大一課件-大一上化學chapter

Chapter

FiveRates

of

Chemical

ReactionChemicalReactionWeather

can

occur

?---chemical

thermodynamicsHow

fast

(reaction

rate)?---chemical

kinetics化學反應速度的快慢——主要決定于反應的內在機理。反應機理：化學反應所經(jīng)歷的途徑或具體步驟，又稱為反應歷程?；瘜W動力學的基本任務就是研究反應的機理。某一反應究竟是經(jīng)過哪些步驟完成的，了解各個步驟的特征和相互聯(lián)系，揭示化學反應速度的本質，使人們能夠自覺控制反應速度。反應機理可以告訴：問題為什么有些口服藥物的服用方法是2次/天？3次/天?

為什么靜脈滴注某些青霉素類藥物要加快滴注的速度？)Reaction

rate:

changes

in

a

concentrationof

aproduct

or

a

reactant

perunit

time.tv

cThe

rate

is

defined

to

be

a

positive

number.The

unit

of

reaction

rate

is

mol·L-1·S-1,mol·L-1·min-1

or

mol·L-1·h

-1

et

al5-1 Rates

and

Mechanisms

of

Chemical

Reactionsv

c

(5

-

4)taverage

reaction

rate

is

obtainedby

dividing

the

change

inconcentration

of

a

reactant

orproduct

by

the

time

interval

overwhich

the

change

occursDefine

reaction

rate

:2N2O5=

4NO2

+

O21.

Average

reaction

rateSAMPLE

EXERCISE

:H2O2

=

H2O

+

1/2

O20〃:

c1(H2O2)

=

15.88×10-3

mol/L5〃

:

c2(H2O2)

=

12.8×10-3

mol/LSo:

2 1

0.6103

(mol

L1

s1)t2

t1c

c

c

3.0103t

5v(H2O2

)

2

1

0.3103

(mol

L1

s1

)t2

t1c

c

c

1.5103t

5v(O2

)

Reaction

Rates

and

StoichiometryH2O2

=

H2O

+

1/2

O2rate

[H

2O2

]

[H

2O]

1

[O2

]t

t

2

t用不同物質濃度變化所表示的反應速率之間存在著一定的關系是：它們之間的速率比正好等于反應式中各物質分子式的系數(shù)之比，To

generalize,

for

the

reactionaA

+

bB

cC+

dDThis

equation

can

be

used

to

establish

therelationship

betweenrate

of

change

of

onereactant

or

product

to

another

reactant

or

product.You

have

to

be

able

to

do

this

on

the

test,

too!2.

Instantaneous

reaction

rateThe

instantaneous

rateofchangeat

a

pointisthe

same

as

theslope

ofthe

tangentline.

That

is,it's

the

slope

of

acurve.=

limΔcΔt0Δtdcdt—

=

cvdtv

d[O2

]dtv

d[H2O2

]Define

reaction

rate

:2N2O5=

4NO2

+

O25-1.2

The

Mechanisms

of

Chemical

ReactionsA

reactionmechanism

is

a

description

of

thepaththatareactiontakes.elementary

reaction？overall

reaction？types

of

elementary

reactions5-1.2

The

Mechanismsof

Chemical

Reactionselementary

reaction:

is

one

that

the

reactants

canconvert

directly

into

the

products

in

a

single

stepwhen

they

act

each

other.

(one-step

reaction)基元反應：反應物間相互作用直接轉化為生成物的反應。（簡稱元反應）。Cl2

M

2Cl

MCl

H

2

HCl

HH

Cl2

HCl

Cl2Cl

M

Cl2

M例如：overall

reaction:Many

reactions

areactually

made

up

of

severalelementary

steps,

which

are

combined

toyieldthe

overall

reaction.NO2

+CO

→ NO

+

CO2NO2

+NO2

→

NO3

+NONO3

+CO →

NO2

+CO2(slow)(fast)This

means

that

the

rate

of

the

overall

reaction

isdominated（控制）

by

the

rate

of

the

reaction,this

is

the

rate-determining

step.H2(g)

+

I2(g)

2HI(g)I2(g)

→

2IH2+

2I→

2HI(fast)(slow)——rate-determining

step.Types

of

Elementary

Reactionsunimolecular

reaction:

an

elementary

reaction

in

whichthe

rearrangement

of

a

single

molecule

produces

one

ormore

molecules

of

product.I2(g)

