Transcript Diapositivo 1

Chemical Thermodynamics
2013/2014
10th Lecture: Phase Equilibria – Pure Substances
Valentim M B Nunes, UD de Engenharia
Introduction
In previous lectures we studied the equilibrium in chemical reactions. Let
us now study the equilibrium in phase transitions or Phase Equilibria. Our
goal is to understand the thermodynamics of phase transitions and
coexistence for single component systems.
Fusion, vaporization, sublimation…
Phase: An homogenous portion of a
given system (for instance, one gas
or a mixture of gases, mixture of
miscible liquids, etc..)
2
Gibbs Phase Rule
Before examining the one component systems, we will derive the Gibbs
Phase Rule. This rule allow us to calculate the number of intensive
variables necessary to determinate the thermodynamic state of a
system with any number of phases and components.
What is a component of the system? Substance whose concentration
varies independently. For instance, for the reaction:
CaCO3(s) = CaO(s) + CO2(g)
we have 2 components, since one of them its not independent! So the
number of components of a system is:
C SR
Where S is the number of substances and R is the number of chemical
equilibriums.
3
Gibbs Phase Rule
Consider a system with several phases and components:
Thermal equilibrium: Tα= Tβ = Tγ = …
α
Mechanical equilibrium: pα = pβ = pγ = …
β
Chemical equilibrium: μ1α = μ1β = μ1γ = …
γ
μ2α = μ2β = μ2γ = …
.
.
……
If we have C components and F phases, then the maximum number of
variables, g, it will be: g = FC+2. But in each phase we have:
x
i
1
i
We can subtract one variable for each phase! Also μ1α = μ1β = μ1γ = …,
then we can subtract F-1 variables for each component.
4
Gibbs Phase Rule
The total number of variables its then:
g  FC  2  F  C ( F  1)  FC  2  F  FC  C
g F C2
This is the final expression for the Gibbs Phase Rule, a very important
result when we are studying phase equlibria.
5
Examples
Systems with one component, C=1:
F
g
Observations
1
2
We have to define temperature and pressure
2
1
For instance liquid/vapor equilibrium: if we set the temperature,
pressure is already known.
3
0
There is only one condition that satisfies this condition: the triple
point
Several components:
CaCO3(s) = CaO(s) + CO2(g)
In this case, F = 3, C = 2, then g = 1. We have to define only
temperature.
6
One Component Systems
Phase transitions are accompanied by a change of entropy and also of
enthalpy. For a given temperature, T, and pressure, p, the more stable
phase is the one with lower chemical potential, that is, the lower molar
Gibbs energy:
phase2  phase1
The chemical potential controls phase transitions and phase equlibria. At
equilibrium μ must be equal throughout the system. When multiple
phases are present μ must be same in all phases.
7
How does μ depends on T?
We know that dG = -SdT + Vdp or per mole dμ = -SmdT + Vmdp , and
  
   Sm
 T  p
The
3rd
Law tell us that Sm>0, so as a consequence:
  
 0
 T 
We also know that Sm,gas > Sm,liq > Sm,sol then:
  gas 

   S m, gas
 T 
 liq 

   S m,liq
 T 
  sol

 T

   S m, sol

This means that the negative slope is steepest in gas phase, less steep
in liquid and even less in the solid phase.
8
How does μ depends on T?
Figure shows the change of chemical potential with temperature for a
pure substance:
μ
μsol = μliq
μliq = μgas
Tfus
Tb
T
At the temperatures Tfus and Tb two phases coexist in equilibrium.
9
Clapeyron,s Equation
As we saw, the equilibrium condition is μα (T,p) = μβ (T,p). Let us suppose
that temperature shifts from T to T+dT and pressure to p+dp. Then the
equilibrium condition it will be μα (T,p) + d μα = μβ (T,p) + d μβ . As a
consequence d μα = d μβ
Since the chemical potential is the Gibbs energy per mole we can write:
 S dT  V dp   S  dT  V dp
S
Rearranging

