• No results found

Chapter 7 : Simple Mixtures

N/A
N/A
Protected

Academic year: 2021

Share "Chapter 7 : Simple Mixtures"

Copied!
25
0
0

Loading.... (view fulltext now)

Full text

(1)

Chapter 7 : Simple Mixtures

Using the concept of chemical potential to describe the physical properties of a mixture.

Outline

1)Partial Molar Quantities

2)Thermodynamics of Mixing

3)Chemical Potentials of Solutions

4)Colligative Properties

(2)

Partial Molar Quantities

Partial Molar Quantities

a) Partial Molar Volume

Add 1 mol of water to 100 moles of water

Add 1 mol of water to 100 moles of EtOH n V

nj V

(3)

Partial Molar Quantities j ~ j T,P,n'

V

V

n

= ⎜

The definition of partial molar volume is V

nj V

nj

At 298 K, the density of a 50%, by mass, EtOH/water mixture is 0.914 g/cm3. The partial molar volume of

water at this composition is known to be 17.4 cm3/mol. What is the partial

(4)

b) Partial Molar Gibbs Free Energy

Partial Molar Quantities

G n G nj Pure substance Mixture

For a binary mixture

a a b b

(5)

Volume

Entropy

Gibbs Free Energy

Partial Molar quantity Molar quantity

Parameter

Partial Molar Quantities

(6)

Gibbs-Duhem Equation

For a pure substance G = G(T,P)

For a mixture G = G(T,P,na,nb ) j j j

d

0

n

μ

=

An equivalent analysis can be done for all partial molar quantities.

For a binary system then….

a = EtOH b = water

(7)

Thermodynamics of Mixing

Thermodynamics of Mixing

Two ideal gases (A and B) are introduced to a common container. We’ve looked at the entropy of mixing, can we determine the Gibbs free energy of mixing ?

o o

R

Tln

P

P

μ μ

=

+

o o

nR

P

G G

Tln

P

=

+

Recall… Where Po is 1 bar.

( )

( )

mix

G

R

T n ln

a a

n ln

b b

Δ

=

χ

+

χ

(8)

Earlier we derived the change in entropy for the mixing of an ideal gas. mix a b

1

1

R

a b

S

n ln

n ln

Δ

χ

χ

=

+

G

H-T S

Δ

=

Δ

Δ

H

G T S

Δ

=

Δ

+

Δ

( )

a

( )

b a b

1

1

R

a b

R

a b

RT

H

T n ln

n ln

T n ln

n

ln

Δ

χ

χ

χ

χ

=

+

+

+

Thermodynamics of Mixing

(9)

Chemical Potentials of Ideal Solutions

Before we develop expressions for the chemical potential of real solutions we first need to develop expressions for the chemical potentials of solutes and solvents.

Chemical Potentials of Solutions

Consider a pure solvent (A).

Gas phase (A)

(10)

Chemical Potentials of Solutions

At Equilibrium

Rate of condensation = k’PA Rate of vaporization

Rate of condensation = k’PA* Pure Liquid (A)

A surface A tot surface , ,

n

k

k

n

χ

=

=

Mixture of Two Liquids (A and B)

Rate of vaporization A surface A tot surface , ,

n

k

k

n

χ

=

=

Mixture of Two Liquids (A and B)

* * o

ln

A A A o

P

RT

P

μ

=

μ

+

(11)

Chemical Potentials of Solutions

Can now determine the chemical potential of a solvent in the presence of a solute for an ideal solution.

*

ln

A A

RT

A

μ

=

μ

+

χ

In an ideal solution both the solute and the solvent should obey Raoult’s Law.

(12)

Henry’s Law Mole Fraction of B B

P

0 1 B = solute

Real solutes don’t obey Raoult’s Law. In the limit of low solute concentration though, the solute’s partial pressure is directly proportional to its mole fraction.

Chemical Potentials of Solutions

* B

P

B B B

P

=

χ

K

Henry’s Law * B B B

P

=

χ

P

Raoult’s Law

(13)

Chemical Potentials of Solutions * *

ln

B B B B

P

RT

P

μ

=

μ

+

P

B

=

χ

B

K

B Henry’s Law A (solvent) B (solute)

In an ideal-dilute solution the solute should obey Henry’s

(14)

A A A

a

=

γ χ

What about non-ideal solutions ?

