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Thermodynamics of chemical process 39
2. THERMODYNAMICS OF
CHEMICAL PROCESSES
Thermodynamics (Greek: therm = heat; dynamys = power) studies the
effects of changes in temperature, pressure and volume on physical systems at the
macroscopic scale by analysing the collective motion of their particles by using
statistical methods.
Chemical thermodynamics has as main objectives:
- the thermal effects of chemical transformation;- the quantity of work, accompanying thermodynamic process which can
include a chemical reaction;- the laws describing the processes in which the energy pass from one
state to another;
- the possibility, the direction and the limits of natural and industrialprocess;
- the organization or disorganization state of thermodynamic systems;- the chemical system and thermal equilibrium after physical chemical
transformations;
- establish the determining conditions for equilibrium.
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40 GENERAL CHEMISTRY
2.1. Thermodynamic systems
Thermodynamic system: a part of the universe separated by a real or
imaginary boundary from the rest of the universe. The surroundingscomprises the
region outside the system, where we make our measurements.
Figure 13. Schematic representation of a thermodynamic system
Phase part of a system having the same properties in all its points and
separated from the rest of the system by a separation surface.
Thermodynamic systems are classified as follows:
- fromphysical point of view homogenous formed from only one phase. It has the
same properties in all its points (gas, solution, mixture of
gases)
heterogenous formed from 2 phases or many (ex.Physical equilibrium H2OL H2OV the properties
vary in at least 2 points)
- from point of view of interaction with the environment isolatedsystem, there is noenergy or mass exchange with
the environment;
System
Surrounding
Boundary
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Thermodynamics of chemical process 41
no isolatedor open system it changes energy and masswith the environment;
closed system it does not exchange mass but it canexchange energy (when the system allows energy to
escape as heat system is calleddiathermic);
adiabatic system there is no change of heat with theenvironment.
The properties of a system could be:
- intensive they do not depend on quantity of substance: pressure,temperature;
- extensive they depend on quantity of substance: internal energy (U),enthalpy (H), Gibbs free energy (G), Helmholtz free energy (A or F).
In order to describe a thermodynamic process state equationsare used:
f(P,V,T) = 0 for a monocomponent system and f(P,V,T,xi) = 0 for multicomponent system.where P is the pressure, T is the temperature, V is the volume and xiis the
molar ratio of the ith
component -n
nx ii= ; ni is the number of moles of i
th
component and nis the total number of moles.
Thermodynamic process the energetic evolution of a thermodynamic
system proceeding from an initial state to a final state. This evolution is determined
by the variation of one or more state parameters.
The thermodynamic processes are classified as follows:
- reversible the process passes spontaneously from the initialstate to the final state and from the final state to the initial state
through the same sequence of intermediate states;
- irreversible the process in which the system can not comeback from the final state to the initial state through the same
sequence of intermediate states.
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42 GENERAL CHEMISTRY
2.2. The Thermodynamics laws
2.2.1 Zeroth law of thermodynamics
If two systems are in thermal equilibrium with a third system, then they
must be in thermal equilibrium with each other.This principle introduces the temperature parameter, T. When two
different systems are in thermal contact, the big body called also thermostat
preserves unchanged its properties while the small body, called also thermometer
varies its properties.
Figure 14. Schematic representation of zeroth law of thermodynamic
For the temperature there are several notations and units:
- t C (Celsius degrees)- T K (Kelvin degrees) (T = t + 273,16)- F = 9/5 C + 32 (Fahrenheit degrees) (anglosaxon system)
2.2.2. First law of thermodynamics
Energy is neither created nor destroyed; it changes from one form to
another.
Thermostatthermometer
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Thermodynamics of chemical process 43
Consequences
- It is impossible to create a perpetual motion machine of I type (machinewhich produces energy from nothing) Helmholtz 1847;
- If a quantity of energy disappears from a system, a new form of energy,in an equivalent quantity appears in its place;
- The internal energy of an isolated system is constant EquivalenceLaw of Joule
calJconstQ
W
Q
W/18,4
2
2
1
1 === [1]
where:
W The Work is done when an object is moved against an opposing force;
Q The Heat: the system gives or receives heat.