─→

2I(g)bimolecular

reaction:

the

collision

and

combinationof

two

reactants

to

give

an

activated

complex

in

anelementary

reaction.NO(g)

+

O3(g)

─→

NO2(g)

+

O2(g)molecularity

--for

ElementaryReactions:_

termolecular

reaction:

an

elementary

reactioninvolving

the

simultaneous

collision

of

anycombination

ofthreemolecules,

ions,oratoms.2

NO+H2

─→

N2

+H2O25-2

Theories

of

ReactionRateOne

is

thecollision

theory:The

collision

theory

is

based

on

the

kinetictheory

and

assumes

a

collision

between

reactantsbefore

a

reaction

can

take

place.1918年Lewis以氣體分子運動論為基礎提出Another

is

the

transition

state

theory:The

transition

state

theory

suggests

that

asreactant

molecules

approach

each

other

closely

theyare

momentarily

in

a

less

stable

state

than

either

thereactants

or

the

products.Contents

of

Collision

Theory:⑴

reacting

molecules

must

comeso

closethat

theycollide.⑵

"

effective"

collisions

:

not

everycollisionbetween

molecules

createsproducts,only

few

collisions between

reactantmolecules

willreact.5-2.1

Collision

Theory

and

ActivationEnergyContents

of

Collision

Theory:According

tothistheory,product

formationcan

onlytake

placewhen

there

are"effective"collisionsbetween

reactant

molecules

involved

inthe

ratedetermining

stepof

theprocess.Br(g)

+

HI(g)

─→

HBr(g)

+

I(g)Straight

on

collision,

hydrogen

facinging

bromine.

Reactionoccurs.Straight

on

collision,

hydrogen

facingawayfrom ing

bromine.

Reaction

does

notoccur.Straight

on

collision,

bromine

at

90degrees.Reaction

does

not

occur.Whatconstitutesaneffectivecollision?發(fā)生有效碰撞的兩個基本前提:enough

energyproper

orientation對HCl

和NH3

的氣相反應Activation

molecule:is

the

molecule

haveenough

energyandcan

produce

effective

collisionthe

average

moleculesmust

absorbsome

energy

to eactivationmolecules活化分子一般只占極少數(shù)，它具有的最低能量為Ec。通常把活化分子具有的平均能量與反應物分子的平均能量之差稱為反應的活化能，用符號Ea表示。Ea

Ec

Ek在一定溫度下，反應的活化能越大，活化分子的分子分數(shù)越小，活化分子越少，有效碰撞次數(shù)就越少，因此化學反應速率越慢；反應的活化能越小，活化分子的分子分數(shù)越大，活化分子越多，有效碰撞次數(shù)就越多，化學反應速率越快。Figure:

As

the

activation

energy

of

a

reaction

decreases,

thenumber

of

moleculeswith

at

least

this

muchenergyincreases,

as

shown

by

the

yellow

shaded

areas.一般認為Ea小于63

kJ·mol-1的為快速反應小于40

kJ·mol-1和大于400

kJ·mol-1的都很難測定出▲每一反應的活化能數(shù)值各異,可以通過實驗和計算得

到。活化能越小,反應速率越快。Ea是動力學參數(shù)?！恍┓磻腅a2SO2(g)+

O2(g)=2SO3(g),N2(g)

+3H2(g)

=2NH3(g),HCl

+

NaOH

→NaCl+H2O,▲離子反應和沉淀反應的Ea都很小Ea=251

kJ·mol-1Ea=175.5

kJ·mol-1Ea≈20

kJ·mol-1的特征活化能Character

of

activation

energy活化能的特征Character

of

activation

energyr

H

=

-199.6

kJ

?

mol-1O3

(g)

+

NO(g)

=

NO2

(g)

+

O2

(g)★反應物的能量必須爬過一個能壘才能轉化為產(chǎn)物★

即使是放熱反應

(△rH為負值)，外界仍必須提供最低限度的能量，這個能量就是反應的活化能energyEa

與△rH

的關系energy5-2.2

The

Transition

State

TheoryTransition

state

theory

(TST)

is

also

calledactivated

complextheory.reactants

pass

through

high-energy

transitionstates

before

forming

products,

they

are

associated

inan

unstable

entity

called

an

activated

complex,then

change

into

products.要點：化學鍵重排、活化絡合物形成Example:NO2(g)

