Chapter 20. Chemical Kinetics - Seoul National University · 2018. 4. 20. · Chapter 20. Chemical Kinetics Concentrations of Reactants and Products = function α, β, γ,… (time)

Chapter 20. Chemical KineticsConcentrations of Reactants and Products = functionα,β,γ,…(time)

The methods of monitoring the concentrations

1. Pressure change

2N2O5(g) → 4NO2(g) + O2(g) - constant-volume container

The pressure of the system increases during the course of reaction.

Inappropriate for the reactions that leave the overall pressure unchanged, and for reactions in solution.

2. Spectroscopy

H2(g) + Br2(g) → 2HBr(g)

By monitoring the intensity of absorption of light hν by the bromine, the progress of the reaction can be monitored.

http://bp.snu.ac.kr 1

toxic toxic

toxicodor

4. Electrochemical MethodsWhen a reaction changes the number or nature of ions present in a solution, its course may be followed by monitoring the electrical conductivity or pH of the solution.

5. Characterization Methodsmass spectroscopy charge/massemission spectroscopy hν, electron, ionnuclear-magnetic resonanceRaman spectroscopy hνinfra-red absorption x-ray absorption hνx-ray diffraction hνetc.

3. PolarimetryWhen the optical activity of a mixture changes in the course of reaction, it can be monitored by measuring the angle of optical rotation.


The Rates of Reactions

A+B → P

where [A], [B], [P]: the concentrations of the species, A, B, and P.

The rate of reaction

Methods of applying these analytical techniques1. Real-Time Analysis: the composition of the system is analyzed while

the reaction is in progress.

2. Quenching: the reaction is frozen after it has been allowed to proceed for a certain time, and then the composition is analyzed by any suitable techniques.

(Product)


[ ] [ ] [ ]dt

Bd

dt

Ad

dt

Pd−=−=

- - - - - - - - - - - - - - - - - - - - - -

Figure 20A.2. Real-Time AnalysisIn the stopped-flow technique, the reagents are driven quickly into the mixingchamber by the driving syringes, and then the time dependence of the concentrations is monitored.


P(t), T(t), V(t), Cα(t), Cβ(t), Cγ(t), …..

in situ in vivo

Steady State Quenching- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Figure 20A.3The definition of (instantaneous) rate as the slope of the tangent drawn to the curve showing the variation of concentration with time. For negative slopes, the sign is changed when reporting the rate, so all reaction rates are positive.

Rate of Reaction


[ ] [ ] [ ] [ ]dt

ddt

ddt

ddt

dv DC31B

21AD3C2BA

==−=−=

+→+

___________________

_________

20A.2(c). Determination of the Rate LawA direct method for determining the rate law from the raw kinetic data giving the concentration as a function of time is from the measurement of slopes.*

The reaction between A and B:

A plot of the rate against various concentrations of A and B gives the orders a and b from the slopes, and log k from the intercept.

A + B → P (Product)

* Only when the rate constant k is independent of concentration.

__________________


[ ] [ ] [ ] [ ]

[ ] [ ] [ ]

[ ] [ ] [ ]BlogAloglogAlog

BAA

PBAA

bakdt

d

kdt

ddt

dkdt

dv

ba

ba

++=

−

=

−

==−= Reaction RateReaction Order

Figure 20A.4The plot of log v0 against (a) log[I]0 for a given [Ar]0, and (b) log[Ar]0 for a given [I]0.

The reaction between A and B


( ) ( ) ( ) ( )[ ] [ ]ArI

ArIArI22

2

kv

gggg

=

+→+

[ ] [ ] [ ]

[ ] [ ] [ ]

[ ] [ ]

[ ]Blog

AloglogAlog

BAA

BAA

b

akdt

d

kdt

d

kdt

d

ba

ba

+

+=

−

=

−

=−

I:slope 2

Ar:slope 1

A+2B → 3C+D

The ambiguity in the definition of Reaction Rate is avoided if we define the rate of reaction v as

where vJ is the stoichiometric constant of substance J.

