8 chemical kinetics

8/8/2019 8 Chemical Kinetics

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8. Chemical Kinetics

objectives of chemical kinetics

1) Determine empirical rate laws

H2+ I2−→ 2 HI

How does the concentration of H2, I2, and HI change

with time?

2) Determine the mechanism of the reaction

H2+ I2−→ 2 HI

is a stoichiometric statement

reactants −→ intermediates −→ products

elementary reaction: the stoichiometric equation

reflects what actually happens to the molecules

mechanism: series of elementary reactions that

constitute the stoichiometric equation

PChem I 8.1


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example

2 H2+O2−→ 2 H2O

some intermediate steps:

H2+O2−→HO2+H

H2+HO2−→OH+H2O

OH+H2−→H2O+H

O2+H−→OH+O

H2+O−→OH+H

3) Empirical study of elementary reactions

important intermediates are often unstable, short-

lived species: H, OH, . . .

4) Predict rates of reactions

theoretical calculations

5) Chemical dynamics

study the dynamics of collisions and reactions of indi-

vidual molecules; high vacuum

PChem I 8.2


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reaction rates

2 NO+O2−→ 2 NO2

time dependence: −d[NO]

dt , −

d[O2]

dt ,

d[NO2]

dt

ambiguity: NO reacts twice as fast as O2

chemical reaction νJJ= 0νJ positive for products

νJ negative for reactants

extent of the reaction ξ

dn J≡ νJdξ

Using numbers of moles in the definition of the extent

of the reaction, we ensure that it is general and that

it applies even to non-constant volume reactions.

dn NO=−2dξ

dn O2 =−dξ

dn NO2 = 2dξ

PChem I 8.3


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reaction rate: ξ̇= dddξdddt

(rate of conversion)

dn J= νJdξ =⇒dn J

dt

=νJξ̇

ξ̇= 1

νJ

dddn J

dddt single rate for reaction

can follow the time-dependence of any component

If V = const, then the reaction rate can be expressed

by the rate of concentration change

v ≡ 1

V ξ̇ for V = const

v = reaction velocity

dn J

dt = νJξ̇

1V PChem I 8.4


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1

V

dn J

dt =

d(n J/V )

dt =νJ

V ξ̇ if V = const

v J≡d[J]

dt =

dc J

dt =νJ

V ξ̇= νJv reaction velocity of species J

ξ̇=V

νJv J

v J=νJv

rate law: v = f (concentrations)

If v depends on the concentration of some species

that does not appear in the stoichiometric equation,

then that species is called a catalyst if

v as [catalyst]

and an inhibitor if

v as [inhibitor]

If v depends on one or more products, then the re-

action is autocatalytic if v as [product], and the

reaction is self-inhibiting if v as [product].

PChem I 8.5


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often v = k [A]x [B] y [C]z · · ·

k : rate constant (depends on T ), k > 0

x , y ,z , . . .: order of the reaction, typically but not al-

ways integers

x + y + z +·· · = overall order

For elementary reactions, the x , y , . . . are inte-

gers.

A −→D v = k [A] unimolecular

A +B−→D v = k [A][B] bimolecular

A +B+C−→D v = k [A][B][C] termolecular (rare)

Reactions that behave kinetically just like a single-

step, elementary reaction are called kinetically sim-

ple: order of reactant Ji =νi . Such reactions are also

said to obey the Law of Mass Action action or to have

PChem I 8.6


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mass action kinetics. That means, the stoichiometric

equation looks like an elementary reaction, but gen-

erally it is not elementary.

Example:

H2+ I2−→ 2 HI

empirical rate law: v = k [H2][I2], i.e., the equationis kinetically simple and looks like a bimolecular el-

ementary reaction but

H2+Br2−→ 2HBr

empirical rate law:

v = k [H2][Br2]

3/2

[Br2]+k [HBr]

The reaction is first order in H2; no order is definablefor Br2 or HBr, and no overall order is definable.

The reaction is self-inhibiting.

