Classical Thermodynamics, the Clapeyron Equation, the Phase Rule, Schreinemakers’ Rules and Solid Solutions
2. Gibbs energy of reaction, equilibrium constant and equilibrium conditions
For a reaction, ΔG°(T) = ΔH° − TΔS° if ΔH° and ΔS° are taken as constant, and ΔG = ΔG° + RT ln Q. At equilibrium ΔG = 0 and Q = K, so ΔG° = −RT ln K, with R = 8.314 J/(mol·K) and T in kelvin. K is dimensionless and built from activities: pure solids and water in excess have a = 1, and a gas enters through its fugacity in bar. At 1 bar the equilibrium temperature follows from ΔG° = 0: T(eq) = ΔH°/ΔS°.
| ΔH | ΔS | Behaviour | Example |
|---|---|---|---|
| + | + | Products favoured above T(eq) | Dehydration, decarbonation, melting |
| − | − | Products favoured below T(eq) | Hydration, crystallisation |
| − | + | Products favoured at all T | Rare for geological reactions |
| + | − | Products never favoured | Rare for geological reactions |
Worked example (temperature). A dehydration reaction has ΔH° = +60 kJ/mol and ΔS° = +100 J/(mol·K). Then T(eq) = 60,000/100 = 600 K, which is 327 °C. At 700 K, ΔG° = 60,000 − 700 × 100 = −10,000 J/mol, so ln K = 10,000/(8.314 × 700) = 1.718 and K = 5.6: the products are favoured.
Pressure enters through (∂ΔG/∂P) at constant T = ΔV. For solids with constant ΔV, ΔG(P) = ΔG(1 bar) + ΔV(P − 1 bar). Worked example (pressure). Calcite → aragonite at 25 °C has ΔG ≈ +1.0 kJ/mol at 1 bar and ΔV ≈ −2.8 cm³/mol = −2.8 × 10⁻⁶ m³/mol. Setting ΔG = 0 gives P ≈ 1.0 × 10³/(2.8 × 10⁻⁶) ≈ 3.6 × 10⁸ Pa, about 0.4 GPa: above this pressure the denser aragonite is the stable polymorph, which is why it forms in high-pressure, low-temperature blueschists.
3. The Clapeyron equation
Along a univariant curve the two sides of a reaction have equal G, so dG is equal on both: −S₁dT + V₁dP = −S₂dT + V₂dP. This gives the Clapeyron equation: dP/dT = ΔS/ΔV = ΔH/(TΔV). If ΔS and ΔV have the same sign the slope is positive; if opposite, negative. With ΔS in J/(mol·K) and ΔV in m³/mol the slope is in Pa/K. Solid-solid reactions typically have slopes of a few MPa/K, so a change of 0.5 GPa in pressure shifts T by tens to hundreds of K. Over a narrow range ΔS and ΔV are nearly constant and the curve is straight.
Worked example. A reaction has ΔS = 25 J/(mol·K) and ΔV = 5 cm³/mol = 5 × 10⁻⁶ m³/mol. Then dP/dT = 25/(5 × 10⁻⁶) = 5 × 10⁶ Pa/K = 5 MPa/K, so a 100 K rise in equilibrium temperature goes with a 500 MPa rise in pressure. The rule of thumb (Le Chatelier): high pressure favours the assemblage of smaller volume, and high temperature the assemblage of larger entropy. Hence aragonite beats calcite, diamond beats graphite, and kyanite, the densest Al₂SiO₅ polymorph, is the high-pressure one.
- Melting: dT/dP = TΔV/ΔH = ΔV/ΔS. For most silicates ΔV(fusion) is positive, so melting temperature rises with pressure, but ever more slowly as the melt compresses. Ice is the exception, with ΔV < 0 and a negative slope.
- Dehydration: at low pressure the fluid has a very large molar volume, so ΔV is large and dP/dT = ΔS/ΔV is small and the curve is shallow. As pressure rises the fluid becomes dense, ΔV shrinks and the curve steepens, and may bend back to a temperature maximum.
- Vaporisation: with an ideal vapour and a negligible liquid volume, d ln P/dT = ΔH/(RT²), so ln P against 1/T is a straight line.
