2012/3/29 1

Chapter 8Thermodynamic Propertiesof Mixtures

2012/3/29 2

Abstract

The thermodynamic description ofmixtures, extended from pure fluids. The equations of change, i.e., energy and entropy balance, for mixtures are developed. The criteria for phase and chemical equilibrium in mixtures

2012/3/29 3

8.1 THE THERMODYNAMIC DESCRIPTION OF MIXTURES

( )( )

( )1 2

Thermodynamic property for pure fluids, = , , where is the number of moles.

= , where the number of mole equals to 1.

Thermodynamic property for mixtures, = , , , , , where c

T P N N

T P

T P N N N

θ θ

θ θ

θ θ L

( )

( ) ( )( )

1 2

1 2 1 2

1 2

is the number of moles of the ith component.

= , , , , , where is the mole fraction of the ith component.

For example= , , , , , or = , , , , ,

= , , , , , o

i

c i

c c

c

N

T P x x x x

U U T P N N N U U T P x x x

V V T P N N N

θ θ L

L L

L ( )1 2r = , , , , ,

cV V T P x x xL

2012/3/29 4

( ) ( )1 2 11

, , , , , ,

where is the molar internal energy, is

(8.

the internal ener

)

g

1-1

Summation of the proper before mixing at ties of pure fluids ( )and C

ic ii

i

U T P x x x x U T P

U U

T P

−=

= ∑L

( ) ( )1 2 11

yof the pure i-th component at and .

ˆ ˆ, , , , , , (8.1-2 )

where is the mass fraction of component i.

C

c i ii

i

T P

U T P w w w wU T P

w

−=

= ∑L

2012/3/29 5

25 cc

At the same T and P

+ 25 cc

H2O H2O

50 cc

48 cc25 cc + 25 cc

BA

52 ccor

- 2 cc + 2 cc

Attractive Repulsive

2012/3/29 6

( ) ( )

( ) ( ) ( )

( ) ( ) ( )

mix1

mix1

mix1

, , ,

Volume change upon mixing

, , , ,

Enthalpy change upon mixing

, , , ,

Property change upon mixing (at constant and )C

ii ii

C

ii ii

C

ii ii

T P x x T P

V T P V T P x x V T P

H T P H T P x x H T P

T P

θ θ θ=

=

=

Δ = −

Δ = −

Δ = −

∑

∑

∑

Experimental data : properties changes upon mixing (H and V)

2012/3/29 8

Figure 8.1-1 Enthalpy-concentrationdiagram for aqueous sulfuric acid at0.1 MPa. The surfuric acid percentageIs by weight. Reference states: The Enthalpies of pure liquids at 0oCAnd their vapor pressures are zero.

2012/3/29 9

Figure 8.1-2 (a) Volume change on mixing at 298.15 K: o methyl formate +methanol, methy formate + ethanol. (b) Enthalpy change on mixing at 298.15 K for mixtures of benzene (C6H6) and aromatic fluorocarbons (C6F5Y), with Y = H, F, Cl, Br, and I.

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Equations relating molar and partial molar properties

( )

( )

( ) ( ) ( ) ( )

( ) ( )

1 2

1 2

1 2

1 21 2, , , , , ,

1 1, ,

, , , , ,At constant and ,

, , ,Total differential of is:

, ,

j j j

j i

c

c

ccT P N T P N T P N c

c c

iii ii T P N

N T P N N NT P

N N N NN

N N Nd N dN dN dN

N N N

NdN T P x d

N

θ φ

θ φθ

θ θ θθ

θθ

≠ ≠

≠

≠

= =

=

=

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

⎛ ⎞∂= =⎜ ⎟∂⎝ ⎠

∑ ∑

L

L

L

( ) ( )

( )

, ,

1

, ,

Definition of the par

tial molar property

(8 .1-12) j i

i

i

i T P N

c

i ii

N

NT P x

N

d N dN

θθ

θ θ

≠

=

⎛ ⎞∂= ⎜ ⎟∂⎝ ⎠

= ∑

2012/3/29 11

At a constant T and P

50 cc

A + B

+

θ

1 mol of A θ + Δθ

Δθ is a partial molar property (at Constant T, P, and NB)

Total property and partial molar property

2012/3/29 12

( )

( )( )

( ) ( )

1

1 1

1 1

1

( b)a) (

(a)

(

b)

c

i ii

i i i i

c c

i ii i ii i

c c

i ii ii i

c

i ii

d N dN

d N Nd dN

dN d Nx Ndx x dN

Nd dN d N dN Ndx x dN

d dx x

d

dN N

N

N

d

N

x

d

θ θ

θ θ θ

θ θ θ θ θ

θ θ θ θ

θ θ θ

=

= =

= =

=

←⎯⎯

=

= +

⎯

= = +

+ = +

+ = +

⎡ ⎤− + −⎢

→

⎥⎣

⎯

⎦

∑

∑ ∑

∑ ∑

∑1

0

The and are the number of moles in system and changing in numberof moles, respectively. T arehus, and arbitrary and not equal to zero .

c

i ii

x dN

N dNN dN

θ=

⎡ ⎤=⎢ ⎥

⎣ ⎦∑

2012/3/29 13

( )

1

1

1

1

(c)

, , (8.1-13)

Euler Theorem def

0; resulted from 1st term of LHS

0; resulted

ined by (8.1-13)

from 2nd term of

means

LHS c

i

c c

i ii ii i

c

i i iii

d dx

T P x x

d dx

x

θ θ θ θ

θ θ θ θ=

= =

=

− = =

− ==

∑

∑ ∑

∑

( )

1

1 1

(c

1 1

)

1

1

Since:

can be c

alculated from

0

, , . c

i ii

c c

i ii ii i

c c c

i i ii i iic

i

i

i

ii

i

x

d dx x d

d

T P x

x

x

d d x d

d

x

θ θ

θ

θ θ θ

θ θ

θ

θ

θ θ

=

= =

= = =

=

=

= +

⎯⎯ =

=

← +

∑

∑ ∑

∑

∑

∑ ∑

(8.2-9b)

Gibbs Duhem equation a

t constant

and

.

T P⇒

2012/3/29 14

2012/3/29 15

25 cc

Property change upon mixingat constant T and P

+ 25 cc

H2O H2O

50 cc

48 cc25 cc + 25 cc

BA

52 ccor

- 2 cc + 2 cc

Attractive Repulsive

2012/3/29 16

N1

N2

θ1

θ2

i

i ix

θ

θ θ= ∑i

i ix

θ

θ θ= ∑

Beforemixing

Aftermixing

Constant T and P

2012/3/29 17

Properties change upon mixing (i)

( )

( )

( )

The total volume and enthalpy of the pure components are

,

,

whereas the v

unmixed

the mixture at the same

ol t

ume empe

and enthrature a

alpy , , of

are, from Eq.nd pressu

r

8.1 3,e

-1

C

iiiC

iii

i

V N V T P

H N H T P

T P x

V T

=

=

∑

∑

( ) ( ) ( )

( ) ( ) ( )

1 2

1 2

, , , , , , , ,

, , , , , , , ,

C

iiiC

iii

P N N NV T P x N V T P x

H T P N N N H T P x N H T P x

= =

= =

∑

∑

L

L

2012/3/29 18

Properties change upon mixing (ii)

( ) ( ) ( )

( ) ( )

( ) ( ) ( )

mix 1 2 1 2

mix 1 2 1 2

The isot

, , ,

(8.1-1

hermal volume change on mixing,

, , , , , ,

, , , 4)

, , , , , , , , ,

C

iii

C

i iii

C

iii

ii

T P TV N N V N N N V

N V T P x V T P

H T P N N H T P N N

P T

N H T P

N H

PΔ = −

⎡ ⎤= −⎣ ⎦

Δ = −

=

∑

∑

∑

L L

L L

( ) ( ), , , (8.1-15) C

ii

T P x H T P⎡ ⎤−⎣ ⎦∑

2012/3/29 19

Thermodynamic relations for partial molar properties

j i j i j i, , , , , ,

Also,

For each relationship among the ththe identical relations

ermodynamic variables in a pure flre exists

id, an

u

i i iT P N T P N T P N

i i i

ii i

A U T S

N A NU NT SN N N

A U T S

G H T S

≠ ≠ ≠

= −

⎛ ⎞ ⎛ ⎞ ⎛ ⎞∂ ∂ ∂= −⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= −

= −

for the partial molar thermodynamic properti in a mixt

hes

ipure.

2012/3/29 20

o

ILLUSTRATION 8.1-lCalculating the Energy Release of an Exothernuc Mixing Process

and are mixed isothermally Three moles of water at . How muchh

oe

ne mole of suat must be ab

lfso

uric arbed o

cr

0 d Ci rele oased to keep the mixture at 0 C?

