Download Lecture 5

Predicting Equilibrium
and
Phases, Components,
Species
Lecture 5
Free Energies
Helmholz and Gibbs
Helmholz Free Energy
• Helmholz Free Energy defined as:
A = U-TS
• Functionally, it is:
dA = –SdT-PdV
• The Helmholz Free Energy is the amount of internal
energy available for work.
• Clearly, this is a valuable piece of knowledge for
engineers.
• It is sometimes used in geochemistry.
• More commonly, we use the …
Gibbs Free Energy
• The Gibbs Free Energy is defined as:
G = H-TS
• Which is the amount of internal energy available for
chemical work.
• As usual, we are interested in changes, not absolute
amounts. The Gibbs Free Energy change for a
reaction is:
dG = VdP-SdT
• Notice that it, like the Helmholz Free Energy,
contains a (negative) entropy term and hence will
help us determine the directions in which reactions
will naturally proceed (lower free energy).
Relationship to Enthalpy
and Entropy
• Since Gibbs Free Energy is defined as:
G = H-TS
dG = dH-TdS-SdT
• For a reaction at constant temperature
∆G = ∆H-T∆S
• Equilibrium states are characterized by minimum energy
and maximum entropy. The Gibbs Free Energy is a
function that decreases with decreasing energy (∆H)
and increasing entropy (∆S) and thereby provides a
criterion for equilibrium.
• (The above equation also lets us calculate the free
energy of reaction from enthalpy and entropy changes).
Criteria for Equilibrium
and Spontaneity
• Products and reactants are at equilibrium when
their Gibbs Free Energies are equal.
• At fixed temperature and pressure, reactions will
proceed in the direction of lower Gibbs Free Energy.
Hess’s Law Again
• We can calculate the Gibbs Free Energy change of reaction
using Hess’s Law:
∆ Gr = ån iG f ,i
i
o Again, ν is the stoichiometric coefficient (by convention negative for reactants,
positive for products) and the sum is over all compounds in the reaction.
• If we ask: in which direction will the reaction below proceed
(i.e., which side is stable)?
2MgO + SiO2 = Mg2SiO4
∆Gr = Gf,Mg2SiO4 – 2Gf,MgO- Gf,SiO2
• The answer will be that it will proceed to the right if ∆Gr is
negative.
• However, ∆Gr is a function of T and P, so that ∆Gr may be
negative under one set of conditions and positive under
another.
Geochemical Example
(finally!)
• At some depth, mantle rock
will transform from
plagioclase peridotite to
spinel peridotite.
• We can represent this
reaction as:
• CaAl2Si2O8 + 2Mg2SiO4 =
CaMgSi2O6 + MgAl2O4 +
2MgSiO3
• For a temperature of
1000˚C, what will be the
pressure at which these two
assemblages are at
equilibrium?
Predicting equilibrium
• The two assemblages will be at equilibrium when
the ∆Gr of reaction is 0.
• We can look up values for standard state ∆Gr in
Table 2.2, but to calculate ∆Gr at 1273K we need to
begin with
d∆Gr = ∆VrdP - ∆SrdT
• and integrate
∆ GT ',P' = ∆ GTref ,Pref + ò ∆ Vr dP - ò ∆ Sr dT
• Since ∂S/∂T)P = CP/T
• The temperature integral becomes:
é
∆ CP
∆b
∆ c∆ T
-∆ Sref (∆ T ) - òò
dT dT = -∆ T ê∆ STref - ∆ a +
∆T T
2
2T 'Tref
êë
T
ù
T'
ú - ∆ aT 'ln
Tref
úû
• This is a general form using Maier-Kelly heat capacities of the
change in ∆Gr with temperature.
• In the example in the book, we are allowed to assume the
phases are incompressible, so the pressure integral is simply:
ò ∆ V dP = ∆ V ∆ P
• Using values in Table 2.2, we predict a pressure of ~1.5 GPa
• For volume pressure dependence expressed by constant ,βthe
integral would be:
ò V dP = V
o
2 P'
éë P - b P ùû P
ref
• Using this approach, our result hardly changes.
How did we do?
• The experimentally
determined phase
boundary is closer to 1
GPa.
• Why are we so far off?
• Real minerals are
solutions; in particular Fe
substitutes for Mg in
olivine and pyroxenes
and Na for Ca in
plagioclase.
• We need to learn to deal
with solutions!
