Komponenten SiO 2 - Al 2 O 3 - H 2 O Phasen KyaniteKy Al 2 SiO 5 SillimaniteSi Al 2 SiO 5...

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Transcript of Komponenten SiO 2 - Al 2 O 3 - H 2 O Phasen KyaniteKy Al 2 SiO 5 SillimaniteSi Al 2 SiO 5...

Komponenten

SiO2 - Al2O3 - H2O

PhasenKyanite Ky

Al2SiO5

Sillimanite SiAl2SiO5

Andalusite AndAl2SiO5

-Quartz aQzSiO2

-Quartz bQzSiO2

Kaolinite KlnAL2Si2O5(OH)4

Pyrophyllite PrlAL2Si4O10(OH)2

Water WH2ONur stabile

Reaktionen

Komponenten

SiO2 - Al2O3 - H2O

PhasenKyanite Ky

Al2SiO5

Sillimanite SiAl2SiO5

Andalusite AndAl2SiO5

-Quartz aQzSiO2

-Quartz bQzSiO2

Nur stabile Reaktionen

Komponenten

SiO2 - Al2O3 - H2O

PhasenKyanite Ky

Al2SiO5

Sillimanite SiAl2SiO5

Andalusite AndAl2SiO5

-Quartz aQzSiO2

-Quartz bQzSiO2

alle Reaktionen

Komponenten

SiO2 - Al2O3 - H2O

PhasenKyanite Ky

Al2SiO5

Sillimanite SiAl2SiO5

Andalusite AndAl2SiO5

-Quartz aQzSiO2

-Quartz bQzSiO2

alle Reaktionen

Al2SiO5

Al2SiO5

T ? T ?

Δ aG = Δ fHT0 ,P0 + Cp dT

T0

T

∫ −T ⋅ST0 ,P0 −T ⋅ CpTT0

T

∫ dT + V dPP0

P

T ? T ?

Gesucht:

für 4000 Bar und 10000 Bar:

∆aG(Sillimanit) = ∆aG(Kyanit)

∆aG(Sillimanit) - ∆aG(Kyanit) = 0

678.5 oC

678.5 K

Δ aG = Δ fHT0 ,P0 + Cp dT

T0

T

∫ −T ⋅ST0 ,P0 −T ⋅ CpTT0

T

∫ dT + V dPP0

P

Sillimanit KyaniteV0 [J/Bar] 4.9900 4.4090

Sillimanit- Kyanite

0.581

ΔV dP1

P

∫ = ΔV ⋅(P −1)

Δ aG = Δ fHT0 ,P0 + Cp dT

T0

T

∫ −T ⋅ST0 ,P0 −T ⋅ CpTT0

T

∫ dT + V dPP0

P

Sillimanit KyaniteV0 [J/Bar] 4.9900 4.4090

Sillimanit- Kyanite

ΔV dP1

4000

ΔV dP1

10000

0.581

= 0.581·(3999) = 2324 J€

ΔV dP1

P

∫ = ΔV ⋅(P −1)

= 0.581·(9999) = 5810 J

Δ aG = Δ fHT0 ,P0 + Cp dT

T0

T

∫ −T ⋅ST0 ,P0 −T ⋅ CpTT0

T

∫ dT + V dPP0

P

Δ aG = Δ fHT0 ,P0 + Cp dT

T0

T

∫ −T ⋅ ST0 ,P0 + T ⋅ CpTT0

T

∫ dT ⎛

⎝ ⎜ ⎜

⎠ ⎟ ⎟+ V dPP0

P

Δ fG = Δ fHT0 ,P0 + Cp dT

T0

T

∫ − mi Cpi dTT0

T

∫i

∑ −T ⋅ ST0 ,P0 + T ⋅ CpTT0

T

∫ dT − mi SiT0 ,P + Cpi

TdT

T0

T

∫ ⎛

⎝ ⎜ ⎜

⎠ ⎟ ⎟

i∑

⎝ ⎜ ⎜

⎠ ⎟ ⎟+ V dPP0

P

Sillimanit Kyanite

∆fG [J/mol]