4. The phase rule, components and degrees of freedom
A phase is a physically distinct, homogeneous, mechanically separable part of the system (quartz, a melt, a fluid, one solid solution). A component is a chemically independent constituent. The number of components is the number of chemical species S minus the number of independent reactions R among them, C = S − R (and minus any compositional restrictions). Components are not minerals: calcite, quartz, wollastonite and CO₂ are four species but, with the reaction CaCO₃ + SiO₂ = CaSiO₃ + CO₂, only three components, CaO, SiO₂ and CO₂.
Gibbs’s phase rule is F = C − P + 2, where F is the number of intensive variables (T, P, compositions) that can change independently without changing the assemblage. At fixed pressure, or in condensed systems where pressure is ignored, F = C − P + 1. The assemblage is invariant if F = 0 (a point), univariant if F = 1 (a curve) and divariant if F = 2 (an area). For calcite + quartz + wollastonite + CO₂ gas, C = 3 and P = 4, so F = 1: the reaction is a curve, and at a given CO₂ pressure there is one temperature. In the one-component Al₂SiO₅ system, one phase is divariant, two are univariant and three (kyanite, andalusite, sillimanite) are invariant at the triple point.
The mineralogical phase rule (Goldschmidt) applies this to rocks: an assemblage in equilibrium over a range of P and T (divariant) has as many phases as components, P = C; one extra phase pins it to a P-T line, and two extra to a point. A rock with more phases than components is either at a special P-T condition or out of equilibrium, which is why reaction rims and zoning are read as disequilibrium. Fewer phases than components is common, since many rocks are monomineralic or lack some components.
5. Mineral stability, phase equilibria and Schreinemakers’ rules
A stability field is the region of P-T (or P-T-X) space where an assemblage has the lowest G. Fields meet along univariant curves, where the G of the two assemblages is equal, and at invariant points. Beyond an invariant point a curve continues as a metastable extension: the reaction can be written there but the assemblages on its two sides are not the stable ones. On every curve the high-temperature side carries the assemblage of larger total entropy, and the high-pressure side the assemblage of smaller total volume.
Schreinemakers’ analysis works out the topology of the curves around an invariant point without data. In a system of C components an invariant point has n = C + 2 phases. Leaving out each phase in turn gives n univariant curves, written [i] for the reaction among all phases except i, so a binary (C = 2) invariant point has four phases and four curves, and a unary (C = 1) point has three phases and three curves. Each curve has a stable half and a metastable half on the other side of the point. The rules:
(1) the metastable extension of each curve lies between two other stable curves, which is the same as saying that adjacent stable curves are separated by less than 180°;
(2) the stable halves are fixed by the entropy and volume of the phases, as above.
| Polymorph | Stable at | Reason |
|---|---|---|
| Kyanite | High pressure | Smallest molar volume (densest) |
| Sillimanite | High temperature | Largest entropy |
| Andalusite | Low pressure, moderate temperature | Largest volume, intermediate entropy |
In the unary Al₂SiO₅ system the three curves meet at a triple point near 0.4 GPa and 500 to 550 °C (published positions differ a little). The metastable extension of each curve runs into the stability field of the phase absent from that reaction: the extension of the andalusite-sillimanite boundary [kyanite] enters the kyanite field, and similarly for the others.
6. Mineral solid solutions and activity-composition relations
Most rock-forming minerals are solid solutions (olivine Mg-Fe, plagioclase Ca-Na, garnet, pyroxene). The chemical potential of an end-member in a phase is μ = μ° + RT ln a, where μ° is that of the pure end-member at the same P and T, and the activity is a = γX, with X the mole fraction and γ the activity coefficient. An equilibrium constant is written with activities, K = Π a^ν, so a diluted end-member (a < 1) is stabilised relative to its pure form. Because composition adds variables, a reaction among solid solutions occurs over a P-T interval (divariant), not along a line.
Ideal mixing takes place on crystallographic sites: ΔH(mix) = 0 and ΔS(mix) = −R Σ(n × X ln X), summed over the mixing sites, with n the number of equivalent sites per formula unit. The activity of an end-member is the product of site fractions, a = X^n. In olivine (Mg,Fe)₂SiO₄ the two M sites are equivalent, so a(forsterite) = X(Mg)² and ΔS(mix) = −2R[X ln X + (1 − X) ln(1 − X)], with a maximum 2R ln 2 = 11.5 J/(mol·K) at X = 0.5. For X(Mg) = 0.9, a(forsterite) = 0.81, not 0.9.
