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Part 1,Equilibrium
Bilingual Program
2,The First Law,the concepts
1
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This chapter introduces the basic concepts of
thermodynamics,It concentrates on the
conservation of energy,The target concept of
the chapter is enthalpy,which is a very useful
book-keeping property for keeping track of the
heat output of physical processes and chemical
reactions at constant pressure.
2,The First Law,the concepts
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The basic concepts
2.1 Work,heat,and energy
2.2 The First Law
Work and heat
2.3 Expansion work
2.4 Heat transactions
2.5 Enthalpy
2.6 Adiabatic changes
Thermochemistry
2.7 Standard enthalpy changes
2.8 Standard enthalpies of formation
2.9 The temperature dependence of reaction enthalpies
2,The First Law,the concepts
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(a),Open system
can exchange matter and
energy with its surroundings.
(b),Closed system
can exchange energy with its
surroundings,but it cannot
exchange matter.
(c),An isolated system can
exchange neither energy nor
matter with its surroundings.
2.1 Work,heat,and energy
*
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(a),A diathermic system is one that
allows energy to escape as heat through
its boundary if there is a difference in
temperature between the system and its
surroundings.
(b),An adiabatic system is one that does
not permit the passage of energy as heat
through its boundary even if there is a
temperature difference between the
system and its surroundings.
2.1 Work,heat,and energy
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Exothermic process,one that releases energy as heat.
Endothermic process,one that absorbs energy as heat.
(a) When an endothermic process occurs
in an adiabatic system,the temperature
falls;(b) if the process is exothermic,then
the temperature rises,(c) When an endo-
thermic process occurs in a diathermic
container,energy enters as heat from the
surroundings,and the system remains at
the same temperature;(d) if the process is
exothermic,then energy leaves as heat,
and the process is isothermal.
2.1 Work,heat,and energy
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Heat-molecular interpretation
Heat is the transfer of energy
that ma kes u se of chaotic
molecular motion,The chaotic
motion of molecules is called
thermal motion,When a system
heats its surroundings,molecules
of the system stimulate the
thermal motion of the molecules
in the surroundings.
2.1 Work,heat,and energy
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Work-molecular interpretation
Work is the transfer of energy
that makes use of organized
motion,When a system does
work,it stimulates orderly
motion in the surroundings
2.1 Work,heat,and energy
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The basic concepts
2.1 Work,heat,and energy
2.2 The First Law
1),The internal energy
2),The conservation of energy
3),The formal statement of the First Law
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The internal energy,U,the total energy of a
system is called its internal energy;it is the total
kinetic and potential energy of the molecules
composing the system,
The internal energy is a state function,It is a
extensive property.
Internal energy,heat,and work are measured
in the same units,the Joule (J),
2.2 The First Law
1),The internal energy
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2),The conservation of energy
w>0 or q>0,energy is transferred to the system as work or heat,
w <0 or q<0,energy is lost from the system as work or heat.
△ U = q + w —the First Law of thermodynamics
If write w for the work done on a system,q for the energy
transferred as heat to a system,and △ U for the resulting
change in internal energy,then it follows that
The internal energy of a system may be changed either
by doing work on the system or by heating it,
2.2 The First Law
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3),The formal statement of the First Law
The work needed to change an adiabatic system from
one specified state to another specified state is the same
however the work is done.
wad =Uf - Ui =△ U
h =Af - Ai =△ A
The same quantity of work must
be done on an adiabatic system to
achieve the same change of state
even though different means of
achieving that work may be used.
2.2 The First Law
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Work and heat
2.3 Expansion work
1),The general expression for work
2),Free expansion
3),Expansion against constant pressure
4),Reversible expansion
5),Isothermal reversible expansion
2.4 Heat transactions
2.5 Enthalpy
2.6 Adiabatic changes
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2.3 Expansion work
wqU ddd
If the work done on a system is dw and the energy
supplied to it as heat is dq,the infinitesimal change
of internal energy dU is:
Infinitesimal form
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1),The general expression for work
zFw dd
The work required to move an object a distance dz
against an opposing force of magnitude F is
The negative sign presents that when the system moves
an object against an opposing force,the internal energy
of the system doing the work will decrease
2.3 Expansion work
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When the system expands through a
distance dz against pex,the work done,
dw = - pex A dz
where Adz is the change in volume,dV,
in the course of the expansion,
1),The general expression for work
the area of the piston,A
the external pressure,pex
the force on the outer face,F =pex A.
2.3 Expansion work
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The work done when the system
expands by dV against a pressure pex is
1),The general expression for work
VpzFw ddd ex
f
i
dexV
V
Vpw
The total work done,w:
2.3 Expansion work
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In general,the work done on a system can be expressed in
the form dw = -Fdz,where F is a 'generalized force'
and dz is a 'generalized displacement',
Type of w dw Comments Units
Expansion -pexdV pex
dV
Pa
m3
Surface expansion?d
d?