PChem I 8.7


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Further examples of empirical rate laws:

C2H4O−→CH4+CO v = k [C2H4O]

CH3CHO−→CH4+CO v = k [CH3CHO]3/2

2 O3N2O5−−−−→ 3 O2 v = k [O3]

2/3[N2O5]2/3

2 SO2+O2Pt−→ 2 SO3 v = k [SO2][SO3]

−1/2

2 N2O Pt−−−−−→741◦C

2 N2+O2 v = k [N2O]

1+k [N2O]

Experimental determination of rate laws

One cannot measure v J directly.

measurements: [J](t); make connection between [J](t)

and ˙[J](t )

two possibilities: differentiate the data or integrate

the rate law

1) Differential method

A −→ P

PChem I 8.8


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d[A]

dt ≈

[A](t 2)− [A](t 1)

t 2− t 1=∆[A]

∆t

v A ([A])=−∆[A]

∆t average velocity

[A]=1

2

[A](t 1)+ [A](t 2)

If the rate law has a simple order, say v A = k [A]n

, thenlnv A = lnk +n ln[A]; so to determine n , plot lnv A ver-

sus ln[A].

Method has a major inherent difficulty:

d[A]

dt ≈∆[A]

∆t if ∆

t is small

but then ∆[A] is small also: division of two small num-

bers =⇒ need very high precision in the concentra-

tion measurement

2) Method of initial velocities (variant of the differen-

tial method)

example: say v A = k [A]x [B] y

PChem I 8.9


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t = 0 : v A,0= k [A]x 0[B]

y

0

lnv A,0= lnk +x ln[A]0+ y ln[B]0

keep [B]0 fixed and vary [A]0: slope of lnv A,0 versus

ln[A]0 is equal to x

keep [A]0 fixed and vary [B]0: slope of lnv A,0 versus

ln[B]0 is equal to y

problems with this method:

H2+Br2−→ 2HBr

v 0= k [H2]0[Br2]

3/20

[Br2]0+k [HBr]0 =0

= k [H2]0[Br2]1/20

•method does not work for multistep mechanisms

•method does not work for complex reactions with ini-

tial transients and induction periods

3) Method of integrated rate laws

d[J]

dt = v J=νJ f (concentrations)

PChem I 8.10


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rate law = differential equation

solve (integrate) the differential equation (analytical-ly or numerically); then fit data

Zero-order reaction

A k −→ P v = k

d[A]

dt = ν A v =−v =−k

d[A]=−k dt [A](t )[A]0

d[A]=−k

t 0

dt

[A](t )− [A]0=−kt

[A](t )= [A]0−kt

For a zero-order reaction, a plot of [A](t ) versus t

yields a straight line.

Note: Reactions cannot be of zero order for all times

— there are no true zero-order reactions — because

the reactant concentration cannot become less than

zero, which would occur at t = [A]0/k . A zero-order is

PChem I 8.11


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an approximation, which is valid only as long as the

reactant is in excess.

First-order reaction

A k −→ P v = k [A]

d[A]

dt = ν A v =−v =−k [A]

d[A]

[A]=−k dt [A](t )

[A]0

d[A]

[A]=−

t 0

k dt

ln[A](t )[A]0 =−kt [A](t )= [A]0 exp(−kt )

For a first-order reaction, a plot of ln([A](t )/[A]0) ver-

sus t results in a straight line with slope −k .

Second-order reaction

A k −→ P v = k [A]2

PChem I 8.12


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d[A]

dt = ν A v =−v =−k [A]

2

d[A]

[A]2 = −k dt

[A](t )[A]0

d[A]

[A]2 = −

t 0

k dt

− 1

[A](t )+

1

[A]0=−kt

[A](t )= 1

1

[A]0+kt

= [A]0

1+kt [A]0

For a second-order reaction, a plot of 1/[A](t ) versus t

results in a straight line with slope k .

Note: If the reaction is

A + A k −→ P v = k [A]2

then

d[A]

dt = ν A v =−2v =−2k [A]

2

PChem I 8.13


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and k has to be replaced by 2k in the expressions

above.