Non-ideal solutions are often modelled as regular solutions: ΔS(mix) is ideal but ΔH(mix) = W X₁X₂, so RT ln γ₁ = W X₂². Where the interaction parameter W is positive, the solution is unstable below a critical temperature, for the symmetric case T(c) = W/(2R), and unmixes into two phases along a solvus: exsolution in alkali feldspar, in augite-pigeonite and in calcite-dolomite. The solvus narrows as T rises, so the compositions of coexisting phases are geothermometers. Two limits matter: Raoult’s law, a → X as X → 1 for the solvent, and Henry’s law, a = hX as X → 0 for a dilute solute such as a trace element.
Key takeaways
- G = H − TS decides stability: at constant T and P a closed system moves to lower G and equilibrium is ΔG = 0. With dG = V dP − S dT, high T favours larger S and high P favours smaller V. Convert kJ to J and °C to K before using ΔH − TΔS.
- ΔG° = −RT ln K and T(eq) = ΔH°/ΔS° at 1 bar; the signs of ΔH and ΔS tell whether products are favoured above or below T(eq). At fixed T, ΔG(P) = ΔG(1 bar) + ΔV(P − 1) gives an equilibrium pressure, as for calcite to aragonite near 0.4 GPa.
- The Clapeyron equation is dP/dT = ΔS/ΔV = ΔH/(TΔV), in Pa/K for SI inputs. High pressure favours the denser assemblage and high temperature the higher-entropy one; melting curves rise for most silicates, and dehydration curves are shallow at low pressure because the fluid volume is large.
- The phase rule is F = C − P + 2 (F = C − P + 1 at fixed P or T), with C = S − R. A divariant rock assemblage has P = C. Components are not minerals: calcite, quartz, wollastonite and CO₂ give C = 3 and a univariant reaction.
- At an invariant point of a C-component system there are C + 2 phases and C + 2 univariant curves, each with a stable and a metastable half. Schreinemakers’ rule: each metastable extension lies between two stable curves. In Al₂SiO₅ it runs into the field of the absent phase; kyanite is dense, sillimanite is high-entropy.
- Activity is a = γX, and ideal mixing is on sites: a = X^n, ΔS(mix) = −R Σ(n X ln X), so a(forsterite) = X(Mg)² and the maximum for olivine is 2R ln 2. Positive W gives a solvus below T(c) = W/(2R); Raoult’s law holds for the solvent and Henry’s law for a dilute solute.
Practice questions (18)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
At constant temperature and pressure, a closed system is at equilibrium when:
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Answer: B — its Gibbs free energy has reached a minimum, so that ΔG = 0 for any small change
At constant T and P the criterion is minimum Gibbs free energy, G = H − TS, so a small change has ΔG = 0. Minimum enthalpy alone is tempting but ignores the entropy term, and the entropy of the system alone is maximised only for an isolated system; the entropy of the system plus its surroundings is what matters here, which G accounts for.A dehydration reaction has ΔH° = +90 kJ/mol and ΔS° = +120 J/(mol·K), both taken as independent of temperature. At 1 bar it proceeds toward the products:
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Answer: D — only above 750 K, where the term TΔS has grown larger than ΔH
T(eq) = ΔH/ΔS = 90,000/120 = 750 K, and with both ΔH and ΔS positive the products are favoured above it, where TΔS exceeds ΔH and ΔG turns negative. Forgetting to convert kJ to J gives the tempting value of 0.75 K, and a positive ΔS alone is not enough, since ΔH opposes the reaction at low T.A reaction, as written, has ΔS > 0 and ΔV < 0. Its Clapeyron slope dP/dT is:
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Answer: A — negative, so raising the pressure lowers the temperature at which the reaction takes place
dP/dT = ΔS/ΔV, a positive number divided by a negative one, is negative. The curve leans backward in P-T space, as for the melting of ice. The signs of ΔS and ΔV do not cancel: they combine as a ratio, and the slope has the sign of the product ΔS × ΔV, not of ΔS alone.At the same temperature, which of two polymorphs of a mineral is the stable one at higher pressure?
