App 1 Thermochemistry

bjc A1.1 12/15/10

A

PPENDIX

1 E

LEMENTS

OF

T

HERMOCHEMISTRY

A1.1 T

HERMOCHEMICAL

T

ABLES

In Chapter 9 we developed the theory for determining the equilibrium composi-tion of a mixture of reacting gases. The procedure requires knowledge of thestandard Gibbs free energy for each molecule in the mixture.

(A1.1)

where is the standard heat of formation of the chemical species

at some reference temperature, is the change in the stan-

dard enthalpy from the reference temperature and is the standard entropy.

Data for these variables for a given chemical species can be found in tabulationsof thermochemical properties that have been developed from experimental mea-surements of material properties over many decades. The purpose of this appendixis to describe the tables, explain the data and connect thermodynamic propertiesof gases to the chemical bonds between the atoms that make up the gas. Perhapsthe most widely used thermochemical data is from the compilation produced bythe JANAF (Joint Army, Navy, Air Force) Committee. Parts of the JANAF tablesfor monatomic and diatomic hydrogen are shown below.

Figure A1.1 JANAF data for diatomic and monatomic hydrogen in the temperature range from to . The full tabulation runs to .

gi° T( ) !h° fi T ref( ) h°i T( ) hi° T ref( )–{ } T si° T( )–+=

!h° fi T ref( ) ith

h°i T( ) hi° T ref( )–{ }

si° T( )

0K 2000K 6000K

Standard Pressure

12/15/10 A1.2 bjc

All quantities are quoted on a per mole basis and in units of Joules and degrees

Kelvin. One mole is an Avagadro’s number of molecules.

In the table headings in Figure A1.1 properties are symbolized using capital letterswhereas we have adopted the convention that all intensive (per unit mass or perunit mole) variables are denoted with lower case letters as in (A1.1) and we willstick to that convention. Full tables for a variety of chemical species are providedin Appendix 2.

A1.2 S

TANDARD

P

RESSURE

Conservation of energy for a system containing some substance is expressed asthe First Law of thermodynamics.

(A1.2)

The enthalpy is defined as

. (A1.3)

If we use (A1.3) to replace the internal energy in (A1.2) the result is the alternateform of the first law

. (A1.4)

The internal energy and enthalpy are related to the temperature, pressure andvolume of the system by the definitions of the specific heats.

. (A1.5)

For a process that takes place at constant pressure, the heat added or removed froma system is given by the change of enthalpy

. (A1.6)

It is probably fair to say that the use of the tables of thermochemical properties isdominated by applications to mixtures of reacting gases typical of combustion atelevated temperatures. As long as the pressure is not extreme, the equation of statefor each gas in the mixture is the ideal gas law

(A1.7)

6.0221415 1023×

"q de PdV+=

h e PV+=

"q dh VdP–=e h

CV#e#T-------

Volume= CP

#h#T-------

Pressure=

"q P constant dh P constant CPdT= =

PiV niRuT=

Standard Pressure

bjc A1.3 12/15/10

where is the number of moles of the component of the mixture, is the

partial pressure and is the volume of the mixture. The universal gas constant is

. (A1.8)

For a general substance the heat capacity, enthalpy, entropy and Gibb potential arefunctions of temperature and pressure.

(A1.9)

For any substance that follows the ideal gas law equation of state the heat capac-ities, internal energy and enthalpy are independent of pressure and the entropy andGibbs free energy depend in a known way on the logarithm of the pressure. Ittherefore makes sense to standardize all thermochemical data at a standard pres-sure rather than, say, at standard volume.

The thermodynamic properties of a substance, whether it is an ideal gas or not,are always tabulated as a function of temperature at standard pressure. That thedata is tabulated this way is indicated by the superscript that appears with eachsymbol in the tables. In various units, the universally accepted value of the stan-dard pressure is

. (A1.10)

The various thermodynamic variables for any species at standard pressure aredenoted by

. (A1.11)

The standard pressure is very close to one atmosphere and indeed before 1982 the standard was atmospheric pressure at sea level. In1982 the International Union of Pure and Applied Chemistry (IUPAC) recom-

ni ith Pi

V

Ru 8.314510 J/mole-K=

C p T P,( )

h T P,( )

s T P,( )

g T P,( )

°

P° 105 N/m2 105 Pascals 100 kPa = 1 bar= = =

C p T 100,( ) C°p T( )=

h T 100,( ) h° T( )=s T 100,( ) s° T( )=g T 100,( ) g° T( )=

1 atm = 1.01325 105× Pa

Standard Pressure

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mended that for the purposes of specifying the physical properties of substancesthe standard pressure should be defined as precisely . This had the imme-diate effect of simplifying thermochemical calculations and the practical effect ofspecifying the standard pressure to be closer to the actual average altitude abovesea level where most people around the world live.

A1.2.1 WHAT ABOUT PRESSURES OTHER THAN STANDARD?For substances in their gaseous state at elevated temperatures where the ideal gaslaw applies and the heat capacities, energy and enthalpy are independent of pres-sure the enthalpy is simply

. (A1.12)

The entropy is

(A1.13)

and the Gibbs free energy is determined from the definition

(A1.14)

In principle the standard entropy requires an integration of the heat capacity fromabsolute zero.

A1.2.2 EQUILIBRIUM BETWEEN PHASES

For a liquid in equilibrium with its vapor the Gibbs free energy of the liquid isequal to the Gibbs free energy of the vapor

(A1.15)

In terms of the definition of the Gibbs free energy

(A1.16)

Differentiate (A1.16).

