The Third Law,
Absolute Entropy
and Free Energy
Lecture 4
The Third Law
• “If the entropy of each element in some crystalline
state may be taken as zero at the absolute zero of
temperature, every substance has a finite positive
entropy, but at absolute zero, the entropy may
become zero, and does so become in the case of
perfectly crystalline substances.” – Lewis & Randall
• In the natural world, perfectly crystalline substances
are non-existent.
o Many naturally crystalline substances, i.e., minerals, are actually solutions,
with substitution of one type of atom for another, e.g., Fe and Mg in
olivine, (Fe,Mg)2SiO4
o This substitution constitutes a randomness or loss of information about the
system; in other words, it constitutes an entropy, called the configurational
entropy.
Configurational Entropy
• Configurational entropy may be calculated as:
Sconf
æ
ö
= -Rå ç m j å Xij ln Xij ÷
ø
j è
i
o (text is missing the parentheses)
• where mj is the total number of atoms in the jth
crystallographic site (in atoms per formula unit) and
Xi,j is the mole fraction of the ith atom (element) in
the jth site.
o But note that since we are using R rather than k, units will be joules-kelvins-1
-moles-1
Absolute Entropy
• The absolute entropy of a substance may be
calculated as the sum of the configurational
entropy, the entropy changes associated with
changes in T and P, and entropy associated with
any phase changes.
• Mathematically, the entropy of a substance at
some temperature, τ, is :
St = Sconfig + ò
t
0
CP
dT + ∆ Sf
T
• where ∆Sϕ is the entropy change of any phase
change (e.g., melting of ice to form water).
Standard State Enthalpies
and Entropies
• We define the enthalpies of elements (or elemental
compounds such as O2) in the standard state of 298.16K
and 0.1 MPa (25˚C, 1 atm) as 0.
o Units of enthalpies are what?
• We can then determine standard states of formation
(from the elements) of compounds such as SiO2 as the
enthalpy of reaction to form the compounds under
standard state conditions:
• ∆Hof,SiO2 = ∆Hr where the reaction is:
Si + O2 = SiO2
• Standard State entropies and enthalpies are available in
compilations such as the Handbook of Chemistry and
Physics (and, for your convenience, Table 2.2 in the text).
Enthalpies and Entropies at
non-standard state T and P
• Given standard state enthalpy and entropy, we
can calculate these at any other T and P by
integrating, e.g.:
T1
ST1 ,P1
P
1
C
= S o + ò P dT + ò -aV dP
T
298
0.1
• The temperature integral would be:
T1
T
cö
c ù1
æa
é
ò298 çè T + b - T 3 ÷ø dT = êëa lnT + b + 2T 2 úû298
The Entropy Pressure
Integral
• If we can consider a substance to be
incompressible (e.g., a solid), the pressure integral is
simply:
P
1
ò -aV dP = -aV (P - 0.1)
1
0.1
• If not, and assuming α and β are constants:
V = V o (1- b P)
• We can substitute
• The pressure integral becomes
P1
ò -aV
0.1
o
2 P1
(1- b P)dP = a V éë P - b P ùû 0.1
o
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 of thermodynamic quantities. 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 (toward 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.