L10 Nonelementary rxns

Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.

L10-1

Review: Multiple Rxns & SelectivityA B Ck1 k2

A

C

B

k2

k1

2) Series rxns

3) Complex rxns

1) Parallel / competing rxns

Desired product

A+B C+D A+C Ek1 k2

DD U

U

F molar flow rate of desired productSF molar flow rate of undesir

ExitE ed prx oductit

DD U

U

N moles of desired productSN moles of undes

Fiir

nalFi ed pr unal od ct

instantaneous rate selectivity, SD/U

DD U

U

rrate of formation of DSrate of formation of U r

overall rate selectivity, D US

DD

A

rrate of formation of DYrate of consumption of A r

instantaneous yield, YD

(at any point or time in reactor)

overall yield, DY

DD

A0 A

FYF F

flow

DD

A0 A

NYN N

batch

at exit

at tfinal

Maximize selectivity / yield to maximize production of desired product


L10-2Review: Maximizing SD/U for Parallel Rxns

E ED UD RT 1 2 1 2

D U A BU

AS e C CA

What reactor conditions and

configuration maximize selectivity?

a) If ED > EU

Specific rate of desired reaction kD increases:

Use higher temperature

b) If ED < EU

less rapidly with increasing TUse lower temperature(not so low

that the reaction rate is tiny)

more rapidly with increasing T

To favor production of the desired product

1 2 1 2a) 0

Now evaluate concentration:

→ Use large CA

1 2 1 2b) 0

→ Use small CA

1 2 1 2c) 0

→ Use large CB

1 2 1 2d) 0

→ Use small CB


L10-3DkD

A+BUkU

1 2

High CB favors

desired product

formation

1 2

High CB favors

undesired product

formation(keep CB

low)

1 2High CA favors desired

product formation1 2

High CA favors undesired product formation

(keep CA low)

PFR/PBR

Batch reactor

When CA & CB are low (end time or position), all rxns will be slow

High P for gas-phase rxn, do not add inert gas (dilutes reactants)

PFR/PBR w/ side streams feeding low CB CB

←High CA

Semi-batch reactor, slowly feed B to large amount of A

CB CB CBCSTRs in seriesB consumed before leaving CSTRn

CA0u0

CB0u0

CAu0

CBu0 CSTR

PFR/PBRPFR/PBR w/ high recycle

• Dilute feed with inerts that are easily separated from product

• Low P if gas phase

PFR/PBRSide streams feed low CA

←High CB

CASemi-batch reactor slowly feed A to large amt of B

CA CA CACSTRs in series


L10-4

The reactor V (for a given u0) and t that maximizes CB occurs when dCB/dt=0

k k1 A0B 1 21 2

2 1

k CdC k e k e 0d k k

t tt

1opt

1 2 2

k1 lnk k k

t 0

V tu

so opt 0 optV u t

AB

C

k1A A0C C e t

k k1 2B 1 A0

2 1

e eC k Ck k

t t

C A0 A BC C C C topt

Review: Reactions in Series: Cj & Yield

(desired) (undesired)A D U

k1 k2Time is the key factor here!!!

Spacetime t for a flow reactorReal time t for a batch reactor

To maximize the production of D, optimize the time (batch) or spacetime (flow)


L10-5

L10: Nonelementary Reaction Kinetics

• Nonelementary reaction kinetics: no direct correspondence between reaction order and stoichiometry– Nonelementary kinetics– Pseudo-steady-state hypothesis (PSSH)– Chain reactions: cracking ethane or polymerizations– Enzymatic reactions (covered in CHBE 471)– Bioreactors (covered in CHBE 471)

In practice, knowledge of the reaction mechanism helps use to design better catalyst, trouble shoot, aid in troubleshooting poor reactor performance

