chapter 13 thermo 2
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Chemical Reaction Equilibria Chapter 13
The Reaction CoordinateApplication of Equilibrium Criteria to Chemical ReactionsThe Standard G and the Equilibrium ConstantEffect on Temperature on the Equilibrium ConstantEvaluation of the Equilibrium ConstantRelation of Equilibrium Constant to CompositionEquilibrium Conversion for Single Reactions
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Chemical Engineering Thermodynamics II ChE 342 12-2
Introduction
Transformation of raw materials into products having commercial use and value is a major industry.
Raw Materials
Products{crude oil, minerals, etc..} {Gasoline, NaOH, H2
SO4
, ammonia, etc..}
This often involves chemical reactions and subsequent separation of the products.
The rate and maximum (or equilibrium) conversion of chemical reaction are of important concern in its commercial development. Both of these aspects depend on the T, P, and composition of the reactants.
We can often increase reaction rates by the use of CATALYST, but we can not change the equilibrium (or ultimate) conversion by using catalyst.
D
e
g
r
e
e
o
f
r
e
a
c
t
i
o
n
Time
With catalyst
Withou
t cataly
st
Uultimateconversion
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Chemical Engineering Thermodynamics II ChE 342 12-3
Chemical Reaction Course concerns on study of reaction rate and
equilibrium, where the following tasks are often considered:
Design and operation of reaction equipment Effect of controllable variables upon conversion Extent of reaction as t
In general:Reaction Equilibrium is governed by THERMODYNAMICS
while,Reaction Rate is governed by REACTION KINETICS
Objectives:
Definitions of reaction stoichiometry
and criterion of reaction equilibrium. Dependence of equilibrium constant
on temperature and its evaluation.
Development of connection between equilibrium constant and composition.
Calculation of equilibrium conversion
for single reactions.
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Chemical Engineering Thermodynamics II ChE 342 12-4
The Reaction CoordinateThe general chemical reaction can be written as:
44332211 +........++ AAAAwhere |i
| are stoichiometric
coefficients and Ai
denotes the chemical formulas of species. i
themselves are stoichiometric
numbers
Sign ConventionFor products: i
is positive (+) For reactants: i
is negative (-)
N.B. The reaction, as written above, does not represent any mechanisms. It represents only the overall macroscopic picture. It does not tell us what is happening at the microscopic level. Example:
CH4
+ H2
O
CO + 3H2CH4
= -1CO
= +1 H2
= +3
H2O
= -1
The stoichiomertric
number of an inert species
is zero.
(I)
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Chemical Engineering Thermodynamics II ChE 342 12-5
In all reactions, the change in the number of moles of the species present are related to each other through the stoichiometric
number.
For example, for the reaction CH4
+ H2
O
CO + 3H2
, if 0.5 mole CH4
reacts, then 1.5 moles H2
and 0.5 mol CO generated, thats ratio of change in number of moles to the stoichiometric
number of CH4
is the same as that for H2
.Thus, in general for reaction (I), we can write:
1
1
2
2
=
dndn
1
1
3
3
=
dndn, , etc
or, in general:
.........=
=
= 3
3
2
2
1
1 dndndn
where d = the amount of reaction.d=
It will be observed that the sign of i
allows for the fact that:dni
> 0 for the product & dni
< 0 for the reactants The preceding equation provides the definition for , the Reaction Coordinate; a quantity characterizes the extent, or degree, to which chemical reaction has taken place.
is also called the Degree of Reaction
(II)
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Chemical Engineering Thermodynamics II ChE 342 12-6
From II, we can write,ddn ii =
moles molesdimensionless
By convention,
= 0 at the initial state (before the reaction occurs). Integration yields:
0
=0
i
n
ni ddn
i
i
= f(t)
nn iii += 0Summing over all N species (reactant and products) yields: +==
0 iii nnnor, nn += 0 where, 0=0 inn = i&
The quantity
tells us whether there is a net increase or decrease in the total number of moles in the system. The mole fraction of species i
at any time, or stage of reaction, is by:
nn
y ii = nn ii
++
=0
0This expression is very useful, since it provides us with a relationship between species concentration and the progress of the reaction.
