Download - 08 Exercise Oxide Activities
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(1) Calculate the distribution of the normative components in the silica-rich part
of the system Na2O-SiO2.
(2) Calculate the activity coefficient (Na2O) ate 1400 C for x(SiO2) = 0.7.Compare the result to the experimental data given below.
Exercise:
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0.0 0.2 0.4 0.6 0.8 1.0
-10
-8
-6
-4
-2
0
non-linear approach
log f(SiO2),
1400 C
log f(Na2O)
1000 C
1200 C
1400 C
log(activitycoefficient)
molar fraction of SiO2
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(1) Calculate the distribution of the normative components in the silica-rich partof the system Na2O-SiO2.
(2) Calculate the activity coefficient (Na2O) ate 1400 C for x(SiO2) = 0.7.
Compare the result to the experimental data given below.
Exercise:
0.0 0.2 0.4 0.6 0.8 1.0
-10
-8
-6
-4
-2
0
non-linear approach
log f(SiO2),
1400 C
log f(Na2O)
1000 C
1200 C
1400 C
log(ac
tivitycoefficient)
molar fraction of SiO2
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Set up a coefficient matrix like this: . Let .
10
21
2
2
2
ONa
SiO
NSS
30.0
70.0
2
2
=
ONa
SiO
( )
==
=
)(
)(
10
21
)(
)(
22
2
NSn
Snk
ONan
SiOnj jk
rr
The relation between oxides and compounds reads:
jk=
The equation is resolved for the vector of coumpounds:
( ) ( )
==
=
)(
)(
)(
)(
2
211
2 ONan
SiOnj
NSn
Snk jkjk
rr
( ) ( )kjT
jk A
uu
uu=
=
+
+=
=
10
21
10
21
2221
1211
1
1
The coefficient matrix jk is inverted like this (better use a computer!):
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Now we start with the inverted matrix .
( ) ( )kjjk A==1
10
21
2
22
NS
S
ONaSiO
( )
=
==
=
)(
)(2)(
)(
)(
10
21
)(
)(
2
22
2
2
2 ONan
ONanSiOn
ONan
SiOnjA
NSn
Snk kj
rr
and calculate
Normalization to nS + nNS2 = nSiO2 - 2nNa2O + nNa2O = nSiO2 - nNa2O yields
Thus, xS = 0.25, xNS2 = 0.75.
=
=
=
=
4/3
4/1
17.02
7.01 17.02
27.03
12
1 12
23
)()(
)( )()(
)(2)(
2
2
2
2
22
2
22
22
2
SiO
SiO
SiO
SiO
NS
S
x
xx
x
ONanSiOn
ONan ONanSiOn
ONanSiOn
x
x
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)0,(
)21(
1
21
),(
1
13
),(
)1(
1
1
12
),1(
)12(
1
12
23
/1
31
2
2
22
222
31
21
2
222
22
21
32
2
222
222
32
2
222
22
2
=
=
=
=
=
SiO
SiOSiO
SiOSN
SiO
SiOSiO
SiONS
SiO
SiOSiO
SiONS
SiO
SiOSiO
SiOS
SiOBAB
x
xx
xxNSN
x
xx
xxSNNS
x
xx
xxNSNS
x
xx
xxNSS
xxxxABrange
We calculated the constitutional structure for range I. The results across the entire
compositional range are compiled below:
I.
II.
III.
IV.
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0 20 40 60 80 100
0.0
0.2
0.4
0.6
0.8
1.0
SiO2
Na2Si2O5
Na2SiO3
Na4SiO4
Na2O
mol % SiO2
fraction
ofspecies
The results for the constitutional structure across the entire compositional range are
shown. The calculated result for xSiO2 = 0.70 is marked.
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0 20 40 60 80 1000.0
0.2
0.4
0.6
0.8
1.0
SiO2
Na2Si2O5
Na2SiO3
Na4SiO4
Na2O
mol % SiO2
fractiono
fspecies
The results for the constitutional structure across the entire compositional range are
shown. The calculated result for xSiO2 = 0.70 is marked.
