Suggested reading:

Garrels, R.M. and Christ, C.M. (1965) Solutions, minerals and equilibria, Freeman and Cooper. 450 pages.

Nordstrom D. K. and Munoz J. L. (1985) *Geochemical
Thermodynamics:*
Benjamin/Cummings Publishing Co., Inc., Menlo Park, CA, Pages 270-281.

**Construction of stability diagrams**.

Graphical representation of mineral phases and the effect of aqueous fluid composition and temperature on their stability's.

Recall Gibb's phase rule.

f = c - p + 2

where:

- f = number of degrees of freedom
- c = number of components in the system
- p = number of phases in the system

variables include; T, P, µ_{}_{i} , µ_{}_{j} ...

Example: K_{}_{2}O
- Na_{}_{2}O
- Al_{}_{2}O3
- SiO_{}_{2}
- H_{}_{2}O
system

In a five component system at constant T and P, description of a single phase requires 4 degrees of freedom

f = 5 - 1 + 2 = 6 - 2 (T, P) = 4

If the 4 variables are taken to be activities and 2 are held constant then we can make a 2 dimensional coordinate system where any reaction between any two pair of phases will produce univariant curve. It follows then that any reaction between three phases will be invariant.

The specific choice of variables is both dictated and represented by the formulas of the reacting species.

In the above system we might choose the aqueous species:

*a*_{}_{K}^{+},
*a*_{}_{Na}^{+}, *a*_{}_{H}^{+},
*a*_{}_{H4SiO4}^{o}

This example is simplified for the purpose of demonstration. The Athens
Gneiss can be represented by mineral assemblage: Quartz, Muscovite,
Albite,
Microcline, Kaolinite and Gibbsite.

Simplifying assumptions:

- Aquesous solution is always present
- Al is always in a solid phase
- Si concentrations is fixed by quartz saturation
- P and T are constant

I. Write reactions for mineral pairs using ions you wish to plot.

Kao + 5 H_{}_{2}O
= 2 Gib + 2 H_{}_{4}SiO_{}_{4}^{o}

Qtz + 2 H_{}_{2}O
= H_{}_{4}SiO_{}_{4}^{o}

2 Mic + 9 H_{}_{2}O
+ 2H^{+} = 2K^{+}
+ H_{}_{4}SiO_{}_{4}^{o} + Kao

2 Alb + 9 H_{}_{2}O
+ 2H^{+} = 2Na^{+}
+ H_{}_{4}SiO_{}_{4}^{o} + Kao

Mus + 9 H_{}_{2}O
+ H^{+} = 3 Gib + 3 H_{}_{4}SiO_{}_{4}^{o}
+ K^{+}

2 Mus + 3 H_{}_{2}O
+ 2 H^{+} = 3 Kao + 2 K^{+}

Mic + 7 H_{}_{2}O
+ H^{+} = Gib + 3 H_{}_{4}SiO_{}_{4}^{o}
+ K^{+}

3 Mic + 12 H_{}_{2}O
+ 2 H^{+} = Mus + 2 K^{+}
+ 6 H_{}_{4}SiO_{}_{4}^{o}

Alb + 7 H_{}_{2}O
+ H^{+} = Gib + 3 H_{}_{4}SiO_{}_{4}^{o}
+ Na

3Alb + K^{+} + 3H^{+}
+ 12 H_{}_{2}O
= Mus + 3Na +H + 6H_{}_{4}SiO_{}_{4}^{o}

Alb + K^{+} + H^{+}
= Mic + Na + H^{+}

II. determine the DG of reaction and solve for K (at equilibrium).

Recall: If DG = 0 , then the reaction will not proceed in either direction (at equilibrium state). In this case,

ΔG^{o}
= - RT ln K

To calculate the ΔG_{reaction} the ΔG^{o}_{formation}
must be obtain from a a data table such as:

Robie R. A., Hemingway B. S., and Fisher J. R.
(1984) *Thermodynamic properties of minerals and related substances
at
298.15 K and 1 bar (10+e5 pascals) pressure and higher temperature:*
Bulletin 1452, U.S. Geological Survey, Washington DC.

For the reaction: Kaolinite + 5 H_{}_{2}O = 2 Gibbsite + 2H_{}_{4}SiO_{}_{4}^{o}

- ΔG
^{o}_{f}Gibbsite = -1155 kJ/mol - ΔG
^{o}_{f}Kaolinite = -3799 kJ/mol - ΔG
^{o}_{f}H_{2}O = -237 kJ/mol - ΔG
^{o}_{f}H_{4}SiO_{4}^{o}= -1308 kJ/mol

ΔG_{reaction} = the products -
reactants

ΔG_{reaction} = (2 x -1155) + (2
x-1308) - (-3799) - (5 x -237)
= -59 kJ/mol

therefore solve for K,

K = 4.2 x 10^{-11
}

K = (*a*H_{}_{4}SiO_{}_{4}^{o})^{2}

or

log K = -10.4

log K = 2 log *a*H_{}_{4}SiO_{}_{4}^{o} =
-10.4

III. Determine the coordinate system. Because the silica activity is fixed by quartz saturation, we wish to display the stability relations for all of the above reactions in terms of K

The coordinate system is therefore define by the ratios:

*a*_{}_{K}^{+}
/ *a*_{}_{H}^{+}

*a*_{}_{Na}^{+}
/*a*_{}_{H}^{+}

IV. Determine the relative stability of the none alkali-bearing phases.

Because kaolinite and gibbsite and quartz do not contain sodium or potassium, their stability's are govern by the silica activity and temperature.

If quartz saturation is always maintained at an activity of *a*H_{}_{4}SiO_{}_{4}^{o} = 10^{-3.95}
the direction of reactions
can be assessed.

For example in step II above, it was shown that for the reaction of kaolinite ---> gibbsite

2 log *a*H_{}_{4}SiO_{}_{4}^{o} =
-10.4

Therefore, under these conditions, the equilibrium activity of
silica
is 10^{-5.4}.

The reaction will be driven to the left and kaolinite is more stable
than gibbsite under these conditions.

V. Assess all the reaction pairs.

Form of equations is y = mx +b

where

y = log (*a*Na+ / *a*H+)

x = log (*a*K+ / *a*H+)

m is the log coefficient

b = zero intercept

VI. Plot lines on coodinates and determine stable forms.