Maxwell Relations
• Maxwell Relations are some additional relationships
between thermodynamic variables that we can derive
from the reciprocity relationship (equality of cross
differentials).
æ ¶G ö
æ ¶G ö
dG = ç
• For example:
÷ dP + ç
÷ dT = VdP - SdT
è ¶P ø T
è ¶T ø P
• Since G is a state function
• Therefore:
æ ¶2 G ö æ ¶2 G ö
çè ¶P ¶T ÷ø = çè ¶T ¶P ÷ø
æ ¶V ö
æ ¶S ö
=
çè ÷ø
çè ÷ø
¶T P
¶P T
• Refer to section 2.12 as necessary.
Chapter 2
Thermodynamics of multi-component systems
The real world is
complicated
• Our attempt to estimate the plagioclase-spinel
phase boundary failed because we assumed the
phases involved had fixed composition. In reality
they do not, they are solutions of several
components or species.
• We need to add a few tools to our thermodynamic
tool box to deal with these complexities.
Some Definitions
• Phase
o Phases are real substances that are homogeneous, physically distinct,
and (in principle) mechanically separable. For example, the phases in a
rock are the minerals present. Amorphous substances are also phases.
o NaCl dissolved in seawater is not a phase, but seawater with all its
dissolved components (but not the particulates) is.
• Species
o A species is a chemical entity, generally an element or compound (which
may or may not be ionized). The term is most useful in the context of gases
and liquids. A single liquid phase, such as an aqueous solution, may
contain a number of species. Na+ in seawater is a species.
• Components
o Components are more carefully defined. But:
• We are free to define the components of our system
• Components need not be real chemical entities.
Minimum Number of
Components
• The minimum number of components of a system is
rigidly defined as the minimum number of
independently variable entities necessary to
describe the composition of each and every phase
of a system.
• The rule is:
c=n–r
o where n is the number of species, and r is the number of independent
chemical reactions possible between these species.
• How many components do we need to describe a
system composed of CO2 dissolved in H2O?
Graphical Approach
• If it can be graphed in 1
dimension, it is a two
component system, in 2
dimensions, a 3 component
system, etc.
• Consider the hydration of
Al2O3 (corundum) to form
boehmite (AlO(OH)) or
gibbsite Al(OH)3. Such a
system would contain four
phases (corundum,
boehmite, gibbsite, water).
• How many components?
The system Al2O3–H2O
Phase diagram
for the system
Al2O3–H2O–SiO2
The lines are called joins because they
join phases. In addition to the endmembers, or components, phases
represented are g: gibbsite, by:
bayerite, n: norstrandite (all
polymorphs of Al(OH)3), d: diaspore,
bo: boehmite (polymorphs of
AlO(OH)), a: andalusite, k: kyanite, s:
sillimanite (all polymorphs of Al2SiO5),
ka: kaolinite, ha: halloysite, di: dickite,
na: nacrite (all polymorphs of
Al2Si2O5(OH)4), and p: pyrophyllite
(Al2Si4O10(OH)2). There are also six
polymorphs of quartz, q (coesite,
stishovite, tridymite, cristobalite, aquartz, and b-quartz).
Degrees of Freedom of a
System
• The number of degrees of freedom in a system is equal
to the sum of the number of independent intensive
variables (generally T and P) and independent
concentrations of components in phases that must be
fixed to define uniquely the state of the system.
• A system that has no degrees of freedom is said to be
invariant, one that has one degree of freedom is
univariant, and so on.
• Thus in a univariant system, for example, we need
specify the value of only one variable, for example,, and
the value of pressure and all other concentrations are
then fixed and can be calculated at equilibrium.
Gibbs Phase Rule
• The phase rule is
ƒ=c - ϕ + 2
o where ƒ is the degrees of freedom, c is the number of components, and f
is the number of phases.
o The mathematical analogy is that the degrees of freedom are equal to
the number of variables minus the number of equations relating those
variables.
• For example, in a system consisting of just H2O, if two
phases coexist, for example, water and steam, then
the system is univariant. Three phases coexist at the
triple point of water, so the system is said to be
invariant, and T and P are uniquely fixed.
Back to Al2O3–H2O-SiO2
• What does our
phase rule (ƒ=c - ϕ
+ 2) tell us about
how many phases
can coexist in this
system over a
range of T and P?
• How many to
uniquely fix the
system?