*

ln

A A

RT

a

A

μ

=

μ

+

Solvents in Real Solutions

Chemical Potentials of Real Solutions

The activity is defined as an effective mole fraction.

*

ln

A A

RT

A A

μ

=

μ

+

γ χ

Chemical Potentials of Solutions

* *

ln

A A A A

P

RT

P

μ

=

μ

+

(15)

Chemical Potentials of Solutions

*

ln

A A

RT

A A

μ

=

μ

+

γ χ

We can now define the standard state to be the pure solvent. i.e. χA → 1 and γA → 1 .

ln

o

A A

RT

a

A

μ

=

μ

+

Chemical potential of the

solvent in a real solution

A A A

a

=

γ χ

* A A A

P

a

P

=

(16)

Chemical Potentials of Solutions * *

ln

B B B B

P

RT

P

μ

=

μ

+

But let’s define the solute

activity to be…

And if we also introduce the solute activity coefficient Solving for PB

We get an eqn. that accounts for deviations to ideal-dilute, Henry’s Law behaviour. * *

ln

B B B B B B

K

RT

P

μ

=

μ

+

γ χ

(17)

All that’s left to do is define a standard state. B B B

a

=

γ χ

χ

B B

P

0 1 * B

P

As χB → 0, Henry’s Law applies

and γB must → 1 . * *

ln

B B B B B B

K

RT

P

μ

=

μ

+

γ χ

( )

( )

* *

ln

ln

ln

o B B B B B B B

K

RT

P

μ

=

μ

=

μ

+

+

γ

+

χ

(18)

Chemical potential of the solute in a real solution

ln

o B B

RT

a

B

μ

=

μ

+

(

)

ln

o B B

RT

B B

μ

=

μ

+

γ χ

Ideal solutions Real solutions Ideal-dilute solution

Solute (B)

Solvent (A)

Chemical Potentials of Solutions

B B B a = γ χ B B B P a = K A A A a = γ χ * A A A P a P =

(19)

Chemical Potentials of Solutions

Activities in Terms of Molaties

Sometimes it is preferable to use concentrations rather than mole fractions.

Molarity is not the best unit of concentration for

thermodynamics. Why not ?

moles of solute

volume of solution

M

=

(20)

The choice of standard state is entirely at our discretion.

and a new definition of the standard state.

ln

o

B B

RT

a

B

μ

=

μ

+

To express the chemical potential of a solute in a solution in terms of molality, …

only requires a slightly different definition of the activity

(21)

Colligative Properties

Colligative Properties

μ

Temperature Gas Liquid Solid

What happens if the

chemical potential of the liquid phase is lowered while maintaining the chemical potentials of the solid and gas

phases ?

Colligative properties stem from the lowering if the pure solvent’s

chemical potential. They depend only on the number of solute particles and not their chemical identity.

(22)

Colligative Properties

The starting point for an analysis of all colligative properties is the fundamental equation of thermodynamics.

A A B B C C

d

G

= −

S T V P

d

+

d

+

μ

d

n

+

μ

d

n

+

μ

d

n ....

For a two component system (A = liquid ; B = solute);

A A B B

d

G

= −

S T V P

d

+

d

+

μ

d

n

+

μ

d

n

(23)

Colligative Property #1 : Boiling Point Elevation

( )

vap 2 B R T* T H Δ χ Δ =

Equation found from experiment. Can we derive it ?

(24)

Colligative Property #2 : Freezing Point Depression

( )

fus 2 B R T* T H Δ χ Δ = Colligative Properties

(25)

Colligative Property #3 : Osmosis Pure Solvent (A) Solvent (A) + Solute (B)

Solvent will flow from the pure solvent side to the mixture’s side. As a result there will be a pressure difference, π.

Membrane is permeable to solvent (A), but not to solute (B).

B RT n V Π = Colligative Properties

References

Related documents