Chapter 8: The Thermodynamics of Multicomponent Mixtures 1.

Chapter 8:The Thermodynamics of

Multicomponent Mixtures

1

Important Notation

2

Learning objectives

Be able to:

• Understand the difference between partial molar properties and pure component properties;

• Use the mass, energy, and entropy balance for mixtures;

• Compute partial molar properties from experimental data;

• Derive the criteria for phase and chemical equilibria in multicomponent systems.

3

Thermodynamic description of mixtures

In previous chapters (which dealt with pure components), it was identified that properties such as volume, internal energy, enthalpy, Helmholtz and Gibbs energies are extensive, i.e., they depend on the amount of substance present.

, ,U U P T N , ,V V P T N

, ,A A P T N , ,G G P T N

, ,H H P T N

4


For mixtures, it is reasonable to expect that these properties will depend on the amount of each of the C components present, i.e.:

1 2, , , ,..., CU U P T N N N

1 2, , , ,..., CV V P T N N N

1 2, , , ,..., CA A P T N N N

1 2, , , ,..., CG G P T N N N

1 2, , , ,..., CH H P T N N N

5


For mixtures, it is reasonable to expect that the corresponding molar properties will depend on the mole fractions of the C components present.

But, there is a subtle point to note:

Consider an open system. Depending on the type of interaction with the surroundings, the number of moles of each component may vary independently: can each be treated as an independent variable.

The same does not happen with mole fractions. Why?

1 2, ,..., CN N N

6


The summation of the mole fractions of all C components is always equal to 1.

1

1C

ii

x

Only (C-1) mole fractions can be treated as independent. One of them, say that of component C, can be calculated as function of the other mole fractions

1

1

1C

C ii

x x

7


For the molar properties:

1 2 1, , , ,..., CV V P T x x x

1 2 1, , , ,..., CU U P T x x x

1 2 1, , , ,..., CH H P T x x x

1 2 1, , , ,..., CA A P T x x x

1 2 1, , , ,..., CG G P T x x x 8


Would a relationship such as this one be true?

1 2 11

, , , ,..., ,C

iC ii

V T P x x x x V T P

9


Would a relationship such as this one be true?

1 2 11

, , , ,..., ,C

iC ii

V T P x x x x V T P

Let us watch a movie to help answer this question.

http://www.youtube.com/watch?v=84k206qaVRU

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http://www.youtube.com/watch?v=84k206qaVRU


We see that:

1 2 11

, , , ,..., ,C

iC ii

V T P x x x x V T P

11


We see that:

1 2 11

, , , ,..., ,C

iC ii

V T P x x x x V T P

Can we quantify this difference, perhaps give it a name?

12


Volume change of mixing:

1

, , ,C

imix ii

V V T P x x V T P

Enthalpy change of mixing:

1

, , ,C

imix ii

H H T P x x H T P

13


Volume change of mixing:

1

ˆ ˆ ˆ, , ,C

mix i ii

V V T P w wV T P

Enthalpy change of mixing:

1

ˆ ˆ ˆ, , ,C

mix i ii

H H T P w w H T P

14


Benzene+C6F5Y at 298.15 K

Methyl formate+ethanol

Methyl formate+methanol

298.15 K

15

Example 1

What is the density of an equimolar liquid solution of methyl formate and ethanol at 298.15 K, 1 atm?

Data

Density of pure ethanol at 298.15 K: 0.789 g/cm3 Density of pure methyl formate at 298.15 K: 0.977 g/cm3

Molar mass of ethanol: 46.07 g/molMolar mass of methyl formate : 60.1 g/mol

16

Example 2

A continuous mixer operating at steady-state blends has two input streams (streams 1 and 2) to produce one output stream (stream 3), all of them liquid binary mixtures of water and sulfuric acid. Stream 1 has a temperature of 21.1oC and a sulfuric acid mass fraction of 0.3 and flow rate of 1 kg/s. Stream 2 is at the same temperature and a water mass fraction of 0.2 and flow rate of 2 kg/s.

a) What is the molar enthalpy change of mixing of stream 1?b) What is the mass fraction of sulfuric acid in stream 3?c) What is the heat transfer rate to/from the mixer if the

temperature of output stream is 21.1oC?d) What is the temperature of the output stream if the mixer

operation is adiabatic?17

f8_1_1

Sulfuric acid + water

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Partial molar properties

The discussion that follows is applicable to V, U, H, S, A, G, but it will be developed based on volume.

