Transcript Diapositivo 1

Chemical Thermodynamics
2013/2014
11th Lecture: Thermodynamics of Simple Mixtures
Valentim M B Nunes, UD de Engenharia
Introduction
In this lecture we will begin the study of simple non-reactive mixtures.
To do that we will introduce the concept of partial molar properties.
At this stage we deal mainly with binary mixtures, that is:
x1  x2  1
We shall also consider mainly non-electrolyte solutions, where the solute
is not present as ions.
2
Partial Molar Volume
To understand the concept of partial molar quantities, the easiest
property to visualize is the “volume”. When we mix two liquids A and B,
we have three possibilities:
•Volume contraction
•Volume dilatation
•No total volume change.
In the last case if we mix, for instance, 20 cm3 of liquid A and 80 cm3
of liquid B, we will obtain 100 cm3 of solution. This mean that A…B
interactions are equal to A…A or A…B molecular interactions!
In most cases that is not going to happen! For instance the molar volume
of water is 18 cm3.mol-1. But if we add 1 mol of water to a huge amount
of ethanol the volume will only increase by 14 cm3. The quantity 14 cm3
is the partial molar volume of water.
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Partial Molar Volume
The partial molar volumes change with composition because molecular
environment also change. For water/ethanol system we have:
The partial molar volume of a substance is
defined as:
 V 

Vi  
 ni  p ,T ,n j  ni
Once we know the partial molar volume of two
components of a binary mixture we can
calculate the total volume:
V  nAVA  nBVB
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Measuring Partial Molar Volume
One method is measuring the dependence of volume on composition and
determine the slope dV/dn:
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Method of Intercepts
Dividing both members of last equation (slide 4) by the total number
of moles, n, we obtain:
Vm  xAVA  xBVB
Differentiating:
See next
part of
lecture!
dVm  VA dxA  VB dxB
dVm
dV
 VB  VA then VB  VA  m
dxB
dxB
Substituting:

dV 
Vm  x AV A  xB V A  m 
dxB 

Vm  x AV A  V A 1  x A   xB
dVm
dxB
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Method of Intercepts
Finally we obtain:
dVm
Vm  VA  xB
dxB
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Partial Molar Gibbs Function
Another partial molar property, already introduced, is the partial molar
Gibbs function or the chemical potential.
 G 

i  
 ni  p ,T ,n j  ni
Then, the total Gibbs function of a mixture is:
G  nA  A  nB B
dG  nAd A  nB dB   AdnA  B dnB
But we have seen before that (recall the fundamental equations for
open systems):
dG   AdnA  B dnB
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Gibbs – Duhem Equation
Since G is a state function the previous equations must be equal, and
then we obtain:
nAd A  nB dB  0
Generalizing:
 n d
i
i
0
i
This is the Gibbs – Duhem Equation. It tell us that partial molar
quantities cannot change independently: in a binary mixture if one
increases the other should decrease! It can also be applied to the
partial molar volume.
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Thermodynamics of Mixing Perfect Gases
We already now that systems evolves spontaneously to lower Gibbs
energy. If we put together two gases they will spontaneously mix, then
G should decrease.
Let us consider two perfect gases in two containers at temperature T
and pressure p. The initial total Gibbs energy is:
Ginitial  n A  A  nB  B



Ginitial  n A  A0  RT ln p  nB  B0  RT ln p

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Gibbs energy of Mixing
After mixing we have:



G final  nA  A0  RT ln pA  nB B0  RT ln pB

The difference is the Gibbs energy of mixing and therefore:
Gmix  n A RT ln
pA
p
 nB RT ln B
p
p
Using Dalton´s Law, yi=pi/p, we finally obtain:
Gmix  nRT y A ln y A  yB ln yB 
Since molar fractions are always lower than 1, the Gibbs energy of
mixing is always negative!
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Other Thermodynamic mixing functions
Since
 G 

