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物理化学电子教案 —第二章（下 ）
U Q W? ? ?
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Adiabatic process
In Fig.PV:
Slope in AB,() TppVV? ???
Slope in AC,()
S
pp
VV?
? ??
?
In reversible process the work
done （ below the line AB） is
big than the work done (below
the line AC) in adiabatic
reversible process.
1? ?
T
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Fig,adiabatic and isothermal rev,process
In an isothermal
expansion heat
continuously flows
into the system,
and so the pressure
does not fall as
much as in a
thermally
isolated,adiabatic
expansion.
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Work in adiabatic process
（ 1） work in pg,a,r,process.
2
1
= dV
V
K V
V ?? ?
11
21
= 11()( 1 )K VV??? ?????
therefore 2 2 1 1=
1
p V p VW
?
?
?
1 1 2 2p V p V K????
because
2
1
dV
V
W p V?? ? ( )p V K? ?
21()
1
n R T T
?
??
?
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A better calculation method
（ 2） another way to calculate
WU??
This equation is true for all adiabatic
expansions or contractions involving a
perfect gas or not,reversible or not,
21 = (( ) V VCTC T T? 设 与 无关）
2
1
dT VT CT? ?
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2.7 real gas
Joule-Thomson experiment
Joule在 1843年所做的气体自由膨胀实验是不够精
确的,1852年 Joule和 Thomson 设计了新的实验，称为
throttling process。
在这个实验中，使人们对实际气体的 U和 H的性质
有所了解，并且在 获得低温和气体液化 工业中有重要
应用。
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throttling process
The Joule-Thomson expansion
consists of allowing a gas to
expand through a porous plug
from a region of higher
pressure to a region of lower
pressure as depicted in Fig.
The process is carried out
steadily and adiabatically.
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throttling process
When the flow is sufficiently slow,the gas has well
defined pressure and temperature on both sides of
the restriction.
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U and H in the throttling process
11W p V? ? ?
Left,Surroundings compresses gaseous system:
work,system got：
Adiabatic Q=0, then,21U U U W? ? ? ?
Right,gaseous system expansion,work,
surr,got:
22W p V? ? ?
1 1 1 1 ( = 0 )p V V V V? ? ? ? ?
2 2 2 2 ( = 0 )p V V V V? ? ? ? ?
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U and H in the throttling process
the sum of work:
1 2 1 1 2 2W W W p V p V? ? ? ?
then
2 1 1 1 2 2U U p V p V? ? ?
Throttling process is const.-enthalpy process!
21HH?
2 2 2 1 1 1U p V U p V? ? ?
It is isenthalpic.
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Joule-Thomson coefficient：
>0 经节流膨胀后，气体温度降低。
T-J?
J - T () H
T
p
? ??
?
- The isenthalpic J-T
coefficient
J-T?
是系统的强度性质。因为节流过程的,
所以当：
d0p ?J-T?
T-J?
<0 经节流膨胀后，气体温度升高。
T-J?
=0 经节流膨胀后，气体温度不变。
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Extent of reaction
20世纪初比利时的 Dekonder引进反应进度 ? 的定义为：
B B,0
B
nn?
?
?? B
B
dd n?
??
D E F GD E F G? ? ? ?? ? ??? ? ? ? ???
?? ??,tt Dn En Fn Gn
0,0 ?? ?t D,0n E,0n F,0n G,0n
单位,mol
?
? is the stoichiometric coefficient
of component B,negative - reactants
and positive - products.
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extent of reaction
引入反应进度的优点：
在反应进行到任意时刻，可以用任一反应物
或生成物来表示反应进行的程度，所得的值都是
相同的，即：
GD E F
D E F G
dd d dd nn n n?
? ? ? ?
? ? ? ? ? ? ? ?
d?=
g
dn
f
dn
e
dn
d
dn GFED
?????
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If the transfer of heat occurs at constant volume,
and if no other forms of work are permitted,
(△ U)v = Qv,
For a specified change of state △ U is independent
of how the change is brought about,therefore the
subscript v can be dropped from U,△ U = Qv,
The significance of this equation is that if we measure the
heat transferred at constant volume we can identify it with
the change in a thermodynamic state function.
The reaction enthalpy
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The analogous result for changes
occurring at constant pressure.
△ H = Qp, When there is no work other than PV-work.
The relation between △ H and △ U,
The enthalpy of a substance differs from its
thermodynamic energy by an amount pV.