400 K500 K600 K700 K800 K900 K

1000 K1100 K

V0 [J/Bar] 4.9900 4.4090

Sillimanit- Kyanite

ΔV dP1

4000

ΔV dP1

10000

-2388660-2339238-2289948-2240847-2191922-2143153-2092987-2042207

-2389682-2339004-2288516-2238279-2188275-2138464-2087312-2035566

1022-234

-1432-2568-3647-4689-5675-6641

0.581

= 0.581·(3999) = 2324 J€

ΔV dP1

P

∫ = ΔV ⋅(P −1)

= 0.581·(9999) = 5810 J

Sillimanit KyaniteSillimanit- Kyanite

Sillimanit Kyanite

∆fG [J/mol]

400 K500 K600 K700 K800 K900 K

1000 K1100 K

V0 [J/Bar] 4.9900 4.4090

Sillimanit- Kyanite

ΔV dP1

4000

ΔV dP1

10000

-2388660-2339238-2289948-2240847-2191922-2143153-2092987-2042207

-2389682-2339004-2288516-2238279-2188275-2138464-2087312-2035566

1022-234

-1432-2568-3647-4689-5675-6641

0.581

= 0.581·(3999) = 2324 J€

ΔV dP1

P

∫ = ΔV ⋅(P −1)

= 0.581·(9999) = 5810 J

+ 2324 + 5810Sillimanit KyaniteSillimanit- Kyanite

Sillimanit Kyanite

∆fG [J/mol]

400 K500 K600 K700 K800 K900 K

1000 K1100 K

V0 [J/Bar] 4.9900 4.4090

Sillimanit- Kyanite

ΔV dP1

4000

ΔV dP1

10000

-2388660-2339238-2289948-2240847-2191922-2143153-2092987-2042207

-2389682-2339004-2288516-2238279-2188275-2138464-2087312-2035566

1022-234

-1432-2568-3647-4689-5675-6641

0.581

= 0.581·(3999) = 2324 J€

ΔV dP1

P

∫ = ΔV ⋅(P −1)

= 0.581·(9999) = 5810 J

+ 2324 + 5810Sillimanit KyaniteSillimanit- Kyanite

892-244

135-831

Sillimanit Kyanite

∆fG [J/mol]

400 K500 K600 K700 K800 K900 K

1000 K1100 K

V0 [J/Bar] 4.9900 4.4090

Sillimanit- Kyanite

ΔV dP1

4000

ΔV dP1

10000

-2388660-2339238-2289948-2240847-2191922-2143153-2092987-2042207

-2389682-2339004-2288516-2238279-2188275-2138464-2087312-2035566

1022-234

-1432-2568-3647-4689-5675-6641

0.581

= 0.581·(3999) = 2324 J€

ΔV dP1

P

∫ = ΔV ⋅(P −1)

= 0.581·(9999) = 5810 J

+ 2324 + 5810Sillimanit KyaniteSillimanit- Kyanite

892-244

135-831

678.5 K

1014.0 K

-244

892

-244

892

600700 K

∆ rG(S

il-K

y)f1 = a·x1 + bf2 = a·x2 + b

P = 4000 Bar

-244

892

600700 K

∆ rG(S

il-K

y)f1 = a·x1 + bf2 = a·x2 + b

Nullpunkt (∆rG = 0) :

x = f1 ⋅ x2 − f2 ⋅ x1

f2 − f1

⎛ ⎝ ⎜

⎞ ⎠ ⎟=

892 ⋅700 − −244( ) ⋅600892 − (−244)

= 678.5

P = 4000 Bar

-244

892

600700 K

∆ rG(S

il-K

y)f1 = a·x1 + ba = (∂∆rG/∂T)P = -∆rS

P = 4000 Bar

∆rS = 11.69 J/K·mol

-244

892

600700 K

∆ rG(S

il-K

y)f1 = a·x1 + ba = (∂∆rG/∂T)P = -∆rS

P = 4000 Bar

∆rS = 11.69 J/K·mol

x = x1 + f1ΔrS

= 600 + 89211.69

= 676.3

-831

135

10001100

K

∆ rG(S

il-K

y)f1 = a·x1 + bf2 = a·x2 + b

Nullpunkt (∆rG = 0) :

x = f1 ⋅ x2 − f2 ⋅ x1

f2 − f1

⎛ ⎝ ⎜

⎞ ⎠ ⎟=

135 ⋅1100 − −831( ) ⋅1000135 − (−831)

=1014.0

P = 10000 Bar

678.5 K 1014.0 K

678.5 K 1014.0 K

678.5 K 1014.0 K

Kyanit

Sillimanit