chpt15sg

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Chapter 15-1

Chapter 15: Chemical and Phase Equilibrium

Consider the combustion of propane with 120 percent theoretical air.

The combustion equation is

C H O N

D CO E CO F O G H O B N

3 8 2 2

2 2 2 2

12 376+ . (5)( + . )

+ + + +

Conservation of mass for each species yields

C D E

H G or G

O A D E F G

N B

: = +

: = =

: (5)( )( ) = + + +

: = ( . )(5)( . )

3

8 2 4

2 2 2

3 76 122

Since the percent theoretical air is known, the N2balance gives B. The H

balance gives G. Then the C and O balances give two equations that relate

the remaining three unknowns D, E, and F. Therefore, we need one more

equation.

To obtain this last equation relating the mole numbers of the products, we

assume that the products are in chemical equilibrium.

To develop the relations for chemical equilibrium, consider placing the

product gases in a system maintained at constant Tand P. The constant T

Combustion

Chamber

FuelC3H8

120%

Th. air

CO2

COO2H2O

N2

QnetReactants

TR, PR

Products

PP, TP

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Chapter 15-2

andPare achieved if the system is placed in direct contact with a heat

reservoir and a work reservoir. In particular, lets consider that CO2, CO,

and O2form a mixture in chemical equilibrium.

CO O CO+ 2 2

Taking the positive direction of heat transfer to be to the system, the increase

of entropy principle for a reacting system is

dsQ

Tsys

sys

If the reaction takes place adiabatically, then dssys 0 . A reaction taking

place adiabatically does so in the direction of increasing entropy.

Heat

Reservoir

At T

System of Product

Gases At T and P

WorkReservoir

At P

QWb

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Chapter 15-3

If we apply both the first law and the second law for the fixed mass system

of reacting gases for fixed Tand Pwhere there is both heat transfer and

work, we obtain

Q W dU

W PdV

dS Q

T

b

b

=

=

T dS Q

T dS dU PdV

dU P dV T dS

or

dU P dV T dS

+

+

+

0

0Now, we define a new (for us anyway) thermodynamic function, the Gibbs

function, as

G H TS =

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Chapter 15-4

The differential of the Gibbs function when Tand Pare constant is

dG dH T dS S dT

dU P dV V dP T dS S dT

dU P dV T dS

T P

=

= + +

= +

( )

0 at constant and

The chemical reaction at constant temperature and pressure will proceed in

the direction of decreasing Gibbs function. The reaction will stop andchemical equilibrium will be established when the Gibbs function attains a

minimum value. An increase in the Gibbs function at constant Tand Pwould

be a violation of the second law.

The criterion for chemical equilibrium is expressed as

( ) ,dG T P = 0

Consider the equilibrium reaction among four reacting components Aand B

as reactants and Cand Das products. These components will have mole

numbers as NA, NB, NC, and ND. As differential amounts of Aand Breact to

0

0

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Chapter 15-5

form differential amounts of Cand Dwhile the temperature and pressure

remain constant, we have the following reaction to consider.

dN A dN B dN C dN DA B C D+ +

For equilibrium the Gibbs function of this mixture must be a minimum. This

yields

( ) ( ) ( ), , ,dG dG g dN

g dN g dN g dN g dN

T P i T P i i T P

D D C C A A B B

= = =

+ + + =

0

0

Here gi is the molar Gibbs function for component i(also called the

chemical potential).

To find a relation among the dNs, we select a corresponding stoichiometric

reaction. For the CO, CO2, O2reaction, the stoichiometric or theoretical

reaction is

CO CO O2 21

2 +

The change in moles of the components is related to their stoichiometric

coefficients, 1, 1, for CO2, CO, and O2, respectively. If 0.01 mole of CO2disassociates, 0.01 mole of CO and 0.005 mole of O2are formed. For the

four general components in equilibrium, A, B, C, and D, the corresponding

stoichiometric equation is

A B C DA B C D+ +

where the s are the stoichiometric coefficients.

The change in mole numbers of the reacting components is proportional to

the stoichiometric coefficients by

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Chapter 15-6

dN dN

dN dN

A A C C

B B D D

= =

= =

where is the proportionality constant and represents the extent of reaction.

A minus sign is added to the dNAand dNBbecause the number of moles of A

andBdecrease as the reaction takes place. Now, we substitute these into the

requirement for equilibrium and cancel the s.

g g g g

g g g g

D D C C A A B B

D D C C A A B B

( ) ( ) ( ) ( )

+ + + =

+ =

0

0

This last result is known as the criterion for chemical equilibrium. It is this

last equation that we use to relate the mole numbers of the reacting

components at equilibrium. Now lets see how the mole numbers are

imbedded in this equation. First we assume that the mixture of reacting

components is an ideal gas. Then we need the Gibbs function for idealgases.

