thermodynamics of reacting mixtures - stanford...

Chapter 9

Thermodynamics of reactingmixtures

9.1 Introduction

For an open system containing several reacting chemical species that can exchange massand work with its surroundings the fundamental Gibbs equation relating equilibrium statesis

TdS = dE + PdV �IX

i=1

µidni +KX

k=1

Fkdlk. (9.1)

The Fk are forces that can act on the system through di↵erential displacements. Ordinarily,lower case letters will be used to denote intensive (per unit mole) quantities (h, s, e, etc)and upper case will designate extensive quantities (H,S,E, etc). Heat capacities, pressureand temperature are symbolized in capital letters. One mole is an Avagadro’s number ofmolecules, 6.0221415⇥ 1023.

The chemical potential energy per unit mole µi is the amount by which the extensive energyof the system is changed when a di↵erential number of moles dni of species i is added orremoved from the system. If the system is closed so no mass can enter or leave and if it isisolated from external forces, the Gibbs equation becomes

TdS = dE + PdV �IX

i=1

µidni (9.2)

9-1

CHAPTER 9. THERMODYNAMICS OF REACTING MIXTURES 9-2

where the di↵erential changes in the number of moles of species i occur through chemicalreactions that may take place within the closed volume. The main di↵erence between(9.1) and (9.2) is that, in the closed system, changes in mole numbers are subject to theconstraint that the number of atoms of each element in the system is strictly constant.The precise expression of the chemical potential in terms of conventional thermodynamicvariables of state will be established shortly. For the present it can be regarded as a new,intensive state variable for the species i. Mathematically, equation (9.2) implies that

µi (E, V, n1

, . . . , nI) = �T

✓

@S

@ni

◆

E,V,nj 6=i

. (9.3)

If no reactions occur then (9.2) reduces to the familiar form

TdS = dE + PdV. (9.4)

According to the second law of thermodynamics, for any process of a closed, isolatedsystem

TdS � dE + PdV. (9.5)

Spatial gradients in any variable of the system can lead to an increase in the entropy.Smoothing out of velocity gradients (kinetic energy dissipation) and temperature gradients(temperature dissipation) constitute the two most important physical mechanisms thatcontribute to the increase in entropy experienced by a non-reacting system during a non-equilibrium process. If the system contains a set of chemical species that can mix, thenchanges in entropy can also occur through the smoothing out of concentration gradientsfor the various species. If the species can react, then entropy changes will occur throughchanges in the chemical binding energy of the various species undergoing reactions.

The inequality (9.5) can be used to establish the direction of a thermodynamic system asit evolves toward a state of equilibrium.

9.2 Ideal mixtures

Consider a mixture of species with mole numbers (n1

, n2

, . . . , nI). The extensive internalenergy of the system is E and the volume is V . The extensive entropy of the system is thefunction, S (E, V, n

1

, n2

, . . . , nI). An ideal mixture is one where all molecules experience thesame intermolecular forces. In an ideal mixture surface e↵ects, (surface energy and surfacetension) can be neglected and the enthalpy change when the constituents are mixed is


zero. Ideal mixtures obey Raoult’s law that states that the vapor pressure of a componentof an ideal mixture is equal to the vapor pressure of the pure component times the molefraction of that component in the mixture. In the ideal approximation the volume ofthe system is the sum of the volumes occupied by the pure species alone. Similarly theinternal energy is the sum of internal energies of the pure species. Most real mixturesapproximate ideal behavior to one degree or another. A mixture of ideal gases is perhapsthe best example of an ideal mixture. Liquid mixtures where the component moleculesare chemically similar, such as a mixture of benzene and toluene, behave nearly ideally.Mixtures of strongly di↵erent molecules such as water and alcohol deviate considerablyfrom ideal behavior.

Let the mole numbers of the mixture be scaled by a common factor ↵.

n1

= ↵n1

, n2

= ↵n2

, n3

= ↵n3

, . . . , nI = ↵nI (9.6)

According to the ideal assumption, the extensive properties of the system will scale by thesame factor.

E = ↵E, V = ↵V (9.7)

Similarly the extensive entropy of the system scales as

S (E, V, n1

, . . . , nI) = ↵S⇣

E, V , n1

, . . . , nI

⌘

. (9.8)

Functions that follow this scaling are said to be homogeneous functions of order one.Di↵erentiate (9.8) with respect to ↵.

E@S

@E+ V

@S

@V+

IX

i=1

ni@S

@ni= S

⇣

E, V , n1

, . . . , nI

⌘

(9.9)

Multiply (9.9) by ↵ and substitute (9.8).

E@S

@E+ V

@S

@V+

IX

i=1

ni@S

@ni= S (E, V, n

1

, . . . , nI) (9.10)

The Gibbs equation is


dS =dE

T+

P

TdV �

IX

i=1

µi

Tdni. (9.11)

According to (9.11) the partial derivatives of the entropy are

@S

@E=

1

T

@S

@V=

P

T

@S

@ni= �µi

T.

(9.12)

Inserting (9.12) into (9.10) leads to a remarkable result for an ideal mixture.

E + PV � TS =IX

i=1

niµi (9.13)

Equation (9.13) is called the Duhem-Gibbs relation. The combination of state variablesthat appears in (9.13) is called the Gibbs free energy.

G = E + PV � TS = H � TS (9.14)

Equation (9.13) expresses the extensive Gibbs free energy of an ideal mixture in terms ofthe mole numbers and chemical potentials.

G =IX

i=1

niµi (9.15)

This important result shows that the chemical potential of species i is not really a newstate variable but is defined in terms of the familiar state variables, enthalpy, temperatureand entropy. The chemical potential of species i is its molar Gibbs free energy.

µi = gi = hi � Tsi (9.16)

The enthalpy in (9.16) includes the chemical enthalpy associated with the formation of thespecies from its constituent elements.


9.3 Criterion for equilibrium

The Gibbs free energy is sometimes described as the ”escaping tendency” of a substance.At low temperatures the enthalpy dominates. A chemical species with a positive enthalpywould like to break apart releasing some of its chemical enthalpy as heat and producingproducts with lower enthalpy. A few examples are ozone (O

3

), hydrogen peroxide (H2

O2

),and nitrous oxide (N

2

O). These are stable chemicals at room temperature but will de-compose readily if their activation energy is exceeded in the presence of a heat source or acatalyst. The entropy of any substance is positive and at high temperatures the entropyterm dominates the Gibbs free energy. In a chemical reaction the Gibbs free energy ofany species or mixture will increasingly tend toward a state of higher entropy and lowerGibbs free energy as the temperature is increased. Take the di↵erential of the Gibbs freeenergy.

dG = dE + PdV + V dP � TdS � SdT (9.17)

For a process that takes place at constant temperature and pressure dT = dP = 0. TheSecond Law (9.5) leads to the result that for such a process

dG = dE + PdV � TdS 0. (9.18)

A spontaneous change of a system at constant temperature and pressure leads to a decreaseof the Gibbs free energy. Equilibrium of the system is established when the Gibbs freeenergy reaches a minimum. This result leads to a complete theory for the equilibrium of areacting system.

9.4 The entropy of mixing

Consider the adiabatic system shown in Figure 9.1 consisting of a set of (n1

, n2

, . . . , ni, . . . , nI)moles of gas species segregated into volumes of various sizes such that the volumes are allat the same temperature and pressure.

Figure 9.1: A system of gases separated by partitions.


The total number of moles in the system is

N =IX

i=1

ni. (9.19)

The entropy per unit mole of an ideal gas is determined using the Gibbs equation

ds = CpdT

T�Ru

dP

P(9.20)

where the units of CP are Joules/(mole�Kelvin). Tabulations of gas properties are alwaysdefined with respect to a reference temperature and standard pressure. The referencetemperature is universally agreed to be Tref = 298.15K and the standard pressure is

P � = 105N/m2 = 105 Pascals = 102 kPa = 1 bar. (9.21)

All pressures are referred to P � and the superscript ’�’ denotes a species property evalu-ated at standard pressure. A cautionary note: In 1999 the International Union of Pure andApplied Chemistry (IUPAC) recommended that for evaluating the properties of all sub-stances, the standard pressure should be taken to be precisely 100kPa. Prior to this date,the standard pressure was taken to be one atmosphere at sea level, which is 101.325kPa.Tabulations prior to 1999 are standardized to this value. The main e↵ect is a small changein the standard entropy of a substance at a given temperature tabulated before and after1999. There are also small di↵erences in heat capacity and enthalpy as well. The IUPACcontinues to provide standards for chemistry calculations and chemical nomenclature.

The pressure has no e↵ect on the heat capacity of ideal gases, and for many condensedspecies the e↵ect of pressure on heat capacity is relatively small. For this reason, tabulationsof thermodynamic properties at standard pressure can be used to analyze a wide varietyof chemical phenomena involving condensed and gas phase mixtures. Inaccuracies occurwhen evaluations of thermodynamic properties involve phase changes or critical phenomenawhere wide deviations from the ideal gas law occur, or condensed phases exhibit significantcompressibility.

