physical chemistry chapter 11 3 atkins
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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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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.
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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 -
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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
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[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
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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]
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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).
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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.
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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
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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 .
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11.13 Enzymes
Ø A Lineweaver-Burk plot is a plot of 1/ υ against 1/[S] .
Slope = K M/υmax
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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 .
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- 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.
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- 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
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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
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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.
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11.13 Enzymes
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11.13 Enzymes
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11.13 Enzymes
(Self-test 11.6)
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11.13 Enzymes
Slope = 40.0
= K M / υ max
I ntercept = 4.0
= 1/ υ max
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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.
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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.
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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.
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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.
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11.14 The Structure of Chain Reactions
- 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.
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11.14 The Structure of Chain Reactions
(Self-test 11.7)
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11.14 The Structure of Chain Reactions
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
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11.14 The Structure of Chain Reactions
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.
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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.
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11.15 The Rate Laws of Chain Reactions
- 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
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- 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.
11.15 The Rate Laws of Chain Reactions
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11.15 The Rate Laws of Chain Reactions
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)
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11.15 The Rate Laws of Chain Reactions
(2) + (3) (4)
(5) (4) " (2)
(1) ! (2) and (5) "
e
e
1/2
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11.15 The Rate Laws of Chain Reactions
e
e
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- 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.
11.15 The Rate Laws of Chain Reactions
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Road map of key equations
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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