Chemical Thermodynamics 2018/2019

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Chemical Thermodynamics2018/2019
10th Lecture Phase Equilibria Pure
Substances Valentim M B Nunes, UD de Engenharia
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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..)
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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
Where S is the number of substances and R is the
number of chemical equilibriums.
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Gibbs Phase Rule
Consider a system with several phases and
components
Thermal equilibrium Ta Tß T? Mechanical
equilibrium pa pß p? Chemical
equilibrium µ1a µ1ß µ1?
µ2a µ2ß µ2?
. .

a
ß
?

If we have C components and F phases, then the
maximum number of variables, g, it will be g
FC2. But in each phase we have
We can subtract one variable for each phase! Also
µ1a µ1ß µ1? , then we can subtract F-1
variables for each component.
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Gibbs Phase Rule
The total number of variables its then
This is the final expression for the Gibbs Phase
Rule, a very important result when we are
studying phase equlibria.
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Examples
Systems with one component, C1
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.
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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
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.
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How does µ depends on T?
We know that dG -SdT Vdp or per mole dµ
-SmdT Vmdp , and
The 3rd Law tell us that Smgt0, so as a
consequence
We also know that Sm,gas gt Sm,liq gt Sm,sol then
This means that the negative slope is steepest in
gas phase, less steep in liquid and even less in
the solid phase.
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How does µ depends on T?
Figure shows the change of chemical potential
with temperature for a pure substance
µsol µliq
µliq µgas
At the temperatures Tfus and Tb two phases
coexist in equilibrium.
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Clapeyron,s Equation
As we saw, the equilibrium condition is µa (T,p)
µß (T,p). Let us suppose that temperature
shifts from T to TdT and pressure to pdp. Then
the equilibrium condition it will be µa (T,p) d
µa µß (T,p) d µß . As a consequence d µa d
µß
Since the chemical potential is the Gibbs energy
per mole we can write
Rearranging
This is the Clapeyrons Equation. From this
equation we can calculate phase diagrams.
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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)
At equilibrium temperature its a reversible
change, so ?Sfus ?Hfus/Tfus
Since ?Hfus gt 0, ?Sfus gt 0. On the other hand
?Vfus can be positive (general case) or negative

water
p
S
L
Typical values are ?Sfus 2 6
cal.K-1.mol-1 ?Vfus 1 10 cm3.mol-1
Melting line is very steep
T
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Liquid - Gas equilibrium
In this case, ?Svap gt 0 and ?Vvap gt 0 (all
substances), then dp/dT gt 0
Near the normal boiling point (p1atm) for most
substances we have ?Svap 20 cal.K-1.mol-1 ?Vvap
20 000 cm3.mol-1
Boiling line is less steep
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.
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Measuring vapor pressures
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Solid - Gas equilibrium
In this case, as for liquid gas equilibrium,
?Ssubl gt 0 and ?Vsubl gt 0 (all substances), then
dp/dT gt 0
This curve as a higher slope, near triple point,
than the liquid gas curve equilibrium, since, at
triple point, ?Hsubl ?Hfus ?Hvap gt ?Hvap and
?Vsubl ?Vvap.
pt
Tt
This is the general aspect of a p,T phase
diagram. We will examine later some phase
diagrams of real substances!
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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!).
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Integration of Clapeyron Equation
From Clapeyrons equation and for the solid
liquid equilibrium we have
Assuming that ?Hfus and ?Vfus are T independent
then
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Clausius-Clapeyron Equation
For the liquid vapor equilibrium ?V its not
constant, but we can make some approximations.
Ignoring the molar volume of the condensed phase,
compared with gas, we have
Then the Clapeyron equation comes
Assuming an ideal gas, then
Substituting in the equation we obtain
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Clausius-Clapeyron Equation
Integration of equation leads to
Assuming now that ?Hvap is constant in the
temperature interval we obtain
If p0 1 atm, the T0 is the normal boiling
point, Tb, then
This is the Clausius Clapeyron equation. It
allow us, for example, to calculate the boiling
point at any pressure.
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Clausius-Clapeyron Equation
Graphically it can be observed that for many
substances the Clausius Clapeyron equation is
obeyed for relatively large temperature
intervals.
?Hvap/RTb
Slope -?Hvap/R
ln p f(1/T)
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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
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Some considerations about the critical point
At critical point we have some thermal
anomalies. For instance if we rearrange the
Clapeyron equation
Now, since ?L ? ?G, ?Vvap ? 0, then
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 (Tc374 ºC and pc 221 bar) readily
dissolve organic molecules and inorganic salts
are nearly insoluble.
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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.
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Phase diagram Carbon
Synthetic diamonds!