Nm-1
m2
Extension f dl f
dl
N
m
Electrical?dq?
dq
V
C
Varieties of work
2.3 Expansion work
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No work is done when a system expands freely,
Expansion of this kind occurs when a system
expands into a vacuum.
2),Free expansion
pex = 0 dw = 0 for each stage of the expansion
w = 0
Free expansion,the expansion against zero opposing force.
2.3 Expansion work
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The magnitude of w,is equal to the
area beneath the horizontal line at
p = pex lying between the initial and
final volumes.
fi dex VV Vpw
3),Expansion against constant pressure
VpVVp exifex )(
For an expansion of the external
pressure constant throughout of
the expansion
2.3 Expansion work
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A reversible change in thermodynamics is a change
that can be reversed by an infinitesimal modification
of a variable,The key word 'infinitesimal' sharpens
the every-day meaning of the word 'reversible' as
something that can change direction.
4),Reversible expansion
2.3 Expansion work
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To achieve a reversible expansion,set the external
pressure pex equal to the pressure of the confined
gas p at each stage of the expansion,pex = p
VpVpw ddd exr e v
4),Reversible expansion
fi dVV Vpw
This equation is for the reversible expansion,The
integral can be evaluated once we know how the
pressure of the confined gas depends on its volume.
2.3 Expansion work
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5),Isothermal reversible expansion
For the isothermal,reversible
expansion of a perfect gas:
i
f
r e v
f
i
d
V
V
n R T
V
V
n R Tw
V
VT
ln
,
2.3 Expansion work
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Example
Calculate the work done when 50 g of iron reacts with
hydrochloric acid in (a) a closed vessel of fixed volume,
(b) an open beaker at 25 ℃,
Method,We need to judge the magnitude of the volume
change,and then to decide how the process occurs,If
there is no change in volume,there is no expansion
work however the process takes place,If the system
expands against a constant external pressure,the work
can be calculated,A general feature of processes in
which a condensed phase changes into a gas is that the
volume of the former may usually be neglected relative
to that of the gas it forms.
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nR TpnR TpVpw
ex
exex
Answer,In (a) the volume cannot change,so no work
is done and w = 0,In (b) the gas drives back the
atmosphere and therefore w =-pex ΔV,If neglect the
initial volume,ΔV = Vf – Vi? Vf = nRT/ pex,where n
is the amount of H2 produced,
Example
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n R Tw
Because 1 mol H2 is generated when 1 mol Fe is
consumed
The reaction is
Fe(s) + 2HCl(aq)? FeCl2 ( aq) + H2 ( g)
2,2 k J
) ( 2 9 8,1 5 K )m o l K J ( 8,3 1 45 5,8 5 g m o lg50 111
Example
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2.4 Heat transactions
The change in internal energy of a system generally is
dU = dq + dwexp + dwe
dU = dq
at constant volume,no additional work,For a
measurable change:
U = qV
where the subscript means a change at constant
volume.
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VV TUC
CV – Extensive properties
CV,m – Intensive properties
Heat capacity
The internal energy of a substance
increases when its temperature is
raised,The slope of the curve at any
T is called the heat capacity,CV,of
the system at that T.
2.4 Heat transactions
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A change in internal energy to a change in temperature of
a constant-volume system follows:
Heat capacity
TCU V dd?
(at constant V )
TCU V
(at constant V )
An infinitesimal change in T brings about an infinitesi-
mal change in U,and the constant of proportionality is
CV at constant V,If CV is independent of T over the
range of temperatures of interest,a measurable change
in internal ΔU
2.4 Heat transactions
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Because a change in U can be identified with the heat
supplied at constant V,the above equation is then
Heat capacity
TCq VV
This relation provides a simple way of measuring CV
of a sample,The ratio of the heat supplied to T rise it
causes is the heat capacity of the sample,
2.4 Heat transactions
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2.5 Enthalpy
1),The definition of enthalpy
The change in internal energy is not equal to the heat
supplied when the system is free to change its volume.
Under these circumstances some of the energy supplied
as heat to the system is returned to the surroundings as
expansion work,However,we shall now show that in
this case the heat supplied at constant pressure is equal
to the change in another thermodynamic property of
the system,the enthalpy,H.
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1),The definition of enthalpy
The enthalpy,a state function,is defined as:
H = U+ pV
pqH
(at constant pressure,no additional work)qH dd?
or
The change in enthalpy between any pair of initial and
final states is independent of the path between them,and
is equal to the heat supplied at constant pressure to
a system:
H?
2.5 Enthalpy
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pp THC
The H of a substance increases as its
T is raised,The relation between the
increase in H and the increase in T
depends on the conditions,The most
important condition is constant p,
and the slope of a graph of H
against T at constant p is called the
h e a t
capacity at constant p,Cp
2),The variation of enthalpy with temperature
2.5 Enthalpy
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2),The variation of enthalpy with temperature
For infinitesimal changes of temperature:
TCH p dd?
