At the eutectic point of a binary system at constant pressure, where liquid coexists with two crystalline phases, the number of degrees of freedom is:

0, because F = C − P + 1 = 2 − 3 + 1, so temperature and composition are fixed

1, because the liquid composition can still slide along the liquidus

3, because C = 2 and each crystalline phase adds one more variable

0, because F = C − P + 1 = 2 − 3 + 1, so temperature and composition are fixed

2, because both temperature and composition remain free at constant pressure

In a simple binary eutectic system A-B a bulk composition lies on the A-rich side of the eutectic. During equilibrium cooling, what are the first crystals to form and the composition of the last liquid?

Crystals of A, and a last liquid of the eutectic composition that solidifies at the eutectic temperature

Crystals of B, and a last liquid of pure A composition that solidifies at the melting point of A

A fine eutectic mixture of A and B, and a last liquid of pure B composition at the lowest temperature

Crystals of A, and a last liquid of pure B composition that persists below the eutectic temperature

In the forsterite-silica system enstatite melts incongruently. Which consequence is correct?

On cooling, forsterite reacts with the liquid at the peritectic to form enstatite, so olivine and quartz do not coexist at equilibrium

Forsterite and quartz can coexist stably at every temperature below the peritectic, since both are crystalline phases of the same system

Enstatite cannot form from a cooling liquid, because incongruent melting is an irreversible process that leaves forsterite behind

At the peritectic only liquid and enstatite are stable, and forsterite is completely absent from the equilibrium assemblage there

On cooling, forsterite reacts with the liquid at the peritectic to form enstatite, so olivine and quartz do not coexist at equilibrium

A plagioclase liquid crystallises with each crystal isolated from the liquid as it grows (fractional crystallisation). Compared with equilibrium crystallisation the crystals are:

zoned from calcic cores to sodic rims, with the final liquid reaching a more sodic composition

homogeneous and calcic, because the composition of the first crystal is preserved throughout

zoned from calcic cores to sodic rims, with the final liquid reaching a more sodic composition

zoned from sodic cores to calcic rims, because the liquid becomes enriched in the anorthite component

identical to the equilibrium crystals, because crystal composition depends only on temperature

In a ternary system a boundary curve has its temperature maximum where it crosses the Alkemade line joining two crystalline phases. A liquid lying on the boundary curve just to one side of this divide, during fractional crystallisation, will:

move along the boundary away from the divide to a lower-temperature eutectic on its own side, never crossing it

cross the divide, because the liquid always moves to the lowest-temperature eutectic in the whole system

move along the boundary toward the divide, because the temperature is highest there and crystallisation is driven uphill

stay at constant composition, because the divide is a barrier to crystallisation itself and no crystals can grow

move along the boundary away from the divide to a lower-temperature eutectic on its own side, never crossing it

Using pK₁ = 6.35 and pK₂ = 10.33 at 25 °C, which species of dissolved inorganic carbon is dominant at pH 8.3?

HCO₃⁻, because pH 8.3 lies between pK₁ = 6.35 and pK₂ = 10.33

CO₃²⁻, because the pH is above 7 and the water is therefore alkaline

H₂CO₃*, because it is the first species formed when CO₂ dissolves in water

an equal mixture of the three species, because pH 8.3 is close to the mean of the two pK values

A dilute water is in equilibrium with solid calcite. Which change increases the amount of calcite that dissolves, that is, the solubility of calcite?

Sodium chloride, which raises the ionic strength and so lowers the activity coefficients of Ca²⁺ and CO₃²⁻

Sodium carbonate, which adds CO₃²⁻ as a common ion and drives the equilibrium toward calcite

Calcium chloride, which adds Ca²⁺ as a common ion and drives the equilibrium toward calcite

Sodium chloride, which raises the ionic strength and so lowers the activity coefficients of Ca²⁺ and CO₃²⁻

A 20 K rise in temperature at fixed P(CO₂), because most solids dissolve more when warm and calcite is no exception

A groundwater has a saturation index SI(calcite) = −0.50. This means that:

IAP is about 0.32 of K(sp), so calcite, if present, would tend to dissolve

calcite is supersaturated and will precipitate from the water immediately

IAP is about 0.32 of K(sp), so calcite, if present, would tend to dissolve

the water is exactly in equilibrium with calcite, because any SI smaller than 1 in magnitude means saturation

the water cannot contain any dissolved calcium or carbonate at all

In an organic-rich groundwater in which dissolved oxygen has been used up, which electron acceptor do microbes use next?

Nitrate, which is reduced to N₂ in denitrification once oxygen is gone

Sulphate, which is reduced to H₂S before any nitrate is touched

Carbon dioxide, which is reduced to CH₄ as the first step after oxygen

Fe(III) oxyhydroxides, which are always reduced before nitrate

For the half-reaction MnO₂ + 4H⁺ + 2e⁻ ⇌ Mn²⁺ + 2H₂O at 25 °C, with the activity of Mn²⁺ held fixed, the boundary on an Eh-pH diagram has a slope of:

−0.118 V per pH unit, from the factor 0.0592 × m/n with m = 4 and n = 2

−0.0592 V per pH unit, as for every redox boundary on the diagram

−0.0296 V per pH unit, taking the number of electrons n = 2 as the divisor only

−0.118 V per pH unit, from the factor 0.0592 × m/n with m = 4 and n = 2

zero, because the boundary is horizontal when a solid phase is involved

Which of the following statements about the thermodynamics of magma crystallisation are correct?

