔ

=

26

Example

AinorderondisreactionM

MAAtt

X

tMAandBAaction

tMAandBAaction

sec

2239.0477.0

734.1log3log

173.03.0log

'5'15log

][][log

log

'15;173.0][:2Re

'5;3.0][:1Re

2

1

1

2

2

1

∴

===⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

=⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛ΔΔ

=

=Δ=→

=Δ=→

27

Practical Application• Reaction rates are proportional to some power of

[reactant]• Determined by using “initial reaction rate method”

28

Compare Reaction 2 with Reaction 1

29


30

• Fuse both effects and the rate equation for this reaction =

Rate = k [Zn] [HCl]2• REMEMBER:• The exponent in a rate equation generally

does NOT match the chemical equation coefficients.

• The exponent MUST be determined experimentally.

31

Another way to do this

32


33


34

Reaction Order Half-Lives

35

11

21

21

1

21

1

21

][1

:

2ln:

2][

:

−−

−

−

==

==

==

sMkF

t

OrderSecond

IONCONCENTRATofTINDEPENDENistOrderFirst

sk

t

OrderFirst

sMk

Ft

OrderZero

o

o

36

Example

• For the reaction:• C → D + E,• Half of the C is used up in 60 seconds.

Calculate the fraction of C used up after 10 minutes – reaction is first order in C.

37

Solution

UpUsedCorUPUSEDCHenceLEFTCC

CCC

CKey

tkCC

ondskandk

t

o

o

o

%9.99999022.0000978.01:000978.0][

93.6][ln)600()01155.0(][ln][ln

1][:][][ln

sec01155.060693.02ln 1

21

−−=−=

−=−=−

=

−=

=== −

38

Example

• 14C is present at about 1.1*10-13 mol% naturally in living matter. A bone dug up showed 9*10-15 mol% 14C. The half life of 14C is 5720 years. How old is the bone?

39

Solution

oldyears

t

tkCC

yrst

k

o

45.2066210*2115.1

503.2

)10*2115.1(10*1.1

10*9ln

][][ln

10*211.15720

693.02ln

4

413

15

14

14

14

21

=−

−

−=

−=

===

−

−−

−

−−

40

Example

• A decomposition reaction occurs in a fixed-volume container at 460°C. Its rate constant is 4.5*10-3 seconds-1. At t = 0, P = 0.75 atm. What is the pressure (P) after 8 minutes?

41

Solution

atmPP

P

P

tkPP

o

0865.016.2288.0ln

)480()10*5.4(75.0lnln

)480()10*5.4(75.0

ln

ln

3

3

=−=+

−=−

−=

−=

−

−

42

Q10 Effect

43

Example

• If the rate of a chemical reaction doubles for every 10°C rise in temperature, how much faster would the reaction proceed at 55°C than at 25°C?

44

Q 10 EffectQ 10 Effect on Reaction Rate

0, 1 10, 2 20, 4 30, 840, 16

50, 32

60, 64

70, 128

0

20

40

60

80

100

120

0 20 40 60

Temperature (Celsius)

Rel

ativ

e R

eact

ion

Rat

e

45

Solution

• Temperature increased 30°C, therefore, reaction rate increases 8-fold

Example:• What if the temperature was increased to

105°C from 25°C?• Temperature increased 80°C, therefore

reaction rate increases 256-fold

46

Example

• How much faster would a reaction go at 100°C than at 25°C?

• For every 10°C increase in temperature, the reaction rate doubles. The change in temperature is 75°C. This is 7.5 10°C increases.

• Hence 27.5 = 181 times faster

47

Example

• In an experiment, a sample of NaOCl was 85% decomposed in 64 minutes. How long would it have taken if the temperature was 50°C higher?

• For every 10°C increase, the reaction rate doubles. 50°C increase is 5 10°C increases.