(III)
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Chemical Engineering Thermodynamics II ChE 342 12-7
Example:
yCO 2+8+1
=
Consider the chemical reaction, CH4
+ H2
O
CO + 3H2Initially there present 2 moles CH4
, 1 mole H2
O, 1 mole CO and 4 moles H2
. Obtain yi
= f(). = i
= -1 -1
+ 1 + 3 = 2 n0
= n0i
= 2 + 1 + 1 + 4 = 8 moles
Application of the equation:
yCH 28-2
4 +=
nn
y iii ++
=0
0
y OH 28-1
2 +=
yH 2834
2 ++
=
Definition of a Mole of Reaction
This means that
has changed by a unit amount (i.e. 1 mol).
When
= 1 mole, the reaction proceeds to such an extent that the change in mole numbers for each reactant and product species is equal to its stoichiometric
numbers, thats: ni = i
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Chemical Engineering Thermodynamics II ChE 342 12-8
Multireaction Stoichiometry
In this case, we allocate a reaction coordinate, j
, to each reaction
(j
= 1, 2, 3, R), where R
= number of separate reactions.
The stoichiometric
numbers are now identified as:
i,j Stoichiometric
number of species i
in reaction j
ni
may change because of the involvement of species i
in several reactions. The general equation analogous to Eq. (III) for single reactions is:
,=j
jjii ddn (i
= 1, 2, .N;
j = 1, 2, . , R)
Integration yields; )(+= ,0j
jjiii nn
Summing over all species: )(+== ,0i j
jjii nnn )(+= ,0j i
jji n
Similarly, ,=i
jij v )(+= 0j
jjnn
Mole fraction,nn
y ii =
+
)(+=
0
,0
jjj
jjjii
n
n(i
= 1, 2, . , N)
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Chemical Engineering Thermodynamics II ChE 342 12-9
Example: Consider the following reactions,CH4
+ H2
O
CO + 3H2 (1) j
= 1CH4
+ 2H2
O
CO2
+ 4H2 (2) j
= 2
There are present 2 mol CH4
and 3 mol H2
O. Obtain expressions for yi
= f(1
, 2
).
Construct the following table for i,j
:
i
= CH4
H2
O CO CO2
H2j
12
-1-1
-1-2
10
01
34
j1
= 22
= 2
)(+= 0j
jjnn = 5 + 21
+ 22
21
21
225--2
4
yCH ++=
21
21
2252--3
2
y OH ++=
21
1
2+2+5=
yCO
21
2
2+2+5=
2
yCO21
21
2+2+54+3
=2
yH
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Chemical Engineering Thermodynamics II ChE 342 12-10
Application of Equilibrium Criterion to Chemical ReactionsAs noted previously, in a closed system at equilibrium (constant
T
&P):
(dGt)T,P
= 0
Recalling that Gt
decreases during an irreversible process (eg. Mixing process, chemical rxn, etc) If a mixture is not in equilibrium, any reaction that occurs at
constant T
and P
must
lead to a decrease in the total energy of the system
is the single variable that characterizes the progress of the reaction.
Recall that
is related to the composition of the system
Criterion for equilibrium:1. (dGt)T,P
= 02. Gt()
is minimim
N.B. Equilibrium expression can also be used for OPEN SYSTEM at constant T
& P.