Crystallisation releases latent heat, with ΔH(cryst) = −ΔH(fus)

The entropy of a pure phase falls on crystallisation at its melting point by ΔH(fus)/T(m)

A second component that dissolves only in the liquid raises its entropy and lowers the liquidus temperature

Crystallisation releases latent heat, with ΔH(cryst) = −ΔH(fus)

The entropy of a pure phase falls on crystallisation at its melting point by ΔH(fus)/T(m)

The latent heat released can raise a magma above its liquidus temperature once enough crystals have formed

A second component that dissolves only in the liquid raises its entropy and lowers the liquidus temperature

Which of the following statements about ternary igneous phase diagrams at constant pressure are correct?

A liquid on a cotectic curve coexists with two crystalline phases and has F = 1

At the ternary eutectic three crystalline phases coexist with liquid and F = 0

A liquid on a cotectic curve coexists with two crystalline phases and has F = 1

At the ternary eutectic three crystalline phases coexist with liquid and F = 0

While a single phase crystallises, the liquid composition moves toward the composition of that phase

Fractional crystallisation can carry a liquid across a thermal divide if enough crystals are removed

Which of the following statements about Eh-pH diagrams are correct?

A boundary between two species that differ only in the number of electrons is horizontal

A boundary between two species that differ only in the number of protons is vertical

The upper stability limit of water at pH 7 and P(O₂) = 1 atm is about +0.82 V at 25 °C

A boundary between two species that differ only in the number of electrons is horizontal

A boundary between two species that differ only in the number of protons is vertical

The upper stability limit of water at pH 7 and P(O₂) = 1 atm is about +0.82 V at 25 °C

An Eh-pH diagram predicts how fast each stable species forms from the others

Crystallisation of Magma, Igneous Phase Diagrams, Speciation in Natural Waters, Saturation Index and Eh-pH

This chapter answers the second half of the Thermodynamics heading of Part B1: “Enthalpy and entropy of magma crystallization, Binary and ternary phase diagrams in igneous systems; Chemical speciation and equilibrium in natural waters, solubility and precipitation of minerals - saturation index calculations, Eh-pH diagrams and redox equilibria”. It teaches the latent heat and entropy of crystallisation, how to read binary eutectic, peritectic and solid-solution diagrams with the lever rule, how ternary diagrams and thermal divides control liquid paths, ionic strength and activity in natural waters, the carbonate system, the saturation index, and Nernst-based Eh-pH diagrams and the redox ladder. The numericals are latent-heat budgets, lever-rule proportions, ionic strength, saturation index and Eh, with every constant stated in the question.

1. Enthalpy and entropy of magma crystallisation

Melting a pure crystal at its melting temperature T(m) is an equilibrium, so ΔG = 0 and ΔS(fus) = ΔH(fus)/T(m). Crystallisation is the reverse: it releases the latent heat ΔH(fus) and lowers the entropy of the magma, since liquid becomes ordered crystal, while the heat given to the surroundings raises their entropy by more, so the total still increases. Latent heats of silicates are large, of order 10² kJ/mol (diopside and anorthite) or a few hundred kJ/kg, about 4 × 10⁵ J/kg for basaltic magma.

Thermal buffering. Crystallising a mass fraction φ releases φL per kilogram, which would otherwise raise the temperature by φL/c(p). For φ = 0.20, L = 4 × 10⁵ J/kg and c(p) = 1,200 J/(kg·K) the release equals 0.20 × 4 × 10⁵/1,200 = 67 K of cooling. A crystallising magma therefore cools more slowly than a liquid would.
Freezing-point depression. A second component adds entropy of mixing to the liquid, stabilising it, so the liquidus falls below T(m). For an ideal dilute liquid ΔT = R·T(m)²·X(B)/ΔH(fus). For T(m) = 1,500 K, ΔH(fus) = 100 kJ/mol and X(B) = 0.05 this is 9.4 K. This is the thermodynamic origin of eutectics and of the fall of the liquidus when water dissolves in a melt.
Melting range. A solid solution has no single T(m): it begins to melt at the solidus and finishes at the liquidus, with the liquid richer in the lower-melting component.

⚠️ Latent heat slows cooling; it does not heat the magma above its liquidus

Crystallisation is exothermic, ΔH(cryst) = −ΔH(fus), and the magma’s entropy falls, ΔS(cryst) = −ΔH(fus)/T(m). The released heat buffers the cooling path, but the magma still cools as heat is lost to the country rock. Do not carry the sign of ΔH(fus) over to crystallisation.