It follows that △ H = △ U +[pV](products)-[pV](reactants)
For reactions involving only solids and liquids,
△ H ? △ U
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The relation between Qp and Qv
pVQ Q n R T? ? ?
When the extent of reaction is unity:
r m r m B
B
H U R T?? ? ? ? ?
rr H U n R T? ? ? ? ?
or
)t an()( tsr e a cnp r o d u ct snn gasgasgas ???
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Const.-p and const.-V heat of reaction
????? ?? ?? pr QH 1 )1( 等压
reactants
111 VpT
products
121 VpT
（ 3）
3r H?
（ 2)cons.V
r2 VUQ??
2r H?
and
pQ VQ
1 1 2T p V
products
Deduce the relation between
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The heat at cons,volume and cons,pressure
????? ?? ?? pr QH 1 )1( 等压
反应物
111 VpT
生成物
121 VpT
（ 3）
3r H?
（ 2）等容
r2 VUQ??
2r H?
1 1 2T p V
生成物
3r2r1r HHH ????? 3r22r )( HpVU ?????
For pg，
r 3
2
0,
()
H
p V n R T
??
? ? ?
rrH U n R T? ? ? ? ?
P2V1-P1V1=P1V2- P1V1
p VQ Q n R T? ? ?
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The equation of thermochemistry
U,H are state functions，
所以方程式中应该注明物态、温度、压力、组成
等。对于固态还应注明结晶状态。
例如,298.15 K时
22H ( g,) I ( g,) 2 H I ( g,)p p p??$ $ $
-1rm ( 2 9 8, 1 5 K ) - 5 1, 8 k J m o lH? ? ?$
式中,表示反应物和生成物都
处于标准态时，在 298.15 K,反应进度为 1 mol 时
的焓变。
rm ( 2 9 8,1 5 K )H? $
p?代表气体的压力处于标准态。
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The standard molar enthalpy of reaction
Enthalpy change
All products and reactants being
in their standard states at the T
Extent of reaction is 1 mol
reaction
rm ( 2 9 8,1 5 K )H? $ Reaction temperature
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压力的标准态
随着学科的发展，压力的标准态有不同的规定：
标准态用符号, ?” 表示,表示压力标准态。p$
最老的标准态为 1 atm
1985年 GB规定为 101.325 kPa
1993年 GB规定为 1?105 Pa。 标准态的变更对凝
聚态影响不大，但对气体的热力学数据有影响，
要使用相应的热力学数据表。
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压力的标准态
气体的标准态,压力为 的理想气体，是假想态。p$
固体、液体的标准态,压力为 的纯固体或纯液体。p$
标准态 不规定温度,每个温度都有一个标准态。
一般 298.15 K时的标准态数据有表可查。为方便
起见,298.15 K用符号 表示。
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2.9 Hess’s law of constant heat summation
1840年，根据大量的实验事实 赫斯 提出了一个定律：
反应热只与起始和终了状态有关，与变化途径无关。
不管反应是一步完成的，还是分几步完成的，其热
相同，当然要保持反应条件（如温度、压力等）不
变。
It states that the standard enthalpy change in any
reaction can be expressed as the sum of the
standard enthalpy changes,at the same
temperature,of a series of reactions into which
the overall reaction may formally be divided.
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赫斯定律
例如：求 C(s)和 生成 CO(g)的反应热。g)(O
2
已知,（ 1）
（ 2）
( g )CO)(OC ( s ) 22 ?? g m,1r H?
2212C O ( g ) O ( g ) C O ( g )??m,2r H?
则 （ 1） -（ 2） 得 （ 3）
（ 3） C O ( g )g)(OC ( s )
221 ?? m,3r H?
m,2rm,1rm,3r HHH ?????
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Enthalpy of formation of a compound
?没有规定温度，一般 298.15 K时的数据有表可查。
?生成焓仅是个相对值，相对于稳定单质的焓值等于零。
standard molar enthalpy of formation
The enthalpy changes that occur when unit
amount of the compound in its standard state is
formed from its elements in their standard states：
fmH? $
（ 物质，相态，温度）
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Enthalpies of formation of compounds
For example,at 298.15 K
221122H ( g,) C l ( g,) H C l ( g,)p p p??$ $ $
-1rm ( 2 9 8, 1 5 K ) - 9 2, 3 1 k J m o lH? ? ?$
This is the standard molar enthalpy of formation
of HCl(g),
-1mf ( H C l,g,2 9 8, 1 5 K ) - 9 2, 3 1 k J m o lH? ? ?$
Reaction
enthalpy：
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Define the reaction enthalpy
? ? ? ?r m f m f m f m f m( C ) 3 ( D ) 2 ( A ) ( E )H H H H H? ? ? ? ? ? ? ? ?$ $ $ $ $
B f m
B
( B )H? ?? ? $
is the stoichiometric coefficient,the negative sign going
with the reactant and the positive with the product,
B?
