化學(xué)基地-2014級(jí)-化學(xué)反應(yīng)速率

化學(xué)反應(yīng)速率Chemical

Kinetics2第二章How

reactions

proceed?What

determines

their

rates?How

to

controlthose

rates?有的化學(xué)反應(yīng)，從熱力學(xué)的角度考慮，進(jìn)行的趨勢(shì)很大，但因其反應(yīng)速率太小，事實(shí)上幾乎不可能發(fā)生。有的化學(xué)反應(yīng)，反應(yīng)速率都很大。能降低不大，但其正逆兩個(gè)方向的Rate

may

be

expressed

in

three

main

ways:Average

reaction

rate:

a

measure

of

thechange

in

concentration

with

timeInstantaneous

rate:

rate

of

change

ofconcentration

atany

particular

instant

during

thereactionInitial

rate:

instantaneous

rate

at

t

=

0-

that

is,

when

the

reactants

aremixed第1節(jié)反應(yīng)速率的定義一、平均反應(yīng)速率(Average

reaction

rate)定義：?jiǎn)挝粫r(shí)間內(nèi)反應(yīng)物濃度的減少或生成物濃度的增加表示。單位：mol·dm-3·s-1

，mol·dm-3·min-1

或

mol·dm-3·h-1。例1、乙酸乙酯的皂化反應(yīng)CH3COOC2H5

＋

OH－＝

CH3COO－

＋

CH3CH2OHc(OH－

)

－c(OH－

)r(OH－

)＝

2 1

＝－t2－t1c(OH－

)t[

]t[

]tReaction

rate:

changes

in

a

concentrationof

a

product

or

a

reactant

per

unit

time.[

]Reaction

rate

=

——

tconcentrationchangeDefine

reaction

rate

and

explainAverage

reaction

rateInstantaneous

reaction

rate(2

tangents

shown)Initial

reaction

rateRate= in

[p_r_od_u_c_ts_]=change

in

timein

[r_e_actan_t_s]change

in

timetreactionatbConsider

the

hypothetical

reactionaA

+

bB

cC

+dDRate

of

=-

1

[A]

=

-

1

[B]

=[C]

=

1t

d

1

c[D]t對(duì)于一般的化學(xué)反應(yīng)

aA

＋

bB

＝gG

＋

hH原則上，用任何一種反應(yīng)物或生成物的濃度變化均可表示化學(xué)反應(yīng)速率，但經(jīng)常采用其濃度變化易于測(cè)量的那種物質(zhì)來(lái)進(jìn)行研究。二、瞬時(shí)反應(yīng)速率(Instantaneous

reaction

rate)瞬時(shí)速率：某一時(shí)刻的化學(xué)反應(yīng)速率稱為瞬時(shí)速率。例3、利用表中反應(yīng)物的濃度對(duì)時(shí)間作圖。AB的斜率表示時(shí)間間隔Δt

＝

tB－tA

內(nèi)反應(yīng)的平均速率。EFr(OH－

)＝

DEdDetermination

of

the

rate

of

deteriorationof

penicillin

during

storageat

two

different

times.

Note

that

the

rate

(the

slope

of

the

tangent

tothe

curve)

at

5

weeks

is

greater

than

the

rate

at

10

weeks,

when

lesspenicillin

is

present.四、如何測(cè)得反應(yīng)速率(Determine

Reaction

Rates)To

measure

reaction

rate,

we

measure

the

concentrationof

either

a

reactant

or

product

at

several

time

intervals.The

concentrations

are

measured

using

spectroscopicmethod

or

pressure

(for

a

gas).

For

example,

the

totalpressure

increases

for

the

reaction:2

N2O5

(g)

4

NO2

(g)

+

O2(g)Because

5

moles

of

gas

products

are

produced

from

2moles

of

gas

reactants.

For

the

reactionCaCO3

(s)

CaO(s)

+CO2

(g)The

increase

in

gaspressure

is

entirelydue

to

CO2formed.barometer三、初始反應(yīng)速率(Initial

rate)Initial

rate:

instantaneous

rate

at

t

=

0 -

that

is,when

the

reactants

aremixed。The

orange

curves

showhow

the

concentration

ofN2O5

changes

withfor

five

differenttimeinitialconcentrations.