SOLUTIONWater has a molecular weight of 18.015, and that of sulfuric acid is 98.078. Therefore, themixture will contain 3 18.0154 + 1 98.078 = 152.12 g, and will h× × ave a ( ) composition of98.078 g 100% 64.5 wt% of sulfuric acid152.12 g

the enthalpy of the mixture is about 315 kJ/kg. Therefore, when 3 mol waterand 1 mol sul

mass

furiFrom Fig. 8.

c acid are m e1

d -1

ix

× =

−

( ) ( )mix

o o1

mix 1 1 2 2

2

isothermally,kJˆ ˆ ˆ 315kg

, so that a total of 315 kJ/kg 0.152 kg= 47.9 kJ

of

ˆ

ˆ ˆsince 0 C = 0 and 0 C

must beenergy to keep the mixture at a constant tem remov peratued re

H w HH

H T H

w H

T

Δ = − − = −

−= = − ×oof 0 C.

2012/3/29 21

Figure 8.1-1 Enthalpy-concentrationdiagram for aqueous sulfuric acid at0.1 MPa. The sulfuric acid percentageIs by weight. Reference states: The Enthalpies of pure liquids at 0oC And their vapor pressures are zero.

-315

64.5

2012/3/29 22

Sulfuric acid and water are said to mix exothermically since energy must be to theenvironment to mix these two components at con

releasedThe temperatstant ure r temperature. that

occi

use

rs when

COMMENT

these two components are mixed is considered in Illustration 8.4-1.Note also that to solve this problem we have, in effect, used an energy balance without explicitlywriting a

adiabalica

detaile

lly

d balance equation.

2012/3/29 23

Partial molar property

A + B

+θ 1 mol of A θ + Δθ

Δθ is a partial molar property (at Constant T, P, and NB)

2012/3/29 24

Physical meaning of partial molar property

( ) ( ) ( )1 21 2 1 2

1

For a binary mixture,

, , , , , , , Now suppose we of species 1, , which is s

(8.1-17a)ad o small comd a sma pared

with the total numberll amoun

of t

moles

T P N N N T P x N T P xN

θ θ θ= +

Δ

( ) ( ) ( ) ( )

1 2

11 1 21 2 1 2

of species 1 and 2 that and are essentially , In this case we have

, , , , , , , Subtracting Eq. 8.1-17a from Eq. 8.1-17b, we find that the change

(8.1-17b)

unchangedx x

T P N N N T P x N T P xN Nθ θ θΔ Δ+ = + +

( ) ( ) ( )1 1 2 1 12 1

in the property is

= , , , , , ,

Therefore, the amount by which a small addition of a species to a mixture changesthe mixture property is equal to the product of the amount

,

a

,T P N N T P N N xNN T Pθθ θ θΔΔ + − = Δ

how the species behaves in a dded and its partial molar

property, that is, , and not its pure componentpro

mixpe

turerty.

2012/3/29 25

8.2 THE PARTIAL MOLAR GIBBS ENERGY AND THE GENERALIZED GIBBS-DUHEM EQUATION

Since the Gibbs energy of a multicomponent mixture is a function of temperature, pres-sure, and each species mole number, the total differential of the Gibbs energy functioncan be written as

GdGT

∂⎛= ⎜ ∂⎝j�

1, , , ,

1

The partial molar Gibbs energ

(8.2-

y

1)

,

,

i i

C

iiP N T N i T P N

C

i ii

ii

G GdT dP dNP N

SdT VdP G d

G

N

μ

=

=

⎛ ⎞∂ ∂⎞ ⎛ ⎞+ + ⎜ ⎟⎟ ⎜ ⎟∂ ∂⎠ ⎝ ⎠ ⎝ ⎠

= − + +

=

∑

∑

j�

j�

1, , , ,

1 , ,

t

he chemic

al potential.

(8.2 -2)

i i

C

iiS N P N i P S N

C

ii i P S N

H H HdH dP dS dNP S N

HVdP TdS dNN

=

=

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

⎛ ⎞∂= + + ⎜ ⎟∂⎝ ⎠

∑

∑

j�, , , ,

i

i iP S N T P N

H H HN N

⎛ ⎞ ⎛ ⎞∂ ∂≠ =⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

2012/3/29 26

1

1

(8

Comparing Eqs. 8.2-2 and 8.2-3 establ

.2

ishes t

-3)

C

iC

i ii

i i

H G TS

dH dG TdS SdT TdS SdT

Vd

SdT VdP

P Td

dN

S N

G

G d

=

=

= +

= + + = + +− + +

= + +

∑

∑

j i j i, , , ,

hat

Using the procedure established here, it is also easily shown that

(8.2-4)

P S N T P N

i

i i

i i

H GGN N

dH TdS PdV G dN

≠ ≠

⎛ ⎞ ⎛ ⎞∂ ∂= =⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

= − +

j i j i j ij i, , , ,

1

1

, , , ,

(8.2-5)

i

C

iC

i ii

i

i P S N T P N S V N Ti Vi i N

dA SdT PdV G dN

H G U AGN N N N

μ≠ ≠ ≠≠

=

=

= − − +

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

∑

∑

Chemical Potential

2012/3/29 27

Generalized Gibbs-Duhem equation

( )( )

( ), , , ,

Any thermodynamic function can be written as

(8

, ,

.2-6)

i i j i

ii

i ii i

i

C

iP N T N T Pi N

N N

d N N d dN

N N T P N

N N Nd N dT d dP N

PT

θ

θ θ

θ θ θ

θ θ

θ θ θθ≠

=

= +

=

⎛ ⎞∂ ∂ ∂⎛ ⎞ ⎛ ⎞=∂ ∂

+ + ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠∂

∑∑ ∑

∑

( ), ,

, ,

Subtracting Eq. 8.2-7 from Eq. 8.2-6 gives the relation

=0

and divi

(8.2-7)

(8.2-8a)

i i

i i

i

C

i iiP N T N

iiP N T N

N

d N dT dP dNT P

N dT N dP N dT

N

P

N θ θθ θ

θ θ θ

∂ ∂⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

∂ ∂⎛ ⎞ ⎛ ⎞− − +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

∑

∑

, ,

(8

ding by the total number of moles yields

=0 .2 8 )-

b

The Generalized Gibbs-Duhem equation.i i

iiP N T N

N

dT dP x dT Pθ θ θ∂ ∂⎛ ⎞ ⎛ ⎞− − +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

∑

2012/3/29 28

, ,

=0

=0

(8.2-8b)=0

(8.2-9a)

Gibbs-Duhem equation at constant and

i i

ii

ii

iiP N T N

N d

x d

dT dP x dT P

T Pθ

θ

θ θ θ∂ ∂⎛ ⎞ ⎛ ⎞− − +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

∑∑

∑

,

Finally, for a change in any property at constant temperatureand pressure, Eq. 8.2-9a can be rewritten as

=0

(8.2-9b

)

ii

T P

Y

YdN θ⎛ ⎞

⎜ ⎟∂⎝ ⎠∑

, ,

whereas for a change in the number of moles of species j at constant temperature,pressure, and all other mole numbers, we have

(8.2-10)

k

ii

T Nj P

dNNθ⎛ ⎞

⎜ ⎟⎜ ⎟∂⎝ ⎠i=0 (8.2 1 1

) -j

C

≠

∑

2012/3/29 29

1

1

The Gibbs-Duhem equations for the , obtained by setting = in Eqs. 8.2-8, are

0

0

Gibbs energy

(8.2-

12a)C

iiiC

iii

G

SdT VdP N dG

SdT VdP x dG

θ

=

=

= − +

= − +

∑

∑

1

1

0

(8.2-12b)

At constant and ,

(8.2-13a)

0

C

iiiC

iii

N dG

x d

T

G

P

=

=

=

=

∑

∑

1 ,

1 , ,

0

(8.2-13b)

(8.2-14)C

ii

i T P

Ci

ii j T P N

GNY

GNN

=

=

⎛ ⎞∂=⎜ ⎟∂⎝ ⎠

⎛ ⎞∂⎜ ⎟⎜ ⎟∂⎝ ⎠

∑

∑ 0 (8.2-15) k j≠

=

2012/3/29 30

2

1

1 21 2

1 21 2

1 1

1 1

, ,

1 21 2

, ,

For a binary mixture,

0 at constant and

0

0

0

(8.2-19b)

=

iii

T P T P

T P T P

x d T P

x d x d

x x

G

x

x x

G Gx

x x

θ

θ θ

θ θ

θ

=

=

+ =

⎛ ⎞ ⎛ ⎞∂ ∂+ =⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞∂ ∂+ =⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

∑

(8. 2-20)