• Rate law is typically determined from experimental data

• Goal: Use the experimental rate law to postulate a reaction mechanism

• Elementary: the reaction orders and stoichiometric coefficients are identical


L10-6

Review (L4): Reversible ReactionskA

k AaA b B c C d D

KC: concentration equilibrium constant (capital K)

a b a bfA A A B fA A A Br k C C r k C C

At equilibrium, the reaction rate is zero, rA=0

Rate of disappearance of A (forward rxn):

c dbA A C Dr k C CRate of generation of A (reverse reaction):

A,net A fA bAr r r r

a b c dA A A B A C Dr 0 k C C k C C

c dC DA

Ca bA A B

C Ck Kk C C

Thermodynamic equilibrium relationship

a b c dA A B A C Dk C C k C C

a b c dA A A B A C Dr k C C k C C

consumed generated


L10-7

Determining Mechanism to Describe Nonelementary Reaction Kinetics

2 2Overall Reaction: 2NO O 2NO • If the reaction were elementary, the reaction kinetics would follow:

22

i NO ONO

ii NO

k C Cr

1 k C

22

NO 1 O NOr k C C

• Instead experiments show that the kinetics are:

• Nonelementary kinetics are the result of multiple elementary reaction steps and reactive intermediates (an intermediate that is so reactive it is consumed as fast as it is formed)

Nonelementary

How do we determine the mechanism?• Postulate a reaction mechanism that is a series of elementary reactions• Derive a rate equation for the postulated mechanism• Is the rate equation for the postulated mechanism consistent with the

experimental results?For example…


L10-8

Postulating a Reaction Mechanism1. If CB appears in the denominator of the rate law, then one elementary rxn

step is probably:*B A Collision products A* is a reactive intermediate

2. If the denominator contains a constant that is not multiplied by a concentration, then one rxn step is probably:

*A Decomposition products

3. If the numerator contains a species concentration, then one step is probably: *

speciesC other species? A other products?

22

i NO ONO

ii NO

k C Cr

1 k C

Apply:

2 2overall reaction : 2NO O 2NO Experimentally observed rate equation for

CNO in denominator: NO collides with reactive intermediate, NO3 3 2NO NO 2NO

CNO & CO2 in numerator: NO and O2 produce NO3 in one reaction step2 3NO O NO

Constant in denominator:NO3 produces NO and O2 (reverse of previous?)3 2NO NO O


L10-9

A Reaction Mechanism for Observed Kinetics?

22

i NO ONO

ii NO

k C Cr

1 k C

Apply:

2 2overall reaction : 2NO O 2NO Experimentally observed rate equation for

CNO in denominator: 3 2NO NO 2NO

CNO & CO2 in numerator: 2 3NO O NO

Constant in numerator: 3 2NO NO O

k12 3k 1

NO O NO

Postulated mechanism:

k23 2NO NO 2NO

Does this add up to the overall reaction?

2 2 2NO O 2NO yes

+

Now derive a rate equation for the postulated mechanism and check if it describes the experimentally observed rate equation

Reactive intermediate


L10-10

Postulated Mechanism for Nonelementary Reaction Kinetics

22

i NO ONO

ii NO

k C Cr

1 k C

Nonelementary kinetics, result of multiple elementary rxns & active intermediates

Postulated mechanism: 1

1

k2 3k

NO O NO

2k3 2NO NO 2NO


3 2 3NO 1 NO 1 NO O 2 NO NOr k C k C C k C C

• -rNO is in terms of CNO3, which is not measurable species because it is a reactive intermediate (so reactive it is consumed as fast as it is formed)

• Need to get CNO3 in terms of measurable species and plug into -rNO

2 3 3NO 1 NO O 1 NO 2 NO NOr k C C k C k C C Consumption of NO:(change signs)

NOr rxns that form NO - rxns that consume NO

Write –rNO for the postulated reaction mechanism


L10-11

Pseudo-Steady State HypothesisPostulated mechanism 1

1

k2 3k

1.) NO O NO

2k3 22.) NO NO 2NO

Reactive intermediate, must replace CNO3 in the rate equation

1) Write rNO3 2) Rearrange to get CNO3 in terms of measurable species3) plug eq for CNO3 back into -rNO