The

initialrate

of

consumption

ofN2O5

can

be

determinedby

drawing

a

tangent(black

line)

to

each

curveat

the

start

of

the

reaction.This

graph

was

obtainedby

plottingthe

five

initialrates

against

the

initialconcentration

of

N2O5.

Theinitial

rate

is

directlyproportional

to

the

initialconcentration.This

graph

also

illustrateshow

we

can

determine

thevalue

of

the

rate

constantk

by

calculating

the

slopeof

the

straight

line

fromtwo

points.Initial

rate

of

consumption

of

N2O5

=

k

[N2O5]initial第2節(jié)反應(yīng)速率與反應(yīng)物濃度的關(guān)系Initial

rate

of

consumption

of

N2O5

=

k

[N2O5]initialThe

constantk

is

called

the

rate

constantfor

the

reaction，速率常數(shù)。At

any

stage

of

the

reaction,

and

provided

products

do

notparticipate

in

the

reaction,Rate

of

consumption

of

N2O5

=

k

[N2O5]把用來(lái)表達(dá)反應(yīng)速率與反應(yīng)物濃度關(guān)系的方程式就叫做反應(yīng)速率定律或質(zhì)量作用定律（Ra

aw）。不同的化學(xué)反應(yīng)，其速率表達(dá)式不同。(a)

When

the

ratesof

disappearance

of

NO2

are

plottedagainst

itsconcentration,

astraight

line

is

not

obtained.

(b)

However,

a

straight

line

is

obtained

when

the

ratesare

plotted

against

the

square

of

the

concentration,

indicating

that

the

rate

isdirectly

proportional

to

the

square

of

theconcentration.Differential

Ra

awsDependence

of

reactionrate

on

the

concentrations

of

reactants

iscalled

the

ra aw,

which

is

unique

for

each

ductsFor

a

general

reaction,the

ra aw

has

ta

A

+b

B+

c

C

eral

formorder

wrt

A,

B,

and

C,

determined

experimentallyreaction

rate

=

k

[A]X

[B]Y

[C]ZFor

example,

the

ra aw

israte

=

k

[Br-]

[BrO3-]

[H+]

for5

Br-

+

BrO3-

+

6H+

3Br2

+

3

H2OThe

reaction

is

1st

order

wrt

all

three

reactants,

total

order

3.Use

differentialsto

express

rates基元反應(yīng)：指反應(yīng)物分子一步直接生成產(chǎn)物的反應(yīng)。質(zhì)量作用定律：基元反應(yīng)的速率與反應(yīng)物濃度以其化學(xué)計(jì)量數(shù)為冪指數(shù)的連乘積成正比。對(duì)于基元反應(yīng)a

A

＋bB

＝

gG

＋hH質(zhì)量作用定律的數(shù)學(xué)表達(dá)式：r

＝

k

c(A)m

c(B)nk

稱為速率常數(shù)m，n

稱反應(yīng)物A，B的反應(yīng)級(jí)數(shù)k，m和n

均可由實(shí)驗(yàn)測(cè)得H2O2

+3

I-

+2

H+

I3-

+

2

H2OExprmt[H2O2][I-][H+]Initial

rate

M

s-110.0100.0100.00501.15e-620.0200.0100.00502.30e-630.0100.0200.00502.30e-640.0100.0100.01001.15e-6例子：Estimatethe

ordersand

rate

constant

k

from

the

resultsobserved

for

thereaction?

