Download - Stoichiometry, Chemical Equilibrium
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Lecture 3
Stoichiometry, Chemical Equilibrium
http://en.wikipedia.org/wiki/Category:Physical_chemistryhttp://en.wikipedia.org/wiki/Category:Thermodynamics
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aj1 ...nelem Eaj
nelem
Ni = aji
j=1
ns
! NReacj = aj
i
j=1
ns
! NProductsj
aji NProducts
j ! NReacj( )
j=1
Ns
" = 0 i=1,Ne
Homogeneous linear system with
Ne eqns & Ns(>Ne) unknowns # $Ns-Ne solutions
Stoichiometry = Constraints for Kinetics
Q?: for a prescribed amount of reactants, how can the composition evolve towards equilibrium?
N ! NReactants = " k ! "Reactantsk( )
k=1
Nr
# Sk
This system has non-trivial (non zero) solution of the form:
Skk = !kj{ } j=1
Ns stoichiometric vector of the k-th reaction
! k
j stoichiometric coefficient of the j-th species in the k-th reaction
N ,NReactants number of moles, of reactants
" k ,"Reactants
k # 0 rate of progress of the k-th reaction, at the initial composition
Sk{ }k=1
Ns$Ne : Basis of Null Space of the matrix A= a ji{ } j=1,Ns
i=1,Ne
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Nkj ! Nk,Reac
j
"kj = # k , k=1,Nr; j=1,Ns
N ! NReac = " k
k=1
nreac
# Sk $ N j ! NReacj = " k
k=1
nreac
# %kj = Nk
j ! Nk,Reacj
k=1
nreac
#
dNdt
= Skk=1,Nr! d" k
dtRate Equation for Composition
d! k
dt= rk (p,T ,N j ) = rf
k " rbk
Constitutive law for rate of progress
Stoichiometry = Constraints for Kinetics
Q?: for a prescribed amount of reactants, how can the composition evolve towards equilibrium?
1!k
j
dNkj
dt=d" k
dt, k=1,Nr; j=1,Ns
Mole change always proportional to !kj
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1 2 3 4 5 6 7 8H2 O2 H O OH H2O HO2 H2O2
NH = 2n1 + n3 + n5 + 2n6 + n7 + 2n8NO = 2n2 + n4 + n5 + n6 + 2n7 + 2n8
2 0 1 0 1 2 1 20 2 0 1 1 1 2 2
!"#
$%&
dn1dn2dn3dn4dn5dn6dn7dn8
'
(
)))))
*
)))))
+
,
)))))
-
)))))
= Ax = 0
2 elements, 8 species
How to find the stoichiometric coefficients
An example: H2/O2 Mixture
dNH = 0 = 2dn1 + dn3 + dn5 + 2dn6 + dn7 + 2dn8dNO = 0 = 2dn2 + dn4 + dn5 + dn6 + 2dn7 + 2dn8
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2 00 2
!"#
$%&dn1dn2
'()
*+,= -
1 0 1 2 1 20 1 1 1 2 2
!"#
$%&
dn3dn4dn5dn6dn7dn8
'
(
.
.
.
)
.
.
.
*
+
.
.
.
,
.
.
.
ek = eki{ } = 1 se i = k
0 se i ! k"#$
%&'
k=1,8-2=6
dn1
dn2
"#$
%&'= (
2 00 2
)*+
,-.
(1 1 0 1 2 1 20 1 1 1 2 2
)*+
,-.dn2+ k
dn2+ k = ekd/k
"
#0
$0
1Ns(Ne=6 solutions
dn1 = !12d"1
dn2 = 0dn3 = +d"1
#
$
%%
&
%%
'12H2 = H
Negative sign = reactantPositive sign = product
8-2=6 linearly independent reactions
How to find the stoichiometric coefficients
An example: H2/O2 Mixture
2 0 1 0 1 2 1 20 2 0 1 1 1 2 2
!"#
$%&
dn1dn2dn3dn4dn5dn6dn7dn8
'
(
)))))
*
)))))
+
,
)))))
-
)))))
= Ax = 0
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07)
Reversible (zero entropy) process dG = µ jj! dN j = 0
Stoichiometric constraint dN j = "kjd# k
k!
