Review1The

two

empiricallaws

of

ideal

dilute

solutionHenry’s

law:pB

kx

xB

,

(

xB

0)pB

kmmB

,(mB

0)pB

kc

cB

,(cB

0)Raoult’s

law:Ap

p*

x

,

(

x

1)A

A

A-3th

Lecture-24. Isothermal

equation

of

of

ideal

dilute

solution1.Definition

of the

ideal

dilute

solution:The

solution

for

which

thesolute

obeys

Henry’s

law

and

thesolvent

obeys

Raoult’s

law

is

calledideal

dilute

solution.For

solvent:p

p*

x

,(

x

1)A

A

A

AFor

solute:pB

kx

xB

kmmB

kc

cB

,(

xB

0,

mB

0,

cB

0)

of

solvent

A

in

ideal

dilute

solution1.

Suppose

the

solvent

A

obey

Raoult’s

law

for

non-electrolyte,then

we

have,A(

xl

(T

,

p,

x

)

*l

(T

,

p)

RT

ln

xlA

A

A

A

1)

(4.4.4.1)A*l

(T

,

p)whereis

chemicalactual

statepotential

of

pure

substance

A,

atthe

temperature

T

and

pressure

pof

the

solution,

which

is

an

actualstate

of

solvent.Al

(T

,

p,

x

)

(T

)

RT

ln

xA

A

Ap

(T

,

p,

x

)

(T

, )

l

g

g(T

)

RTlnBB

of

solute

B

in

ideal

dilute

solution2.

Suppose

the

species

in

gas

phase

being

in

gas-solutionequilibrium

are

mixing

ideal

gases,

and

the

solute

B

obeyHenry’s

law

for

non-electrolyte,

then:(1)

xB

expresses

the

concentration

of

soluteBBBB

BBk

gBp

(T

)

RT

lnp

RT

ln

xwhere*gat

the

temperature

T

and

pressure

p=k

of

the

solution.4B

*

g

(T

,

p

k)

RT

ln

x

,B

B

(T

)

RT

ln

x

,B

BT

,

p

k

is

the

gas

chemical

potential

of

pure

solute

B

k)

RT

lnl

(T

,

p,

x

)

*

g

(T

,B

B

BBBl

(T

,

p)Supposeis the

liquid

chemical

potential

of

puresolute

B

at

the

temperature

T

and

pressure

p=k

of

the

solution,and

the

pure

solute

B

in

ideal

dilute

solution

has

a

equilibriumwith

ideal

gas

in

T,

=

,

then:*

g

(T

,

p

k

)

l

(T

,

p)

l

(hyp,T

,

p)B

B

BB(x

0)Bl

(T

,

p,

x

)

l

(hyp,T

,

p)+RTlnxB

B

B(4.4.4.2)5B(x

0)Bl

(T

,

p,

x

)

l

(hyp,

T

,

p)+RTlnxB

B

Bwhereis

thechemical

potential

of

pure

soluteBl

(hyp,

T

,

p)hypothetical

stateB

at

the

temperature

T

andpressure

p=k

of

the

solution.It

is

defined

to

be

the

fictitiousstate

at

the

temperature

T

andpressure

p=k

of

the

solution,where

this

is

an

extrapolationof

the

properties

of

solute

B

inthe

very

dilute

solution

to

thelimit

xB

16(2)

mB

expresses

the

concentration

of

soluteB

B

B

B

Bl

(T

,

p,

m

)

g

(T

,

p

)

g

(T

)

RTln

pBk m,B

m

m

gBp

Bm

(T

)

RT

lnp

RT

ln

B

m

'l

(T,

p)

RTln

mB(4.4.4.3)BB

Bl

(T,

p,m

)

'l(hyp,T,

p)

RTlnmBm

mB

0'l

(T,

p)is

chemical

potential

of

standard

state

(hypotheticalBstate)

for

solute

B,

which

illustrates

that

solute

B

obeys

Henry’slaw

( )

in

a

dilute

solution,

where

this

hypotheticalpB

km,BmB7the

limit

standard

molality

m

.state

is

an

extrapolation

of

the

properties

of

solute

B

from

mB

toB

B

BBl

(T,

p,m

)

'l

(hyp,T,

p)

