physical chemistry chapter 11 3 atkins

8/16/2019 Physical Chemistry Chapter 11 3 Atkins

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1

Homogeneous Catalysis

Ø A catalyst ( "

,#$

) is a substance that accelerates a reaction

but undergoes no net chemical change .

- The catalyst lowers the activation energy of the reaction by

providing an alternative path that avoids the slow, rate-determining

step of the uncatalyzed reaction.

- The decomposition of H 2 O 2 insolution, E a : 76 KJmol -1

I - added , E a : 57 KJmol -1

ó The rate constant increases by

order of 2000.

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2

Homogeneous Catalysis

Ø Enzymes (%&

), which are biological catalysts , are very selective

and can have a dramatic effect on the reactions they control .

- H 2 O 2 decomposition wi th enzyme catalyst, E a = 8 KJmol

-1

;acceleration of the reaction by a factor of 10 15 at 298K.

Ø A homogeneous catalyst( '()#$ ) is a catalyst in the same

phase as the reaction mixture .

- H 2 O 2 decomposition by Br -

or catalase

Ø A heterogeneous catalyst ( *'()#$

) is a catalyst in a

dif ferent phase from the reaction mixture .

- The hydrogenation of ethene to ethane, a gas phase reaction, a

solid catalyst (Pd, Pt, Ni) - The metal provides a sur face upon which the reactants bind; this

binding facil i tates encounters between reactants and increasing the

rate of the reaction.






3

11.12 Acid and Base Catalysis

Ø I n acid catalysis the crucial step is the transfer of a proton to the

substrate.

X + HA HX + + A -

HX + products

- Keto-enol toutomerism

CH 3 COCH 2 CH 3 CH 3 C(OH)=CHCH 3

Ø I n base catalysis a proton is transferr ed from the substrate to a base.

HX + B X - + HB +

X -

products

- The hydrolysis of esters

CH 3 COOCH 2 CH 3 + H 2 O CH 3 COOH (as CH 3 CO 2 - )

+ CH 3 CH 2 OH

H +

OH -






4

11.13 Enzymes

Ø The M ichaeli s-Menten mechanism

- The action of enzyme in an aqueous environment.

(1 ) E + S ES Rate of formation of ES = k a [E][S]

(2 ) ES E + S Rate of decomposition of ES = k a [ES]

(3) ES P + E Rate of formation of P = k b [ES]

Rate of consumption of ES = k b [ES]

- Rate of formation of P = k b [ES]

- Net rate of formation of ES

k a

[E][S] - k a

[ES] - k b

[ES] = 0

k a [E][S] ! (k a + k b ) [ES] = 0






5

[E ] : the molar concentration of the fr ee enzyme

[S] : the molar concentration of the substrate

[E] 0 : the total concentration of the enzyme

[E] + [ES] = [E] 0

11.13 Enzymes






6

11.13 Enzymes

k a [ES] + k b [ES] = k a [E] 0 [S] - k a [ES][S]

(k a + k b + k a [S]) [ES] = k a [E] 0 [S]

(K M + [S]) [ES] = [E ] 0 [S]






7

11.13 Enzymes

- The M ichaeli s-Menten rate law

Rate of formation of P = k b [ES] = k b [E] 0 [S]/([S]+ K M )

- The M ichaeli s constant, K M

- [E] 0 is the total concentration of enzyme (both bound and

unbound).






8

11.13 Enzymes

Ø The rate of enzymolysis is f irst-order in the added enzyme

concentration , but the effective rate constant k r depends on the

concentration of substrate [S].

- When [S] << K M , , k r =k b [S]/K M , the rate constant increases

l inearl y with [S] at low concentration.

Rate of formation of P = k b [E] 0 [S]/K M

- When [S] >> K M , , k r ~ k b , Rate of formation of P = k r [E] 0 = k b [E] 0.






9

Ø When [S] >> K M , , the rate is independent of the concentration of S

because there is so much substance present that it remains aneffectively the same concentration even though products are being

formed.