100 kPa

h T( ) h° T( ) h f 298.15( ) C°pT TdT ref

T

$+!= =

s T P,( ) C°p T( )TdT------ RuLog P

100---------% &' (–

0

T$ s° T( ) RuLog P

100---------% &' (–= =

g T P,( ) h Ts– h° T( ) T s° T( ) RuTLog P100---------% &' (+– = = =

g° T( ) RuTLog P100---------% &' (+

gliquid T P,( ) ggas T P,( )=

hliquid T sliquid– hgas T sgas–=

Standard Pressure

bjc A1.5 12/15/10

(A1.17)

The Gibbs equation is

(A1.18)

where is the molar density of the substance. Substitute (A1.18)

into (A1.17). The result is

(A1.19)

with a little bit of rearrangement (A1.19) becomes

. (A1.20)

The heat required to convert a mole of liquid to gas at constant temperature iscalled the latent heat of vaporization. From (A1.16)

. (A1.21)

If we substitute (A1.21) into (A1.20) the result is the famous Clausius-Clapeyronequation

(A1.22)

that relates the vapor pressure of a gas-liquid system to the temperature. Gener-ally the temperature of the system is specified and the vapor pressure is to bedetermined. In some cases the condensed phase is a solid such as the graphite formof carbon or solid iodine sublimating directly from the solid to the vapor phase.

In the case of a solid in equilibrium with its liquid such as water and ice the Gibbsfree energies of the solid and liquid phases are equal and a similar relation governsthe transition between phases. The enthalpy change is called the latent heat offusion. In a solid-liquid system usually the pressure is the specified condition andthe temperature of melting is the property to be determined.

The heat of vaporization depends on the temperature according to

dhliquid T dsliquid– sliquiddT– dhgas T dsgas– sgasdT–=

Tdsliquid dhliquid vliquiddP–=

Tdsgas dhgas vgasdP–=

v V n=

vliquiddP sliquiddT– vgasdP sgasdT–=

dPdT-------

sgas sliquid–vgas vliquid–---------------------------------=

!hvap hgas hliquid– T sgas sliquid–( )= =

dPdT-------

!hvapT vgas vliquid–( )------------------------------------------=

Standard Pressure

12/15/10 A1.6 bjc

. (A1.23)

Generally speaking the heat capacity term is relatively small for modest changesin temperature. For example, for water between its freezing and boiling point

. (A1.24)

If the effect of heat capacity differences on the latent heat of vaporization isneglected, and the liquid molar volume is much smaller than the gas, and the gasfollows the ideal gas law then (A1.22) can be integrated to give

. (A1.25)

The reference temperature and pressure are taken at the boiling point where thevapor pressure equals the gas pressure.

The Clausius -Clapeyron equation provides a direct connection between the tem-perature of a mixture of gases and the partial pressure of any constituent that is inequilibrium with its condensed phase.

In order to determine the heat capacity and enthalpy of a substance at conditionswhere pressure effects do apply either in its condensed state or in a gaseous stateclose to the vaporization temperature at some given pressure, an equation of stateis needed in order to evaluate . The heat capacities aredetermined using the definitions in (A1.5). Generally the effect of pressure on theheat capacities of most liquids and solids are well known.

hgas T( ) hgas T ref( ) C pgasT T ref–( )+=

hliquid T( ) hliquid T ref( ) C pliquidT T ref–( )+=

!hvap T( ) !hvap T ref( ) C pgasC pliquid

–( ) T T ref–( )+=

!hvap 273.15K( ) !hvap 373.15K( ) 33 75–( ) 100–( )+=

!hvap 273.15K( ) 41 kJ/mole 4.2+ 45 kJ/mole= =

PPref----------- e

!hvap T ref( )

Ru------------------------------- 1

T--- 1

T ref----------–% &

' (–

=

h T P,( ) e T P,( ) PV+=

Reference temperature

bjc A1.7 12/15/10

A1.3 REFERENCE TEMPERATUREThe standard enthalpy of a species is usually tabulated as the difference betweenthe standard enthalpy at a given temperature and the standard enthalpy at a refer-ence temperature. By wide agreement the reference temperature is taken to be

and this temperature is indicated in the headings of the tables shown inFigure A1.1. The enthalpy of a species is expressed as

. (A1.26)

This seemingly trivial representation of the enthalpy in (A1.26) is explained in thenext two sections.

A1.4 REFERENCE REACTION, AND REFERENCE STATE FOR ELEMENTSThe thermochemical tables are designed to enable one to analyze systems of react-ing gases and condensed materials. To facilitate this, the enthalpy of a substanceis defined to include the energy contained in the chemical bonds that hold togetherthe various atoms that compose the substance. This leads to the concept of a ref-erence reaction for a substance, the corresponding reference state for the elementsthat make up the substance and the heat of formation for the substance.

The reference reaction for a given substance is the reaction in which the substanceis created from its elements when those elements are in their reference state. Thereference state for elements is always taken to be the thermodynamically stablestate of the element at the standard temperature . For elements that aregases at and the reference state is taken by convention to begaseous over the entire range of temperatures from up to the highest temper-ature tabulated (typically ). For elements that are condensed at such as a metal like aluminum or titanium, the reference state is taken to be theelement in its condensed state over the entire temperature range.

The great advantage of using a reference temperature of close to room tem-perature is that for applications at elevated temperature the enthalpy can bedetermined by integrating the heat capacity from the reference temperature with-out having to address the complexities of a given substance at low temperature. Achoice of, say, for the reference temperature would not offer this same con-

298.15K

h° T( ) h° 298.15( ) h° T( ) h° 298.15( )–{ }+=

298.15K298.15K 100kPa

0K6000K 298.15K

25C

0K

The heat of formation

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venience. For the analysis of systems at low temperature such as the cryogenicliquids used in rocket propulsion applications the tables are not very useful sincethe heat capacity data and enthalpy are not provided below .