A +B k −→ P v = k [A][B]

d[A]

dt

= ν A v =−v =−k [A][B]

[A](t )= [A]0− z (t ), [B](t )= [B]0− z (t )

d[A]

dt =−

dz

dt

dz

dt = k ([A]0− z ) · ([B]0− z )

dz

([A]0− z ) · ([B]0− z )= k dt

z (t )0

dz

([A]0− z ) · ([B]0− z )= kt

z (t )

0

1

[B]0− [A]0 1

[A]0− z −

1

[B]0− z dz = kt 1

[B]0− [A]0

− ln

[A]0− z (t )

[A]0

+ ln

[B]0− z (t )

[B]0

= kt

PChem I 8.14


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1

[B]0− [A]0ln

[A]0 · [B](t)

[A](t ) · [B]0= kt If [B]0= [A]0, then

d[A]

dt =−k [A]2

Reactions of other orders

A k −→ P v = k [A]x

d[A]

dt = ν A v =−v =−k [A]

x

d[A]

[A]x = −k dt

[A](t )[A]0

d[A][A]x

= −t

0

k dt

[A](t )[A]0

d[A][A]−x =−

t 0

k dt

1

1−x [A](t )1−x − [A]1−x 0 =−kt , x = 1

1

x −1

1

[A](t )x −1−

1

[A]x −10

= kt , x = 1

PChem I 8.15


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For more complicated rate laws, use the method

of isolation or method of “flooding” . Focus on one

species; prepare all other species in large excess.

Example:

A +B k −→ P v = k [A][B]3/2

[B]0 [A]0 :

d[A]

dt =−k [A][B]3/2

−k [B]3/20 [A]

=−k eff [A] pseudo-first order

[A]0 [B]0 :

d[A]

dt =−k [A][B]3/2

−k [A]0[B]

3/2

=−k eff [B]3/2

pseudo-one-and-a-half order

PChem I 8.16


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Information on rate laws from half-life data

half-life t 1/2 : [A](t 1/2)= 12

[A]0

First-order reaction, A −→ P:

ln

12[A]0

[A]0 =−kt 1/2

t 1/2=ln 2

k = 0.693k −1

Second-order reaction, A −→ P:

1

2

[A]0= [A]0

1+kt 1/2[A]0

kt 1/2[A]0= 1

t 1/2= 1

k [A]0

Reactions of order x (x = 1), A −→ P:

t 1/2= 2x −1−1

(x −1)k [A]x −10

PChem I 8.17


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Temperature dependence of reaction rates

k = A exp− E

a

RT

Arrhenius law

E a activation energy; A pre-exponential factor

In the Arrhenius law, E a and A are considered to be

constants, though in applications they are often not

strictly constant but vary slowly with temperature.

if E a is large, then k is very sensitive to changes in

temperature

if E a is zero, e.g., radical recombination reactions,

then the rate is largely independent of temperature

Include backward reactions

Reversible first-order reaction

A k 1−−−−k −

1

B

Reversible has a different meaning in kinetics than in

thermodynamics!

PChem I 8.18


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All chemical reactions are reversible to some extent,

but if K > 100, this fact can usually be ignored for pur-

poses of chemical kinetics.

d[A]

dt = ν

f

A v f +ν

b A v b

ν f

A =−1, v f = k 1[A]

νb

A =+1, v b = k −1[B]d[A]

dt =−k 1[A]+k −1[B]

d[B]

dt =ν

f

Bv f +νb Bv b

d[B]

dt = k 1[A]−k −1[B]

Add the two rate equations:

d[A]

dt +

d[B]

dt =−k 1[A]+k −1[B]+k 1[A]−k −1[B]

ddt

[A]+ [B]

= 0

[A](t )+ [B](t )= const= [A](0)+ [B](0)=C

PChem I 8.19


20/49

[B](t )=C − [A](t )

d[A]dt

=−k 1[A]+k −1{C − [A](t )}

d[A]

dt = k −1C − (k 1+k −1)[A]

equilibrium:

d[A]eq

dt = 0

0=−k 1[A]eq+k −1[B]eq

[B]eq

[A]eq= k 1

k −1=K c

C − [A]eq

[A]eq= k 1

k −1=K c

[A]eq= k −1

k 1+k −1C

d[A]

dt = k −1C − (k 1+k −1)[A]

=−(k 1+k −1)

[A]−

k −1

k 1+k −1C

PChem I 8.20


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d[A]

dt =−(k 1+k −1)[A]− [A]eq[A](t )

[A]0

d[A]

[A]− [A]eq=−(k 1+k −1)t

ln

[A](t )− [A]eq

[A]0− [A]eq

=−(k 1+k −1)t

[A](t )− [A]eq

[A]0− [A]eq= exp−(k 1+k −1)t

[A](t )= [A]eq+[A]0− [A]eq

exp

−(k 1+k −1)t

relaxation methods: relaxation = return of a system

to equilibrium

Relaxation methods are used to measure rate con-

stants. Apply a sudden perturbation to the equilib-

rium, e.g., a temperature jump or a pressure jump,

and monitor the relaxation.