STOP HERE

2012/3/29 31

8.3 A NOTATION FOR CHEMICAL REACTIONS

i(8.3-1

0I 0

is the stoichiometric coefficient of species I, so defined that is positive for reaction

)

pro

i

i i

A B RR A B

α β ρρ α β

ν

ν ν

+ + → ++ − − − =

=∑

L L

L L

i,0

, negative for reactants, and equal to zerofor . Using to represent the number of moles of species in a closed system at

du

anytime , and for the initial

ctsi

numb

nert spec

er of mol

ies

es of siN i

t N i,0

i,0

pecies , then and are relatedthrough the reaction variable , and the stoichiometriccoefficient by

, the molar extent of rea

ctio

n

i

i

i i

i N N

N N X

Xνν= +

i,0

(8.3-2a)

(8.3-2 b) i

i

N NX

ν−

=

2012/3/29 32

i,0

i,0

rxn

(8.3-2a)

(8.3-2b)

i i

i

i

ii i

N N XN N

X

dN dX Xdt dt

ν

ν

ν ν

= +

−=

⎛ ⎞ = =⎜ ⎟⎝ ⎠

&

( ) ( ) ( )i,0 i,0

The total number of moles in a closed system at any time is

(8.3-3)

(8.3-4)

C C C C

i i ii i i i

N N N X N Xν ν= = + = +∑ ∑ ∑ ∑

2012/3/29 33

the smallest collection of reactio that, on forming Indepenlent reactions= various linear combinations, includes all possible

chemical reactions

n

amon e

s

g th

independent chemicaThe con l ce rept o acf tions

{ }

2 2

2

2 2

species present

For example, three reactions between carbon and ox

No reaction in the set = a linear combinatio

ygen:C + O = CO2C + O = 2CO2CO + O =

n of the other

2COIf we

s

add the seco

⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪⎩ ⎭

nn

od

t and

indethird of

pendent these equations, we get twice the first, so these three

reactions are . any of the three reactions form In anin

tde

thispend

woen

case, t set.

2012/3/29 34

Remarks on the chemical reacting system

(1) To describe a chemically reacting system it is necessary to consider all the chemical reactions that can occur between the reactant species

no

t,

the independent reactionsthe molar extent of reaction for chemical react

. (2) Furthermore, among the species can be computed from an appropriate linear combination of the

on

ion

lyany

known extents of reaction for the set of chemical independe reactiont ns.

2012/3/29 35

In the Denbigh's method, the procedure to get the independent chemical reactionsare as follows: (1) w equations for the formatirites the stoichiometric , from its constituent , of

oneac

atom

sh of present in the chemical reaction system.

(2) One of the equations that contains an in

the molecular

the atomatomic spe

ic state acies actually pr

t the reaction co

sp

ndi

eci

tio

esnot

ns isesent

then used, by and/or , that atomic species from the remaining equations. (3) In this way the number of stoichiometric equations is red

additionsubtrac

uced by one.

t

io

to

T

elimi

he pr o

naten

ceduThe equations that remain form a set of indep

until all atomic species pendent chemical reactions

resent have been eliminated. (4) ; the molar extents of reactio

n

n

re is repeate

for these

otd

reactions are the variables to be used for the description of the multiple reaction system and to follow the composition changes in the mixture.

Denbigh’s method (1981)

2012/3/29 36

An example using Denbigh’s method

2 2

2

Considering the reaction: the oxidation of carbonMolecular species: O , CO, CO , and atomic species

the reaction of formation for each mo

Cwere found in

lecular speciesthe system,

write

2 = O O 2

2

2

2 2

= O

C = COC 2 = COSince a free oxygen atom does not occur, the first equation is used from the other two equations to obtain2C + O =

12

to eliminat 2CO

C +

e

O = C

O

O

l

O

O

OA

++

l the remaining atomic species in these equations (here only carbon) are present inthe reaction system, so no further reduction of the equations is pos thesetwo equations form a se

sible, t of i

and ndependent reactions.

Pop-quiz

A system contains the following substances: N2(g), O2(g), NO(g) , NO2(g), N2O(g), N2O4(g), and N2O5(g), find an set of independent reactions for the system.

2012/3/29 38

For multiple-reactions (j reactions)M

i,0 ij jj=1

M Mj

ij ij jj=1 j=1rxn

j j

ˆFinally, with = / defin

(8.3-5)

(8.3

ed to be

-6)

i

i

N N X

dXdN Xdt dt

X X V

ν

ν ν

= +

⎛ ⎞ = =⎜ ⎟⎝ ⎠

∑

∑ ∑ &

( )M M Mj

ij ij j ijj=1 j=1 j=1rxn

M M

ij j ij jj=1 j=1

j

the molar extent of reaction

,

ˆˆ +

ˆ = +

per unit v

olumeˆ

(8. 3

id XdN V X V

dt dt dt dt

dV X V r

dX dV

dt

ν ν ν

ν ν

⎛ ⎞ = =⎜ ⎟⎝ ⎠

∑ ∑ ∑

∑ ∑

jj

ˆ , , is the reaction variable most

frequently used by chemists and chem

-7)

t

ical engineers in chemical reactor analysi

he rate of reaction per unit volume

s.

dXr

dt=

2012/3/29 39

8.4 THE EQUATIONS OF CHANGE FOR A MULTICOMPONENT SYSTEM

( )

( )

( )

i1

ii

1 rxn

ii

Rate at which species i is being is entry port

produced by chemical reaction

Species mass balance for a reacting systemK

ik

k

K

kk

k

dN N k kthdt

dNNdt

dN Ndt

=

=

⎛ ⎞= + ⎜ ⎟

⎝ ⎠⎛ ⎞= + ⎜ ⎟⎝ ⎠

=

∑

∑

&

&

&ij

11j

We can obtain a balance equation on in the system by sumt -he

mi

tot

ng E

al numbe

q. 8.4-l

(8.4-1a)

r of moles

over all specia es , reco gni i

M

j

K

kXν

==

+ ∑∑ &

( ) ( )

( )

i i1 1 1 i 1

i1

1

ii 1

zing that = is the total number

of moles, and that

where is the total molar flow rate at the kth entr y port.

C C

k ki k k

K K

C

i

kk

k

K

C

k

N N

N

NN

N

N= = = = =

=

=

= =

=

∑∑ ∑∑

∑

∑

∑& & &

&&

2012/3/29 40

Total mass balance fora reacting system

C

ij j1 i=1 j=1

the number of mole

Since is aconserved quantity, we need inf

(8.4-1b)

the total number of moless of speciormat

neither noro

esi

Molar basisK M

kk

dN N Xdt

ν=

= +∑ ∑∑& &

n on the at which all chemical reactionsoccur to use Eqs. 8.4

rates-1.

2012/3/29 41

Total mass balance fora reacting system

Although we will be interested in the equations of change mainly on a ,for completeness and for several illustrations that follow, some of the equations ofchan

molar basis

ge will also be give

Mass Basis

i

i i ii k i i

n on a . To obtain a balance equation for the mass ofspecies , we need only multiply Eq. 8.4-1a by the molecular weight of species . ,

and use the notation = and (

mass basis

) = (

i i m

M N M m Nm & &k

ii k i ij j

1 j=1

k1

k i k1

) to get

( )

(8.4-2a)

(8.4-2b)

where ( ) .

In Eq.

K M

k

K

kC

i

dM M m Xdt

dM Mdt

M M

ν=

=

=

= +

=

=

∑ ∑

∑

∑

& &

&

& &

C

i ij ji=1 j=1

the chemical reaction term vanishes since total m8.4-2b

ass is a conserved quantity.

0

M

m Xν =∑∑ &

2012/3/29 42

Energy balance for a reacting system

( )

( )

( ) ( )

1

1

i i1

(molar basis)

(mass basis)

(8.4-

3)

ˆ

K

SkkK

SkkC

k ki

dU dVN Q W Pdt dtdU dVM Q W Pdt dt

N H N H

H

H

=

=

=

= + + −

= + + −

=

∑

∑

∑

&& &

&& &

& &

i i1

(8.4-4)

C

iU N U

=

= ∑

2012/3/29 43

Figure 8.4-1 A simple stirred-tank reactor.

