CHAPTER 5: REACTION THERMODYNAMICS Chemical reactions result in a rearrangement of atoms among molecules. Some of the examples of chemical reaction are (1) Photosynthesis, (2) Metabolism, (3) Combustion, (4) Petroleum refining, and (5) Plastics manufacturing. Energy cannot be extracted from a system that is constrained unless the thermodynamic constraint is removed. Fossil fuels such as coal, oil and natural gas are stored below the earth’s surface and are in thermal, mechanical and chemical equilibrium. If the chemical constraint is removed (allow the fuel to come in physical contact with air), it is possible to release the chemical energy. 5.1 CHEMICAL REACTIONS AND COMBUSTION 5.1.1 STOICHIOMETRIC OR THEORETICAL REACTION A stoichiometric or theoretical reaction results in the complete combustion of fuel. For example, consider the reaction

4 2 2 2CH aO bCO cH O+ → + (5.1)

involving one kmole of fuel and “a” kmoles of molecular oxygen. Species on the left hand side of the reaction equation (Eq. (5.1)) are usually called the reactants as they react and are consumed during the overall chemical reaction. Species on the right hand side of Eq. (5.1), produced as a result of chemical reaction are called products. In case of a steady flow reactor, the input and output streams contain the following quantity in kmole of the elements.

Input Output C 1 = b H 4 = 2c O 2a = 2b+c

We deduce from the carbon atom balance that b=1, from the H atom balance that c=2, and from O balance that a=(2b+c)/2. Consequently, Eq. (5.1) can be written in the form of the stoichiometric reaction

4 2 2 22 2CH O CO H O+ → + (5.2)

The coefficients (1, 2, 1 and 2) in Eq. (5.2) are called the stoichiometric coefficients. The minimum amount of air which provides sufficient oxygen for the complete combustion of all the element like carbon, hydrogen, etc., which may oxidize is called the theoretical or stoichiometric air. There is no oxygen in the products when complete combustion (oxidation) is achieved with this theoretical air.

86

Let us consider again the combustion of 10 kmole of CH4, 25 kmoles of O2, 5 kmoles of CO2 and 3 kmole of H2O during which the fuel completely burns. Thus only 10 kmole of CH4 are consumed in the reactor. Applying Eq. (5.2) we readily determine that 20 kmole of O2 are utilized during the process, and hence 5 kmole of O2, 15 (=10+5) kmole of CO2 and 23 (=20+3) kmole of H2O leave the combustor. The stoichiometric coefficients do not necessarily represent the amounts of species entering and leaving a reactor. From Eq. (5.2), the stoichiometric air fuel ratio is

( ): 2 2 3.76 1 9.52mole

A F = + × ÷ = kmole of air to 1 kmole of fuel.

On mass basis, the A:F ratio is ( ) ( ): 2 32 2 3.76 28 1 16 17.16

moleA F = × + × × ÷ × = kg of air/kg of fuel.

Atmospheric air contains 21% oxygen, 78% nitrogen, and 1% argon by volume. In combustion calculations, however, the argon is usually neglected, and air is assumed to consist of 21% oxygen and 79% nitrogen by volume (or molar basis). On a mass basis, air contains 23% oxygen and 77% nitrogen. If we desire more accuracy, air is a mixture of 78% , 1% and 21% . In this case, 2N Ar 2O

( ): 2 2 3.76 1 9.52mole

A F = + × ÷ = kmole of air to 1 kmole of fuel. ( ) ( ): 2 2 3.76 0.1 40 1 16 17.24

massA F = + × + × ÷ × = kg of air/kg of fuel.

5.1.2 REACTION WITH EXCESS AIR (LEAN COMBUSTION) Fuel and air are often introduced separately (without premixing) into a combustor, e.g., boiler, CI engine etc. Due to the large flow rates and short residence time there is no assurance that each molecule of fuel is surrounded by the appropriate number of oxygen molecules required for the stoichiometric combustion. Therefore, it is customary to supply excess air in order to facilitate better mixing and thereby ensure complete combustion. The excess oxygen remains unburned and appears in the product. 5.1.3 REACTION WITH EXCESS FUEL (RICH MIXTURE) Incomplete combustion occurs when the air supplied is less than the stoichiometric amount required. For this condition, the products of incomplete oxidation may contain a mixture of and 2, ,CO CO H 2 2H O . 5.1.4 EQUIVALENT RATIO AND STOICHIOMETRIC RATIO The equivalent ratio is defined as the actual fuel-air ratio to stoichimetric fuel-air ratio.

87

( )( )

2

2

:: :: : :

actual stoichiometric stoichiometric

stoichiometric actual actual

O FF A A FF A A F O F

φ = = = (5.3)

The stoichiometric ratio is given by

1SRφ

= (5.4)

For example, if 0.5φ = , for methane-air combustion, this implies that the excess air is supplied for every kmole of fuel that is burned. In such case SR=2. In otherwords, air supplied is two times as large as the stoichiometric or theoretical air demand of the fuel. Eq. (5.2) can be written under such condition as

( )4 2 2 2 2 2

2 13.76 2 2 1 3.76CH O N CO H O O Nφ φ

⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + → + + − + ×⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠2

2φ

(5.5)

5.2 DEGREE OF REACTION Let us suppose that we have a mixture of four substances, 1, 2, 3A A A and

4A capable of undergoing a reaction of the type

1 1 2 2 3 3 4 4v A v A v A v A+ + (5.6)

where the are the stoichimetric coefficients. 'v Starting with arbitrary amounts of both initial and final constituents, let us imagine that the reaction proceeds completely to the right with the disappearance of at least one of the initial constituents, say, 1A . Then the original number of moles of the initial constituents is given in the form

1n (original) 0 1n v= (5.7)

2n (original) 0 2 2n v N= + (5.8)

where is an arbitrary positive number, and is the residue(or excess) of 0n 2N2A , i.e., the number of moles of 2A which cannot combine. If the reaction is

assumed to proceed complete it to the left with the disappearance of the final constituent 3A then

3n (original) 0 3n v′= (5.9)

4n (original) 0 4 4n v N′= + (5.10)

88

where is an arbitrary positive number and is the excess number of moles of

0n′ 4N4A left after the reaction is complete from right to left. For a reaction that

occurred completely to the left, there is a maximum possible amount of each initial constituent, and a minimum possible amount of each final constituent, so that

1n (max) 0 1n v= + 0 1n v′ 0 0( )n n v1′= + (5.11)

where Original number of moles of 0 1n v = 1A Number of moles of 0 1n v′ = 1A formed by chemical reaction 0 3 3 0 4 4 0 1 1 0 2 2( )n v A n v A n v A n v A′ ′ ′ ′+ → +