  S we easily obtain:
 T  p
Smix  nR y A ln y A  yB ln yB 
Now since ΔH = ΔG + TΔS:
Purely
entropic
H mix  0
And since for perfect gases ΔGmix is independent of pressure, then:
Vmix  0
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Summary
S mix
H mix
Gmix
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Chemical potential of liquids
In order now to study the equilibrium properties of liquid solutions we
need to calculate the chemical potential of a liquid.
To do this, we will use the fact that the chemical potential of a
substance present as a dilute vapor must be equal to the chemical
potential of the liquid, at equilibrium.
Remember also that it is usual, in order to characterize a given
solution, to distingue between the solvent (usually the substance in
bigger quantity or in the same physical state of solution) and solute (in
lower quantity or in a different state of aggregation)
Important note: We will denote quantities relating to pure substances
by the superscript “*” so the chemical potential of a pure liquid A it
will be written as μ*A(l).
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Ideal Solutions
For a pure liquid in equilibrium with the vapor we can write:
 *A (l )   A0  RT ln p*A
If another substance is present (solution!) then the chemical potential
is:
 A (l )   A0  RT ln pA
Combining the two equations leads to:
pA
 A (l )   (l )  RT ln *
pA
*
A
?
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Ideal Solutions
French chemist François Raoult found that the ratio pA/pA* is equal to
mole fraction of A in the liquid, that is:
pA  xA p
*
A
This equation is known as the Raoult´s Law.
It follows that for the chemical potential of liquid:
 A (l )   (l )  RT ln xA
*
A
This important equation is the definition of ideal solutions. Alternatively
we may also state that an ideal solution is the on in which all the
components obey to the Raoult´s Law.
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Raoult`s Law
p  p A  pB
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Examples
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Ideal Dilute Solutions
In ideal solutions the solute, as well as the solvent, obeys Raoult´s Law.
But in some cases, in non-ideal solutions, the partial pressure of solute
is proportional to the mole fraction but the constant of proportionality
is not the pure component vapor pressure but a constant, known as the
Henry´s constant, KB:
pB  xB K B
This is known as the Henry`s Law.
Mixtures obeying to the Henry`s
Law are called ideal dilute
solutions.
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Example
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Ideal Mixture of Liquids
When two liquids are separated we have:
Ginitial  nA *A (l )  nB B* (l )
When they are mixed we have:



G final  nA  *A (l )  RT ln xA  nB B* (l )  RT ln xB

As a consequence:
Gmix  nRTxA ln xA  xB ln xB 
This equation is the same as that for two perfect gases! All the other
conclusions are equally valid. The “driving” force for mixing is the
increase of entropy.
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Excess Functions
Real solutions have different molecular interactions between A…A, A…B
and B…B particles. As a consequence ΔH≠0 and ΔV≠0. For instance if
ΔH is positive (endothermic) and ΔS negative (“clustering”) ΔG is
positive and the liquids are immiscible. Alternatively the liquids can be
only partially miscible.
Thermodynamic properties of real solutions can be expressed in terms
of excess functions (GE, SE, etc.). For example for entropy:


S E  Smix real   nR xA ln xA  xB ln xB

In this context the concept of regular solutions is very important. In
this solutions HE≠0 but the SE=0.
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Examples
These figure shows experimental excess
functions at 25 ºC. HE values for benzene/
cyclohexane show that the mixing is
endothermic (since ΔH=0 for an ideal
solution).
VE for tetrachloroethene/cyclopentane
shows that there is a contraction at lower
mole fractions of tetrachloroethene but
expansion at higher mole fractions (because
ΔV=0 for an ideal solution)
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Solutions containing non-volatile solutes
Let us now consider a solution in which the solute is non-volatile, so it
does not contribute to the vapor pressure, and it does not dissolve in
solid sovent. Recalling that the chemical potential of liquid A in solution
is:
 A (l )   (l )  RT ln xA
*
A
We can conclude that the chemical potential in the liquid solvent is
lower that the chemical potential of pure liquid (remember that xA<1).
These give origin to a set of properties of this solutions that are
called colligative properties. For instance the vapor pressure lowering:
pA  xA p*A  (1  xB ) p*A
or
pA  p  xA p
*
A
*
A
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Colligative Properties
All colligative properties have in common that they only depend on the
amount of solute but not on the particular substance. The other
colligative properties are: Boiling point elevation, Freezing point
depression and Osmotic pressure.
Graphically we can observe
the first two properties:
Effect of
the solute
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The elevation of boiling point
The heterogeneous equilibrium that maters when considering boiling is
the solvent vapor and the solvent in the solution.
The equilibrium condition is:
 *A ( g )   *A (l ) * RT ln xA
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The elevation of boiling point
This rearranges to:
ln1  xB  
 *A ( g )   *A (l )
RT