Integrating (9.20) from the reference temperature at standard pressure, the entropy perunit mole of the ith gas species is

si(T, P )� s�i (Tref ) =

Z T

Tref

C�pi (T )

dT

T�Ru ln

✓

P

P �

◆

. (9.22)


where the molar heat capacity C�pi is tabulated as a function of temperature at standard

pressure. The standard entropy of a gas species at the reference temperature is

s�i (Tref ) =

Z Tref

0

C�pi (T )

dT

T+ si

�(0) (9.23)

where the integration is carried out at P = P �. To evaluate the standard entropy, heatcapacity data is required down to absolute zero. For virtually all substances, with theexception of superfluid helium (II), the heat capacity falls o↵ rapidly as T ! 0 so thatthe integral in (9.23) converges despite the apparent singularity at T = 0. From the thirdlaw, the entropy constant at absolute zero, s�i (0), is generally taken to be zero for a puresubstance in its simplest crystalline state. For alloys and pure substances such as COwhere more than one crystalline structure is possible, the entropy at absolute zero may benonzero and tabulated entropy data for a substance may be revised from time to time asnew research results become available. Generally the entropy constant is very small.

The entropy per unit mole of the ith gas species is

si(T, P ) = s�i (T )�Ru ln

✓

P

P �

◆

. (9.24)

The entire e↵ect of pressure on the system is in the logarithmic term of the entropy. Theextensive entropy of the whole system before mixing is

Sbefore =IX

i=1

nisi (T, P ) =IX

i=1

nis�i (T )�

IX

i=1

niRu ln

✓

P

P �

◆

. (9.25)

Figure 9.2: System of gases with the partitions removed at the same pressure, temperatureand total volume as in Figure 9.1.

If the partitions are removed as shown in Figure 9.2 then, after complete mixing, each gastakes up the entire volume and the entropy of the ith species is

si(T, Pi) = s�i (T )�Ru ln

✓

Pi

P �

◆

. (9.26)


where Pi is the partial pressure of the ith species. The mixture is ideal so there is noenthalpy change during the mixing. If the pressure was so high that the potential energyassociated with inter-molecular forces was significant then the enthalpy of mixing wouldbe non-zero.

The entropy of the system after mixing is

Safter =IX

i=1

nisi (T, P ) =IX

i=1

nis�i (T )�

IX

i=1

niRu ln

✓

Pi

P �

◆

. (9.27)

The change of entropy due to mixing is

Safter�Sbefore =

IX

i=1

nis�i (T )�

IX

i=1

niRu ln

✓

Pi

P �

◆

!

�

IX

i=1

nis�i (T )�

IX

i=1

niRu ln

✓

P

P �

◆

!

.

(9.28)

Cancel common terms in (9.28).

Safter � Sbefore = Ru

IX

i=1

ni ln

✓

P

Pi

◆

> 0 (9.29)

Mixing clearly leads to an increase in entropy. To determine the law that governs thepartial pressure let’s use the method of Lagrange multipliers to seek a maximum in theentropy after mixing subject to the constraint that

P =IX

i=1

Pi. (9.30)

That is, we seek a maximum in the function

W (T, n1

, n2

, ..., nI , P1

, P2

, ..., PI ,�) =IX

i=1

nis�i (T )�

IX

i=1

niRu ln

✓

Pi

P �

◆

+ �

IX

i=1

Pi � P

!

(9.31)

where � is an, as yet unknown, Lagrange multiplier. The temperature of the system andnumber of moles of each species in the mixture are constant. Di↵erentiate (9.31) and setthe di↵erential to zero for an extremum.


dW =@W

@P1

dP1

+@W

@P2

dP2

+ .....+@W

@PIdPI +

@W

@�d� = 0 (9.32)

Now

dW = �IX

i=1

niRu

✓

dPi

Pi

◆

+ �

IX

i=1

dPi

!

+ d�

IX

i=1

Pi � P

!

= 0. (9.33)

The last term in (9.33) is zero by the constraint and the maximum entropy conditionbecomes.

IX

i=1

✓

�niRu

Pi+ �

◆

dPi = 0 (9.34)

Since the dPi are completely independent, the only way (9.34) can be satisfied is if theLagrange multiplier satisfies

� =niRu

Pi(9.35)

for all i. In the original, unmixed, system each species satisfies the ideal gas law.

PVi = niRuT (9.36)

Using (9.35) and (9.36) we can form the sum

�IX

i=1

Pi =IX

i=1

PVi

T. (9.37)

Finally the Lagrange multiplier is

� =V

T(9.38)

where, V =IX

i=1

Vi. Using (9.35) and (9.38) the partial pressure satisfies

PiV = niRuT (9.39)


which is Dalton’s law of partial pressures. What we learn from this exercise is, not onlythat the entropy increases when the gases mix, but that the equilibrium state is one wherethe entropy is a maximum. Using Dalton’s law, the mole fraction of the ith gas species isrelated to the partial pressure as follows

xi =ni

N=

Pi

P(9.40)

The entropy of a mixture of ideal gases expressed in terms of mole fractions is

Sgas =IX

i=1

ni�

s�igas (T )�Ru ln (xi)�

�NRu ln

✓

P

P �

◆

(9.41)

and the entropy change due to mixing, (9.29), is expressed as

Safter � Sbefore = �NRu

IX

i=1

xi ln (xi) > 0. (9.42)

9.5 Entropy of an ideal mixture of condensed species

The extensive entropy of an ideal mixture of condensed species (liquid or solid) with molenumbers n

1

, . . . , nI is S (T, P, n1

, . . . , nI). If sipure (T, P ) is the entropy of the pure form ofthe ith component, then in a system where the mole numbers are fixed, the di↵erential ofthe entropy is

dS (T, P, n1

, . . . , nI) =IX

i=1

nidsipure (T, P ) (9.43)

since dn1

= 0, . . . , dnI = 0. If we integrate (9.43) the result is

S (T, P, n1

, . . . , nI) =IX

i=1

nisipure (T, P ) + C (n1

, . . . , nI) (9.44)

with a constant of integration that is at most a function of the mole numbers.

In order to determine this constant we will use an argument first put forth by Max Planckin 1932 and also described by Enrico Fermi in his 1956 book Thermodynamics (page 114).


Since the temperature and pressure in (9.44) are arbitrary let the pressure be reduced andthe temperature be increased until all of the condensed species in the system are fullyvaporized and behave as ideal gases as shown in Figure 9.3.

Figure 9.3: System of condensed species with the temperature increased and pressure de-creased to fully vaporize the mixture.

Since the number of moles of each constituent has not changed, the constant of integrationmust be the same. The entropy of a system of ideal gases was discussed in the previoussection and is given by equation (9.41).

Sgas =IX

i=1

ni

✓

s�igas (T )�Ru ln

✓

P

P �

◆◆

�IX

i=1

niRu ln⇣ni

N

⌘

(9.45)

Comparing (9.44) with (9.45) we can conclude that the constant of integration mustbe

C (n1

, . . . , nI) = �Ru

IX

i=1

ni ln⇣ni

N

⌘

. (9.46)

Therefore, the entropy of a mixture of condensed species is

S (T, P, n1

, . . . , nI) =IX

i=1

ni�

sipure (T, P )�Ru ln (xi)�

(9.47)

where the mole fraction, xi = ni/N is used. Finally, the contribution of each componentto the extensive entropy of the mixture of condensed species is

si (T, P ) = sipure (T, P )�Ru ln (xi) . (9.48)


The form of this equation is very similar to that for gases and comes as something of asurprise since it involves the ideal gas constant which would seem to have no particularrelevance to a liquid or a solid. According to Kestin (A Course in Thermodynamics,Ginn Blaisdell 1966) when (9.48) was first introduced it was ”met with general incredulity,but its validity has since been confirmed experimentally beyond any reasonable doubtwhatever.”

It can be noted that if the gases in the mixing problem described in the last section werereplaced by ideal liquids, and those liquids were allowed to evolve from the unmixed to themixed state, the entropy change would be given by an expression identical to (9.29).

(Safter � Sbefore)ideal liquids = �NRu

IX

i=1

xi ln (xi) (9.49)

9.6 Thermodynamics of incompressible liquids and solids

For a single homogeneous substance the Gibbs equation is

dS =dE

T+

P

TdV. (9.50)

If the substance is an incompressible solid or liquid the Gibbs equation reduces to

dS = Cv(T )dT

T(9.51)

and the entropy is

S � Sref =

Z T

Tref

Cv(T )dT

T. (9.52)

The entropy and internal energy of an incompressible substance depend only on its temper-ature. The constant volume and constant pressure heat capacities are essentially equivalentCp (T ) = Cv (T ) = C (T ).

Since liquids and solids tend to be nearly incompressible, the entropy sipure (T, P ) tends tobe independent of pressure and in most circumstances one can use

sipure (T, P ) = s�i (T ) . (9.53)


Now the entropy of a condensed species in a mixture really does resemble that of an idealgas, but without the dependence on mixture pressure.

si (T, P ) = s�i (T )�Ru ln (xi) (9.54)

For a general substance, the di↵erential of the Gibbs function is

dG = dE + PdV + V dP � TdS � SdT = �SdT + V dP (9.55)

where the Gibbs equation (9.4) has been used. The cross derivative test applied to theright side of (9.55) produces the Maxwell relation

✓

@S

@P

◆

T

= �✓

@V

@T

◆

P

. (9.56)

For an incompressible material, the entropy is independent of the pressure and (9.56)implies that the coe�cient of thermal expansion of the material is also zero. Therefore, foran incompressible substance

✓

@V

@P

◆

T

= 0✓

@V

@T

◆

P

= 0

V = constantCP = CV = CdE = CdT

dS =C

TdT

dH = CdT + V dP.