3DCEA2 ???
The standard enthalpy of reaction
is the difference H(products)-H(reactants),all substances
being in their standard states at the temperature T.
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Enthalpy of combustion
The subscript ―c‖ denotes combustion。
The superscript ―?‖ denotes all substances being at
standard pressure.
The subscript ―m‖ denotes the ?being 1 mol.
At standard pressure and reaction temperature，
the change of enthalpy accompanying total oxidation
of a material is called Standard molar enthalpy of
combustion.
cmH? $
It is denoted (substance,phase,temperature)
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燃烧焓
指定产物通常规定为：
g)(COC 2? O ( l)HH 2?
g)(SOS 2? g)(NN
2?
H C l( aq )Cl ? 金属
? 游离态
显然，规定的指定产物不同，焓变值也不同，查表
时应注意。 298.15 K时的燃烧焓值有表可查。
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燃烧焓
例如：在 298.15 K及标准压力下：
O ( l )2Hg)(2 C Og)(2OC O O H ( l )CH 2223 ???
-1rm 8 7 0,3 k J m o lH? ? ? ?$
-1c m 3( C H C O O H,l,298.15 K ) - 870.3 kJ m olH? ? ?$
则
显然，根据标准摩尔燃烧焓的定义，所指定产物如
等的标准摩尔燃烧焓，在任何温度 T时，
其值均为零。
OHg ),(CO 22
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Use of the enthalpy of combustion
r m B c m
B
( 298.15 K) - ( B,298.15 K)HH ?? ? ??$$
例如：在 298.15 K和标准压力下，有反应：
l)(O2Hs)()( C O O C HO H ( l )2 C Hs)(C O O H )( 22332 ???
（ A） （ B） （ C） (D)
则
r m c m c m c m( A ) 2 ( B ) - (C)H H H H? ? ? ? ? ?$ $ $ $
The reaction enthalpy is given by：
in calculating the enthalpy of a reaction.
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Calculate the enthalpy of formation
O H ( l )CHg)(Og)(2HC ( s ) 322 21 ???
f m 3 c m c m 2( CH O H,l ) ( C,s) 2 ( H,g )H H H? ? ? ? ?$ $ $
该反应的反应焓变就是 的生成焓，则：
3C H OH(l )
例如：在 298.15 K和标准压力下：
c m 3- ( CH O H,l )H? $
using enthalpies of combustion.
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Kirchhoff`s law
2
1r m 2 r m 1
2, ( ) ( ) d T pTH T H T C T? ? ? ? ??积分式
1858年 Kirchhoff提出了 The temperature –
dependence of reaction enthalpies
()1, [ ]
pp
H C
T
?? ??
?微分式 B,mB ( B )ppCC??? ?
也是温度的函数，只要将 Cp - T的关系式代
入，就可从一个温度时的焓变求另一个温度下的焓变。 pC?
如有物质发生 相变,就要进行 分段积分 。
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Adiabatic reaction
求终态温度的示意图
设反应物起始温度均为 T1,产物温度为 T2,整个
过程保持压力不变：
rm
d = 0,0
0
12
D + E F + G
( ) ( )
ppQ
Hd e f g
T T x
?
???????
?已知
? rm(1)H? rm(2)H??
rm ( 2 9 8, 1 5 K )
D + E F + G
( 2 9 8, 1 5 K ) ( 2 9 8, 1 5 K )
Hd e f g????????
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2.12 绝热反应
根据状态函数的性质
r m r m r m( 1 ) ( 2 9 8, 1 5 K ) ( 2 ) 0H H H? ? ? ? ? ?
可由 表值计算
rm ( 2 9 8, 1 5 K )H? fm ( 2 9 8, 1 5 K )H? $
1
298
rm ( 1 ) (pTH C T?? ?? 反应物) d
可求出
2
rm 298( 2 ) (
T
pH C T?? ?? 生成物) d
从而可求出 T2值
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Phase transitions
During a change of phase of a pure substance,
such as vaporization,melting,and sublimation,
its temperature and pressure remain constant
while its entropy and volume undergo changes.
The temperature and pressure of a pure
substance consisting of two phases in
equilibrium are not independent variables.