20B.1. The First-Order Reactions

(20A.3b)


[ ] [ ] [ ] [ ]dt

ddt

ddt

ddt

dv DC31B

21A

==−=−=

[ ]dt

dv

vJ

J1=

[ ] [ ][ ][ ]

[ ][ ][ ]

[ ]dtkd

dtkd

kdt

d

tt

∫∫ =

−

=−

=−

0 1

A

A

1

1

0 AA

AA

AA[ ] [ ]( )

[ ][ ][ ] [ ] ( )tk

tk

tk

t

t

t

10

10

10

expAAAAln

AlnAln

−=∴

−=

=−−

k1 = rate constant = independent of concentration

A → P____

______

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

The concentration of A falls exponentially with time with a rate determined by k1.

20B.2. The Second-Order Reactions A + A → P

the slope is second-order rate coefficient k2.


[ ] [ ] ( )tkt 10 expAA −=

[ ] [ ]

[ ][ ][ ]

[ ]dtkd

kdt

d

tt

∫∫ =

−

=−

0 2

A

A 2

22

0 AA

AA[ ] [ ]

[ ] [ ][ ]02

0

20

A1AA

A1

A1

tk

tk

t

t

+=

=−∴

[ ] and obained be should linestraight a ,against plotted isA1When t

t

First-Order Reactions

____- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Figure 20B.1The exponential decay of the reactant in a first-order reaction. The larger the rate constant, the more rapid the decay: here klarge = 3ksmall.

First-Order Reaction


log [A] vs. time

[ ] [ ] ( )tkt 10 expAA −=A → P

(IQ 50)

Figure 20B.2The determination of the rate constant of a first-order reaction: a straight line is obtained when ln [A] (or, as here, ln p) is plotted against t, and the slope gives k.

First-Order Reaction

= First-Order Rate Constant


1k

[ ] [ ] ( )[ ] ( )τ/expA

expAA

0

10

ttkt

−=−=

= Time Constant= Life Time= Relaxation Time

1

1

k=τ

Figure 20B.3The variation with time of the concentration of a reactant in a second-order reaction. The dotted line is the corresponding decay in a first-order reaction with the same initial rate. For this illustration, klarge = 3ksmall.

Second-Order Reaction

First-Order (dotted lines)

Second-Order


[ ]A1

[ ] [ ]reactionorder 2nd

A1

A1

20

−

=− tk

t

A+A → P [ ] [ ]22 AA

kdt

d=−

(IQ 50)

Table 20B.3 Integrated rate laws





20B.1. Half-Life and Time ConstantA simple indication of the rate of a chemical reaction is the time it takes for the concentration of a reagent to fall to half of its initial value: this is called the half-life of the reaction, and is denoted t1/2.

First-Order Reaction:

- Time Constant- Life Time- Relaxation Time

15http://bp.snu.ac.kr

[ ] [ ][ ][ ]

[ ][ ]

12/1

100

021

2/11

2/100

2lnAAln 2ln

AAln

at A21A

kt

tktk

t

t

=∴

−=−==−

→

1

1k

=τ

For a first-order reaction, one-half of the reactant disappears in t1/2 independent of the initial concentration.


12/1

2ln k

t =

[ ] [ ] ( )[ ] ( )τ/expA

expAA

0

10

ttkt

−=−=

-161017(월)-161019(수) 휴강

where A and Ea are Pre-Exponential Factor: Independent of temperatureActivation Energy: Determined from a plot of ln kr against 1/T

Approximately independent of temperature

20D. Temperature Dependence of Reaction RatesIn many experiments, the temperature dependence of the rate has been found to fit the expression proposed by Arrhenius:


__________

−=

RTEAk a

r exp

RT

EAk a

r −= lnln

Rate Constant

Activation Enthalpy≈ Activation Energy

[ ] [ ]barkv BA−= A + B → P Reaction Rate

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Rate Constant

Figure 20D.1A plot of ln k against 1/T is a straight line when the reaction follows the behavior described by the

Arrhenius-Type Temperature Dependence.


RTEAk a

r −= lnln

Figure 20D.3. A potential energyprofile for anexothermic (not always)reaction. The height of the barrierbetween the reactants and productsis the activation energy(activation enthalpy)of the reaction.