Let x be the deviation from equilibrium. Then for a re-

versible first-order reaction A −− B, our results show

PChem I 8.21


22/49

that

x (t )= [A](t )− [A]eq= x (0)exp(−t /τ)

with

τ= 1

k 1+k −1

i.e., after a temperature jump at t = 0, the system re-

laxes exponentially with a rate 1/τ = k 1+k −1. Com-

bine the measurement of the equilibrium constant

K c = k 1/k −1 with a measurement of the relaxation

time τ to obtain the forward and backward rate con-

stants k 1 and k −1.

Parallel first-order reactions

Assume the reactions are irreversible for the moment.

A k 1−→B

A k

2−→C

d[A]

dt =−k 1[A]−k 2[A]

PChem I 8.22


23/49

d[B]

dt = k 1[A]

d[C]

dt = k 2[A]

add the three equations:

d

dt [A]+ [B]+ [C]= 0[A](t )+ [B](t )+ [C](t )= const

= [A](0)+ [B](0)+ [C](0)= [A]0

if initially only the reactant A is present

conversion of reactant into products:

[A](t )= [A]0 exp{−(k 1+k 2)t }

d[B]

dt = k 1[A]= k 1[A]0 exp{−(k 1+k 2)t }

[B](t )= k 1[A]0

k 1+k 2 1−exp[−(k 1+k 2)t ][C](t )=

k 2[A]0

k 1+k 2

1−exp[−(k 1+k 2)t ]

PChem I 8.23


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•yield (t →∞)

Y B

irr=[B](∞)

[A]0=

k 1

k 1+k 2

Y C

irr=[C](∞)

[A]0=

k 2

k 1+k 2

•selectivity

S irr=[B](t )

[C](t )=k 1

k 2for all times

What happens if the reactions are reversible?

A k 1−−k −1

B

A k 2−−k −2

C

d[A]

dt =−k 1[A]+k −1[B]−k 2[A]+k −2[C]

d[B]

dt = k 1[A]−k −1[B]

PChem I 8.24


25/49

d[C]

dt = k 2[A]−k −2[C]

[A](t )+ [B](t )+ [C](t )= const= [A]0

can eliminate on variable, say [C]:

[C](t )= [A]0− [A](t )− [B](t )

Consider the case t →∞. The reaction reaches equi-

librium, i.e., the concentrations no longer change with

time:

d[A]

dt =−k 1[A]+k −1[B]−k 2[A]+k −2[C]= 0

d[B]dt

= k 1[A]−k −1[B]= 0

d[C]

dt = k 2[A]−k −2[C]= 0

k 1[A]eq= k −1[B]eq

k 2[A]eq= k −2[C]eq[B]eq

[A]eq= k 1

k −1=K c ,1

PChem I 8.25


26/49

[C]eq

[A]eq= k 2

k −2=K c ,2

[A]eq+ [B]eq+ [C]eq= [A]0

[A]eq+K c ,1[A]eq+K c ,2[A]eq= [A]0

[A]eq= [A]0

1+K c ,1+K c ,2

[B]eq= K c ,1[A]0

1+K c ,1+K c ,2

[C]eq=K c ,2[A]0

1+K c ,1+K c ,2

•yield (t →∞)

Y B

rev=[B]eq

[A]0=

K c ,1

1+K c ,1+K c ,2

Y C

rev=[C]eq

[A]0=

K c ,2

1+K c ,1+K c ,2

PChem I 8.26


27/49

•selectivity

S rev=[B]eq

[C]eq=K c ,1

K c ,2= k 1

k −1·k −2

k 2=k 1

k 2·k −2

k −1

S rev=S irr ·k −2

k −1

The selectivity

S =[B]

[C]=k 1

k 2=S irr

will be observed for short times, as along as the back-

ward reactions are negligible: The reaction is un-der kinetic control .

For long times, the selectivity approaches the equilib-

rium value

S =

[B]

[C] −−−−−→t →∞

K c ,1

K c ,2 =S rev

The reaction is under thermodynamic control .