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Reversible phase transitions
l
g
P*
T
??
H2O(l) ? H2O(g)
T,P*
enthalpy of phase transition
?vap Hm
enthalpy vaporization
Qp = ?vap Hm = Qr
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The temperature –dependence of enthalpy
H2O(l) ? H2O(g)
H2O(l) ? H2O(g)
? ?
T2,P*2
T1,P*1
?vap Hm,2 = ?vap Hm,1 + ? ?dTgmCpLmCpT
T
? ?2
1
)(,)(,
Take it as const,P
The end
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JAMES PRESCOTT JOULE
JAMES PRESCOTT JOULE (1818-1889)
English physicist,had the strength of mind to put
science ahead of beer.He owned a large brewery but
neglected its management to devote himself to scientific
research.His name is associated with Joule’s law,which states
that the rate at which heat is dissipated by a resistor is given
by I2R.He was the first to carry out precise measurements of
the mechanical equivalent of heat;and the firmly established
that work can be quantitatively converted heat.
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JOSEPH LOUIS GAY-LUSSAC
JOSEPH LOUIS GAY-LUSSAC (1778-1850)
French chemist,was a pioneer in balloon ascensions,
In 1804,Gay-Lussac made several balloon ascensions to
altitudes as high as 7000 m,where he made observations on
magnetism,temperature,humidity,and the composition of
air.He could not find any variation of compositions with
height.In 1809,he pointed out that gases combine in simple
proportions by volume;and this is still called Gay-Lussac’s
work on chlorine brought the scientist into controversy
with Sir Humphry Davy.
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JOSEPH LOUIS GAY-LUSSAC
Gay-Lussac assumed chlorine to be an oxygen-containing
compound,while Davy correctly considered it an element,a
view that Gay-Lussac eventually accepted,He showed that
prussic acid contained hydrogen but no oxygen.Lavoisier
had insisted that oxygen was the critical constituent of
acids,and Gay-Lussac,Gay-Lussac was one of the
tubing,all of which had to be imported from German,and
the French had an import duty on glass tubing.He
instructed his German supplier to seal both ends of each
piece of tubing and label the tubes ―German air.‖ The
French government had no duty listed for ―German air‖,
and he was able to import his tubing duty free.
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WILLIAM THOMSON,Lord Kelvin
WILLIAM THOMSON,Lord Kelvin (1824-1907)
Irish-born British physicist,proposed his absolute
scale of temperature,which is independent of the
thermometric substance in 1848.In one of his earliest papers
dealing with heat conduction of the earth,Thomson showed
that about 100 million years ago,the physical condition of
the earth must have been quite different from that of
today.He did fundamental work in telegraphy,and
navigation.For his services in trans-Atlantic
telegraphy,Thomson was raised to the peerage,with the title
Baron Kelvin of Larg.There was no heir to the title,and it is
now extinct.
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HESS
HESS (1802-1852)
俄国化学家,1802年出生于德国。在 1836年提出
了著名的赫斯定律。赫斯定律是热化学的最基本规律。
根据这个定律，热化学公式可以互相加减，从一些反应
的反应热可求出另一些反应的反应热。这个定律的发现
以及当时所采用的实验方法，为以后热力学第一定律的
确立奠定了实验基础。
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LINUS CARL PAULING
LINUS CARL PAULING (born 1901)
American chemist,did his earliest work in crystal
structure determinations,using X-ray diffraction.The early
years of his career coincided with the development of
quantum mechanics,and his interest in structural chemistry
led him to a variety of quantum mechanical investigations
concerned with the solid and nonsolid states of matter.After
the war,his interests turned partly to biochemistry,and
Pauling discovered the cause of sickle-cell anemia.
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LINUS CARL PAULING
He received the Nobel Prize in chemistry in 1954 for his
research into the natrue of the chemical bond and the
structure of complex molecules.In the late 1950s and early
1960s,he was one of the most vocal opponents of atomic
bomb testing,and received the Nobel Peace Prize in 1963 for
his efforts on behalf of the nuclear ban treaty,thereby
becoming the only person to win two individual Nobel
awards.
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KIRCHHOFF,GUSTER ROBERT
KIRCHHOFF,GUSTER ROBERT(1824-1887)
德国物理化学家。 1858年发表了著名的基尔霍夫
定律。该定律描述了反应的等压热效应和温度之间的关
系。根据基尔霍夫公式，可以从一个温度时的反应热求
得另一个温度时的反应热。
物理化学电子教案 —第二章（下 ）
U Q W? ? ?