∆G = Gactivated – Greactant= ∆H – T ∆S= ∆E – T ∆S + P ∆V


***∆G

20C.1. First-Order Reactions Approaching Equilibrium- considering Backward Reaction

If the initial amount of A is [A]0, and if initially there is no B present, then [A]+[B]=[A]0 at all times:

The solution of this first-order differential equation:

If (no reverse reaction), the equation becomes

(20C.2)

(20C.4)


[ ] [ ] [ ]BAAbf kk

dtd

+−=

[ ] [ ] [ ] [ ]( ) ( )[ ] [ ]00 AAAAAAbbfbf kkkkk

dtd

++−=−+−=

[ ] [ ] ( )[ ]

++−+

=bf

bffbt

kk

tkkkkAA

exp0

0=bk[ ] [ ] ( )tkAA ft −= exp0

→ fk

←bk

B A

__________________________________________

ppt 20-9

Figure 20C.1The approach of concentrations to their equilibrium values as predicted by the equation for a reaction that is first-order in each direction, and for which .


A B

Sec. 20C.1

bf kk 2=

B A → fk←

bk

[A]+[B]=[A]0

What is the final state of the system?

(20C.6)Equilibrium Constant


________

Sec. 20C.1

∞→t[ ] [ ] [ ] [ ] [ ] [ ]

[ ][ ] b

f

bf

f

bf

b

kk

K

kkk

kkk

==

+=−=

+=

∞

∞

∞∞∞

AB

AAAB ,AA 0

00

(20C.4)

→ fk←

bk

A

B

[ ] [ ] ( )[ ]

++−+

=bf

bffbt kk

tkkkkAA

exp0

First-Order Reactions Approaching Equilibrium

The term relaxation denotes the return of a system to equilibrium. It is used in chemical kinetics to indicate that an externally applied influence has shifted the equilibrium position of a reaction, normally suddenly, and that the reaction is adjusting to the equilibrium composition characteristic of new conditions.

where x0 is the departure from equilibrium immediately after the temperature jump, and x is the departure from equilibrium at the new temperature after a time t.

20C.2. Relaxation Method

Consider


(First-Order)Temperature jump Suddenly from T1 to T2

at T2

→ fk

←bk

B A

bf

t

kkexx +==−

ττ 1 ,0

[ ] [ ] ( )[ ]

++−+

=bf

bffbt kk

tkkkkAA

exp0at T1

We write the deviation of [A] from its new equilibrium value as x, so and at T2. The concentration of A then changes as:

Since , its value may be combined with the relaxation-time measurement to find the individual kf and kb.http://bp.snu.ac.kr 24

[ ] [ ]eqbeqf kk BA =

[ ] [ ] xeq += AA [ ] [ ] xeq −= BB

[ ] [ ] [ ][ ]( ) [ ]( )

( )xkkxkxk

kkdt

d

bf

eqbeqf

bf

+−=

−++−=

+−=

BA

BAA

[ ]( )

( ) ( )

( )bf

ttkk

bfbf

bf

kkexexx

tkkx

xdtkk

x

dx

xkkdt

dx

dtdxdtAd

bf +===

+−=⇒+−=

+−=

=

−+−

ττ 1

where ,

ln

,// of Because

00

0

bf kkK ≈

[ ][ ] b

f

kk

K ==∞

∞

AB

Equilibrium Constantat T1

at T2 kf (T2) kb (T2)perturbation

________________________________________________________________________________at T2

Figure 20C.2The relaxation to the new equilibrium composition

when a reaction initially at equilibrium at a temperature T1 is subjected to a sudden change to temperature T2.


[A][A]eq at T2

at T1

Temperature Jumping Up

20E.2. Consecutive Reactions and the Steady State

23.5 min 2.35dayshalf-lives

We suppose that only A is present initially, and that its concentration is then [A]0.