RTln

mB

,(m

0)mB'l

(T,

p)

ischemical

potential

ofhypothetical

statestandard

state

(hypothetical

state)for

solute

B,

which

illustrates

thatkmBp*solute

B

obeys

Henry’s

lawPressure,

pi(

pB

kmmB

)

in

a

dilute

solution.m0Molality

of

species

B,

m

BThis

hypothetical

state

is

an

extrapolation

of

the

properties

ofsolute

B

from

mB

to

the

limit

standard

molality

m

.83

cB

expresses

the

concentration

of

soluteFrom

the

same

method,

we

have:BBlB

Bc,(c

"

lBc

(T

,

p,

c

)

(hyp,T

,

p)

RTlnB(T,

p)''lis

chemical

0)

(4.4.4.4)potential

ofstandard

state

(hypothetical

state)for

solute

B,

which

illustrates

thathypothetical

statekcsolute

B

obeys

Henry’s

law(

pB

kccB

)

in

a

dilute

solution,Pressure,

piBp*where

this

hypothetical

state

is

anextrapolation

of

the

properties

of9csolute

B

from

cB

to

the

limit

standardconcentration

.0molar

concentration

of

cspecies

B,

cBStandard

state

of

the

solute

in

ideal

dilute

solution

is

thefictitious

stateB

:(

pB

k

)B

(k)BC

:

(c

c

)C

(kc)p*BD

(km)D

:

(mB

m

)Pressure,

pi01cmThe

k

is

the

slope

of

tangents

of

experimental

curve

of

thevapour

pressure

of

B

against

its

mole

fraction

at

xB=

010Another

definition

of the

ideal

dilute

solution:Thethermodynamic

definition:

onein

which

the

chemicalpotentials

of

solvent

and

solute

are

given

respectively

by

(4.4.4.1)and

(4.4.4.2

~

4.4.4.4)

for

a

range

of

composition

with

xA

close

to1

and

for

a

range

of

T

and

p.l

*lA

A

AT

,

p,

x T

,

p

RTxl,

x

(4.4.4.1)A

ABl

(T

,

p,

x

)

l

(hyp,

T

,

p)+RTlnx

,

(

x

0)B

B

B

B(4.4.4.2)(4.4.4.3)l'lmB

0)BlcBBB

"

lB,(cc

(T

,

p,

c

)

(hyp,T

,

p)

RTln

0)

(4.4.4.4)11NOITICE：Isothermal

equation

of

chemical

potential

between

the

idealdilute

solution

and

the

ideal

solution

is

different

in

nature.x

i

:

0

1

ideal

solutionx

A

1

ideal

dilute

solution(1).

Application

range

is

different,

and

formula

(4.4.4.1)

and(4.4.4.2

~

4.4.4.4)

only

suit

for

dilute

solutionActual

state

of

the

pure

solvent2

.

The

meanin

ofhypothetical

state

of

the

pure

solute(3).

From

molecular

lever,

only

fA-B

and

fA-A

are

taken

into

account.12is

different*iComparison

of ideal

solution

and

ideal

dilute

solutionIdeal

liquidmixture

oideal

solutionideal

dilute

solutionsolvent

soluteApplication

*

RT

ln

x

*

RT

ln

x

*

(hyp)

RT

ln

xi

i

i

A

A

A

B

B

Brange

x

:0

1

xA

1

xB

0ithe

Actual

state: Actual

state:

Hypothetical

state:meaning

of

*iPure

liquid

ithat

has

thesame

T,

p

withmixturePure

solvent

Athat

has

thesame

T,

p

withsolutionChemicalpotential

of

puresolute

B

in

T,

p=k134.4.4.2.

Application:

Duhem-Margules

equationApply

Gibbs-Duhem

equation

and

isothermal

equation

ofchemical

potential

of

each

substance

i

to

the

binary

solution

ofgas-liquid

equilibrium,

show

that:T

,T

,)(

p

1

p1p

2

(

p

2)Justification:

For

binary

solution,

we

have:x1

d1+

x2

d2

=

0,1

1

22

xx

x

xor1

)T

,

p

l

l

x

2

( 2

)T

,

p

0x

1

(

x1

x11l

gpg

(T

)

RTlnSuppose

vapour

is

ideal

gas,

then:1

1

1p2pl

g

g

(T

)