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Chemical Engineering Thermodynamics II ChE 342 12-11
The Standard Gibbs Energy Change and the Equilibrium Constant
Recall the fundamental property relation for single-phase system:
)(-)()(i
iidndTnSdPnVnGd +=
ddn ii =If changes in ni
occur as a result of single reaction in a closed system, then:
)(-)()(i
ii ddTnSdPnVnGd +=
=]
)([ , iiPT
nG= 0.0 At equilibrium state, according to the
previous figure. The criterion for chemical reaction equilibrium can be expressed as:
0.0= iiSince ii G =Recall that
ln+)(= iii fRTT
Write this equation for pure species in its standard state at the same temperature
(A)
oii
oi fRTTG ln+)(= (B)
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Chemical Engineering Thermodynamics II ChE 342 12-12
See section 4.3 (p. 133) for the definition of standard state:Gases: pure species in ideal gas state at 1 bar (or 1 atm)Liquids or solids: actual pure liquid or solid at 1 bar (or 1 atm)
Subtracting (B) from (A):o
iioii ffRTG /ln=_
ln= iaRTwhere is the activity of species i
in solution.ia
N.B. the activity coefficientia )/=(
iiii fxfln+= i
oii aRTGNow, & 0.0= ii
0=]ln+[ 0 iiii aRTG 0=)ln(+][ 0 i
iii aRTGor
RTG
a iivi i 0 -)(ln =or where product over all species
]/Exp[-)( 0 RTGa iivi i = = KGoi
= f(T) only, independent of pressure; K
= the equilibrium constant
= f(T) only
ln- ooii GGKRT == where G o
the Standard Gibbs Energy Change of Reaction.
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Chemical Engineering Thermodynamics II ChE 342 12-13
Note: For a specific species the standard state represented by Gio
must be the same state represented by fio
upon which the activity coefficient is specified
iaCase of Gases
The standard state is the ideal gas state of pure species at a
pressure of 1 bar. fio
= 1 bar for each species in has phase reaction
= =
iii o
i
fa f
f = ( ) i viK f must be measured in bar
+ for product, -
for reactant
Example: A + 2B + 3C
D + 4 E
-2 -3-1 4= ( ) = ( ) ( ) ( ) ( ) ( )i vi A B C D EK f f f f f for
4
2 3
( ) ( )=
( ) ( ) ( )
D E
A B C
f fK
f f fCase of liquids and solids
The usual standard state is pure liquid or pure solid at 1 bar and T
of the system. for this state, fio
is not generally equal to 1 bar, thus =
ii o
i
fa
f
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Chemical Engineering Thermodynamics II ChE 342 12-14
Other Standard Property Change of ReactionFor the general property, M, we can write
Mo
= i
Mio
For example: Ho
= i
Hio
CPo
= i
CPiof(T) only
Relationship between Hio and GioPreviously, we have derived:
Therefore, in the standard state for species i:
= PT
RTGT
RTH
])/(
- [
=dT
RTGdRTH
oio
i)/(
- 2 = f(T)Why we removed the partial derivative?
Summing over all species:
= ][- 2
RTG
dTd
RTHoiio
ii
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Chemical Engineering Thermodynamics II ChE 342 12-15
= ][- 2RT
GdTd
RTHo
o Relationship between Ho
and Go
But, RTGK o /-ln = 2lnRT
HdT
Kd o= Highly important
relation
Note: If Ho
is negative (exothermic reactions) K
as T
Special case: Assuming Ho
is independent of T,
]1
-1
[
-)/ln(1
1 TTRH
KKo
= A plot of lnK
vs. 1/T
should be a straight line
See the next slide for plot of K
versus 1/T
for several reactions.
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Chemical Engineering Thermodynamics II ChE 342 12-16
This figure allows determination of K at any temperature.
If K1
is known at T1
K
can be determined at any temperature for a given Ho.
Important Remarks:
Slope of the lines are ve
or +ve; -ve
for endothermic reactions
+ve
for exothermic reactions
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Chemical Engineering Thermodynamics II ChE 342 12-17
Determination of Go
K
can be determined from the previous figure at a given temperature for certain
reactions if Ho
is constant (
f(T)) Another alternative method, when Ho
= f(T),
for determination of K
is by using:
RTG
Ko
- ln = = f(T)
If an estimate of Go
is available K
can be determined Recall that Go
= f(T)
Go
can be calculated by using the general relation: ooo STHG - =
Go
= i
GioHo
= i
HioSo
= i
SioCPo/R = i
GPio/RRecalling (from Chapter 4 and 5):
0
0
T
T
oPoo dT
RC
RHH += or: 0
0
T
T
oPoo
TdT
RC
RSS +=
Where H0o
and S0o is the standard enthalpy and entropy changes at a reference temperature T0
, respectively.