- The rate of formation of product is a maximum , and k b [E] 0 , is

called the maximum velocity , υ max , of the enzymolysis .

υ max = k b [E] 0

- The rate-determining step is ES P + E, because there is ample

ES present, and the rate is determined by the rate at which ES

reacts to form the products.

11.13 Enzymes






10

Ø The reaction rate υ at a general substrate composition i s related to

the maximum velocity, υ max = k b [E] 0

11.13 Enzymes

- The graph gives a clue to the

signif icance of K M , the rate of

enzymolysis reaches 1/2 υ max

when [S] = K M ; broadly

speaking, K M is a measur e of

the substrate concentration at

an above which the enzyme is

effective .






11

11.13 Enzymes

Ø A Lineweaver-Burk plot is a plot of 1/ υ against 1/[S] .

Slope = K M/υmax






12

11.13 Enzymes

- 1/ υ vs. 1/[S] -- a straight line

- The slope of the straight li ne : K M / υ max .

- The extrapolated intercept at 1/[S] = 0 : 1/ υ max .

- The intercept can be used to fi nd υ max , and the value combined

with the slope to find the value of K M .

- The extr apolated in tercept with the horizontal axis (where 1/ υ = 0)

occurs at 1/[S] = -1/K M .






13

- The time constant f or the first-order decomposition of ES is 1/k b .

- The number of catalytic events in an interval "t is

"t /(1/k b ) = k b "t

- The frequency of such event is this number divided by the interval

"t , which i s k b itself .

k cat = k b

11.13 Enzymes

Ø The turnover frequency , or catalytic constant , of an enzyme, k cat , is

the number of catalytic cycles (turnovers) per formed by the active

site in a given interval divided by the duration of the interval.






14

- An alternative interpretation of k cat is a measure of theeffectiveness of an active site , as it is the maximum rate of

enzymolysis that can be achieved divided by the concentrations of

active sites .

11.13 Enzymes






15

Ø The catalytic eff iciency , η , of an enzyme is the ratio.

11.13 Enzymes

- k cat is a number of the effectiveness of an enzyme and K M is a

measure of the concentration at which the enzyme becomes

effective .

- The higher the value of η , the more eff icient is the enzyme.

- We can think of the catalytic eff iciency as the effective rate

constant of the enzymatic reaction.

K M = (k a + k b )/k a






16

11.13 Enzymes

- The eff iciency reaches its maximum value of k a when k b >> k a .

- Because k a is the rate constant for the formation of a complexfrom two species that are dif fusing freely in solution, the

maximum eff iciency is related to the maximum rate of di ffusion of

E and S in soluti on.

- I n th is limit, rate constants are about 10 8

~ 10 9

dm 3

mol -1

s -1

formolecules as large as enzyme at room temperature.

- The enzyme catalase has η = 4.0 x 10 8 dm 3 mol -1 s -1 and as said to

have attained catalytic perfection.

- The rate of the reaction i t catalyzes is controlled only by diffusion :

i t acts as soon as a substrate makes contact.






17

11.13 Enzymes






18

11.13 Enzymes






19

11.13 Enzymes

(Self-test 11.6)






20

11.13 Enzymes

Slope = 40.0

= K M / υ max

I ntercept = 4.0

= 1/ υ max






21

11.13 Enzymes

Ø The action of an enzyme may be partial ly suppressed by the present

of a foreign substance , whi ch is call ed an inhibitor .

- An inhibitor may be a poison that has been admini stered to the organi sm, or it may be a substance that is naturally present in a

cel l and involved in its regulatory mechanism.

Ø I n competiti ve inhibition , the inhibitor competes for the active site

and reduces the abil ity of enzyme to bind the substance.






22

11.13 Enzymes

Ø I n non-competi tive inhibition , the inh ibitor attaches to another part

of the enzyme molecule, thereby distorting i t and reducing i ts abil i tyto bind the substrate.






23

Chain Reactions

+,-

Ø I n chain reaction, an intermediate produced in one step generates

a reactive intermediate in a subsequent step, then that intermediate

generates another reactive intermediate and so on.