The first law of thermodynamic postulate that U the change in internal
energy of the system is given by:
WQU += [2]
Or in other words:
The change in internal energy of a closed system is equal to the energy that
passes through its boundary as heat or/and work.
All these dimensions are expressed in J (Joule) or cal (calorie):
1J = 4,186 cal
The internal energy is a state function. It is depending on the initial and
final state and not on the way, while the heat and the work are functions which
depend on the followed way. So we can write for the change in internal energy wecan write:
12 UUU = [3]
12
2
1
UUdUU == [4]
= 0dU for a reversible cycle [5]
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44 GENERAL CHEMISTRY
Sign convention:
(+) any quantity which increases the energy of the system (the heat
absorbed by the system, the work done on the system);
(-) any energy lost by the system.
Examples:
Heat:
(+) received by the system (evaporation, endothermic reaction)
(-) lost by the system (solidification, exothermic reaction)
Work:
(+) performed by the surrounding on the system (compression)
(-) performed by the system (expansion)
In these conditions the internal energy is given by:
WQU += - for a finite process
And:
WQdU += - for an elemental process [6]
Q , W are infinite small quantities.
Exemplification for expansion work:
pdVSdlS
FFdlW ===
pdVW = [7]
That means that the system receives work when its volume is reduced.
In these equations Fis the force, lis the distance, Sand Vare the surface and the
volume respectively.
Examples Reversible processes(perfect gas)
a) isotherm process(constant temperature)12
UUU = [8]
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Thermodynamics of chemical process 45
==
2
1
)(
T
T
vdTCUTfU [9]
(where Cvis the molar heat capacity at constant volum)
0,12 == UUU [10]
WQ = [11]
1
2
2
1
2
1
lnV
VnRTdV
V
nRTpdVW === [12]
nRTpV= [13]
V
nRTp= [14]
constant temperature
2211 VpVp = [15]
2
1
1
2
p
p
V
V= [16]
2
1lnp
pnRTW = [17]
b) isochoric process (constant volum)00 == WdV [18]
QU= [19]
c) adiabatic process 0=Q WU=
)( 12 TTCW V =
=
2
1
T
T
VdTCU [20]
This means that the temperature will increase when the system is
compressed and the temperature will decrease when the system will undergoes an
expansion.
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46 GENERAL CHEMISTRY
d) isobar process(constant pressure)A new state function is introduced in order to characterize the isobar
process:
H Enthalpy, calorific function of Gibbs
PVUH + [21]
HHHdH ==2
1
12 [22]
0= dH for a reversible cycle [23]
Because the enthalpy is a state function its elemental change is an exact
differential.
VdPPdVdUdH ++= [23]
If the system is in mechanical equilibrium with its surroundings at a given
pressure Pand undergoes only expansion work, we can write: PdVdW =
VdPPdVPdVQdH ++= [25]
VdPQdH += ,0=dP (constant pressure)
( )p
QH = [26]
Qp= H [27]
The thermal effects of all process which take place at constant pressure
(chemical reactions, phase transitions, solubilization) represent the change in
enthalpy for that system.
2.2.2.1. Applications of the first law to the material constants
Heat capacity represents the amount of heat necessary to increase the
temperature of the system by 1C.
Specific heat capacity(specific heat) of a substance represents the heat
capacity divided by the mass, usually in grams or, in other words, it represents the
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48 GENERAL CHEMISTRY
calculations of quantities such as the heat capacity, heat of absorption, heat of
formation, etc.
Considering the following chemical equation:
HAAAAAAA r
iiii ++++++++ .............'''
2
'
2
'
1
'
1332211
where Hr is (+) for endothermic reactions and () for exothermic
reactions
At a given temperature and pressure:
=react
ii
prod
iiPT
rHHH ''
, [31]
where vi is the number of moles for ith
reactant; Hi is the molar enthalpy of ith
reactant; viis the number of moles for ith
product;Hiis the molar enthalpy of ith
product.
Example:322
2NH3HN +
22332
, NHNHPT
r HHHH =
a) Heat of formationThe enthalpy change for a reaction at a given temperature and pressure can
be derived from heat of formations of the reactants and products in the following
way:
=react
i
f
i
prod
i
f
iPT
rHHH ''
, [32]
The values for standard enthalpy0
298Hf (standard conditions, 1 atm and
298 K (reference state)) are collected in thermodynamic tables.
The standard enthalpy of formation0
Hf of a substance is the standard
reaction enthalpy for the formation of the compound from its elements in their
reference states.
