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 SiO 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.

Configurational Entropy Configurational entropy may be calculated as: o (text is missing the parentheses) where m j is the total number of atoms in the j th crystallographic site (in atoms per formula unit) and X i,j is the mole fraction of the i th atom (element) in the j th 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 ∆S ϕ 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 O 2 ) in the standard state of 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 SiO 2 as the enthalpy of reaction to form the compounds under standard state conditions: ∆H o f,SiO2 = ∆H r where the reaction is: Si + O 2 = SiO 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 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 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: 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 + SiO 2 = Mg 2 SiO 4 ∆G r = G f,Mg2SiO4 – 2G f,MgO - G f,SiO2 The answer will be that it will proceed to the right if ∆G r is negative. However, ∆G r is a function of T and P, so that ∆G r 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: CaAl 2 Si 2 O 8 + 2Mg 2 SiO 4 = CaMgSi 2 O 6 + MgAl 2 O 4 + 2MgSiO 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 ∆G r of reaction is 0. We can look up values for standard state ∆G r in Table 2.2, but to calculate ∆G r at 1273K we need to begin with d∆G r = ∆V r dP - ∆S r dT and integrate Since ∂S/∂T) P = C P /T The temperature integral becomes:

This is a general form using Maier-Kelly heat capacities of the change in ∆G r 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.