RTln

214pApplyto

relation

nature

of

partial

molarties,1ll22

2

and

Duhem-Margules

equationWe

have:}T

,

px1{ppln

p1ln

p2

x2

{}T

,

p

0xxx1

(

p1

)

x2

(

p2

)112

1T

,

pT

,

pp1

x1

p

xthenfrom

dx1

dx2(

p

1

(

p

2)

)T

,

pT

,

p

x

x

1

p

1x

1

2

p

2x

2The

formula

comes

intoexistence

when

vapour

isideal

gas.15Three

deductions

of

Duhem-Marulese

uation1T

,

p(1)

if: (

p1

)

0,x1T

,

pshow

that:

(

p2

)

0,x1

1

11(2)

if:

p

p*

x

,

(

x

:

0

1),*2

2

2show

that:

p

p

x

,2(

x

:

0

1)*1(

x

:

1

1

x),1

1

1(3)

if:

p

p x

,2

2162sho

hat

p

kx

,

(

x

:

0

x)(1)p1p21,

s

ohat:T

,

px1

,T

,

pxJustification:T

,

pbecause:

(

p1

)

0,x1From

Duhem-Margules

equation:,,x1

(

p1

)

x2

(

p2

)2

1p1

x1p

x117T

,

pThen:

(

p2

)

0x1

1

11(2)

if:

p

p*

x

,

(

x

:

0

1),showJustification:p

p*

x22

22(

x

:

0

1),1111we

have:

(T

,

pxp)

p1

1,(

x

:

0

1),x*1111,T

,

ppxfrom:

(

p1

)

p

x22

2T

,

pp2xthen:

(

p2

)

1,(

x

:

1

0),xAt

constant

T,

p,

above

formula

change

to:d

ln

p2

d

ln

x2

,1822

22from:

x

:1

x

,

p

:

p*

p

,Integral

above

formula,

we

have:2p*ln

p2

ln

x

,Then:*2

2

2p

p

x

,2(

x

:

1

0),1(

x

:

1

1

x),*1

1

1(3)

if:

p

p x

,p

pJustification:show

that:

p2

kx2

,(

x2

:

0

x),

1x1

1,(

x1

:

1

1

x),x1from:

(

1

)T

,

pT

,

px(

p2

)p2

1,xwe

have:22d

ln

p2

d

ln

x2

,At

constant

T

above

formula

chan e

to:Integral

above

formula,

we

have:ln

p2

ln

x2

ln

k,then:

p2

kx2

,

(x2

:0x),195.5(b)

The

vapour

pressure

of

2-propanol

is

50.00

kPa

at338.8

C,

but

it

fell

to

49.62

kPa

when

8.69

g

of

an

involatileorganic

compound

was

dissolved

in

250

g

of

2-propanol.Calculate

the

molar

mass

of

the

compound.We

assume

that

the

solvent,

2-propanol,

is

ideal

and

obeysRaoult’s

law.AAAx

(solvent)=

pp*

49.62

50.00

0.9924_An

(solvent)=

250

g 60.096

g

mol1

4.1600

molnB

(solute)=

wB

MB

8.69

g

MB20-BM

27

3

g

mol-1

270

g

mol-1例1.

水

A

和乙酸乙酯

B

不完全互溶，在37.55℃時(shí)兩液相呈平衡。一相中含質(zhì)量分?jǐn)?shù)wB=0.0675的酯，另一相中含wA=0.0379的水。假定 定律對每相中的溶劑都能適用，已知37.55℃時(shí)，純乙酸乙

酯的蒸氣壓是22.13kPa，純水的蒸氣壓是6.399kPa，試計(jì)算：氣相中酯和水蒸氣的分壓；總的蒸氣壓力。(乙酸乙酯的摩爾質(zhì)量為88.10g?mol-1，水的摩爾質(zhì)量為18.02g?mol-1

。)解(1)p

p*

xB

B

B

22.13kPa()

18.56kPa0.9621/88.100.0379

/

18.02

0.9621

/

88.100.9325/18.02

6.399kPa(p

p*

xA

A

A)

6.306kPa0.9325

/

18.02

0.0675

/

88.1021（2）p

=

pA+pB

=（6.306+18.56）kPa

=

24.86

kPa例

2.