Example:
)(2)()()(32323
gsgs COFeCOOFe ++
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Thermodynamics of chemical process 49
molkcalHsOFe
f /2.1960)(,298 32
=
molkcalHgCO
f /4.260)(,298 =
molkcalHgCO
f/968.930 )(,298 2 =
00,298
= FefH (generally 0
298Hf is equal with zero for elements)
( ) ( ) ( ) kcalHr 504.64.2632.196968.9330298
==
b)Heat of combustion
The heat of combustion is the energy released as heat when one mol of a
compound undergoes complete combustion to upper stable oxides.
molkcalHc
/79.212O2HCO2OCH 0298(l)2(g)22(g)4(g)
=++
='
''
,
i
ic
i
i
i
c
iPT
r HHH [33]
Other heat type:
- heat of solubilization- heat of decomposition- heat of neutralization- heat from bond energy
2.2.2.3 The laws of thermochemistry
a) Hesss law (1840)The heat exchange accompanying a transformation is the same,
independently whether the process occurs in one or several steps.
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.
The thermal effects of one reaction depend only on the nature and the state
of the reactants and products and not on the way in which takes place the reaction.
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50 GENERAL CHEMISTRY
Example:
calHCOOCO
calHCOOC
calHCOOC
ggg
ggs
ggs
676402/1
263902/1
94030
3)(2)(2)(
2)()(2)(
1)(2)(2)(
=+
=+
=+
321 HHH +=
b) Kirchhoff s lawThe variation with temperature of the thermal effects of a chemical
reaction is equal to the stoichiometric molar heat capacity of the system (constant
pressure).
+=2
1
12 ,,
T
T
P
r
PT
r
PT
r dTCHH [34]
Equation [34] represent the integral form of Kirchhoff equation
Usually:
...2
2
+++=
++=
cTbTaC
cTbTaC
rrr
P
r
P
Generally:
T1= 298K and P = 1 atm. (standard heat of formation)
( ) ++++=2
2
298
0
298
0 ...
T
rrrr
T
rdTcTbTaHH [35]
HCp > 0 increases when the temperature increases
HCp
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Thermodynamics of chemical process 51
If phase transitions take place in the temperature range considered, the
enthalpies of transition should be taken into account.
++=2
2
1
112
00
T
T
P
rtr
T
T
P
r
T
r
T
r
tr
tr
dTCHdTCHH [36]
where represents the mole numbers of the substance undergoing the phase
transition. The transition enthalpy will be positive when a product is phase
transformed and will be negative when a reactant undergoes a phase transition.
1PC and
2PC are the change in heat capacity calculated with the values of heat
capacity function of the phase.
Example:
Calculate the thermal effect for the reaction:
Al2O3+ 3CO 2Al + 3CO2
at 1000 K and 1 atm.
It is known: calHr 1770802980 = , melting temperature of aluminium is
930 K, molcalHAl
melt /25000 = and the values of heat capacities at constant
pressure, given in the following table:
Substance Cp, cal/molK
Al2O3 22 + 810-3T
CO 6.6 + 1,210-3T
CO2 10.3 + 2.810-3T
Al(s) 4.8 + 3.2 10-3
Al(l) 7
Solution:
322)(2
332OAlCOCOsAl
PPPPP
r CCCCC +=
TTT
TTCPr
333
33
102.33.1)10822()102.16.6(3
)108.26.10(3)102.38.4(21
+=++
+++=
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52 GENERAL CHEMISTRY
322)(2
332OAlCOCOlAl
PPPPP
r CCCCC +=
T
TTTCpr
3
333
2
102.31.3
)10822()102.16.6(3)108.26.10(372
=
=++++=
+=+
++++=
)298930(3.1177080)102.31.3(
25002)102.33.1(177080
3
1000
930
3
930
298
0
1000
dTT
dTTHr
++
+
2
9301000102.3)9301000(1.325002
2
298930102.3
22
3
22
3
calHr
18250101000
=
c) Lavoisier and Laplaces law (1782)The heat exchange accompanying a transformation is equal in absolute
value and opposite in sign to the heat exchange accompanying the reverse
transformation.