(9.57)

On a per unit mole basis, the enthalpy of an incompressible material referenced to thestandard state is

h(T, P ) = h� (T ) + (P � P �) v� (9.58)

where v� = constant is the molar volume of the material. The superscript on v� is notreally necessary since it assumed to be a constant at all pressures and temperatures butas a practical matter, when using (9.58) as an approximation for a real condensed solid orliquid, the value of the molar density will be taken to be at the standard pressure 105Pa


and the reference temperature 298.15K. Often the pressure term in (9.58) is neglectedwhen dealing with reacting systems of gases and condensed materials. The reason is that,while the molar enthalpy of the condensed and gaseous form of a species are of the sameorder, the molar volume of a solid or liquid is generally two to three orders of magnitudesmaller than the gaseous molar volume.

9.7 Enthalpy

The enthalpy per unit mole of a gas is determined from

dh = Cp(T )dT. (9.59)

The enthalpy of a gas species is

hi (T )� hi (Tref ) = h�i (T )� h�i (Tref ) =

Z T

TrefC�pi (T )dT. (9.60)

In principle the standard enthalpy of the ith gas species at the reference temperature couldbe taken as

h�i (Tref ) =

Z Tref

0

C�pidT + h�i (0). (9.61)

In this approach the enthalpy constant is the enthalpy change associated with chemicalbond breaking and making that occurs when the atoms composing the species are broughttogether from infinity to form the molecule at absolute zero. Note that even for an atomicspecies, the enthalpy constant is not exactly zero. A quantum mechanical system containsenergy or enthalpy arising from ground state motions that cannot be removed completelyeven at absolute zero temperature. In practice, enthalpies for most substances are tabulatedas di↵erences from the enthalpy at the reference temperature of 298.15K which is muchmore easily accessible than absolute zero so the question of the zero point enthalpy rarelycomes up. Thus the standard enthalpy of a species is

h�i (T ) =

Z T

TrefC�pi (T )dT +�h�fi (Tref ) (9.62)

where �h�fi (T ) is the enthalpy change that occurs when the atoms of the species arebrought together from at infinity at the finite temperature T . The enthalpy including the


heat of formation (9.62) is sometimes called the complete enthalpy. In practice certain con-ventions are used to facilitate the tabulation of the heat of formation of a substance.

9.7.1 Enthalpy of formation and the reference reaction

The enthalpy of formation of a substance, denoted �h�f (Tref ), is defined as the enthalpychange that occurs when one mole of the substance is formed from its elements in theirreference state at the given temperature T and standard pressure P �. The reference statefor an element is generally taken to be its most stable state at the given temperature andstandard pressure. The reference reaction for a substance is one where the substance is thesingle product of a chemical reaction between its elements in their most stable state.

This convention for defining the heat of formation of a substance is useful even if thereference reaction is physically unlikely to ever actually occur. A consequence of thisdefinition is that the heat of formation of a pure element in its reference state at anytemperature is always zero. For example, the enthalpy of formation of any of the diatomicgases is zero at all temperatures. This is clear when we write the trivial reaction to form,for example hydrogen, from its elements in their reference state.

H2

! H2

(9.63)

The enthalpy change is clearly zero. In fact the change in any thermodynamic variablefor any element in its reference state is zero at all temperatures. A similar referencereaction applies to any of the other diatomic species O

2

, N2

, F2

, Cl2

, Br2

, I2

, and the heatof formation of these substances is zero at all temperatures. The most stable form of carbonis solid carbon or graphite and the reference reaction is

C(s) ! C

(s) (9.64)

with zero heat of formation at all temperatures.

The reference reaction for carbon dioxide at 298.15K is

C(s) +O

2

! CO2

�h�fCO2

(298.15) = �393.522 kJ/mole. (9.65)

Here the carbon is taken to be in the solid (graphite) form and the oxygen is taken to bethe diatomic form. Both are the most stable forms over a wide range of temperatures.Even if the temperature is well above the point where carbon sublimates to a gas (3915K)and significant oxygen is dissociated, the heats of formation of C

(s) and O2

remain zeroeven though the most stable form of carbon at this temperature is carbon gas.


The heats of formation of metal elements are treated a little di↵erently. The heat offormation of crystalline aluminum is zero at temperatures below the melting point and theheat of formation of liquid aluminum is zero at temperatures above the melting point. Thesame applies to boron, magnesium, sulfur, titanium and other metals.

The enthalpy (9.62) is usually expressed in terms of tabulated data as

hi (T ) = h�i (T ) = �h�fi (Tref ) + {h�i (T )� h�i (Tref )} . (9.66)

For a general reaction the enthalpy balance is

�h� (Tfinal) =

IproductX

iproduct

niproducth�iproduct (Tfinal)�

IreactantX

ireactant

nireactanth�ireactant

⇣

Tinitialireactant

⌘

.(9.67)

So for example, to determine the heat of formation of CO2

at 1000K where the initialreactants are also at 1000K, the calculation would be

�h�fCO2

(1000) =

⇣

�h�fCO2

(298.15) +n

h�fCO2

(1000)� h�fCO2

(298.15)o⌘

�

⇣

�h�fC(s)

(298.15) +n

h�fC(s)

(1000)� h�fC(s)

(298.15)o⌘

�

⇣

�h�fO2

(298.15) +n

h�fO2

(1000)� h�fO2

(298.15)o⌘

.

(9.68)

Putting in the numbers from tabulated data (See Appendix 2) gives

�h�fCO2

(1000) =

[�393.522 + 33.397]� [0 + 11.795]� [0 + 22.703] = �394.623(9.69)

which is the tabulated value of the heat of formation of carbon dioxide at 1000K. Note thatthe enthalpy of the reference reactants at the reaction temperature must be included in thecalculation of the heat of formation calculation. Further discussion of heats of formationcan be found in Appendix 1 and tables of thermo-chemical data for selected species can befound in Appendix 2.


9.8 Condensed phase equilibrium

The pressure and temperature of the system may be such that one or more or all of thespecies may be evolving with their condensed phase. In this case it may be necessary tovary the volume to keep the temperature and pressure constant until the system reachesequilibrium.

Figure 9.4 depicts the various species of the system in several phases all of which are in con-tact with each other. Let the total number of moles in each phase beN

1

, N2

, . . . , Np, . . . , NP .Generally only a subset of the species will be evolving with their condensed phase and sothere is a di↵erent maximal index Ip for each phase. E↵ects of surface tensions betweenphases are ignored.

Figure 9.4: System of molecular species with several phases.

Note the total numbers of moles in each phase are related to the species mole numbersby

N1

=I1

X

i=1

n1i N

2

=I2

X

i=1

n2i · · · NP =

IPX

i=1

nPi. (9.70)

The mole fractions in each phase are

xpi =npi

Np. (9.71)

In the gas phase, the partial pressures of the gas species add up to the mixture pres-sure

P =I1

X

i=1

Pi. (9.72)


The mole fractions of the species in the gas phase are related to the partial pressuresby

x1i =

Pi

P. (9.73)

Finally the mole fractions in each phase add to one.

IpX

i=1

xpi = 1 (9.74)

The extensive Gibbs free energies in each phase are

G1

(T, P, n11

, n12

, n13

, ..., n1I

1

) = N1

(

I1

X

i=1

x1i (g

�1i(T ) +RuT ln (x

1i)) +RuT ln

✓

P

P �

◆

)

G2

(T, P, n21

, n22

, n23

, ..., n2I

2

) = N2

I2

X

i2

=1

x2i (g2i(T, P ) +RuT ln (x

2i))

...

GP (T, P, nP1

, nP2

, nP3

, ..., nPIP ) = NP

IPX

iP=1

xpi (gpi(T, P ) +RuT ln (xpi))

(9.75)

where phase 1 is assumed to be the gas phase. In (9.75) we have assumed that all condensedphases are ideal mixtures and the entropy per mole of the condensed phase of species idoes not depend on the pressure. This is a reasonable assumption for a condensed phasethat is approximately incompressible as was argued earlier in conjunction with equation(9.58). There are a few examples of liquids such as liquid helium and liquid nitrous oxidethat are quite compressible and the assumption would break down. This treatment of thecondensed phases is similar to that used by Bill Reynolds in the development of STANJAN.The NASA Glenn code CEA is a little di↵erent. In CEA, each condensed phase is treatedas a pure substance. For example, if there are two species that condense out as liquids, eachis treated as a separate, distinct phase. In the approach used by CEA the mole fractionsof each condensed species are one by definition. In CEA the pressure term in (9.58) isneglected. See equation 2.11 in NASA Reference Publication 1311 (1994) by Gordon andMcBride.

In the case of a gas, the Gibbs free energy does depend on pressure through the dependenceof entropy on pressure. This is connected to the fact that, as quantum mechanics tells us,


the number of energy states that a gas can occupy (and therefore the entropy of the gas)increases with the volume containing the gas (see Appendix 1 of the AA210a notes).

The Gibbs free energy of the whole system is

G (T, P, n11

, n12

, . . . , n1I

1

, n21

, n22

, . . . , n2I

2

, ..., nP1

, nP2

, ..., nPIP ) =

PX

p=1

IPX

i=1

npigpi (T, P, npi) .(9.76)

With the gas phase written separately Equation (9.76) is

G (T, P, n11

, n12

, . . . , n1I

1

, n21

, n22

, . . . , n2I

2

, ..., nP1

, nP2

, ..., nPIP ) =

PX

p=1

IPX

i=1

n1i

0

@g�1i (T ) +RuT ln (n

1i)�RuT ln

0

@

I1

X

î=1

n1

î

1

A

1

A+

PX

p=2

IPX

i=1

npi

0

@gpi (T, P ) +RuT ln (npi)�RuT ln

0

@

IPX

î=1

npî

1

A

1

A .

