Magmalogy assignment

Different Mineral Reaction Equilibirum Question Solved
1.Al2SiO5 (Kyanite) ↔ Al2SiO5 (Sillimanite)
2.NaAlSi3O8 (Low-Albite) ↔ NaAlSi2O6 (Jadeite) + SiO2 (Quartz)
3.3CaAl2Si2O8 (Anorthite) ↔ Ca3Al2Si3O12 (grossular-garnet) + 2Al2SiO5 (Kyanite) + SiO2 (Quartz)

1. EXPERIMENTAL MAGMALOGY ASSIGNMENT
By Sanjay Dubey
5TH MAY 2017
Problem No.1
Reaction Equilibria:
Al2SiO5 (Kyanite) ↔ Al2SiO5 (Sillimanite)
Al2SiO5 (Sillimanite) ↔ Al2SiO5 (Andalusite)
Al2SiO5 (Andalusite) ↔ Al2SiO5 (Kyanite)
Objective:
A. Illustrate the Reaction Boundaries
B. Label the Mineral Phase at Each Boundaries
Assumption
∆Cp =0 and ∆V is independent of Temperature and Pressure
Data:
Phase H1bar, 298
[KJ/mol]
S1bar,298
[J/mol.k]
V1bar,298
[cm
3
/mol]
Kyanite
Al2SiO5
-2596.200 82.3000 44.150
Sillimanite
Al2SiO5
-2587.770 95.690 49.860
Andalusite
Al2SiO5
-2591.700 91.420 51.520

Solution:
Reaction 1.
Al2SiO5 (Kyanite) ↔ Al2SiO5 (Sillimanite)

Reaction are equilibrium hence ∆G=0
∆G=∆H-T∆S=0
∆H= HProduct - HReactant = {Hsillimanite- HKyanite }
= {-2587.770 – (-2596.200)}
= +8.43 KJ/mol = 8430J/mol
∆S= SProduct - SReactant = {Ssillimanite - Skyanite}
= {95.690-82.300} = 13.39J/mol. K
∆V= Vproduc t- Vreactant = {Vsillimanite-VKyanite}
= {49.860 - 44.150} = 5.71cm
3
/mol= 0.571J/bar.mol

Making the use of assumption that ∆Cp =0 and ∆Vis independent of P and T. The equilibrium temperature at
1bar is given by:
∆G= 0 = ∆H-T∆S
T(Equilibrium) = ∆H/∆S= 8430/13.39 = 629.5K

The Clausius-Clapeyron equation enables the slope of the reaction boundary at any pressure and temperature
of interest to be calculated from ∆S and ∆V.

dP
dT
=
∆S
∆V
= 13.39/0.571=23.45bar/K

2. Notes: Before calculating dP/dT from the Clausius-Clapeyron equation, it is necessary to make the units of ∆V
consistent with those of ∆S. If ∆S is in entropy units, the consistent units of ∆V are Joules per bar (J/bar).

Alternatively, the 1bar result may be extrapolated to higher pressure using the Clausius-Clapeyron equation and
making the assumption that ∆S and ∆V are constant at 1bar, 298K values. The value obtained from the pressure
will be used to plot reaction boundary. This is generally reasonable for solid-solid reaction. Now, if you add 200K
to the equilibrium temperature, what is equilibrium pressure and the resultant pressure can be calculated by using
Clausius-Clapeyron Equation.

dP/dT= ∆S/∆V
dP = (∆S/∆V) * dT
P=1+23.45(829.5-629.5) =4691bar=4.691Kbar is equilibrium pressure at 829.5k

Similarly,
Now, if you add 400K to the equilibrium temperature,
P=1+23.45(1029.5-629.5) =9381bar=9.381Kbar is equilibrium pressure at 1029.5k
SIMILARLY
Reaction 2.
Al2SiO5 (Sillimanite) ↔ Al2SiO5 (Andalusite)

Reaction are equilibrium hence ∆G=0
∆G=∆H-T∆S=0
∆H= -3930J/mol and ∆S = -4.37J/mol. K and ∆V= +0.166J/bar.mol
T(Equilibrium) = 899K
slope of the reaction boundary
dP/dT= -26.32 bar/K

If you subtract 100K from the equilibrium temperature, the reason behind the subtraction is if you take any value
more than 899K then because of slope the pressure values becomes negative. Now, if you subtract 100K to the
equilibrium temperature, what is equilibrium pressure and the resultant pressure can be calculated by using
Clausius-Clapeyron Equation.

