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The Third Law, Absolute Entropy and Free Energy

The Third Law, Absolute Entropy and Free Energy. Lecture 4 . The Third Law.

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The Third Law, Absolute Entropy and Free Energy

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  1. The Third Law, Absolute Entropy and Free Energy Lecture 4

  2. 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.” • In the natural world, perfectly crystalline substances are rare if not non-existent. • 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 • This substitution constitutes a randomness or loss of information about the system; in other words, it constitutes an entropy, called the configurational entropy.

  3. Configurational Entropy • Configurational entropy may be calculated as: • (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 thejthsite.

  4. 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 ∆Sϕis the entropy change of any phase change (e.g., melting of ice to form water).

  5. 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. • 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).

  6. 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:

  7. 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

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