(9.77)

Di↵erentiate (9.77).

dG =@G

@TdT +

@G

@PdP+

@G

@n11

dn11

+@G

@n12

dn12

+ · · ·+ @G

@n1I

1

dn1I

1

+

@G

@n21

dn21

+@G

@n22

dn22

+ · · ·+ @G

@n2I

2

dn2I

2

+ . . .+

@G

@nP1

dnP1

+@G

@nP2

dnP2

+ · · ·+ @G

@nPIPdnPIP .

(9.78)

Written out fully (9.78) is


dG =I1

X

i=1

dn1i

0

@g�1i(T ) +RuT ln (n

1i)�RuT ln

0

@

I1

X

_i =1

n1

_i

1

A

1

A+

I1

X

i=1

n1i

0

B

B

B

B

@

RuT

✓

dn1i

n1i

◆

�RuT

0

B

B

B

B

@

I1

P

_i =1

dn1

_i

I1

P

_i =1

n1

_i

1

C

C

C

C

A

1

C

C

C

C

A

+RuT ln

✓

P

P �

◆ I1

X

i=1

dn1i+

I1

X

i=1

n1i

0

@

@g�1i(T )

@T+Ru ln (n1i)�Ru ln

0

@

I1

X

_i =1

n1

_i

1

A

1

A dT+

I1

X

i=1

n1i

!

RuT

✓

dP

P

◆

+

I1

X

i=1

n1i

!

Ru ln

✓

P

P �

◆

dT+

PX

p=2

IpX

i=1

dnpi

0

@gpi(T, P ) +RuT ln (npi)�RuT ln

0

@

IpX

_i =1

np_i

1

A

1

A+

PX

p=2

IpX

i=1

npi

0

B

B

B

B

B

@

RuT

✓

dnpi

npi

◆

�RuT

0

B

B

B

B

B

@

IpP

_i =1

dnp_i

IpP

i=1

np_i

1

C

C

C

C

C

A

1

C

C

C

C

C

A

+

PX

p=2

IpX

i=1

npi

0

@

@gpi(T, P )

@T+Ru ln (n1i)�Ru ln

0

@

IpX

_i =1

n1

_i

1

A

1

A dT +PX

p=2

IpX

i=1

npi

✓

@gpi(T, P )

@T

◆

dP .

(9.79)

Note that

IpX

i=1

npi

0

B

B

B

B

B

@

dnpi

npi�

IpP

_i =1

dnp_i

IpP

_i =1

np_i

1

C

C

C

C

C

A

= dNp � dNp = 0. (9.80)

At constant temperature and pressure dT = dP = 0 and(9.79) becomes


dG =I1

X

i=1

dn1i

0

@g�1i(T ) +RuT ln (n

1i)�RuT ln

0

@

I1

X

_i =1

n1

_i

1

A+RuT ln

✓

P

P �

◆

1

A+

PX

p=2

IpX

i=1

dnpi

0

@gpi(T, P ) +RuT ln (npi)�RuT ln

0

@

IpX

_i =1

np_i

1

A

1

A .

(9.81)

In terms of mole fractions (9.81) reads

dG =I1

X

i=1

dn1i

✓

g�1i(T ) +RuT ln (x

1i) +RuT ln

✓

P

P �

◆◆

+

PX

p=2

IpX

i=1

dnpi (gpi(T, P ) +RuT ln (xpi)).

. (9.82)

At equilibrium dG = 0. For species that are only present in one phase, dnpi = 0 atequilibrium. For those species that are in equilibrium with another phase, dnphase1i =�dnphase pi and equation dG = 0 can only be satisfied if

gphase1i = gphase pi. (9.83)

If phase 1 is a gas species (9.83) is

g�phase1i(T )+RuT ln (xphase1i)+RuT ln

✓

P

P �

◆

= gphase pi(T, P )+RuT ln (xphase pi) . (9.84)

At equilibrium, the Gibbs free energy (or chemical potential) of species i is the sameregardless of its phase. For example if gas species 1 is in equilibrium with its liquid phase,then ggas1 (T, P ) = gliquid1 (T, P ). There can also be more than one solid phase and so thetotal number of phases can exceed three. For example, in helium at very low temperaturethere can be multiple liquid phases.

Suppose phase p in is a pure liquid so the mole fraction is one. Also assume the liquid isincompressible so that the Gibbs function of the liquid is given by

g�gasi(T ) +RuT ln (xgasi) +RuT ln

✓

P

P �

◆

= g�liquidi(T ) + (P � P �) v�liquid. (9.85)


Solve for the partial pressure of the species in the gas phase.

Pi

P � = e(

P�P�)

v�liquidiRuT

eg�liquidi

(T )�g�gasi (T )

RuT

!

(9.86)

The term in brackets is the classical form of the Clausius-Clapeyron equation that relatesthe vapor pressure of a gas in equilibrium with its condensed phase to the temperature ofthe system.

9.9 Chemical equilibrium, the method of element poten-tials

If the species are allowed to react at constant temperature and pressure, the mole fractionswill evolve toward values that minimize the extensive Gibbs free energy of the systemsubject to the constraint that the number of moles of each element in the mixture remainsfixed. The number of moles of each atom in the system is given by

aj =PX

p=1

IpX

i=1

npiApij (9.87)

where Apij is the number of atoms of the jth element in the ith molecular species of thepth phase. The appropriate picture of our system is shown in Figure 9.5.

Figure 9.5: System of reacting molecular species in several phases at constant temperatureand pressure with fixed number of moles of each element.

Generally the number of species in each phase will be di↵erent. This may be the case evenin a situation where, at a given temperature and pressure, there is phase equilibrium fora given species. Consider graphite in equilibrium with its vapor at low temperature, themole fraction in the vapor phase may be so small as to be essentially zero and the species


may be rightly excluded from the vapor mixture. The volume required to maintain thesystem at constant pressure and temperature must be allowed to vary.

The Gibbs free energy of the system is

G (T, P, n11

, n12

, ..., np1, ..., nPIP ) =PX

p=1

Np

IpX

i=1

xpigpi (T, P, xpi) . (9.88)

With the gas phase written separately

G (T, P, n11

, n12

, ..., n21

, ..., nPIP ) = N1

I1

X

i=1

x1i (g

�1i(T ) +RuT ln (x

1i)) +N1

RuT ln

✓

P

P �

◆

+

PX

p=2

Np

IpX

i=1

xpi (gpi(T, P ) +RuT ln (xpi))

.

(9.89)

We will use the method of Lagrange multipliers to minimize the Gibbs free energy subjectto the atom constraints. Minimize the function

W�

T, P, n11

, n12

, ..., n1I

1

, n21

, n22

, ..., n2I

2

, ...., np1, np2, ..., npIp ,�1

, ...,�J�

=

G�

T, P, n11

, n12

, ..., n1I

1

, n21

, n22

, ..., n2I

2

, ...., np1, np2, ..., npIp

�

�

RuTJX

j=1

�j

0

@

PX

p=1

IpX

i=1

npiApij � aj

1

A .

(9.90)

where the J unknown Lagrange multipliers, �j , are dimensionless. Our modified equilib-rium condition is

dW =@W

@TdT+

@W

@PdP+

@W

@n11

dn11

+....+@W

@npIdnpIP +

@W

@�1

d�1

+....+@W

@�Jd�J = 0. (9.91)

Substitute (9.76) into (9.91) and impose dP = dT = 0.

dW =PX

p=1

IpX

i=1

npidgpi (T, P, xpi) + gpi (T, P, xpi) dnpi �RuTjX

j=1

�j

PX

p=1

IpX

i=1

dnpiApij (9.92)


The order of the sums can be rearranged so (9.92) can be written as

dW =PX

p=1

IpX

i=1

npidgpi (T, P, xpi) +PX

p=1

IpX

i=1

0

@gpi (T, P, xpi)�RuTJX

j=1

�jApij

1

A dnpi = 0.