2n (max) 0 2 2( )n v N= + + 0 2n v′ 0 0 2( )n n v N2′= + + (5.12)

where Original number of moles of 0 2 2(n v N+ ) = 2A Number of moles of 0 2n v′ = 2A formed by chemical reaction The constituent 3A completely disappears by reaction, hence

3n (min)= 0 (5.13)

The excess number of moles of 4A that are left after the reaction is complete to left

4n (min)= 0 (5.14)Similarly, if the reaction is imagined to proceed completely to the right, there is a minimum amount of each initial constituent, and a maximum amount of each final constituent, so that

1n (min) = 0 (5.15)

2n (min) = 2N (5.16)

3n (max) = 0 3n v′ + 0 3n v 0 0( )n n v3′= + (5.17) where = original amount 0 3n v′ = amount formed by chemical reaction 0 3n v 0 1 1 0 2 2 0 3 3 0 4 4( )n v A n v A n v A n v A+ → +

4n (max) = 0 0 4( )n n v N4′+ + (5.18)

Let us suppose that the reaction proceeds partially either to the right or to the left to the extent that there are moles of 1n 1, 2A n moles of 2, 3A n moles of 3A and

moles of 4n 4A . The degree (or advancement) of reaction ε is defined in terms of any one of the initial constituents, say, 1A as the fraction

89

1 1

1 1

(max)(max) (min)n n

n nε −

=−

(5.19)

It is seen that when 1 1(max),n n= 0ε = , the reaction will start from left to right. When 1 1(min),n n= 1,ε = reaction is complete from left to right. The degree of reaction can thus be written in the form

0 0 1

0 0 1

( )( )

n n v nn n v

ε 1′+ −=

′+

(5.20)

Therefore 1 0 0 1 0 0 0( ) ( )n n n v n n v ε′ ′ n= + − + = (at start) n− (consumed) (5.21)

or,

1n = Number of moles of 1A at start − number of moles of 1A

consumed in the reaction = 0 0 1( ) (1n n v )ε′+ −

(5.22)

2n = n (at start) n− (consumed)

0 0 2 2 0 0 2( ) ( )n n v N n n v ε′ ′= + + − + 0 0 2( ) (1 )n n v N2ε′= + − +

(5.23)

3n n= (at start) n+ (formed)

0 0 30 ( )n n v ε′= + + 0 0 3( )n n v ε′= +

(5.24)

4n = n (at start) n+ (formed)

4 0 0 4( )N n n v ε′= + +

0 0 4( )n n v N4ε′= + +

(5.25)

The number of moles of the constituents change during a chemical reaction, not independently but restricted by the above relations (Eqs. (5.22-5.25)). These equations are the equations of constraint. The are functions of 'n s ε only. In a homogeneous system, in a given reaction, the mole fraction 'x s is also function of ε only, as illustrated below. Let us take the reaction

2 2

1 12 2

NO N O+ (5.26)

90

in which moles of dissociates to produce 0n NO 0

2n

moles of and 2N 0

2n

moles of . The s and 2O 'n 'x s as function of ε are shown in the Table-5.1 given below Table 5.1 Values of , nν and x (Eq. (5.26)) A v n x

1A NO= 1 1v = 1 0 (1 )n n ε= − 11

11

nxn

εε

−= =

+∑

3 2A N= 3

12

v = 03 2

nn ε= 3 1x ε

ε=

+

4 2A O= 4

12

v = 04 2

nn ε= 4 1x ε

ε=

+

( )0 1n n ε= +∑ If the reaction is imagined to advanced to an infinitesimal extent, the degree of reaction changes from ε to ,dε ε+ and the various s will change by the amounts (Upon differentiation of Eqs. (5.22-5.24) w.r.t.

'nε )

1 0 0 1( )dn n n v dε′= − +

2 0 0 2( )dn n n v dε′= − +

3 0 0 3( )dn n n v dε′= +

4 0 0 4( )dn n n v dε′= + or,

1 2 3 40 0

1 2 3 4

( )dn dn dn dn n n dv v v v

ε′= = = = +− −

(5.27)

which shows that the s are proportional to the 's. 'dn v

5.3 REACTION EQUILIBRIUM

Let us consider a homogeneous phase having arbitrary amounts of the constituents, 1, 2, 3A A A and 4A , capable of undergoing the reaction

1 1 2 2 3 3 4 4v A v A v A v A+ + (5.28)

91

The phase is at uniform temperature T and pressure p . The Gibbs function of the mixture is

1 1 2 2 3 3 4 4G n n n nµ µ µ µ= + + + (5.29)

where the s are the number of moles of the constituents at ant moment, and the

'n'µ s are the chemical potentials. Let us imagine that the reaction is allowed to

take place at constant T and p . The degree of reaction changes by an infinitesimal amount from ε to dε ε+ . The change in the Gibbs function is

,T p k kdG dnµ= ∑ 1 1 2 2 3 3 4 4dn dn dn dnµ µ µ µ= + + + (5.30)

The equations of constraint in differential form are

1 0 0 1( )dn n n v dε′= − + (5.31)

2 0 0 2( )dn n n v dε′= − + (5.32)

3 0 0 3( )dn n n v dε′= + (5.33)

4 0 0 4( )dn n n v dε′= + (5.34)

On substitution of Eqs. (5.31-5.34) in Eq. (5.30),

, 0 0 1 1 2 2 3 3 4 4( )( )T pdG n n v v v v dµ µ µ µ ε′= + − − + + (5.35)

From Eq. (5.35), following interpretations can be made, (1) When the reaction proceeds spontaneously to the right,dε is positive, and since , 0T pdG <

1 1 2 2 3 3 4 4( ) (v v v v )µ µ µ µ+ > + (5.36)

(2) When the reaction proceeds spontaneously to the left,dε is negative

1 1 2 2 3 3 4 4( ) (v v v v )µ µ µ µ+ < + (5.37)

i.e., k kv µ∑ is positive. (3) At equilibrium, the Gibbs function will be minimum, hence

1 1 2 2 3 3 4 4v v v vµ µ µ µ+ = + (5.38)

which is called the equation of reaction equilibrium. Therefore, it is the value of k kv µ∑ which causes or forces the spontaneous reaction and is called the

“chemical affinity”.