(at constant p )
If the heat capacity is constant over the range of
temperatures of interest:
TCH p
(at constant p )
TCq pp
2.5 Enthalpy
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3),The relation between heat capacities
nRCC Vp
In most cases the heat capacity at constant p of a
system is larger than its heat capacity at constant V.
For the two heat capacities of a perfect gas:
RCC mVmp,,
2.5 Enthalpy
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Example - Relating ΔH and ΔU
The internal energy change when 1.0 mol CaCO3 in
the form of calcite converts to aragonite is +0.21 kJ.
Calculate the difference between the enthalpy change
and the change in internal energy when the pressure
is 1.0 bar given that the densities of the solids are 2.71
gcm-3 and 2.93 gcm-3,respectively.
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Method,The starting point for the calculation is the
relation between the enthalpy of a substance and its
internal energy,The difference between the two
quantities can be expressed in terms of the pressure
and the difference of their molar volumes calculated
by ρ=M/Vm,
Example - Relating ΔH and ΔU
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( c a l c i t e ))( a r a g o n i t e HHH
Answer,The change in enthalpy when the transfor-
mation occurs is
( c )( c )( a )( a ) pVUpVU
( c )( a ) VVpU
VpU
Example - Relating ΔH and ΔU
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For the volume of 1.0 mol CaCO3 (100 g) and ρ=M/Vm:
33 cm34)2,9 3 ( g c m1 0 0 ( g )( a )V
33 cm37)2,7 1 ( g c m1 0 0 ( g )( c )V
J30, UH
365 m103 7 )( 3 4Pa1001,Vp
J30P a m30 3,,
Example - Relating ΔH and ΔU
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Water is heated to boiling under a pressure of 1.0
atm,When an electric current of 0.50 A from a 12
V supply is passed for 300 s through a resistance
in thermal contact with it,it is found that 0.798 g
of water is vaporized,Calculate the molar
internal energy and enthalpy changes at the
boiling point (373.15 K).
Example - Calculating a change in enthalpy
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Method,Because the vaporization occurs at constant
pressure,the enthalpy change is equal to the heat
supplied by the heater,Therefore,calculate the heat
supplied and express that as an enthalpy change; then
convert the result to a molar enthalpy change by
division by the amount of H2O molecules vaporized.
To convert from enthalpy change to internal energy
change,we assume that the vapour is a perfect gas.
Example - Calculating a change in enthalpy
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tIq p V?
Answer,The enthalpy change is
1,8 J( 3 0 0 s )( 1 2 V )( 0,5 0 A )
1,8 J pqH
Because 0.798 g of water is 0.0443 mol H2O,the enthalpy
of vaporization per mole of H2O is
1k J m o l41
0,0 4 4 3 m o l
1,8 J
pm qH
In the process H2O (l)? H2O(g) the change in the
amount of gas molecules Δn = +1 mol
RTHU mm 1k J m o l38
Example - Calculating a change in enthalpy
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1),The work of adiabatic change
2.6 Adiabatic changes
The change in internal energy:
TCTTCU VV )( if
For adiabatic expansion
q=0,U= wad
TCw Vad
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1),The work of adiabatic change
For the reversible:
R
C
cTVTV mVcc,iiff,
c
V
V
TT
1
f
i
if
2.6 Adiabatic changes
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2),Heat capacity ratio and adiabats
The change in pressure resulting from an adiabatic,
reversible expansion of a perfect gas:
c o n s a n t?γpV
m,
m,
V
p
C
C
γ?
The heat capacity ratio is defined as:
2.6 Adiabatic changes
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2),Heat capacity ratio and adiabats
Since the heat capacity at constant pressure is greater
than the heat capacity at constant volume
γ?1
m,
m,
V
V
C
RC
γ
For a perfect gas,it follows that:
2.6 Adiabatic changes
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Thermochemistry
2.7 Standard enthalpy changes
1),The standard enthalpy change
2),Enthalpies of physical change
3),Enthalpies of chemical change
4),Hess's law
2.8 Standard enthalpies of formation
2.9 The T dependence of reaction enthalpies
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Thermochemistry,a branch of thermodynamics,
concentrating on the heat produced or required
by chemical reactions.
Process releasing heat－ exothermic:
H < 0.
Process absorbing heat－ endothermic,
H > 0.
Thermochemistry
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2.7 Standard enthalpy changes
The standard enthalpy change,,the change
in enthalpy for a process in which the initial and
final substances are in their standard states.
H
The standard state,of the pure form a substance
at 1 bar and at a specified temperature.
Conventionally,T=298.15 K (25 ℃ )
1),The standard enthalpy change
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2),Enthalpies of physical change
The standard enthalpy of transition,,the
standard enthalpy change accompaning a change of
physical state,
trsH
The standard enthalpy of vaporization - enthalpy change per
mole when a pure liquid at 1 bar vaporizes to a gas at 1 bar:
O ( g )HO ( l )H 22 vapH? (373 K) = +40.66 kJmol
-1?vapH
O ( l )HO ( s )H 22
The standard enthalpy of fusion - enthalpy change
accompanying the conversion of a solid to a liquid
fusH?