What

is

the

rate

when

[H2O2]

=

[I-]

=

[H+]

=

1.0

M?Learn

the

strategy

to

determine

the

ra awfromthis

example.Figure

out

the

answer

without

writing

down

anything.Estimate

the

orders

from

the

results

observed

for

the

reactionH2O2

+3

I-

+

2

H+

I3-

+2

H2OExprmt[H2O2][I-][H+]Initialrate

M

s-110.0100.0100.00501.15e-620.0200.0100.00502.30e-6

1

forH2O230.0100.0200.00502.30e-6

1

forI-40.0100.0100.01001.15e-6

0

for

H+1.15e-6

=

k

[H2O2]x

[I-]y

[H+]z1.15e-6k

(0.010)x(0.010)y(0.0050)z

exprmt

111-----------=2.30e-6-------------------------------------k

(0.020)x(0.010)y(0.0050)z

exprmt

2----

=2---2x

=

1Other

orders

are

determined

in

a

similar

way

as

shown

before.Now,

lets

find

k

and

the

rateThurs,

rate

=

1.15e-6

=

k

(0.010)(0.010)

fromexprmt

1k

=

1.15e-6

M

s-1

/

(0.010)(0.010)

M3

=

0.0115

M-1

s-1And

the

ra aw

istherefore,–

d

[H2O2]

krate

=

—————

=

0.0115

[H2O2]

[I-]d

tThe

rate

when

[H2O2]

=

[I-]

=[H+]

=1.0

M:a

differential

ra

awtotal

order

2The

rate

is

the

same

as

the

rate

constant

k,

when

concentrations

ofreactants

are

all

unity(exactly1),

doesn’tmatter

what

the

orders

are.當(dāng)反應(yīng)物的濃度都為1.0M時(shí)，其反應(yīng)速率的值和反應(yīng)級(jí)數(shù)無(wú)關(guān)！?。he

reaction

rate–d[H2O2]/dt

=

0.0115

[H2O2]

[I–],

forH2O2

+

3

I-

+

2

H+

I3-

+

2

H2OWhat

is

–

d[I–]/dt

when

[H2O2]

=

[I–]

=

0.5?Solution:

Please

note

the

stoichiometry

of

equation

and

how

the

rate

changes.–

d[I–]/dt

=–

3

d[H2O2]/dt

=

3*

0.0115

[H2O2]

[I–]=

0.0345

*

0.5

*

0.5=

0.0086

Ms-1In

order

to

get

a

unique

rate

constant

k,

we

evaluate

k

forthereactiona

A

+

bB

productthis

wayrate

=-1/a

d[A]/dt

=

-1/b

d[B]/dt

=

k

[A]x

[B]y第3節(jié)反應(yīng)物濃度與時(shí)間的關(guān)系一、0級(jí)反應(yīng)The

concentration

of

the

reactant

in

a

zero-order

reaction

falls

onstantrate

until

thereactant

is

exhausted.The

rate

of

a

zero-order

reaction

is

independent

of

the

concentration

of

thereactant

and

remains

constant

until

all

the

reactant

has

been

consumed,

whenthe

ratefalls

abruptly

tozero.速率常數(shù)的單位與反應(yīng)級(jí)數(shù)有關(guān)：一級(jí)反應(yīng)s－1二級(jí)反應(yīng)dm3·mol－1·s－1n級(jí)反應(yīng)dm3(n－1)·mol－1(n－1)·s

－1化學(xué)反應(yīng)速率常數(shù)k是在給定溫度下，各反應(yīng)物濃度皆為1mol·dm-3時(shí)的反應(yīng)速率，因此也稱比速率常數(shù)。速率常數(shù)是溫度的函數(shù)。poses

in

1st

or2nd

order

ra

aw.Integratedra

aw[A]

=

[A]o

–

k

tOnereactant

ADifferentialra

awd[A]

/

dt

=

kd[A]——

=

k

[A]d

td[A]——

=

k

[A]2d

t[A]

=

[A]o

e

–

k

t

or ln

[A]

=

ln

[A]o

–

k

t1

1

[A]

conc

at

t——

–

——=

k

t[A]

[A]o

[A]o

conc

at

t=0二、1級(jí)反應(yīng)[A]tln[A]t[A]

=[A]o

e

–

k

tln

[A]

=

ln

[A]o

–k

tt?d[A]–

——

=

k

[A]d

t[A]

=

[A]o

e

–

k

t

or ln

[A]

=

ln

[A]o

–

k

tHalf

life

&

k

of

Order半衰期和速率常數(shù)positionThe

time

required

for

half

of

A

to pose

is

called

half

life

t1/2.Since[A]

=

[A]o

e

–

k

t

orln

[A]

=

ln

[A]o

–

k

tWhenThust

=

t1/2,

[A]

=

?