P=1-26.32*(799-899) =2633bar=2.633Kbar is equilibrium pressure at 799K

Reaction 3.
Al2SiO5 (Andalusite) ↔ Al2SiO5 (Kyanite)

Reaction are equilibrium hence ∆G=0
∆G=∆H-T∆S=0
∆H= -4500J/mol and ∆S = -9.12J/mol. K and ∆V= - 0.737J/bar.mol
T(Equilibrium) = 493.42K
slope of the reaction boundary
dP/dT= 12.37bar/K
Now, if you add 200K to the equilibrium temperature, what is equilibrium pressure and the resultant pressure can
be calculated by using Clausius-Clapeyron Equation.

P=1+12.37*(693.42-493.42) =2475bar=2.475Kbar is equilibrium pressure at 694.42K

3. To plot the graph:
a. Plot the point Teq calculated of Reaction 1 and Join it with the points {i.e. (829.5K, 4.691Kbar) and
(1029.5K, 9.381Kbar)} calculated by using C.C. equation in the case of Kyanite ↔ Sillimanite. It will
provide the reaction boundary for the Kyanite and Sillimanite.
b. Plot the Point Teq calculated of Reaction 2 and Join it with the point with lower temperature (i.e.
799K,2.633Kbar) calculated by using C.C. equation in the case of Sillimanite ↔ Andalusite. It will
provide the reaction boundary for the Sillimanite and Andalusite.
c. Plot the Point Teq calculated of Reaction 3 and Join it with the point with higher temperature (i.e.
694.42K,2.475Kbar) calculated by using C.C. equation in the case of Andalusite ↔ Kyanite. It will
provide the reaction boundary for the Andalusite and Kyanite.

























Figure 1: Pressure-Temperature phase diagram for Andalusite-Sillimanite-Kyanite
Results:
a. In reaction 1: ∆V positive and ∆H positive it means increasing P favors the reactant and Increasing T
favors the product since the reaction is endothermic. In reaction 2: ∆V positive and ∆H negative it means
Increasing P favors reactant and increasing temperature favors the reactant since the reaction is exothermic.
In reaction 3: ∆V is negative and ∆H is negative it means increase in pressure will favors the product and
increase in temperature favors the product.
b. Above figure 1 show the stability region for Andalusite-Kyanite-Sillimanite. Three reaction boundaries
not meet at single point in my case. Expected triple point from the figure 1 is 3.5Kbar and 780K and
(refer to figure 2 to check). My reaction boundary is not close and miss the triple point by 10K and 10bar.
c. From the figure.1 it is clear that low temperature and pressure favors the Kyanite and region below the
triple point and 900K provides the stable region for the Andalusite. Figure 1 explain that Sillimanite forms
at high temperature.

4. Figure: 2 Phase diagram for –Andalusite-Sillimanite-Kyanite from Kieffer, S.W. (1982): My triple point
was quite similar to the one I found in the research paper.

Problem No.2
Reaction Equilibria:
NaAlSi3O8 (Low-Albite) ↔ NaAlSi2O6 (Jadeite) + SiO2 (Quartz)
Objective:
A. Calculate the equilibrium pressure at 800K with ∆Cp=0 and ∆V is independent of P and T
B. Calculate the equilibrium pressure at 800K with ∆Cp is not 0 and ∆V is independent of P and T
C. Compare the equilibrium pressure in A and B
Data:
Phase H1bar, 298
[KJ/mol]
S1bar,298
[J/mol.k]
V1bar,298
[cm
3
/mol]
Low Albite
NaAlSi3O8
-3935.120 207.40 100.070
Jadeite
NaAlSi2O6
-3029.400 133.47 60.400
Quartz
SiO2
-910.700 41.46 22.688

Solution:
Part 1.
Reaction are equilibrium hence ∆G=0 and ∆Cp=0 and ∆V is independent of P and T
∆G=∆H-T∆S=0
∆H= -4980 J/mol and ∆S = -32.47J/mol. K and ∆V= -1.6982 J/bar.mol
T(Equilibrium) = 153.3K
The Clausius-Clapeyron equation enables the slope of the reaction boundary at any pressure and temperature of
interest to be calculated from ∆S and ∆V.
dP/dT= 19.210bar/K
the equilibrium pressure obtained is 12.2Kbar and by using Clausius-Clapeyron equation i.e. dP/dT= ∆S/∆V; dP
= (∆S/∆V) * dT; P=1+19.210 (800-153) =12429 bar=12.429 K-bar is equilibrium pressure at 800K