If this result is inserted into the equation for B and the condition [B]0=0 imposed, we arrived at


(Assume no backward reaction)

(20E.4a)

(20E.4b)

(20E.4c)

CBA'11 →→ kk

PuNpU e.g. 23994

23993

23992 →→ ββ

[ ] [ ] [ ] [ ] [ ] [ ] [ ]BC ,B -AB ,AA '1

'111 k

dtdkk

dtdk

dtd

==−=

[ ] [ ] ( )tkt 10 expAA −=

[ ] [ ] ( )[ ] [ ] [ ] [ ]

[ ] [ ] [ ] [ ] [ ] ( )tktkttt

ttt

tktkt

ekekkk

eekk

k

'11

'11

1'1'

1100

0

1'1

10

11ABAAC

ACBA

AB

−−

−−

−

−

+=−−=

=++

−

−=

The solution for [B]

(1)

(2)

(3)

(4)

(4) → (2)

(5)


CBA'11 →→ kk

[ ] [ ][ ] [ ] [ ][ ] [ ]

[ ] [ ] ( )tk

kdt

d

kkdt

d

kdt

d

10

'1

'11

expAA

BC

BAB

AA

−=

=

−=

−=

[ ] [ ] ( ) [ ]BexpAB '1101 ktkk

dtd

−−=

(skip)

Rearranging eq (5)

(7)

From eqs (6) and (7)

(6)


[ ][ ] [ ]

[ ] [ ]BB

B

BLet

'1

'1

'1

'1

'1

'1

kdt

ddtdYe

ekdtBde

dtdY

eY

tk

tktk

tk

+=∴

+=

=

−

[ ] [ ] [ ] ( )tkkkdt

d101

'1 expABB

−=+

[ ] ( )

[ ] ( )

[ ] ( ) dtekdY

ekdtdY

tkkdtdYe

tkk

tkk

tk

1'1

1'1

'1

01

01

101

A

A

expA

−

−

−

=

=

−=[ ] ( ) [ ]

[ ] [ ] ( ) [ ]

[ ] [ ] ( )tktk

tkkk

tkkt

eekk

kkk

kekk

ke

kkke

kkkYY

'11

1'1

'1

1'1

1'1

01

1'1

01

1'1

01

1'1

01

1'1

010

AB

AA0B

AA

−−

−

−

−−

=

−−

−=−

−−

−=−

(skip)

(i) The three equations for [A], [B] and [C] indicate how to analyze a reaction scheme consisting of two consecutive first-order reactions.

(ii) “Rate-Determining Step” {Bottle-Neck Effect}If k1

’>>k1, whenever B molecules is formed, it decays quickly into C.

The formation of C depends only on the smaller rate coefficient, as anticipated. For this reason, the step with the slowest rate is called the rate-determining step of the reaction.


[ ] [ ] ( )

[ ] [ ] ( )tktk

tktkt

eekk

k

ekekkk

C

11

'11

1A 1A

1

1A

0'11

'1

0

1'1'

11

0

−−

−−

−≈

−+≅

−

−+=

[ ] [ ] ( )tkt e

kk'11AC

, If

0

1'1

−−≈

<<

CBA'11 →→ kk

병목현상

Sec. 20E.4

________________________________________________________________________________

____________________________________________________________________

Figure 20E.4(a) The first step is rate determining.(b) The second step is rate-determining.(c) Although one step is slow, it is not

rate-determining step because there is a fast route that circumvents it. Figure 20E.5. Reaction profile for a

mechanism in which the first step is rate-determining step.

http://bp.snu.ac.kr

SlowFast

FasterG

낙하산인사

x-axis:High (1023) Dimensional Space- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

in Series

The steady-state approximation assumes that during the major part of the reaction,

the concentrations of all reaction intermediates are approximately constant, and

the rates of changes of all reaction intermediates are negligibly small.

20E.3. Steady-State Approximation


A couple of kinetic steps can inducemathematical complexity.

Figure 20E.1The curves are plots with k1 = 10k1`. If the intermediate B is in fact the desired product, it is important to be able to predict when its concentration is greatest.


CBA '11 →→ kk

[A]

[B]

[C]

intermediates

Fig. 20E.2Steady-State ApproximationThe concentrations of intermediates remain small and hardly change during most of the course of the reaction.