PChem I 8.27


28/49

Example: k 1 = 1 s−1

, k −1 = 0.01s−1

, k 2 = 0.1s−1

, k −2 =

0.0005s−1

Initially B is formed ten times faster than C, however

K c ,1= 1 s−1

0.01s−1= 100

K c ,2= 0.1s−1

0.0005s−1= 200

and C is the more stable product.

1000 2000 3000 4000 5000 6000 7000

t

0.2

0.4

0.6

0.8

1.

conc

Figure 1: Parallel first-order reversible reactions. [A]:

solid line, [B]: dashed line, [C]: dot-dash line.

PChem I 8.28


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1 2 3 4 5 6 7 8 9 10

t

0.2

0.4

0.6

0.8

1.

conc

Figure 2: Parallel first-order reversible reactions: Ini-

tial stage. [A]: solid line, [B]: dashed line, [C]: dot–

dash line.

PChem I 8.29


30/49

0.01 0.1 1 10 100 1000 10000

t

0.2

0.4

0.6

0.8

1.

conc

Figure 3: Parallel first-order reversible reactions: con-

centrations versus log t . [A]: solid line, [B]: dashed

line, [C]: dot-dash line.

In the above parallel-reaction mechanism C can be

converted into B only via A . What happens if there is

the additional reaction

Ck 3−−k −3

B

(An example are the isomerization reactions be-

tween ortho-xylene, meta-xylene, and para-xylene:

PChem I 8.30


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o-xylene−− m-xylene, m-xylene−− p-xylene, p-xylene

−− o-xylene.) Let us assume that the three steps are

the mechanism of the reaction, i.e., all reactions are

elementary.

Is it then possible to establish equilibrium by the

cyclic reaction mechanism below?

A

B

C

d[A]

dt =−k 1[A]+k −2[C]

d[B]

dt = k

1

[A]−k −3

[B]

d[C]

dt = k −3[B]−k −2[C]

[A](t )+ [B](t )+ [C](t )= const= [A]0

stationary state:

k 1[A]= k −2[C] =⇒ [C]

[A]= k 1

k −2

PChem I 8.31


32/49

k 1[A]= k −3[B] =⇒ [B]

[A]= k 1

k −3

k −3[B]= k −2[C], which follows from the previous two equations

[B]= k 1

k −3[A]

[C]= k 1

k −2[A]

[A]+ k 1

k −3[A]+

k 1

k −2[A]= [A]0

[A]= [A]0

1+ k 1k −3+

k 1k −2

We have determined the stationary concentrations

[A], [B], and [C] in terms of [A]0 and the three rate

constants.

This stationary state is not an equilibrium state!! It

violates the principle of detailed balance.

The principle of microscopic reversibility :

In a system at equilibrium, any molecular pro-

PChem I 8.32


33/49

cess and the reverse of that process occur, on

the average, at the same rate.

implies the principle of detailed balance:

In a system at equilibrium, each collision has its

exact counterpart in the reverse direction, so

that the rate of every chemical process is ex-

actly balanced by that of the reverse process.

=⇒

In a system at equilibrium, every elementary forward

reaction is balanced by its own backward reaction.

In other words, in a system at equilibrium each ele-

mentary reaction must be individually at equilibrium.

Principle of detailed balance implies that all elemen-

tary reactions must be reversible.

Application to the scheme

A k 1−−k −1

B

PChem I 8.33


34/49

A k 2−−k −2

C

Ck 3−−k −3

B

equilibrium:

k 1[A]eq= k −1[B]eq =⇒ K c ,1=[B]eq

[A]eq= k 1

k −1

k 2[A]eq= k −2[C]eq =⇒ K c ,2=[C]eq

[A]eq= k 2

k −2

k 3

[C]eq= k

−3[B]

eq =⇒ K

c ,3=

[B]eq

[C]eq= k 3

k −3

The product of the second and third equilibrium con-

stants is

K c ,2K c ,3= [C]eq[A]eq

[B]eq[C]eq

= [B]eq[A]eq

=K c ,1

=⇒

PChem I 8.34


35/49

k 1

k −1= k 2

k −2

k 3

k −3

or

k 1k −2k −3= k −1k 2k 3

This condition on the six rate constants has been de-

rived by considering chemical equilibrium, but rate

constants are concentration independent. Therefore,

this relation is general for this mechanism; it also

holds for all nonequilibrium conditions. We see that

the rate constants can never be chosen independent-

ly, and the cyclic mechanism is impossible, since it

requires k −1 = k 2 = k 3 = 0, which is forbidden by the

relation.