Example:
1(l)2(g)2(g)2(g)4OHCO2OCH Hr++
2(g)2(g)4(l)2(g)22OCHO2HCO Hr++
molkcalH
molkcalH
r
r
/7.212
/7.212
2
1
=
=
Consequence
When one mole of a chemical compound is decomposed into chemical
elements quantity of heat is absorbed or evolved a quantity of heat equal to the heat
evolved or absorbed when one mole is formed from the constituent elements.
HH df = [37]
where Hf represents heat of formation and Hd heat of decomposition.
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Thermodynamics of chemical process 53
2.2.3 The second law of thermodynamics
Similarly to the Ist law, the II
nd law of thermodynamics is the result of
generalization of the huge human experience. It cannot be demonstrated. At the
earth scale it was not infirmed by any particular case.
Clausius formulation (1850)
Heat cannot spontaneously flow from a colder to a hotter material.
Kelvin formulation
No process is possible in which the sole result is the absorption of heat
from a reservoir and its complete conversion into work. Such a kind of machine
will be a perpetual Motion Machine of IInd
type.
A heat engineis a mechanical device that provides useful work based on
the difference in temperature of two bodies. In other words, a heat engine is a
mechanical device that continuously transforms heat into an useful work. Therefore
in order to function such an engine necessarily has to be endowed with two heat
reservoirs: a hot source (T1) and a cold source (sink) (T2) (figure 15)
Figure 15. Schematic representation of the functioning of heat engine
During the functioning of this engine some heat Q1 is absorbed from the
hot source, a part is transformed into work and the rest Q2is transferred to the sink.
Sign convention
(+) for heat received by the system
(-) for heat lost by the system
2121)( QQQQW +== [38]
Hot body
T1
EngineQ1
W
Q2 Cold body
T2
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54 GENERAL CHEMISTRY
The efficiency is given by:
1
21
1 Q
Q
L +== [39]
Sadi Carnot (1824) established the ideal efficiency of heat engines for an
engine which work through cycles formed from two isotherm transformations
(constant temperature) and two adiabatic transformations (Q = ct). Such a cycle is
called Carnot cycle (figure 16).
1
21
T
TT= [40]
Figure 16. Carnot Cycle
Homework: demonstrate the eq. [40] knowing the heat for the isotherm
transformations:1
2
11ln
V
VRTQ = and
1
2
22ln
V
VRTQ =
Consequence
The maximum efficiency of a heat engine does not depend on the nature of
substance which follows the transformations. It is determined only by the
temperature of the two sources.
For an irreversible cycle the efficiency will be lower than the ideal
efficiency.
=
+processreversiblefor
processleirreversibfor
1
21
1
21
T
TT
Q
QQ [41]
V
AB
CD
T1
T2
Q1
Q2
P
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Thermodynamics of chemical process 55
2.2.3.1. State functions introduced by the IInd
Law of Thermodynamics
a) The Entropy
1
21
1
21
T
TT
Q
+ [42]
1
2
1
2 11T
T
Q
Q+ [43]
1
1
2
2
TQ
TQ [44]
02
2
1
1 +T
Q
T
Q [45]
The ratioT
Q between the heat exchanged in an isotherm and reversible
process and the temperature corresponding to this exchange is called reduced heat.
For any elemental cycle it can be written in the following way:
02
2
1
1 +TQ
TQ [46]
By summing all the reduced heats one obtaines:
0= TQrev 0
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56 GENERAL CHEMISTRY
For anirreversible process
dST
Qirev VUdS exists. So the IInd
law of
thermodynamics can be formulate in terms of entropy as follows:
In one isolated system the entropy of the system can not decrease, it is
constant ( )0, =VUdS if in the system only reversible processes take place, and it
increases (dS>0) in any irreversible process. This is why the IInd
law of
thermodynamics is called also the entropy law.
Entropy for a chemical reaction
For a general chemical reaction like:
.......................... '''2
'
2
'
1
'
12211 +++++++ iiii AAAAAA
the variation of the entropy is:
=react
ii
prod
ii
r SSS '' [55]
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Thermodynamics of chemical process 57
where vi is the number of moles for ith reactant; Si is the molar entropy of i
th
reactant; vi is the number of moles for ithproduct; Si is the molar entropy of i
th
product.