92

5.4 LAW OF MASS ACTION For a homogeneous phase chemical reaction at constant temperature and pressure, when the constituents are ideal gases, the chemical potential are given by the expressions of the type

( ln lnk k )kRT pµ φ= + + x (5.39)

where the 'φ s are functions of temperature only. Substituting in the equation of reaction equilibrium (Eq. (5.38))

1 1 1 2 2 2( ln ln ) ( ln lnv p x v p )xφ φ+ + + + +

3 3 3 4 4 4( ln ln ) ( ln lnv p x v p )xφ φ= + + + + + (5.40)

On rearranging

3 3 4 4 1 1 2 2 3 4 1 2ln ln ln ln ( ) lnv x v x v x v x v v v v+ − − + + − − p

3 3 4 4 1 1 2 2( )v v v vφ φ φ φ= − + − − (5.41)

∴

3 4

3 4 1 2

1 2

3 4

1 2

.ln

.

v vv v v v

v v

x x px x

+ − −3 3 4 4 1 1 2 2( )v v v vφ φ φ φ= − + − − (5.42)

Denoting

3 3 4 4 1 1 2 2ln ( )K v v v vφ φ φ φ= − + − − (5.43)

where K , is known as the equilibrium constant, is a function of temperature only 3 4

3 4 1 2

1 2

3 4

1 2.e

v vv v v v

v v

x x p Kx x

ε ε

+ − −

=

⎡ ⎤=⎢ ⎥

⎣ ⎦

(5.44)

This equation is called the law of mass action. K has the dimension of pressure raised to the th power. Here the 3 4 1 2(v v v v+ − − ) 'x s is the values of mole fractions at equilibrium when the degree of reaction is eε . The law of mass action can also be written in this form

3 4

1 2

3 4

1 2

.

.

v v

v v

p p Kp p

= (5.45)

where the 'p s are the partial pressures.

5.5 HEAT OF REACTION The equilibrium constant K is defined by the expression

3 3 4 4 1 1 2 2ln ( )K v v v vφ φ φ φ= − + − − (5.46)

Differentiating ln K with respect to T

93

3 4 13 4 1 2ln ( )d d d d dK v v v v

dT dT dT dT dT2φ φ φ φ

= − + − − (5.47)

Let,

0 02

1 pc dTh sRT R T R

φ = − −∫∫ (5.48)

Therefore,

02 2

pc dTd hdT RT RT

φ= − − ∫

02

1 ( )ph c dTRT

= + ∫ 2

hRT

= −

(5.49)

Therefore,

3 3 4 4 1 1 2 22

1ln ( )d K v h v h v h v hdT RT

= + − − (5.50)

where the 's refer to the same temperature T and the same pressure h p . If moles of

1v1A and moles of 2v 2A combine to form moles of 3v 3A and moles of 4v

4A at constant temperature and pressure, the heat transferred would be, as shown in Fig.5.1, equal to the final enthalpy 3 3 4 4(v h v h )+ minus the initial enthalpy This is known as the heat of reaction, and is denoted by 1 1 2 2(v h v h+ ).

H∆ .

H

T,P

V3V1

V2

A1

A2 V4

A3

A4

Fig.5.1 Heat of Reaction Hence, heat of reaction is

3 3 4 4 1 1 2 2H v h v h v h v h∆ = + − − (5.51)

94

Therefore,

2lnd HK

dT RT∆

= (5.52)

This is known as the van’t Hoff equation. This equation can be used to calculate the heat reaction at any desired temperature or within a certain temperature range. By rearranging Eq. (5.52)

2

lnd K HdT RT

∆= (5.53)

or, ln

(1/ )d K Hd T R

∆= − (5.54)

Therefore, log log2.303 19.148 /1 1( ) ( )

d K d KH R kJ kgmold d

T T

∆ = − = − (5.55)

If 1K and 2K are the equilibrium constants evaluated at temperatures and respectively

1T 2T

1 2

1 2

log log19.148 1 1K KH

T T

−∆ = −

− (5.56)

or, 1 2 1

1 2

19.148 logTT KHT T K

∆ = −− 2

(5.57)

If H∆ is positive, the reaction is said to be endothermic. If H∆ is negative, the reaction is exothermic.

5.6 TEMPERATURE DEPENDENCE OF THE HEAT OF REACTION

3 3 4 4 1 1 2 2H v h v h v h v h∆ = + − − (5.58)

0 ph h c dT= + ∫ (5.59)

Therefore,

3 03 4 04 1 01 2 02 3 3 4 4 1 1 2 2( )p p p pH v h v h v h v h v c v c v c v c dT∆ = + − − + + − −∫ (5.60)

Denoting 3 03 4 04 1 01 2 02H v h v h v h v h∆ = + − −

0 3 3 4 4 1 1 2 2( )H H v c v c v c v c d∆ = ∆ + + − −∫ T (5.61)

95

If is known as a function of temperature and if at any temperature pc H∆ is known, then at any other temperature, H∆ can be determined for a certain chemical reaction from the above relation. Some chemical reactions may be expressed as the result of two or more reactions. If 0H∆ is known for each of the separate reactions, then 0H∆ of the resultant reaction may be calculated. For example,

2 2

12 2H O H + O 0 239,250 /H J gmol∆ =

2

12

CO CO O+ 2 0 279,890 /H J gmol∆ =

2 2CO H O CO H+ 2+ 0 40640 /H J gmol∆ = − 5.7 TEMPERATURE DEPENDENCE OF THE EQUILIBRIUM CONSTANT

3 3 4 4 1 1 2 2ln ( )K v v v vφ φ φ φ= − + − − (5.62)

where

0 02

1 pc dTh sdTTRT R R

φ = − −∫∫ (5.63)

On substitution

3 03 4 04 1 01 2 02

1ln ( )K v h v h v h v hRT

= + − −

3 3 4 4 1 1 2 2

2

( )1 .p p p pv c v c v c v c dTdT

TR+ − −

+ ∫∫

3 03 4 04 1 01 2 02

1 ( )v s v s v s v sR

+ + − −

(5.64)

If 0 3 03 4 04 1 01 2 02

0 3 03 4 04 1 01 2 02

H v h v h v h v hS v s v s v s v s

∆ = + − −∆ = + − −

(5.65)

Then,

3 3 4 4 1 1 2 20 02

( )1ln p p p pv c v c v c v c dTH SK dT

TRT R R

+ − −∆ ∆= − + +∫∫

(5.66)

This equation is sometimes called as the Nernst’s equation.

96

5.8 THERMAL IONIZATION OF A MONATOMIC GAS One interesting application of Nernst’s equation was made by Dr. M.N. Saha to the thermal ionization of a monatomic gas. If a monatomic gas is heated to a high enough temperature (5000 K and above), some ionization occurs, with the electrons in the outermost orbit being shed off, and the atoms, ions, and electrons may be regarded as a mixture of three ideal monatomic gases, undergoing the reaction

A A e+ + (5.67)

Starting with moles of atoms, 0nA v n x

1A A= 1 1v = 1 0 (1 )en n ε= − 1

11

e

e

x εε

−=

+

3A A+= 3 1v = 3 0 en n ε= 3 1e

e

x εε

=+

4A e= 4 1v = 4 0n n 0ε= 4 1e

e

x εε

=+

0 (1 )en n ε= +∑ The equilibrium constant is given by

3 4

3 4 1 2

1 2

3 4

1 2

.ln ln .