Consecutive first-order reactions (series reactions)

A k 1−−→B

k 2−−→C

Assume the reactions are first order and irreversible.

d[A]

dt =−k 1[A]

PChem I 8.35


36/49

d[B]

dt = k 1[A]−k 2[B]

d[C]

dt = k 2[B]

Assume again [A](0) = [A]0, [B](0) = 0, [C](0) = 0, and

add the three equations:

[A](t )+ [B](t )+ [C](t )= const= [A]0

[A](t )= [A]0 exp(−k 1t )

d[B]

dt = k 1[A]0 exp(−k 1t )−k 2[B]

[B](t )= k 1[A]0

k 2−k 1{exp(−k 1t )−exp(−k 2t )}

[C]= [A]0− [A](t )− [B](t )

= [A]0

1+

1

k 2−k 1[k 1 exp(−k 2t )−k 2 exp(−k 1t )]

Consider the case: k 2 k 1.

Then

exp(−k 2t ) exp(−k 1t )

PChem I 8.36


37/49

and

k 2−k 1≈ k 2

[C]≈ [A]0

1+

1

k 2· (−k 2)exp(−k 1t )

= [A]0{1−exp(−k 1t )}

The formation of C depends only on the rate constantof the first step A −→B.

A −→B is the rate-determining step.

Motivation for the

Steady-State Approximation: After an initial induction

period, the concentration and the rate of change of all reaction intermediates are negligibly small.

Illustrate for the consecutive scheme:

[B](t )

[A](t )=

k 1[A]0

k 2−k 1{exp(−k 1t )−exp(−k 2t )}

[A]0 exp(−k 1t )

= k 1

k 2−k 1{1−exp[−(k 2−k 1)t ]}

PChem I 8.37


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For k 2 k 1:

[B](t )

[A](t )≈k 1

k 2{1−exp[−k 2t ]

≈0 for t 1/k 2

}=k 1

k 2

[B](t )=k 1

k 2[A](t ) [A](t )

d[B]

dt =

k 1

k 2

d[A]

dt

d[A]

dt

If we apply directly the steady-state approximation

(SSA), also known as the quasisteady-state approxi-

mation (QSSA), to the intermediate B, we obtain

0≈d[B]

dt = k 1[A]−k 2[B]

=⇒

[B]=k 1

k 2[A]

d[C]

dt = k 2[B]= k 2

k 1

k 2[A]= k 1[A]

PChem I 8.38


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= k 1[A]0 exp(−k 1t )

[C](t )= [A]0{1−exp(−k 1t )} (same as above)

Application: decomposition of ozone

2 O3−→ 3 O2

empirical rate law (M is an inert gas):

v =−1

2

d[O3]

dt

=1

3

d[O2]

dt

= k [O3]

2[M]

k

[O2][M]+ [O3]

proposed mechanism:

stoichiometric number

O3+Mk 1−−→O2+O+M 2

O2+O+Mk −1−−→O3+M 1

O+O3k 2−−→ 2 O2 1

PChem I 8.39


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2 O3+2 M+O2+O+M+O+O3−→

2 O2+2 O+2 M+O3+M+2 O2

2 O3−→ 3 O2

intermediate O: steady-state approximation

d[O]

dt = 0

Intermediates do not appear in the rate law!!

d[O]

dt = k 1[O3][M]−k −1[O2][O][M]−k 2[O][O3]= 0

[O]= k 1[O3][M]

k −1[O2][M]+k 2[O3]

d[O3]

dt =−k 1[O3][M]+k −1[O2][O][M]−k 2[O][O3]

=−k 1[O3][M]+k −1[O2][M]−k 2[O3]

[O]

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d[O3]

dt =−k 1[O3][M]+

+k −1[O2][M]−k 2[O3]

k 1[O3][M]k −1[O2][M]+k 2[O3]

= 1

k −1[O2][M]+k 2[O3]×

×−k 1[O3][M]k −1[O2][M]−k 1[O3][M]k 2[O3]++k −1[O2][M]k 1[O3][M]−k 2[O3]k 1[O3][M]