+

CO(g)

─→

NO(g)

+

CO2(g)ΔH

=Ea,f

-

Ea,r=358

kJ·mol-1

-132

kJ·mol-1ΔH

=

–226kJ·mol-1.E

+

S

=E-S

P

+

E許多實驗事實證明了E－S復合物的存在。理論受限活化絡合物的結構無法在實驗中加以確定計算過于復雜酶作用的機制——中間產(chǎn)物學說酶與底物形成酶－底物中間復合物，中間復合物再分解成產(chǎn)物和酶。Factors

That

Affect

Reaction

RatesConcentration

of

ReactantsTemperatureCatalysts

Speed bychangingmechanism5-3 Reaction

Rates

andConcentrationsChemical

reactionsare

fasterwhentheconcentrations

of

the

reactants

are

increased

.Because

more

molecules

will

exist

in

a

givenvolume.

More

collisions

will

occur

and

the

rateofa

reaction

will

increase.Concentration

and

RateEach

reaction

has

its

own

equationthat

gives

itsrateasafunctionofreactantconcentrations.this

is

called

its

Ra

awA

ra aw

shows

the

relationship

between

thereaction

rate

and

the

concentrations

of

reactants.說明在一定溫度下，反應速度與反應物濃度之間的定量關系——質量作用定律5-3.1

The

Ra

awaA

+

bB

→

cC

+

dDv∝[A]m[B]nv

=

k[A]m[B]n反應速率方程或稱質量作用定律The

rate

of

a

reaction

is

proportional

to

theproduct

of

the

concentrations

of

the

reactantsraised

to

some

power.v=

k[A]m[B]nk

is

a

rate

constant

that

has

a

specific

valuefor

each

reaction.The

value

of

k

isdetermined

experimentally.“Constant”

is

relative

here

-k

changes

with

T，the

unit(量綱)

depend

on

m

+

n①

when

[A]=[B]=1mol·L-1,

v

=k②

the

greater

the

k

,

the

faster

the

rate參見中文P101質量作用定律（law

of

mass

action）對于基元反應，反應速率與反應物濃度的冪乘積成正比。冪指數(shù)就是基元反應方程中各反應物的系數(shù)，這就是質量作用定律。v

k[Cl2

][M]v

k[Cl][H

2

]v

k[H][Cl

2

]v

k[Cl]2[M]Cl2

M

2Cl

MCl

H2

HCl

HH

Cl2

HCl

Cl2Cl

M

Cl2

M對于復雜反應，它適用于每一步的基元反應，它的反應速度取決于定速步驟。2N2O5（g）→4NO2（g）＋

O2

(g)實驗證明N2O5

→NO3＋NO2

（慢）NO2＋NO3

→NO2＋O2＋NO

（快）NO＋NO3

→2NO2（快）v

k[N2O5

]質量作用定律僅適用于基元反應。速率方程應為in

general,

m

and

n

are

not

equal

to

thestoichiometric

coefficients

a

and

b[A],

[B]

are

the

concentration

of

A

and

B;m

and

n

are

themselves

constants

for

agiven

reaction,

it

must

be

determinedexperimentallyv=

k[A]m[B]naA

+

bB

→

cC

+

dDThe

orderof

a

reactionwith

respect

to

one

ofthe

reactants

is

equaltothe

power

to

whichthe

concentrationof

thatreactant

is

raised

intherate

equation.The

sum

of

the

powers

to

which

allreactant

concentrations

appearing

in

thera aw

are

raised

is

called

the

overallreaction

order.5-3.2

Order

of

A

ReactionFor

rate

equationv

=

k[A]m[B]ntruethat

fora

A

+

b

B→

c

C+

dD,m

≠aand

n≠bis

the

order

of

the

reaction

with

respectto

A,is

the

order

of

the

reactionwith

respectto

B.m+n

is

overall

order

of

the

reactionthe

exponents

m

and

n

are

not

necessarilyrelated

to

the

stoichiometric

coefficients

in

thebalanced equation,

thatis,

in

general

it

is

notThera aw

forthethermal

position（熱分解）

of

acetaldehyde

(CH3CHO)CH3CHO(g)

→CH4(g)