A1.5 THE HEAT OF FORMATIONThe enthalpy (heat) of formation of a substance is defined as the enthalpy changethat occurs when the substance is formed from its elements in their reference stateat temperature and standard pressure and is denoted

(A1.27)

A consequence of this definition is that the heat of formation of a pure element inits reference state at any temperature is always zero.

For example, the enthalpy of formation of any of the diatomic gases is zero at alltemperatures. This is clear when we write the (trivial) reaction to form from

its elements in their reference state as

. (A1.28)

The enthalpy change is clearly zero. In fact the change in any thermodynamicvariable for any element in its reference state is zero at all temperatures. This isthe reason for all the zeros in the last three columns in the left data set in FigureA1.1. A similar reference reaction applies to any of the other diatomic species

, etc and the heat of formation of these elements is zero

at all temperatures.

The most stable form of carbon is solid carbon or graphite and the reference reac-tion is

. (A1.29)

with zero heat of formation at all temperatures. The heat of formation of crystal-line aluminum is zero at temperatures below the melting point and the heat offormation of liquid aluminum is zero at temperatures above the melting point. Thesame applies to Boron, Magnesium, Sulfur, Titanium and other metals.

The enthalpy of a substance is tied to its heat of formation by the equality

. (A1.30)

100K

T

!h° f T( )

H2

H2 H2)

O2, N2, F2, Cl2, Br2, I2

C s( ) C s( ))

h° 298.15( ) !h° f 298.15( )=


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Using (A1.30) in (A1.26), the standard enthalpy of a substance is expressed as

. (A1.31)

The two terms in (A1.31) are the quantities that are tabulated.

A1.5.1 EXAMPLE - HEAT OF FORMATION OF MONATOMIC HYDROGEN AT AND AT

.

The heat of formation of atomic hydrogen at is by definition theenthalpy change of the reference reaction

(A1.32)

Using the data in Figure A1.1, the enthalpy change for (A1.32) is

(A1.33)

The positive heat of formation of (A1.33) indicates that the heat is absorbed in theprocess and the reaction is said to be endothermic. The conceptual physical pro-cess that is invisioned by these calculations is illustrated in Figure A1.2. Theemphasis here is on the word “conceptual”. This is not an experiment that couldbe actually performed since the reactivity of monatomic hydrogen is so strong thatit would be impossible to stabilize the gas depicted in state 2 at room temperature.However that does not stop us from analyzing the energetics of such a reaction.

The appearance of the term on both sides of (A1.33) should not

cause too much concern. It is just a reminder of the fact that for any reference reac-tion the heats of formation of the reactants are always zero by definition.

h° T( ) !h° f 298.15( ) h° T( ) h° 298.15( )–{ }+=

298.15K1000K

298.15K

12---H

2 gas @ 298.15KH gas @298.15K)

!h° f 298.15( )H

!h° f 298.15( )H

12---!h° f 298.15( )

H2 in its reference state– = =

217.999 12--- 0( )– 217.999 kJ/mole of H formed=

!h° f 298.15( )H


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Figure A1.2 Dissociation of diatomic hydrogen to produce monatomic hydrogen by the addition of heat.

Now let’s use the heat capacity data in Figure A1.1 to determine the heat of for-mation of atomic hydrogen at . We need the data from the tables in FigureA1.1 for the enthalpy change from the reference temperature to carry out the cal-culation at the new temperature. The reference reaction is still

(A1.34)

and the enthalpy balance is

. (A1.35)

The somewhat higher value than that calculated in (A1.33) comes from the largervolume change that occurs when the number of moles is doubled at versusdoubling the number of moles at .

0.5 moles of

T 298.15K=

H

State 1 State 2

P 105 N/m2=P 105 N/m2=

H2

1 mole of

T 298.15K=

218 kJadded

1000K

12---H

2 gas @ 1000KH gas @1000K)

!h° f 1000( )H

!h° f 298.15( )H

h° 1000( ) h° 298.15( )–{ } H+% &' ( –=

12--- !h° f 298.15( )

H2

h° 1000( ) h° 298.15( )–{ } H2+% &

' (

217.999 14.589+( )12--- 0 20.680+( )– 222.248 kJ/mole of H formed=

1000K298.15K


bjc A1.11 12/15/10

A1.5.2 EXAMPLE - HEAT OF FORMATION OF GASEOUS AND LIQUID WATER

In this case the reference reaction is

(A1.36)

Using the data for gaseous water in Appendix 2, the enthalpy balance is

(A1.37)

Figure A1.3 below depicts one possible experiment that would be used to measurethis heat of formation.

Figure A1.3 Reference reaction for (gaseous) water.

Half a mole each of diatomic oxygen and hydrogen are placed in an adiabatic pis-ton-cylinder combination. The two will not react spontaneously and so a smallsource of ignition is needed to exceed the threshold energy to start the reaction.Once the reaction proceeds, the chemical energy contained in the chemical bonds

H212---O2 H2O g( ))+

!h° f 298.15( )H2O

!h° f 298.15( )H2O

–=

12---!h° f 298.15( )

O2 in its reference state

12---!h° f 298.15( )

H2 in its reference state–

241.826– 12--- 0( )– 1

2--- 0( )– 241.826 kJ/mole of H2O formed–=

T 298.15K=

H H2,

T 3078K=T 298.15K=1 mole of H2O

removedState 1 State 2 State 3

P 105 N/m2= P 105 N/m2= P 105 N/m2=

1.0 mole of H20.5 mole of O2

O O2 O3,,

OH HO2,

241.826 kJ

H2O2 H, 2O


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of the reactants is released and a mixture of gases at a temperature of results. The piston is withdrawn to keep the pressure constant. The mixture con-tains atoms and molecules that essentially represent all the reasonablecombinations of and that one could conceive although the more complexmolecules such as and would only be present in extremely low con-

centrations. More complex molecules, though possible in principle, are toounlikely to be worth considering.