(273 K) = +6.01 kJmol-1?fusH
2.7 Standard enthalpy changes
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2),Enthalpies of physical change
:O ( g )HO ( s )H 2nS u b l i m a t i o2 Hsub?
:O ( g )HO ( l )H 2onV a p o r i z a t i2
:O ( l )HO ( s )H 2F u s i o n2
Hvap?
Hfus?
Hfus?
Hsub?
Hvap?
= +H
f us?Hsub? Hva p?
2.7 Standard enthalpy changes
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2),Enthalpies of physical change
A
B
H?
H?
H?H?
2.7 Standard enthalpy changes
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3),Enthalpies of chemical change
e.g.
O ( l )H)(CO2O( g )CH 224 g2
= -890 kJmol-1H
r?
The standard reaction enthalpy,,the change in
enthalpy when reactants in their standard states change
to products in their standard states.
Hr?
2.7 Standard enthalpy changes
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3),Enthalpies of chemical change
e.g.
O ( l )H)g(CO( s )OHC 26126 66 2
= -2808kJmol-1H
c?
Hc?
The standard enthalpy of combustion,,the
standard reaction enthalpy for the complete oxidation of
an organic compound to CO2 and H2O.
2.7 Standard enthalpy changes
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The standard molar enthalpy,
mH
ν
— the stoichiometric coefficients;
m
R e a c t a n t s
m
P r o d u c t s
r HνHνH
)J(m
J
Jr
HνH
)J(m?H — the standard molar enthalpy of species J
3),Enthalpies of chemical change
2.7 Standard enthalpy changes
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The standard enthalpy of an overall
reaction is the sum of the standard
enthalpies of the individual reactions
into which a reaction may be divided.
4),Hess's law
2.7 Standard enthalpy changes
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This reaction can be recreated from the following sum
CH2=CHCH3(g) + 5O2(g)? 3CO2 (g) +4H2O (l),-2220 kJmol-1
CH2=CHCH3(g) + H2(g)? CH3CH2CH3 (g),-124 kJmol-1
e.g.
Calculate the standard enthalpy of combustion of propane.
286
2 2 2 05
124
2
1
)g(O)g(H)l(OH
)l(OH)g(CO)g(O)g(HC
)g(HC)g(H)g(HC
222
22283
83263
2 0 5 83329 )l(OH)g(CO)g(O)g(HC 22263
1 k J m o l /r H
2.7 Standard enthalpy changes
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The standard enthalpy of formation,:the
standard reaction enthalpy for the formation of
the compound from its elements in their
reference states,
The reference state of an element is its most
stable state at the specified temperature and 1
bar.
2.8 Standard enthalpy of formation
Hf?
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Products
Reactants
Hr
Elements
The reaction enthalpy in terms of enthalpies of formation
f
R e a c t a n t s
f
P r o d u c t s
r HνHνH
)J(f
J
Jr
HνH
2.8 Standard enthalpy of formation
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Illustration
The standard reaction enthalpy of
2HN3(l) + 2NO(g)?H2O2,(l) + 4N2(g)
is calculated as follows,
= {-187.78 + 4(0)} - {2(264.0) + 2(90.25)}
= -892.3 kJmol-1
gN O,l,HN
g,Nl,OH
f3f
2f22fr
HH
HHH
22
4
2.8 Standard enthalpy of formation
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2.9 The temperature dependence of reaction enthalpies
TCTHTH T p d2
1
T
12
Standard reaction enthalpies at different T can be estimated
from heat capacities and reaction enthalpy at some other T,
If a substance is heated from T1toT2,
The standard enthalpy changes:
TCTHTH pTT d2
1 r1r2r
From
TCH p dd?
(at constant p )
□ Kirchhoff ’ law
The estimation from Cp and at some other T H
r
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m,
R e a c t a n t s
m,
P r o d u c t s
r ppp CνCνC
Where is the difference of the molar heat capacities of
products and reactants under standard conditions weighted
by the the stoichiometric coefficients:
pCr?
More generally,
J
mr ( J )
,pJp CνC
2.9 The temperature dependence of reaction enthalpies
□ Kirchhoff ’ law
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m,
R e a c t a n t s
m,
P r o d u c t s
r ppp CνCνC
The change in reaction enthalpy reflects the difference
in the changes of the enthalpies.
TCTHTH pTT d2
1 r1r2r
2.9 The temperature dependence of reaction enthalpies
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1,The properties of gases
2,The First Law,the concepts
3,The First Law,the machinery
4,The Second Law,the concepts
5,The Second Law,the machinery
6,Physical transformations of pure substances
7,Simple mixtures
8,Phase diagrams
9,Chemical equilibrium
10,Electrochemistry
Part 1,Equilibrium
Part 1,Equilibrium
Bilingual Program
2,The First Law,the concepts
1
版权所有：华东理工大学物理化学教研室 2
This chapter introduces the basic concepts of
thermodynamics,It concentrates on the
conservation of energy,The target concept of
the chapter is enthalpy,which is a very useful
book-keeping property for keeping track of the
heat output of physical processes and chemical
reactions at constant pressure.