• Hence: 25 = 32 times faster• So: (64 minutes)/(32 times faster) = 2 minutes

48

There are 4 types of temperature dependence for reaction rates -- 1

• Rate increases with increasing temperature

• NORMAL

49


• Rate increases to a point, then reduces with increasing temperature

• E.g., enzymes being denatured

50


• Rate decreases with increasing temperature

• VERY RARE• Known only for a

few reactions that are multi-step reactions:– A → B Fast step– B → C Rate limiting step

51


• Rate increases with increasing temperature

• Odd behavior• Explosive reaction

when temperature shoots up

• Gradual rise in temperature due to chain reactions

52

• The Q 10 effect is “not entirely perfect”

• There is another way to study temperature dependence:

• Mathematically with Energy of Activation

53

etemperaturabsoluteTtconsGasR

ActivationofEnergyEmoleculestreacbetweencollisionsof

frequencytotalfactorfrequencyAtconsratek

a

TRaEeAk

==

=

==

−=

tan

)tan(

tan

/

54

Take Natural Log (ln) of Above Equation

TRE

Ak a−= lnln

55

Previous Equation Can Be Manipulated

• If you know the rate constants for reactions at 2 different temperatures you can calculate the Ea:

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

121

2 11lnTTR

Ekk a

56

Example

• At T1 of 308 K, k1 = 0.326 s-1; at T2 of 318 K, k2 = 1.15 s-1. R = 8.314 J/mol-K. Determine Ea

molkJ

molJE

E

E

TTRE

kk

a

a

a

a

6.1028.601,10210*021.1

)314.8()26.1(

)10*021.1(314.8

26.1

3081

3181

314.8326.015.1ln

11ln

4

4

121

2

⇔==−−

−−=

⎟⎠⎞

⎜⎝⎛ −−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

−

−

57

Ea -- Transition

• Top of Energy curve (“hump”) = transition state

• The smaller the Ea, the easier it is for the reaction “to go”

58

Ea – “thermic”

Ea is related to ΔG

59

Reaction Mechanisms

60

Unimolecular Steps

• A single molecule or atom breaks down or rearranges itself into another species.

61

Bimolecular Steps

• Collision of two molecules or atoms; relatively common

62

Termolecular Steps

• 3 molecules or atoms collide simultaneously; relatively rare and SLOW

63

Elementary Steps

• = each step in a mechanism; a specific occurrence in the reaction sequence– The rate of elementary steps are proportional

to the number of collisions per step– The number of collisions per step are

proportional to every reactant concentration– Hence the rate of elementary steps are

proportional to the concentration of every reactant

64

The exponents of an elementary step rate equation are equal to the coefficients in the step’s equation

65

E.g., Bi Elementary Steps

OrderstSOkStepforequationRate

OrderdOSOkStepforequationRate

StepOSOSOStepSOOSO

RxnOverallOSOOSO

1][:2

2][][:1

21

51

321

235

532

2332

+→→+

+→+

66

E.g., Termolecular Elementary Steps

OrderstClHkStepforequationRate

OrderdHClkStepforequationRate

OrderstClkStepforequationRate

StepHClClHStepClHHCl

StepClCl

RxnOverallHClHCl

1][:3

2][][:2

1][:1

3222

12

2

223

22

2

21

22

222

2

22

=

=

=

→→+

→

→+

−

−

−

67

Reaction MechanismsTWO requirements of postulated mechanisms:

1) MUST account for the postulated mechanism2) MUST explain the experimental rate equation

1) – INHERENT in this is the assumption that2) 1 step is incredibly SLOWER than the others,

hence, it determines the speed of the reaction.3) This step is called the rate limiting step.4) Therefore, the overall rate equation is determined

by rate equations for the rate limiting step.5) ALSO: when a catalyst is present, there MUST BE

at least 2 steps in the mechanism!

