第十節(jié)

DG的計(jì)算第十節(jié)

DG計(jì)算一般方法（1）熱力學(xué)關(guān)系式dG

=

–SdT

+

Vdp定義式

DG

=

DH

–

D

(TS)=

DH

–

(

T2S2

–

T1S1

)恒溫：DG

=DH

–TDS恒熵：DG

=DH

–SDT物理意義DG

=–W

'max1．恒溫過(guò)程DG

=

Vdpp1DG

=

nRTln

p22．恒壓過(guò)程3．非恒溫過(guò)程DG

=

DH

–

D

(TS)一、簡(jiǎn)單pVT變化dG

=–SdT

+VdpdG

=

Vdp理想氣體固，液dG

=

–SdT

+DGVd=pV

(p2

–

p1

)dG

=–SdT

不易積分，不常用，按3處理一、簡(jiǎn)單pVT變化例求1

mol

雙原子理想氣體，經(jīng)不同過(guò)程后的DGS1

=

200

J

K–1T1

=

300

Kp1

=

2

pq恒溫可逆真空膨脹恒外壓恒溫絕熱可逆絕熱恒外壓p2

=

pq解1．恒溫可逆p1DG

=

nRTln

p2＝–1729

J2．真空膨脹3．恒外壓恒溫終態(tài)同1DG

=

–1729

J一、簡(jiǎn)單pVT變化S

=

200

J

K–11T1

=

300

Kp1

=

2

pq．真空膨脹34．恒外壓恒溫．絕熱可逆p

=

pq4．絕熱可逆1-γ=

T2

p2

γ等熵過(guò)程1-γ終態(tài)溫度：T1

p1

γ7g=

CC

p,m=

5

\

T2

=

246.14

KV

,m焓變：

DH

=

nCp,

m

(T2

–

T1

)

=–1567

J恒熵過(guò)程：DG

=

DH

–

SDT

=

DH

–

S

(T2

–

T1

)

=

9204

J1．恒溫可逆2

TV

g-1

=常數(shù)T

p

=常數(shù)1-γγ5．p絕V熱g

恒=外常壓數(shù)恒溫：DG

=DH

–TDS恒熵：DG

=DH

–SDT一、簡(jiǎn)單pVT變化S1

=

200

J

K–1T1

=

300

Kp1

=

2

pq恒溫可逆真空膨脹恒外壓恒溫4．絕熱可逆p

=

pq3．非恒溫過(guò)程DG

5=．絕H

熱–

恒(外TS壓)D

D=

DH

–

(

T2S2

–

T1S1

)

-112Tpp2=

-R

T(T

-

T

)5．絕熱恒外壓終態(tài)溫度：CV

,m

2

1T2

=

257.1K終態(tài)熵：p,mln

T2

-

nRln

p2T1

p1S2

=

S1

+

DS

=

S1

+

nC=

201.3

JK–1焓變：DH

=

nCp,

m

(T2

–

T1

)

＝–

1248.3

J—、簡(jiǎn)單pVT變S1

=

200

J

K–1T1

=

300

Kp1

=

2

pq化1．恒溫可逆真空膨脹恒外壓恒溫絕熱可逆絕熱恒外壓p

=

pq5．絕熱恒外壓終態(tài)溫度：終態(tài)熵：

焓變：T2

=

257.1KS2

=

201.3

J

K–1DH

=

–

1248.3

J非恒溫過(guò)程：DG

=DH

–(T2S2

–T1S1

)所有計(jì)算的DG結(jié)=

果–1，24能8

–否(

用25來(lái)7.1判2斷01過(guò).3程–的30方0

向20？0

) =

6997

J·

·此處DG不能判斷過(guò)程的方向，因?yàn)椴环虾銣睾銐簵l件。二、恒溫恒壓混合過(guò)程nA

mol

A

氣體

nK

mol

K

氣體壓力

p

壓力

p混合氣體，壓力為p氣體混合混合前：pA

=pB

=p混合后：AnAAp

=

x

pnA

+

nKp

=pK

=

xK

p1AAp=

n RTln

p2計(jì)算DG：DGA=

n RTln

xADG

=

n RTln

xK

K

KDG

=

DGA

+

DGB多種氣體混合：DG

=RTSnBln

xB=nRTSxB

ln

xB此式也適合液體混合時(shí)DG的計(jì)算。計(jì)算表明，DG

<0，混合過(guò)程是自發(fā)過(guò)程。三、相變過(guò)程恒溫恒壓相變，符合DG判據(jù)條件，可作過(guò)程方向的判斷?？赡嫦嘧儯篋G

=0不可逆相變

設(shè)計(jì)可逆過(guò)程方法一：設(shè)計(jì)恒溫變壓可逆過(guò)程（已知飽和蒸汽壓數(shù)據(jù)）氣體

凝聚相pq(s或l)pq

(

g

)p飽和(s或l)p飽和(g)DG2

=

0DG

=

?DG

3DG

1三、相變過(guò)程凝聚相

凝聚相例

求1

mol

過(guò)冷

CO2

在

–59

C凝固時(shí)的DG。已知該溫度時(shí)的蒸氣壓，pl

=465.96

kPa，ps

=439.24

kPa。設(shè)計(jì)可逆過(guò)程：101.325

kPa

(

l

)101.325

kPa

(

s

)465.96

kPa

(

l

)439.24

kPa

(

s

)DG

=

?DG

5DG

1465.96

kPa

(

g

)439.24

kPa

(

g

)DG3DG

4DG2三、相變過(guò)程101.325

kPa

(

l

)101.325

kPa

(

s

)465.96

kPa

(

l

)439.24

kPa

(

s

)DG

=

?DG

5DG

1465.96

kPa

(

g

)439.24

kPa

(

g

)DG3DG

4DG2DG2

=

DG4

=

0DG1

=

Vl

(p2

–

p1

)DG5

=

Vs

(p2

–

p1

)3p1DG

=

S

DGi=

–105.2

JDG

<0，為自發(fā)過(guò)程。DG

=

nRTln

p2

=

–105.2

JT

=

214

K?

0(2)方法二：設(shè)計(jì)恒壓變溫可逆過(guò)程（已知熱容和相變熱數(shù)據(jù)）先求DS和DH，再求DG四、化學(xué)反應(yīng)已知DfHqmSqm52.28

–241.8219.45

188.75–235.31282kJ

mol–1J

K–1

mol–1mmDrGq

=

D

Hq

–TD

Sqm

r

m

r

m=

–45.76

–

298·

(

–126.17

)·10

–3

=

–8.16

kJ解

DrHq

=

–235.31

–

(

52.28

–

241.83

)

=

–45.76

kJDrSq

=

282

–

(219.45

+

188.75

)

=

–126.17

J

K–1用DfGq

求。m恒溫過(guò)程，DrG

=DrH–TDrS例

求反應(yīng)

C2H4

(g)

+ H2O

(g)

= C2H5OH

(g)

在298K時(shí)的DrGq。五、溫度對(duì)DG的影響Gibbs–Helmholtz公式G與T的關(guān)系=

-DS

=

-S

?T

p

?T

p

?G

?DG

即T將DS表示為DH，若溫度為T(mén)，DS

=DH

-DGT

?T

p

?DG

=

-

DH

-

DG-

DG

=

-DH

?T

p合并DG項(xiàng)：T

?DG

兩邊除以T2：T

2T

2=

-

DH-

DG

?T

pT

?DG

五、溫度對(duì)DG的影響Gibbs–Helmholtz公式兩邊除以T2：T

2T

2=

-

DH-

DGT

?T

p

?DG

對(duì)照商的導(dǎo)數(shù)公式：2