Predicting Equilibrium and Phases, Components, Species Lecture 5 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. 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 (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. Chapter 2 Thermodynamics of multi-component systems The real world is complicated • Our attempt to estimate the plagioclase-spinel phase boundary failed because we assumed the phases involved had fixed composition. In reality they do not, they are solutions of several components or species. • We need to add a few tools to our thermodynamic tool box to deal with these complexities. Some Definitions • Phase o Phases are real substances that are homogeneous, physically distinct, and (in principle) mechanically separable. For example, the phases in a rock are the minerals present. Amorphous substances are also phases. o NaCl dissolved in seawater is not a phase, but seawater with all its dissolved components (but not the particulates) is. • Species o A species is a chemical entity, generally an element or compound (which may or may not be ionized). The term is most useful in the context of gases and liquids. A single liquid phase, such as an aqueous solution, may contain a number of species. Na+ in seawater is a species. • Components o Components are more carefully defined. But: • We are free to define the components of our system • Components need not be real chemical entities. Minimum Number of Components • The minimum number of components of a system is rigidly defined as the minimum number of independently variable entities necessary to describe the composition of each and every phase of a system. • The rule is: c=n–r o where n is the number of species, and r is the number of independent chemical reactions possible between these species. • How many components do we need to describe a system composed of CO2 dissolved in H2O? Graphical Approach • If it can be graphed in 1 dimension, it is a two component system, in 2 dimensions, a 3 component system, etc. • Consider the hydration of Al2O3 (corundum) to form boehmite (AlO(OH)) or gibbsite Al(OH)3. Such a system would contain four phases (corundum, boehmite, gibbsite, water). • How many components? The system Al2O3–H2O Phase diagram for the system Al2O3–H2O–SiO2 The lines are called joins because they join phases. In addition to the endmembers, or components, phases represented are g: gibbsite, by: bayerite, n: norstrandite (all polymorphs of Al(OH)3), d: diaspore, bo: boehmite (polymorphs of AlO(OH)), a: andalusite, k: kyanite, s: sillimanite (all polymorphs of Al2SiO5), ka: kaolinite, ha: halloysite, di: dickite, na: nacrite (all polymorphs of Al2Si2O5(OH)4), and p: pyrophyllite (Al2Si4O10(OH)2). There are also six polymorphs of quartz, q (coesite, stishovite, tridymite, cristobalite, aquartz, and b-quartz). Degrees of Freedom of a System • The number of degrees of freedom in a system is equal to the sum of the number of independent intensive variables (generally T and P) and independent concentrations of components in phases that must be fixed to define uniquely the state of the system. • A system that has no degrees of freedom is said to be invariant, one that has one degree of freedom is univariant, and so on. • Thus in a univariant system, for example, we need specify the value of only one variable, for example,, and the value of pressure and all other concentrations are then fixed and can be calculated at equilibrium. Gibbs Phase Rule • The phase rule is ƒ=c - ϕ + 2 o where ƒ is the degrees of freedom, c is the number of components, and f is the number of phases. o The mathematical analogy is that the degrees of freedom are equal to the number of variables minus the number of equations relating those variables. • For example, in a system consisting of just H2O, if two phases coexist, for example, water and steam, then the system is univariant. Three phases coexist at the triple point of water, so the system is said to be invariant, and T and P are uniquely fixed. Back to Al2O3–H2O-SiO2 • What does our phase rule (ƒ=c - ϕ + 2) tell us about how many phases can coexist in this system over a range of T and P? • How many to uniquely fix the system?