Sufficient heat is removed and the piston is compressed to bring the mixture backto the original temperature and pressure. In the process the various molecules andatoms in the mixture combine to form gaseous water which is the most stable mol-ecule with the lowest energy. State 3 in Figure A1.3 is assumed to be pure watervapor.

The water vapor at state 3 depicted in Figure A1.3 is not stable but will tend tocondense to form liquid water in equilibrium with its vapor. This is really the low-est energy state of the system.

Figure A1.4 Condensation of water vapor to liquid at constant pressure

The last step in the process is illustrated in Figure A1.4. The water vapor con-denses to liquid water. At a temperature of the vapor pressure of theliquid is much lower than the pressure of the system and since the system is closedwith no possible mixing of outside gas with the vapor all the water vapor mustcondense to form liquid.

The reference reaction to form liquid water at the same temperature and pressureis

3078K

H OH2O2 O3

T 298.15K=1 mole of H2O g( )

State 3

P 105 N/m2=

T 298.15K=1 mole of H2O l( )

P 105 N/m2=

State 4 41.85 kJremoved

298.15K


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(A1.38)

Subtraction of (A1.36) and (A1.38) gives the heat released when steam condensesto form liquid water at the reference temperature.

(A1.39)

In practice the experimental measurement of the heat of formation would proba-bly not be carried out in such a complex apparatus as a piston and cylinder butwould be carried out at constant volume in a device called a bomb calorimeter.The amount of heat removed in the experiment would then be used to work outthe heat that would have been released if the process were at constant pressure.

A1.5.3 EXAMPLE - COMBUSTION OF HYDROGEN AND OXYGEN DILUTED BY NITROGEN

Now let’s look at an example where liquid water is produced in equilibrium withits vapor. We start with a mixture in state 1 of diatomic hydrogen, oxygen andnitrogen. A source of ignition is used to initiate the reaction and the mixture comesto state 2 at equilibrium at . A fair amount of the nitrogen reacts to form

and . As the mixture temperature is lowered these gases react leaving

only trace amounts in the final mixture which is composed of approximately onemole of and one mole of water. The partial pressure of water vapor in the gas

mixture is determined by the vapor pressure equation (A1.25) repeated here forconvenience.

(A1.40)

H212---O2 H2O l( ))+ !h° f 298.15( ) 283.666 kJ/mole–=

H212---O2 H2O l( ))+ !h° f 298.15( ) 283.666 kJ/mole–=

minus

H212---O2 H2O g( ))+ !h° f 298.15( ) 241.826 kJ/mole–=

equalsH2O g( ) H2O l( )) !h° f 298.15( ) 41.84 kJ/mole–=

2709KNO NO2

N2

PPref----------- e

!h°vap T ref( )

Ru---------------------------------- 1

T--- 1

T ref----------–% &

' (–

=


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Figure A1.5 Hydrogen-Oxygen combustion in the presence of Nitrogen diluent.

The heat of vaporization at the pressure and temperature of interest is, from(A1.39), . The reference temperature is the

boiling temperature at the reference pressure of one atmo-

sphere, . The simplified solution of the Clausius-

Clapeyron, Equation (A1.40) gives the vapor pressure as

(A1.41)

corresponding to which is reasonably close to the more

precise value shown in Figure A1.5 derived from the tables.

T 298.15K=

H H2,

T 2709K= T 298.15K=


P 105 N/m2= P 105 N/m2= P 105 N/m2=

1.0 mole H20.5 mole O2

O O2 O3,,

OH HO2, H2O2 H, 2O1.0 mole N2

N N2,NO NO2,

1.0 mole N2

traces of NO, NO2

0.968 mole H2O l( )

0.032 mole H2O g( )

284.2 kJ

!h°vap T ref( )H2O 41.84 kJ/mole=

T ref 373.15K=

Pref 1.0132 105 Pa×=

P

1.0132 105×

-------------------------------- e

41.84 103×

8.314510----------------------------- 1

298.15---------------- 1

373.15----------------–% &

' (–0.03363= =

nH2O g( ) 0.035 moles=


bjc A1.15 12/15/10

A1.5.4 EXAMPLE - COMBUSTION OF METHANE

The heats of formation of various species can be used to evaluate the enthalpyreleased or absorbed during a reaction involving those species. From the data inAppendix 2, the reference reactions to produce carbon dioxide and methane are

(A1.42)

and

. (A1.43)

This data together with (A1.36) and (A1.38) can be used to evaluate the heatreleased by the combustion of methane with oxygen to form carbon dioxide andliquid water at . The reaction is

. (A1.44)

To produce (A1.44) the algebra for the various reactions can be viewed as follows.