(A)
Substituting in (A) gives:
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Chemical Engineering Thermodynamics II ChE 342 12-18
00
- -
00
T
T
oPo
T
T
oPoo
TdT
RC
RTSTdTRC
RHG +=
0
000
-
TGH
Soo
o =
According to (A):
00
-)-( -
000
0
T
T
oPoo
T
T
oPoo
TdT
RC
RTGHTT
dTRC
RHG +=whence:
00
- 1-
0
000T
T
oP
T
T
oP
oooo
TdT
RC
dTRC
TRTHG
RTH
RTG
++=
Division by RT:
CPio/R = f(T)
presented in Tables C.1-C.3 (pp. 657/658) Go
and Ho
values are tabulated for many formation reactions at standard conditions
Gof298
and Gof298
(G0o
and H0o) for a number of chemical compounds are listed in Table C.4 (pp. 569/660)
Go/RT
can be determined after determining the two integrals in the above equations
Hence, K can be determined
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Chemical Engineering Thermodynamics II ChE 342 12-19
Example: calculate the equilibrium constant for the vapor-phase hydration of ethylene at 145 & 320oC from data given in Appendix C
C2
H4
(g) + H2
O(g)
C2
H5
OH(g)
Data: SpeciesC2
H4
(g)H2O(g)C2
H5
OH(g)
i-1-1+1
Ho29852510-241818-235100
Go29868460
-228572-168490
Ai1.4243.473.518
Bi
10314.3941.45
20.001
Ci
105-
4.392-----
-
6.002
where Cpi
/R = Ai
+ Bi
T
+ Ci
T2
298,o298 oii HH = = (-1)(52510) + (-1)(-241818)+(1)(-235100) = -
45792 J/mol
298,o298 oiiGG = = (-1)(52510) + (-1)(-241818)+(1)(-235100) = -
45792 J/mol2
,o )()()/(/ TBTBARCRC iiiiiio iPiP ++==
= [(-1)(1.424) + (-1)(3.47) + (1)(3.518)] + [(-1)(14.39) + (-1)(1.45) + (1)(20.0)]10-3T+ [(-1)(-4.392) + (-1)(0.0) + (1)(-6.002)]
10-5T2
= -1.376 + 4.157
10-3T -
1.61
10-6T2
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Chemical Engineering Thermodynamics II ChE 342 12-20
418
298
26-3-
418
298
26-3-
]101.61-104.157-1.376[-
]101.61-104.157-1.376[418
1)298)(314.8(
457928378-)418)(314.8(
45792-
TdT
T
dTTRTG o
+
+++
+=
@ T
= 145oC = 418 K:
936.1
=RTG o@ T
= 145oC = 418 K:
RTG
Ko
- ln = K
= exp(-1.936) = 0.1443
Similarly @ T = 320oC = 593 K K
= 0.00294
K
with TExothermic rxn
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Chemical Engineering Thermodynamics II ChE 342 12-21
Relationship between Equilibrium Constant and Composition
Gas-Phase ReactionsFor the gas phase reactions: ( where fio
= Po
= 1 bar)= =
iii o
i
fa f
f i
)( iifK =
),,( ii yPTff =K
= f(T) only
Reflects nonideality
of the equilibrium mixture
But
&This means that for a given T, the composition at eqm
must change
with P
in such away that remains constant iif )(
Now, Pyf iii = i
)( iii PyK =
Separate P
term, i ][)( ii PyK i= where = i
Or, i- )( iii
yPK =(recalling that yi
can be related to e
) Relates K
to e
for a given T
and P
Complicated problem because: ),,( ii yPTf = iterative solution is needed