Continuous-flow stirred-tank reactor

2012/3/29 44

Mass/Energy balances for asteady state process

( ) ( )

( ) ( )in out

C C

in outi=1 i=1

(8.4-5a)

(8.4-5b)

the very s

For a stead

imple isothe

y-state process

0 (singl

0

e reactio

rmal

F

n)

r co

ii i i

i i i i

dN N N Xdt

dU N H N H Qdt

ν= = − +

= = − +∑ ∑

& & &

&& &

asewith the ass

in which thumptio

e inlet and outlet streams and thereactor contents are all at temperature , and that the partial molarenthalpy of each species is just equal t

pure

n-compoo its nent e

T

( ) ( )( ) ( )

C

out ini=1

out in

, we obtain

=

Now if there were and

should b

the heat flow rate

to maintain the constant temperatur

no chemical re

n

e . However, when e equal to zer

thalpy

actio

o

n

ii i

i i

Q N N H

N N Q

T

⎡ ⎤−⎣ ⎦

=

∑& & &

&& &

( ) ( )

( ) ( )

out in

C C

out ini=1 i=1 i

a chemical

reaction occurs, and , and the steady heat

flow required to keep the reactor at constant te

are not

mperature is

equal in magnit

=

udei i

i i ii i ii

N N

Q

Q N N H X H X Hνν⎡ ⎤− = =⎣ ⎦∑ ∑

& &

&

& & & &&C

rxn=

rx

1

n can be calculated by and .H Q X

H X

Δ

= Δ∑& &

&

2012/3/29 45

Entropy balances for asteady state process

( )

( ) ( )

( )

gen=1

1

gen

2 22

1 1

For a steady-state process

0 +

where

1

(8.4-6)

(8.4- 7)

K

kk

C

i ik ki

s

M C

s ij i jj i

dS QNS Sdt T

NS N S

S dV

T G XT T T

σ

λ μσ φ ν

=

= =

= = +

=

=

= Δ + −

∑

∑

∫

∑∑

&&&

& &

& &

&&

2012/3/29 46

2012/3/29 47

Table 8.4-1 continued

2012/3/29 48

2012/3/29 49

Table 8.4-2 continued

2012/3/29 50

o

ILLUSTRATION 8.4-1 on Adiabatic Mixing of an Acid and Water

Three moles of water and one mole of sulfuric acid, each at 0 C, a adiabaticalre mixed . Usethe data in

Temperature Change

Fig. 8.1- an1ly

d the

2 2 4H O H SO

information in Illustration 8.1-1 to of the mixture.SOLUTIONFrom the c

estimate the final temperature

losed-system mass balance, we have3 18.015 1 98.078 152.2 kg

and from the energy bala

fM M M= + = × + × =

2 2 2 4 2 4mix H O H O H SO H SO

nce, we haveˆ ˆ ˆ 3 1 0 kJ

Thus finally we have a mixture of 64 that has an enthalpy of 0 kJ/kg.5 wt % sulfuric acid , (Notethat we have used the fact that for li

0

q

0f f i fU H H M H M H M H= = = = + = × + × =

uids and solids at low pressure, ). From Fig. 8.1-1 we see thal a mixture containing

the internal energy andenthaipy are essentially equa 64.5 wt %sulfuric acid has an enthalpy of 0 kJ/kg at b

l,ou

a

U H=o

othe mixture will act . Therefore,

hieve a temperaif water and sulfuric acid are

adiabalically mixed in the ratio of 3:1, , just below the boiling poin

ture of 150t of the mi

whichxtureis . T

15

h

0 CC

is lm

aa

rgke

e tes mi

mperaturxing sul

e rise, and the furic acid and w

fact that themixture is just bel ater an operation thatmust be do

ow its boiling pne with extreme

oint, care.

2012/3/29 51

Figure 8.1-1 Enthalpy-concentrationdiagram for aqueous sulfuric acid at0.1 MPa. The sulfuric acid percentageIs by weight. Reference states: The Enthalpies of pure liquids at 0oC And their vapor pressures are zero.

-315

≈150 oC

2012/3/29 52

o

o

COMMENTIf instead of starting with refrigerated sulfuric acid and water (at 0 C), one started with thesecomponents and mixed them adiabatically, the resulting 3:1 mixture would be in at 21 the.2 Cliquid + vapor region; that is, the mixture would boil (and splatter). Also note that because ofthe shape of the curves on the enthalpy-concentration diagram, adding sulfuric acid to wateradiabalically (i.e., moving to the right from the pure water edge of the diagram) results in amore gradual temperature rise than adding water to sulfuric acid (i.e., moving to the left fromthe pure-sulfuric acid edge sulfuric acid should be added tow

). Therefore, whenever possible, , and not vice versa. ater

2012/3/29 53

o

ILLUSTRATION 8.4-2Mass and Energy Balances on a Nonreacting SystemA continuous-flow steam-heated mixing kettle will be used to produce a 20 wt % sulfuric acidsolution at 65.6 C from a solution of 90 w o ot % sulfuric acid at 0 C and pure water at 21.1 C.Estimate

a. needed per kilogram of initial sulfuric acid solution to produce a mixture of th

The kilogre desired

ams of pure wateconcentratio

rn

b. per kilogram of initial sulfuric acid solution to heat the mixture to

The amount of heat needed

The temperature of the kettle effluent the desired temperature

c. if the mixing process is carried out adiabatically

2012/3/29 54

SOLUTIONWe choose . The difference form of the equationsof change will be used for a time interval in which

the contents of the mixing 1 kg of concentrated sulf

kettle as the systuric acid enter

ems the

( )2 4

3 3

H SO1 1

kettle.a. Since there is no chemical reaction, and the mixing tank operates continuously, the total and sulfuric acid mass balances reduce to

0 = and 0 =

Denoting th

k kk k

M M= =

∑ ∑&

1

1

e 90 wt % acid stream by the subscript 1 and its mass flow by , the water stream by the subscript 2, and the dilute acid stream by subscript 3, we have, from the total mass balance, 0 =

M

M +

&

&

( )2 3 1 1 3

3 1

1 2

1 used per kilogram of the 90 wt %

acid. Alwhere Z is equal to

so from the mass balthe number of kilograms

ance on sulfuric acid, we have0 = 0.9

of wate

0. 0

r

0 0 2

M M M ZM M

M Z M

M M

+ = + +

= − +

+ +

& & & & &

& &

& & & ( )3 1 10.90 0.20Therefore,

0.901+Z = 4.5 or Z = 3.50.20

so that 3.5 kg of water must be added to each 1 kg of 90 wt % acid solution to produce a20 wt % soluti .

1

on

Z MM M += −

=

&&

2012/3/29 55

( )s

2 1 3 1

1

b. The steady-state energy balance is

ˆ0

since = 0 and = 0. From the mass balance of part (a).

3.5 and 4.5From the enthalpy-concentration chart, Fig. 8.1-1, we haveˆ

k k

MH Q

W PdV

M M M M

H

= +

= = −

∑

∫

&&

& & & &

( )( )( )

( )( ) ( )

o2 4

o2 2

o3 2 4

ˆ 90 wt % H SO , 0 C 183 kJ/kg

ˆ ˆ pure H O, 21.1 C 91 kJ/kg

ˆ ˆ 20 wt % H SO , 65.56 C 87 kJ/kg

4.5 87 3.5 91 1 183 391.5 318.5 1 256 kJ/kg of initial aci83

c. For adi

d solution

abatic oper

H T

H H T

H T

Q

H

= = = −

= = =

= = =

× − ×= − × − = − + =

( )( )( ) 33 3

o

ation, the enerey balance is

ˆ0

ˆ ˆ0 4.5 3.5 91 1 183 ; 4.5 135.5 and

Referring to the enthalpy-conce

ˆ 30.1 kJ/

ntration diagram, we find that ~ 50 C

k

g

.

k k

MH

HH H

T

=

=

= × − × − × − × =

∑ &

2012/3/29 56

8.5 THE HEAT OF REACTION AND A CONVENTION FOR THE THERMODYNAMICPROPERTIES OF REACTING MIXTURES

( ) ( ) ( )12 2 22

In the ideal gas-phase reactionH g + O g H O git is observed that 241.82 kJ are liberated for each mole of water vapor producedwhen this reaction is run in an , isothermal constant-pressure calori

→

( ) ( )2

2

o

o

H Orxno

H

at 25 C and1 bar with all species in the vapor phase. Clearly, then, the enthalpies of the reac

25 C,

tingmolecules must be related as follows:

vapor,

= 1 bar 25 C, = 1 bar

2

m ter

e

T PH

H

T P

T

HΔ == =

=− ( ) ( )2

1O2

o o

2

kJ241.82mol of H O produced

so that we are free to choose the values of the enthalpy of hydrogen, oxyge

5 C, = 1 bar 25 C,

n, andwater vapor

=

a

1

ll arbitrarily. o

bar

n t

P T PH =− = −

Enthalpy of formation

2012/3/29 57

( ) ( )( ) ( )

2

2 2

2

2 2

o oH Of f

oH O

o o

o o

H Of

1H O

O f

2

H

The enthalpy of formation is

25 C, = 1 bar 25 C, = 1 bar

and the Gibbs energy of formation is . B

25 C, = 1 bar 25 C, =

y definition

1 b

,

v arapor,

T P T P

T P

H

H G

H H

H H T P

Δ Δ

Δ =

− =−

=

=

= Δ

=

2 2 2

o 1H O H O

of

2

of

o

contains a listing of and for a large collection of sub-

stances in at 25

Appendi

C and

x A.IV

1 bartheir nor .Isotherm

mal states of aggral heats (enthalpi

egationes) and Gibbs e

H G

H H H

Δ Δ

+ +

o

nergies of formation of species may besummed to compute the enthalpy change and Gibbs free energy chang

25 C, e that

1 bar would

occur if the molecular species at , and the state of aggregation listed ino oo o

rxn rxnA (25 C, 1 ppendix A.IV. We will denote these changes by and bar) (25 C,1 bar , respective y) l .