2. Binary phase diagrams and the lever rule

A binary diagram plots temperature against composition at constant pressure. Above the liquidus everything is liquid, below the solidus everything is solid, and between them crystals and liquid coexist. With C = 2 and fixed P, F = 3 − P: a one-phase field has F = 2, a two-phase field has F = 1 (at a given T the compositions are fixed), and an invariant point with three phases has F = 0.

The lever rule gives the proportions of the two phases. Draw the horizontal tie-line at the temperature; the end points are the phase compositions X₁ and X₂ and the bulk composition X₀ lies between. The fraction of phase 1 is (X₂ − X₀)/(X₂ − X₁): the arm on the far side of the bulk composition, over the whole tie-line. Worked example. In a system A-B with eutectic liquid E at 40 wt% B and pure B crystals, take a bulk of 70 wt% B just above the eutectic temperature. The fraction of liquid is (100 − 70)/(100 − 40) = 0.50. Check by mass balance: 0.50 × 40 + 0.50 × 100 = 70.

Binary diagram types
Type	Reaction on cooling	F at fixed P	Example
Eutectic	L → A + B at one temperature, an isothermal halt	0	Diopside-anorthite: near 1270 °C and about 40 wt% anorthite
Peritectic	L + A → C: crystal A is consumed	0	Forsterite-silica: forsterite + liquid → enstatite
Solid-solution loop	Continuous: crystal and liquid compositions slide along solidus and liquidus	1 inside the loop	Albite-anorthite plagioclase, about 1100 to 1550 °C

Eutectic path. For a bulk on the A side of the eutectic, A crystallises first at the liquidus, the liquid slides down the liquidus to E, and at the eutectic temperature A and B crystallise together at constant T until the liquid is gone. Solid solution. In plagioclase the first crystal is calcic; with equilibrium the crystal re-equilibrates along the solidus, but when it does not (fractional crystallisation) the crystals are zoned, calcic cores to sodic rims, and the liquid reaches more sodic compositions. Peritectic. Enstatite melts incongruently to forsterite plus a silica-rich liquid, so on cooling olivine reacts with the liquid to form orthopyroxene. This reaction relationship is why olivine and quartz do not coexist in equilibrium, and incomplete reaction leaves olivine with pyroxene rims.

⚠️ The lever rule works on a tie-line in a two-phase field only

Apply it only at one temperature inside a two-phase region, and measure the arm on the opposite side of the bulk composition from the phase you want. At exactly the eutectic temperature the three phases (A, B, liquid E) coexist, so the lever rule is applied just above or just below it.

3. Ternary diagrams and thermal divides

A ternary composition is plotted in an equilateral triangle: each corner is 100 percent of one component, a point on an edge is binary, and a line from a corner keeps the ratio of the other two constant. Percentages sum to 100, so two read-offs fix the third. Temperature is shown by contours on the liquidus surface or by arrows pointing to falling temperature.

Primary phase field: liquid plus one crystalline phase, F = 3 − 2 + 1 = 2 at fixed P.
Cotectic (boundary) curve: liquid plus two crystalline phases, F = 1; the liquid moves along it toward falling temperature.
Ternary eutectic: liquid plus three crystalline phases, F = 0, the lowest temperature of the system.

Liquid path. While one phase crystallises, the liquid moves along the straight line directly away from the composition of that phase; mass balance puts the bulk composition on the line joining the liquid and the crystals, and the lever rule applies along it. On reaching a cotectic the liquid follows the curve while two phases crystallise, and ends at the ternary eutectic. Any bulk composition inside the system finishes at that eutectic unless a thermal divide intervenes.

An Alkemade line joins the compositions of two phases whose primary fields share a boundary curve. Where that curve crosses the line, the temperature is a maximum along the curve: a thermal divide. Liquids cannot cross a divide by fractional crystallisation, so compositions on opposite sides end at different eutectics. The Alkemade lines also cut the diagram into compatibility triangles, and the final assemblage is the three phases at the corners of the triangle that encloses the bulk composition. A thermal divide of this kind at low pressure separates silica-undersaturated (nepheline-normative) from silica-saturated basalt liquids, so alkali basalt and tholeiite cannot be derived from one another by low-pressure fractional crystallisation.

🎯 Why a divide matters

Fractional crystallisation always drives the liquid down in temperature. At a thermal divide the boundary curve has its temperature maximum, so the liquid on either side is driven away from the divide, never across it. This is a thermodynamic barrier that sets which magma series can be related.

4. Chemical speciation, activity and the carbonate system

A dissolved element is spread over free ions and complexes (ion pairs): calcium occurs as Ca²⁺, CaSO₄⁰ and CaHCO₃⁺, so total concentration is not free-ion concentration. Speciation is found by solving mass balance, charge balance and the mass-action equations with an activity model. Complexation raises the solubility of metals such as Al (fluoride complexes), Fe (hydroxy complexes) and Au (chloride and bisulphide complexes).