5. Part 2
∆Cp is not zero and ∆V is independent of pressure and temperature.
Additional Data

Low Albite
Cp= 5.8394E
2
– 0.092852T + 2.2722E
-5
T
2
-6.4242

E
3
T
-0.5
+1.6780

E
6
T
-2

Jadeite
Cp=3.0113

E
2
+ 1.0143

E
-2
T – 2.0551

E
3
T
-0.5
-2.2393

E
6
T
-2

Alpha quartz
Cp=44.603 + 3.7754

E
-2
T -1.0018

E
6
T
-2


∆Cp is not zero and ∆V is independent of pressure and temperature.
Calculation for the Entropy
∆S800= ∆S298 + ∫∆Cp.T
-1
dT and ∆V= -1.6982 J/bar.mol
For above equation ∆Cp can be written as:
∆Cp= a + bT + cT
2
+ dT
-0.5
+ eT
-2

∆Cp = ∆Cp(product) -∆Cp(reactant)
= -238.207 + 0.140749T -2.272E
-5
T
2
+ 4369.1T
-0.5
- 4.919E
6
T
-2

Now Integrate the Equation and
∫∆Cp.T
-1
dT= ∫∆Cp.1/TdT
800
298

=∫-238.20T
-1
+0.1407-2.272E
-5
T +4369.1T
-1.5
-4.91E
6
T
-3
. dT
After integration
=-238.20lnT+0.1407T -1.135E
-5
T
2
-8738.2T
-0.5
-2.45E
6
T
-2
Put the finite limit on integration
=-235.23+70.6314-6.262+197.24-23.80= 2.5794J/mol.K
∆S800= ∆S298 + ∫∆Cp.T
-1
dT=-32.47+2.5794= - 29.8906J/mol.K…. Equation (1)
Calculation for the Enthalpy
∆H800= ∆H298 + ∫∆Cp.dT
∫∆Cp.dT= ∫∆Cp.dT
800
298

=∫-238.207 + 0.140749T -2.272E
-5
T
2
+ 4369.1T
-0.5
- 4.919E6T
-2
.dT
After integration
=-238.207T+0.07035T
2
-0.7573

E
-5
T
3
+8738.2T
0.5
+4.91

E
6
T
-1
Putting the finite limit
=-119579.9+38776.6-3677.12+96308.8-10339.0= 1489.38J/mol
∆H800= ∆H298 + ∫∆Cp.dT= -4980+1489.3= - 3490.9J/mol…..Equation(2)
Reaction at equilibrium
∆G=∆H-T∆S=0
So Tequilibrium = ∆H/∆S = -3490.95/-29.8906 = 116.79K …… (Eq.1 and 2)
Use of Clausius-Clapeyron equation:
dP/dT= 17.60bar/K
dP/dT= ∆S/∆V; dP = (∆S/∆V) * dT; P=1+17.60 (800-116.79) =12025.4 bar=12.0 K-bar is equilibrium pressure when ∆Cp
≠ 0
Results
 Equilibrium pressure calculated with ∆Cp=0 was 12.429Kbar and if ∆Cp ≠ 0 is 12.025 Kbar. Error between
two calculations is very small. This explain the necessity of performing the heat capacity integration for
reactions involving in the solids. Figure A: show the graph for the equilibrium pressure at ∆Cp = 0 and ∆Cp
≠ 0. Also the value for the pressure matches to the reference in figure 4.
 Equilibrium temp in the case of ∆Cp=0 was T=153.3K and ∆Cp ≠ 0 was T=116.79. This may because of
the relation provided between ∆Cp and Temperature.

6. In this corrected case the error is also not so large in the calculation ∆G with both above cases (∆G∆Cp=0= 20.996KJ/mol
and ∆G∆Cp≠0=20.422K. This explain that ∆H and ∆S are effectively constant over wide range of temperature





















Figure A : Phase Diagram: Low Albite ↔ Jadeite + Quartz





Figure 4: Phase diagram for –Albite-Jd-Qz from Kieffer, S.W. (1982): Equilibrium pressure from the
research paper at 800K is around 12Kbars which is similar to the value obtained by keeping ∆Cp=0.