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Adiabatic process
In Fig.PV:
Slope in AB,() TppVV? ???
Slope in AC,()
S
pp
VV?
? ??
?
In reversible process the work
done （ below the line AB） is
big than the work done (below
the line AC) in adiabatic
reversible process.
1? ?
T
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Fig,adiabatic and isothermal rev,process
In an isothermal
expansion heat
continuously flows
into the system,
and so the pressure
does not fall as
much as in a
thermally
isolated,adiabatic
expansion.
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Work in adiabatic process
（ 1） work in pg,a,r,process.
2
1
= dV
V
K V
V ?? ?
11
21
= 11()( 1 )K VV??? ?????
therefore 2 2 1 1=
1
p V p VW
?
?
?
1 1 2 2p V p V K????
because
2
1
dV
V
W p V?? ? ( )p V K? ?
21()
1
n R T T
?
??
?
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A better calculation method
（ 2） another way to calculate
WU??
This equation is true for all adiabatic
expansions or contractions involving a
perfect gas or not,reversible or not,
21 = (( ) V VCTC T T? 设 与 无关）
2
1
dT VT CT? ?
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2.7 real gas
Joule-Thomson experiment
Joule在 1843年所做的气体自由膨胀实验是不够精
确的,1852年 Joule和 Thomson 设计了新的实验，称为
throttling process。
在这个实验中，使人们对实际气体的 U和 H的性质
有所了解，并且在 获得低温和气体液化 工业中有重要
应用。
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throttling process
The Joule-Thomson expansion
consists of allowing a gas to
expand through a porous plug
from a region of higher
pressure to a region of lower
pressure as depicted in Fig.
The process is carried out
steadily and adiabatically.
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throttling process
When the flow is sufficiently slow,the gas has well
defined pressure and temperature on both sides of
the restriction.
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U and H in the throttling process
11W p V? ? ?
Left,Surroundings compresses gaseous system:
work,system got：
Adiabatic Q=0, then,21U U U W? ? ? ?
Right,gaseous system expansion,work,
surr,got:
22W p V? ? ?
1 1 1 1 ( = 0 )p V V V V? ? ? ? ?
2 2 2 2 ( = 0 )p V V V V? ? ? ? ?
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U and H in the throttling process
the sum of work:
1 2 1 1 2 2W W W p V p V? ? ? ?
then
2 1 1 1 2 2U U p V p V? ? ?
Throttling process is const.-enthalpy process!
21HH?
2 2 2 1 1 1U p V U p V? ? ?
It is isenthalpic.
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Joule-Thomson coefficient：
>0 经节流膨胀后，气体温度降低。
T-J?
J - T () H
T
p
? ??
?
- The isenthalpic J-T
coefficient
J-T?
是系统的强度性质。因为节流过程的,
所以当：
d0p ?J-T?
T-J?
<0 经节流膨胀后，气体温度升高。
T-J?
=0 经节流膨胀后，气体温度不变。
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Extent of reaction
20世纪初比利时的 Dekonder引进反应进度 ? 的定义为：
B B,0
B
nn?
?
?? B
B
dd n?
??
D E F GD E F G? ? ? ?? ? ??? ? ? ? ???
?? ??,tt Dn En Fn Gn
0,0 ?? ?t D,0n E,0n F,0n G,0n
单位,mol
?
? is the stoichiometric coefficient
of component B,negative - reactants
and positive - products.
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extent of reaction
引入反应进度的优点：
在反应进行到任意时刻，可以用任一反应物
或生成物来表示反应进行的程度，所得的值都是
相同的，即：
GD E F
D E F G
dd d dd nn n n?
? ? ? ?
? ? ? ? ? ? ? ?
d?=
g
dn
f
dn
e
dn
d
dn GFED
?????
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If the transfer of heat occurs at constant volume,
and if no other forms of work are permitted,
(△ U)v = Qv,
For a specified change of state △ U is independent
of how the change is brought about,therefore the
subscript v can be dropped from U,△ U = Qv,
The significance of this equation is that if we measure the
heat transferred at constant volume we can identify it with
the change in a thermodynamic state function.
The reaction enthalpy
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The analogous result for changes
occurring at constant pressure.
△ H = Qp, When there is no work other than PV-work.
The relation between △ H and △ U,
The enthalpy of a substance differs from its
thermodynamic energy by an amount pV.