(A1.45)

and the algebra for the heats of formation is

. (A1.46)

C s( ) O2 CO2)+ !h° f 298.15( ) 393.510 kJ/mole–=

C s( ) 2H2 CH4)+ !h° f 298.15( ) 74.600 kJ/mole–=

298.15K

CH4 2O2 CO2 2H2O l( )+)+

C s( ) O2 CO2)+

plus

2 H212---O2 H2O l( ))+% &

' (×

minusC s( ) 2H2 CH4)+

minus2 O2 O2)( )×

equalsCH4 2O2 CO2 2H2O l( )+)+

!h° 298.15( ) reaction !h° f 298.15( )CO2

2 !h° f 298.15( )H2O l( )

–×+=

!h° f 298.15( )CH4

2 !h° f 298.15( )O2

×– =

393.510– 2 283.666–( ) 74.600–( )– 2 0( )–+ =886.242 kJ per mole of CH4 burned–


12/15/10 A1.16 bjc

In general the reactants can be at different temperatures. Note that the reaction(A1.45) is somewhat artificial in that, at equilibrium, there would actually be asmall amount of water vapor mixed with the carbon dioxide and in equilibriumwith the liquid water.

The general enthalpy balance for a reaction is

(A1.47)

where it is recognized that the products are mixed at the final temperature of thereaction. Equation (A1.47) fully written out is

(A1.48)

For example, suppose in the methane-oxygen combustion problem above themethane is initially at and the oxygen is at while the products areassumed to be at . The heat of reaction is determined from

(A1.49)

The data is

(A1.50)

!h° T final( ) nih°i T i( ) n jh° j T final( )

j reactant*–

i product*=

!h° T final( ) ni !h° f 298.15( ) h° T final( ) h° 298.15( )–( )+{ }i product

–i*=

n j !h° f 298.15( ) h° T j( ) h° 298.15( )–( )+{ }j reactantj

*

600K 800K1500K

!h° 1500( ) =!h° f 298.15( ) h° 1500( ) h° 298.15( )–( )+{ }CO2

+

2 !h° f 298.15( ) h° 1500( ) h° 298.15( )–( )+{ }H2O g( ) –

!h° f 298.15( ) h° 600( ) h° 298.15( )–( )+{ }CH4

–

2 0 h° 800( ) h° 298.15( )–( )+{ } O2

!h° 1500( ) =393.510– 61.705+{ }CO2

2 241.826– 48.151+{ }H2O g( ) –+

74.600– 13.130+{ } CH4 2 0 15.835+{ } O2

– =

689.355 kJ per mole of CH4 burned–


bjc A1.17 12/15/10

Somewhat less enthalpy is evolved compared to the enthalpy change at the refer-ence temperature.

A1.5.5 EXAMPLE - THE HEATING VALUE OF JP-4In Chapter 2 a typical value of the fuel enthalpy for JP-4 jet fuel was given as

(A1.51)

although we never stated explicitly that this refers to the fuel and its products ofcombustion at a temperature of . Where does this number come fromand how does it relate to the heat of formation of JP-4? According to CEA themolecular formula and heat of formation of JP-4 are

. (A1.52)

The reaction of this fuel with oxygen (the oxidizer contained in air) is

(A1.53)

and the enthalpy released by the reaction is

(A1.54)

On a per mass basis the heating value is

(A1.55)

which agrees closely with the accepted value (A1.51). Notice the relatively smallcontribution of the actual enthalpy of formation of JP-4 to the calculation of theheat of reaction (A1.54). Note also that the water formed in the reaction is takento be in the gas phase and the values given in (A1.51) and (A1.55) are what wouldbe called the lower heating value of JP-4. If the water is assumed to condense toits liquid phase in equilibrium with its vapor at (see Figure A1.5) , whichof course it does, the enthalpy change is

h f JP 4 @ 298.15K–4.28 107J kg×=

298.15K

CH1.94 !h° f 298.15( ) 22.723 kJ/mole–=

CH1.94 1.485O2 CO2 0.97H2O+)+

!h° 298.15( ) !h° f 298.15( )CO2

0.97 !h° f 298.15( )H2O g( )

–×+=

!h° f 298.15( )CH1.94

1.485 !h° f 298.15( )O2

×– =

393.510– 0.97 241.826–( ) 22.723–( )– 1.485 0( )–+ =605.358 kJ per mole of CH1.94 burned–

h f JP 4 @ 298.15K–

605.358 kJ per mole–

13.9664 10 3– kg/mole×---------------------------------------------------------- 4.33– 107J kg×= =

298.15K

Heat capacity

12/15/10 A1.18 bjc

(A1.56)

which would be called the higher heating value.

A1.6 HEAT CAPACITYThe key piece of data needed for any species is its heat capacity. One of the mostcomplete collections of such data can be found in the well known reference.

NASA/TP - 2002-211556 - NASA Glenn Coefficients for Calculating Thermody-namic Properties of Individual Species by Bonnie J. McBride, Michael J. Zehe,and Sanford Gordon, Glenn Research Center, Cleveland, Ohio.

These authors provide accurate curve fits for the standard heat capacities, enthal-pies and entropies of a vast range of species over the temperature range to

.

The enthalpy change terms in (A1.35) are

(A1.57)

The molar heat gas capacities of both monatomic hydrogen and gaseous diatomichydrogen are consistent with the classic formula from kinetic theory

(A1.58)

where is the number of degrees of freedom of the gas molecular model.