2 3 33NO 1 NO O 1 2 N ONO NOr k Ck C C k CC

CNO3 is very small, and NO3 is assumed to be so reactive that it is consumed as fast as it is formed, so NO3r 0

Pseudo-Steady State Hypothesis: Net formation of reactive intermediate ≈0

323 31 NO O 1 2 NONO NONO k C C k CkC Cr 0

2 3 3NO 1 NO O 1 2 NONO NOr k C C k Ck CC Factor out to simplify

32NO 1 NO O 2 NONO 1r k C C k C k C

Solve for concentration of reactive intermediate NO3 in terms of other species


L10-12

Concentration of Reactive IntermediatePostulated mechanism 1

1

k2 3k

1.) NO O NO

2k3 22.) NO NO 2NO


323 31 NO O 1 2 NONO NONO k C C k CkC Cr 0

2

31 NO O

NO1 2 NO

k C CC

k k C

32NO 1 NO O 2 NONO 1r k C C k C k C

22

1 NO O

1 2 NNO 1 NO O 2 NO 1

O

kr k C C k C

C

k k Ck

C

Now we will rearrange and simplify to see if it matches the experimental data

22

i NO ONO

ii NO

k C Cr

1 k C

Observed rate equation (nonelementary)

Plug CNO3 into -rNO

3 23NO NO1 2 NO 1 NO Ok k C k C CC C

23 1 2 N NNO O 1 O Ok k C k C CC

Solve for CNO3 in terms of other species


L10-13

Rearranging the Postulated Rate Eq.

Plug in CNO3

22NO 2 NO 1

1

1 NO O1 N

OO

NO

2

kr k C k

k k

C Ck

CC C

Factor out

2

2 NO1 NO

1NO

1 2 NOO

k C kr 1

k kk C

CC

21 NO Ok C C

21 2 NO 2 NO 1

NO 1 NO O1 2 NO 1 2 NO

k k C k C kr k C C

k k C k k C

22 NO

NO 1 NO O1 2 NO

2k Cr k C C

k k C

Common denominator

Add fraction

22

1 2 NO ONO

1 2 NO

2k k C Cr

k k C

Multiply

Postulated mechanism 1

1

k2 3k

1.) NO O NO

2k3 22.) NO NO 2NO


Conventional to reduce the additive constant in the numerator to 1

22

1 2 1 NO ONO

2 1 NO

2k k k C Cr

1 k k C


L10-14

Comparison Between Postulated and Experimental Rate Equation

22

i NO ONO

ii NO

k C Cr

1 k C

Compare rate eq for postulated mechanism to the experimental rate eq

Postulated mechanism 1

1

k2 3k

1.) NO O NO

2k3 22.) NO NO 2NO


Rate equation for postulated mechanism

Experimentally observed rate equation

22

1 2 1 NO ONO

2 1 NO

2k k k C Cr

1 k k C

Clicker Q: Are these rate equations the same?a) Yesb) No 1 2

i1

2k kk

k 2

ii1

kk

k

Yes, these are the same → postulated rate law explains the experimental data


L10-15

Chain Reaction

• A chain reaction consists of the following sequence:– Initiation

• formation of an active intermediate (radicals)– Propagation or chain transfer

• interaction of an active intermediate with the reactant or product to produce another active intermediate (a radical species)

– Termination• deactivation of the active intermediate

Common in radical polymerizations and cracking of ethane


L10-16

Free Radical Polymerizations (4 steps)

M MM

MM

MM-M-M

nmonomer polymer

1. Initiation:

2. Propagation:

3. Chain transfer:

4a. Termination by addition:

k02I 2I Initiator (I) decomposes to 2 free radicals

ki1I M R Radical (1)

kp1 2R M R

kpj j 1R M R

Chain elongation, new monomers add to chain

kmj j 1R M P R

“Live” polymer chain transfers radical to monomer. Polymer chain is no longer reactive (dead). Can also transfer to solvent or other species

kaddj k j kR R R

4b. Termination by disproportionation: kdisj k j kR R R R


L10-17

PSSH Applied to Thermal Cracking of Ethane

The thermal decomposition of ethane to ethylene, methane, butane andhydrogen is believed to proceed in the following sequence:

Initiation: k12 6 3C H 2CH • 1,C H 1 C H2 6 2 6r k C

Propagation: k23 2 6 4 2 5CH C H CH C H 2,C H 2 CH C H2 6 3 2 6r k C C

k32 5 2 4C H • C H H• 3,C H 3 C H2 4 2 5r k C

k42 6 2 5 2H• C H C H • H 4,C H 4 H C H2 6 2 6r k C C

Termination: k52 5 4 102C H C H 25,C H 5 C H2 5 2 5r k C

(a) Use the PSSH to derive a rate law for the rate of formation of ethylene(b) Compare the PSSH solution in Part (a) to that obtained by solving the complete setof ODE mole balance


L10-18

2,C H 2 CH C H2 6 3 2 6r k C C

The rate of formation of ethylene C H 3 C4 H2 52 Cr k

The net rates of reaction of reactive intermediates C2H5•, CH3•, H• are:

CH 1,C H 2,C H3 2 6 2 6r 2r r

H 3,C H 4,C H2 4 2 6r r r

Goal: replace CC2H5•

1,C H 1 C H2 6 2 6r k C

3,C H 32 4-r -k C H •2 5C4,C H 4 H C H2 6 2 6r k C C

25,C H 5 C H2 5 2 5r k C

C2H5• was formed in rxns 2 & 4, and consumed in rxns 3 & 5:

CH3• was formed in rxn 1 and consumed in rxn 2:

H• was formed in rxn 3 and consumed in rxn 4:

Plug rate eqs into eq above, assume rate = 0 (PSSH) & solve for reactive species

2

2 6

5C H3H

4 C H

kC

k C

C

C H 2,C H 3,C H 4,C H 5,C H2 5 2 6 2 4 2 6 2 5r r r r r

31

CH2

2kC

k

CH 1 C H 2 C H3 2 6 2CH 63r 0 2k C k C C

C H C H2 52

C 2H 2 C H 3 4 C H 52 5 2 6 2 6CH3 H 5r C C C Ck C k k C k 0

Need to replace CCH3• and CH•

1 C H 2 C C HH6 632 22k C k CC

1 C H2 6CH3

2 C H2 6

k C

k CC Do the same for CH•


L10-19

2,C H 2 CH C H2 6 3 2 6r k C C


The net rates of reaction of active intermediates C2H5•, CH3•, H• are (PSSH):

C H C H2 52


CH 1,C H 2,C H3 2 6 2 6r 2r r

H 3,C H 4,C H2 4 2 6r r r


1,C H 1 C H2 6 2 6r k C


25,C H 5 C H2 5 2 5r k C




Plug rate eqs into eq above, assume rate=0 (PSSH) & solve for reactive species

2

2 6

5C H3H

4 C H

kC

k C

C


Plug in expressions for CCH3• and CH• into C2H5• rate eq for C2H5• & solve for C2H5•