The activity of an ion is a = γm, with m the molality and γ the activity coefficient. The ionic strength is I = ½ Σ m(i) z(i)² in mol/kg. At 25 °C the Debye-Hückel limiting law is log γ = −0.509 z² √I (for I below about 0.005 m), the extended form adds a denominator 1 + B·a₀·√I (to about 0.1 m), and the Davies equation is log γ = −0.509 z² [√I/(1 + √I) − 0.3 I] (to about 0.5 m). Seawater, with I of about 0.7 m, needs ion-pairing or Pitzer models; neutral species have γ close to 1. Worked example: 0.010 m CaCl₂ has m(Ca²⁺) = 0.010 and m(Cl⁻) = 0.020, so I = ½(0.010 × 4 + 0.020 × 1) = 0.030 m. Rising I lowers γ, which is why salts of non-common ions raise the solubility of sparingly soluble minerals (the salt effect).

The carbonate system controls the pH of most natural waters: CO₂(aq) + H₂O ⇌ H₂CO₃* ⇌ H⁺ + HCO₃⁻ ⇌ 2H⁺ + CO₃²⁻, with pK₁ = 6.35 and pK₂ = 10.33 at 25 °C. By the Henderson-Hasselbalch relation pH = pK + log([base]/[acid]), H₂CO₃* dominates below pH 6.35, HCO₃⁻ between 6.35 and 10.33 and CO₃²⁻ above 10.33; the two neighbours are equal at pH = pK. In waters of pH 6 to 9, alkalinity [HCO₃⁻] + 2[CO₃²⁻] + [OH⁻] − [H⁺] is close to [HCO₃⁻].

⚠️ Total concentration is not activity

Equilibrium constants use activities of the free species. Putting total concentration, or molality without γ, into an ion activity product overstates it, most strongly in saline waters, where the activity coefficients of divalent ions fall far below 1. Always correct for complexation and ionic strength first.

5. Solubility, precipitation and the saturation index

For calcite, K(sp) = a(Ca²⁺)·a(CO₃²⁻) = 10^−8.48 at 25 °C (gypsum 10^−4.58). The ion activity product IAP is the same expression evaluated for the water, and the saturation index is SI = log(IAP/K(sp)). SI = 0 is equilibrium; SI > 0 is supersaturation, so precipitation is thermodynamically possible; SI < 0 is undersaturation, so a mineral that is present would dissolve. Because the free energy of dissolution is ΔG(r) = RT ln(IAP/K), SI is a free-energy measure: ΔG(r) = 2.303RT·SI = 5.71 kJ/mol × SI at 25 °C.

Worked example. A spring water has m(Ca²⁺) = m(SO₄²⁻) = 8.0 × 10⁻³ and γ = 0.55 for each, with K(sp) of gypsum 10^−4.58 and a(H₂O) taken as 1. IAP = (0.55 × 8.0 × 10⁻³)² = 1.94 × 10⁻⁵, log IAP = −4.71, and SI = −4.71 + 4.58 = −0.13: slightly undersaturated, so gypsum, if present, would dissolve.

CO₂ pressure and degassing. Ca²⁺ + 2HCO₃⁻ ⇌ CaCO₃ + CO₂ + H₂O: higher P(CO₂) dissolves calcite, and loss of CO₂ from a spring or cave drip raises pH and SI and precipitates travertine, tufa and speleothems.
Common-ion and salt effects. Adding Ca²⁺ (gypsum dissolution) lowers the CO₃²⁻ that calcite can hold; adding NaCl lowers γ and raises calcite solubility. Calcite is retrograde: its solubility falls as temperature rises.
Kinetics. SI > 0 does not guarantee precipitation: nucleation barriers and inhibitors such as Mg²⁺ and phosphate keep surface seawater supersaturated with calcite and dolomite, and dissolved silica precipitates as opal or chalcedony long before quartz.

6. Redox equilibria and Eh-pH diagrams

For a half-reaction Ox + m H⁺ + n e⁻ ⇌ Red the Nernst equation at 25 °C is Eh = E° − (0.0592 m/n) pH + (0.0592/n) log(a(Ox)/a(Red)), where 0.0592 V = 2.303RT/F. The free energy is ΔG = −nFE, and the electron activity is expressed as pe = Eh/0.0592. A high Eh means an oxidising environment.

The stability field of water is bounded by two lines. The upper one, O₂ + 4H⁺ + 4e⁻ = 2H₂O, is Eh = 1.23 − 0.0592 pH (at P(O₂) = 1 atm) and the lower one, 2H⁺ + 2e⁻ = H₂, is Eh = 0.00 − 0.0592 pH. At pH 7 these are +0.82 V and −0.41 V, and natural waters lie between them. On an Eh-pH diagram a boundary between species that differ only by electrons is horizontal (m = 0, as for Fe³⁺/Fe²⁺ with E° = +0.77 V), a boundary with only protons is vertical (an acid-base or hydrolysis equilibrium), and a boundary with both slopes at −0.0592 m/n volts per pH unit. The diagram shows the species that are thermodynamically stable for chosen activities, and says nothing about rates.