.e

v vv v v v

v v

x xK px x

ε

+ − −⎧ ⎫= ⎨ ⎬

⎩ ⎭

(5.68)

or,

3 4 1

.1 1ln .1

1

e e

v v ve e

e

e

K p

ε εε ε

εε

+ −+ +=

−+

2

2ln .1

e

e

pεε

=−

(5.69)

Since the three gases are monatomic,52pc = R which, on being substituted in the

Nernst’s equation, gives 2

02

5ln . ln ln1 2

e

e

Hp T BRT

εε

∆= − + +

− (5.70)

where 0 lnS BR

∆=

97

Now, 2

02 5/ 2ln .

1 .e

e

p HT B RT

εε

∆= −

− (5.71)

∴ 0

2 5/

2 .1

H RTe

e

TBe/ 2

pε

ε−∆=

− (5.72)

where eε is the equilibrium value of the degree of ionization. This is known as the Saha’s equation. For a particular gas the degree of ionization increases with an increase in temperature and a decrease in pressure. It can be shown that

0H∆ is the amount of energy necessary to ionize 1 gmole of atoms. If we denote the ionization potential 0H∆ of the atom in volts by E , then

0H E∆ = ( )volt 191.59 10−× × 236.06 10Coulomb electronelectron gmol

× ×

49.6354 10 /E J gmol= ×

(5.73)

Equation (5.70) becomes

2

96354 5ln ln ln1 2

e

e

Ep T BRT

εε

= − + +−

(5.74)

Expressing p in atmospheres, changing to common logarithms and introducing the values of B from statistical mechanics, Saha finally obtained the equation

2

2

96,354 5log ( ) log log 6.4911 19.148 2

e i

e a

Ep atm TT

ε ωε ω

= − + + −−

eω

e

(5.75)

where ,iω ω and aω are constants that refer to the ion, electron and atom respectively. The value eω for an electron is 2 . The value of , i aE andω ω for a few elements are given below Table 5.2 Values for E and ω Element ,E volts aω iω

NaCaCdZn

5.123.876.098.969.36

22111

11222

98

For alkali metals like etc., the ionization potential is less. It means less energy is required to ionize one gmole of atoms. So these are used as seed for magnetohydrodynamic power generation. Saha applied Eq. (5.70) to the determination of temperature of stellar atmosphere. The spectrum of a star contains lines which originate from atoms (arc lines) and those which originate from ions (spark lines). A comparison of the intensities of an arc line and a spark line from the same element gives a measure of

, ,Cs Na K

eε . Considering a star as a sphere of ideal gas, the pressure of a star can be estimated. Thus, knowing

, ,e p Eε and the ' ,sω temperature of star can be calculated. 5.9 GIBBS FUNCTION CHANGE Molar Gibbs function of an ideal gas at temperature T and pressure p is given by

( lng RT pφ= + ) (5.76)

For the reaction of the type

1 1 2 2 3 3 4 4v A v A v A v A+ + (5.77)

The Gibbs function change of the reaction G∆ is defined by the expression

3 3 4 4 1 1 2 2G v g v g v g v g∆ = + − − (5.78)

where the refer to the gases completely separated at is also known as the free energy change. Substituting the values of the

'g s , , T p G∆'g s

∆ 3 4 1 2

3 3 4 4 1 1 2 2( ) l v v v vG RT v v v v RT pφ φ φ φ n + − −= + − − + (5.79)

But

3 3 4 4 1 1 2 2ln ( )K v v v vφ φ φ φ= − + − − (5.80)

∴ 3 4 1 2ln ln v v v vG RT K RT p + − −∆ = − + (5.81)

If p is expressed in atmospheres and G∆ is calculated from each gas is at a pressure of 1 atm, the second term on the right drops out. Under these conditions is known as the standard Gibbs function change and denoted by

G∆

0G∆0 lnG RT K∆ = (5.82)

This is an important equation which relates the standard Gibbs function change with temperature and the equilibrium constant. From this the equilibrium constant can be calculated from changes in the changes in the standard Gibbs function, or vice versa. For dissociation of water vapour,

2 2

12 2H O H + O (5.83)

298ln K = 93.7.−

99

Therefore, 0298 8.3143 298 ( 93.7) 232,157 /G J∆ = − × × − = gmol (5.84)

Substituting ln K from Nernst’s equation 3 3 4 4 1 1 2 20

0 02

( p p p pv c v c v c v cG H T dT T

T)

S+ − −

∆ = ∆ − − ∆ (5.85)

from which also may be calculated directly. Values of 0G∆ 00 0 2, 98H S and G∆ ∆ ∆

for fundamental ideal gas reactions are given in Table 5.3 + Table 5.3 Fundamental ideal gas reactions Reaction 0, /H J gmol∆ 0 /S J gmol K∆ − 0

298 /G J gmol∆

2 2H H 4 2 7 , 3 8 0 4.90 404,335

2 2

1 12 2

HCL H Cl+ 91,760 22.25− 95 ,110

2 2

1 12 2

HBr H Br+ 50,280 24.05− 54 ,050

2

1 12 2 2HI H + I 5, 320 21.00− 8, 380

2 2

1 12 2

NO N O+ 90,500− 10.48− 87,570−

2 2 2

12

H O H O+ 239,250 14.70 232 ,545

2 2

12 2H S H + S 6.90 80,450 71,230

2 2

12

CO CO O+ 279,890 18.69 258 ,940

2 2

12

NO NO O+ ,500 59 11.44 37,290

2 2

12

SO S O+ 2 349 3.77 329 ,025 ,330

3 2 2

1 32 2

NH N H+ 39,800 16 40.27 ,380

3 2

12

SO SO O+ 2 94 89.50 ,690 67,880

2 4 22N O NO 174.30 5, 56,980 030

100

Both and may be added and subtracted in the same manner as 0S∆ 0

298G∆ 0H∆ . For example,

2 2

12 2H O H + O gmol 0

298 232,545 /G J∆ =

2

12

CO CO O+ 2 gmol 0 258,940 /G J∆ =

2 2CO H O CO H+ 2+ gmol0

298 26,395 /G J∆ = −

0298

298

26,395ln 10.6538.3143 298

GKRT

∆= = =

×

298 42,330K = From this, the value of the degree of reaction at equilibrium eε may be

calculated. From Eq.(5.35), if 0 1n = and 0 0n′ =

3 3 4 4 1 1 2 2,

( )T p

G v v v vµ µ µ µε

∂⎛ ⎞ = + − −⎜ ⎟∂⎝ ⎠

Since ( ln lnk k )kRT pµ φ= + + x

And ( lnk kg RT pφ= + )

So lnk kg RT xµ = + k Therefore

3 4

1 2

3 43 3 4 4 1 1 2 2

, 1 2

.ln

.

v v

v vT p

G xv g v g v g v g RT xx xε

∂⎛ ⎞ = + − − +⎜ ⎟∂⎝ ⎠

3 4

1 2

3 4

1 2

.ln

.