H GΔ Δ

Enthalpy of reaction and enthalpy of formation

2012/3/29 58

( ) ( ) ( )( ) ( )

3 H

3

2 2

22

2 2 3

o o o oHNO NOrxn

o oNO H O

o 1HNOf

3O2 2

For the gas-phase reaction 3NO +H O = 2HNO +NO, we have

25 C,1 bar 2 25 C,1 bar 25 C,1 bar

3 25 C,1 bar 25 C,1 bar

2 H H

H H H

H H

H

Δ = +

− −

Δ + += +

2

2

o

o2

o2 H2

o2 H2

3 2

2

2

2

25 C,1 bar

o 1NO Nf 2 25 C,1 bar

o 1NOf 2 25 C,1 bar

oH Of 25 C,1 bar

o o o oHNO NO NO H Of f f f

N

1O2

O

1O2

3

2 3

H H

H H

H H

H H H

H

H

H

H

H

⎡ ⎤⎣ ⎦

⎡ ⎤+ Δ + +⎣ ⎦

⎡ ⎤− Δ + +⎣ ⎦

⎡ ⎤− Δ + +⎣ ⎦

= Δ + Δ − Δ − Δ

( )( ) ( )( ) ( )

o25 C,1 bar

o o

oo o orxn f

oo o orxn

f

f

25 C,1 bar

25 C,1 bar 25 C,1 bar

25 C,1 ba

r 25 C,1 bar

(8.

5-1)ii

ii

ii

H

G G

H

Hν

ν

ν

⎡ ⎤⎣ ⎦

= Δ

Δ = Δ

Δ = Δ

∑∑

∑

(8.5-2 )

At T and 1 bar

2012/3/29 59

( ) ( )( ) ( )

oo o orxn f

oo o orxn f

25 C,1 bar 25 C,1 bar

25 C,1 bar 25 C,1 bar

(8.5-1)

(8.5-

The standard heat of reactio

2)

n

ii

ii G

H H

G

ν

ν

Δ = Δ

Δ = Δ

∑∑

( ) ( )( ) ( )

oorxn f

oorxn f

at any temperature

,1 bar ,

1 ba

(8.5-

r

,1 bar ,

3)

(8.1 5 4)bar -ii

iiG

H T H T

T TG

T

ν

ν

Δ = Δ

Δ = Δ

∑∑

( ) ( ) ( )

( ) ( )( )

o

o

o

o o oP,i

oP,i2

oo orxn f 5 C

oP,i25

o

C

o

P

r

o,i

oxn (

, =1 bar , =1 bar , =1 bar

,1 bar 25 C,1 bar

25 C,1

is

ba 8. -5r 5 )

T

i i T

T

TT

T

ii i

i

H T P H P C T P dT

C dTH T

T

C dT

C

H

H

ν ν

ν

=

=

Δ = Δ

Δ

+

+

+

=

=

∫

∫∑∑ ∫

∑

the heat capacity of species i in its standard state.

2012/3/29 60

Atomic species at 25 oC and 1 bar. O2, H2, N2

H(25 oC, 1 bar)

Molecular species at 25 oC and 1 bar. H2O, HNO3, NO2

Enthalpy of formation

Heat of reaction at 25 oC and 1 bar.

Heat of reaction at T and 1 bar.

o

oP,i25 C

T

Ti C dTν=

+ ∑ ∫

Heat of combustion

2012/3/29 61

o

6 6 2 6 12

ILLUSTRATION 8.5-1Calculation of the Standard Heat of Reaction at 25 CCompute the standard heat of reaction for the hydrogenation of benzene to cyclohexane,C H + 3H C H from t standard-heahe t-of→ .SOLUTIONThe standard heat of reaction can, in principle, be computed from Eq. 8.5-3;

-combustion data

we will use the heat of combustion for cyclohowever, for

illustration, From the stanh dexane. ard-h

( ) ( )

6 12 2 2 26 12

oc

6 12 2 2 2

oC H CO H O C Oc H

eat-of-

combustion data in Appendix A.V, we have = 3,919,906 J/mol of forthe following reaction:C H l 9O 6CO + 6H O l

Thus

cycloh

6 6

exan

9 3 1990

e

9H

H

H H H H

Δ −

+ →

Δ = + − − = −

6 12 2 2 2

6 6 6 6 2 2 2

2 2 2

6 1

2

2

oC H O CO H Oc

o 1C H C H O CO H Oc 2

o 1H H O H Oc 2

C H

6 J/mol

or 9 6 6

Similarly, 7 6 3

and 3 1

3

H H H H

H H H H H

H H

H

H H

= −Δ − + +

= −Δ − + +

= −Δ − +

2012/3/29 62

( )

6 12 6 6 2

6 12

6

2

2

2

2

226

2 H O

H

oC H C H Hrxn

oC H O

1O2

1

c

oC Hc

oHc

CO

OCO

Therefore, 3

9

7

1

6

36

6

H H H H

H H

H

H

H

H

H

HH

Δ = − −

= −Δ − + +

− −Δ + +

−Δ −−

−

( )

6 12 6

2

6 2

2

o o oC H C H Hc c c

H OO2

oic

i

J 3,919,906 3,267,620 3 285,840 205,234mole of benzene

kJ 205.

3

23mo

i

H

H

H

H H H ν

+

= −Δ + Δ + Δ =

= − − × =

Δ

=

−

−

−

∑

le of benzene

2012/3/29 63

o orxn c

COMMENT

Note that in the final equation, = , the enthalpies of the reference-stateatomic species cancel, as they must due to conservation of atomic species on chemical reaction.

The equati

iiH Hν Δ−Δ ∑

( ) ( )o oo orx cn

on developed in this illustration,

25 C,1 bar = 25 C,1 bar

is always valid and provides a way of computing t

(

he standa

8.5-

rd

)

a

6

heiiH Hν−Δ Δ∑

t of reaction fromstandard-heat-of-combustion data.

2012/3/29 64

2 4 2

ILLUSTRATION 8.5-2Calculation of the Standard Heat of Reaction Compute the heat of reaction for the gas-phase reaction

as a FunctioN O = 2NO

n of Temperaturestanda over the

temperaturerd-

rastate

nge

( ) ( )o orxn f

o

f f

of .Data: See Appendices A.II( ) and A. IV( ).SOLUTIONThe heat of reaction at a temperature can be computed

200 to 600

from

At = 2

ideal Cp a

5 C we find, from the data in A

d K

n

i

o o

i

T

H T

G

T

H

H Tν

Δ Δ

Δ = Δ∑

( ) [ ]2 4

o orxn

ppendix A.IV, that for each mole of N O reacted,

25 C 2 33.18 9.16 kJ/mol=57.20 kJ/mol

To compute the heat of reaction at any temperature , we start from Eq. 8.5-5 and note that sincethe stand

H T

T

Δ = = × −

( ) ( )

2 2 4

o *p,i p,i

o o o *rxn rxn p,i298.2K

* * *p,i p,NO p,N O

ard state for each species is a low-pressure gas, = . Therefore,

25 C

For the case here we have, from Appendix A.II,

2

12.804

T

iTi

ii

C C

H T H T C dT

C dT C C

ν

ν

=Δ = Δ = +

= −

= −

∑∫

∑2 5 2 8 3 J7.239 10 4.301 10 1.5732 10

mol KT T T− − −× + × + ×

2012/3/29 65

( ) ( )

( ) ( )

( ) ( )

o 2 5 2 8 3rxn 298.2

22 2

5 83 3 4 4

2 2 5 3

57200 12.804 7.239 10 4.301 10 1.5732 10

7.239 1057200 12.804 298.15 298.152

4.301 10 1.5732 10298.15 298.153 4

56189 121.804 3.619 10 1.4337 10

TH T T T T dT

T T

T T

T T T

− − −

−

− −

− −

Δ = + − × + × + ×

×= + − − −

× ×+ − + −

= + − × + × +

+

∫

( )

( )