68

Example

• The empirically derived rate equation for:2NO2 (g) + Cl2 (g) → 2NO2Cl ↑ is

k [NO2] [Cl2]

• Write a two-step mechanism with a slow first step

69

stepslowtheisStepHenceEquationRateEmpiricalStepforEquationRate

BUTClClNOk

EquationRateStepClNOClClNO

ClNOkEquationRate

StepClClNOClNORxnOverallClNOClNO

1,1

][][

22][][

122

2

22

22

222

222

=

→+

+→+→+

70

Comments

• Any steps after Step 1 will NOT effect the rate because step 1 is the rate limiting step

• Add up the 2 steps (just like in thermodynamics) and you get the overall reaction

71

Illustrative Example

)1,Re(2

2][][

2][][

1][][

22

2

2

2

22

purposesveillustratiforwasthisSTEPisalityInLimitingRateisStep

EquationStepEqualsEquationRateEmpiricalthatNoteHIIClk

EquationRateStepHClIHIICl

HIClkEquationRate

StepHClHIHIClEquationRateDerivedyEmpiricallHIIClkRxnOverallIHClHICl

−

+→+

+→+

+→+

72

Example

• The empirical rate equation for

2ClO2 + F2 → 2FClO2

Is: k [ClO2] [F2]

• Write the reaction mechanism (2-step) consistent with the rate equation.

73

LimitingRateisStepHenceEquationRatedDtnyEmpiricall

MatchesEquationRateStepFClOk

EquationRateStepFClOFClO

FClOkEquationRate

StepFFClOFClORxnOverallFClOFClO

1,'

1][][

2][][

122

2

22

22

222

222

→+

+→+→+

74

Steady State Approximation• The majority of mechanisms are NOT as easy as

previous examples.• Let’s return to our Bimolecular Reaction Step Example:

75

Focus• SO5• This is an “intermediate” – a “transition”• An intermediate (or transition) is a compound

that is neither reactant nor product, rather in between the two.

• Intermediates are very difficult to isolate in many cases.

• They are highly reactive and, more often than not, are used up as quickly as they are formed. It is because of this characteristic that we have the Steady State Approximation

76

Steady State Approximation• The concentration of intermediate is a constant• I.e., the rate of formation of the intermediate is

equal to the rate of disappearance of the intermediate:

tdtermediateind

tdtermediateind ][][

−=

77

Example• 2O3 → 3O2

• Empirically derived rate equation = k [O3]2/[O2]

SLOWStepOOO

clearlyreactionbackwardstheforsitthatyoudsreitaskpreferIkitcallsourcessomereactionbackwardsforkratek

FASTStepcatalystOcatalystOO

reactionforwardforkratekFASTStepcatalystOOcatalystO

MECHANISM

k

k

k

32

'min)(

2

1:

23

1

21

32

1

23

3

1

1

⎯→⎯+

=+⎯→⎯++

=++⎯→⎯+

−

−

−

78

• The constant of a reverse reaction is indicated by (-), e.g., k-1

• The intermediate in this mechanism is the oxygen ATOM, hence:

tdOd

tdOd ][][

−=

79

• Knowing this, we need to show that the empirical rate equation and mechanism are consistent with each other.

80

][][

][][

][][][][

][][][][][

][][][][][][][''

21

31

2131

2131

2131

OOkOk

OforSolveOOkOk

catalystoutCancelcatalystOOkcatalystOk

TwoTheEquateNow

catalystOOktdOdcatalystOk

tdOd

RxndBckwofRateRxndForofRate

=

=

=

=−=

−

−

−

−

81

• Stop temporarily• REMEMBER: the rate limiting step is #3;

the rate equation for this step = k3 [O] [O3]• So, substitute the solution for [O] from the

first for’d and bkw’d reactions in to the rate equation for Step 3:

82

equationratederivedyempiricalltoidenticalIsOO

kequation

substituteandk

kkkLet

OOkOk

k

][][

:

][*][

][*

2

23

1

13

321

313

=

=−

−

83

• This series of kinetic equations fulfills the 2 requirements:

• 1) Accounts for the products, and,• 2) Explains the observed rate law

84

Second Example• For 2NO + O2 → 2NO2, the empirical rate equation is:

k [NO]2 [O2]

• The postulated mechanism follows:

SLOWStepNONONO

FASTStepNOONO

k

k

k

22

1

223

32

2

1

1

⎯→⎯+

⇔+−

85

• Show that the empirical rate equation and mechanism are consistent with each other.

• Use the methodology we used in the previous example.