Worked example. For Fe³⁺ + e⁻ ⇌ Fe²⁺ with E° = +0.77 V and a(Fe³⁺)/a(Fe²⁺) = 10⁻², Eh = 0.77 + 0.0592 × log(10⁻²) = 0.77 − 0.118 = +0.65 V. In natural waters iron is mobile as Fe²⁺ under anoxic conditions and precipitates as ferric hydroxide, Fe(OH)₃, once oxidised at near-neutral pH, giving the rust-coloured deposits where anoxic groundwater meets air.

The redox ladder: electron acceptors used as Eh falls
Order	Electron acceptor	Product and consequence
1	O₂	H₂O; oxic water
2	NO₃⁻	N₂; nitrate removed
3	Mn(IV) oxides	Mn²⁺ dissolved
4	Fe(III) oxyhydroxides	Fe²⁺ dissolved; sorbed arsenic released
5	SO₄²⁻	H₂S; sulphide minerals such as pyrite precipitate
6	CO₂	CH₄; methanogenesis

⚠️ Measured Eh is rarely an equilibrium Eh

A platinum electrode reads a mixed potential, because couples such as O₂/H₂O, SO₄²⁻/H₂S and N₂/NO₃⁻ do not equilibrate at the electrode. Use measured Eh as an indicator of the redox zone, and the Fe²⁺/Fe³⁺ ratio or the dominant electron acceptor, not a single Eh number, to interpret a groundwater.

Key takeaways

Crystallisation releases latent heat (ΔH(cryst) = −ΔH(fus), about 4 × 10⁵ J/kg for basalt) and lowers the magma’s entropy by ΔH(fus)/T(m); the heat buffers cooling, and a second component lowers the liquidus by about R·T(m)²·X(B)/ΔH(fus).
In binary diagrams F = 3 − P at fixed P; the lever rule gives (far arm)/(whole tie-line). A eutectic crystallises isothermally with F = 0, a peritectic is a reaction relationship (olivine + liquid → pyroxene), and solid solutions give zoning under fractional crystallisation.
In ternary systems the liquid moves directly away from the crystallising phase, then along a cotectic (F = 1) to the ternary eutectic (F = 0). A thermal divide, the temperature maximum where a boundary crosses an Alkemade line, cannot be crossed by fractional crystallisation.
Activity is a = γm and I = ½ Σ m z². The carbonate system has pK₁ = 6.35 and pK₂ = 10.33, with HCO₃⁻ dominant between them. Complexation raises solubility, and the salt effect raises the solubility of minerals when non-common ions are added.
SI = log(IAP/K(sp)): zero at equilibrium, positive when supersaturated, negative when undersaturated, with the free energy of dissolution ΔG(r) = 5.71 kJ/mol × SI at 25 °C. Degassing of CO₂ precipitates calcite, and SI > 0 does not guarantee precipitation because of kinetics.
Eh = E° − (0.0592 m/n) pH + (0.0592/n) log(a(Ox)/a(Red)). Water is stable between 1.23 − 0.0592 pH and 0 − 0.0592 pH; horizontal Eh-pH boundaries are pure redox, vertical ones pure pH, and slopes are −0.0592 m/n. Microbes use O₂, NO₃⁻, Mn(IV), Fe(III), SO₄²⁻ and CO₂ in turn as Eh falls.