(9.93)

The di↵erential of the molar Gibbs free energy is

dgpi =@gpi@T

dT +@gpi@P

dP +RuTdxpixpi

. (9.94)

For a process that takes place at constant temperature and pressure

dW = RuTPX

p=1

IpX

i=1

npidxpixpi

+PX

p=1

IpX

i=1

0


j=1

�jApij

1

A dnpi = 0. (9.95)

The first sum in (9.95) can be re-written as follows

dW = RuTPX

p=1

Np

IpX

i=1

xpidxpixpi

+PX

p=1

IpX

i=1

0


j=1

�jApij

1

A dnpi = 0

(9.96)

or

dW = RuTPX

p=1

Np

IpX

i=1

dxpi +PX

p=1

IpX

i=1

0


j=1

�jApij

1

A dnpi = 0. (9.97)

But the normalization conditions for the mole fractions of each phase imply that

IpX

i=1

dxpi = d

0

@

IpX

i=1

xpi

1

A = d (1) = 0. (9.98)

Finally our modified equilibrium condition is


dW =PX

p=1

IpX

i=1

0


j=1

�jApij

1

A dnpi = 0. (9.99)

Since the dnpi are completely free, the condition (9.99) can only be satisfied if

gpi (T, P, xpi) = RuTJX

j=1

�jApij . (9.100)

The Gibbs free energy of the system is

G (T, P, n11

, n12

, ..., np1, ..., nPIP ) = RuTPX

p=1

IpX

i=1

JX

j=1

npi�jApij . (9.101)

Each atom in the mixture contributes equally to the extensive Gibbs free energy regardlessof which molecule it is in or which phase it is in. The molar Gibbs free energy of the ithgas phase species is

g1i(T, P ) = g�

1i(T ) +RuT ln (x1i) +RuT ln

✓

P

P �

◆

. (9.102)

Insert into (9.100). For the gas phase species

g�1i(T )

RuT+ ln (x

1i) + ln

✓

P

P �

◆

=JX

j=1

�jA1ij . (9.103)

For the condensed phase species

gpi(T, P )

RuT+ ln (xpi) =

JX

j=1

�jApij p = 2, . . . , P . (9.104)

Solve for the mole fraction of the ith species in the pth phase.


x1i = Exp

8

<

:

�g�1i(T )

RuT� ln

✓

P

P �

◆

+JX

j=1

�jA1ij

9

=

;

xpi = Exp

8

<

:

�gpi(T, P )

RuT+

JX

j=1

�jApij

9

=

;

p = 2, . . . , P

. (9.105)

The constraints on the atoms are

aj =PX

p=1

Np

IpX

i=1

xpiApij . (9.106)

Substitute (9.105) into (9.106).

aj = N1

I1

X

i=1

A1ijExp

8

<

:

�g�1i(T )

RuT� ln

✓

P

P �

◆

+JX

j1

=1

�j1

A1ij

1

9

=

;

+

PX

p=2

Np

IpX

i=1

ApijExp

8

<

:

�gpi(T, P )

RuT+

JX

j1

=1

�j1

Apij1

9

=

;

j = 1, ..., J.

(9.107)

Note that we have to introduce the dummy index j1

in the formula for xpi when we makethe substitution. The normalization conditions on the mole fractions give

8

<

:

�g�1i(T )

RuT� ln

✓

P

P �

◆

+JX

j=1

�jA1ij

9

=

;

= 1 (9.108)

and

IpX

i=1

Exp

8

<

:

�gpi(T, P )

RuT+

JX

j=1

�jApij

9

=

;

= 1 p = 2, ..., P . (9.109)

The total number of moles in the mixture is

PX

p=1

Np = N. (9.110)


Equations, (9.107), (9.108), (9.109) and (9.110) are J + P + 1 equations in the unknowns�1

, ....,�J , N1

, ..., NP and N .

As a practical matter, it is easier to compute the solution to equations (9.107), (9.108),(9.109) and (9.110) by reformulating the equations to get rid of the exponentials. De-fine

B1i(T ) ⌘ Exp

⇢

�g�1i(T )

RuT

�

(9.111)

and for the condensed species, p > 1

Bpi(T, P ) ⌘ Exp

⇢

�gpi(T, P )

RuT

�

. (9.112)

In addition, define

yj = Exp (�j) . (9.113)

The mole fractions become

x1i =

✓

P �

P

◆

B1i

JY

j=1

(yj)A

1ij

xpi = Bpi

JY

j=1

(yj)Apij p = 2, . . . , P.

(9.114)

The system of equations that needs to be solved now becomes


✓

P �

P

◆

N1

I1

X

i=1

A1ijB1i

JY

j1

=1

(yj1

)A1ij1 +

PX

p=2

Np

IpX

i=1

ApijBpi

JY

j1

=1

(yj1

)Apij1 = aj , j = 1, . . . , J

✓

P �

P

◆ I1

X

i=1

B1i

JY

j=1

(yj)A

1ij = 1

IpX

i=1

Bpi

JY

j=1

(yj)Apij = 1, p = 2, . . . , P

PX

p=1

Np = N.

(9.115)

Note that in this formulation only the always positive yj are needed to determine the molefractions. The element potentials �j , which are the logarithm of the yj , never actually needto be calculated.

A key advantage of this formulation of the problem is that the equations that need tobe solved for the unknown yi and Np are multivariate polynomials, and algorithms areavailable that enable the roots to be determined without requiring an initial guess of thesolution. Typically a number of real and complex roots are returned. The correct root isthe one with all positive real values of the yi and Np. In general, there is only one suchroot.

9.9.1 Rescaled equations

The equations (9.115) admit an interesting scaling invariance where any constants, say ↵j ,j = 1, . . . , J can be added to the normalized Gibbs free energies as long as the coe�cientsand unknowns are transformed as

Bpi = Bpi

JY

j=1

(↵j)Apij

yj =yj↵j

.

(9.116)

If the transformations (9.116) are substituted into (9.115) the system of equations remainsthe same but expressed in tildaed variables. This leads to the following reformulation ofthe problem.


Rewrite (9.111) and (9.112) by adding and subtracting the standard Gibbs free energies ofthe elements that make up the given species.

B1i = Exp

8

<

:

�g�1i(T )

RuT+

JX

j=1

A1ij

g�j(T )

RuT�

JX

j=1

A1ij

g�j (T )

RuT

9

=

;

Bpi = Exp

8

<

:

�gpi(T, P )

RuT+

JX

j=1

Apijgj(T, P )

RuT�

JX

j=1

Apijgj(T, P )

RuT

9

=

;

p > 1

(9.117)

The first two terms in the bracket in equation (9.117) constitute the Gibbs free energy offormation of the given species from its individual elements

B1i = Exp

8

<

:

��g�1i(T )

RuT�

JX

j=1

A1ij

g�j (T )

RuT

9

=

;

Bpi = Exp

8

<

:

��gpi(T, P )

RuT�

JX

j=1

Apijgj(T, P )

RuT

9

=

;

p > 1

(9.118)

where

�g�1i(T ) = g�

1i(T )�JX

j=1

A1ijg

�j (T )

�gpi(T, P ) = gpi(T, P )�JX

j=1

Apijgj(T, P ) p > 1.

(9.119)

We can write (9.118) as

B1i = B

1i

JY

j=1

Exp

✓

�A1ij

g�j (T )

RuT

◆

Bpi = Bpi

JY

j=1

Exp

✓

�Apijgj(T, P )

RuT

◆

p > 1

(9.120)

where


B1i = Exp

⇢

��g�1i(T )

RuT

�

Bpi = Exp

⇢

��gpi(T, P )

RuT

�

p > 1.(9.121)

The system of equations (9.115) can now be written as

✓

P �

P

◆

N1

I1

X

i=1

A1ijB1i

JY

j1

=1

(yj1

)A1ij1 +

PX

p=2

Np

IpX

i=1

ApijBpi

JY

j1

=1

(yj1

)Apij1 = aj , j = 1, . . . , J

✓

P �

P

◆ I1

X

i=1

B1i

JY

j=1

(yj)A

1ij = 1

IpX

i=1

Bpi

JY

j=1

(yj)Apij = 1 p = 2, . . . , P

PX

p=1

Np = N

(9.122)

where

yj = yjExp

✓

�g�j (T )

RuT

◆

j = 1, . . . , J. (9.123)

The mole fractions in scaled variables are

x1i =

✓

P �

P

◆

B1i

JY

j=1

(yj)A

1ij

xpi = Bpi

JY

j=1

(yj)Apij p = 2, . . . , P.

(9.124)

Note that the governing equations (9.115) and (9.122) have exactly the same form and theyj determine the mole fractions. The implication of this result is that the calculation ofmole fractions can be carried out either in terms of the standard Gibbs free energy of aspecies or the Gibbs free energy of formation of the species.


Generally the Gibbs free energy of formation of a substance is less than the Gibbs freeenergy itself and so usually Bpi < Bpi making the numerical solution of (9.122) simplerthan (9.115). This is especially important for gas calculations at low temperature whereg�j/RuT can be quite large producing values of the Bpi that can range over many ordersof magnitude. For many calculations, the Gibbs free energy of formation of a substance isthe only data tabulated. This is especially true for reactions in aqueous solution.

9.10 Example - combustion of carbon monoxide

If we mix carbon monoxide (CO) and oxygen (O2

) at 105Pa and 298.15K then ignite themixture, the result is a strongly exothermic reaction. The simplest model of such a reactiontakes one mole of CO plus half a mole of O

2

to produce one mole of carbon dioxide.

CO +1

2O

2

! CO2

(9.125)

But this model is not very meaningful without some information about the temperatureof the process. If the reaction occurs in an adiabatic system at constant pressure, thefinal temperature is very high and at that temperature the hot gas consists of a mixtureof a number of species beside CO

2

. A more realistic model assumes that the compositionincludes virtually all of the combinations of carbon and oxygen that one can think ofincluding

C,CO,CO2

, O,O2

. (9.126)

Other more complex molecules are possible such as C2

and O3

but are only present inextraordinarily low concentrations. For the composition (9.125) the temperature of themixture at one atmosphere turns out to be 2975.34K. This is called the adiabatic flametemperature.

Let’s use the minimization of the Gibbs free energy to determine the relative concentrationsof each molecular species for the mixture (9.126) at the equilibrium temperature 2975.34K.We will order the species as in (9.126). In this case all the species are in the gas phase andthe matrix of element coe�cients Aij is shown in Figure 9.6.

For this system of molecular species, (9.107) and (9.108) lead to the following equationsgoverning the mole fractions and the total number of moles. Note that P = P � in thiscase.


