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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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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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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