2,The First Law,the concepts
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The basic concepts
2.1 Work,heat,and energy
2.2 The First Law
Work and heat
2.3 Expansion work
2.4 Heat transactions
2.5 Enthalpy
2.6 Adiabatic changes
Thermochemistry
2.7 Standard enthalpy changes
2.8 Standard enthalpies of formation
2.9 The temperature dependence of reaction enthalpies
2,The First Law,the concepts
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(a),Open system
can exchange matter and
energy with its surroundings.
(b),Closed system
can exchange energy with its
surroundings,but it cannot
exchange matter.
(c),An isolated system can
exchange neither energy nor
matter with its surroundings.
2.1 Work,heat,and energy
*
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(a),A diathermic system is one that
allows energy to escape as heat through
its boundary if there is a difference in
temperature between the system and its
surroundings.
(b),An adiabatic system is one that does
not permit the passage of energy as heat
through its boundary even if there is a
temperature difference between the
system and its surroundings.
2.1 Work,heat,and energy
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Exothermic process,one that releases energy as heat.
Endothermic process,one that absorbs energy as heat.
(a) When an endothermic process occurs
in an adiabatic system,the temperature
falls;(b) if the process is exothermic,then
the temperature rises,(c) When an endo-
thermic process occurs in a diathermic
container,energy enters as heat from the
surroundings,and the system remains at
the same temperature;(d) if the process is
exothermic,then energy leaves as heat,
and the process is isothermal.
2.1 Work,heat,and energy
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Heat-molecular interpretation
Heat is the transfer of energy
that ma kes u se of chaotic
molecular motion,The chaotic
motion of molecules is called
thermal motion,When a system
heats its surroundings,molecules
of the system stimulate the
thermal motion of the molecules
in the surroundings.
2.1 Work,heat,and energy
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Work-molecular interpretation
Work is the transfer of energy
that makes use of organized
motion,When a system does
work,it stimulates orderly
motion in the surroundings
2.1 Work,heat,and energy
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The basic concepts
2.1 Work,heat,and energy
2.2 The First Law
1),The internal energy
2),The conservation of energy
3),The formal statement of the First Law
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The internal energy,U,the total energy of a
system is called its internal energy;it is the total
kinetic and potential energy of the molecules
composing the system,
The internal energy is a state function,It is a
extensive property.
Internal energy,heat,and work are measured
in the same units,the Joule (J),
2.2 The First Law
1),The internal energy
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2),The conservation of energy
w>0 or q>0,energy is transferred to the system as work or heat,
w <0 or q<0,energy is lost from the system as work or heat.
△ U = q + w —the First Law of thermodynamics
If write w for the work done on a system,q for the energy
transferred as heat to a system,and △ U for the resulting
change in internal energy,then it follows that
The internal energy of a system may be changed either
by doing work on the system or by heating it,
2.2 The First Law
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3),The formal statement of the First Law
The work needed to change an adiabatic system from
one specified state to another specified state is the same
however the work is done.
wad =Uf - Ui =△ U
h =Af - Ai =△ A
The same quantity of work must
be done on an adiabatic system to
achieve the same change of state
even though different means of
achieving that work may be used.
2.2 The First Law
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Work and heat
2.3 Expansion work
1),The general expression for work
2),Free expansion
3),Expansion against constant pressure
4),Reversible expansion
5),Isothermal reversible expansion
2.4 Heat transactions
2.5 Enthalpy
2.6 Adiabatic changes
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2.3 Expansion work
wqU ddd
If the work done on a system is dw and the energy
supplied to it as heat is dq,the infinitesimal change
of internal energy dU is:
Infinitesimal form
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1),The general expression for work
zFw dd
The work required to move an object a distance dz
against an opposing force of magnitude F is
The negative sign presents that when the system moves
an object against an opposing force,the internal energy
of the system doing the work will decrease
2.3 Expansion work
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When the system expands through a
distance dz against pex,the work done,
dw = - pex A dz
where Adz is the change in volume,dV,
in the course of the expansion,
1),The general expression for work
the area of the piston,A
the external pressure,pex
the force on the outer face,F =pex A.
2.3 Expansion work
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The work done when the system
expands by dV against a pressure pex is
1),The general expression for work
VpzFw ddd ex
f
i
dexV
V
Vpw
The total work done,w:
2.3 Expansion work
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In general,the work done on a system can be expressed in
the form dw = -Fdz,where F is a 'generalized force'
and dz is a 'generalized displacement',
Type of w dw Comments Units
Expansion -pexdV pex
dV
Pa
m3
Surface expansion?d
d?