9 4

2 4

orxn

2 4

J3.933 10mol N O

K 200 300 400 500 600

kJ 57.423 57.192 56.538 55.580 54.448mol N O

T

T

H T

−×

⎛ ⎞Δ ⎜ ⎟

⎝ ⎠

2012/3/29 66

of

In some databases for example, in the very extensive NIST Chemistry Web-Book - the data reported for each substance are the the standard state heat of

formation and the ab

Third law reference state

HΔ o o Here by absolute entropy ismeant entropy based on the third law of t

solute entropy , both at 25 C. as defined in Sec. 6.8. The

reason for reporting these two quantities is he

thrmodyna

at they emics ar

S

the Gibbs energy of formation, which is obtainedby measuring chem

determined directly by thermalor calorimetric measurements, unli

If the stanical equ

dard stailibrium constan

te heat of formats.

ke

tion

o orxn f

1

, the as follows. First, and the absolute entropy of each

the heat ofreaction is computed u

Gibbs ensing

substanceergy of reaction a

and then the entropy cha

re know

nge

can be compu dn te

C

iii

H Hν=

Δ = Δ∑

oorxn

1o

o oo o orxn rxn rxn f rxn

1

for the reaction is computed from

The standard-State Gibbs energy change on reaction at = 25 C can then be computedfrom

298.15 298.15

C

ii

C

ii ii

S S

T

G H S H S

ν

ν

=

=

Δ =

⎡ ⎤Δ = Δ − ⋅Δ = Δ − ⋅ Δ⎣ ⎦

∑

∑

2012/3/29 67

the Gibbs free energy of reaction aThen using the heat capacities repo

t any other temperrted in the NIST Chemistry WebBook and Eq. 8.5-5

can be obtained. As an example of the use of data

aturein this form, we return to the gas-phase reaction of

hydrogen and oxygen to form water considered at the beginning of this section. Usingthe NIST Chemistry WebBook, we obtain the following data.

o of

2

2

2

f rxn rxn

kJ JSpeciesmol mol K

H 0 130.680 O 0 205.150

H O -241.826 188.835

This results in = 241.826 kJ/mol, = 44.43 J/(mol K), an 228.579 kJ/

d =. This last v agralul eemo

H S

H S G

Δ

Δ − Δ Δ−

( )with the Gibbs energy of formation for water

as a vapor in228.6 kJ/ Appendixe

A.IV.s

mol

−

2012/3/29 68

mixture mix i

mixture m

8.6 THE EXPERIMENTAL DETERMINATION OF THE PARTIAL MOLAR VOLUMEAND ENTHALPY

Experimental data

(1) PVT measurement,

(2) Heat capacity (enthalpy) measurement,

V V V

H H

→ Δ →

→ Δ

( )

ix i

i i

i i i

(3) VLE measurement VLE data = ln

= / .

i

H

G RT

S G H T

γ γ

→

→ →

−

2012/3/29 69

2012/3/29 70

2012/3/29 71

Equation for the calculation of thepartial molar volume

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

1 2 1 2mix 1 1 2 2 1 2 1 1 2 2

mix 1 21 21 1 2 2

1 1 1, ,,

1 21 2

1 1, ,

mix

Due to Gibbs-Duhem

(8.6-1)

(8.6-2)

equation,

0

T P T PT P

T P T P

V xV x V x V x V x V V x V V

V V VV V x V V xx x x

V Vx xx x

Δ = + − − = − + −

⎛ ⎞∂ Δ ⎛ ⎞ ⎛ ⎞∂ ∂= − + − − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎝ ⎠

⎛ ⎞ ⎛ ⎞∂ ∂+ =⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

∂ Δ( ) ( ) ( )

( ) ( )

( )

1 21 21 ,

mix2mix 1 2

1 ,

mixmix 2 mix 2

2 ,

(8.6-

3)

(8.6 -4a )

T P

T P

T P

VV V V V

x

VV x V V

x

VV x V x

x

⎛ ⎞= − − −⎜ ⎟∂⎝ ⎠

⎛ ⎞∂ ΔΔ − = −⎜ ⎟∂⎝ ⎠

⎛ ⎞∂ ΔΔ − = Δ +⎜ ⎟∂⎝ ⎠

( ) ( )mix11

1 ,

(8.6-4 b) T P

VV V

x⎛ ⎞∂ Δ

= −⎜ ⎟∂⎝ ⎠

2012/3/29 72

2012/3/29 73

( )

( )

mix 1 2 1 2=0

mix 1 2 1 2

Redlich-Kister expansion (can be used for the change or mixing of )

any mol

ar

or

property

(8.6-5)n

ii

i

ii

x x a x x

x x a x x

V

H

Δ = −

Δ = −

∑

( )

( ) ( )

mix 1 2 1 2=0 =0

mix 1=0

mi

1

1

1

x

1 2 1

n ni

ii i

ni

ii

x x a x x

V x a

V

x x

G

x

Δ = −

Δ =

⎛ ∂Δ⎜ ∂⎝

− −

∑ ∑

∑

( ) ( ) ( )

( ) ( )

1 11

=0 =0,

mix22 mix 1

1 ,

121 1 2 2 1 2

=0

m11 mix

1

2

1 1+2

2

2 1 1 2 1 (8.6-5)

(8.6-6a )

n ni i

i ii iT P

T P

ni i

ii

a x a i

VV V V x

x

x

x a x x ix x x

V V V x

x x+ −

−

⎞= −⎟

⎠

⎛ ⎞∂Δ− = Δ − ⎜ ⎟∂⎝ ⎠

⎡ ⎤= − − −⎣ ⎦

∂Δ−

− − −

= Δ +

∑ ∑

∑

( ) ( )

ix

1 ,

122 1 2 1 1 2

=02 (8 . 6-6b)

T P

ni i

ii

Vx

x a x x ix x x −

⎛ ⎞⎜ ⎟∂⎝ ⎠

⎡ ⎤= − − −⎣ ⎦∑

2012/3/29 74

( )

( ) ( )

mix 1 2 1 2=0

12mix11 mix 2 2 1 2 1 1 2

=01 ,

mix22 mix 1

Redlich-Kister expansion

(8.6-5)

(8.6-6b)

2

ni

ii

ni i

iiT P

V x x a x x

VV V V x x a x x ix x xx

V V V x

−

Δ = −

⎛ ⎞∂Δ ⎡ ⎤− Δ + − − −⎜ ⎟ ⎣ ⎦∂⎝ ⎠

∂Δ− Δ −

∑

∑

( ) ( ) 121 1 2 2 1 2

=01 ,

(8.6-6a)2n

i ii

iT P

V x a x x ix x xx

−⎛ ⎞ ⎡ ⎤− − −⎜ ⎟ ⎣ ⎦∂⎝ ⎠∑

2012/3/29 75

( )

3

mix 1 2 1 2=0

An accurate representation of the has been obtained using Eq 8.6-5 with (in units of m /mol)

water-methanol da

ta

(8.6-5

)n

ii

iV x x a x xΔ = −∑

60

61

62

63

a -4.0034 10a 0.17756 10

a 0.54139 10a 0.60481 10

and Table 8. the partial molar volumes in have been computed using these constant6 s.-2

−

−

−

−

×− ×

××

2012/3/29 76

The partial molar of a species in a bA steady-st

inaate

ry mix flow

ec

ntal

to

halrim

uy repeter

2012/3/29 77

Two streams, one of pure fluid 1 and the second of pure fluid 2, both at a temperature and a pressure , enter this steady-state mixing device,and a single mixed stream, also at and , leaves. Hea

T PT P

1 2 3

Taking the contents of the calorimeter to be the syst

t is added or removed to maintain the temperature of the outlet stream.

, the mass and energy balances are

0=

em

N N N+ +& & &

3 1 2

1 2 mix1 2 3

mix 1 21 2 1 2 1 2

0 = +

(8.6-7)

(8.6-8)

=

N N N

N H N H N H Q

Q N N H N H N H N N

= +

+ +

⎡ ⎤ ⎡ ⎤+ − − = + Δ⎣ ⎦ ⎣ ⎦

− & & &

&& & &

& & & & & & &mix

mix 1 2

mix

/

can be calculated from .