+

CO(g)hasbeen

determinedexperimentallyto

bev

=k[CH3CHO]3/2and

notrate

=

k[CH3CHO]×Example5-1:Given

the

following

data,

what

is

the

rate

expressionfor

the

reaction

between

hydroxide

ion

and

chlorinedioxide?2ClO2(aq)

+

2OH-(aq)

→ClO3

(aq)

+

ClO2

(aq)

+

H2O-

-[ClO2]

(mol.L-1)[OH-]

(mol.L-1)Rate

(mol.L-1

s-1)0.0100.0306.00×10-4

v10.0100.0751.50×10-3

v20.0550.0301.82×10-2

v3[ClO2

]0.0100.0306.00

10-4v10.0100.0751.50

10-3v20.0550.0301.82

10-2v3rate(mol

L1s1

)[OH

]m0.055[ClO

]mv

[ClO

]m[ClO2

12

3

v31

2

130.3

(5.5)m]

([

ClO2

]3

)m1.82

102

6.00

104

(

0.010

)By

inspection,

m=

2.

The

reaction

is

2nd

order

inClO2By

inspection,

n

=

132[ClO2

]

[OH

]0.0100.0306.00

10-4v10.0100.0751.50

10-3v0.055

0.030 1.82

10-2

vrate(mol

L1s1

)nnvv0.0751(

2

)1

2

(

)0.0301.50

1036.00

1042.5

(2.5)n[OH

][OH

]The

overall

rate

expression

isthereforev

=

k[ClO2]2[OH-]2v=

k[ClO

]m[OH-]nn=?想

：反應速度與速率方程？

反應分子數(shù)和反應級數(shù)？-order

reactionsA

-order

reaction

is

a

reactionwhoseratedependsonthe

reactant

concentration

raised

to

thepower.A

→Bthe

rate

equation

isv

k[

A]

k1cDifferential

form:dt

dt1v

d[

A]

dc

k

cIntegrate

theleft

sidefrom

c=

c0

to

c

andtherightfromt

=0to

t.dt1

dc

k

cc1

dc

k

dtThustcck

dt

dcc010速度方程的積分表達式Can

be

rearranged

to

give:10ln

ct

k

tc-

{lnc

–

lnc0}=

k1(t-0)c0

is

the

initial

concentration

of

c

(t

=0).c

istheconcentrationof

c

atsometimet.(5

-

8)1k

2.303

log

c0t

c(5

-

9)0k12.303t

log

clog

c

orThe

characteristics

of-order

reactions：A

ploto

c

versus

t(time)gives

astraightline

with

a

slope

of-k1/2.303.The

rate

constant,

k,has

units

of[time]-1.1/23.

half-life

(t

)is

thetimeittakesfor

theconcentrationof

areactant

A

tofallto

one

half

of

itsoriginal

value.(5

-

8)log012.303

ct

ck

1/2By

definition,

whent

=

t

,,

socck1log

02.303t

21112kc0kt

2.303

lg2

2.303

log

c02c

c0(5

-10)11/

2t

0.693k——一級反應的半衰期與速率常數(shù)k1成反比，與反應物的起始濃度無關。恒溫下，一級反應的半衰期是一個常數(shù)。半衰期越長，k1就越小，反應速率就越慢。可以用半衰期衡量反應速度，t1/2越大，反應越慢。常見藥物的半衰期：藥物名稱t1/2（h）青霉素鏈霉素安定0.52～320～40放射性同位素14C半衰期為5580年一級反應的特點是：（1）1gc對時間t作圖可得一條直線，直線的斜率為－k1/2.303。因此可由斜率求得反應速率常數(shù)k1。（2）k1的單位為[時間]－1。這說明k1的數(shù)值與時間單位有關，而與濃度無關。（3）恒溫下，一級反應的半衰期是與速率常數(shù)k

成反比，與反應物的起始濃度1無關的一個常數(shù)。log01c2.303

ctk

=11/

2t

0.693kExample

5-2(a)

What

is

the

rate

constant

k

for

the

-orderposition

of

N2O5(g)

at

25℃

if

the

half-life

ofN2O5(g)

at

that

temperature

is

4.03×104

seconds?Solution:

(a)k1/

2t

0.693

4.034.0k

=

0.693

=

0t1/

2-Solution:

(b)Putting

in

the

value

for

k

andsubstitutingt

=

8.64×104

seconds(one

day

has

86,400seconds)givesc0

2.303log

c

kt

0.6452.303log

1.72105

8.64104c0c(b)

Under

these

conditions,

what

percent

of

the

N2O5molecules

will

not

have

reacted

after

one

day?HenceTherefore,

22.6%

of

the

N2O5

molecules

will

nothave reactedafter

one

day

at25℃.