Figure 9.6: Matrix of element coe�cients for the CO, O2

system.

a1

= N

✓

A11

Exp

✓

� g�1

RuT+ �

1

A11

+ �2

A12

◆

+A21

Exp

✓

� g�2

RuT+ �

1

A21

+ �2

A22

◆

+

A31

Exp

✓

� g�3

RuT+ �

1

A31

+ �2

A32

◆

+A41

Exp

✓

� g�4

RuT+ �

1

A41

+ �2

A42

◆

+

A51

Exp

✓

� g�5

RuT+ �

1

A51

+ �2

A52

◆◆

(9.127)

a2

= N

✓

A12

Exp

✓

� g�1

RuT+ �

1

A11

+ �2

A12

◆

+A22

Exp

✓

� g�2

RuT+ �

1

A21

+ �2

A22

◆

+

A32

Exp

✓

� g�3

RuT+ �

1

A31

+ �2

A32

◆

+A42

Exp

✓

� g�4

RuT+ �

1

A41

+ �2

A42

◆

+

A52

Exp

✓

� g�5

RuT+ �

1

A51

+ �2

A52

◆◆

(9.128)

1 = Exp

✓

� g�1

RuT+ �

1

A11

+ �2

A12

◆

+ Exp

✓

� g�2

RuT+ �

1

A21

+ �2

A22

◆

+

Exp

✓

� g�3

RuT+ �

1

A31

+ �2

A32

◆

+ Exp

✓

� g�4

RuT+ �

1

A41

+ �2

A42

◆

+

Exp

✓

� g�5

RuT+ �

1

A51

+ �2

A52

◆

(9.129)

The unknowns in this system are the two Lagrange multipliers �1

and �2

correspondingto each element in the mixture and the total number of moles . The number of moles ofcarbon atoms is a

1

= 1 and the number of moles of oxygen atoms is a2

= 2. The greatadvantage of this method is that the number of unknowns is limited to the number ofelements in the mixture not the number of molecular species. Use (9.111) and (9.113) torewrite (9.127), (9.128) and (9.129) in the form of (9.115).


a1

/N = A11

B1

y1

A11y

2

A12 +A

21

B2

y1

A21y

2

A22 +A

31

B3

y1

A31y

2

A32+

A41

B4

y1

A41y

2

A42 +A

51

B5

y1

A51y

2

A52

(9.130)

a2

/N = A12

B1

y1

A11y

2

A12 +A

22

B2

y1

A21y

2

A22 +A

32

B3

y1

A31y

2

A32+

A42

B4

y1

A41y

2

A42 +A

52

B5

y1

A51y

2

A52

(9.131)

1 = B1

y1

A11y

2

A12 +B

2

y1

A21y

2

A22 +B

3

y1

A31y

2

A32+

B4

y1

A41y

2

A42 +B

5

y1

A51y

2

A52

(9.132)

According to (9.116) these equations can be rescaled as follows. Let

B1

= B1

↵1

A11↵

2

A12

B2

= B2

↵1

A21↵

2

A22

B3

= B3

↵1

A31↵

2

A32

B4

= B4

↵1

A41↵

2

A42

B5

= B5

↵1

A51↵

2

A52

(9.133)

and

y1

=y1

↵1

y2

=y2

↵2

.(9.134)

If (9.133) and (9.134) are substituted into (9.130), (9.131) and (9.132) the resulting equa-tions read exactly the same except in terms of tildaed variables. This invariance can beexploited to reduce the range of magnitudes of the coe�cients in the equation if needed.Now

1 = N�

B1

y1

+B2

y1

y2

+B3

y1

y2

2

�

(9.135)

2 = N�

B2

y1

y2

+ 2B3

y1

y2

2 +B4

y2

+ 2B5

y2

2

�

(9.136)

1 = B1

y1

+B2

y1

y2

+B3

y1

y2

2 +B4

y2

+B5

y2

2 (9.137)

where the coe�cients are


B1

= e�g�CRuT

B2

= e�g�CORuT

B3

= e�g�CO

2

RuT

B4

= e�g�ORuT

B5

= e�g�O

2

RuT .

(9.138)

At this point we need to use tabulated thermodynamic data to evaluate the coe�cients.The Gibbs free energy of the ith molecular species is

g�i (T ) = �h�fi (Tref ) + {h�i (T )� h�i (Tref )}� Ts�i (T ) . (9.139)

Data for each species is as follows (See appendix 2). In the same order as (9.126)

g�C (2975.34) = 715.004 + 56.208� 2975.34 (0.206054) = 158.131 kJ/moleg�CO (2975.34) = �110.541 + 92.705� 2975.34 (0.273228) = �830.782 kJ/moleg�CO

2

(2975.34) = �393.522 + 151.465� 2975.34 (0.333615) = �1234.68 kJ/moleg�O (2975.34) = 249.195 + 56.1033� 2975.34 (0.209443) = �317.866 kJ/moleg�O

2

(2975.34) = 0.00 + 97.1985� 2975.34 (0.284098) = �748.09 kJ/mole.(9.140)

The universal gas constant in appropriate units is

Ru = 8.314472⇥ 10�3 kJ/mole�K (9.141)

and RuT = 24.7384 kJ/mole. Now the coe�cients are

B1

= Exp

✓

� g�CRuT

◆

= Exp

✓

�158.131

24.7382

◆

= 1.67346⇥ 10�3

B2

= Exp

✓

� g�CO

RuT

◆

= Exp

✓

830.782

24.7382

◆

= 3.84498⇥ 1014

B3

= Exp

✓

�g�CO

2

RuT

◆

= Exp

✓

1234.68

24.7382

◆

= 4.73778⇥ 1021

B4

= Exp

✓

� g�ORuT

◆

= Exp

✓

317.866

24.7382

◆

= 3.80483⇥ 105

B5

= Exp

✓

�g�O

2

RuT

◆

= Exp

✓

748.090

24.7382

◆

= 1.35889⇥ 1013.

(9.142)


Notice the ill conditioned nature of this problem. The constants in (9.142) vary over manyorders of magnitude requiring high accuracy and very careful numerical analysis. Thescaling (9.133) is

B1

= B1

↵1

B2

= B2

↵1

↵2

B3

= B3

↵1

↵2

2

B4

= B4

↵2

B5

= B5

↵2

2.

(9.143)

Choose ↵2

=p

1.35889⇥ 1013 so that B5

= 1 and ↵1

= 1.67346 ⇥ 10�3 so that B1

= 1.The scaled coe�cients are

B1

= B1

/↵1

= 1.67346⇥ 10�3/1.67346⇥ 10�3 = 1

B2

= B2

/↵1

↵2

= 3.84498⇥ 1014/⇣

1.67346⇥ 10�3

p

1.35889⇥ 1013⌘

= 6.23285⇥ 1010

B3

= B3

/↵1

↵2

2 = 4.73778⇥ 1021/�

1.67346⇥ 10�3

�

1.35889⇥ 1013��

= 2.08341⇥ 1011

B4

= B4

/↵2

= 3.80483⇥ 105/p

1.35889⇥ 1013 = 0.103215B

5

= B5

/↵2

2 = 1.35889⇥ 1013/1.35889⇥ 1013 = 1.(9.144)

Using this procedure we have reduced the range of the coe�cients from 24 down to 11orders of magnitude, a very significant reduction. The equations we need to solve are asfollows.

1 = N⇣

B1

y1

+ B2

y1

y2

+ B3

y1

y22

⌘

(9.145)

2 = N⇣

B2

y1

y2

+ 2B3

y1

y22

+ B4

y2

+ 2B5

y22

⌘

(9.146)

1 = B1

y1

+ B2

y1

y2

+ B3

y1

y22

+ B4

y2

+ B5

y22

(9.147)

I used Mathematica to solve the system, (9.145), (9.146), and (9.147). The result is

y1

= 1.42474⇥ 10�11

y2

= 0.392402n = 1.24144

(9.148)


At the mixture temperature T = 2975.34K, the mole fractions of the various speciesare

xC = B1

y1

= 1.42474⇥ 10�11

xCO = B2

y1

y2

= 6.23285⇥ 1010 ⇥ 1.42474⇥ 10�11 ⇥ 0.392402 = 0.34846xCO

2

= B3

y1

y22

= 2.08341⇥ 1011 ⇥ 1.42474⇥ 10�11 ⇥ 0.3924022 = 0.45706xO = B

4

y2

= 0.103215⇥ 0.392402 = 0.0405018xO

2

= B5

y22

= 0.3924022 = 0.153979.

(9.149)

Note that there is almost no free carbon at this temperature. We could have dropped Cfrom the mixture (9.126) and still gotten practically the same result.

9.10.1 CO Combustion at 2975.34K using Gibbs free energy of forma-tion.

The Gibbs free energies of formation of the various species are

�g�C (2975.34) = 250.012 kJ/mole�g�CO (2975.34) = �365.761 kJ/mole�g�CO

2

(2975.34) = �395.141 kJ/mole�g�O (2975.34) = 55.8281 kJ/mole�g�O

2

(2975.34) = 0.0 kJ/mole.

(9.150)

These values can be found in the JANAF tables in Appendix 2. Using the Gibbs freeenergies of formation, the coe�cients of the element potential equations are

B1

= Exp

��g�C (2975.34)

RuT

�

= Exp

�250.012

24.7382

�

= 4.0821⇥ 10�5

B2

= Exp

��g�CO (2975.34)

RuT

�

= Exp

365.761

24.7382

�

= 2.63731⇥ 106

B3

= Exp

��g�CO

2

(2975.34)

RuT

�

= Exp

395.141

24.7382

�

= 8.6486⇥ 106

B4

= Exp

��g�O (2975.34)

RuT

�

= Exp

�55.8281

24.7382

�

= 0.104689

B5

= Exp

��g�O

2

(2975.34)

RuT

�

= Exp

� 0.0

24.7382

�

= 1.00.