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 the eutectic point of a binary system at constant pressure, where liquid coexists with two crystalline phases, the number of degrees of freedom is:
1. 1, because the liquid composition can still slide along the liquidus
2. 3, because C = 2 and each crystalline phase adds one more variable
3. 0, because F = C − P + 1 = 2 − 3 + 1, so temperature and composition are fixed
4. 2, because both temperature and composition remain free at constant pressure
Show answer
Answer: C — 0, because F = C − P + 1 = 2 − 3 + 1, so temperature and composition are fixed
At fixed pressure F = C − P + 1. With C = 2 and P = 3 (liquid plus two crystals), F = 0: the eutectic is an invariant point, so the temperature and the liquid composition are both fixed and cooling halts there until the liquid is gone. The liquid can slide along the liquidus only while a single crystalline phase is present, where F = 1.
In a simple binary eutectic system A-B a bulk composition lies on the A-rich side of the eutectic. During equilibrium cooling, what are the first crystals to form and the composition of the last liquid?
1. Crystals of A, and a last liquid of the eutectic composition that solidifies at the eutectic temperature
2. Crystals of B, and a last liquid of pure A composition that solidifies at the melting point of A
3. A fine eutectic mixture of A and B, and a last liquid of pure B composition at the lowest temperature
4. Crystals of A, and a last liquid of pure B composition that persists below the eutectic temperature
Show answer
Answer: A — Crystals of A, and a last liquid of the eutectic composition that solidifies at the eutectic temperature
A bulk on the A side first meets the liquidus of A, so A crystallises first and the liquid moves along the liquidus to the eutectic. The last liquid always has the eutectic composition and freezes isothermally at the eutectic temperature into A + B. B cannot crystallise first for an A-rich bulk, and no liquid survives below the eutectic temperature in equilibrium.
In the forsterite-silica system enstatite melts incongruently. Which consequence is correct?
1. Forsterite and quartz can coexist stably at every temperature below the peritectic, since both are crystalline phases of the same system
2. Enstatite cannot form from a cooling liquid, because incongruent melting is an irreversible process that leaves forsterite behind
3. At the peritectic only liquid and enstatite are stable, and forsterite is completely absent from the equilibrium assemblage there
4. On cooling, forsterite reacts with the liquid at the peritectic to form enstatite, so olivine and quartz do not coexist at equilibrium
Show answer
Answer: D — On cooling, forsterite reacts with the liquid at the peritectic to form enstatite, so olivine and quartz do not coexist at equilibrium
The peritectic is a reaction point, forsterite + liquid ⇌ enstatite, with three phases and F = 0 at fixed pressure. Cooling drives the reaction toward enstatite, so olivine and quartz are separated by enstatite in the equilibrium assemblage. Incongruent melting is reversible, and forsterite is one of the three phases at the peritectic, not absent.
A plagioclase liquid crystallises with each crystal isolated from the liquid as it grows (fractional crystallisation). Compared with equilibrium crystallisation the crystals are:
1. homogeneous and calcic, because the composition of the first crystal is preserved throughout
2. zoned from calcic cores to sodic rims, with the final liquid reaching a more sodic composition
3. zoned from sodic cores to calcic rims, because the liquid becomes enriched in the anorthite component
4. identical to the equilibrium crystals, because crystal composition depends only on temperature
Show answer
Answer: B — zoned from calcic cores to sodic rims, with the final liquid reaching a more sodic composition
The first plagioclase is calcic. If crystals are removed from contact with the liquid, the early calcic material is not re-equilibrated and later growth is progressively more sodic, so crystals are normally zoned and the liquid gets more sodic than it would at equilibrium. Homogeneous or reverse-zoned crystals do not follow, and composition depends on temperature only when the crystals re-equilibrate.
In a ternary system a boundary curve has its temperature maximum where it crosses the Alkemade line joining two crystalline phases. A liquid lying on the boundary curve just to one side of this divide, during fractional crystallisation, will:
1. cross the divide, because the liquid always moves to the lowest-temperature eutectic in the whole system
2. move along the boundary toward the divide, because the temperature is highest there and crystallisation is driven uphill
3. stay at constant composition, because the divide is a barrier to crystallisation itself and no crystals can grow
4. move along the boundary away from the divide to a lower-temperature eutectic on its own side, never crossing it
Show answer
Answer: D — move along the boundary away from the divide to a lower-temperature eutectic on its own side, never crossing it
Fractional crystallisation always drives the liquid to lower temperature along the boundary. At a thermal divide the temperature is a maximum along the curve, so liquids on either side move away from the divide toward different eutectics. The idea that the liquid can reach the global minimum ignores the divide, and the divide does not stop crystallisation, only the crossing.