0.226cc0[N2O5

]0[N2O5

]

=

0.226Example

5-3SO2Cl2（磺酰氯

）posesto

sulfur

dioxideandchlorine

gas.

The

reaction

is order.

If

it

takes13.7

hours

for

a

0.250mol/L

solution

of

SO2Cl2

topose

into

a

0.117mol/L

solution,

what

is

therate

constantfor

the

reaction

and

what

is

the

half-lifeofSO2Cl2

position?Solution:2.3030log

c

log

c

ktlog

0.250

log

0.K

=

0.0554

h-1

0.693

0.693

12.5(h)k

0.05541/

2t一級反應很多，許多熱分解反應、分子重排反應、放射性元素的蛻變等都是一級反應，許多化合物的水解反應在低濃度的水溶液中進行時也表現(xiàn)為一級反應，許多藥物在生物體內的吸收、分布、代謝和排泄過程，也常近似地看作是一級反應。[例5-2]已知四環(huán)素在內的代謝服從一級反應規(guī)律。設給注射0.5g四環(huán)素，然后在不同時間測定血液中四環(huán)素的含量，得如下數(shù)據(jù)：服藥后時間t/h4

6

8

10

12

14

16血中四環(huán)素含量c/mg·L-1

4.6

3.9試求：①四環(huán)素代謝的半衰期；3.2

2.8

2.5

2.0

1.6②若血液中四環(huán)素的最低有效量相當于3.

7mg·L-1，則需幾小時后注射第二次？解：①先求速率常數(shù)k1，一級反應以logc對t作圖，得直線，見圖5-6。由前后兩點或由直線回歸得:斜率

lg

1.6

lg

4.6

0.038）16

4k

2.303

(0.038)

0.08（8

h-1∴(h)

0.693

7.90.0881

0.693k1/

2t一般控制計算表明：要使血液中四環(huán)素含量不低于3.7mg·L-1，應于第一次注射后6.3h之前注射第二次，臨在6h后注射第二次，即4次/天。②由圖5-6知，t＝0時，lgc0＝0.81，最低有效量c＝3.7mg·L-1lgc＝0.57,將（5-8）重排得：t

2.303(lg

c0

lg

c)k1

2.303(0.81

0.57)

6.3(h)0.088log (5

-

8)01c2.303

ctk

另外，衡量藥物分解的速率時，常用分解10％所需的時間，稱為十分之一衰期，用t0.9表示，恒溫下t0.9也是與濃度無關的常數(shù)。10.9k1

90

kt

2.303

lg

100

0.1054了解藥物的半衰期，對于合理用藥有著重要意義。常見藥物的半衰期，如磺胺甲惡唑（新諾明），半衰期12小時左右；頭孢氨芐，半衰期為1～2小時；為20～40h，保泰松為48～120h。Second

-orderreactions二級反應是一類常見的反應，溶液中的許多有機反應像加成、取代及消去反應等都是二級反應。22dtdcv

=

k2[A]2

,

v

=

k2

[A][B]

[A]=[B]v

k

c2c2

dc

k

dt(5

-11)20c

c1

1

k

t02c

c1

k

t

102t

c

ck

1

(1

1

)Theintegrated

raawfor

a2ndorder

reactioncan

be

easily

shown

to

betcck

dtc

dc

0

2

20The

characteristics

of

second-order

reactions：

A

graphof1/c

against

time

is

a

straight

line

,the

slopeof

whichgivesthe

rateconstantforthereaction；The

rate

constant,

k,has

units

of

[c]-1[t]-1；The

half-life

of

2th-02t

c

ck

1

(1

1

)order

reactions(c

=2/c0代入上式)t1/2=1

/

kc0.Note

that

the

half-life

of

a

second-orderreaction

is

not

independent

of

the

initialconcentration,

as

in

the

case

of

a

-orderreaction.