Atomic hydrogen is a gas at all temperatures with three translational degrees offreedom, and one can expect that

(A1.59)

h f JP 4 @ 298.15K–

605.358– 0.97 0.968 41.85××–

13.9664 10 3–×

---------------------------------------------------------------------------------- 644.653–

13.9664 10 3– kg/mole×----------------------------------------------------------= =

644.653 kJ/mole–

13.9664 10 3– kg/mole×---------------------------------------------------------- 4.62– 107J kg×=

200K6000K

h° 1000( ) h° 298.15( )–{ } H C°p T( ) H Td298.15

1000

$=

h° 1000( ) h° 298.15( )–{ } H2C°p T( ) H2

Td298.15

1000

$=

C p+ 2+

2------------% &' ( Ru=

+

+ 3=

C°p H

52---Ru 20.786( ) Joules/mole-K= =

Heat capacity

bjc A1.19 12/15/10

over a wide temperature range. That this is the case is clear from the right-handtable in Figure A1.1. In fact this value of the heat capacity of monatomic hydrogenis valid up to which is the temperature at which the tables stop. Above thistemperature the heat capacity increases as electronic degrees of freedom begin tobe excited. The enthalpy change for monatomic hydrogen is

(A1.60)

which reproduces the tabulated data very well.

The number of degrees of freedom for diatomic hydrogen at room temperature isnominally with three translational degrees of freedom and two rotationaldegrees. Kinetic theory (A1.58) predicts

. (A1.61)

Actually the tabulated heat capacity is slightly less than this since the rotationaldegrees are not fully excited at room temperature.

As the gas is heated, vibrational degrees of freedom begin to come into play andthe number increases to at high combustion temperatures. Quantum sta-tistical mechanics can be used to develop a theory for the onset of vibrationalexcitation. According to this theory, the specific heat of a diatomic gas from roomtemperature up to high combustion temperatures is accurately predicted by

(A1.62)

where is the vibrational transition temperature for a given gas. Values of the

vibrational transition temperature for several common diatomic species are pre-sented in Table A1.1.

6000K

h° T( ) h° 298.15( )–{ } H52---Ru T 298.15–( )=

+ 5=

C°p H2

29.101 J/mole-K=

+ 7=

C°pR

---------- 72---

,v 2T( )

Sinh ,v 2T( )----------------------------------- ./ 01 2

2+=

,v

Heat capacity

12/15/10 A1.20 bjc

The vibrational transition temperatures for common diatomic molecules are all athigh combustion temperatures. Using the results of this theory, the standardenthalpy change of a diatomic gas heated to temperatures above room temperaturecan be expressed as

(A1.63)

which integrates to

. (A1.64)

plotted in Figure A1.6 for hydrogen. This equation produces good approximateresults for the enthalpy changes in hydrogen and other diatomic molecules. Forhydrogen (A1.64) gives which compareswell with the value given in the left table in Figure A1.1.

Figure A1.6 Enthalpy change for diatomic hydrogen at elevated temperature.

H2 6297

N2 3354

O2 2238

CO 3087

Table A1.1 Vibrational transition temperature for several gases.

,v K

h° T( ) h° 298.15( )– C°p Td298.15

T$ Ru

72---

,v 2T( )

Sinh ,v 2T( )----------------------------------

- ./ 01 2

2+ Td

298.15

T$= =

h° T( ) h° 298.15( )–Ru T 298.15–( )

------------------------------------------------- 72---

,v2 T 298.15–( )----------------------------------- Coth

,v2T-------% &' ( Coth

,v2 298.15( )------------------------% &' (–+=

h° 2000( ) h° 298.15( )– 51872.9 J/mole=52951 J/mole

0.5 1 1.5 2

3.6

3.8

4.2

h° T( ) h° 298.15( )–Ru T 298.15–( )

------------------------------------------------- T ,v

,v 6297K=

Chemical Bonds and the Heat of Formation

bjc A1.21 12/15/10

A1.7 CHEMICAL BONDS AND THE HEAT OF FORMATION

A1.7.1 POTENTIAL ENERGY OF TWO HYDROGEN ATOMS

The figure below depicts the variation in potential energy as two hydrogen atomsat zero energy are brought together from infinity.

Figure A1.7 Potential energy of two hydrogen atoms as a function of bond distance.

The molecule is stable at a bond length of 74 picometers. Noting the existence ofminimum energy quantum vibrations, this distance can be viewed as an approxi-mate measure of the bond length between the two nuclei.

The potential energy of the system at this spacing is where

. This is the bond energy that isradiated away when the two atoms are allowed to approach each other throughmutual attraction and react to form the chemical bond that holds them together ina stable molecule. It is also the energy that would have to be added to the moleculein order to dissociate the two atoms and return them to their zero energy state atinfinity. The hydrogen molecule is a classic example of a covalent bond where the

50 100 150 200 250 300

-5

-4

-3

-2

-1

1

electronvolts

meters 1012×

- 4.476 ev

74 pm

h3 = 4.476 ev

4.476 electron volts–

1 electron volt = 1.60217646 10 19– Joules×


12/15/10 A1.22 bjc

mutual attraction of the two nuclei arises from the attraction of each nucleus tothe electron pair that is concentrated in the space between them. Some other bondlengths and zero Kelvin bond energies are given in Table A1.2 below.

Covalent bonds are the strongest chemical bonds found in nature and are alwaysformed from the sharing of one or more pairs of electrons. Typical covalent bondenergies range from for the single bond in diatomic iodine to

for the double bond between two carbon atoms to for thetriple bond that can occur between two atoms of nitrogen. For comparison the

Molecule Bond lengthBond

energy at 0K

electron volts

0.7415 4.476

1.20739 5.080

1.0976 9.756

1.442 1.59

1.988 2.475

2.283 1.971

2.666 1.5417

0.9706 4.35

1.1198 3.47

1.1281 11.11

Table A1.2 Bond lengths and bond energies for several diatomic molecules at . Reference: Moelwyn-Hughes, Physical Chemistry page 427

108cm×

H2

O2

N2

F2

Cl2

Br2

I2

OH

CH

CO

0K

1.5417 ev–6.364 ev– 9.756 ev–


bjc A1.23 12/15/10

thermal energy per molecule of air at is only about . This isone of the reasons why covalently bonded molecules tend be so difficult to breakby heating.