100℃

時(shí)

，

純

CCl4

及

純

SnCl4

的

蒸

氣

壓

分

別

為1.933×105

Pa及0.666×105

Pa。這兩種液體可組成理想液態(tài)混合物。假定以某種配比混

的這種液態(tài)混合物，

在外壓力為1.013×105Pa的條件下，加熱到100℃時(shí)開始沸騰。計(jì)算：12該液態(tài)混合物的組成；該液態(tài)混合物開始沸騰時(shí)的第一個(gè)氣泡的組成。解：(1)設(shè)A：CCl4，B：SnCl4，則p

*=1.933×105Pa；p

*=0.666×105PaA

B

*

*AB

A

Bxp

p*

p

pBB

Axp*p

p*A

0.7261.013105

Pa

1.933105

Pa

p*

0.666105

Pa-1.933105

PaB(2)設(shè)上述溶液平衡時(shí)氣相組成為yB，則yB

p

=pB

=xB

p

**x

p

0.726

0.666

105

Pa22ByB

B

0.477p

1.013

105

PayA=1－yB=0.523例3.

液體

A

和

B

可形成理想液態(tài)混合物。把組成為

yA

=*

*0.400的蒸氣混合物放入一帶有活塞的汽缸中進(jìn)行定溫定壓壓縮

(溫度為t

)，已知溫度t

時(shí)pA

、pB

分別為40530Pa

和121590Pa。(1)計(jì)算剛開始出現(xiàn)液相時(shí)的蒸氣總壓；(2)求組分A和B的液態(tài)混合物在101325Pa下沸騰時(shí)液相的組成。23解：（1）剛開始凝結(jié)時(shí)氣相組成仍為yA

=0.400，yB=0.600，而

pB=

pyB，

故

p=pB/yB=

pB*xB/yB

①又

p

=

pA*+（

pB*－

pA*）xB

②聯(lián)立式①和式②，代入yB=0.6，pA*

=40530Pa，pB*

=121590Pa解得xB

=0.333

。再代入①，解得

p

=

67583.8Pa

。（2）由式②：101325Pa=40530Pa＋(121590Pa－40530Pa)xB解得

xB

=

0.750例4.

35℃時(shí)，純

的蒸氣壓力為43.063kPa。今測得氯仿摩爾分?jǐn)?shù)為0.3的氯仿—

溶液中，

的蒸氣分壓力為26.77kPa

，問此混合物是否為理想液態(tài)混合物？為什么？解：由定律可知，的蒸氣分壓力為：p()=

p*( )

x

(

)=43.063

×0.7kPa=30.14

kPa顯然與實(shí)驗(yàn)測得的結(jié)果不符，所以該混合物不是理想液態(tài)混合物。244.5.

The

colligative

properties

of

dilute

solution25Ideal

solution

and

ideal

dilute

solution

have

four

colligativeproperties:(1)The

lowering

of

vapour

pressure,(2)the

elevationof

boiling

point,(3)

the

depression

of

freezing

point,

and(4)

theosmotic

pressure

arising

from

the

presence

of

a

solute.In

dilutesolutions

these

properties

depend

only

on

the

number

of

soluteparticles

present,

not

their

identity.(“依數(shù)性”:“depending

onthe

collection”)We

assume

throughout

the

following

that

the

solute

is

notvolatile,

so

it

does

not

contribute

to

the

vapour.

We

also

assumethat

the

solute

does

not

dissolve

in

the

solid

solvent:

that

is,

thepure

solid

solvent

separates

when

the

solution

is

frozen.26Constant

T,

and

p,

hase

e

uilibrium,

then:(A

)*

S*

,p,ni

ATFor

the

same

substance

i,

we

have:S*

g

S*l

S*

sA

A

A

hase

e

uilibrium

then:

,A

AAdd

solute

to

pure

solvent,

then:A27AAl*l

(T

,

p)

RT

ln

xThe

chemical

potential

of

a

solvent

inthe

presence

of

a

solute.

The

loweringof

the

liquid’s

chemical

potential

has

agreater

effect

on

the

freezing

pointthan

on

the

boiling

point

.Atherefore,

lA

*l4.5.1.

The

lowering

of

vapour

pressurep

p*

p

p*

x1

1

1

21)

p*

(n2]n1

n2W2

p*[1M2W1

W2M1

M21

2

12

1M

Wp

p*

x

p*

(

M1W2

)284.5.2.