$
dG = µ jj! "k
jd# k
k! = d# k
k! µ j"k
j
j! = 0 %&d# k ' 0, µ j"k
j
j! = 0
()*
+*
,-*
.*k=1
Nreactions
$ Free Enthalpy (Gibbs) is stationary % Chemical equilibrium
Condition of Chemical Equilibrium (2)
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Equilibrium Condition µ j T , pj( )!kj
j" = 0 k=1,Nreactions
µ j (T , pj ) := !H j (T ) # T !Sj (T , pi ) !Sj (T , pj ) := !Sj0 (T ) #$ Log(
pjpref
)
µ j (T , pj ) := !H j (T ) # T !Sj0 (T ) #$ Log(
pjpref
)%
&''
(
)**= !H j (T ) # T !Sj
0 (T )%& () +$T Log(pjpref
) = µ j0 (T ) +$T Log(
pjpref
)
µ j0 (T ) +$T Log(
pjpref
)%
&''
(
)**!k
j
j" = 0 + µ j
0 (T )%& ()!kj
j" = # $T Log(
pjpref
)%
&''
(
)**!k
j
j"
!H j (T ) # T !Sj0 (T )%& ()!k
j
j" = #$T !k
jLog(pjpref
) = #$T Log(pjpref
)!kj
j"
j"
Exp #1$T
!H j (T ) # T !Sj0 (T )%& ()!k
j
j"
,-.
/.
01.
2.Kp (T )
" #$$$$$$ %$$$$$$= Log(
pjpref
)!kj
j" = Exp Log (
pjpref
)!kj
j3
,-.
/.
01.
2.= (
pjpref
)!kj
j3
Law of Mass Action Kpk (T ) = (
pjpref
)!kj
j3
Definition Kpk (T ) 4 Exp #
1$T
!H j (T ) # T !Sj0 (T )%& ()!k
j
j"
,-.
/.
01.
2.
Law of Mass Action / Equilibrium Constant
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Law of Mass Action Kpk (T ) = (
pjpref
)!kj
j"
with Kpk (T ) # Exp $
1%T
!H j (T ) $ T !Sj0 (T )&' ()!k
j
j*
+,-
.-
/0-
1-= Exp $
2Gko(T )
%T+,.
/01
2Gko(T ) # !H j (T ) $ T !Sj
0 (T )&' ()!kj
j*
Alternative form of law of mass actions:
Kck = c
j
!kj
j" ; KX
k = Xj
!kj
j"
cj = pj / (%T ) ; Xj = pj / p Relations among different form of equilibrium constants:
Kpk (T ) = Kc
k (%T )!"kj $! '
kj( )
j*
= KXk ( ppref
)!"kj $! '
kj( )
j*
2!k # ! "kj $! '
kj( )
j*
Kpk (T ) = F(T ) ; Kc
k = F(T ) ; KXk = Kp (T )( p
pref)$ !"
kj $! '
kj( )
j*
= F(T , ppref
$2!k
)
Kc, Kp are dimensional
Kx is nondimensional
Alternative Form of Equilibrium Constants
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Val
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i (20
07)Equilibrium Condition µ j (T , pj )!k
j
j" = 0
Define Equilibrium Constant Kpk (T ) # Exp $ %Gk
o(T )&T
'()
*+,
Law of Mass Action KXk (T , p) # Kp
k (T ) / ( ppref
)!"kj $! '
kj( )
j"
= Xj!kj
j-
The equilibrium constant of a reaction is a pure thermodynamic function !!It depends on T, p and on thermochemical properties of the species
Equilibrium state depends on thermodynamics only !!
Equilibrium is a Thermodynamic Condition !
NB: pressure affects equilibrium only if !"kj "! 'k
j#$%
&'(j
) * 0
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Log(Kpk ) = ! "Gk
o
#T
"Gko = 0 $ Kp
k = 1 reactants & products in equilibrium pj%kj
j=1,Nreac& = pj
%kj
j=1,Nprod&
"Gko > 0 $ Kp
k <1 reaction favors reactants pj%kj
j=1,Nreac& > pj
%kj
j=1,Nprod&
"Gko < 0 $ Kp
k >1 reaction favors products pj%kj
j=1,Nreac& < pj
%kj
j=1,Nprod&
Small Displacement from Equilibrium
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1) Method of equilibrium constants:Easy to grasp but of difficult convergence
2) Method of minimization of Gibbs free enthalpy:Complex formulation, but fast and robust
How to find the equilibrium state
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1 2 3 4 5 6 7 8H2 O2 H O OH H2O HO2 H2O2
{Nj} j=1,8, N, T: 8+1+1 unknown values need 10 equations to be uniquely identified
2 Atomic Mass ConservationEquilibrium Condition for 6
lin. indep. reactions
Kpm (T *) =
pj*
pref
!
"#$
%&
' jm
j( m=1,6 ; j=1,8
pj* =
N j*
Np
Kpm (T *) =
N j*
Nppref
!
"#$
%&
' jm
j( =
ppref
!
"#$
%&
' jm
j) N j
*
N!