When some phase transitions take place:
T
QdS
= ; at constant pressure QP =H so for elemental processes
dHQ= .
T
dHdS= [56]
For a finite process:
tr
tr
tr
T
HSSS
==
12 [57]
Ex.v
vap
vap
T
HS
=
For a chemical reaction:
+
+=2
2
1
1
12
T
T
P
tr
tr
i
T
T
P
T
r
T
r
tr
tr
dTT
C
T
HdT
T
CSS [58]
Example:
Calculate the entropy for the reaction:
3CO2Al3COAl32
++
At 1000 K and 1 atm. It is known grdcalSr /88,1380
298 = , the melting
temperature for aluminium 930K, molcalHAl
melt /2500= and the heat capacities
listed in the table below:
Substance Cp, cal/mol K
Al(s) 4.8 + 3.2 10-2T
Al(l) 7
CO(g) 6.6 + 1,2 10-3T
Al2O3(s) 22 + 9 10-3T
C(s) 2.7 + 2.6 10-3T
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58 GENERAL CHEMISTRY
Solution:
+
++=1000
930
930
298
2981000 212
T
dTC
T
H
T
dTCSS
p
r
m
m
p
rrr
)3(32)((32)()(1
) sCsOAlgCOsAl ppppp
r CCCCC ++=
TCpr 3
1108.67.0 =
)3(32)()(32)()(2 sCsOAlgCOlAl
pppppr CCCCC ++=
TCp
r 3102.137.3
2
=
+
+
+=1000
930
3930
298
3
1000
102.137.3
930
25002108.67.088.138 dT
T
TdT
T
TSr
grdcalSr /51.138
1000=
b) Thermodynamic potentials Helmholtz Free Energy (A or F), GibbsFree Energy (G)
According to the correlation between Istand II
ndlaws:
LdUTdS [59]
LTdSdU [60]
( ) LT TS-Ud [61]
= reversible process
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Thermodynamics of chemical process 59
( )VTAA ,= - state function
= 0dA [63]
==2
1
12 AAAdA [64]
( ) LdA T [65]
The Helmholtz free energy is a thermodynamic potential which measuresthe useful work obtainable from a closed thermodynamic system at a constant
temperature.
Similarly:
pdVdUTdS + [66]
PVUH += [67]
VdPPdVdUdH ++= [68]
VdPPdVdHdU = [69]
PdVVdPPdVdHTdS + [70]
VdPdHTdS [71]
VdPTdSdH [72]
( ) VdPTSHd T [73]
By definition:
TSHG [74]
G Gibbs free energy
( )PTGG ,= state function [75]
= 0dG [76]
GGGdG == 122
1
[77]
( ) VdPdG T [78]
if the pressure is constant then
( ) 0, PTdG [79]
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60 GENERAL CHEMISTRY
In a system, at constant temperature and pressure can take place
spontaneously only the processes in which the Gibbs free energy is
decreasing ( ) 0,
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Thermodynamics of chemical process 61
2.2.4. Third Law of thermodynamics
The Nernst heat Theorem
The entropy change accompanying any physical or chemical
transformation approaches zero as the temperature approaches zero: S0 as T0
provided all the substances involved are perfectly ordered.
Planck affirmation:
At 0 K not only S is zero but also the molar entropy (S) of any crystalline
pure element is zero.
00 =S [87]
S0= 0 [88]
Equation [88] represents the mathematical expression of the IIIrd
law of
thermodynamic.
Summarized Third Law of thermodynamics is: The entropy of all perfect
crystalline substances is zero at T = 0 K.
2.2.5. Chemical Potential
In open systems formed from many substances, (so the composition is
varying due to the changes with the surroundings), the expressions of the
thermodynamics functions established in the frame of Istand II
ndLaws should be
completed with composition variables:
.....),......,,,,(21 innnSVUU= [89]
.....),......,,,,(21 i
nnnSPHH= [90]
.....),......,,,,(21 i
nnnSVFF= [91]
.....),......,,,,(21 i
nnnPTGG= [92]
The elementary variation for the Gibbs free energy will be:
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62 GENERAL CHEMISTRY
...