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Answer: C — The one with the smaller molar volume, because (∂G/∂P) = V and its G rises more slowly
Since the slope of G against P is the molar volume, the denser phase gains Gibbs energy more slowly as P rises and eventually has the lower G. Aragonite over calcite and kyanite over andalusite are examples. Entropy governs the temperature response, not the pressure response, and enthalpy alone cannot decide stability.Why does the P-T curve of a dehydration reaction rise only gently (dP/dT is small) at low pressure?
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Answer: B — Because ΔV is large there, the released fluid having a very large molar volume, so ΔS/ΔV is small
A free fluid at low pressure occupies an enormous volume, so ΔV of the reaction is large while ΔS is also large, and the ratio dP/dT = ΔS/ΔV is small. At high pressure the fluid is dense, ΔV shrinks and the curve steepens. The idea of a solid-like fluid reverses this: it would make ΔV small and the slope steep, which is the high-pressure behaviour.Calcite, quartz, wollastonite and CO₂ gas coexist in equilibrium through the reaction CaCO₃ + SiO₂ = CaSiO₃ + CO₂, with T and P free to vary. The variance (degrees of freedom) of this assemblage is:
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Answer: D — 1, univariant, because S = 4 and R = 1 give C = 3 and F = 3 − 4 + 2
The four species are linked by one independent reaction, so C = S − R = 4 − 1 = 3, and with P = 4 phases, F = C − P + 2 = 1: the assemblage is univariant, a curve in P-T space, and at a given P(CO₂) it fixes the temperature. Counting species as components gives the tempting F = 2, which is wrong because the reaction already ties the compositions.How many univariant curves meet at an invariant point of a two-component system with four phases, and what does each curve represent?
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Answer: A — Four, each being the reaction among the three phases that remain when one phase is omitted
With n = C + 2 = 4 phases, omitting each phase in turn leaves a three-phase reaction, and each such reaction is a univariant curve [i]: four curves in all, each with a stable and a metastable half. The tempting count of six is the number of pairs of phases, but a curve is a reaction among all but one phase, not a pair boundary.At the Al₂SiO₅ triple point, the metastable extension of the andalusite-sillimanite boundary runs into the stability field of:
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Answer: C — kyanite, the phase that takes no part in the andalusite-sillimanite reaction
In a unary system the metastable extension of the boundary between two polymorphs lies in the field of the third phase, the one absent from the reaction: here kyanite. This follows from Schreinemakers’ rule that each extension lies between two stable curves, so the curves cannot be collinear. The extension never continues into either of its own two fields, which is the tempting error.A binary solid solution with a positive interaction parameter W (a regular solution) is cooled below its critical temperature. What happens?
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Answer: D — It unmixes into two coexisting phases of different composition along a solvus
With W > 0 the enthalpy of mixing is positive, and below T(c) = W/(2R) (symmetric case) the −TΔS(mix) term is too small to compensate it, so a single phase has a G curve with a hump and the solution exsolves into two compositions: a solvus, as in alkali feldspar. Cooling lowers TΔS and so makes unmixing more likely, not less.Olivine with X(Mg) = 0.90 mixes ideally on two equivalent M sites per formula unit, (Mg,Fe)₂SiO₄. The activity of the forsterite end-member, Mg₂SiO₄, is:
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Answer: B — 0.81, because each of the two M sites contributes a factor of 0.90
For ideal mixing on n equivalent sites the end-member activity is the product of the site fractions, a = X^n. Here n = 2, so a = 0.90² = 0.81. Setting a equal to the mole fraction (0.90) is the tempting slip, and ideal mixing does not make a = 1: a is 1 only for the pure end-member.Which of the following statements about Gibbs free energy and equilibrium are correct?
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Answer: A — A reaction with ΔG < 0 at constant T and P is thermodynamically favoured, yet it may be extremely slow; B — At constant pressure the slope of G against T is −S, so heating lowers the G of a high-entropy phase faster; D — ΔG° = −RT ln K, so K = 1 when ΔG° = 0
ΔG gives direction, not rate, so a favoured reaction can be kinetically blocked. Since (∂G/∂T) = −S, the high-entropy phase becomes stable at higher T. ΔG° = 0 means ln K = 0 and K = 1. A reaction releasing fluid creates disorder and has a large positive ΔS, so the statement of a negative ΔS is wrong.Which of the following statements about the Clapeyron equation are correct?