H

H Q N N

H Q

⎡ ⎤Δ = +⎣ ⎦Δ

& & &

&

2012/3/29 78

2012/3/29 79

Figure 8.6-3 Heat of mixing data for the water(1)-methanol(2) systemAt T = 19.69 oC

2012/3/29 80

1 2 in a manner completely Once the composition depe

analogous to the procedur

ndence of the heat of mixing is known, and may be computed

used for the partial molar volumes, inparticular, it is eeasi

H H

( ) ( )

( ) ( )

mix22 mix 1

1 ,

mix11 mix 2

1 ,

ly established that

(8.6-9a)

(8

.6-9b)

T P

T P

HH H H x

x

HH H H x

x

⎛ ⎞∂ Δ− = Δ − ⎜ ⎟∂⎝ ⎠

⎛ ⎞∂ Δ− = Δ − ⎜ ⎟∂⎝ ⎠

2012/3/29 81

2012/3/29 82

General equation relating the partial molar property to the pure component property and the property change on mixing

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( )

mix11 mix 2

2 ,

mix22 mix 1

1 ,

1 22 ,

mixmix

, , , , ,

, , , , ,

, , , ,

(8.6-10a)

(8.6-1 )

(

0b

8

T P

T P

T P

T P x T P T P x xx

T P x T P T P x xx

T P x T P x xx

θθ θ θ

θθ θ

θ θ

θ

θ

⎛ ⎞∂ Δ− = Δ − ⎜ ⎟∂⎝ ⎠

⎛ ⎞∂ Δ− = Δ − ⎜ ⎟∂⎝ ⎠

⎛ ⎞∂= − ⎜ ⎟∂⎝ ⎠

( ) ( ) ( )mixm

1 ,

ix2 1, , , ,

.6-11a)

(8.6- 11b)T P

T P x T P x xx

θθθ

⎛ ⎞∂= − ⎜ ⎟∂⎝ ⎠

2012/3/29 83

( ) ( ) ( ) ( ) ( )

( ) ( )

( ) ( ) ( )

mix11 2

2 ,

mixmix1 2

2 2, ,

mix mi

mix

mix 1 2mix 1 2

xmix1 22 2 2

2 2 2,

mi

,

x, ,, , ,

=

(8.6-10a)

+

+

, ,

=

T P

T P T P

T P T P

T P x

x

TT P x T P xx

x

x

x

x x xx

x

x

P

x

θθ θ

θ θθ θ

θ θ θ

θ

θ θ θ

θ

θ

θ

θ

⎛ ⎞∂ Δ− = − ⎜ ⎟∂⎝ ⎠

⎛ ⎞ ⎛ ⎞∂ Δ ∂−⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

⎧ ⎫⎛ ⎞ ⎛ ⎞∂ Δ ∂ ∂⎪ ⎪− =⎨ ⎬⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠

Δ

Δ = −

⎠

−

⎝⎪ ⎪⎩

Δ

⎭

( ) ( ) ( ) ( )

( )

( )

( ) ( ) ( )

1 22 2,

mix1 mix 2

2 ,

mixmix 1 11 2 2

2 ,

mixmix 2

2 ,

mixmix1 2

2 ,

1

2 22

1

2

, , , , ,

Thus,

, , , ,

T P

T P

T P

T P

T P

x x

T P x T P T P x xx

x x xx

xx

T P x T P x x

x

x

x

θ θ

θθ θ

θθ θ θ

θθ

θ

θ θ

θθ θ

θ

⎛ ⎞+ −⎜ ⎟

⎝ ⎠

⎛ ⎞∂ Δ− = Δ − ⎜ ⎟∂⎝ ⎠

⎧ ⎫⎛ ⎞∂⎪ ⎪= − − +⎨ ⎬⎜ ⎟∂⎝ ⎠⎪ ⎪⎩ ⎭⎛ ⎞∂

= − −⎜ ⎟∂⎝ ⎠

⎛ ⎞∂= − ⎜ ⎟∂⎝

−

⎠

−

mix(8.6-11a) θ

2012/3/29 84

ILLUSTRATION 8.6-1Calculations of Partial Molar from Experimental DataUsing the data in Fig. , determine the partial molar enthalpy of sulfuric acid and water a

Enthalpies8.1-1

50 mol % sulfuric at

cid

o

2 4

o

2

and .SOLUTIONFirst we must obtain values of enthalpy versus concentration at 65.6 C. The values

converted to a molar baread from

this figure and are give

65

n below.(MW(H SO ) 98.07 = ; MW(H8

si

. C

s

6

O) = 18.015)

2012/3/29 85

Figure 8.1-1 Enthalpy-concentrationdiagram for aqueous sulfuric acid at0.1 MPa. The sulfuric acid percentageIs by weight. Reference states: The Enthalpies of pure liquids at 0oC And their vapor pressures are zero.

278

85

-78

-175

-153

-60

92

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0-16

-14

-12

-10

-8

-6

-4

-2

0

2

Δ m

ixΗ

, kJ

/mol

x1

Plot of Example 8.6-1

Slope=-82.8

Slope=+26.11

2 4 2 4

2 4

2mixH SO H SO

H SO82.795 278.965 170.049d H x x

dxΔ

= − + −

Slope=+14.17

2012/3/29 87

( )( )( )

2 4 2 2 4

2 4 2 4 2 4

2 4

mix H SO H O H SO

H SO H SO H SO

H SO

These data with the simple expressionkJ 82.79 56.683

mol

1 8

are fit re

2.79 56.683

asona

b

82.795 1

ly w

.

e

39

ll

H x x x

x x x

x

⎛ ⎞Δ = − +⎜ ⎟⎝ ⎠

= − − +

= − +

( )

2 4 2 4

2 4 2 4

2 4

2 4

2 42 4 2

2 4

2 3H SO H SO

2mixH SO H SO

H SO

H SOH O

mix mix mixH SO

H SO

mix H SO

478 56.683

and 82.795 278.965 170.049

kJ kJFor 0.5; 14.17 ; 14.17mol mol

kJ0.5 13.61mol

x x

d H x xdx

d H d H d Hxdx dx dx

H x

−

Δ= − + −

Δ Δ Δ= = = − = −

Δ = = −

2012/3/29 88

( )2 4 2 4

2 4

2

2 4

oH SO

H SO

H O

H S

H O

O

S

kJ 98.708 g kJ9.02 1000 g mol mol

kJ 18.015

from Fig. 8.1-1

from Fig. g kJ5.01 1000 g mol mol

Finally, from Eq. 8.6-9b, we have

kJ92kgkJ278kg

0.5, 65.

.1-1

6 C

8

H

H

H T

x T

H = × × =

= × × =

=

=

=

LLL

LLL

( ) ( )

( )

( )

( )

( ) ( )

2 4 2

2 4 2

2 4H SO2 4

2 2 4 2 4

2

4

4

2 2

H SO H S

o mixmix H SO H O

H SO 0.5

o mixH O

oH O

mix H SO H SO

H

O

SO

H S

kJ13.61 0.5

65.6 C 0.5

kJ13.61 0.5 14.7 11.68mol

65.6 C 0

14.7 0.7mol

9.02

0.5, 65.6 C

.5

x

HH x xx

HH T H

H

H x

x

T

x

H

x

=

=

⎛ ⎞∂Δ= + Δ = + ⎜ ⎟⎜ ⎟∂⎝ ⎠

= − + − = −

∂Δ

− =

= = + Δ = +∂

− + − =

=

−

( )

( )

4H SO2 4

2 2

O 0.5

H O H O

kJ13.61 0.5 14.7 1.52mol

kJ13.61 0.5 14.7 6.5m l

5.01

o

x

H H

=

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

= − + = −

− = − + = −

2012/3/29 89

At infinite dilution

( ) ( )

( )

( )

2 1

1 2

mix11 1

2 ,

mix mix11

2 1

mix mix22

mix 2

11 0

21 21 0

(8.6-12a)

and

(8.6-12b)

The largest differenc

, , , ,

0

e b t

, ,

,

e

, 0

T

x x

P

x x

T P x T P xxx

T P xx x

T P xx x

θθ θ

θ θθ θ

θ θ

θ

θ θ= =

= =

⎛ ⎞∂Δ− = − ⎜ ⎟∂⎝ ⎠

⎛ ⎞ ⎛ ⎞∂Δ ∂Δ− = − = +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞∂Δ ∂Δ→ − = − = +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠

→

Δ

⎝ ⎠

i iween is at infinite dilution.θ θ−

2012/3/29 90

ILLUSTRATION 8.6-2Calcutation of from Experimental DataCompute the difference between the infinite

Inf di

inite Dilution ( = 0) Partial Molalution partial molar enthalpy and t

r Enthalpiehe pur

se

compo

x

2 4

2 4

o

mixH SO

H SO ,

using the information in the previous illus

nent molar enthalpy for sulfuric acid and water at 65.6 C

SOLUTION

From the previous illustration 82.795 278.965 170

tration.