"#$
%&
' jm
j(
Kpm (T *) p
pref
!
"#$
%&
* ' jm
j)
=N j
*
N!
"#$
%&
' jm
j(
NH = 2N1 + N3 + N5 + 2N6 + N7 + 2N8
NO = 2N2 + N4 + N5 + N6 + 2N7 + 2N8
N = N jj=1,8!
Xj :=N j
N
Total number of Moles
Conservation of AbsoluteEntalpy
!H (T ,N j ) := !H j (T )j! N j
!H (T0 ,N j ,0 )reactants = !H (T *,N j*)products
Method of the Equilibrium Constants
Q.?: Find Equilibrium Condition of an Adiabatic (H=const), Isobaric (p=const) System
Equilibrium StateT *, N j
*,N *
p = given
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Stoichiometric constraint Ni = ajiN j
j! i=1, Ne ; j=1, Ns
Condition of Minimal Gibbs Free Enthalpy "G p,T =" µ jNj
j!
#
$%&
'(p,T
= 0
Adjoint Form (Lagrange Multipliers) "G*p,T
:= "G p,T + ) ii=1,Ne! dNi
Constrained Minimum
"G*p,T
= 0 * "G p,T = 0 and dNi =0 for i=1, Ne
After some manipulations the system is written in the unknowns ) i , log N j, log N , p , T 1) A Ne x Ne system is solved to find the Ne multipliers2) The remaining unknowns are found as a funtion of the multipliers
PROs: the system to solve is only Ne x Ne instead of Ns x Ns as for the method of the equlibrium constants The convergence properties of this method are excellent
Method of Minimization of Gibbs Free Enthalpy
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Enthalpy of Reaction & Heat of Combustion
For an isobaric, isothermal system, but NOT adiabatic, wherein one reactiontransforms the composition from X0 to X*, the first principle reads
! !Q = d !H " !Vdp =dp=0"
d !H with !H (T ,Xj ) = !H jref
j# Xj + Xj !cj , p (T )dT
T ref
T
$j#
! !Q = d !H (T ,Xj ) = !H jref
j# dX j + Xj !cj , p (T )dT =
dT =0"
j# !H j
ref
j# dX j
Enthalpy of Reaction % !HReaction = !H (T ,Xj*) " !H (T ,Xj
0 ) = !H jref
j# Xj
* " !H jref
j# Xj
0
Heat of Combustion !QComb = " % !HReaction Jmol
&'(
)*+
% !HReaction < 0 !QComb > 0 Exothermic Reactions% !HReaction > 0 !QComb < 0 Endothermic Reactions
%HReaction,Fuel =% !HReaction
WFuel
KJKg-Fuel
&
'(
)
*+ %HReaction,Mix =
% !HReaction
WMix
KJKg-Mix
&
'(
)
*+
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Equilibrium (Adiabatic, Isobaric) Temperature
For an isobaric and adiabatic system, wherein many reactions transform thecomposition from X0 to X*, the first principle reads
! !Q = d !H " !Vdp #! !Q,dp=0"
d !H = 0 with !H (T ,Xj ) = !H jref
j$ Xj + Xj !cj , p (T )dT
T ref
T
%j$
0 = !H (T *,Xj*) " !H (T 0 ,Xj
0 ) = !H jref
j$ Xj
* + Xj* !cj , p (T )dTT ref
T *
%j$
&
'(
)
*+ " !H j
ref
j$ Xj
0 + Xj0 !cj , p (T )dTT ref
T 0
%j$
&
'(
)
*+
Xj* !cj , p (T )dTT ref
T *
% " Xj0 !cj , p (T )dTT ref
T 0
%j$
j$
&
'(
)
*+
Change in sensible enthalpy# $%%%%%%%% &%%%%%%%%
= !H jref
j$ Xj
0 " !H jref
j$ Xj
*&
'()
*+
Heat of Combustion# $%%%% &%%%%
T *, p Adiabatic (Isobaric) Temperature
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Equilibrium (Adiabatic, Isocoric) Temperature
For an isocoric and adiabatic system, wherein many reactions transform thecomposition from X0 to X*, the first principle reads
! !Q = d !U + pd !V "! !Q,d !V =0"
d !U = 0
0 = !U(T *,Xj*) # !U(T 0 ,Xj
0 )
= !H (T *,Xj*) #$T *( ) # !H (T 0 ,Xj
0 ) #$T 0( )= !H (T *,Xj
*) # !H (T 0 ,Xj0 )( )
Enthalpy of Combustion# $%%%% &%%%%
#$ T * # T 0( )
T *,v Adiabatic (Isocoric) Temperature