v v

v v

x xG RTx x

= ∆ +

At 3 40, 0, 0x xε = = =

,T p

Gε

∂⎛ ⎞ = −∞⎜ ⎟∂⎝ ⎠

and at 1 21, 0, 0x xε = = =

101

,T p

Gε

∂⎛ ⎞ = +∞⎜ ⎟∂⎝ ⎠

At 1 21 2

1 , ,2

v vx xv v

ε = = =∑ ∑

3 43 4,v vx x

v v= =

∑ ∑

where 3 4 1v v v v v2= + + +∑ 3 4

1 2

3 4

,1 1 22

ln

v v

v vT p

v vv vG G RT

v vv v

εε

=

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟

∂⎛ ⎞ ⎝ ⎠ ⎝ ⎠= ∆ +⎜ ⎟∂⎝ ⎠ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∑ ∑

∑ ∑

If 1 , 298p atm T K= =

02981

29812

p atmT K

G G

ε

ε ==

=

∂⎛ ⎞ = ∆⎜ ⎟∂⎝ ⎠

The slope ,T p

Gε

∂⎛ ⎞⎜ ⎟∂⎝ ⎠

at 12

ε = is called the “affinity” of the reaction, and it is equal

to at the standard reference state. The magnitude of the slope at 0G∆12

ε = (Fig.5.2) indicates the direction in which the reaction will proceed.

ε = 0 ε = ε ε = 1 /2

T ,p

G

ε = 1

( δG /δε ) = − αT ,p

( δG /δε ) = + αT ,p

e

Fig. 5.2 Plot of against G ε at constants T and p

102

For water vapour reaction, is a large positive number, which indicates the

equilibrium point is far to the left of

0298G∆

1 ,2

ε = and therefore, eε is very small. Again

for the reaction, 02 2

1 1 , 2 2

NO N O G+ ∆ 298 is a large negative value, which

shows that the equilibrium point is far to the right of 1 ,2

ε = and so eε is close to

unity. 5.10 FUGACITY AND ACTIVITY The differential of Gibbs function of an ideal gas undergoing an isothermal process is

nRTdG Vdp dpp

= = (ln )nRTd p= (5.86)

Analogously, the differential of Gibbs function for a real gas is

(ln )dG nRTd f= (5.87)

where f is called the fugacity, first used by Lewis. The value of fugacity approaches the value of pressure, as the letter tends to zero, i.e., when ideal gas conditions apply. Therefore,

0lim 1p

fp→

= (5.88)

For an ideal gas .f p= Fugacity has the same dimension as pressure. Integrating Eq. (5.87)

00ln fG G nRT

f− = (5.89)

where and 0G 0f refer to the reference state when 0 1p = atm. The ratio

0/f f is called the activity. Therefore,

0 lnG G nRT a− = (5.90)

For ideal gases, the equilibrium constant is given by 3 4

1 2

3 4

1 2

.

.

v v

v v

P PKP P

= (5.91)

For real gases

103

3 4

1 23 4

1 2

.

.

v v

real v v

f fKf f

= (5.92)

Similarly, it can show that

0 ln realG nRT K∆ = − (5.93)

and

0

2

ln reald K HdT RT

∆= (5.94)

5. 11 ENTHALPY OF FORMATION Let us consider the steady state steady flow combustion of carbon and oxygen to from (Fig.5.3). Let the carbon and oxygen each enter the control volume at

and 1atm. Pressure, and the heat transfer be such that the product

leaves at ,1atm Pressure. The measured value of heat transfer is of formed. If

2CO025 C 2CO

025 Cl393,522− /kJ kgmo 2CO RH and PH refer to the total enthalpy

of the reactants and products respectively, then the first law applied to the reaction gives 2 2C O CO+ →

C . V .

C

O 2

2 5 C , 1 a t m

C O 2

Q C . V .= - 3 9 3 , 5 2 2 k j / k g m o l o f C O 2

2 5 C , 1 a t m

Fig.5.3 Enthalpy of formation

For all the reactants and products in a reaction, the equation may be written as

. ei C V eR Pi

n h Q n h+ =∑ ∑ (5.95)

104

where R and refer to the reactants and products respectively. The enthalpy of all the elements at the standard reference state of atm is assigned the value of zero. In the carbon-oxygen reaction,

P025 ,C 10.RH = Hence, the energy

equation gives

. 393,522 /C V PQ H kJ kg= = − mol (5.96)

This is known as the enthalpy of formation of at and

designated by the symbol,

2CO 025 ,C 1 .atm ,

( )2

0 393,522fCO

h = − . In most cases,

however, the reactants and products are not at , therefore, the change of enthalpy (in case of constant pressure or S.S.S.F. process) between

and the given state must be known. Thus the enthalpy at any

temperature and pressure,

/kJ kgmol025 ,C 1atm

025 ,C 1atm.

,T ph is

( ) ( )0, 298 ,1 ,298 ,1

fT p K atm T pK atmh h h

→= + ∆ (5.97)

For convenience, the subscripts are usually dropped, and 0

,T p fh h h= + ∆ (5.98)

where h∆ represents the difference in enthalpy between any given state and the enthalpy at 298.15 , 1K atm . Table 5.4 gives the values of the enthalpy of formation of a number of substances in . /kJ kgmol

105

Table 5.4 Enthalpy of Formation, Gibbs Function of Formation, and Absolute Entropy of Various Substance at pressure 025 , 1C atm

Substance ,

MolecularWeight M

0

/fh

kJ kgmol

0

/fg

kJ kgmol

0

/skJ kgmolK

28.01144.00118.01518.01516.04326.03828.05430.07044.09758.124114.23114.23

110529393522241827285838748732267315228384667103847126148208447249952

−−−−−++−−−−−

1371503943742285832371785075120923468207327772331616914168596940

−−−−−++−−−++

197.653213.795188.83370.049186.256200.956219.548229.602270.019310.227466.835360.896

Table 5.5 give the values of 0 0298h h h∆ = − /kJ kg mol for various substances at

different temperature. Table 5.5 Enthalpy of formation at , ideal gas enthalpy, and Absolute Entropy at 0.1 MPa (1bar) Pressure

025 C

Nitrogen, Diatomic ( )2N Oxygen, Diatomic ( )2O

( )0

2980 /fh kJ k= mol ( )0

2980 /fh kJ k= mol

31.999M = 28.013M =

Temperature ( )0 0298h h−

0s ( )0 0

298h h− 0

s

/kJ kmol /kJ kmolK /kJ kmol /kJ kmolKK 0 -8669 0 -8682 0 100 -5770 159.813 -5778 173.306 200 -2858 179.988 -2866 193.486

2

2

2

4

2 2

2 4

2 6

3 8

4 10

8 18

8 18

( )( )( )( )( )( )( )( )( )( )( )( )