.T P

d H xdx

⎛ ⎞Δ= − + −⎜ ⎟⎜ ⎟

⎝ ⎠

( ) ( )

2 4

2 4 2 4H

2 4 2 4

SO H

2

SO4

4

2 2 4

2H SO

mix mix

H SO H S1 O

o oH SO H SO H SO

0

049

kJ kJTherefore, 26.11 and 82.80mol mol

kJso that , 65.6 C 65.6 C 8 8m l

0 2. 0o

x x

x

d H d Hdx dx

H x T H T

= =

⎛ ⎞ ⎛ ⎞Δ Δ= + = −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= = − = = −

2012/3/29 91

( )

( ) ( )

( )

2 4 2 4

2 2 2

2 2

oH SO H SO

o oH O H O H O

oH O H O

ILLUSTRATION 8.6-2

in which casekJ, 65.6 C 9.02 82.8 73.8

molkJ, 65.6 C 65.6 C 26.11

molkJand , 65.6 C 5.01 26.11 21.1

molNote that for the sulfuric acid + w

0

0

0

H x T

H x T H T

H x T

= = = − = −

= = − = = −

= = = − = −

o

at infinater syst

ite dilutem at = 65.6 C the differences between the pure

component and partial molar properties are considerably at the

mole fraction of 0.5 in the pr

greater

evious i

t

l

ion

lu a

han

str

T

kJtion. ( ; -20.7, --82.8, -26.11 ).m6. ol 5

2012/3/29 92

( )

( ) ( )1 1 2

There is for partial molar properties .

In general, we have from Eq. 8.1-13 for any total property

a simple physi at inf

,

cal interpreta

, in a binarymixture that

, ,

inite dilutiont

,

ion

,

T P x

T P x N T P x N

θ

θ θ= + ( )

( ) ( )2 1 2 1

2

11 1

(8.6-13) = 1 and >> so th

, , Now consider the case when , in which case

, , ~ 1 , , since species 1 is essentially at that ~

e pu e 1

r

T P x

T PN N

xN

T Px

θ

θ θ≅

( )

( ) ( ) ( )

2 1

21 1 21

1 2 2

component

limit. Also, , , ~ 1 is the partial molar property of species 2 at infinitedilution, s >> and = o that at in this limit ( )

, , ~ 1 , , ~ 0

1

,

T P x

T P x TN T P

N N

P

N

xθ

θ

θ θ= +

( )2 2the infinite dilution partial molar proper

From this equation we see that is the amount by which the total property changes as a result

ty , , ~ of the

(8.6

addi

-14

t

)

ion of o0T P xθ

θ

2

nemole of species 2 to (so that remains aboutzero). Note that if the solution were ideal, the total pro

an infinitely laperty would change by an

am

rge amount of

ount equal t

o

specie

the pure com

s 1 xθ

2

2

ponent molar property : however, since most solutions

are nonideal, the change is instead equal to .

θ

θ

2012/3/29 93

o

ILLUSTRATION 8.6-3Calculation of the Isothermal Enthalpy Change of Mixing

of sulfuric acid at 65.6 C is added to of water at the same temperature. Ifthe mixing is dOne mole 1000 mole

one isothennas

lly,

( ) ( )( )

2 2 4 2 2 4

2 4 2 4

o oH O H SO H O H SO

oH SO H SO

mix 2

the change in enthalpy of the mixture

Eq. 8.6-14

1

estimate .SOLUTIONFrom ,

65.6 C, 1000, 65.6 C, 1000,

6

0

kJ735.6 C, 0.001 0

1000 mol H O + 1 mol

.8mo

l

H T N N H T N N

H T x

H

= = = = = =

= = = ≈ =

Δ

−

−

= ( )

2 4 2 4

2 4

oH SO H SO

H SO

[The numerical value for ( = 65.6 C, 0) was obtained from the previous

illustration.]

H T x ≈

2012/3/29 94

8.7 CRITERIA FOR PHASE EQUILIBRIUM IN MULTICOMPONENT SYSTEMS

gen

For a closed system, we have for both the pure component and multicomponent cases

(8.7-1)

SdU dVQ W Pdt dtdS Q Sdt T

= + −

= +

& &

&&

( )( )

( )

( )

ii=1

ii=1

For the pure component system (molar basis),

,and for a multicomponent system (molar basis)

, ,

, ,

The equi

(8.7-2)

l

C

i

C

i

U NU T P

S N S T P

U N U T P x

S N S T P x

=

=

=

=

∑

∑ibrium criteria for a closed multicomponent mixture are

= maximum for equilibrium at coinstant , , and = minimum for equilibrium at constant , , and (8.7- = minimum f

4o

)S M U VA M T VG r equilibrium at constant , , and M T P

2012/3/29 95

The equilibrium criterion

i

The equilibrium criterion for a closed, adiabatic, constant-volumesystem is

= maximumsubject to the constraints of constant , , and total number of moles of each species

. For the two-phase system

SU V

N

I IIi i i

I II

, each extensive property (e.g., , , , is the sum ofthe properties for the individual phases, for example,

iN S U V

N N N

dS dS dS

= +

= +

i i1

I II

I II

I IIi i

I II I III I Ii i

iI II I II I IIi 1

1 1

1 1 0

(8.7-5)C

i

C

PdS dU dV G dNT T T

dU dUdV dVdN dN

P P G GdS dU dV dNT T T T T T

=

=

= + −

= −

= −

= −

⎛ ⎞ ⎛ ⎞⎛ ⎞= − + − − − =⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

∑

∑

2012/3/29 96

I II

I II

First criteriaon for phase equilibrium

Second criteriaon for phase equilibrium

(8

.7

-7a)

T T

P P

=

=

I II I IIi i i i

Third criteriaon for phase equilibrium

or

(8.7

-7b)

(8.7-7 c)G G μ μ= =

2012/3/29 97

8.8 CRITERIA FOR CHEMICAL EQUILIBRIUM, AND COMBINED CHEMICAL ANDPHASE EQUILIBRIUM

( )C C

i ii i,0 ii=1 i=1

at constant temperature and A single chemical reaction occurring in a single phase in a closed system

the only variati

pressure,

Since possible in a one-phao se, closed systemn

G N G N X Gν= = +∑ ∑

,

C

i=1

at constant temperature and pressure is in , the equilibrium criterion the extent of reaction

Criterion for chemical equilibrium of a single reactio

is

0

0=

n

T P

ii

GX

G

X

ν

∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠

∑ ( 8 . 8-1)

2012/3/29 98

M

i i,0 ij jj=1

For the multiple chemical reactions in a closed, single-phase, constanl-temperalure and constant-pressure system.

(8.3-5)N N X

G N

ν= +

=

∑C C M

ii i i,0 ij ji=1 i=1 j=1

C C M

ii,0 ij ji=1 i=1 j=1

The condition for chemical equilibrium in this multireaction system is = m

(8.8-

in

)

i

2i

G N X G

N G X G

G

ν

ν

⎛ ⎞= +⎜ ⎟

⎝ ⎠

= +

∑ ∑ ∑

∑ ∑∑

i�j , ,

mum or = 0 for all variations consistent with the stoichiometry at constant temperature, pressure, and total mass. For the present case this implies

0 j = 1, 2, 3, , T P X

dG

G MX

⎛ ⎞∂=⎜ ⎟⎜ ⎟∂⎝ ⎠

L

i�

i�

C C

iij ii=1 i=1j j, , , ,

C

ii=1 j , ,

0

for all independent reactions j=1,

(8.8-3)

2, 3, , . 0 due to the Gibb

i

T P X T P X

i

T P X

G GG NX X

GM NX

ν⎛ ⎞ ⎛ ⎞∂ ∂

= = +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

⎛ ⎞∂=⎜ ⎟⎜ ⎟∂⎝ ⎠

∑ ∑

∑L

C

iiji=1

s-Duhem equation

0= j=1, 2, 3, , (8.8 - 4)

Chemical equilibrium criteria for multiple reactions

G Mν∑ L

2012/3/29 99

8.9 SPECIFICATION OF THE EQUILIBRIUM THERMO-DYNAMIC STATE OF A MULTICOMPONENT, MULTIPHASE SYSTEM; THE GIBBS PHASE RULE

I II III

I II III

I II III

ij

(8.9-1)(8.9-

For in

2

dependent chemica

)

(8.9-4)l reactions,

0

P

P

Pi i i i

i

T T T TP P P P

G G G GM

Gν

= = = =

= = = =

= = = =

=

L

L

L

C

i=1 (8.9-5)∑

2012/3/29 100

Gibbs phase rule

( ) ( ) ( )

Number of unknowsNumber of independent relations

thermodynamicamong the unknows parameters

parameters

1 2 1 1

2

F

P C P C P M

C M P

⎛ ⎞⎛ ⎞⎜ ⎟= − ⎜ ⎟⎜ ⎟ ⎝ ⎠⎜ ⎟

⎝ ⎠⎡ ⎤= + − − + − +⎣ ⎦

= − − + where

number of phasesnumber of componentsnumber of independent reactions

degrees of freedom

(8.9-6)

PCMF

===

=

2012/3/29 101

Problems

8.7, 8.9, 8.17, 8.26, 8.28, 8.31