The

depression

of

freezing

pointA(pure

solid,

T,

p) A(soln,

T,

p,

xA)*

s

(T

,

p)

l

(T

,

p,

x

)

*l

(T

,

p)+RTlnxA

A

A

AAEquation

rearranges

into:*s

(T

,

p)-*l

(T

,

p)lnx

A

A

ARTGfusRT(

Gfus

)[lnxA]

1

[]

1(Hfus

)HfusTpTR

TpRT

2RT

2xAdlnx

TfHfusdT1AT*

2fRTIgnore

the

small

temperature

dependent

of

H

fuslnx

H

fus

(

1

1

)ART

*Tff29H

fusT

*

TlnxA

=ln

1

xB

H

fus

(

1

1T

*R

T

R

TT

*We

now

suppose

that

the

amount

of

solute

present

is

so

small

that

xB

1.We

can

then

wrin(1

xB

)

xBandhence

obtainusHHT

*RTT

*R

T

us

(

T

)T

*2xB

( )

mB(

1BB

A

m

M

,

it

also

follows

thatAM

mB

)Becaus

,T

T*and

x

B

BRT

*2

MRT

*2T

T

*

T

x

A

m

K

f

mBH

Hfus30Where

T

is

the

freezfiunsg

point

depresfsuison,

and

H

is

the

enthalpy

ofmelting

of

the

solvent.Above

formula

illustrates

that

the

value

of

depression

offreezing

point

of

a

solvent

is

proportional

to

molality

of

solute

mB.NOTICES:(1).

When

solution

is

frozen,

solvent

must

be

pure

solvent

solid.RT

*2

Mf

Afs

ml

H

*(2).

Kis

the

constant

of

depression

of

molar

freezingf

1

s*

lpoint

at

p,

it

can

be

calculated

if

T M

,

Δown,

and

it

canm0(

T

)mofalso

be

obtained

by

experiment.

The

k

is

the

valuecurve

ΔT/m m

at

m

=

0.(3).

The

depression

of

freezing

point

may

be

used

to

measurethe

molar

massof

a

solute.314.5.3.

The

elevation

of

boiling

pointA(soln,

T,

p,

xA) A(pure

gas,

T,

p)Al

(T

,

p,

x

)

g

(T

,

p)

*l

(T

,

p)+RTlnxA

A

A

AEquation

rearranges

into:AlnxRTRTg

*l

(T

,

p)-

(T

,

p)GVapA

A

Vap(

GVap

)H[lnxA]ppRT

2

1

[

T

]

T*2R

TTbbA1

THVap

dTRTxA

dlnx

I nore

the

small

tem

erature

de

endent

of321

1(bABbbbbRT

*TRT

*2RT

*2HVapHVap

(Tb

T

*

)HVap

Tlnx

)

xWe

may

obtain

the

expression

of elevation

of

boiling:BbxRT

*2

HVap

TBBRT

*2

MRT

*2xT

T

T

*

A

mHHvapvapT

KmmBNOTICES:The

rule

only

suits

the

solution

with

non-volatile

solute.The

elevation

of

boiling

point

may

be

used

to

measure

themolar

mass

of

a

solute.33344.5.4.

The

osmosis35The

equilibrium

involved

in

thecalculation

of

osmotic

pressure,

,

is

between

pure

solvent

A

at

apressure

p

on

one

side

of

thesemipermeable

membrane

and

Aas

a

component

of

the

mixtureon

the

other

side

of

the

embrane,where

the

pressure

is

p

+

AAAl∵

*l(T

,

p)

RT

ln

x4.5.4.

The

osmosis4.5.4.1. The

equilibrium

of

osmosishp0p

*l(T

,

p)>AlASolvent

molecules

flow

spontaneously

by

aap0pk h

mbA

A,B121

decreasesemipermeable

from

1

to

2,

and

then:

left：descend

of

solution

interface,right：ascent

of

solution

interface,

2

increaseDifference

of

two

side

height,

h,

is

the

osmoticpressure

,

which

is

the pressure

that

must

beof

the

osmosisapplied

to

the

solution

to

stop

the

influx

of

solvent.Equilibrium

is

reached

when

the

hydrostatic pressure

of

thecolumn

of

solution

matches

the

osmotic

pressure

at

h.364.5.4.2. The

osmotic

law,

x

),*

(T

,

p

)

(T

,

pA

0

AAA37A,mppV

dpor

*

(T

,

p

)