CO gCO gH O gH O lCH gC H gC H gC H gC H gC H gC H gC H l

106

298 0 191.611 0 205.142 300 54 191.791 54 205.322 400 2971 200.180 3029 213.874 500 5912 206.740 6088 220.698 600 8891 212.175 9247 226.455 700 11937 216.866 12502 231.272 800 15046 221.016 15841 235.924 900 18221 224.757 19246 239.936 1000 21406 288.167 22707 243.585 1100 24757 231.309 26217 246.928 1200 28108 234.225 29765 250.016 1300 31501 236.941 33351 252.886 1400 34936 239.484 36966 255.564 1500 38405 241.878 40610 258.078 1600 41903 244.137 44279 260.446 1700 45430 246.275 47970 262.685 1800 48982 248.304 51689 264.810 1900 52551 250.237 55434 266.835 2000 56141 252.078 59199 268.764

Carbon Dioxide ( )2CO Carbon Monoxide ( )CO

( )0

298393522 /fh kJ km= − ol ( )0

298110529 /fh kJ kmol= −

44.01M = 28.01M =

Temperature ( )0 0298h h−

0s ( )0 0

298h h− 0

s

K /kJ kmol /kJ kmolK /kJ kmol /kJ kmolK 0 -9364 0 -8669 0 100 -6456 179.109 -5770 165.850 200 -3414 199.975 -2858 186.025 298 0 213.795 0 197.653 300 67 214.025 54 197.833 400 4008 225.334 2975 206.234 500 8314 234.924 5929 212.828 600 12916 243.309 8941 218.313 700 17765 250.773 12021 223.062 800 22815 257.517 15175 227.271 900 28041 263.668 18397 231.006 1000 33405 269.325 21686 234.531 1100 38894 274.555 25033 237.719 1200 44484 279.417 28426 240.673 1300 50158 283.956 31865 243.426

107

1400 55907 288.216 35338 245.999 1500 61714 292.224 38848 248.421 1600 67580 296.010 42384 250.702 1700 73492 299.592 45940 252.861 1800 79442 302.993 49522 254.907 1900 85429 306.232 53124 256.852 2000 91450 309.320 56739 258.710

Water ( )2H O Hydrogen,Diatomic ( )2H

( )0

298241827 /fh kJ= − kmol ( )0

2980 /fh kJ kmol=

18.015M = 2.016M =

Temperature ( )0 0298h h−

0s ( )0 0

298h h− 0

s

K /kJ kmol /kJ kmolK /kJ kmol /kJ kmolK 0 -9904 0 -8468 0 100 -6615 152.390 -5293 102.145 200 -3280 175.486 -2770 119.437 298 0 188.833 0 130.684 300 63 189.038 54 130.864 400 3452 198.783 2958 139.215 500 6920 206.523 5883 145.738 600 10498 213.037 8812 151.077 700 14184 218.719 11749 155.608 800 17991 223.803 14703 159.549 900 21924 228.430 17628 163.060 1000 25978 232.706 20686 166.223 1100 30167 236.694 23723 169.118 1200 34476 240.443 26794 171.792 1300 38903 243.986 29907 174.281 1400 43447 247.350 33062 176.620 1500 48095 250.560 36267 178.833 1600 52844 253.622 39522 180.929 1700 57685 256.559 42815 182.929 1800 62609 259.372 46150 184.833 1900 67613 262.078 49522 186.657 2000 72689 264.681 52932 188.406

5.12 FIRST LAW FOR REACTIVE SYSTEMS For the S.S.S.F. process as shown in Fig.5.4 the first law gives

108

. . . .R C V P C VH Q H W+ = + (5.99)

or,

. . . .i i C V e e C VR p

n h Q n h W+ = +∑ ∑ (5.100)

or,

0 0. . . .i f C V e f C V

eR pin h h Q n h h W⎡ ⎤ ⎡ ⎤+ ∆ + = + ∆ +⎣ ⎦ ⎣ ⎦∑ ∑ (5.101)

n1

n2

C.V

Qc.v W c.v

n3

n4Rea

ctan

ts Products

Fig.5.4 First law for a reaction system When the states of reactants and products are not in the standard reference state (298 , 1 ),K atm then, as shown in Fig.5.5,

109

R

R0R0

P0

PhRP

298 K T

0

H

Fig.5.5 Enthalpy of reactants and products varying with temperature

0 0 0 0. . ( ) ( ) (C V P R P P P R R RQ H H H H H H H H= − = − + − + − ) (5.102)

0 0

0( ) ( )p p RP R Re i

P Rn h h h n h h= − + ∆ − −∑ ∑ (5.103)

where 0PRh∆ is the enthalpy of reaction at the standard temperature (298 )K . The

variation of enthalpy with pressure is not significant.

0 0 0 0. . ( ) ( ) (C V P R P P P R R RQ U U U U U U U U= − = − + − + − )

0 0

0( ) ( )p p RP R Re i

p Rn u u u n u u= − + ∆ − −∑ ∑

(5.104)

where 0RPu∆ is the internal energy of reaction at 298 .K

5.13 ADIABATIC FLAME TEMPERATURE If a combustion process occurs adiabatically in the absence of work transfer or changes in K.E. and P.E., then the energy equation becomes

R PH H= (5.105)

or, i ei e

R pn h n h=∑ ∑ (5.106)

or, i ei e

R pn h n h=∑ ∑ (5.107)

or,

110

0 0f fi e

eR Pin h h n h h⎡ ⎤ ⎡ ⎤+ ∆ = + ∆⎣ ⎦ ⎣ ⎦∑ ∑ (5.108)

For such a process, the temperature of the products is called the adiabatic flame temperature which is the maximum temperature achieved for the given reactants. The adiabatic flame temperature can be controlled by the amount of excess air supplied; it is the maximum with a stoichiometric mixture. Since the maximum permissible temperature in a gas turbine is fixed from metallurgical considerations, close control of the temperature of the products is achieved by controlling the excess air. For a given reaction the adiabatic flame temperature is computed by trial and error. The energy of the reactants RH being known, a suitable temperature is chosen for the products so that the energy of products at that temperature becomes equal to the energy of the reactants. 5.14 ENTHALPY AND INTERNAL ENERGY OF COMBUSTION: HEATING

VALUE The enthalpy of combustion is defined as the difference between the enthalpy of the product and the enthalpy of the reactants when complete combustion occurs at a given temperature and pressure. Therefore,

RP P Rh H H= − (5.109)

or,

0 0RP f fe i

e iP Rh n h h n h⎡ ⎤ ⎡ ⎤= + ∆ − + ∆⎣ ⎦ ⎣ ⎦∑ ∑ h (5.110)

where RPh is the enthalpy of combustion or of fuel. The value of the enthalpy of combustion of different hydrocarbon fuels at

are given in Table 5.6.The internal energy of combustion,

( /kJ kg /kJ kgmol)

025 , 1C atm ,RPµ is defined similar way.