Although, in principle, it is possible to analyze the energetics of all chemical reac-tions as a process of breaking down reactants and assembling products to and fromtheir constituent atoms at zero Kelvin it is not particularly convenient to do so.The reason is that virtually all chemical reactions of interest take place in a sur-rounding gas at some pressure above vacuum and in accounting for all the energychanges it is necessary to include the work associated with the changes in volumethat occur during the reaction. Consider the reaction depicted in Figure A1.8which is the opposite of the reaction shown in Figure A1.2.

.

Figure A1.8 Reaction of atomic hydrogen to form diatomic hydrogen at standard pressure and temperature.

Two moles of hydrogen atoms are placed in an adiabatic piston-cylinder combi-

nation at a temperature and pressure . The atomsin state 1 are allowed to react and the piston is withdrawn to produce a mixture of

and in state 2 at very high temperature and at the original pressure. Heat

is removed and the piston is compressed to bring the gas back to the original tem-

298.15K 0.064 ev

2 moles of H

T 298.15K=

H H2,

T 4062.79K=

T 298.15K=1 mole of H2


P 105 N/m2=P 105 N/m2=

P 105 N/m2=

436 kJ

T 298.15K= P 105 N/m2=

H H2


12/15/10 A1.24 bjc

perature and pressure at state 3. At the relatively low final temperature of virtually all of the atomic hydrogen has reacted to produce one mole of

diatomic hydrogen.

A1.7.2 ATOMIC HYDROGEN

Taking the reference energy of the hydrogen atoms to be zero at absolute zero, theatomic hydrogen is brought to state 1 through the addition of heat and the produc-tion of work as the collection of atoms expands from essentially zero volumeagainst the surrounding constant pressure.

Any phase changes that may occur in atomic hydrogen are ignored and it seemsthat little is known about the nature or even the existence of a stable solid or liquidstate of atomic hydrogen in the neighborhood of absolute zero. There are paperson the subject that describe theoretical models of atomic hydrogen in a face-cen-tered cubic crystal but there is no definitive experimental evidence of an ignitionthreshold near absolute zero. If such a condensed state does exist the temperatureof any phase change would be extremely low and would make only a very smallcontribution to the enthalpy.

The total enthalpy change needed to bring the two moles of atomic hydrogen fromzero energy at to state 1 at is therefore determined from the gasphase heat capacity.

(A1.65)

The enthalpy required per mole of is

. (A1.66)

This is the enthalpy change given in the tables in Figure A1.1 but with the oppositesign.

A1.7.3 DIATOMIC HYDROGEN

An accurate determination of the enthalpy of state 3 requires a knowledge of theheat capacity of diatomic hydrogen at very low temperatures as well as data forthe heat of fusion and vaporization.

298.15K

0K 298.15K

h° 298.15( ) h° 0( )– n C pdT 2( )20.785 298.15( )40

298.15$ 12394 J= =

H

h° 298.15( ) h° 0( )– 123942

--------------- 6197 J/mole of H= =


bjc A1.25 12/15/10

The number of degrees of freedom for diatomic hydrogen is at room tem-perature and drops to near the vaporization point as rotational degrees offreedom “freeze out”. Hydrogen is somewhat unusual in this respect. Because ofthe strong bond reflected in the short bond length between the two hydrogen atomsand the correspondingly low vaporization point, the freezing out of the rotationaldegrees of freedom is evident in the gas heat capacity with a transition tempera-ture at about . For other diatomic gases the theoretical transition temperaturetends to be far below the vaporization point and is therefore less evident in theheat capacity function.

The heats of fusion and vaporization of diatomic hydrogen are well known. Here

is some data at .

(A1.67)

The heat capacity of all materials is zero at absolute zero. An approximation tothe heat capacity of gaseous that works reasonably well between the temper-

atures of and is

(A1.68)

This model for the heat capacity is plotted below. For simplicity we assume theheat capacities of the solid and liquid states vary linearly between the values givenin (A1.67).

+ 5=+ 3=

87K

P 102kPa=

Heat capacity, Cp @ 0.0K 0.0 J/mole-K=Heat capacity, Cp @ 13.15K 5.729 J/mole-K=Enthalpy of fusion = 117.25 J/mole @ 13.95K

Heat capacity, Cp @ 20K 19.4895 J/mole-K=Enthalpy of vaporization = 912 J/mole @ 20.28KHeat capacity, Cp @ 175K 26.4499 J/mole-K=Heat capacity, Cp @ 250K 28.3248 J/mole-K=

Heat capacity, Cp @ 298.15K 28.836 J/mole-K=

H2

20.28K 298.15K

C pH2T[ ] 18.834 1.30487 103× T 2 1.14829 105× T 3– + +=

3.98205 10 8– T 4 4.9369 10 11– T 5×–×


12/15/10 A1.26 bjc

Figure A1.9 Approximation to the heat capacity of gaseous at low temperatures

and .

The enthalpy of a species including the absolute zero energy of the chemical bondis

(A1.69)

The pressure dependence comes primarily from the effect of pressure on the heatof vaporization and to a lesser extent on the heat of fusion as well as the possiblepressure dependence of the heat capacities of the condensed phases particularlynear phase transition points. The gas phase heat capacity becomes independent ofpressure as the gas approaches ideal gas behavior at increasing temperaturesabove the vaporization temperature.