Using pK₁ = 6.35 and pK₂ = 10.33 at 25 °C, which species of dissolved inorganic carbon is dominant at pH 8.3?
1. HCO₃⁻, because pH 8.3 lies between pK₁ = 6.35 and pK₂ = 10.33
2. CO₃²⁻, because the pH is above 7 and the water is therefore alkaline
3. H₂CO₃*, because it is the first species formed when CO₂ dissolves in water
4. an equal mixture of the three species, because pH 8.3 is close to the mean of the two pK values
Show answer
Answer: A — HCO₃⁻, because pH 8.3 lies between pK₁ = 6.35 and pK₂ = 10.33
H₂CO₃* dominates below pH = pK₁ = 6.35, HCO₃⁻ between 6.35 and 10.33 and CO₃²⁻ above 10.33. At pH 8.3 the ratio [HCO₃⁻]/[H₂CO₃*] is 10^(8.3 − 6.35) ≈ 90, and [CO₃²⁻]/[HCO₃⁻] is 10^(8.3 − 10.33) ≈ 0.01, so HCO₃⁻ is about 98 percent of the total. Being alkaline (pH above 7) does not make carbonate dominant; that needs pH above 10.33.
A dilute water is in equilibrium with solid calcite. Which change increases the amount of calcite that dissolves, that is, the solubility of calcite?
1. Sodium carbonate, which adds CO₃²⁻ as a common ion and drives the equilibrium toward calcite
2. Calcium chloride, which adds Ca²⁺ as a common ion and drives the equilibrium toward calcite
3. Sodium chloride, which raises the ionic strength and so lowers the activity coefficients of Ca²⁺ and CO₃²⁻
4. A 20 K rise in temperature at fixed P(CO₂), because most solids dissolve more when warm and calcite is no exception
Show answer
Answer: C — Sodium chloride, which raises the ionic strength and so lowers the activity coefficients of Ca²⁺ and CO₃²⁻
K(sp) fixes the product of activities a = γm. Added NaCl supplies no common ion but raises I and lowers γ, so more Ca²⁺ and CO₃²⁻ must dissolve to reach the same activity product: the salt effect. Common ions (Na₂CO₃, CaCl₂) push calcite toward precipitation and reduce how much calcite can dissolve, and calcite is retrograde, becoming less soluble when warmed.
A groundwater has a saturation index SI(calcite) = −0.50. This means that:
1. calcite is supersaturated and will precipitate from the water immediately
2. IAP is about 0.32 of K(sp), so calcite, if present, would tend to dissolve
3. the water is exactly in equilibrium with calcite, because any SI smaller than 1 in magnitude means saturation
4. the water cannot contain any dissolved calcium or carbonate at all
Show answer
Answer: B — IAP is about 0.32 of K(sp), so calcite, if present, would tend to dissolve
SI = log(IAP/K(sp)) = −0.50 gives IAP/K(sp) = 10^−0.50 = 0.32, so the water is undersaturated and calcite, if present, would dissolve (for dissolution, ΔG(r) = 5.71 × (−0.50) = −2.9 kJ/mol, which is negative, so dissolution is favoured). SI > 0 is the supersaturated case, and equilibrium means SI = 0, not merely a small |SI|: SI = −0.50 puts IAP at only about a third of K(sp).
In an organic-rich groundwater in which dissolved oxygen has been used up, which electron acceptor do microbes use next?
1. Nitrate, which is reduced to N₂ in denitrification once oxygen is gone
2. Sulphate, which is reduced to H₂S before any nitrate is touched
3. Carbon dioxide, which is reduced to CH₄ as the first step after oxygen
4. Fe(III) oxyhydroxides, which are always reduced before nitrate
Show answer
Answer: A — Nitrate, which is reduced to N₂ in denitrification once oxygen is gone
Microbes use the electron acceptor that yields most energy first, giving the sequence O₂, NO₃⁻, Mn(IV), Fe(III), SO₄²⁻, CO₂. After oxygen the next acceptor is nitrate, so denitrification occurs and nitrate is lost. Sulphate reduction and methanogenesis come much later, at lower Eh, and Fe(III) reduction follows nitrate and manganese.
For the half-reaction MnO₂ + 4H⁺ + 2e⁻ ⇌ Mn²⁺ + 2H₂O at 25 °C, with the activity of Mn²⁺ held fixed, the boundary on an Eh-pH diagram has a slope of:
1. −0.0592 V per pH unit, as for every redox boundary on the diagram
2. −0.0296 V per pH unit, taking the number of electrons n = 2 as the divisor only
3. −0.118 V per pH unit, from the factor 0.0592 × m/n with m = 4 and n = 2
4. zero, because the boundary is horizontal when a solid phase is involved
Show answer
Answer: C — −0.118 V per pH unit, from the factor 0.0592 × m/n with m = 4 and n = 2
The Nernst equation gives Eh = E° − (0.0592 m/n) pH + (0.0592/n) log(a(Ox)/a(Red)). With m = 4 protons and n = 2 electrons the slope is −0.0592 × 4/2 = −0.118 V per pH unit. The value 0.0592 applies only when m = n, as for the water lines; a horizontal boundary needs m = 0, which is not so here since protons take part.
Which of the following statements about the thermodynamics of magma crystallisation are correct?
1. Crystallisation releases latent heat, with ΔH(cryst) = −ΔH(fus)
2. The entropy of a pure phase falls on crystallisation at its melting point by ΔH(fus)/T(m)
3. The latent heat released can raise a magma above its liquidus temperature once enough crystals have formed
4. A second component that dissolves only in the liquid raises its entropy and lowers the liquidus temperature
Show answer
Answer: A — Crystallisation releases latent heat, with ΔH(cryst) = −ΔH(fus); B — The entropy of a pure phase falls on crystallisation at its melting point by ΔH(fus)/T(m); D — A second component that dissolves only in the liquid raises its entropy and lowers the liquidus temperature