[A]oln

? [A]o

=

ln

[A]o

–

k

t1/2–ln

2

=

–

k

t1/2k

t1/2

=

ln2

=

0.693

relationship

between

k

and

t1/2Radioactive

decay

usually

follow

1st

order

kinetics,

and

half

life

of

anisotope

is

used

to

indicate

its

stability.Evaluate

t?

from

k

or

k

from

t?1

1——

–

——

=

k

t[A]

[A]o[A]

conc

att[A]o

conc

at

t

=

0三、2級(jí)反應(yīng)Dimerization

of

butadiene

is

secondorder:2C4H6(g)

=

C8H12(g).The

rate

constant

k

at

some

temperature

is

0.100

/min.

The

initialconcentration

of

butadiene

[B]

is

2.0

M.Calculate

the

time

required

for

[B]

=

1.0

and

0.5

MCalculate

concentration

of

butadiene

when

t

=

1,

5,

10,

and

30.d[A]–——

=

k

[A]2d

t例子：A

2nd

Order

ExampleDimerization

of

butadiene

is

second

order:2

C4H6(g)

=

C8H12(g).The

rate

constant

k

at

some

temperatureis

0.100

/min.

The

initial

concentration

ofbutadiene

[B]

is

2.0

M.Calculate

the

time

t

required

for

[B]

=1.0and

0.5

MCalculate

concentration

of

butadienewhen

t

=

1,

5,

10,

and

30.1

1——

–

—— =

kt[B]

[B]o1

1——

–

——[B]

[B]ot

=

————————k[B]o[B]

=

——————[B]o

k

t+

1t

=

1

5[B]

=

1.67

1.010

15

30

350.67

0.50

0.29

0.25Work

out

the

formulas

andthen

evaluate

values第

4

節(jié)

反應(yīng)機(jī)理一、基本概念所謂基元反應(yīng)是指反應(yīng)物分子一步直接轉(zhuǎn)化為產(chǎn)物的反應(yīng)。

如：NO2

＋

CO

＝

NO

＋

CO2反應(yīng)物NO2

分子和CO分子經(jīng)過(guò)一次碰撞就轉(zhuǎn)變成為產(chǎn)物

NO分子和CO2

?；磻?yīng)是動(dòng)力學(xué)研究中的最簡(jiǎn)單的反應(yīng)，反應(yīng)過(guò)程中沒(méi)有任何中間產(chǎn)物。Elementary

reactions

are

steps

of

molecular

events

showing

howreactions

proceed.

This

type

of

description

is

a

mechanism.The

mechanism

for

the

reaction

between

CO

and

NO2

is

proposed

to

beStep

1 NO2

+

NO2

NO3

+

NOStep

2 NO3

+

CO

NO2

+

CO2(an

elementary

reaction)(an

elementary

reaction)Addthese

two

equations

led

tothe

overall

reactionNO2

+

CO

=NO+

CO2

(overall

reaction)A

mechanism

is

a

proposal

to

explain

thera aw,

and

it

has

to

satisfythe

ra

aw.

A

satisfactory

explanation

is

not

a

proof.例如：

H2

(

g

)

＋

I2

(

g

)

＝

2

HI

(

g

)實(shí)驗(yàn)上或理論上都證明，它并不是一步完成的基元反應(yīng)，它的反應(yīng)歷程可能是如下兩步基元反應(yīng)：①

I2

＝

I

＋I(xiàn)

（快）②

H2

＋2I

＝

2HI

（慢）化學(xué)反應(yīng)的速率由反應(yīng)速率慢的基元反應(yīng)決定。基元反應(yīng)或復(fù)雜反應(yīng)的基元步驟中發(fā)生反應(yīng)所需要的微粒（分子、原子、離子）的數(shù)目一般稱為反應(yīng)的分子數(shù)。分子數(shù)

Molecularity

of

Elementary

ReactionsThe

total

order

of

ra aw

in

an

elementary

reaction

is

molecularity.The

ra aw

of

elementary

reaction

is

derived

from

the

equation.