Specific Gibbs Function for Ideal Gases

Recall that g= h- Ts. Then

dg dh T ds s dT and dh T ds v dP

dg T ds v dP T ds s dT

v dP s dT

= - - = +

= - -( )+

=

Consider an isothermal process with T= constant and on a mole basis

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Chapter 15-7

dg v dP =

For an ideal-gas mixture undergoing an isothermal process, the Gibbs

function for the ith

component, gi , on a mole basis is found by

dg v dP R T

PdP R T

P

P

g T P g T P R T P

P

iT P

T P

i iP

Pu

i

iP

P

ui

o

i i i o ui

o

o

i

o

i

o

i

,

,'

'

' ln

( , ) ( , ) ln

z z z= = =

= +

Lets take Poto be one atmosphere and measure the component partial

pressure Piin atmospheres. Let gi* be the Gibbs function for any component

at the temperature Tand a pressure of 1 atm. Then the Gibbs function

becomes

g T P g T R T Pi i i u i( , ) ( ) ln*

= +

We substitute g T Pi i( , ) into the criterion for chemical equilibrium.

D D C C A A B Bg g g g+ = 0

C C u C D D u D

A A u A B B u B

g T R T P g T R T P

g T R T P g T R T P

[ ( ) ln ] [ ( ) ln ]

[ ( ) ln ] [ ( ) ln ]

* *

* *

+ + +

+ + = 0

Remember that we are trying to find a way to calculate the mole numbers of

components in the product gases for equilibrium at fixed Tand P. Do you

see where the mole numbers are hidden in this equation?

The mole numbers are contained in the expression for the partial pressures.

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Chapter 15-8

P y P N

NPi i

i

total

= =

Now, lets put the last result into a workable form. We define the standard-state Gibbs function change as

G T g T g T g T g T C C D D A A B B* * * * *( ) ( ) ( ) ( ) ( )= +

Substituting we get

G T R T P P P P

R T P P

P P

u C C D D A A B B

uC D

A B

C D

A B

*

( ) ( ln ln ln ln )

ln

= +

=

We define the equilibrium constantKP(T) as

ln ( )

( )

ln

( ) exp( )

( )

*

*

K T

G T

R T

P P

P P

K T G T

R T

or

K T P P

P P

Pu

C D

A B

P

u

PC D

A B

C D

A B

C D

A B

= =

= L

NM

O

QP

=

Table A-28 gives lnKP(T) for several equilibrium reactions as a function of

temperature.

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Chapter 15-9

Now, lets writeKP(T) in terms of the mole numbers of the reacting

components in the real product gas mixture. Using the definition of partial

pressure given above,KP(T) becomes

K T N N

N N

P

NP

C D

A B total

C D

A B( )= F

HGIKJ

where

= + C D A B

Clearly the equilibrium constant is a function of the mole numbers,

temperature, and pressure at equilibrium as well as the stoichiometric

coefficients in the assumed equilibrium reaction. Ntotalis the total moles of

all components present in the equilibrium reaction, including any inert gases.

For values of ln KP(T) given in Table A-28 for several equilibrium

reactions, we make the following observations. Again consider the typical

equilibrium reaction

A B C DA B C D+ +

If

ln ( ) ( )

ln ( ) ( )

K T then K T

K T then K T

P P

P P

< >

0 1

0 1

When ln KP(T) < -7, ( KP(T) 7, ( KP(T) >> 1), the components are so active that the

reaction will proceed to completion (left to right).

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Chapter 15-10

Review the discussion of the equilibrium constant given in Section 15-3.

Example 15-1

Consider the disassociation of H2

H xH yH2 2 +

The equilibrium reaction is assumed to be

H H2 2

For T < 2400 K, ln KP(T) < -8.276, and the reaction does not occur. x 1

andy 0.

For T > 6000 K, ln KP(T) > 5.590, and the reaction occurs so rapidly

that the products will be mostly H with traces of H2. x 0 and y 2.

Example 15-2

Determine the amount of N2that dissociates into monatomic N at 1 atm

when the temperature is 3000 K and 6000 K.

The reaction equation is

N xN zN2 2 +

The nitrogen balance yields

N: 2 = 2X+ Z

or

Z= 2 - 2X

The balanced reaction equation becomes

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Chapter 15-11

N xN x N2 2 2 2 + ( )

The total moles of products is

= + = + ( - ) = - xN x z x xtotal 2 2 2

To solve for x (or to get a second equation that relatesxand z), we assume

the product gases N2and N to be in chemical equilibrium. The equilibrium

reaction is assumed to be

N N2

2

Assuming ideal gas behavior, the equilibrium constantKP(T) is defined by

K T P P

P PP

C D

A B

C D

A B( )=

or

K T N N

N N

P

NP

C D

A B total

C D

A B( )=

FHG

IKJ

K T z

x

P

x

x

x

P

x

P( )

( )

( )

( )

=

F

HG

I

KJ

=

FHG

IKJ

2

1

2 1

2

1

2 1

2

2 2

2

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Chapter 15-12

This becomes

(( )

) (8( )

)42

4 02+ + + =K T

Px

K T

PxP P

The solution to the above equation is the well-known quadratic formula; or,

the solution can be found by trial and error. Note bothxand zmust be

greater than zero or negative moles will result. For z 0, x 1. So a trial-

and-error solution would be sought in the range 0 < x < 1. Recall P = 1

atm.