7. Problem No.3
Reaction Equilibria:
3CaAl2Si2O8 (Anorthite) ↔ Ca3Al2Si3O12 (grossular-garnet) + 2Al2SiO5 (Kyanite) + SiO2 (Quartz)
Objective :
A. Define the system (components suitable for expressing the chemical compositions for these four mineral
phases)
B. Plot the chemical composition of these mineral phases onto the system you defined in A.
C. Calculate Reaction-boundary by keeping assumption that ∆Cp=0 and ∆V is independent of P and T
D. Describe the mineral assemblage is stable at each side of the reaction boundary.

Data:
Phase H1bar, 298
[KJ/mol]
S1bar,298
[J/mol.k]
V1bar,298
[cm
3
/mol]
Anorthite
CaAl2Si2O8
-4232.500 203.00 100.070
Grossular-garnet
Ca3Al2Si3O12
--6634.308 256.118 125.310
Kyanite
Al2SiO5
-2596.200 82.300 44.150

Quartz
SiO2
-910.700 41.460 22.688

Solution:
Part A and B

Reaction:
3CaAl2Si2O8 (Anorthite) ↔ Ca3Al2Si3O12 (grossular-garnet) + 2Al2SiO5 (Kyanite) + SiO2 (Quartz)

The system(component) for mineral phase can be express it as follow:
Anorthite (CaAl2Si2O8 ) : Al2O3 +2SiO2 + CaO
Grossular-garnet (Ca3Al2Si3O12 ) : Al2O3 +3SiO2 +3CaO
Kyanite (Al2SiO5 ) : Al2O3 +SiO2 +0.CaO
Quartz(SiO2): 0. Al2O3 +0. SiO2 +0. CaO

Mineral Moles
CaO
Moles
Al2O3
Moles o
SiO2
%CaO %Al2O3 %SiO2
Anorthite 1 1 2 25 25 50
Grossular-
Garnet
3 1 3 43 14 43
Kyanite 1 1 0 0 50 50
Quartz 0 0 0 0 0 100

Mineral Moles CaO Moles
Al2O3
Moles
SiO2
% CaO % Al2O3 %SiO2
anorthite 1 1 2 25 25 50
grossular 3 1 3 43 14 43
kyanite 1 1
Quartz

8. Above chemical composition of four mineral phase can be plotted in the form of ternary diagram where the
chemical components are SiO2, CaO, and Al2O3. This ternary is important for the study of many metamorphic
rocks




Trinary Diagram shown Below:

Figure 5: Ternary Diagram of SiO2-Al2O3-CaO


Part C and D
Reaction are equilibrium hence ∆G=0
∆G=∆H-T∆S=0
∆H= -39908 J/mol and ∆S = -146.822J/mol. K and ∆V= -6.4302 J/bar.mol
T(Equilibrium) = 271K
The Clausius-Clapeyron equation enables the slope of the reaction boundary at any pressure and temperature
of interest to be calculated from ∆S and ∆V. dP/dT= 22.83bar/K

9. Figure 6: Phase Diagram for Anorthite ↔ Grossular Garnet +Kyanite + Quartz



Figure 7: Value of Equilibrium pressure is around 19Kbar from Newton, R. C. (1983). My diagram
explains the similar result

10. Results:

 Part A and B defines the suitable expression for the mineral phases
 This mineral phases are represented in the form of ternary diagram figure no.5
 Phase diagram shown in the figure no.6. Here ∆H negative and ∆V negative explains that the reaction is
exothermic and increase in temperature will favors the reactant i.e Anorthite will form more and
Increase in pressure will form more product i.e. GG+ KY+QZ.
 Figure 7 show the similar result prediction of my work compare to the research paper.



Reference
 Kieffer, S. W. (1982). Thermodynamics and lattice vibrations of minerals: 5. Applications to phase equilibria, isotopic
fractionation, and high‐pressure thermodynamic properties. Reviews of Geophysics, 20(4), 827-849.
 Newton, R. C. (1983). Geobarometry of high-grade metamorphic rocks. American Journal of Science, 283, 1-28.