assuming rate = 0 (PSSH)3

1CH •

2

2kC

k

2C H 2 C H 3 4 C H 52 5 2 6C H2 5

C H C H2 53

4 C H2 62 52 6

1

2

kr k C k k C k

CC C

2kk

0Ck


L10-20

2,C H 2 CH C H2 6 3 2 6r k C C



C H C H2 52


CH 1,C H 2,C H3 2 6 2 6r 2r r

H 3,C H 4,C H2 4 2 6r r r


1,C H 1 C H2 6 2 6r k C


25,C H 5 C H2 5 2 5r k C




Plug rate eqs into eq above, assume rate=0 (PSSH) & solve for reactive species2

2 6

5C H3H

4 C H

kC

k C

C


31

CH •2

2kC

k

C H2 22

51C H C H2 5 2 5C H 3 4

3C H 5

4 C H6

62

22 6

2

2 kk C k k

CC

k Ck

Ck

k 0C

C H C H C H2 5 2 5 2 52

1 C H 3 3 52 62k C k kC k 0C C 21 C H 52 6 C H2 5C2k C k

1 C H 22 6C H2 5

5

2k C

kC 1 C H2 6

C H2 55

2k CC

k


L10-21

2,C H 2 CH C H2 6 3 2 6r k C C



CH 1,C H 2,C H3 2 6 2 6r 2r r

H 3,C H 4,C H2 4 2 6r r r


1,C H 1 C H2 6 2 6r k C


25,C H 5 C H2 5 2 5r k C




Plug expressions for reactive species into eq

for rC2H4 above

2

2 6

5C H3H

4 C H

kC

k C

C


31

CH •2

2kC

k1 C H2 6

C H2 55

2k CC

k

C H 3 C4 H2 52 Cr k 2 62 4C H 3

0.51

C H5

2kCk

kr =

The CC2H4 depends on CC2H6. How does CC2H6 change with time?

2 6 2 2 6 2 636 CHC H 1 C H 2 C H 4 •• H C H-r = k C + k C + k CC C Need to eliminate the reactive species


L10-22Rate of disappearance of ethane: 2 6 2 6 3 2 6 2 6C H 1 C H 2 CH • C H 4 H• C H-r = k C + k C C + k C C

The net rates of reaction of active intermediates CH3•, C2H5•, H• are (PSSH):

2 6

121

5

3H• 0.5

4 C H

k 1C2

kk

k C

2 6

2 62 6 2 6 2 6

C HC H 1 C H 2 C H

30.5

0.

41

5

C

1

2 H54

k C2k-r = k C + k C

k Ck+

2kk

k

Purpose: replace [CH3•] and [H•]

31

CH •2

2kC

k

C H 2,C H 3,C H 4,C H 5,C H2 5 2 6 2 4 2 6 2 5r r r r r 0

CH 1,C H 2,C H3 2 6 2 6r 2r r 0 H 3,C H 4,C H2 4 2 6r r r 0

3CH • 1 2C 2k k 2

2 6

5C H3H

4 C H

kC

k C

C

2 5 2 6

121

C H • C H5

2kC C

k

From the previous slide:

2 6 2 2 6 2 636 CHC H 1 C H 2 C H 4 •• H C H-r = k C + k C + k CC C

2 6 2 6 2 6

0.51

C H 1 C H 3 C H5

2k-r = 3k C + k C

k

2 6 2 6 2 6 2 6

0.51

C H 1 C H 1 C H 3 C H5

2k-r = k C + 2k C + k C

k


L10-23

2 4 2 6

121

C H 3 C H5

2kr = k C

k

For a constant-volume batch reactor :

For given initial concentration of C2H6 and temperature, these two equation can be solved simultaneously ( i.e., use Polymath)

Providing the concentration - time relationship using the PSSH

Other methods can also be used - solve the complete set of ODE mole balances

2 6 2 6 2 6

121

C H 1 C H 3 C H5

2k-r = 3k C + k C

k

42 6

2

12C H 1

35

C HdC 2k

= kdt k

C

26

62 6 2

121

1 C H 3 C H5

C H 2k= 3k C + k

d

k

C-

dtC


L10-24

1. Mole balances:

C2H6

11

dC= r

dt

CH3•

CH4

C2H5•

C2H4

2. Rate laws for each species:

1 1 1 2 1 2 4 1 6r = -k C - k C C - k C C

(Batch)

k12 6 3C H 2CH •

k23 2 6 4 2 5CH • C H CH C H •

k32 5 2 4C H • C H H•

k42 6 2 5 2H• C H C H • H

k52 5 4 102C H • C H

H2

33

dC= r

dt4

4dC

= rdt

55

dC= r

dt

77

dC= r

dt

22

dC= r

dt 2 1 1 2 2 1r = 2k C - k C C

24 2 1 2 3 4 4 1 6 5 4r = k C C - k C + k C C - k C

5 3 4r = k C

6 3 4 4 1 6r = k C - k C CH• 66

dC= r

dt

C4H108

8dC

= rdt

7 4 1 6r = k C C

28 5 4

1r = k C2

All these O.D.Es can besolved simultaneously

The comparisons of the results obtained from the two methods are shown on page 391 of textbook.The two results are identical, indicating the validity of the PSSH under these conditions