This

is

one

way

to

distinguish

a-order

reaction

from

a

second-orderreaction.（1）1/c對時間t作圖可得一條直線，直線的斜率即為反應速率常數(shù)k2。k2的單位為[濃度]-1[時間]-1。k2的數(shù)值與時間和濃度單位有關。二級反應的

。由此可見：二級反應的半衰期與反應物的初始濃度成反比。反應物初始濃度越大，半衰期越短。二級反應有以下特點是：Solution:t

=

3.6

(min)c

c2

k

t01

1

0.84

t10.200

0.5001Example

5-4Butadiene(C4H6丁二烯)dimerizes(聚合)toformC8H12（二聚物）.Thisreaction

is

2nd

order

in

butadiene.If

the

rateconstant

for

the

reaction

is

0.84

Lmol-1min-1,how

longwillit

take

for

a

0.500

mol/L

sample

of

butadiene

todimerizeuntil

the

butadiene

concentration

is

0.200

mol/L?（5-11）［例5-3］乙酸乙酯在25℃時的皂化反應為二級反應：CH３COOC２H５十NaOH→CH３COONa十C２H５OH乙酸乙酯和氫氧化鈉的起始濃度均為0.0100mol·L－１，反應20min后，氫氧化鈉的濃度消耗了0.00566mol·L－１，求：①反應速率常數(shù)；②反應的半衰期。解：1

1

1t

c

c0(

)1

6.52mol1

L

min10.0100

1

120

0.0100

0.00566k2

2

01/

21k

ct

15.3min6.52

0.01001①②Zero

-

orderreactions——

is

one

wherethe

rate

does

not

dependon

theconcentration

of

thespecies.00

0dtdcv

k

cv

=

k0

[A]0c0

c

k0tintegrated：coc

tdc

k

dt00固體界面上的分解反應c

=

-

k0

t

+

c0c

=

-k t+

cThe

characteristics

of

zero-orderreactions：0

0

A

graph

of

c

against

tis

a

straight

line

withaslopeof

-k0.

The

rate

constant,

k,has

units

of[c][

t

]-1；Thehalf-life

of

a

zero-order

reaction

is0

01/

22k

kt

c0

0.5c0c對t作圖得一直線，斜率為－k0k0的單位為［濃度］［時間］－１所以k0與時間單位和濃度單位有關（3）零級反應的半衰期：0

c0

0.5c001/

2k2kt所以零級反應的半衰期與最初濃度成正比。反應物的初始濃度越大，半衰期越長。近年來發(fā)展的一些緩解長效藥，其釋藥速率在相當長的時間范圍內比較恒定，即屬零級反應。如國際上應用較廣的一種皮下植入劑，內含女性

左旋18－甲基炔諾酮，每天約釋藥30μg，可一直維持5年左右。零級反應的特點：Solution:[HI] =

0.500－0.050×5=

0.250(mol·L-1)0c

kt

c-1c0=

0.500mol·L

k

=0.050mol·L

t=5sExample

5-5The position

of

HI

into

hydrogen

and

iodineon

a

gold

surface

is

0th

order

in

HI.

The

rateconstant

for

the

reaction

is

0.050mol·L-1·s

-1.

If

youbegin

with

a

0.500mol/L

concentration

of

HI,whatis

the

concentration

of

HI

after

5

seconds?-1·

s-15-4Effect

of

Temperature

on

Reaction

RatesChemical

reactions

arefaster

whenthetemperature

is

increased.

Why?Figure:

At

a

higher

temperature,T2,

more

molecules

have

anenergy

greater

than

E

a

,

as

shown

by

the

yellow

shaded

area.When

we

increase

the

temperature,

kineticenergies

and

speeds

will

increase

and

sothe

average

energy

of

a

collision

will

alsoincrease.As

a

result,

the

energy

of

any

particularcollision

will

be

more

likely

to

exceed

theactivation

energy.Thus,

chemical

reactions

are

faster

at

highertemperatures

than

at

lower

temperatures.The

dependence

of

reaction

rate

ontemperature

has

led

to

a

common

ruleofthumb:the

rate

of

a

chemical

reaction

willdouble

for

each

10℃

increase

in

thetemperature.Rule

of

thumb：5-4.1

Rule

of

Thumb經(jīng)驗法則(Van’t

Hoff

Law)v

(T+10)k

(T+10)γ===

2～4v

Tk

TThe

rate

ofa

chemical

reactionwill

double

foreach

10℃

increase

in

the

temperature.T

:

v

T

=

k

T[A]a

[B]baA

+

b

B=

c

C

+

d

D(T

+

10)

