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Lecture 9Kjemisk reaksjonsteknikk

Chemical Reaction Engineering

o Review pervious lectures

o Pseudo Steady State Hypothesis (PSSH)o Hall of Fame Reaction (2NO+O2=2NO2)o Enzymes: Michealis-Menten Kineticso Enzyme Inhibition

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Quasi-equilibrium / rate determining step

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Elementary processes contd.Steady state assumption

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An active intermediate is a molecule that is in a highly energetic and reactive state It is short lived as it disappears virtually as fast as it is formed. That is, the net rate of reaction of an active intermediate, A*, is zero.

The assumption that the net rate of reaction is zero is called the Pseudo Steady State Hypothesis (PSSH)

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Example

CBA

The rate law for the gas phase reaction (without catalysts)

rA kC A

2

1 k C A

is found from experiment to be

How did this rate law come about? Suggest a mechanism consistent with the rate law.

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Nonelementary Rate Laws

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For reactions with active intermediates, the reaction coordinated now has trough in it and the active intermediate, A*, sits in this trough

A*k1

k2

k3

A+A B+C

A+A

+A

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The activation occurs when translational kinetic energy is transferred into internal energy (vibrational and rotational)

A*: free radical (unpaired electrons CH3·, ionic intermediate, enzyme-substrate complex, etc…..

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AAA*A 2 2k AAA CCkr *2*2

CB*A 3 3k *3*3 AA Ckr

A*AAA 1 1k 21*1 AA Ckr

Solution

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*3*3 AAB Ckrr

Need to find CA*

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C A* k1C A

2

k 3 k 2C A

*3*2*1** AAAAA rrrrr (5)

k1C A2 k 2C AC A* k 3C A* 0 (6)

C A*Solving for

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Pseudo Steady State Hypothesis r*A = 0

Substituting for in Equation of the rate of formation of B is

*AC

ACkkA

Ck

ACkkACkk

Br3/21

21

23

231

(8)

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Searching for a mechanism

Species having the concentration(s) appearing in the denominator of the rate law probably collide with the active intermediate,

A+A*→collision products

If a constant appears in the denominator, one of the reaction step is probably the decomposition of the intermediate

A* → decomposition product

Species having the concentration(s) appearing in the numerator of the rate law probably produce the active intermediate in one of the reaction step,

reactant → A*

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22 O2NONO CkCr

The reaction 2NO +O2 2NO2

has an elementary rate law

However… Look what happens to the rate as the temperature is increased.

Active Intermediates / Free Radicals (PSSH)

-rNO2

T11

Hall of Fame Reaction

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Why does the rate law decrease with increasing temperature?

Step 1 propose an active intermediate, NO3*Sept 2: Mechanism:

*3

k2 NOONO 1 (1)

2k*

3 ONONO 2 (2)

2k*

3 NO2NONO 3 (3)

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Write Rate of formation of Product rNO2

Note: k3 is defined w.r.t. NO2

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NONOkCCkr *33NONO33 *

3

21ONO11 ONOkCCkr2

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Step 3: write rate lawAssume that all reactions are elementary reactions, such that:

Step 4: Write rate of formation of product

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NONOkNOkONOkrNO*33

*3221*

3

321*3

rrrrNO

The net reaction rate for NO3* is the sum of the individual reaction rates for NO3*:

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Step 5 Write rate of formation of intermediate and use PSSH

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rNO3

* 0

NONOkNOkONOk *33

*32210

NOkkNOONOk 32*3210

NOkkONOkNO

32

21*3

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Pseudo Steady State Hypothesis (PSSH)The PSSH assumes that the net rate of species A* (in this case NO3

*) is zero.

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NOkk

ONOkNO32

21*3

Pseudo Steady State Hypothesis (PSSH)

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][32

22

312 NOkk

ONOkkrNO

Step 6: eliminate the intermediate concentration from the rate law

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Pseudo Steady State Hypothesis (PSSH)

22RT

EEE

2

312

2

2

31NO ONOe

AAAONO

kkkr

312

2

312 EEE

NOkk 32

The result shows why the rate decreases as temperature increases.17

-rNO2

T

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Enzymes Michealis-Menten Kinetics Lineweaver-Burk PlotEnzyme Inhibition Competitive UncompetitiveNon-competitive

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Kinetics of Bioreaction

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Enzymes

Michaelis-Menten Kinetics. Enzymes are protein like substances with catalytic properties.

Enzyme unease. [From Biochemistry, 3/E by Stryer, copywrited 1988 by Lubert Stryer. Used with permission of W.H. Freeman and Company.]