Using Table A-28 at T = 3000 K, ln KP(T) = -22.359 and

KP(T) = 1.948x10-10

.

Solving for xand then z, we find

x= 1.0 and z= 0.0

The balanced combustion equation when the product temperature is 3000 K

is

N N N2 2

1 0 +

At T = 6000 K, ln KP(T) = -2.865 and KP(T) = 5.698x10-2

.

Solving for xand then z, we find

x= 0.881 and z= 0.238

The balanced combustion equation when the product temperature is 6000 K

is

N N N2 20 881 0 238 + . .

Notice that the product gas mixture at equilibrium has the following

composition

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Chapter 15-13

y or

y or

N

N

2

0881

0 881 0 2380 787 78 7%

0238

0 881 0 2380 213 213%

=+

=

=

+

=

.

. .. .

.

. .. .

Example 15-3

Determine the product temperature for the following when the product

pressure is 1 MPa.

The product pressure in atmospheres is 1 MPa(1 atm/0.1 MPa) = 10 atm.

The balanced reaction equation is

H H H2 20 9 0 2 + . .

The total moles of products are

= 0.9 + 0.2 = 1.1Ntotal

ReactionChamber

H20.9 H2

0.2 H

Qnet= ?

Reactants

TR, PR

Products

PP= 1 MPa

TP= ?

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Chapter 15-14


H H2 2


K T N N

N N

P

NP

C D

A B total

C D

A B( )=

FHG

IKJ

K T

K T

P

P

( ) ( . )( . ) .

.

ln ( ) .

( )

= FHG IKJ

=

=

0 20 9

1011

0404

0 9063

2

1

2 1

For this value of the equilibrium constant, the product temperature is foundin Table A-28 as

TP= 3535 K

If we were to consider the influence of product temperature on the

disassociation of H2at 10 atm, the following results for the mole fraction of

monatomic hydrogen in the products are found.

T(K) KP(T) Mole Fraction of H1000 5.17x10-18

0.00

2000 2.65 x10-6

0.16

3000 2.5 x10-1

14.63

3535 4.04 x10-1

18.18

6000 2.68 x10+2

99.63

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Chapter 15-15

These data show that the larger the product temperature, the larger the

equilibrium constant, and the more complete the reaction for the

disassociation of H2.

Effect of Inert Gases on Equilibrium

Example 15-4

One kmol of CO is combusted with one-half kmol O2and one kmol of N2.

The products are assumed to consist of CO2, CO, O2, and N2when the

product temperature is 2600 K and the product pressure is 1 atm. Write the

balanced combustion equation.

CO O N

X CO W CO Z O N

+ 0.5 +

+ + +

2 2

2 2 2

1

1

The nitrogen does not react with the other components and is inert. Note that

the moles of a given component must be 0, so X, W, and Zall must be 0.

C X W W X X

O X W Z Z X

: = + , = - ,

: + . ( ) = + (2) + , = 0.5X,

1 1 1

1 0 5 2 2 0

( )

The moles of CO at equilibrium must be such that 0 X 1.

Combustion

Chamber

1 CO

O21 N2

XCOWCO2ZO21 N2

QnetReactants

TR, PR

Products

PP= 1 atm

TP= 2600 K

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Chapter 15-16

The total number of moles of product gases at equilibrium is

N X W Z X X X Xtotal = + + + = + ( - ) + . + = . +1 1 0 5 1 05 2


CO CO21

2 O2 +


K T N N

N N

P

NP

C D

A B total

C D

A B( )=

FHG

IKJ

K T X Z

W XX X

X X

P( )

.( . )

( ) .

/ ( )

/ ( / )

=

+

F

H

G I

K

J

= +

FHG

IKJ

+ 1 1 2

1

11

21

1 1 2

1

1 2

1

05 205

1

1

05 2

At T= 2600 K, ln KP(T) = -2.801 and KP(T) = 0.06075.

0 0607505

1

1

05 2

1 1 2

1

1 2

.( . )

( ) .

/ ( / )

= +

FHG

IKJ

X X

X X

Using a trial-and-error method to solve for Xand then Wand Z, we find

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Chapter 15-17

X = 0.212

W = 0.788

Z = 0.106

The balanced reaction equation becomes

CO O N

CO CO O N

+ 0.5 +

+ + +

2 2

2 2 2

1

0 212 0 788 0106 1

. . .