3 2 1 2r k C C


L10-25

Ozone (O3 ) Example Problem3 2Overall Reaction for Ozone Decomposition: 2O 3O

When ozone decomposes in the presence of an inert gas, M, the following kinetics are observed:

33

2 3

2O M

OO M O

kC Cr

C C k 'C

Postulate a reaction mechanism that is consistent with this rate law

This is provided as an extra example that can be skipped if there isn’t enough time to cover it in class.


L10-26

Since one oxygen atom is lost from the ozone molecule (O3), oxygen radicals (O•) are likely the reactive intermediate:

Postulating Mechanism for O3 Decomposition (Step 1)

1. If CB appears in the denominator of the rate law, then one elementary rxn step is probably: *B A Collision products

2. If the denominator contains a constant term, then one rxn step is probably:*A Decomposition products

3. If the numerator contains a species concentration, then one rxn step is probably: *


33

2 3

2O M

OO M O

kC Cr

C C k 'CApply:

3 2overall reaction : 2O 3OExperimentally observed rate equation for

CO2, CM & CO3 in denominator, so O2, M, and O3 must each collide a with reactive intermediate.

2 3 (Possible RxOOO n 1)

What is the reactive intermediate?

3 2O 2OO (PR2)

reactive intermediate

M M ?O (PR3)


L10-27

CM is multiplied by CO2 in the denominator, so combine PR3 with PR1

Postulating Mechanism for O3 Decomposition (Step 1 continued)


33

2 3

2O M

OO M O

kC Cr

C C k 'CApply:


CO2, CM & CO3 in denominator; they must collide with reactive intermediate O• 2 3 (possible rxOOO n 1) 3 2O 2OO (PR2)


M M ?O (PR3)

Wait, M is inert, so it cannot react with O• to create a new chemical species• An inert molecule can provide kinetic energy in another reaction→ M must participate in one of the other reactions Which one?

2 3 (PM OO O M R1b)


L10-28Postulating Mechanism for O3 Decomposition (Steps 2 & 3)


2. If the denominator contains a constant term, then one rxn step is probably:*A Decomposition products

3. If the numerator contains a species concentration, then one rxn step is probably: *


33

2 3

2O M

OO M O

kC Cr

C C k 'CApply:



2 3 (PM OO O M R1b) 3 2 (O 2O O PR2)Denominator doesn’t contain a constant, so step 2 isn’t in our mechanismBoth CO3 and CM appear in the numerator, so one reaction step may be:

3 2 (O M M O O PR4)Note that PR4 is the reverse of PR1


L10-29

Postulated Mechanism for Ozone Decomposition

33

2 3

2O M

OO M O

kC Cr

C C k 'CApply: 3 2overall reaction : 2O 3O

Experimentally observed rate equation for

CM and CO2 in denominator: 2 3M O OO M

CO3 in denominator: 3 2OO 2O

CO3 and CM in numerator: 3 2O M O OM

k13 2k 1

O M O OM Postulated mechanism:

k23 2O O 2O

Does this add up to the overall reaction?

3 2 2O 3O yes!

