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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Komponenten SiO 2 - Al 2 O 3 - H 2 O Phasen Kyanite Ky Al 2 SiO 5 Sillimanite Si Al 2 SiO 5 Andalusite And Al 2 SiO 5 -Quartz aQz SiO 2 -Quartz bQz SiO 2 Kaolinite Kln AL 2 Si 2 O 5 (OH) 4 Pyrophyllite Prl AL 2 Si 4 O 10 (OH) 2 Water W H O Nur stabile Reaktionen
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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