24

11.14 The Structure of Chain Reactions

Ø The intermediates responsible for the propagation of a chain

reaction are called chain carr iers .

- radicals , ions , neutrons

- Radicals are marked with a dot to signify the unpaired electron .

Ø A radical chain reaction typicall y involves some but not necessari ly

all of the fol lowing steps.

- Initiation , in which radicals are formed from non-radical

molecules.

- Propagation , in which a radical attacks a molecule and forms a

new radical.

- Branching , in which the attack by a radical produces two radicals.






25


- Retardation , in which a product molecule is destroyed by a radical.

- Inhibition , in which radicals are removed other than by

termination; perhaps by reaction with the wall s of the reaction

vessel or an added agent.

- Termination , in which two radicals combine to form a molecule

and so end the chain.






26


(Self-test 11.7)






27


I nitiation : CH 3 CH 3 2 CH 3

Propagation : CH 3 + CH 3 CH 3 CH 4 + CH 2 CH 3

Termination : CH 2 CH 3 + CH 2 CH 3 CH 3 CH 2 CH 2 CH 3

I nhibition :

CH 2 CH 3 to react with the wall of the reaction vessel.

Retardation : CH 3 CH 2 CH 2 CH 3 + CH 2 CH 3 CH 2 CH 2 CH 2 CH 3

+ CH 3 CH 3






28


Ø The NO molecule has an unpaired electron and is a very eff icient

chain inh ibitor .

- The observation that a gas-phase reaction i s quenched when NO is

introduced is a good indication that a radical chain mechanism is

in operation.






29

11.15 The Rate Laws of Chain Reactions

Ø The thermal r eaction of H 2 with Br 2 .

- The overall reaction and the observed rate law

H 2 (g) + Br 2 (g) 2 HBr

- The complexity of the rate law suggests that a compl icated

mechanism is involved.






30


- The radical chain mechanism

I nitiation : Br 2 Br + Br

Rate of consumption of Br 2 = k a [Br 2 ]

Propagation : Br + H 2 HBr + H υ = k b [Br] [H 2 ]

H + Br 2 HBr + Br υ = k c [H] [Br 2 ]

Retardation : H + HBr H 2 + Br υ = k d [H] [HBr]

Termination : Br + Br + M

Br 2 + M υ = k e [Br] 2

k a

k b

k c

k d

k e






31

- The # thi rd body $ , M , a molecule of an inert gas removes the

energy of recombination.

- The constant concentration of M has been absorbed in to the rate

constant k e .

- Other possible termination steps include the recombination of H

atoms to form H 2 and the combination of H and Br atoms; however,

i t turns out that only Br atom recombination is important.







32


Ø The rate law of a chain reaction

- The exper imental rate law is expressed in terms of the rate of

formation of products, HBr .

Net rate of formation of HBr

= k b [Br] [H 2 ] + k c [H] [Br 2 ] ! k d [H] [HBr] (1)

- Set up the expressions for the net rates of formation of [H ] and [Br]and apply the steady-state assumpti on to both.

Net rate of formation of H

= k b [Br] [H 2 ] - k c [H][Br 2 ] ! k d [H ][HBr] = 0 (2)

Net rate of formation of Br

= 2k a [Br 2 ] - k b [Br] [H 2 ] + k c [H] [Br 2 ] + k d [H] [HBr] ! 2k e [Br] 2 = 0

(3)






33


(2) + (3) (4)

(5) (4) " (2)

(1) ! (2) and (5) "

e

e

1/2






34


e

e






35

- The proposed mechanism is at least consistent wi th the observedrate law.

- Additional support for the mechanism would come from the

detection of the proposed intermediate (by spectroscopy), and the

measurement of individual rate constants for the elementary stepsand conf irming that they correctly r eproduced the observed

composite rate constants.







36

Road map of key equations






37

Homework (Chapter 11)

11.2

11.4

11.5

11.8

11.9

11.12

11.20

11.21

11.23

11.24

11.26

11.27

11.29

11.30

11.31

physical chemistry chapter 11 3 atkins

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