 S dT  V  V dp
dp S

dT V
This is the Clapeyron’s Equation. From this equation we can calculate
phase diagrams.
10
Phase Diagrams
Phase diagrams describe the phase properties as a function of state
variables, like pressure and temperature (p,T)
Solid – Liquid equilibrium (melting line)
dp S fus

dT V fus
At equilibrium temperature it’s a reversible change, so ΔSfus = ΔHfus/Tfus
Since ΔHfus > 0, ΔSfus > 0. On the other hand ΔVfus can be positive
(general case) or negative
water
p
S
L
Typical values are:
ΔSfus : 2 ~ 6 cal.K-1.mol-1
ΔVfus : ± 1 ~ 10 cm3.mol-1
Melting line is very steep
T
11
Liquid - Gas equilibrium
In this case, ΔSvap > 0 and ΔVvap > 0 (all substances), then dp/dT > 0
Near the normal boiling point (p=1atm) for most substances we have:
ΔSvap ≈ 20 cal.K-1.mol-1
ΔVvap ≈ 20 000 cm3.mol-1
p
S
L
G
Boiling line is less steep
T
This line intersects the melting line in a point for which there are
three phases in equilibrium: the triple point. Accordingly with the phase
rule g = 0.
12
Measuring vapor pressures
13
Solid - Gas equilibrium
In this case, as for liquid gas equilibrium, ΔSsubl > 0 and ΔVsubl > 0 (all
substances), then dp/dT > 0
This curve as a higher slope, near triple point, than the liquid gas curve
equilibrium, since, at triple point, ΔHsubl= ΔHfus+ ΔHvap > ΔHvap and
ΔVsubl ≈ ΔVvap.
p
S
L
G
pt
Tt
T
This is the general aspect of a p,T phase diagram. We will examine
later some phase diagrams of real substances!
14
Critical point
At the critical point the gas-liquid curve stops. Beyond the critical
point the liquid and gas phase become indistinguishable, and we have a
supercritical fluid (no phase change!).
15
Integration of Clapeyron Equation
From Clapeyron’s equation and for the solid liquid equilibrium we have:
dp 
S fus
V fus
p2
dT
 dp 
p1
H fus 1
T V fus T dT
1
T2
Assuming that ΔHfus and ΔVfus are T independent then:
p2  p1 
H fus
V fus
T2
ln
T1
16
Clausius-Clapeyron Equation
For the liquid – vapor equilibrium ΔV its not constant, but we can make
some approximations. Ignoring the molar volume of the condensed phase,
compared with gas, we have:
Vvap  Vgas  Vliq  Vgas
Then the Clapeyron equation comes:
Assuming an ideal gas, then
Vgas
dp H vap

dT TVgas
RT

p
Substituting in the equation we obtain:
dp pH vap

dT
RT 2
or
H vap
1
dp 
dT
2
p
RT
17
Clausius-Clapeyron Equation
H vap
1
dT
Integration of equation leads to:  dp  
2
p
RT
p0
T0
p
T
Assuming now that ΔHvap is constant in the temperature interval we
obtain:
p H vap  1 1 
  
ln

p0
R  T0 T 
If p0 = 1 atm, the T0 is the normal boiling point, Tb, then:
p H vap H vap 1
ln 


1
RTb
R
T
This is the Clausius – Clapeyron equation. It allow us, for example, to
calculate the boiling point at any pressure.
18
Clausius-Clapeyron Equation
Graphically it can be observed that for many substances the Clausius
Clapeyron equation is obeyed for relatively large temperature intervals.
ΔHvap/RTb
Slope =-ΔHvap/R
ln p = f(1/T)
19
Example of phase diagram: Water
The figure shows the simplified phase diagram for water:
Negative slope!
Critical point
Triple point
Water sublimes bellow 0.006 atm (~4.58 mmHg): lyophilization
20
Some considerations about the critical point
At critical point we have some “thermal anomalies”. For instance if we
rearrange the Clapeyron equation:
H vap  TVvap
dp
dT
Now, since ρL  ρG, ΔVvap  0, then:
lim
H vap
T  Tc
0
Above Tc an pc, liquid and gas become indistinguishable, a single fluid
phase known as supercritical fluid. These fluids are finding remarkably
practical applications. Supercritical water (Tc=374 ºC and pc = 221 bar)
readily dissolve organic molecules and inorganic salts are nearly
insoluble.
21
Phase diagram: Carbon Dioxide
Some remarkable features of
the CO2 phase diagram are:
 Solid liquid line has a
positive slope
 CO2 sublimates bellow 5.1
atm, for instance at
atmospheric pressure
CO2 is a supercritical fluid
just above room temperature!
Supercritical CO2 is a reaction solvent. It can replace chlorinated and
volatile organic compounds. It is a “dry cleaning” solvent.
22
Phase diagram: Carbon
Synthetic
diamonds!
23