At T(m) the crystal and liquid are in equilibrium, so ΔS(fus) = ΔH(fus)/T(m); crystallisation reverses both, releasing heat and lowering entropy. Mixing in a second component that stays out of the crystals adds entropy to the liquid, stabilising it and depressing the liquidus (a component that enters the crystals as a solid solution can instead raise it). Crystallisation happens only below the liquidus, and the latent heat merely slows the cooling; it cannot raise the magma above it.
Which of the following statements about ternary igneous phase diagrams at constant pressure are correct?
1. A liquid on a cotectic curve coexists with two crystalline phases and has F = 1
2. At the ternary eutectic three crystalline phases coexist with liquid and F = 0
3. While a single phase crystallises, the liquid composition moves toward the composition of that phase
4. Fractional crystallisation can carry a liquid across a thermal divide if enough crystals are removed
Show answer
Answer: A — A liquid on a cotectic curve coexists with two crystalline phases and has F = 1; B — At the ternary eutectic three crystalline phases coexist with liquid and F = 0
With C = 3 at fixed P, F = 4 − P: liquid plus two crystals (P = 3) gives F = 1 on a cotectic, and liquid plus three crystals (P = 4) gives F = 0 at the eutectic. While one phase crystallises the liquid moves directly away from its composition, since that component is being removed, and a thermal divide cannot be crossed by fractional crystallisation however many crystals are removed.
Which of the following statements about Eh-pH diagrams are correct?
1. A boundary between two species that differ only in the number of electrons is horizontal
2. A boundary between two species that differ only in the number of protons is vertical
3. The upper stability limit of water at pH 7 and P(O₂) = 1 atm is about +0.82 V at 25 °C
4. An Eh-pH diagram predicts how fast each stable species forms from the others
Show answer
Answer: A — A boundary between two species that differ only in the number of electrons is horizontal; B — A boundary between two species that differ only in the number of protons is vertical; C — The upper stability limit of water at pH 7 and P(O₂) = 1 atm is about +0.82 V at 25 °C
A pure redox couple has m = 0 and so no pH term, giving a horizontal line; a pure acid-base equilibrium has no electrons and so no Eh dependence, giving a vertical line. The water line is 1.23 − 0.0592 × 7 = 0.82 V at pH 7. The diagram shows thermodynamic stability only and says nothing about reaction rates, which is why many predicted equilibria are not reached in nature.
In a binary system A-B the eutectic liquid E contains 36 wt% B and the other solid phase is pure A (0 wt% B). A bulk composition of 27 wt% B is held just above the eutectic temperature, where liquid E coexists with crystals of A. By the lever rule, the percentage of liquid by mass is ____.
Numerical answer — type the value.
Show answer
Answer: 75
The tie-line runs from 0 wt% B (crystals of A) to 36 wt% B (liquid E). The fraction of liquid is (27 − 0)/(36 − 0) = 0.75, which is 75 percent. Check: 0.75 × 36 + 0.25 × 0 = 27. Measuring the wrong arm would give 25 percent, the fraction of crystals.
A basaltic magma crystallises 30 percent of its mass (φ = 0.30). The latent heat of crystallisation is L = 3.6 × 10⁵ J/kg of crystals and the specific heat of the magma is c(p) = 1,200 J/(kg·K). The latent heat released is equivalent to how many kelvin of sensible cooling of the magma, φL/c(p)? Give the answer to the nearest whole number.
Numerical answer — type the value.
Show answer
Answer: 90
Heat released per kilogram of magma = φL = 0.30 × 3.6 × 10⁵ = 1.08 × 10⁵ J/kg. Dividing by c(p): 1.08 × 10⁵/1,200 = 90 K. The magma must lose this heat in addition to the sensible heat before it cools by the same amount, which is why crystallisation buffers cooling.
A water contains 0.005 mol/kg Na₂SO₄ and 0.002 mol/kg MgSO₄, both fully dissociated, and no other solutes. Its ionic strength I = ½ Σ m z², in mol/kg to three decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 0.023
The ions are Na⁺ 0.010 (z = 1), Mg²⁺ 0.002 (z = 2) and SO₄²⁻ 0.005 + 0.002 = 0.007 (z = 2). I = ½[0.010 × 1 + 0.002 × 4 + 0.007 × 4] = ½[0.010 + 0.008 + 0.028] = ½ × 0.046 = 0.023 mol/kg. Forgetting to double the sodium (two Na⁺ per formula unit) or the z² weighting are the usual slips.
A water has m(Ca²⁺) = 2.0 × 10⁻³ mol/kg and m(CO₃²⁻) = 1.0 × 10⁻⁵ mol/kg, with activity coefficients γ = 0.70 for each ion. For calcite take K(sp) = 10^−8.48 at 25 °C. The saturation index SI = log(IAP/K(sp)), to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 0.47
a(Ca²⁺) = 0.70 × 2.0 × 10⁻³ = 1.4 × 10⁻³ and a(CO₃²⁻) = 0.70 × 1.0 × 10⁻⁵ = 7.0 × 10⁻⁶. IAP = 9.8 × 10⁻⁹ and log IAP = −8.009. SI = −8.009 − (−8.48) = +0.47: supersaturated. Using molalities without γ gives IAP = 2.0 × 10⁻⁸ and SI = +0.78, which is the tempting error.
For MnO₂ + 4H⁺ + 2e⁻ ⇌ Mn²⁺ + 2H₂O take E° = +1.23 V at 25 °C and use 0.0592 V for 2.303RT/F. At pH 7 with a(Mn²⁺) = 10⁻⁶ and a(MnO₂) = a(H₂O) = 1, the value of Eh in volts, to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 0.58
Eh = E° − (0.0592/2) × log[a(Mn²⁺)/a(H⁺)⁴] = 1.23 − 0.0296 × (log 10⁻⁶ + 4 × 7) = 1.23 − 0.0296 × (−6 + 28) = 1.23 − 0.0296 × 22 = 1.23 − 0.651 = 0.579, so 0.58 V. Forgetting the factor of 4 on pH, or using 0.0592 instead of 0.0592/2, gives other values.