*

(T

,

p)+RT

ln

x

A

0

AVA,mdpA

RT

ln

x

ppA,m,when

x

1,

V

lA

A,m

V

*lA,mAV

*ldp

RTd

ln

xA,m

AV

*ldp

RTd

ln

xA,mA

BV

*l

RT

ln

x

RTx

RT

nB

,We

assures

that

the

pressure

range

in

the

integration

is

so

small

thatthe

molar

volume

of

the

solvent

is

a

constantAnn

V

*l

n

RT

,A

A,m

BxV

*lor

B

RTV

nB

RT

,A,mBV

nB

RT

c

RTThe

equality

is

the

osmotic

law

of

the

Van’t

Hoff

equation:It

implies

that

for

dilute

solution

the

osmotic

pressure

is

given

bythe

Van’t

Hoff

equation.384.5.4.3. The

application

(determine

the

molar

masses

ofmacromecules):

ghh

RT

BcRT

RT2

1...

c...

gM

gMc

gM

M

The

plot

involveddetermination

of39in

themolarmass

by

osmometry.

Themolar

mass

is

calculatedfrom

the

intercept

at

c=0;4.5.5.

The

relations

of

four

colligative

propertiesp

g

H

*

l

H

*

V

*l

1

2b

f1bfp*

RT

*2

RT

*2

ln

x

x

1

l m

T

s m

T

1

RT幾點(diǎn)注意：(1).這種關(guān)系存在，反映了稀溶液的四種依數(shù)性都是溶液內(nèi)在性質(zhì)在不同條件下的表現(xiàn)，它們都是由熵效應(yīng)決定的，只與溶液中溶質(zhì)的濃度有關(guān)，而與其性質(zhì)無關(guān)。2.下面給出293K時(shí)，0.001mol?kg-1溶液的依數(shù)性數(shù)值比較：Tf

0.002

K不易測量；p

5

102

Pa很難測量；

2.4kPa容易測量；(3).提供了由易測性質(zhì)求算難測性質(zhì)的途徑，從而擴(kuò)大了每個(gè)規(guī)律的應(yīng)用范圍。40例1.測定化合物的相對分子質(zhì)量：W2

M2

n

WTf

K

f

m2

KfW1K

fTf 2

K

f

2

WW

M1

1

2式中W1，W2分別為溶液中溶劑和溶質(zhì)的質(zhì)量。例2.CaCl2

~

H2O作為冷凍劑(最低共

218K)：例3.可用以鑒定物質(zhì)的純度：只有例4.

易熔合金的理論基礎(chǔ)，如：保險(xiǎn)絲為Bi,

Pb,Sn,Cd四種金屬組成的易熔合金，

343K，比Sn還要低(Tf

*(Sn)=505K)例5.指導(dǎo)冶金上調(diào)劑造渣材料的選用，如：選用SiO2~

CaO材料，可顯著降低爐渣。41例6.凝固點(diǎn)為271.3K的海水，在293.2K利用反滲透使其淡化，問需要最少加多大壓力？(水的l

H

*

6004J

mol

1

,V

*l

0.018dm

3

mol

1

)s

m

1解：設(shè)海水為稀溶液，則有：fl

H

*

V

*l

s m

T

1

,fT

l

H

*

s m

TfRT

*2RT1

fV

*lT

*2293.2K

6004J

mol1

(273.2K

271.3K)

24.9105

Pa0.018103

m3

mol1

(273.2)2因而最少需加壓力為：p

p

26

105

Pa：首先，海水濃度并不太大，接近稀溶液，這是基本點(diǎn)。其次，盡管近似估算會有一定誤差，但毛估仍有實(shí)際意義。即可使海水淡化可行與否有量化依據(jù)。第三，此法體現(xiàn)了用“直線”近似“曲線”的常用科學(xué)方法應(yīng)用。42.

. .

Distribution

rule

ofdilute

solutionBecause:x

,T

,

p,

xT

,

p

k

RTAAAAx

,T

,

p,

xT

,

p

k

RTAAAAARTxx,

(T

,

p

k

)

,

(T

,

p

k

) A

exp[

A

A

K

(T

,p)xThen:

K

T

,

pA43

AxThree

deductions

of

Distribution

rule(1)

If molecule

in

is

not

same

as

that

in

phase,

and

then:xA2則：K

(T

,p)

A

x若：2A

2，K

(T

,p)

A