Rate-Limiting Step↓

(20E.5)

____

CBA'11 →→ kk

[ ] [ ] [ ] [ ] ( )[ ][ ]

( )( )

[ ][ ]

( )[ ][ ] [ ] [ ] [ ]

[ ] [ ] [ ]

[ ] 0B

AAB

AAB ,AB ,1

1AB , If

1AB

AB ,AA

'1

1

'1

1'1

1'1

'1

11

'1

1'1

1

1'1

010

'1

'11

'111

≈∴

<<=

<<=∴=>>

−=>>

−−

=

−−

==

−

−

−−−

dtd

dtd

dtd

kk

dtd

kk

kktk

kekkk

ekk

k

eekk

ke

tk

tkk

tktktk

(ppt 20-27) [A]

[C]

[B]

-161024(월)-161026(수) 학회

20E.5. Pre-Equilibrium

Steady-state approximation

Therefore, the reaction has overall second-order kinetics.http://bp.snu.ac.kr 34

much faster too slow

The intermediate is in equilibrium with the reactants (approximately).

intermediate

(20E.12)

( ) C AB BA + → 2k←

−1k→ 1k

( )[ ] [ ][ ] ( )[ ]ABBAAB12 −−= kk

dtd

[ ][ ] ( )[ ]

( )[ ] [ ][ ]

( )[ ] [ ][ ][ ] [ ][ ]BA ]AB[ C

BAAB

BAAB

0ABBA

11

1

2

12

Kkkdt

dKkk

kk

==

=

=

≈−

−

− [ ][ ]

[ ][ ] b

f

k

kK ===

∞

∞

A

B

Reactant

Product

20F.1. UnimolecularReactions

Lindemann-Hinshelwood mechanism

If the unimolecular step is slow enough to be the rate-determining step, the overall reaction will have the first-order kinetics (by steady-state).

Rate=k[cyclo-propane]

(*)http://bp.snu.ac.kr 35

(not first-order yet) (20F.4)

22

2

CHCHCH

23 CH CHCH

]A[]A[ ],A[[P] PA

]A][A[]A[ 2A A A

]A[]A[ A AAA

'

2

∗∗

∗∗

∗∗

∗

∗∗

−==→

−=→+

=+→+

bbk

ak

ak

kdt

dkdt

d

kdt

d

kdt

d

b

'a

a

[ ]]A[

]A[]A[P]A[

]A[]A[ 0]A[]A][A[]A[]A[

'

2

'

2'2

ab

bab

ab

abaa

kkkkk

dtd

kkkkkk

dtd

+==∴

+=∴≅−−=

∗

∗∗∗∗

반응속도가단일반응물의 1차반응으로표시되는반응. 그러나실제로는타분자와의충돌에인한에너지이동을포함한복잡한과정이며,외견상간단한속도식으로표시되는경우가많다.________________________________________________________________________________

(skip---)

http://terms.naver.com/entry.nhn?docId=605500

http://terms.naver.com/entry.nhn?docId=607679

Figure 20F.1Unimolecular Reactions.

The species A is excited by collision with A, and the excited A molecule (A*) may either be deactivated by a collision with A, or go on to decay by a unimolecular process to form products.


(skip)

Lindemann-Hinshelwood mechanism:

As the concentration (and therefore the partial pressure) of A is reduced, the reaction should switch to overall second-order kinetics. The physical reason for the change of order is that at low pressures the rate-determining step is the bimolecular formation of A*.

(*)


_______________

______

(20F.7)

(20F.6)

]A[]A[]P[

'

2

ab

ba

kkkk

dtd

+=

],A[]P[ then ,]A[or ,]A[]A][A[ If '' kdt

dkkkk baba =>>>> ∗∗

' wherea

ba

kkkk =

( )∗=∴

<<

eq from ]A[]P[,]A[ If

2

'

a

ba

kdt

dkk

(skip)

38http://bp.snu.ac.kr

Problems from Chap. 20

E 20A.2(b) 20A.5(b)

D 20B.2

E 20B.4(b)

P 20B.16

D 20D.2

P 20E.3 (2017-)

Chapter 20. Chemical Kinetics - Seoul National University · 2018. 4. 20. · Chapter 20. Chemical Kinetics Concentrations of Reactants and Products = function α, β, γ,… (time)

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