Nm-1
m2
Extension f dl f
dl
N
m
Electrical?dq?
dq
V
C
Varieties of work
2.3 Expansion work
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No work is done when a system expands freely,
Expansion of this kind occurs when a system
expands into a vacuum.
2),Free expansion
pex = 0 dw = 0 for each stage of the expansion
w = 0
Free expansion,the expansion against zero opposing force.
2.3 Expansion work
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The magnitude of w,is equal to the
area beneath the horizontal line at
p = pex lying between the initial and
final volumes.
fi dex VV Vpw
3),Expansion against constant pressure
VpVVp exifex )(
For an expansion of the external
pressure constant throughout of
the expansion
2.3 Expansion work
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A reversible change in thermodynamics is a change
that can be reversed by an infinitesimal modification
of a variable,The key word 'infinitesimal' sharpens
the every-day meaning of the word 'reversible' as
something that can change direction.
4),Reversible expansion
2.3 Expansion work
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To achieve a reversible expansion,set the external
pressure pex equal to the pressure of the confined
gas p at each stage of the expansion,pex = p
VpVpw ddd exr e v
4),Reversible expansion
fi dVV Vpw
This equation is for the reversible expansion,The
integral can be evaluated once we know how the
pressure of the confined gas depends on its volume.
2.3 Expansion work
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5),Isothermal reversible expansion
For the isothermal,reversible
expansion of a perfect gas:
i
f
r e v
f
i
d
V
V
n R T
V
V
n R Tw
V
VT
ln
,
2.3 Expansion work
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Example
Calculate the work done when 50 g of iron reacts with
hydrochloric acid in (a) a closed vessel of fixed volume,
(b) an open beaker at 25 ℃,
Method,We need to judge the magnitude of the volume
change,and then to decide how the process occurs,If
there is no change in volume,there is no expansion
work however the process takes place,If the system
expands against a constant external pressure,the work
can be calculated,A general feature of processes in
which a condensed phase changes into a gas is that the
volume of the former may usually be neglected relative
to that of the gas it forms.
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nR TpnR TpVpw
ex
exex
Answer,In (a) the volume cannot change,so no work
is done and w = 0,In (b) the gas drives back the
atmosphere and therefore w =-pex ΔV,If neglect the
initial volume,ΔV = Vf – Vi? Vf = nRT/ pex,where n
is the amount of H2 produced,
Example
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n R Tw
Because 1 mol H2 is generated when 1 mol Fe is
consumed
The reaction is
Fe(s) + 2HCl(aq)? FeCl2 ( aq) + H2 ( g)
2,2 k J
) ( 2 9 8,1 5 K )m o l K J ( 8,3 1 45 5,8 5 g m o lg50 111
Example
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2.4 Heat transactions
The change in internal energy of a system generally is
dU = dq + dwexp + dwe
dU = dq
at constant volume,no additional work,For a
measurable change:
U = qV
where the subscript means a change at constant
volume.
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VV TUC
CV – Extensive properties
CV,m – Intensive properties
Heat capacity
The internal energy of a substance
increases when its temperature is
raised,The slope of the curve at any
T is called the heat capacity,CV,of
the system at that T.
2.4 Heat transactions
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A change in internal energy to a change in temperature of
a constant-volume system follows:
Heat capacity
TCU V dd?
(at constant V )
TCU V
(at constant V )
An infinitesimal change in T brings about an infinitesi-
mal change in U,and the constant of proportionality is
CV at constant V,If CV is independent of T over the
range of temperatures of interest,a measurable change
in internal ΔU
2.4 Heat transactions
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Because a change in U can be identified with the heat
supplied at constant V,the above equation is then
Heat capacity
TCq VV
This relation provides a simple way of measuring CV
of a sample,The ratio of the heat supplied to T rise it
causes is the heat capacity of the sample,
2.4 Heat transactions
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2.5 Enthalpy
1),The definition of enthalpy
The change in internal energy is not equal to the heat
supplied when the system is free to change its volume.
Under these circumstances some of the energy supplied
as heat to the system is returned to the surroundings as
expansion work,However,we shall now show that in
this case the heat supplied at constant pressure is equal
to the change in another thermodynamic property of
the system,the enthalpy,H.
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1),The definition of enthalpy
The enthalpy,a state function,is defined as:
H = U+ pV
pqH
(at constant pressure,no additional work)qH dd?
or
The change in enthalpy between any pair of initial and
final states is independent of the path between them,and
is equal to the heat supplied at constant pressure to
a system:
H?
2.5 Enthalpy
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pp THC
The H of a substance increases as its
T is raised,The relation between the
increase in H and the increase in T
depends on the conditions,The most
important condition is constant p,
and the slope of a graph of H
against T at constant p is called the
h e a t
capacity at constant p,Cp
2),The variation of enthalpy with temperature
2.5 Enthalpy
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2),The variation of enthalpy with temperature
For infinitesimal changes of temperature:
TCH p dd?