Gvap
RT

H vap
RT

Svap
R
When xB=0 , the boiling point is that of pure liquid, Tb, and:
ln 1 
H vap
RTb

S
R
The difference between the two equations is
H vap  1 1 
  
ln(1  xB ) 
R  T Tb 
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The elevation of boiling point
Assuming now dilute solutions, that is, xB<<1, then ln(1-xB) ≈ -xB, so:
H vap  1 1 
  
R  Tb T 
1 1 T  Tb T
 
 2
Since T ≈ Tb we can also write:
Tb T
TTb
Tb
xB 
Which gives:
 RTb2 
 xB
T  
 H 
vap 

This means that the elevation of boiling point, ΔT, depends on xB (no
reference to the identity of solute, the other properties are related to
the solvent!)
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The elevation of boiling point
We can also calculate the elevation of boiling point in terms of the
molality of the solute, mB. Since the mole fraction of solute is small
nA >> nB, so:
nB
n
1 kg
xB 
 B and nA 
nA  nB nA
M
Therefore we obtain:
nB nB M __
xB 

 mB M
n A 1 kg
Boling point elevation can now be calculated by:
 RTb2 M
T  
 H
vap

or
 __
 mB


__
T  K b m B
Kb is the molal ebullioscopic
constant.
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The depression of freezing point
The equilibrium of interest now is between the pure solid solvent and
the solution with dissolved solute:
At freezing point the equilibrium condition is:
 *A (s)   *A (l )  RT ln xA
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The depression of freezing point
All the previous calculations are equal, so we can write directly:
 RTf2
T  
 H
f


 xB


or
___
T  K f m B
Kf is the molal cryoscopic constant.
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Constants of Common Liquids
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Osmosis
We have shown that boiling point and freezing point depends on the
equilibrium between the solvent in solution and in the solid or vapor
state. The last possibility is to have an equilibrium between the solvent
in solution and the pure solvent. This is the basis of the phenomenon of
osmosis. This consists in the passage of a pure solvent into the solution
separated from it by a semipermeable membrane.
h
Π = ρhg is the
osmotic pressure
μA*(p,T)
μA(p+π,T)
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Osmotic pressure
The equilibrium condition is:
 ( p)   A ( p   , xA )   ( p   )  RT ln xA   ( p)  
*
A
*
A
From these equation we can easily obtain:
*
A
p 
p
Vmdp  RT ln xA
 RT ln xA  
p 
p
Vm dp
For dilute solutions ln xA = ln (1-xB) ≈ -xB, so, assuming that Vm is
constant:
RTxB  Vm
Now, since xB ≈ nB/nA and nAVm = V, we finally obtain:
V  nB RT
This is the van`t Hoff equation for osmotic pressure. Note the
remarkable similitude with the perfect gas equation!
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Applications
Van`t Hoff equation can be rewritten as:
  BRT
This is a very useful equation for the determination of molar mass of
polymers and other macromolecules, assuming dilute solutions and
incompressible solutions.
Other application is the reverse or
inverse osmosis, If we apply from the
solution side a pressure bigger than the
osmotic pressure we will reverse the
passage of water molecules. This can be
used to obtain fresh water from sea
water ( π≈ 30 to 50 atm)
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Ideal solubility of solids
Although is not strictly a colligative property the solubility of a solid
can be estimated by the same techniques of previous slides. In contact
with a liquid the solid will dissolve until the solution is saturated. The
equilibrium condition is:
 (s)   (l )  RT ln xB
*
B
*
B
From this starting point and using similar deduction we will obtain:
H fus  1
1 

ln xB 

*

R  T fus T 
Being the properties related only to the solute.
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