86

tdNOd

tdNOd

andNONONO

ONONO

NOONO

k

k

k

][][

2

33

23

23

32

2

1

1

−=

⎯→⎯+

+⎯→⎯

⎯→⎯+−

87

ytemporarilhereStop

NOk

ONOk

NOforSolveNOkONOk

twotheEquate

NOktd

NOdONOk

tdNOd

EquationRatedBkwEquationRatedFor

][][][:][

][][][:

][][

][][][

''

31

21

3

3121

313

213

=

=

=−=

−

−

−

88

• The rate limiting step is step 3.• The rate equation for step 3 is: k2 [NO3]

[NO]• Substitute the solution for [NO3] into the

rate equation for step 3.

89

][][][][][ 22

21

21 ONOkNOONOk

kk=⎟⎟

⎠

⎞⎜⎜⎝

⎛

−

90

Apply This to Enzymes• Enzymes are, with a couple of exceptions,

proteins• Enzymes are biological catalysts• Enzymes speed up biological reactions

incredibly• For this discussion:

– E = enzyme, – S = substrate and – P = product

91

tdESd

tdESd

ANDSLOWPEESFASTSEESFASTESSE

k

k

k

][][

2

1

1

−=

+⎯→⎯

+⎯→⎯

⎯→⎯+−

92

ytemporarilStop

ESSEkk

ESforSolveESkSEk

MethodShort

][][][

:][][][][

1

1

11

=

=

−

−

93

• Rate limiting step is step 3• Rate equation is: k2 [ES]• Substitute as before:

RxnUniUniaisThisPEESSE

reactionoveralltheWrite

SEkSEk

kk

−+→⇔+

=⎟⎟⎠

⎞⎜⎜⎝

⎛

−

:

][][][][1

21

94

Uni-Uni Reaction – Cleland Plot

95

Bi-molecular Reactions

• An enzyme catalyzed reaction may utilize 2 substrates.

• This reaction is always SEQUENTIAL, however,

• May be – ORDERED or – RANDOM

96

E.g., Ordered Sequential Reaction

E + X + Y → E + R + SE is still enzyme

X and Y are substratesR and S are products

97

Putative MechanismE + X → EX

EX + Y → EXYEXY → ERS SLOW STEP

ERS → ER + SER → E + S

And:

tdEXYd

tdERSd ][][

−=

98

ytemporarilhereStop

EXYEXYXEkkk

EXYforSolve

EXYkYEXkXEk

Equate

EXYktd

EXYdYEXkXEktd

ERSd

][][][][][

][

][][][][][

][][][][][][][

3

21

321

321

=⎟⎟⎠

⎞⎜⎜⎝

⎛

=

=−=

99

• Rate Limiting Step is: k3 [EXY]• Substitute:

][][][][

][][][][][][][][ 213

213

EXYXEk

EXYXEkkEXYXEkkkk

=

=

100

Kinetic Data Tells us:

• The following sequence MUST be taking place:

E + X + Y → EX (FIRST!) → EXY (SECOND!) → ERS → ES + R → E + S

• And is an Ordered Bi Bi Reaction

101

Ordered Bi Bi Reaction –Cleland Plot

102

• In the case where separate experiments about the same system give 2 different rate equations, e.g.,

k [E] [X] [Y] [EX]And

k [E] [X] [Y] [EY]

103

Mechanism = Random Sequential – Cleland Plot

104

• What, though, if an enzyme catalyzed a reaction that bound one substrate, released its product, then binds a SECOND substrate and releases ITS product?

Overall Reaction is: E + X + Y → E + R + S

105

Mechanism

E + X → EXEX → ER 1st rate limiting step

ER → E + RE + Y → EY

EY → ES 2d rate limiting stepES → E + S

106

• Note: while written unidirectionally, in many cases the reactions are reversible

• With 2 rate limiting steps, this reaction and its kinetics get ugly fast.

• This sort of reaction between 2 substrates and the 1 enzyme act like a ping pong game.

107

Ping Pong Mechanism –Cleland Plot

108

Function of Kinetics

• To Determine Reaction Mechanisms

reaction kineticsreaction kinetics. an introduction. 2 • a condition of equilibrium is reached in...

Documents