(at constant p )
If the heat capacity is constant over the range of
temperatures of interest:
TCH p
(at constant p )
TCq pp
2.5 Enthalpy
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3),The relation between heat capacities
nRCC Vp
In most cases the heat capacity at constant p of a
system is larger than its heat capacity at constant V.
For the two heat capacities of a perfect gas:
RCC mVmp,,
2.5 Enthalpy
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Example - Relating ΔH and ΔU
The internal energy change when 1.0 mol CaCO3 in
the form of calcite converts to aragonite is +0.21 kJ.
Calculate the difference between the enthalpy change
and the change in internal energy when the pressure
is 1.0 bar given that the densities of the solids are 2.71
gcm-3 and 2.93 gcm-3,respectively.
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Method,The starting point for the calculation is the
relation between the enthalpy of a substance and its
internal energy,The difference between the two
quantities can be expressed in terms of the pressure
and the difference of their molar volumes calculated
by ρ=M/Vm,
Example - Relating ΔH and ΔU
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( c a l c i t e ))( a r a g o n i t e HHH
Answer,The change in enthalpy when the transfor-
mation occurs is
( c )( c )( a )( a ) pVUpVU
( c )( a ) VVpU
VpU
Example - Relating ΔH and ΔU
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For the volume of 1.0 mol CaCO3 (100 g) and ρ=M/Vm:
33 cm34)2,9 3 ( g c m1 0 0 ( g )( a )V
33 cm37)2,7 1 ( g c m1 0 0 ( g )( c )V
J30, UH
365 m103 7 )( 3 4Pa1001,Vp
J30P a m30 3,,
Example - Relating ΔH and ΔU
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Water is heated to boiling under a pressure of 1.0
atm,When an electric current of 0.50 A from a 12
V supply is passed for 300 s through a resistance
in thermal contact with it,it is found that 0.798 g
of water is vaporized,Calculate the molar
internal energy and enthalpy changes at the
boiling point (373.15 K).
Example - Calculating a change in enthalpy
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Method,Because the vaporization occurs at constant
pressure,the enthalpy change is equal to the heat
supplied by the heater,Therefore,calculate the heat
supplied and express that as an enthalpy change; then
convert the result to a molar enthalpy change by
division by the amount of H2O molecules vaporized.
To convert from enthalpy change to internal energy
change,we assume that the vapour is a perfect gas.
Example - Calculating a change in enthalpy
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tIq p V?
Answer,The enthalpy change is
1,8 J( 3 0 0 s )( 1 2 V )( 0,5 0 A )
1,8 J pqH
Because 0.798 g of water is 0.0443 mol H2O,the enthalpy
of vaporization per mole of H2O is
1k J m o l41
0,0 4 4 3 m o l
1,8 J
pm qH
In the process H2O (l)? H2O(g) the change in the
amount of gas molecules Δn = +1 mol
RTHU mm 1k J m o l38
Example - Calculating a change in enthalpy
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1),The work of adiabatic change
2.6 Adiabatic changes
The change in internal energy:
TCTTCU VV )( if
For adiabatic expansion
q=0,U= wad
TCw Vad
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1),The work of adiabatic change
For the reversible:
R
C
cTVTV mVcc,iiff,
c
V
V
TT
1
f
i
if
2.6 Adiabatic changes
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2),Heat capacity ratio and adiabats
The change in pressure resulting from an adiabatic,
reversible expansion of a perfect gas:
c o n s a n t?γpV
m,
m,
V
p
C
C
γ?
The heat capacity ratio is defined as:
2.6 Adiabatic changes
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2),Heat capacity ratio and adiabats
Since the heat capacity at constant pressure is greater
than the heat capacity at constant volume
γ?1
m,
m,
V
V
C
RC
γ
For a perfect gas,it follows that:
2.6 Adiabatic changes
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Thermochemistry
2.7 Standard enthalpy changes
1),The standard enthalpy change
2),Enthalpies of physical change
3),Enthalpies of chemical change
4),Hess's law
2.8 Standard enthalpies of formation
2.9 The T dependence of reaction enthalpies
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Thermochemistry,a branch of thermodynamics,
concentrating on the heat produced or required
by chemical reactions.
Process releasing heat－ exothermic:
H < 0.
Process absorbing heat－ endothermic,
H > 0.
Thermochemistry
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2.7 Standard enthalpy changes
The standard enthalpy change,,the change
in enthalpy for a process in which the initial and
final substances are in their standard states.
H
The standard state,of the pure form a substance
at 1 bar and at a specified temperature.
Conventionally,T=298.15 K (25 ℃ )
1),The standard enthalpy change
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2),Enthalpies of physical change
The standard enthalpy of transition,,the
standard enthalpy change accompaning a change of
physical state,
trsH
The standard enthalpy of vaporization - enthalpy change per
mole when a pure liquid at 1 bar vaporizes to a gas at 1 bar:
O ( g )HO ( l )H 22 vapH? (373 K) = +40.66 kJmol
-1?vapH
O ( l )HO ( s )H 22
The standard enthalpy of fusion - enthalpy change
accompanying the conversion of a solid to a liquid
fusH?