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Chemical Engineering Thermodynamics II ChE 342 12-22
Special cases: For equilibrium mixture as an ideal solution ii =
whence, i- )( iii yPK =
For sufficiently low pressure, equilibrium mixture behaves as an ideal gas
1 =i
whence, i- )( ii yPK = T, P, and composition-dependent terms are distinct are separate, thus solution for any one of
e
, T
or P
given the other two is straightforward Conclusion remarks
-
Dependence of T
can be seen through K: If Ho
+ve, ENDO
rxn, K
with T
(yi
)i
for a fixed P
e
rxn
shifts to the right
If Ho
-ve, EXO
rxn, K
with T
(yi
)i
for a fixed P
e
rxn
shifts to the left
P
@ constant T
(yi
)i
rxn
shifts to the right, e
If
(i
) -ve,-
P
@ constant T
(yi
)i
rxn
shifts to the left, e
If
(i
) +ve,
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Chemical Engineering Thermodynamics II ChE 342 12-23
Example: Consider the water-gas shift reactionCO(g) + H2
O(g)
CO2
(g) + H2
(g)Assuming ideal gas mixture, what is the fraction of the steam reacted in each of the following cases: (Hint: fraction of steam converted = (ni0
nie
)/ni0
= -i
e
/ni0
)(a) n0,H2O
= 1, n0, CO
= 1 mol, T
= 1100 K, P
= 1 bar
n0
= 2 mol & = 0.0From Fig. 13.2: @ 1/T
= 1/(1100 K) = 9.09
10-4
for the above rxn
lnK
= 0 or K
= 1.0
1 = i i0 )( )1( iiyK = = 1
14/)-1(4 2
2
=e
e
,202
02
eeCO
y =
++
= ,2-1
yCO = ,22e
H
y =2-1
2
y OH =
Substituting:
1y
2
2CO =OHCO
H
yyy
e
= 0.5 fraction of steam reacts = 0.5
(b) n0,H2O
= n0, CO
= 1 mol, T
= 1100 K, P
= 10 bar Since
= 0 P
has no effect; e
= 0.5
(c) n0,H2O
= n0, CO
= 1 mol, n0,N2
= 2 mol, T
= 1100 K, P
= 1 bar
As an inert N2
= 0 Does not affect the calculations, e
remains the same.
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Chemical Engineering Thermodynamics II ChE 342 12-24
(d) n0,H2O
= 2, n0, CO
= 1 mol, T
= 1100 K, P
= 1 bar
n0
= 3 mol
1)-1)(-2(
2
=ee
e
,32e
CO
y = ,3-1
yCO = ,32e
H
y =3-1
2
y OH =
Substituting: e
= 0.667 fraction of steam reacts = 0.667/2 = 0.33
(e) n0,H2O
= 1, n0, CO
= 2 mol, T
= 1100 K, P
= 1 bar
n0
= 3 mol
1)-1)(-2(
2
=ee
e
,22e
CO
y = ,3-2
yCO = ,32e
H
y =3-1
2
y OH =
Substituting: e
= 0.667 fraction of steam reacts = 0.667
(f) n0,H2O
= 2, n0, CO
= 1 mol, n0, CO2
= 1 mol, T
= 1100 K, P
= 1 bar
n0
= 3 mol
1)-1(
)1(2 =
+
e
ee
,
31
2
eCO
y
+= ,3
-1 yCO = ,32
eH
y =
3-1
2
y OH =
Substituting: e
= 0.0.33 fraction of steam reacts = 0.33(g) n0,H2O
= 1, n0, CO
= 1 mol, T
= 1650 K, P
= 1 bar From Fig. 13.2: @ 1/T
= 1/(1650 K)
lnK
= -1.15 or K
= 0.316
316.0)-1( 2
2
=e
e
e
= 0.36
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Chemical Engineering Thermodynamics II ChE 342 12-25
Liquid-Phase ReactionsFor the liquid phase reactions: o
i
ii f
fa
= where fi
o
0
iviaK )(
=Then K
remains as
For liquid: iiii fxf =
oi