Consider a mixture:

1 2, , , ,..., CNV V T P N N N

1 2 ... CN N N N

1 2 1, , , ,..., CV V T P x x x

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If the amounts of all components double:

1 22 , , 2 ,2 ,..., 2 2CNV V T P N N N V

1 2 1, , , ,..., CV V T P x x x

20



If the amounts of all components are multiplied by a positive constant :

1 2, , , ,..., CNV V T P N N N V

1 2 1, , , ,..., CV V T P x x x

The total volume is a homogeneous function of degree 1.

21



Now differentiate this expression with respect to :

1 , ,

1 , ,

j i

j i

Ci

i i T P N

C

ii i T P N

d V V d NV

d N d

VN

N

22



1 , , j i

C

ii i T P N

VV N

N

1 , , j i

Ci

i i T P N

NV V

N N N

1 , , j i

C

ii i T P N

VV x

N

23


The partial molar volume is defined as:

, , j i

i

i T P N

VV

N

1

C

i ii

V x V

The molar volume of the mixture is the average weighted by the mole fractions of the partial molar volumes (and not of the pure component molar volumes).

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t8_1_1

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Gibbs-Duhem equation

The Gibbs-Duhem equation can be developed in generalized form (see the textbook). Here, it is developed based on the Gibbs energy:

1 2, , , ..., CG G T P N N N

1, , , , j i

C

iiP N T N i T P N

G G GdG dT dP dN

T P N

1

C

i ii

dG SdT VdP G dN

26


Side note:

i iG

Chemical potential of component i.

27


From the discussion of partial molar properties, we also have that:

1

C

i ii

G N G

1 1

C C

i i i ii i

dG N dG G dN

28


The previous slides have two expressions for changes in Gibbs energy:

1 1

C C

i i i ii i

dG N dG G dN

1

C

i ii

dG SdT VdP G dN

Subtracting them:

1

0C

iii

SdT VdP N dG

29


This is the Gibbs-Duhem equation:

Its interpretation is that the changes in temperature, pressure, and chemical potentials are interrelated. In other words, they cannot all change independently.

1

0C

iii

SdT VdP N dG

1

0C

iii

SdT VdP x dG

30

Generalized Gibbs-Duhem equation

For a molar thermodynamic property (other than temperature, pressure, and mole numbers):

Please refer to the proof in the textbook.

1, ,

0C

iiiP x T x

dT dP x dT P

31

Experimental determination of partial molar volume and enthalpy

The next slide shows a table of experimental density data for liquid mixtures of water and methanol at 298.15 K.

32

t8_6_1

33


How to use the information available in this table to determine the partial molar volumes?

34

Binary solutions

112

121

dx

dMxMM

dx

dMxMM

Obtain dM/dx1 from (a)

Example 11.3

• We need 2,000 cm3 of antifreeze solution: 30 mol% methanol in water.

• What volumes of methanol and water (at 25oC) need to be mixed to obtain 2,000 cm3 of antifreeze solution at 25oC

• Data:

watermolcmV molcmV

methanol molcmV molcmV

/07.18/77.17

/73.40/63.383

23

2

31

31

solution• Calculate total molar volume of the 30% mixture

• We know the total volume, calculate the number of moles required, n

• Calculate n1 and n2

• Calculate the total volume of each pure species needed to make that mixture

Note curves for partial molar volumes

t8_6_1

40


How to use the information available in this table to determine the partial molar volumes?