)(3132 ,,
2
,..,,.,2
1
,...,,,,1
+
+
+
++=
i
nPTinnnPTnnnPT
dnn
Gdn
n
Gdn
n
GVdPSdTdG
ijjii
[93]
The partial molar function:
( )
i
nPTi
i
ijj
n
GG =
=
,,
[94]
is called chemical potential (Gibbs)
++=i
iidnVdPSdTdG [95]
2.3. Chemical equilibrium
The experience showed that the chemical reactions occur in forward
direction as well as in opposite direction.
Example:
H2+ I2 2 HI
In a chemical process, chemical equilibrium is the state in which the
chemical activities or concentrations of the reactants and products have no net
charge over time. Usually, this state results when the forward chemical process
proceeds at the same rate as their reverse reaction. The reactions rates of the
forward and reverse reactions are generally not zero but, being equal, there are no
net charges in any of the reactant or product concentrations.
2.3.1. Thermodynamical condition of chemical equilibrium.
Reaction isotherm
Let take a chemical reaction in general form:
......2211 ++++ iiAvAvAv ......''
2'
2'
1'
1' ++++ iiAvAvAv
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Thermodynamics of chemical process 63
The chemical potentials (free partial molar Gibbs energy) are:
,,21
,..., i for reactants ,''
2
'
1...,,,lyrespective
i for products.
The expression:
( ) 0'', == react
ii
prod
iiPT
rvvG [96]
is the thermodynamical condition of the chemical equilibrium.
For p = 1 atm
=react
ii
prod
iiTr
vvG ''0 [97]
representing the standard Gibbs energy of the reaction.
The expression:
aT
rKRTG ln0 = [98]
links thermodynamic data with chemical important equilibrium constant K.
Ka is the equilibrium constant, calculated with activities:
=react
v
i
prod
v
i
ai
i
a
a
K
'
'
[99]
where : ai is the activity of ith component at equilibrium. It is a kind of
effective mole fraction:
*
i
i
ip
pa = ; [100]
wherepiis the vapor pressure of ithcomponent (at equilibrium) when it is a
component of solution and*
ip vapor pressure of pure ith
component at
equilibrium;
- notation for product operator.
For gases instead of activities fugacities f are used which have pressure
dimensions: 1lim0
= p
f
p.
For an ideal gas mixture pf = :
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64 GENERAL CHEMISTRY
=
react
v
i
prod
v
i
pi
i
p
p
K
'
'
[101]
For very diluted solutions activities can be replaced by corresponding
concentrations:
xa KK or ma KK or ca KK
=react
v
i
prod
v
i
xi
i
x
x
K
'
'
=react
v
i
prod
v
i
mi
i
m
m
K
'
'
=react
v
i
prod
v
i
ai
i
c
c
K
'
'
[102]
where: xi molar ratio at equilibrium
t
i
in
nx = [103]
mi molal concentration (moles of substance in 1 kg of solvent) at
equilibrium
ci molar concentration (mols of substance in 1L of solution) atequilibrium
In this case the free Gibbs energies of reaction will be:
( ) ( ) ( )cT
r
mT
r
T
r GGG 000 [104]
wherexT
rKRTG ln0 =
mmT
r KRTG ln)( 0 =
ccT
r KRTG ln)( 0 =
2.3.2 Expression of equilibrium constants for reactions among
gases
i
i
v
i
vv
v
i
vv
p
ppp
pppK
...
...
21
''2
'1
21
''
2
'
1
= [105]
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Thermodynamics of chemical process 65
a) ii Pxp = ,
where: ip is partial pressure of ithcomponent at equilibrium
Pis the total pressure
xi is molar ratio of ithcomponent at equilibrium
v
xv
v
v
i
vvv
v
i
vvv
p PK
P
P
xxxx
xxxxK
i
i
i
i
i
i
=
=
'
321
''3
'2
'1
......
......
321
''
3
'
2
'
1 [106]
= ii vvv' [107]
b) RTnVp ii = [108]
RTcRTV
np
i
i
i == [109]
( )
( )( )
=
= RTKRT
RT
cccc
ccccK
c
i
i
pi
i
i
i '
321
''3
'2
'1
......