It follows that △ H = △ U +[pV](products)-[pV](reactants)
For reactions involving only solids and liquids,
△ H ? △ U
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The relation between Qp and Qv
pVQ Q n R T? ? ?
When the extent of reaction is unity:
r m r m B
B
H U R T?? ? ? ? ?
rr H U n R T? ? ? ? ?
or
)t an()( tsr e a cnp r o d u ct snn gasgasgas ???
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Const.-p and const.-V heat of reaction
????? ?? ?? pr QH 1 )1( 等压
reactants
111 VpT
products
121 VpT
（ 3）
3r H?
（ 2)cons.V
r2 VUQ??
2r H?
and
pQ VQ
1 1 2T p V
products
Deduce the relation between
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The heat at cons,volume and cons,pressure
????? ?? ?? pr QH 1 )1( 等压
反应物
111 VpT
生成物
121 VpT
（ 3）
3r H?
（ 2）等容
r2 VUQ??
2r H?
1 1 2T p V
生成物
3r2r1r HHH ????? 3r22r )( HpVU ?????
For pg，
r 3
2
0,
()
H
p V n R T
??
? ? ?
rrH U n R T? ? ? ? ?
P2V1-P1V1=P1V2- P1V1
p VQ Q n R T? ? ?
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The equation of thermochemistry
U,H are state functions，
所以方程式中应该注明物态、温度、压力、组成
等。对于固态还应注明结晶状态。
例如,298.15 K时
22H ( g,) I ( g,) 2 H I ( g,)p p p??$ $ $
-1rm ( 2 9 8, 1 5 K ) - 5 1, 8 k J m o lH? ? ?$
式中,表示反应物和生成物都
处于标准态时，在 298.15 K,反应进度为 1 mol 时
的焓变。
rm ( 2 9 8,1 5 K )H? $
p?代表气体的压力处于标准态。
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The standard molar enthalpy of reaction
Enthalpy change
All products and reactants being
in their standard states at the T
Extent of reaction is 1 mol
reaction
rm ( 2 9 8,1 5 K )H? $ Reaction temperature
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压力的标准态
随着学科的发展，压力的标准态有不同的规定：
标准态用符号, ?” 表示,表示压力标准态。p$
最老的标准态为 1 atm
1985年 GB规定为 101.325 kPa
1993年 GB规定为 1?105 Pa。 标准态的变更对凝
聚态影响不大，但对气体的热力学数据有影响，
要使用相应的热力学数据表。
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压力的标准态
气体的标准态,压力为 的理想气体，是假想态。p$
固体、液体的标准态,压力为 的纯固体或纯液体。p$
标准态 不规定温度,每个温度都有一个标准态。
一般 298.15 K时的标准态数据有表可查。为方便
起见,298.15 K用符号 表示。
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2.9 Hess’s law of constant heat summation
1840年，根据大量的实验事实 赫斯 提出了一个定律：
反应热只与起始和终了状态有关，与变化途径无关。
不管反应是一步完成的，还是分几步完成的，其热
相同，当然要保持反应条件（如温度、压力等）不
变。
It states that the standard enthalpy change in any
reaction can be expressed as the sum of the
standard enthalpy changes,at the same
temperature,of a series of reactions into which
the overall reaction may formally be divided.
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赫斯定律
例如：求 C(s)和 生成 CO(g)的反应热。g)(O
2
已知,（ 1）
（ 2）
( g )CO)(OC ( s ) 22 ?? g m,1r H?
2212C O ( g ) O ( g ) C O ( g )??m,2r H?
则 （ 1） -（ 2） 得 （ 3）
（ 3） C O ( g )g)(OC ( s )
221 ?? m,3r H?
m,2rm,1rm,3r HHH ?????
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Enthalpy of formation of a compound
?没有规定温度，一般 298.15 K时的数据有表可查。
?生成焓仅是个相对值，相对于稳定单质的焓值等于零。
standard molar enthalpy of formation
The enthalpy changes that occur when unit
amount of the compound in its standard state is
formed from its elements in their standard states：
fmH? $
（ 物质，相态，温度）
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Enthalpies of formation of compounds
For example,at 298.15 K
221122H ( g,) C l ( g,) H C l ( g,)p p p??$ $ $
-1rm ( 2 9 8, 1 5 K ) - 9 2, 3 1 k J m o lH? ? ?$
This is the standard molar enthalpy of formation
of HCl(g),
-1mf ( H C l,g,2 9 8, 1 5 K ) - 9 2, 3 1 k J m o lH? ? ?$
Reaction
enthalpy：
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Define the reaction enthalpy
? ? ? ?r m f m f m f m f m( C ) 3 ( D ) 2 ( A ) ( E )H H H H H? ? ? ? ? ? ? ? ?$ $ $ $ $
B f m
B
( B )H? ?? ? $
is the stoichiometric coefficient,the negative sign going
with the reactant and the positive with the product,
B?