Theorder

is

the

number

of

reacting

molecules

because

they

must

collide

toreact.A

molecule poses

by

itself

is

a

unimolecular

reaction

(step);twomoleculescollide

and

react

isa

bimolecular

reaction

(step);

&three

molecules

collide

and

react

is

a

termolecular

reaction

(step).O3

O2

+

ONO2

+

NO2

Br+

Br

+

Ar

rate

=

k[O3]rate

=

k[NO2]2rate

=

k

[Br]2[Ar]NO3

+

NOBr2

+

Ar*Caution:

Derive

ra aws

this

way

only

for

elementary

reactions.單分子反應(yīng)SO2Cl2

的分解反應(yīng)SO2Cl2

＝

SO2

＋Cl2雙分子反應(yīng)NO2

的分解反應(yīng)2

NO2

＝

2NO

＋

O2三分子反應(yīng)HI

的生成反應(yīng)

H2

＋

2

I

＝

2

HI四分子或 分子碰撞而發(fā)生的反應(yīng)尚未發(fā)現(xiàn)。Elementary

Reactions

are

Molecular

EventsN2O5

NO2

+NO3NO

+

O2

+

NO2

NO2

+

NO3A

mechanism

is

a

collection

of

elementary

steps

devise

to

explain

the

thereaction

in

view

of

the

observed

ra aw.

You

need

the

skill

to

derive

ara awfrom

a

mechanism,

but

proposing

a

mechanism

istask

afteryouawis,have

learned

more

chemistryForthe

reaction,

2

NO2

(g)

+

F2

(g)

2

NO2F

(g),

the

rarate

=

k[NO2]

[F2]

.Can

the

elementary

reaction

be

the

same

as

the

overall

reaction?If

they

were

the

same

the

ra aw

would

have

beenrate

=

k

[NO2]2

[F2],Therefore,

they

the

overallreaction

is

not

an

elementary

reaction.

Itsmechanism

is

proposed

next.The

rate

determining

step

is

the

slowesand

the

ra aw

for

this

step

isthe

raementary

step

in

a

mechanism,aw

for

the

overall

reaction.The

(determined)

ra aw

is,

rate

=

k

[NO2]

[F2],for

the

reaction, 2

NO2

(g)

+F2

(g)

2

NO2F

(g),and

a

two-step

mechanism

is

proposed:NO2

(g)

+F2

(g)

NO2F(g)+F(g)iii NO2

(g)

+

F(g)

NO2F(g)Which

is

the

rate

determining

step?Answer:The

rate

for

step

iis rate

=k

[NO2]

[F2],

which

is

the

ra aw,

thissuggests

that

step

i

is

the

rate-determiningorthe

s-l-o-w

step.反應(yīng)機(jī)理中的慢反應(yīng)步驟決定總反應(yīng)的速率！二、如何由給出的反應(yīng)機(jī)理推導(dǎo)出速率方程例1、The position

of

H2O2

in

the

presence

of

I–

follow

thismechanism,k1

H2O

+IO–

slowi H2O2

+

I–

iiH2O2

+

IO–

k2

H2O

+O2

+

I–

fastWhat

is

the

ra

aw?Solutionaw

is:The

slow

step

determines

the

rate,

and

the

rarate

=

k1[H2O2]

[I

–]Since

both

[H2O2]

and

[I

–]

are

measurable

in

the

system,

this

isthera

aw.fast

equilibrium

(k1,

k-1)Br2

2BrH2

+

BrH+Brk2

HBr

+

Hk3

HBrslowfastSolution:The

fast

equilibrium

condition

simply

says

thatk1

[Br2]

=

k-1

[Br]2and

[Br]

=

(k1/k-1

[Br2])?The

slow

step

determines

the

ra

aw,rate

=

k2

[H2]

[Br]=

k2

[H2]

(k1/k-1

[Br2])?=

k

[H2]

[Br2]

?