RP P RU Uµ = − (5.111)

0 0f fe i

ip Re

n h h pv n h h pv⎡ ⎤ ⎡= + ∆ − − + ∆ −⎣ ⎦ ⎣ ⎦∑ ∑ ⎤ (5.112)

If all the gaseous constituents are considered ideal gases and the volume of liquid and solid considered is assumed to be negligible compared to gaseous volume.

(RP RP gaseousu u RT n= − products gaseousn− tan )reac ts (5.113)

111

In the case of a constant pressure or steady flow process, the negative of the enthalpy of combustion is frequently called the heating value at constant pressure, which represents the heat transferred from the chamber during combustion at constant pressure. Similarly, the negative of the internal energy of combustion is sometimes designated as the heating value at constant volume in the case of combustion, because it represents the amount of heat transfer in the constant volume process. The higher heating value (HHV) or higher calorific value (HCV) is the heat transferred when 2H O in the products is in the liquid state. The lower heating value (LHV) or lower calorific value (LCV) is the heat transferred in the reaction when 2H O in the products is in the vapour state. Therefore,

2.H O fgLHV HHV m h= − (5.114)

where 2H O

m is the mass of water formed in the reaction.

5.15 ABSOLUTE ENTROPY AND THE THIRD LAW OF THERMODYNAMICS So far only first law aspects of chemical reactions have been discussed. The second law analysis of chemical reactions needs a base for the entropy of various substances. The entropy of substances at the absolute zero of temperature, called absolute entropy, is dealt with by the third law of thermodynamics formulated in the early twentieth century primarily by Nernst (1864-1941) and Max Planck (1858-1947). The third law states that the entropy of a perfect crystal is zero at the absolute zero of temperature ant it represents the maximum degree of order. A substance not having a perfect crystalline structure and processing a degree of randomness such as a solid solution or a glassy solid has a finite value of entropy at absolute zero. The third law provides an absolute base from which the entropy of each substance can be measured. The entropy relative to this base is referred to as the absolute entropy. Table 5.5 gives the absolute entropy of various substances at the standard state

For any other state, 025 , 1 .C atm

( )0,

,1 , ,T p T

T atm T ps s s

→= + ∆ (5.115)

where 0Ts refers to the absolute entropy at 1 atm, and temperature T, and

( ),1 , ,T atm T p

s→

refers to the change of entropy for an isothermal change of pressure

from 1atm to pressure p (Fig.5.6).

112

1 atm

T .p

A bso lu te en tropy is know a long th is isoba r, assum ing idea l gas behav iour a t 1 a tm

( s) T ,1 a tm T ,p

T

s Fig. 5.6 Absolute entropy

Table 5.5 give the value of 0

s for various substances at 1 atm. and at different

temperatures. Assuming ideal gas behaviour ( ),1 , ,T atm T p

s→

∆ can be determined

(Fig 5.6). 2

2 1

1

ln ps s Rp

− = − (5.116)

or,

( ),1 , ,

lnT atm T p

s R p→

∆ = − (5.117)

where p is in atm. Table 5.6 Enthalpy of Combustion of Some Hydrocarbons at 025 C

Liquid 2H O in Vapour 2H O in Products Products ( of ( Negative Negative of Higher Heating Lower Heating Value )Value

Hydrocarbon Formula

/

LiquidHydrocarbonkJ kgfuel

−

/

GaseousHydrocarbonkJ kgfuel

−

/

LiquidHydrocarbonkJ kgfuel

−

/

GaseousHydrocarbonkJ kgfuel

−

113

(1 (2 (3 ) ) ) (4) (5) (6)Paraffin Family

MethaneEthanePropaneButanePentaneHexaneHexaneOctaneDecane

4

2 6

3 8

4 10

5 12

6 14

7 16

8 18

10 22

12 26

CHC HC HC HC HC HC HC HC HC H

4997549130486434830848071478934764147470

−−−−−−−−

55496518755034549500490114867648436482564800047828

−−−−−−−−−−

4598345344449834473344557444254423944100

−−−−−−−−

50010474844635345714459514510144922447884459844467

−−−−−−−−−−

Olefin FamilyEthenePropeneButenePenteneHexeneHepteneOcteneNoneneDecene

2 4

3 6

4 8

5 10

6 12

7 14

8 16

9 18

16 20

C HC HC HC HC HC HC HC HC H

502964891748453481344793747800476934761247547

−−−−−−−−−

471584578045316449964480044662445564447544410

−−−−−−−−−

Alkylbenzene FamilyBenzeneMethylbenezeneEthylbenzenePropylbenzeneButylbenzene

6 6

7 8

8 10

9 12

10 14

C HC HC HC HC H

4183142473429974341643748

−−−−−

4226642847433954380044123

−−−−−

4014140527409244121941453

−−−−−

4057640937413224160341828

−−−−−

114

5.16 SECOND LAW ANALYSIS OF REACTIVE SYSTEMS The reversible work for a steady state flow process, in the absence of changes in K.E. and P.E., is given by

0 0( ) (rev i i i e e eW n h T s n h T= − − −∑ ∑ )s (5.118)

For an S.S.S.F. process involving a chemical reaction

0

0

0

0

frev iR i

feeP

W n h h T

n h h T s

s⎡ ⎤= + ∆ −⎣ ⎦

⎡ ⎤− + ∆ −⎣ ⎦

∑

∑

(5.119)

The irreversibility for such a process is

0 0e ie iP R

. .C VI n T s nT s Q= − −∑ ∑ (5.120)

The availability, ,ψ in the absence of K.E. and P.E. changes, for an S.S.S.F. process is

0 0 0( ) (h T s h T s0 )ψ = − − − (5.121)

When an S.S.S.F. chemical reaction takes place in such a way that both the reactants and products are in temperature equilibrium with the surroundings, the reversible work is given by

i erev i eR P

W n g n= − g∑ ∑ (5.122)

where the g′s refer to the Gibbs function. The Gibbs function for formation,0,fg is

defined similar to the enthalpy of formation,0.fh The Gibbs function of each of the

elements at and 1 is assumed to be zero, and the Gibbs function of

each substance is found relative to this base. Table 5.4 gives

025 C .atm0

fg for some

substance at 1 . 025 ,C atm 5.17 SECOND LAW EFFICIENCY OF A REACTIVE SYSTEM For a fuel at the chemical exergy is the maximum theoretical work that could be obtained through reaction with environmental substances. However, due to various irreversibilities like friction and heat loss, the actual work obtained is only a fraction of this maximum theoretical work. The second law efficiency may thus be defined as the ratio of

0, 0,T p

115

Actual work doneMaximum theoretical workIIη = (5.143)

IIη cV

fuel ch

Wm xa

= (5.144)

The associated irreversibility and the consequent exergy loses require to be reduced to enhance the second law efficiency, which in turn, reduce the fuel consumption and also increases the cost. The trade off between the fuel savings and the additional costs must be carefully weighed. 5.18 CHEMICAL EXERGY A system is in thermal and mechanical equilibrium with the environment when the system attains the dead state. Under such conditions, exergy of the system is zero. However, the contents of a system even at the dead state may undergo chemical reaction with environmental components and produce additional work. Consider a combined system formed by an environment and a system having a certain amount of fuel at Work is obtainable by allowing the fuel to react with oxygen from the environment to produce the components of and

0 0,T p .2CO 2H O .