(273 K) = +6.01 kJmol-1?fusH
2.7 Standard enthalpy changes
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2),Enthalpies of physical change
:O ( g )HO ( s )H 2nS u b l i m a t i o2 Hsub?
:O ( g )HO ( l )H 2onV a p o r i z a t i2
:O ( l )HO ( s )H 2F u s i o n2
Hvap?
Hfus?
Hfus?
Hsub?
Hvap?
= +H
f us?Hsub? Hva p?
2.7 Standard enthalpy changes
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2),Enthalpies of physical change
A
B
H?
H?
H?H?
2.7 Standard enthalpy changes
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3),Enthalpies of chemical change
e.g.
O ( l )H)(CO2O( g )CH 224 g2
= -890 kJmol-1H
r?
The standard reaction enthalpy,,the change in
enthalpy when reactants in their standard states change
to products in their standard states.
Hr?
2.7 Standard enthalpy changes
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3),Enthalpies of chemical change
e.g.
O ( l )H)g(CO( s )OHC 26126 66 2
= -2808kJmol-1H
c?
Hc?
The standard enthalpy of combustion,,the
standard reaction enthalpy for the complete oxidation of
an organic compound to CO2 and H2O.
2.7 Standard enthalpy changes
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The standard molar enthalpy,
mH
ν
— the stoichiometric coefficients;
m
R e a c t a n t s
m
P r o d u c t s
r HνHνH
)J(m
J
Jr
HνH
)J(m?H — the standard molar enthalpy of species J
3),Enthalpies of chemical change
2.7 Standard enthalpy changes
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The standard enthalpy of an overall
reaction is the sum of the standard
enthalpies of the individual reactions
into which a reaction may be divided.
4),Hess's law
2.7 Standard enthalpy changes
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This reaction can be recreated from the following sum
CH2=CHCH3(g) + 5O2(g)? 3CO2 (g) +4H2O (l),-2220 kJmol-1
CH2=CHCH3(g) + H2(g)? CH3CH2CH3 (g),-124 kJmol-1
e.g.
Calculate the standard enthalpy of combustion of propane.
286
2 2 2 05
124
2
1
)g(O)g(H)l(OH
)l(OH)g(CO)g(O)g(HC
)g(HC)g(H)g(HC
222
22283
83263
2 0 5 83329 )l(OH)g(CO)g(O)g(HC 22263
1 k J m o l /r H
2.7 Standard enthalpy changes
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The standard enthalpy of formation,:the
standard reaction enthalpy for the formation of
the compound from its elements in their
reference states,
The reference state of an element is its most
stable state at the specified temperature and 1
bar.
2.8 Standard enthalpy of formation
Hf?
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Products
Reactants
Hr
Elements
The reaction enthalpy in terms of enthalpies of formation
f
R e a c t a n t s
f
P r o d u c t s
r HνHνH
)J(f
J
Jr
HνH
2.8 Standard enthalpy of formation
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Illustration
The standard reaction enthalpy of
2HN3(l) + 2NO(g)?H2O2,(l) + 4N2(g)
is calculated as follows,
= {-187.78 + 4(0)} - {2(264.0) + 2(90.25)}
= -892.3 kJmol-1
gN O,l,HN
g,Nl,OH
f3f
2f22fr
HH
HHH
22
4
2.8 Standard enthalpy of formation
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2.9 The temperature dependence of reaction enthalpies
TCTHTH T p d2
1
T
12
Standard reaction enthalpies at different T can be estimated
from heat capacities and reaction enthalpy at some other T,
If a substance is heated from T1toT2,
The standard enthalpy changes:
TCTHTH pTT d2
1 r1r2r
From
TCH p dd?
(at constant p )
□ Kirchhoff ’ law
The estimation from Cp and at some other T H
r
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m,
R e a c t a n t s
m,
P r o d u c t s
r ppp CνCνC
Where is the difference of the molar heat capacities of
products and reactants under standard conditions weighted
by the the stoichiometric coefficients:
pCr?
More generally,
J
mr ( J )
,pJp CνC
2.9 The temperature dependence of reaction enthalpies
□ Kirchhoff ’ law
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m,
R e a c t a n t s
m,
P r o d u c t s
r ppp CνCνC
The change in reaction enthalpy reflects the difference
in the changes of the enthalpies.
TCTHTH pTT d2
1 r1r2r
2.9 The temperature dependence of reaction enthalpies
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1,The properties of gases
2,The First Law,the concepts
3,The First Law,the machinery
4,The Second Law,the concepts
5,The Second Law,the machinery
6,Physical transformations of pure substances
7,Simple mixtures
8,Phase diagrams
9,Chemical equilibrium
10,Electrochemistry
Part 1,Equilibrium