;total

order

1.5Brisan

intermediatek

=

k2

(k1/k-1)?M-?

s-1explain快速平衡假設(shè)法?。。±?、Derive

the

ra aw

for

the

reaction,

H2

+

Br2

=

2

HBr,from

theproposed

mechanism:例3、Theposition

of

N2O5

follows

the

mechanism:fast

equilibriumslowN2O5

NO2

+

NO3NO2

+

NO3

—k2

NO

+

O2

+

NO2NO3

+

NO

—k3

NO2

+NO2

fastDerive

the

ra

aw.Solution:NO2

&

NO3

are

intermediateK,

equilibrium

constantKdiffer

fromkThe

slow

step

determines

the

rate,rate

=

k2

[NO2]

[NO3]From

1,

we

have[NO2]

[NO3]——————

=

K[N2O5]Thus,

rate

=

K

k2

[N2O5]穩(wěn)態(tài)近似法?。?！以假設(shè)中間產(chǎn)物的濃度恒定不變?yōu)榛A(chǔ)！即、中間產(chǎn)物的生成速率與其消耗速率相等。Rate

of

producing

theintermediate,

Rprod,

isthe

same

asits

rate

ofconsumption,

Rcons.Rprod

=

Rcons[Intermediate]timeRprod

<

RconsRprod

>

RconsBe

able

to

apply

the

steady-state

approximation

to

derivera

aws 2

HI的反應(yīng)機(jī)理如下：I2

—k1 2

I2I

—k-1

I2H2

+ 2I

—k2 2HIaw.假設(shè)

H2

+

I2Step

(1)

Step(1)

Step

(2)Derive

the

raDerivation:rate

=

k2

[H2]

[I]

2

(‘cause

this

step

gives

products?。?！)but

I

is

an

intermediate,

thisis

not

a

ra aw

yet.Since

k1

[I2](=

rate

of

producingI)I)g

Steady

stateThus,=

k-1

[I]2

+

k2

[H2]

[I]2

(=

rate

of

consumink1

[I2][I]2

=

——————k-1

+

k2

[H2]

rate

=

k1

k2

[H2]

[I2]

/

{k-1

+

k2

[H2]

}From

the

previous

result:k1

k2

[H2]

[I2]rate

=

———————{k-1

+k2

[H2]}Discussion:Ifk-1

<<

k2

[H2] then

{k-1

+

k2

[H2]}=k2

[H2]

,then

rate

=

k1

k2

[H2][I2]

/ {k2

[H2]

}=

k1

[I2]

(pseudo

1st

order

wrt

I2)using

large

concentration

of

H2

or

step

2

is

fast

(will

meet

this

condition).If

step

(2)

is

slow,

then

k2

<<

k1,and

if

[H2]

is

not

large,

we

have

{k-1

+

k2

[H2]}

=

k-1and

rate

=

k1

k2

[H2]

[I2]

/

k1

=

k2

[H2]

[I2]第5節(jié)

反應(yīng)速率理論簡(jiǎn)介上20世紀(jì)，反應(yīng)速率理論的研究取得了進(jìn)展；1918年 （Lewis）在氣體分子運(yùn)動(dòng)論的基礎(chǔ)化學(xué)反應(yīng)速率的碰撞理論；30年代基礎(chǔ)上（Eyring）等在量子力學(xué)和統(tǒng)計(jì)力學(xué)的化學(xué)反應(yīng)速率的過(guò)渡狀態(tài)理論。一、碰撞理論碰撞理論認(rèn)為：反應(yīng)物分子間的相互碰撞是反應(yīng)進(jìn)行的先決條件。反應(yīng)物分子能量高；碰撞頻率越大；碰撞方向有利；有效碰撞次數(shù)多，反應(yīng)速率越大。即