:v

(T+10)

=

k

T+10

[A]a

[B]b5-4.2

The

Arrhenius

EquationIn

1889

Svante

Arrhenius

showed

that

thedependence

oftheconstant

ofa

reaction

ontemperaturecan

be

expressed

by

thefollowingequation,

now

known

as

the

Arrheniusequation通過實驗制作速率常數(shù)k隨溫度T升高而變大的曲線圖，并分析曲線，可以得出速率常數(shù)隨溫度變化函數(shù)的經(jīng)驗方程:k

Ae

Ea

/

RTEa

—is

the

activation

energy

of

the

reaction

(in

kJ/mol)R—

is

thegas

constant

(8.314

JK-1mol-1)T

—

is

the

absolute

temperaturee

—

is

the

base

of

the

natural

logarithm

scaleA

—represents

the

collision

frequency,and

is

calledthe

frequency

factor

(指前因子).The

Arrhenius

Equation:烏斯方程式的重要假設：A和Ea是不隨溫度改變的特征參數(shù)。大多數(shù)反應在一定溫度區(qū)間內的這種近似是完全允許的，這種近似方法被稱為線性化。對某一反應：Ea基本不變，溫度升高，e

Ea

/(RT

)

增大，k值也增大，反應速率加快，說明了溫度對反應速率的影

響。對不同反應：當溫度一定時，活化能Ea值越小，則e

Ea

/(RT

)越大，k值愈大，反應速率

。反之，Ea愈大，反應速率愈慢。烏斯方程式把速率常數(shù)k、活化能Ea和溫度T三者聯(lián)系起來，由此關系式可以說明：k

Ae

Ea

/

RTEa2.303R

T(

1

)

lg

Alg

k

Taking

the

natural

logarithm

of

both

sides,

the

equationesorinterms

ofcommon

logarithms

:k

Ae

Ea

/

RTExample

5-6The

rateconstantsforthe position

of

acetaldehydeCH3CHO(g)

─→

CH4(g)

+

CO(g)were

measured

at

five

different

temperatures.

The

data

areshow

below.

Plot

lgk

versus

1/T

and

determine

theactivation

energy

(in

KJ/mol)

for

thereaction.──────────────────────────────T(K)

700

730

760

790k

[1/(mol/L)1/2s]

0.011

0.035

0.105

0.3438100.789──────────────────────────────Answer:

We

need

to

plot

lg

k

versus

1/T

. From

thegiven

data

we

obtain1/T(1/K)log

k1.43×10-3

1.37×10-3

1.32×10-3

1.27×10-3

1.23×10-3-1.96

-1.46

-0.979

-0.465

-0.103The

slope

of

the

line

iscalculated

from

two

pairs

ofcoordinates:

0.30;

1.25103lg

k

Ea

(

1

)

lg

A2.303R

Tlg

k1.80;1.411031/TEa

=

2.303×(8.314J/K·mol)×

(9.38×103K)=

1.80×105

J/mol=

1.80×102

KJ/molThus,

a

ploto

k

versus

1/T

gives

a

straightlinewhose

slope

is

equalto

-E

a

/2.303Rand

whoseintercept（截距）

with

the

ordinate(縱坐標)is

log

A.5-4.3

Application

of Arrhenius

EquationT1

→

k1According

to

this

equation, we

can

calculate

E

a

and

kT2

→

k2(5

-13)11Ea2.303RTlg

k

log

A

(5

-14)22Ea2.303RTlg

k

log

A

5

15

1

1

T1

lg

k1

k2

2.303R

T2Ea(5-14)

-

(5-13)：Example

5-7Therateconstant

of

a -orderreaction

is3.46×10-2

/s

at

298

K.

What

is

the

rate

constantat

350

K

if

the

activationenergy

for

the

reactionis50.2

KJ/mol?Answer:298

2.3038.314

35050.2103

13.461021logk225

15

2

1

T

k

2.303R

Tlog

k1

Ea

1

1

3.46102log

1.31k2k

0.71(s1

)25-5

Effect

of

Catalyst

on

ReactionRatesCatalyst:a

substance

that

increase

therate

of

achemical

reaction,

but

isitselfneither

consumednor

produced

in

thereactionMnO2Catalysts

increase

the

rate

of

a

reaction

bydecreasing

the

activation

energy

of

the

reaction.Catalysts