41

f8_6_1

+

42


The molar volume change of mixing is:

1

, , ,C

imix ii

V V T P x x V T P

1

C

i imix ii

V x V V

For a binary mixture:

1 21 21 2mixV x V V x V V 43


Let’s evaluate the slope of the curve in the binary diagram (at constant temperature and pressure):

111 1 1

1 1 1, , ,

222 2 2

1 1, ,

mix

T P T P T P

T P T P

V x VV V x

x x x

x VV V x

x x

44


1 21 21 ,

1 2

1 21 1, ,

mix

T P

T P T P

VV V V V

x

V Vx x

x x

0: Why?

45


1 21 21 ,

1 2

1 21 1, ,

mix

T P

T P T P

VV V V V

x

V Vx x

x x

Generalized Gibbs-Duhem equation

46


1 21 21 ,

mix

T P

VV V V V

x

47


11 ,

1 2 1 21 2 1 21 2 1 1

2 2

mixmix

T P

VV x

x

x V V x V V x V V x V V

V V A

Note that:

48


21 ,

1 2 1 21 2 1 21 2 2 2

1 1

mixmix

T P

VV x

x

x V V x V V x V V x V V

V V B

Note that:

49

f8_6_1

+

50


The experimental data are often fitted to a polynomial form, known as Redlich-Kister expansion:

1 2 1 20

ni

mix ii

V x x a x x

For water(1) + methanol(2) at 298.15K with volume in m3/mol:

60

61

62

63

4.0034 10

0.17756 10

0.54139 10

0.60481 10

a

a

a

a

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Equilibrium in multicomponent systems

The discussion extends that of Chapter 7, which dealt with pure components.

All the steps are very, very similar.

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Equilibrium in an isolated systemNo mass in or out; neglect changes in kinetic and potential energies. The energy and entropy balances are:

dU dVQ P

dt dt

gen

dS QS

dt T

with because of the 2nd law of thermodynamics.0genS

53

Equilibrium in an isolated system

0 constantdU dV

Q P Udt dt

0gen gen

dS QS S

dt T

0, no heat transfer to an isolated system

0, no change in volume in an isolated system

0, no heat transfer to an isolated system

54

Equilibrium in an isolated system

0gen

dSS

dt for systems in which M, U, V are constant.

Away from equilibrium, the system conditions change, increasing its entropy.

After a long enough wait (from a fraction of a second to many years, depending on the system), the system attains a condition in which its state properties no longer change, including the entropy. At this state, the equilibrium state, the entropy has a maximum value.

55


Chemical reactions may occur in a multicomponent system.

Let us begin with a situation in which chemical reactions do not occur.

56


Consider an isolated system composed of two subsystems containing a non-reactive multicomponent mixture:

1,...,I IIi i iN N N i C

I IIU U U

I IIV V V

I IIS S S

57




I IIU U U

I IIV V V

I IIS S S Find the equilibrium condition if the internal wall is diathermal, rigid, and impermeable.

Find the equilibrium condition if the internal wall is diathermal, moveable, and impermeable. 58




I IIU U U

I IIV V V

I IIS S S

Find the equilibrium condition if the internal wall is diathermal, moveable, and permeable.

59


I IIS S S

I IIdS dS dS

1

1

1

1

II CiI I I

iI I Ii

IIII CiII II II

iII II IIi

P GdS dU dV dN

T T T

P GdU dV dN

T T T

II IdU dU II IdV dV II Ii idN dN

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At constant U, V, Ni


1

1 1I III II Ci iI I I

iI II I II I IIi

P P G GdS dU dV dN

T T T T T T

At equilibrium:

1 1I IIT T

I II

I II

P P

T T 1,...,

I IIi i

I II

G Gi C

T T

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In summary, at equilibrium:

I IIT T

I IIP P

1,...,I IIi iG G i C

62


Find the two-phase equilibrium conditions of a non-reactive multicomponent system with specified temperature, pressure, and amounts of each component.

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Closed system, constant T and P

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0 III dGdGdG

At constant T and P

i

C

i

iPT dNGdG

1

,|

01

Ii

C

i

IIi

Ii dNGGdG

IIi

Ii GG

Recommendation

Read chapter 8 and review the corresponding examples.

65

Chapter 8: The Thermodynamics of Multicomponent Mixtures 1.

Documents

Transcript of Chapter 8: The Thermodynamics of Multicomponent Mixtures 1.