Z**＝Z

f

Pf

為能量因子；P為取向因子；

P

取值在1～10－9之間。Ea

稱活化能，一般的化學(xué)反應(yīng)

Ea

為每摩爾幾十到幾百千焦。Not

all

collisions

leads

to

a

reactionFor

effective

collisions

proper

orientation

ofthe

molecules

must

be

possible二、過(guò)渡狀態(tài)理論過(guò)渡狀態(tài)理論認(rèn)為：當(dāng)兩個(gè)具有足夠能量的反應(yīng)物分子相互接近時(shí)，分子中的化學(xué)鍵要發(fā)生重排，即反應(yīng)物分子先形成活化配合物的中間過(guò)渡狀態(tài)，活化配合物能量很高，不穩(wěn)定，它將分解部分形成反應(yīng)產(chǎn)物。該理論認(rèn)為，活化配合物的濃度；活化配合物分解成產(chǎn)物的概率；活化配合物分解成產(chǎn)物的速率均將影響化學(xué)反應(yīng)的速率。例如

反應(yīng)

NO2

＋

CO

＝

NO

＋

CO2正反應(yīng)活化能Ea

＝活化配合物的勢(shì)能－反應(yīng)物平均勢(shì)能逆反應(yīng)活化能Ea′＝活化配合物的勢(shì)能－產(chǎn)物平均勢(shì)能反應(yīng)的熱效應(yīng)

ΔrHm＝

Ea

－Ea′結(jié)論：①若正反應(yīng)是放熱反應(yīng)，其逆反應(yīng)必定吸熱。不論是放熱還是吸熱反應(yīng)，反應(yīng)物必須先爬過(guò)一個(gè)能壘反應(yīng)才能進(jìn)行。②如果正反應(yīng)是經(jīng)過(guò)一步即可完成的反應(yīng)，則其逆反應(yīng)也可經(jīng)過(guò)一步完成，而且正逆兩個(gè)反應(yīng)經(jīng)過(guò)同一個(gè)活化配合物

。這就是微觀可逆性原理。③化學(xué)反應(yīng)的熱效應(yīng)

ΔrHm

＝

Ea

－

Ea′當(dāng)Ea

＞Ea′時(shí)，ΔrHm

＞0當(dāng)Ea

＜Ea′時(shí)，ΔrHm

＜0反應(yīng)吸熱；反應(yīng)放熱。三、溫度對(duì)化學(xué)反應(yīng)速率的影響過(guò)渡狀態(tài)理論認(rèn)為：在反應(yīng)過(guò)程中反應(yīng)物必須爬過(guò)一個(gè)能壘才能進(jìn)行。升高溫度，反應(yīng)物分子的平均能量提高，減小了活化能的值，反應(yīng)速率加快。碰撞理論認(rèn)為：溫度高時(shí)分子運(yùn)動(dòng)速率增大，活化分子的百分?jǐn)?shù)增加，有效碰撞的百分?jǐn)?shù)增加，反應(yīng)速率增大。1899年Arrhenius總結(jié)了大量實(shí)驗(yàn)事實(shí)，歸納出反應(yīng)速率常數(shù)和溫度的定量關(guān)系式中k為反應(yīng)速率常數(shù)，Ea為活化能，A為指前因子。通常溫度升高，化學(xué)反應(yīng)速率常數(shù)增加。1T2

k2

Ea

1

1

lg

k

2.303R

T1七、催化劑與催化反應(yīng)簡(jiǎn)介?

－1①

H2

(g)

＋

1/2

O2

(g)

＝

H2O

(l)

ΔrGm

＝－237.1

kJ·mol②

H2

(g)

＋

O2

(g)＝

H2O2

(l)?

－1ΔrGm

＝－120.4

kJ·mol③3/2H2

(g)

＋

1/2

N2

(g)

＝

NH3

(g)?

－1ΔrGm

＝－16.4

kJ·mol在熱力學(xué)上看，均為常溫常壓下可以自發(fā)進(jìn)行的反應(yīng)。但是由于反應(yīng)速