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Enzymes

It provides a pathway for the substrate to proceed at a faster rate. The substrate, S, reacts to form a product P.

A given enzyme can only catalyze only one reaction. Example, Urea is decomposed by the enzyme urease.

E S

SlowS P

Fast

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A given enzyme can catalyze only one reaction. Urea is decomposed by the enzyme urease, as shown below.

UREASECONHOHUREASECONHNH k 232*

22 23

EPES OH 2

22

*22221 UreaseCONHNHUREASECONHNH k

UreaseCONHNHUREASECONHNH k 22*

222

SESE k 1

SESE k 2

EPWSE 3k

The corresponding mechanism is:

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Michaelis-Menten Kinetics

WSEkr 3P

SEWkSEkSEk0r 321SE

WkkSk1

EE

32

1

t

WkkSEkSE

32

1

SEEE t

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Michaelis-Menten Kinetics

SKSEk

Sk

WkkSEWkWSEkr

m

V

tcat

K

1

32

t

k

33P

max

m

cat

SK

SVWSEkrm

max3P

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Vmax=kcat* Et

Turnover Number: kcatNumber of substrate molecules (moles) converted to productin a given time (s) on a single enzyme molecule

(molecules/molecule/time)For the reaction

40,000,000 molecules of H2O2 converted to product per second on a single enzyme molecule.

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H2O2 + E →H2O + O + Ekcat

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Summary

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(Michaelis-Menten plot)Vmax

-rs

S1/2

Solving:

KM=S1/2

therefore KM is the concentration at which the rate is half the maximum rateCS

Michaelis-Menten Equation

SKSVrr

M

maxSP

2/1M

2/1maxmax

SKSV

2V

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S1

VK

V1

r1

max

M

maxS

Inverting yields

Lineweaver-Burk Plot

slope = KM/Vmax

Intercept1/Vmax

1/S

1/-rS

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Types of Enzyme Inhibition

)inactive( EIIE

Competitive

)inactive( SEIISE

Uncompetitive

)inactive( SEIISE

)inactive( SEISEI

Non-competitive

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Competitive Inhibition

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k

kk

PESESE

SE3P CkrIEIE

IEIE EPSESESE SESE

1) Mechanisms:

Competitive Inhibition

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4

k

k

)inactive(IEI E

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2) Rates:

SE3SE2ES1SE CkCkCCk0r

4

5I

I

EIEI k

kK KCCC

m

ES

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ES1SE K

CCkkCCkC

m

ES3P K

CCkr

EI5EI4EI CkCCk0r

Competitive Inhibition

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I

I

m

S

EtotEEISEEEtot

KC

KC1

CC CCCC

SI

I

max

m

maxS C1

KC1

Vk

V1

r1

I

mISm

SEtot3P

KKCCK

CCkr

Competitive Inhibition

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I

ImS

SmaxS

KC1KC

CVr

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From before (no competition):Smax

m

maxS C1

VK

V1

r1

Intercept does not change, slope increases as inhibitor concentration increases

Competitive Inhibition

max

m

VKslope

max

1InterceptV

Sr1

SC1

No Inhibition

Competitive Increasing CI

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Competitive

SI

I

max

m

maxS C1

KC1

VK

V1

r1

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Uncompetitive Inhibition

PSESE 3

2

1k

k

k

SEkrr catSP

Inhibition only has affinity for enzyme-substrate complex

Developing the rate law

SEIkSEIkSEkSEkSEk0r 54cat21SE (1)

0r SEI SEIkSEIk 54 (2)

)inactive(SEISEI5

4

k

k

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Adding (1) and (2)

Mcat2

1

cat21

KSE

kkSEkSE

0SEkSEkSEk

M

catcatp

4

5I

MII5

4

KSEkSEkr

kkK

KKSEI

KSEISEI

kkSEI

From (2)

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E t E E S IE S

E 1S

K M

I S K IK M

rp kcat E t S

K M 1S

K M

I S K IK M

Total enzyme

IM

maxPS

KI1SK

SVrr

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Slope remains the same but intercept changes as inhibitor concentration is increased

Lineweaver-Burk Plot for uncompetitive inhibition

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1rS

1

Vmax S K M S 1I

K I

1rS

K M

Vmax

1S

1Vmax

1I

K I

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Both slope and intercept changes

SC1

Sr1

Increasing I

No Inhibition

E + S E·S P + E

(inactive)I.E + S I.E.S (inactive)+I +I-I -I

Noncompetitive Inhibition (Mixed)

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I

I

Smax

m

I

I

maxS

I

ISm

SmaxS

kC1

C1

Vk

kC1

V1

r1

kC1Ck

CVr

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