+

Now derive a rate equation for the postulated mechanism and check if it describes the experimentally observed rate equation



L10-30

Postulated Mechanism for Nonelementary Reaction Kinetics

33

2 3

2O M

OO M O

kC Cr

C C k 'C

Nonelementary kinetics, result of multiple elementary rxns & active intermediates


k1

3 2k 1O M M O O

k23 2O O 2O


3 3 2 3O 1 O M 1 M O O 2 O Or k C C k C C C k C C

• -rO3 is in terms of CO•, which is not measurable species because it is a reactive intermediate (so reactive it is consumed as fast as it is formed)

• Need to get CO• in terms of measurable species and plug into –rO3

3O 3 3r rxns that consume O - rxns that form O

Write –rO3 for the postulated reaction mechanism


L10-31

Pseudo-Steady State HypothesisPostulated mechanism:

1

1

k3 2k

1.) O M M O O 2k

3 22.) O O 2O

Reactive intermediate, must replace CO• in the rate equation

1) Write rO• 2) Rearrange to get CO• in terms of measurable species3) plug eq for CO• back into –rO3

3 2 3O 1 O M 1 M OO O O2r k C C k C C k CCC

CO• is very small, and O• is so reactive that it is consumed as fast as it is formed, so apply pseudo-steady state hypothesis: Or 0

3 2 3O1 O M 1 M O 2O Ok C C k C C k0 CCr

3 3 2 3O 1 O M 1 M O 2 OO Or k C C k C k CC CC Factor out to simplify

3 3 3 2O 1 O M 2 O 1 M O Or k C C k C k C CC

Put concentration of reactive intermediate O• in terms of other species


L10-32

Concentration of Reactive O•Postulated mechanism:

3 2 3O1 O M 1 M O 2O Ok C C k C C k0 CCr

3

2 3

1 O MO

1 M O 2 O

k C CC

k C C k C

3 3 3 2O 1 O M 2 O 1 M OOr k C C k C C Ck C

3 3 3 2

3

2 3O 1 O M 2 O 1 M

1 O M

1 M O 2 OOr k C C

k C C

k C C kk C C

Ck C

Now we will rearrange and simplify to see if it matches the experimental data

33

2 3

2O M

OO M O

kC Cr

C C k 'CObserved rate equation (nonelementary)

Plug CO• into –rO3

2 3 31 M O 2 O 1 O MOk C C k C k C CC

Solve for CO• in terms of other species

1

1

k3 2k

1.) O M M O O 2k

3 22.) O O 2O



L10-33

Rearranging the Postulated Rate Eq.

Plug in CO•

33 3 3 2

2 3

1 O MO 1 O M 2 O 1 M O

1 M O 2 O

k C Cr k C C k C k C C

k C C k C

Multiply outbrackets

3 2 3

3 32 3

2 21 2 O M 1 1 M O O

O 1 O M1 M O 2 O

k k C C k k C C Cr k C C

k C C k C

3 2 3

32 3

3

3

2

2

32

1 O2

1 2 O M 1 1 M O OO

1 M O

1 M 2 OM

2 1 M O O

O

O 2

k k C C k k C C Cr

k C C k C

kk C C

k C C k

C C

C

k C

3 23 33

2

23

3

2 21 2 O M 1 1 M

21 O O

O1 M

21

O

2 O1 O O

O

M M

2

k k C k k C C k k C C Cr

k C C k

C CC C

C

k k

Get common denominator

3

32 3

21 2 O M

OM O 2 O1

2k k C C

C C k Ckr


1

1

k3 2k

1.) O M M O O 2k

3 22.) O O 2O


Conventional to remove constant from 1st term

3

32 3

21 2 O M

OM

1

O 2 1 O

2k k C Cr

C C kk C

k


L10-34

Comparison Between Postulated and Experimental Rate Equation

33

2 3

2O M

OO M O

kC Cr

C C k 'C

Compare rate eq for postulated mechanism to the experimental rate eq

Rate equation for postulated mechanism

Experimentally observed rate equation

3

32 3

21 2 1 O M

OM O 2 1 O

2k k k C Cr

C C k k C

1 2

1

2k kk

k

2

1

kk

k '

These are the same → postulated rate law explains the experimental data


1

1

k3 2k

1.) O M M O O 2k

3 22.) O O 2O

Reactive intermediate, cannot appear in the rate equation

L10 Nonelementary rxns

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