On a per unit mole basis the chemical bond energy of at zero Kelvin is

(A1.70)

C pJ/mole-K

liquid

solid

gas

T K50 100 150 200 250 300

5

10

15

20

25

30

H2

P 100kPa=

h T P,( ) Echemical bond energy/mole C p T P,( ) Td !h fusion T P,( ) + +0

melting point$+=

C p T P,( ) Td !hvaporiz T P,( ) C p T P,( ) Tdvaporization point

T$+ +

melting point

vaporization point$

H2

Echemical bond energy/mole =

4.476 evmolecule--------------------- 6.0221415 1023molecules

mole------------------------ 1.60217646 10 19– J

ev-----××××– =

431.87 103× J/mole–

Heats of Formation computed from bond Energies

bjc A1.27 12/15/10

Using the data in (A1.67), the enthalpy at and including the chemical bond energy is

(A1.71)

Use (A1.68) to carry out the last integral in (A1.71). The result is

(A1.72)

The heat released by the reaction to form diatomic hydrogen from its atoms at and 1 atmosphere is

(A1.73)

The result (A1.73) is the enthalpy released during the reaction of atomic hydrogen

to form diatomic hydrogen at a temperature of and pressure .

This result can be found in the right table in Figure A1.1 as twice the heat of for-mation of one mole of monatomic hydrogen. We worked this out in equation(A1.33).

Note that, because the heat input and work needed to bring two moles of atomichydrogen to the initial temperature is greater that that needed to bring one moleof diatomic hydrogen to the same temperature, the reaction releases four morekilo-Joules of enthalpy than it would if the reaction was carried out at zero Kelvin.

A1.8 HEATS OF FORMATION COMPUTED FROM BOND ENERGIESWe began this appendix by discussing tabulations of the heats of formation of var-ious species from their elements in their reference state. In this last section we willdescribe how heats of formation can be determined from bond energies. The

T 298.15K= P 100kPa=

h°H2298.15 100,( ) 431.87 103

×– 5.72913.95-------------% &' ( T Td 117.25 + +

0

13.95$+=

5.729 19.49 5.729–20.28 13.95–---------------------------------% &' ( T 13.95–( )+% &

' ( Td 912 C p Td20.28

298.15$+ +

13.95

20.28$

h°H2298.15 100,( ) 431.87 103

×– 5.7292

------------- 13.95( ) 117.25 + + +=

19.49 5.729–2

---------------------------------% &' ( 20.28 13.95–( ) 912 7057+ + =

431.87 103×– 8170 J/mole+

298.15K

!h° h° 298.15 102,( ) H2

2h° 298.15 102,( ) H–=

8170 431.87 103×–( ) 2 6197.35×– 436 103 J/mole of H2×–=

298.15K 105 N/m2

Heats of Formation computed from bond Energies

12/15/10 A1.28 bjc

enthalpy changes involved in chemical reactions come from the breaking andmaking of chemical bonds to decompose the reactants and then compose theproducts.

Table A1.3 gives average bond energies for a variety of diatomic molecules.

Consider the reference reaction for water

(A1.74)

MoleculeBond

energy at 298.15K

MoleculeBond

energy at 298.15K

kJoules per mole

kJoules per mole

436 615

145 418

170 498

158 615

243 725

193 477

151 812

463 946

416 890

350 308

Table A1.3 Average single, and multiple bond energies for several diatomic molecules at .

H H– C C=

O O– N N=

N N– O O=

F F– C N=

Cl Cl– C O=

Br Br– C S=

I I– C C5

O H– N N5

C H– C N5

C O– C N–

298.15K

H212---O2 H2O g( ))+ !h° f 298.15( ) 241.826 kJ/mole–=

References

bjc A1.29 12/15/10

To break down the reactants we have to break the 0.5 mole of double bonds

as well as one mole of single bonds. When we compose the products we will

make two single bonds. All the bond changes are evaluated at . Theheat of the reaction is

(A1.75)

Using the data in Table A1.3 we get

(A1.76)

which is very close to the tabulated heat of formation.

A certain amount of caution is always required when computing heats of forma-tion from bond energies. The energy of a chemical bond always depends on theparticular atomic configuration of the molecule where the bond appears. Forexample to remove one of the hydrogens from a molecule of water requires

whereas the energy required to break the remaining bond isonly . Therefore the average bond energy is .

For methane is required to break a single whereas it takes to break all four bonds therefore the average bond energy is

not .

A1.9 REFERENCES1) Linus Pauling, General Chemistry, Dover 1970.

2) Irvin Glassman, Combustion, Third Edition, Academic Press 1996.

3) NASA/TP - 2002-211556 - NASA Glenn Coefficients for Calculating Thermo-dynamic Properties of Individual Species by Bonnie J. McBride, Michael J. Zehe,and Sanford Gordon, Glenn Research Center, Cleveland, Ohio.

In addition to providing curve fits for the heat capacity, enthalpy and entropy fora wide range of species, a number of variables are also tabulated in this referenceincluding , and .

O2

H2

OH 298.15K

!h° !h° bonds broken( )

* !h° bonds made( )

*–4

!h° 436 12--- 498( ) 2 463( )–+ 241 kJ/mole of H2O–= =

502 kJ/mole O H–424 kJ/mole 463 kJ/mole

435 kJ/mole C H–1662 kJ/mole416 kJ/mole 435 kJ/mole

!h° f 298.15( ) h° 0( ) !h° f 0( )

References

12/15/10 A1.30 bjc

Documents

App 1 Thermochemistry