the third law, absolute entropy and free energy lecture 4

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:

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 :

• 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.:

• The temperature integral would be:

The Entropy Pressure Integral

• If we can consider a substance to be incompressible (e.g., a solid), the pressure integral is simply:

• If not, and assuming α and β are constants:

• We can substitute

• The pressure integral becomes

Free EnergiesHelmholz 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:

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

• Since ∂S/∂T)P = CP/T

• The temperature integral becomes:

• 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:

• 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:

• 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).

• For example:

• Since G is a state function

• Therefore:

• Refer to section 2.12 as necessary.