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Answer: A — Along a univariant curve dP/dT = ΔS/ΔV = ΔH/(TΔV); B — On a univariant curve the high-pressure side carries the assemblage with the smaller total volume
The Clapeyron relation, and the Le Chatelier statement that high pressure favours the smaller-volume assemblage, are correct. For most silicates the melt is less dense than the crystals, so ΔV of fusion is positive and the melting temperature rises with pressure. The slope dP/dT has units of J/(mol·K) divided by m³/mol, which is Pa/K, not K/Pa.Which of the following statements about the phase rule and components are correct?
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Answer: A — C = S − R, so calcite, quartz, wollastonite and CO₂ related by one independent reaction form three components; C — A divariant rock assemblage stable over an area of P-T space has as many phases as components
Components are species minus independent reactions, and a divariant assemblage has P = C (F = 2). At fixed pressure the rule becomes F = C − P + 1. Five phases in three components give F = 3 − 5 + 2 = 0, which is invariant rather than univariant, and such a rock is usually either at a special P-T point or out of equilibrium.A hydration reaction has ΔH° = −40.0 kJ/mol and ΔS° = −100 J/(mol·K), both independent of temperature. Take R = 8.314 J/(mol·K). The equilibrium constant K at 300 K, to the nearest whole number, is ____.
Numerical answer — type the value.
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Answer: 55
ΔG° = ΔH° − TΔS° = −40,000 − 300 × (−100) = −40,000 + 30,000 = −10,000 J/mol. Then ln K = −ΔG°/(RT) = 10,000/(8.314 × 300) = 4.009 and K = e^4.009 = 55.1, so 55. A sign slip on the negative ΔS would give ΔG° = −70,000 J/mol instead.A solid-solid reaction has ΔS = 21.0 J/(mol·K) and ΔV = 3.5 cm³/mol (both positive). Assuming that the Clapeyron slope stays constant, by how many kelvin does the equilibrium temperature rise when the pressure is raised by 0.45 GPa? (1 cm³ = 10⁻⁶ m³.)
Numerical answer — type the value.
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Answer: 75
dP/dT = ΔS/ΔV = 21.0/(3.5 × 10⁻⁶) = 6.0 × 10⁶ Pa/K = 6.0 MPa/K. For ΔP = 0.45 GPa = 450 MPa, ΔT = 450/6.0 = 75 K. Leaving ΔV in cm³ (a factor of 10⁶ error) is the usual slip.Olivine (Mg,Fe)₂SiO₄ with X(Mg) = 0.75 and X(Fe) = 0.25 mixes ideally on two equivalent M sites per formula unit. Using R = 8.314 J/(mol·K), the molar entropy of mixing, in J/(mol·K) to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 9.35
ΔS(mix) = −2R[X ln X + (1 − X) ln(1 − X)] = −2 × 8.314 × [0.75 ln 0.75 + 0.25 ln 0.25] = −16.628 × [−0.21576 − 0.34657] = 16.628 × 0.56233 = 9.35 J/(mol·K). Forgetting the factor 2 for the two sites gives 4.68.Graphite → diamond at 298 K has ΔG° = +2900 J/mol at 1 bar (10⁵ Pa) and ΔV = −1.9 × 10⁻⁶ m³/mol, taken as constant. The pressure at which graphite and diamond are in equilibrium at 298 K, in GPa to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 1.5
ΔG(P) = ΔG°(1 bar) + ΔV(P − 10⁵) = 0, so P − 10⁵ = 2900/(1.9 × 10⁻⁶) = 1.526 × 10⁹ Pa. Then P ≈ 1.526 × 10⁹ Pa = 1.5 GPa. Above this pressure ΔG < 0 and diamond is the stable form at 298 K.A closed system contains the species MgCO₃, MgO, CO₂, H₂O and Mg(OH)₂, which are related by two independent reactions, MgCO₃ = MgO + CO₂ and Mg(OH)₂ = MgO + H₂O. An assemblage of magnesite, brucite and one homogeneous H₂O-CO₂ fluid coexists in equilibrium, with T and P both free to vary. The number of degrees of freedom is ____.
Numerical answer — type the value.
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Answer: 2
C = S − R = 5 − 2 = 3 (the components are MgO, CO₂ and H₂O). The phases are magnesite, brucite and the fluid, so P = 3, and F = C − P + 2 = 3 − 3 + 2 = 2: the assemblage is divariant. Counting species as components would give 5 − 3 + 2 = 4, which is wrong.