=−

2k 1k 2[O3]2[M]

k −1[O2][M]+k 2[O3]

=−2

k 1[O3]2[M]

k −1k 2

[O2][M]+ [O3]

=−2v empirical rate law:k = k 1, k = k −1/k 2

If initially we have pure ozone, then

k −1[O2][M] k 2[O3]

PChem I 8.41


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and

d[O3]dt

=−2k 1[O3][M]

=⇒ the first step is rate-determining

The reaction is self-inhibiting in O2, v as [O2]

Another approximation method

Pre-equilibrium or fast-equilibrium approximation

A +Bk 1−−−−k −1

X k 2−−→C

k 1 k 2, k −1 k 2

A , B, and X are in equilibrium:

k 1[A][B]= k −1[X]

[X]= k 1k −1

[A][B]=K [A][B]

(Note: K =K c !!)

PChem I 8.42


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d[C]

dt = k 2[X]= k 2K [A][B]= k [A][B]

k = k 2K =k 2k 1

k −1

enzyme reactions

Michaelis-Menten mechanism

E+Sk 1−−−−k −1

(ES)k 2−−→ P+E

d[P]

dt = v = k 2[(ES)]

Apply the steady-state approximation:

d[(ES)]

dt = k 1[E][S]−k −1[(ES)]−k 2[(ES)]= 0

[(ES)]=k 1[E][S]

k −1+k 2=

[E][S]

K M

K M=k −1+k 2

k 1Michaelis constant

PChem I 8.43


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d[P]

dt = k 2

K M[E][S]

conservation of total enzyme concentration:

[E]0= [E]+ [(ES)]

= [E]+

[E][S]

K M

= [E]

1+

[S]

K M

[E]= [E]0

1+ [S]K M

d[P]

dt = k 2

K M[E][S]

= k 2

K M·

[E]0

1+ [S]K M

[S]

d[P]

dt = v =

k 2[E]0[S]

K M+ [S]

Michaelis-Menten rate law

PChem I 8.44


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At high substrate concentrations, [S]K M,

d[P]dt

k 2[E]0[S]

[S]= k 2[E]0,

the reaction is zero order, and the rate reaches its

maximum:

v max= k 2[E]0

[Figure: rate versus substrate concentration; Atkins 9th ed., Fig. 23.3]

PChem I 8.45


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The Michaelis-Menten rate law can be written in terms

of v max:

d[P]

dt = v =

v max

1+K M/[S]

We rearrange this expression into a form that is

amenable to data analysis by linear regression:

1

v =

1

v max+

K M

v max

1

[S]

A Lineweaver-Burk plot is a plot of the reciprocal

of the velocity v versus the reciprocal of the sub-strate concentration [S] at fixed enzyme concentra-

tion. If the enzyme obeys Michaelis-Menten kinetics,

then the plot should yield a straight line, with slope

K M/v max, that intercepts the ordinate axis at 1/v maxand the abscissa at −1/K M.

PChem I 8.46


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[Figure: Lineweaver-Burk plot; Atkins 9th ed., Fig. 23.4]

At low substrate concentrations, [S]K M,

d[P]

dt

k 2[E]0[S]

K M= k [S],

the reaction is first order.

The turnover number or catalytic constant of an

enzyme, k cat, is the number of catalytic cycles

(turnovers) performed by the active site in a given

PChem I 8.47


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time interval divided by the duration of that inter-

val. It has units of a first-order rate constant, and for

the Michaelis-Menten mechanism is given by k 2, therate constant for release of product from the enzyme-

substrate complex.

k cat= k 2=v max

[E]0

The catalytic efficiency , η, of an enzyme is given by

η=k cat

K M

The higher the value of η, the more efficient is the

enzyme. For the Michaelis-Menten mechanism

η= k 2k −1+k 2

k 1

= k 1k 2

k −1+k 2

η→ ηmax= k 1 if k 2 k −1

= rate of formation of the enzyme-substrate complex;

diffusion-limited: < 108 – 109 M−1 s−1 for enzyme-sized

molecules at room temperature

PChem I 8.48


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catalase (decomposition of hydrogen peroxide) η =

4.0×108 M−1 s−1: “catalytic perfection”

8 chemical kinetics

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