3DCEA2 ???
The standard enthalpy of reaction
is the difference H(products)-H(reactants),all substances
being in their standard states at the temperature T.
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Enthalpy of combustion
The subscript ―c‖ denotes combustion。
The superscript ―?‖ denotes all substances being at
standard pressure.
The subscript ―m‖ denotes the ?being 1 mol.
At standard pressure and reaction temperature，
the change of enthalpy accompanying total oxidation
of a material is called Standard molar enthalpy of
combustion.
cmH? $
It is denoted (substance,phase,temperature)
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燃烧焓
指定产物通常规定为：
g)(COC 2? O ( l)HH 2?
g)(SOS 2? g)(NN
2?
H C l( aq )Cl ? 金属
? 游离态
显然，规定的指定产物不同，焓变值也不同，查表
时应注意。 298.15 K时的燃烧焓值有表可查。
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燃烧焓
例如：在 298.15 K及标准压力下：
O ( l )2Hg)(2 C Og)(2OC O O H ( l )CH 2223 ???
-1rm 8 7 0,3 k J m o lH? ? ? ?$
-1c m 3( C H C O O H,l,298.15 K ) - 870.3 kJ m olH? ? ?$
则
显然，根据标准摩尔燃烧焓的定义，所指定产物如
等的标准摩尔燃烧焓，在任何温度 T时，
其值均为零。
OHg ),(CO 22
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Use of the enthalpy of combustion
r m B c m
B
( 298.15 K) - ( B,298.15 K)HH ?? ? ??$$
例如：在 298.15 K和标准压力下，有反应：
l)(O2Hs)()( C O O C HO H ( l )2 C Hs)(C O O H )( 22332 ???
（ A） （ B） （ C） (D)
则
r m c m c m c m( A ) 2 ( B ) - (C)H H H H? ? ? ? ? ?$ $ $ $
The reaction enthalpy is given by：
in calculating the enthalpy of a reaction.
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Calculate the enthalpy of formation
O H ( l )CHg)(Og)(2HC ( s ) 322 21 ???
f m 3 c m c m 2( CH O H,l ) ( C,s) 2 ( H,g )H H H? ? ? ? ?$ $ $
该反应的反应焓变就是 的生成焓，则：
3C H OH(l )
例如：在 298.15 K和标准压力下：
c m 3- ( CH O H,l )H? $
using enthalpies of combustion.
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Kirchhoff`s law
2
1r m 2 r m 1
2, ( ) ( ) d T pTH T H T C T? ? ? ? ??积分式
1858年 Kirchhoff提出了 The temperature –
dependence of reaction enthalpies
()1, [ ]
pp
H C
T
?? ??
?微分式 B,mB ( B )ppCC??? ?
也是温度的函数，只要将 Cp - T的关系式代
入，就可从一个温度时的焓变求另一个温度下的焓变。 pC?
如有物质发生 相变,就要进行 分段积分 。
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Adiabatic reaction
求终态温度的示意图
设反应物起始温度均为 T1,产物温度为 T2,整个
过程保持压力不变：
rm
d = 0,0
0
12
D + E F + G
( ) ( )
ppQ
Hd e f g
T T x
?
???????
?已知
? rm(1)H? rm(2)H??
rm ( 2 9 8, 1 5 K )
D + E F + G
( 2 9 8, 1 5 K ) ( 2 9 8, 1 5 K )
Hd e f g????????
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2.12 绝热反应
根据状态函数的性质
r m r m r m( 1 ) ( 2 9 8, 1 5 K ) ( 2 ) 0H H H? ? ? ? ? ?
可由 表值计算
rm ( 2 9 8, 1 5 K )H? fm ( 2 9 8, 1 5 K )H? $
1
298
rm ( 1 ) (pTH C T?? ?? 反应物) d
可求出
2
rm 298( 2 ) (
T
pH C T?? ?? 生成物) d
从而可求出 T2值
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Phase transitions
During a change of phase of a pure substance,
such as vaporization,melting,and sublimation,
its temperature and pressure remain constant
while its entropy and volume undergo changes.
The temperature and pressure of a pure
substance consisting of two phases in
equilibrium are not independent variables.