The chemical exergy is thus defined as the maximum theoretical work that could be developed by the combined system. Thus for a given system at a specific state

Total exergy = Thermomechanical exergy + Chemical exergy

Let us consider a hydrocarbon fuel at reacting with oxygen from the environment (Fig.5.7) which is assumed to be consisting of an ideal gas mixture at

( )a bC H 0 0,T p

0, 0.T p

116

T0

H2O at T0,XH20 p0

CO2at T0

Wor

k

Ca H b

p0, T0

O2 at T0

XO2 p0

XCO2, p0

Heat transfer environment at p0 , T0

Boundary of the combined system

Fig. 5.7 Fuel exergy concept

The oxygen that reacts with the fuel is at a partial pressure of

20x 0,p where 20x is

the mole fraction of oxygen in the environment. The fuel and oxygen react completely to produce and 2CO 2 ,H O which exit in separate steams at oτ and respective partial pressures of

2coχ oρ and2OHχ .o

ρ . The reaction is given by

2 24 2a b

b bC H a O aCO H O⎛ ⎞+ + → +⎜ ⎟⎝ ⎠

2

P

(5.123)

At steady state, the energy balance gives

R cVcVnH W nHQ

⋅

+ = + (5.124)

or,

cV cVR P

W Q H Hn n

= + − 2 2

00( )

4 2cV

2f CO H Oa b

Q bh h C H a h ah hn

⎛ ⎞= + + ∆ + + − −⎜ ⎟⎝ ⎠

b (5.125)

where is the rate of fuel flow in moles, and K.E. and P.E. effects are neglected. An entropy balance for the control volume gives

n

2

2 2

0

/04

2

a bcV

C H o

genCO H O

Q n bs a sT

sbas sn

⎛ ⎞= + + +⎜ ⎟⎝ ⎠

− − +

(5.126)

Eliminating between Eq. (5.125) and Eq. (5.126) cVQ

117

2 24 2a bcV

C H o CO H OW bh a h ah hn

⎡ ⎤⎛ ⎞= + + − −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦2

b

2 2 20 4 2b

a bC H o CO H ObT s a s as s⎡ ⎤⎛ ⎞− + + − −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

0 genT sn

−

(5.127)

The specific enthalpies in Eq. (5.127) can be determined knowing only the temperature and the specific entropies can be determined knowing and the composition of the environment. For maximum work,

0T 0 0,T p

0 0genT s

In

= = (5.128)

Therefore, the chemical exergy cha can be expressed as

2 24 2a bch C H O CO H Ob ba h a h ah h⎡ ⎤⎛ ⎞= + + − −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

2

2 20 4 2a bC H O CO H Ob bT s a s as s⎡ ⎤⎛ ⎞− + + − −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

2

2

(5.129)

The specific entropies for and 2 ,O CO 2H O are written from Eq. (5.104),

0, 0 0, 0( ) ( ) lni ii is T x p s T p R x= − (5.130)

where the first term on the right is the absolute entropy at and 0T 0,p and ix is the mole fraction of component in the environment .Therefore ,Eq.(5.129) becomes,

2 2 2 0, 0( )at T p4 2a bch C H o CO H Ob ba h a h ah h⎡ ⎤⎛ ⎞= + + − −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

2 2 20 4 2a bC H O CO H Ob bT s a s as s⎡ ⎤⎛ ⎞− + + − −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

( )0, 0 at T p

( )( ) ( )

2

2 2

/ 4

/ 20 lna b

Oa b

CO H O

xRT

x x

+

+

(5.131)

In terms of Gibbs function of respective substances,

2 2 2 0, 0( )at T p4 2a b

ch C H O CO H O

b ba g a g ag g⎡ ⎤⎛ ⎞= + + − −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

(5.132)

118

+( )

( ) ( )2

2 2

/ 4

/ 20 lna b

Oa b

CO H O

xRT

x x

+

where

( )0 , 0 ,

0

0, 0 T pref reff T pg T p g g→

= + ∆ (5.133)

For the special case when and 0T 0p are the same as and refT ,refp g∆ will be

zero. The chemical exergy of pure at where the reaction is given by CO 0 0,T p

2 2

12

CO O CO+ → (5.134)

( ) 2 2

12

ch CO O COCOa g g g⎡ ⎤= + −⎢ ⎥⎣ ⎦

( )0, 0 at T p

( )2

2

1/ 2

0 ln O

CO

xRT

x+

(5.135)

Water is present as a vapour within the environment, but normally is a liquid at The chemical exergy of liquid water is, 0, 0.T p

2 2( ) ( )H O l H O g→ (5.136)

( ) 2 22

( ) ( )( )ch H O l H O gH O l

a g g⎡ ⎤= −⎣ ⎦ ( )0, 0atT p (5.137)

The specific exergy of a system is

thermo mech chema a a−= + 2

0 0 0 0 0( ) ( ) ( )2 ch

Vu u p v v T s s gz a= − + − − − + + +

(5.138)

and the specific flow exergy is given by

( )2

0 0 0( )2 ch

Va h h T s s gz a= − − − + + + (5.139)

Between two states of system a constant composition, cancels, leaving just the thermo-mechanical contribution. For example, to find the chemical exergy of liquid octane, the reaction is

cha

8 18 2 2 2( ) 12.5 8 9 ( )C H l O CO H O g+ → + (5.140) Assume the composition of the environment (on molar basis):

2 2 2 275.67%, 20.35%, 3.12%, 0.03%, N O H O CO others 0.83%. Then,

8 18 2 2 2( ) 12.5 8 9ch C H l O CO H Oa g g g g⎡ ⎤= + − −⎣ ⎦ ( )0, 0atT p (5.141)

119

( )( ) ( )

2

2 2

12.5

8 90 ln O

CO H O

xRT

x x+

Using the values given in Table 5.4,

( ) 8 18 ( ) 6940 12.45(0) 8( 394,374) 9( 228,583)cha C H l = + − − − −

( )( ) ( )

12.5

8 9

0.20358.3143(298.15)ln

0.003 0.0312+

5,408,068.72 /kJ kgmol= 47,346 /kJ kg=

(5.142)

120