The Third Law Absolute Entropy and Free Energy

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)2 Si. O 4 o This substitution constitutes a randomness or loss of information about the system; in other words, it constitutes an entropy, called the configurational entropy. o “perfectly crystalline substances” are nonetheless another useful fiction.

Configurational Entropy • Configurational entropy may be calculated as: o (text may be 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

Olivine Configurational Entropy • • (Mg, Fe)2 Si. O 4 is an example of solid solution. Three crystallographic sites: o o o • • • i= 1 Anion or O site (m=4) i=2 Si site (tetrahedral (m=1) i= 3 cation or M site (m=2) mj is the total number of atoms in the jth crystallographic and Xi, j is the mole fraction of the ith atom (element) in the jth site. Substitution for O and Si is generally insignificant. Mixing of Fe and Mg in the M site is usually all we are concerned with,

Pyroxene Configurational Entropy • • • Pyroxenes are more complex. For ”Clinopyroxenes” such as diopside (Ca. Mg. Si 2 O 6) We have four crystallographic sites: o o I c= 1 Anion or O site (m=6) I =2 Si site (tetrahedral (m=2) I = 3 M 1 site (normally occupied by Mg) I = 4 M 2 site (normally occupied by Ca) There is commonly substitution of Fe for Mg in the M 1 site and substitution of Mg or Fe in the M 2 site. Furthermore, there can be other substitutions in the M 2 site, such as Na. Since Na has a valence of 1 and Ca 2, this requires a coupled substitution in the tetrahedral site, such as Al or Ti for Si.

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, t, is : • where ∆Sϕ is the entropy change of any phase change (e. g. , melting of ice to form water).

Standard State Enthalpies and Entropies • By convention, we define the enthalpies of elements (or elemental compounds such as O 2) in the standard state of 298. 16 K 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 Si. O 2 as the enthalpy of reaction to form the compounds under standard state conditions: • ∆Hof, Si. O 2 = ∆Hr where the reaction is: Si + O 2 = Si. O 2 • 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 • Enthalpy change as a function of T and P is: • And • Entropy change as a function of T and P is: • And

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 substituting into these expressions and integrating, e. g. : • The temperature integral* would be: *using Maier-Kelley heat capacities

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, then: o We can substitute • The pressure integral becomes

Free Energies Helmholz and Gibbs

Helmholz Free Energy • Helmholz Free Energy defined as: A = U-TS • Functionally, it is: d. A = –Sd. T-Pd. V • 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, for example when we want to use T and V as independent variables. • More commonly, we use the …

Gibbs Free Energy Don’t worry about this • 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. By differentiating the above, we can derive the Gibbs Free Energy change for a reaction ss: Do d. G = Vd. P-Sd. T memorize • Notice that it, like the Helmholz Free Energy, contains this 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: and this G = H-TS d. G = d. H-Td. S-Sd. T • 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.

Temperature and Pressure dependence • Since: d∆Gr = ∆Vrd. P-∆Srd. T o it is apparent that the dependencies of Gibb Free Energy on temperature and pressure are: • A reaction occurring as a result of increase in pressure at constant T will proceed in the direction of lower volume. • A reaction occurring as a result of increase in temperature at constant P will proceed in the direction of higher entropy.

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)? 2 Mg. O + Si. O 2 = Mg 2 Si. O 4 ∆Gr = Gf, Mg 2 Si. O 4 – 2 Gf, Mg. O- Gf, Si. O 2 • 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: • Ca. Al 2 Si 2 O 8 + 2 Mg 2 Si. O 4 = Ca. Mg. Si 2 O 6 + Mg. Al 2 O 4 + 2 Mg. Si. O 3 • 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 1273 K we need to begin with d∆Gr = ∆Vrd. P - ∆Srd. T • 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 , β, the 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.