(9.151)

The results using Mathematica to solve this system are


y1

= 3.382049⇥ 10�7

y2

= 0.3935536n = 1.24391

(9.152)

and the mole fractions are

xC = B1

y1

= 1.380743⇥ 10�11

xCO = B2

y1

y2

= 0.3509709xCO

2

= B3

y1

y2

2 = 0.4529426xO = B

4

y2

= 0.04120202xO

2

= B5

y2

2 = 0.15488445.

(9.153)

These results agree closely with the results in (9.149) and with calculations using CEA.Now transfer heat out of the gas mixture to bring the temperature back to 298.15K at oneatmosphere. The standard Gibbs free energies of the species at this temperature are

g�C (298.15) = 669.54219 kJ/moleg�CO (298.15) = �169.46747 kJ/moleg�CO

2

(298.15) = �457.25071 kJ/moleg�O (298.15) = 201.15482 kJ/moleg�O

2

(298.15) = �61.16531 kJ/mole

(9.154)

and RuT = 2.47897 kJ/mole. The coe�cients are

B1

= Exp

✓

� g�CRuT

◆

= Exp

✓

�669.54219

2.47897

◆

= 5.03442⇥ 10�118

B2

= Exp

✓

� g�CO

RuT

◆

= Exp

✓

169.46747

2.47897

◆

= 4.88930⇥ 1029

B3

= Exp

✓

�g�CO

2

RuT

◆

= Exp

✓

457.25071

2.47897

◆

= 1.277622⇥ 1080

B4

= Exp

✓

� g�ORuT

◆

= Exp

✓

�201.15482

2.47897

◆

= 5.74645⇥ 10�36

B5

= Exp

✓

�g�O

2

RuT

◆

= Exp

✓

61.16531

2.47897

◆

= 5.19563⇥ 1010.

(9.155)

At this relatively low temperature the coe�cients range over 198 orders of magnitude! De-spite this a good modern solver should be able to find the solution of (9.135), (9.136) and(9.137) for this di�cult case even without the scaling used earlier. For example, Mathemat-ica has a feature where the user can specify an arbitrary number of digits of precision forthe calculation. The solution of (9.135), (9.136) and (9.137) at this temperature is


y1

= 7.07098⇥ 10�40

y2

= 3.32704⇥ 10�21

n = 1.0000.(9.156)

The resulting mole fractions at T = 298.15K are

xC = B1

y1

= 3.55983⇥ 10�157

xCO = B2

y1

y2

= 1.15023⇥ 10�30

xCO2

= B3

y1

y2

2 = 1.00000xO = B

4

y2

= 1.91187⇥ 10�56

xO2

= B5

y2

2 = 5.75116⇥ 10�31.

(9.157)

Only when the mixture is brought back to low temperature is the reaction model (9.125)valid. The enthalpy change for this last step is

h�|mixture at 298.15K � h�|mixture at 2975.34K = �8.94⇥ 106 + 2.51⇥ 106 = �6.43⇥ 106 J/kg.(9.158)

This is the chemical energy released by the reaction (9.125) and is called the heat ofreaction.

9.10.2 Adiabatic flame temperature

In the example in the previous section the products of combustion were evaluated at theadiabatic flame temperature. This can be defined at constant volume or constant pressure.For our purposes we will use the adiabatic flame temperature at constant pressure. Imaginethe reactants brought together in a piston-cylinder combination permitting the volume tobe adjusted to keep the pressure constant as the reaction proceeds. A source of ignition isused to start the reaction that evolves to the equilibrium state defined by the equilibriumspecies concentrations at the original pressure and at an elevated temperature called theadiabatic flame temperature. In the process the Gibbs function is minimized and since theprocess is adiabatic, the enthalpy before and after the reaction is the same.

The general enthalpy balance for a reaction is given in (9.67). Fully written out the balanceis


�h� (Tfinal) =

IproductX

iproduct

niproduct

n

�h�fiproduct (298.15) +⇣

h�iproduct (Tfinal)� h�iproduct (298.15)⌘o

�

IreactantX

ireactant

nireactant

�

�h�fireactant(298.15) +

�

h�ireactant(Tireactant)� h�ireactant

(298.15)�

.

(9.159)

If the reaction takes place adiabatically then �h� (Tfinal) = 0 and

IproductX

iproduct

niproduct

n

�h�fiproduct (298.15) +⇣

h�iproduct (Tfinal)� h�iproduct (298.15)⌘o

=

IreactantX

ireactant

nireactant

�

�h�fireactant(298.15) +

�

h�ireactant(Tireactant)� h�ireactant

(298.15)�

.

(9.160)

Equation (9.160) can be solved along with (9.135), (9.136) and (9.137) to determine thefinal temperature of the mixture along with the mole fractions and total number of moles.In the carbon monoxide combustion example of the previous section we would write

nC�

�h�fC (298.15) + (h�C (Tfinal)� h�C (298.15))

+nCO

�

�h�fCO (298.15) + (h�CO (Tfinal)� h�CO (298.15))

+nCO

2

�

�h�fCO2

(298.15) +�

h�CO2

(Tfinal)� h�CO2

(298.15)�

+nO

�

�h�fO (298.15) + (h�O (Tfinal)� h�O (298.15))

+nO

2

�

�h�fO2

(298.15) +�

h�O2

(Tfinal)� h�O2

(298.15)�

=nCO

�

�h�fCO (298.15)

+ nO2

�

�h�fO2

(298.15)

.

(9.161)

The enthalpy of the reactants is

nCO�

�h�fCO (298.15)

+ nO2

�

�h�fO2

(298.15)

=1.0 kgmole

�

�110.527⇥ 103 kJ/kgmole

+ 0.5 kgmole {0 kJ/kgmole} =�110.527⇥ 103 kJ.

(9.162)

On a per unit mass basis the enthalpy of the reactant mixture is

�110.527⇥ 106 J

1⇥ 28.014 + 0.5⇥ 31.98= �2.5117⇥ 106 J/kg. (9.163)


The enthalpy per unit mass of the product mixture at various temperatures is plotted inFigure 9.7.

Figure 9.7: Enthalpy of the product mixture as a function of temperature.

As the products of combustion are cooled from 4000K the enthalpy decreases monitonically.The only temperature where the enthalpy of the product mixture matches that of theoriginal reactants is the adiabatic flame temperature, 2975.34K.

9.10.3 Isentropic expansion

Now consider an isentropic expansion from a known initial state, (Tinitial, Pinitial) to a finalstate (Tfinal, Pfinal) with the final pressure known. The condition that determines thetemperature of the final state is

S(Tfinal, Pfinal, n1final , n2final , n3final , ..., nIfinal) =S(Tinitial, Pinitial, n1initial , n2initial , n3initial , ..., nIinitial)

(9.164)

or

IX

i=1

nifinals�i (Tfinal)�NfinalRu

IX

i=1

xifinal ln�

xifinal

�

�NfinalRu ln

✓

Pfinal

P �

◆

=

IX

i=1

niinitials�i (Tinitial)�NinitialRu

IX

i=1

xiinitial ln (xiinitial)�NinitialRu ln

✓

Pinitial

P �

◆

.

(9.165)

Equation (9.165) can be solved along with (9.135), (9.136) and (9.137) to determine thefinal temperature of the mixture after isentropic expansion along with the mole fractionsand total number of moles.


For example, take the mixture from the previous section at the initial state, Tinitial = 2975.34Kand Pinitial = 1 bar. The entropy of the system on a per unit mass basis is 8.7357kJ/kg�Kwhich is essentially equivalent to the extensive entropy. Now expand the mixture toPfinal = 0.1 bar. If we calculate the entropy of the system at this pressure and varioustemperatures, the result is the following plot.

Figure 9.8: Entropy of the product mixture as a function of temperature.

The temperature at which the entropy of the final state is the same as the initial state isTfinal = 2566.13K.

9.10.4 Nozzle expansion

If we interpret the expansion just described as an adiabatic, isentropic expansion in anozzle we can use the conservation of stagnation enthalpy to determine the speed of thegas mixture at the end of the expansion.

Hinitial = Hfinal +1

2U2 (9.166)

The initial enthalpy is taken to be the reservoir value. For this example the numbersare

U =q

2 (Hinitial �Hfinal) =q

2 (�2.51162 + 3.94733)⇥ 106 J/kg = 1694.53m/ sec .

(9.167)

Ordinarily we are given the geometric area ratio of the nozzle rather than the pressureratio. Determining the exit velocity in this case is a little more involved. Here we needto carry out a series of calculations at constant entropy and varying final pressure. Foreach calculation we need to determine the density and velocity of the mixture and plot the


product ⇢U as a function of the pressure ratio. Beginning with the mixture at the adiabaticflame temperature as the reservoir condition, the results are plotted below.

Figure 9.9: Mass flux in a converging-diverging nozzle as a function of nozzle static pressureratio.

The maximum mass flux occurs at the nozzle throat. Equate the mass flow at the throatand the nozzle exit. For Pinitial/Pfinal = 10.0 the nozzle area ratio is

Ae

At=

⇢tUt

⇢eUe=

73.6768

29.8354= 2.46944. (9.168)

This completes the specification of the nozzle flow. The case we have considered here iscalled the shifting equilibrium case where the gas mixture is at equilibrium at every pointin the nozzle. One can also consider the case of frozen flow where the composition of thegas mixture is held fixed at the reservoir condition.