iiio
i
ii f
fx
ff
a ==
Determination of fi
/fio:
@ T
and P: iii fRTTG ln)( +=@ T
and Po: oii
oi fRTTG ln)( +=
Subtracting:
Integration at constant T
from Po
to P:
oii
oii ffRTGG /ln- =
But: dTSdPVdG ii i-=-P
Pi
oii
odPVGG =
(A)
(B)
Substitute (B) in (A): lnP
Pio
i
i
odPV
ff
RT =
For liquids Vi
is weak function of P
upon integration )-(ln oioi
i PPVff
RT =
or ]/)-(exp[ RTPPVff o
ioi
i =
-
Chemical Engineering Thermodynamics II ChE 342 12-26
]/)-(exp[ RTPPVxa oiiii =
ioiii RTPPVxK }]/)-(exp{[=and
io
ii RT
PPVii ex ][)(
)-(=
RTPPV
ii
oii
i ex)-(
)(=
])-(
exp[])([ i
ii
oi
ii VRTPPV
xK i=
Special cases At low P: exp[{Vi
(P-Po)/RT}{i
Vi
}]
1.0 (or when P
Po)i
iiixK )(= Requires determination i
, but i
= f(xi
) xi
can be found in this case iteratively.
For the case of ideal solution: i
= 1.0
iii
xK )(= Law of Mass Action
xi
= f(e
)
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Chemical Engineering Thermodynamics II ChE 342 12-27
Example: Esterification
of acetic acid in the liquid phase at 100oC and 1 atm:
CH3
COOH(l) + C2
H5
OH(l)
CH3
COOC2
H5
(l) + H2
O(l)
Species
i
CH3
COOH(l), i=1C2
H5
OH(l), i=2CH3
COOC2
H5
(l), i=3H2
O(l), i=4
i-1-1+1+1
Hof298 (J)-484,500-277690-480,000-285,830
Gof298 (J)-389,900-174,780-332,200-273,129
n0,acetic acid
= n0,C2H5OH
= 1 molEstimate the mole fraction of ethyl acetate (x3
) at equilibrium. Assuming ideal solution and that Ho
does not change significantly with from 25oC to 100oC. i
iixK )(=
First, determine K: 298,o298 oii HH = = 484,500 + 277,690
480,00
285,830 = -
3640 J
298,o298 oiiGG = = 389,990 + 174,780 -332,200
273,129 = -
4650 J
lnK
= -Go/RT K298
= exp[-4650/(8.314)(298)] K298
= 6.527
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Chemical Engineering Thermodynamics II ChE 342 12-28
For constant Ho, apply the following,]
1-
1[
-)/ln(
11 TTR
HKK
o
= ]2981
-373
1[
314.83640
-)527.6/ln( 373 =K
K373
= 4.859
21
34
xxxx
K =Now:
K
= 4.859, n0
= 2, = 0
0
0
nn
x eiii+
= ,2-1
1ex = ,23
ex = ,24ex =,
2-1
2ex =
859.44/)-1(
4/2
2
=e
e
e
= 0.6879
x3
= 0.344
(see other examples in the textbook)
Chemical Reaction EquilibriaChapter 13IntroductionSlide Number 3The Reaction CoordinateSlide Number 5Slide Number 6Example:Multireaction StoichiometryExample:Application of Equilibrium Criterion to Chemical ReactionsThe Standard Gibbs Energy Change and the Equilibrium ConstantSlide Number 12Slide Number 13Other Standard Property Change of ReactionSlide Number 15Slide Number 16Determination of GoSlide Number 18Slide Number 19Slide Number 20Relationship between Equilibrium Constant and CompositionSlide Number 22Slide Number 23Slide Number 24Slide Number 25Slide Number 26Slide Number 27Slide Number 28
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