......
321
''
3
'
2
'
1 [110]
R = 0.082 Latm / molK
c) = )(RTKPK cx [111]
=
RT
PKK
Xc [112]
If there is no change of mole numbers in a reaction, 0=v ,
cxp KKK == [113]
These equilibrium constants are dimensionless.
Example
For the reaction COCl2 CO + Cl2, at equilibrium, at 550C and 1 atm
77% of COCl2is dissociated. Calculate Kx, Kpand Kc.
Solution:
COCl2 CO + Cl2
Initial: 1 0 0 moles
Equilibrium: 1 moles Total 1 + + = 1 + moles
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66 GENERAL CHEMISTRY
1100
77= moles of COCl2dissociated.
+=
1COx
+=
12Clx
+
=
1
12COCl
x
( )( )
+=
+
+
+=
11
1
111
2
xK
( )( )( )
456.177.0177.01
77.02
=+
=xK
= PKK xp 1111 =+=
456.11456.1 ==pK
( ) = RTKK pc
0215.0823082.0
1456.1 =
=cK
2.3.3 Estimation of the equilibrium constant from
thermodynamic data
It is known that aTr KRTG ln0 =
But
000
T
r
T
r
T
r STHG = [114]
dTT
CSS
Tp
rr
T
r +=298
0
298
0 [115]
+=T
p
rr
T
r dTcHH298
0
298
0 [116]
SoRT
GK T
r
a
0
ln
= [117]
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Thermodynamics of chemical process 67
orR
S
RT
HK T
r
T
r
a
00
ln
+
= [118]
or
+
=
Tp
rT
p
rT
rr
a dT
T
cdTc
TRR
S
RT
HK
298298
00
298 11ln [119]
The equation [119] solves the most important problem of chemical
equilibrium: the integral calculus of equilibrium constant from thermodynamical
data - reaction heat, molar heat capacities, and entropies.
2.3.4. The shift of chemical equilibrium; Le Chatelier Brawn
principle
If a dynamic equilibrium is disturbed by changing the conditions, the
position of equilibrium shifts in order to counteract the change.
A system at equilibrium, when subjected to a disturbance, responds in a
way that tends to minimize its the effect.
2.3.4.1 Temperature influence on the equilibrium constant. Equation
Vant Hoff
T
G
RK
r
a
01ln
= [120]
Differentiation of aKln with respect to temperature gives:
dTT
Gd
RdTKd
r
a
=
0
1ln [121]
From the Gibbs Helmholtz equation we have:
2
0
0
RT
H
dT
T
Gd
r
r
=
[122]
So:
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68 GENERAL CHEMISTRY
2
ln
RT
H
dT
Kd ra = [123]
The equation [123] is called vant Hoff equation
From the equation [123] one can see the following:
- for endothermic reactions ar KH > 00 increases with theincreasing temperature. So the increase the temperature will have as
effect the shift of the reaction to the right side;
- for exothermic reaction ar KH 0 decreases with the increases of pressureExample: CO2 CO +
2
1O2
2
1=v
xKv
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Thermodynamics of chemical process 69
kcalGr
72.440
298 = . Calculate 0
400G
r and Kp at 400 K, supposing that
0= pC cal/Kfor the temperature range 298 - 400K, and an ideal gases behavior.
Solution:
2
0ln
RT
H
dT
Kd rp =
=2
1
2
1
2
0
ln
T
T
rK
K
pTdT
RHKd
P
P
=
21
0 11ln
1
2
TTR
H
K
K r
P
P
0
2981298 GKT
r=
0
4002400 GKT
r=
pT
r KRTG ln0 =
2
0400
1
0298
21lnln
RT
GK
RT
GK
r
P
r
P==
=
+
21
0
1
0
298
2
0
400 11
TTR
H
RT
G
RT
G rrr
=
211
0
298
2
0
400
11
TTH
T
GTG
r
r
+=
400
1
298
114.44
298
72.444000
400Gr
kcalGr 450
400 =
2
0
400lnRT
GK
r
p
=
51.56400987.1
45000ln
400 ==pK
24105.3 =pK .
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70 GENERAL CHEMISTRY