Note: If in the last example, the initial moles of CO were changed from 1 to

2, then the solution for Xwould be in the range 1 X 2.

What is the heat transfer per unit kmol of CO for this reaction?

How do we find the adiabatic flame temperature of a reacting system when

chemical equilibrium is required?

Simultaneous Reactions

When all the products that are not inert but present in the reacting mixture

cannot be expressed in terms of one equilibrium reaction, we must consider

simultaneous equilibrium reactions are occurring.

Example 15-5

Find the equilibrium composition when CO2disassociates into CO, O2, and

O at 3000 K, 1 atm, according to the following reaction equation:

CO a CO b CO c O d O2 2 2 + + +

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Chapter 15-18

We assume the equilibrium reactions to be

CO CO

O

2

1

2

2

O

O

2

2

+

We note that there are four unknowns, a, b, c, and dso we need four

equations relating them. These equations are the species balance equations

for C and O and the equilibrium constant equations for the two assumed

equilibrium reactions.

C a b b a a

O a b c d a c d b

: = + , = -

: , = - - ,

1 1 1

2 2 2 1 2 0

,

= + + +

The moles of each component at equilibrium must be such that the mole

numbers are greater than or equal to zero.

The total number of moles of product gases at equilibrium is

N a b c d c d c d c d

c d

total = + + + = + + + +

= + +

( ) ( )1 2 2

1

For the first assumed stoichiometric reaction

CO CO21

2 O2 +

and assuming ideal-gas behavior, the equilibrium constantKP1(T) is defined

by

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Chapter 15-19

K T N N

N N

P

NP

C D

A B total

C D

A B1( )=

FHG

IKJ

K T b c

a a b c d P1

1 1 2

1

11

21

1( )

/ ( )

=+ + +

FHG

IKJ

+

For the second assumed stoichiometric reaction

O O2 2

and assuming ideal-gas behavior, the equilibrium constantKP2(T) is defined

by

K T N N

N N

P

NPC D

A B total

C D

A B2( )=

F

HG

I

KJ

K T

d

c a b c d P2

2

1

11

21

1( )

( )

=+ + +

FHG

IKJ

+

Therefore, we have four unknowns (a, b, c, d) and four equations that may

be solved by trial and error.

So,

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Chapter 15-20

a c d

b a c d c d

N a b c d c dtotal

= 1 2

1 1 1 2 2

1

= = = +

= + + + = + +

( )

K T c d c

c d c d

K T d

c c d

P

P

1

1 1 2

1

1 1 2 1

2

2

1

2 1

2

1 2

1

1

1

1

( )( )

( )

( )

/ ( / )

( )

= +

+ +

FHG

IKJ

=+ +

FHG

IKJ

+

At 3000 K, KP1(T) = 0.327 and KP2(T) = 0.0126.

SinceKP2(T) is small, we expect d to be small; and since a 1 , c 1/2 .

So guess c, solve for d from KP2(T), and check both values in KP1(T) .

Repeat until KP1(T) is satisfied within some small error. The results are

a= 0.557

b= 0.443

c= 0.210

d= 0.0246

and the balanced reaction equation is

CO CO CO O O2 2 20557 0 443 0 210 0 0246 . . . .+ + +

How do we find the adiabatic flame temperature of a reacting system whensimultaneous chemical equilibrium is required as in the following reaction?

CO O a CO b CO c O d O2 2 2 2+ + + +

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Chapter 15-21

Phase Equilibrium

The criterion that the Gibbs be a minimum also applies to equilibrium

between phases. Consider the equilibrium of saturated liquid water and

saturated water vapor at 100oC. Let gbe the specific Gibbs function per unit

mass, G/m.

G m g m g f f g g= +

At a fixed T, the gfand ggare constant, and the Gibbs function changes

because mass is changing between the liquid and the vapor states.

dG g dm g dmf f g g= +

and, by conservation of mass,

dm dm dm

dm dm

f g

g f

= + =

=

0

At equilibrium for constant Tand P, dG= 0

dG g g dm

g g

f g f

f g

=

=

( )

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Example 15-6

Show that the Gibbs functions for saturated liquid water and saturated water

vapor at 100oC are equal.

g h Ts kJ

kgK

kJ

kg K

kJ

kg

f f f= =

=

419 04 37315 13069

68 6

. . ( . )

.

g h Ts kJ

kgK

kJ

kg K

kJ

kg

g g g= =

=

267614 37315 7 3549

68 4

. . ( . )

.

The saturated liquid and saturated vapor specific Gibbs functions are in close

agreement. Some property table preparers use the equality of the saturatedliquid and saturated vapor Gibbs functions as another property relation in the

required calculations for generating the table values.

chpt15sg

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