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Reversible phase transitions
l
g
P*
T
??
H2O(l) ? H2O(g)
T,P*
enthalpy of phase transition
?vap Hm
enthalpy vaporization
Qp = ?vap Hm = Qr
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The temperature –dependence of enthalpy
H2O(l) ? H2O(g)
H2O(l) ? H2O(g)
? ?
T2,P*2
T1,P*1
?vap Hm,2 = ?vap Hm,1 + ? ?dTgmCpLmCpT
T
? ?2
1
)(,)(,
Take it as const,P
The end
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JAMES PRESCOTT JOULE
JAMES PRESCOTT JOULE (1818-1889)
English physicist,had the strength of mind to put
science ahead of beer.He owned a large brewery but
neglected its management to devote himself to scientific
research.His name is associated with Joule’s law,which states
that the rate at which heat is dissipated by a resistor is given
by I2R.He was the first to carry out precise measurements of
the mechanical equivalent of heat;and the firmly established
that work can be quantitatively converted heat.
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JOSEPH LOUIS GAY-LUSSAC
JOSEPH LOUIS GAY-LUSSAC (1778-1850)
French chemist,was a pioneer in balloon ascensions,
In 1804,Gay-Lussac made several balloon ascensions to
altitudes as high as 7000 m,where he made observations on
magnetism,temperature,humidity,and the composition of
air.He could not find any variation of compositions with
height.In 1809,he pointed out that gases combine in simple
proportions by volume;and this is still called Gay-Lussac’s
work on chlorine brought the scientist into controversy
with Sir Humphry Davy.
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JOSEPH LOUIS GAY-LUSSAC
Gay-Lussac assumed chlorine to be an oxygen-containing
compound,while Davy correctly considered it an element,a
view that Gay-Lussac eventually accepted,He showed that
prussic acid contained hydrogen but no oxygen.Lavoisier
had insisted that oxygen was the critical constituent of
acids,and Gay-Lussac,Gay-Lussac was one of the
tubing,all of which had to be imported from German,and
the French had an import duty on glass tubing.He
instructed his German supplier to seal both ends of each
piece of tubing and label the tubes ―German air.‖ The
French government had no duty listed for ―German air‖,
and he was able to import his tubing duty free.
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WILLIAM THOMSON,Lord Kelvin
WILLIAM THOMSON,Lord Kelvin (1824-1907)
Irish-born British physicist,proposed his absolute
scale of temperature,which is independent of the
thermometric substance in 1848.In one of his earliest papers
dealing with heat conduction of the earth,Thomson showed
that about 100 million years ago,the physical condition of
the earth must have been quite different from that of
today.He did fundamental work in telegraphy,and
navigation.For his services in trans-Atlantic
telegraphy,Thomson was raised to the peerage,with the title
Baron Kelvin of Larg.There was no heir to the title,and it is
now extinct.
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HESS
HESS (1802-1852)
俄国化学家,1802年出生于德国。在 1836年提出
了著名的赫斯定律。赫斯定律是热化学的最基本规律。
根据这个定律，热化学公式可以互相加减，从一些反应
的反应热可求出另一些反应的反应热。这个定律的发现
以及当时所采用的实验方法，为以后热力学第一定律的
确立奠定了实验基础。
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LINUS CARL PAULING
LINUS CARL PAULING (born 1901)
American chemist,did his earliest work in crystal
structure determinations,using X-ray diffraction.The early
years of his career coincided with the development of
quantum mechanics,and his interest in structural chemistry
led him to a variety of quantum mechanical investigations
concerned with the solid and nonsolid states of matter.After
the war,his interests turned partly to biochemistry,and
Pauling discovered the cause of sickle-cell anemia.
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LINUS CARL PAULING
He received the Nobel Prize in chemistry in 1954 for his
research into the natrue of the chemical bond and the
structure of complex molecules.In the late 1950s and early
1960s,he was one of the most vocal opponents of atomic
bomb testing,and received the Nobel Peace Prize in 1963 for
his efforts on behalf of the nuclear ban treaty,thereby
becoming the only person to win two individual Nobel
awards.
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KIRCHHOFF,GUSTER ROBERT
KIRCHHOFF,GUSTER ROBERT(1824-1887)
德国物理化学家。 1858年发表了著名的基尔霍夫
定律。该定律描述了反应的等压热效应和温度之间的关
系。根据基尔霍夫公式，可以从一个温度时的反应热求
得另一个温度时的反应热。