9.10.5 Fuel-rich combustion, multiple phases

Suppose we choose a mixture of CO and O2

that has an excess of CO. In this case theoverall balance of carbon and oxygen can be satisfied in more than one way when theproducts of the reaction are brought back to low temperature. For example, if we mix 2moles of CO with 0.5 moles of O

2

at a pressure of 105N/m2 we could end up with eitherof the following.

2CO +1

2O

2

! CO2

+ CO

2CO +1

2O

2

! 3

2CO

2

+1

2C(gr)

(9.169)


Which balance actually occurs is of course determined by which one minimizes the Gibbsfree energy at the given temperature. The set of species in the mixture is now

C,CO,CO2

, O,O2

, C(gr) (9.170)

where we have allowed for the possible condensation of solid carbon. The matrices ofelement coe�cients Apij for the two phases are shown in Figure 9.10.

Figure 9.10: Matrix of element coe�cients for the CO, O2

system with condensed species(graphite).

The governing equations (9.122) become the following.

2 = N1

�

B11

y1

+B12

y1

y2

+B13

y1

y2

2

�

+N2

(B21

y1

) (9.171)

3 = N1

�

B12

y1

y2

+ 2B13

y1

y2

2 +B14

y2

+ 2B15

y2

2

�

(9.172)

1 = B11

y1

+B12

y1

y2

+B13

y1

y2

2 +B14

y2

+B15

y2

2 (9.173)

1 = B21

y1

(9.174)

N = N1

+N2

(9.175)

The coe�cients are


B11

= e�g�CRuT

B12

= e�g�CORuT

B13

= e�g�CO

2

RuT

B14

= e�g�ORuT

B15

= e�g�O

2

RuT

(9.176)

and

B21

= e�g�C

(gr)RuT . (9.177)

As an example, let’s solve for the various mole fractions at a mixture temperature of 940K.From tables, the standard Gibbs free energies at this temperature are

g�C (940) = 558.95895 kJ/moleg�CO (940) = �309.37696 kJ/moleg�CO

2

(940) = �613.34861 kJ/moleg�O (940) = 88.41203 kJ/moleg�O

2

(940) = �206.32876 kJ/moleg�C

(gr)(940) = �11.22955 kJ/mole.

(9.178)

The number of moles of each species in the mixture at T = 940K are

nC = N1

B11

y1

= 4.11775⇥ 10�32

nCO = N1

B12

y1

y2

= 0.977065nCO

2

= N1

B13

y1

y2

2 = 1.011467nO = N

1

B14

y2

= 3.23469⇥ 10�22

nO2

= N1

B15

y2

2 = 1.027845⇥ 10�22

nC(gr)

= N2

B21

y1

= 0.011467.

(9.179)

At this temperature, the mixture is predominately CO and CO2

with some C(gr). Figure

9.11 shows the mole fraction of CO and C(gr) at a pressure of 105N/m2 based on the total

number of moles in the mixture for a variety of mixture temperatures.

The excess CO begins to react and solid carbon begins to condense out at about T =942.2K. Below 600K there is virtually no CO in the mixture.


Figure 9.11: Condensation of graphite in the CO, O2

system.

9.11 Rocket performance using CEA

The equilibrium combustion package CEA (Chemical Equilibrium with Applications) fromNASA Glenn can also be used to perform equilibrium chemistry calculations and has acapability similar to STANJAN but with a much wider range of chemicals with data basedon the current standard pressure. Some typical performance parameters for several pro-pellant combinations at two chamber pressures are shown in Figure 9.12. The propellantsare taken to be at an equivalence ratio of one (complete consumption of fuel and oxidizer)and so the exhaust velocity is not optimized. The numbers correspond to the e↵ectiveexhaust velocity for the given chamber pressure and area ratio assuming vacuum ambientpressure. The maximum e↵ective exhaust velocity generally occurs with a somewhat fuelrich mixture that produces more low molecular weight species in the exhaust stream.

9.12 Problems

Problem 1 - A nuclear reactor is used to heat hydrogen gas in a rocket chamber to atemperature of 4000K. The pressure in the chamber is 100 atm. At these conditions asignificant fraction of the H

2

is dissociated to form atomic H. What are the mole fractionsof H, H

2

in the mixture? Relevant thermochemical data is provided in Figure 9.13 . Thereference temperature is 298.15K.

Work the problem by hand and compare with CEA.

Problem 2 - Hydrogen gas heated to 4000K is fed at a mass flow rate of 200 kg/sec intoa rocket chamber with a throat area of 0.2m2. The gas is exhausted adiabatically and


Figure 9.12: Some typical performance parameters for several propellant combinations attwo chamber pressures.

Figure 9.13: Hydrogen dissociation at 4000K.


isentropically through a nozzle with an area ratio of 40. Determine the exhaust velocityfor the following cases.

i) Expansion of undissociated H2

.

ii) Frozen flow expansion of dissociated H2

at the rocket chamber composition of H andH

2

.

(III) Shifting equilibrium expansion.

Problem 3 - An exotic concept for chemical rocket propulsion is to try to harness theenergy released when two atoms of hydrogen combine to form H

2

. The idea is to storethe hydrogen atoms in solid helium at extremely low temperatures. Suppose a space en-gine is designed with a very large area ratio nozzle. Let a 50-50 mixture by mass ofatomic hydrogen and helium be introduced into the combustion chamber at low temper-ature. The propellant vaporizes and the hydrogen atoms react releasing heat. The gasexhausts through the nozzle to the vacuum of space. Estimate the exhaust velocity of thisrocket.

Problem 4 - I would like you to consider the Space Shuttle Main Engine (SSME). Usethe thermochemical calculator CEA or an equivalent application to help solve the problem.The specifications of the SSME are

Pt2 = 3000 psiaTt2 = 3250KAe/At = 77.5

(9.180)

and

mH2

= 69 kg/ secmO

2

= 400 kg/ sec .(9.181)

i) Determine the adiabatic flame temperature and equilibrium mass fractions of the system(H,H

2

, HO,H2

O,O,O2

) at the given chamber pressure.

ii) You will find that the temperature in (1) is higher than the specified chamber temper-ature. Bring your mixture to the specified chamber pressure and temperature (in e↵ectaccounting for some heat loss from the engine).

Determine the exhaust velocity and sea level thrust assuming that the mixture remains atequilibrium during the expansion process. Suppose the mass flow rates are changed to thestoichiometric ratio


mH2

= 52.5 kg/ secmO

2

= 416.5 kg/ sec.(9.182)

How much does the exhaust velocity change?

Problem 5 - A monopropellant thruster for space applications uses nitrous oxide N2

O ata low initial temperature as a monopropellant. The gas is passed through a catalyst bedwhere it is decomposed releasing heat. The hot gas is expelled through a large area rationozzle to the vacuum of space. The exhaust gas is composed of O

2

and N2

. The enthalpyof formation of N

2

O at the initial temperature is 1.86 ⇥ 106 J/kg. Estimate the exhaustvelocity of this rocket.

Problem 6 - Element number three in the periodic table is Lithium which is a soft,silvery white, highly reactive metal at room temperature. Because of its low atomic weightit has been considered as a propellant for propulsion applications. Above 1615K it is amonatomic gas. At high temperatures and pressures the diatomic gas begins to form. Atthe super extreme conditions of 7000K and 105 bar almost 92% of the mixture is Li

2

andonly 8% is Li. By the way, at these conditions the density of the mixture is 2.3⇥104 kg/m3,greater than Uranium. Perhaps this is the di-Lithium crystal propellant that, according toGene Roddenberry, will in the future power generations of starships. Suppose this mixtureat an enthalpy hinitial = 38.4⇥106 J/kg expands isentropically from a rocket chamber to anozzle exit where the enthalpy is hfinal = 9.3⇥ 106 J/kg. What is the nozzle exit velocity?What is the composition of the mixture at the nozzle exit?

Problem 7 - One mole per second of methane CH4

reacts with 2 moles per second ofO

2

in a rocket chamber. The reaction produces products at 3000K and 100 bar. Assumethe mixture contains (CO,CO

2

, H,H2

, H2

O,O,OH,O2

). Set up the system of equations(9.115) for this problem. Use data from Appendix 2 to evaluate the coe�cients in theseequations and solve for the mole fractions. Compare your results with CEA. The gasexhausts isentropically to the vacuum of space through a nozzle with an area ratio of 80.Determine the thrust.

Problem 8 - Ozone (O3

) releases energy when it decomposes and can be used as a mono-propellant for a space thruster. Let ozone be decomposed across a manganese oxide catalystbed and introduced into a thrust chamber. The pressure and temperature in the cham-ber are 105N/m2 and 2688K. At these conditions the mixture is composed of only twoconstituents, O

2

and O. Relevant thermo-chemical data is given in Figure 9.14.

In the units used in this table the ideal gas constant isRu = 0.00831451 kJ/ (mole�K).

a) Determine the mole fractions of O2

and O in the mixture.

b) Determine the total number of moles in the mixture.


Figure 9.14: Ozone decomposition at 2688K.

c) Determine the enthalpy per unit mass of the mixture.

d) If the gas is exhausted to the vacuum of space what is the maximum gas speed thatcould be reached.

thermodynamics of reacting mixtures - stanford...

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