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Now we write down the Gibbs energy G of our mixture. It can be presented equally well
in terms of oxide components j or in terms of constitutional compounds k:
Since nj is replaced by (Akj)nk, we may write the sum term by term for each oxide j:
Resolving this equation for log j yields:
Note that the activity aj is given by jxj.
( )
( ) ( )
+=
+=+
+=+=
+=
+=
+=
j
j
k
kk
kjj
k
kk
kjj
j
k
kkjkj
k
kkk
j
jjj
kkk
jjj
xRT
Gx
RT
GA
xRT
GAa
RT
G
xRTGnAxRTGn
aRTGnG
xRTGG
aRTGG
lnln303.2
1log
lnln
lnln
ln
ln
ln
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+=
jj
k
kk
kjj xRTGx
RTGA lnln
303.21log
When evaluating the equation
we need to employ the coefficients from the Na2O column of the matrix Akj:
This results in the following equation:
The following basic data are given for T = 1400 C:
GSiO2 = GS = -1073.8 kJ/mol, GNS2 = -3067.6 kJ/mol, GNa2O = -681.5 kJ/mol,
RT = 8.3141673.1510-3 kJ/mol = 13.91 kJ/mol. We started with xSiO2 = 0.7
xNa2O = 1 - xSiO2 = 0.3 and calculated xS = 0.75. The result reads log Na2O = -6.8.
10
21
2
22
NS
S
ONaSiO
ONaNSSONaNSS
ONa xxxRT
GGG22
222 loglog1log2
12
303.2
1log +
+=
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0.0 0.2 0.4 0.6 0.8 1.0
-10
-8
-6
-4
-2
0
1000 C1200 C
1400 C
log f(Na2
O)
log(activitycoefficie
nt)
molar fraction of SiO2
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0.0 0.2 0.4 0.6 0.8 1.0
-10
-8
-6
-4
-2
0
1000 C1200 C
1400 C
log f(Na2
O)
log(activitycoefficie
nt)
molar fraction of SiO2
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][][][1
][][][1
][][][1
][][
][
;][][
][
;][][
][
1442
22
13222
1522
222
22
2
44
3
22
322
222
5221
2
2
2
2
=
=
=
==
==
==
SiONaSiOONaK
SiONaSiOONaK
OSiNaSiOONaK
SiOONa
SiONa
KK
SiOONa
SiONaKK
SiOONa
OSiNaKK
SN
NS
NS
SN
NS
NS
Example:
system Na2O-SiO
2with compounds
Na2Si2O5, Na2SiO3, Na4SiO4
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( )
( )
.1
)1(;
1
;1
)1(;
1
22
2
22
2
2
2
2
2
2
2
2
2
SNSSNS
SSSN
SNS
SNSSNS
NNSNSN
NNNS
NNS
NNSNSN
xKxK
xxKs
xK
xKxKr
xKxK
xxKq
xK
xKxKp
++
=
++
=
++
=
++=
This is a simple second degree equation system which can be readily solved:
Either xN od xS is used as independent input variable:
( ) ( )
.323
2)(1)(
;;;
;1412
1412
22
22
2222
22
22
NSNNSNSS
SNNSNSS
SNNSNSSNNSNSSNSNSN
NSSN
xxxxx
xxxxONaxSiOx
xxKxxxKxxxKx
sr
xxorqp
xx
++++
+++==
===
====
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0.0 0.2 0.4 0.6 0.8 1.00.0
0.2
0.4
0.6
0.8
1.0T = 1400 C
Na2O Na
4SiO
4
Na2SiO
3
NaSi2O
5
SiO2
molarfractionofspecies
molar fraction of SiO2
constitutional model &
associated species model
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0.4 0.6 0.8 1.0
T in C =
400
800
1450
molar fraction of SiO2
0.4 0.6 0.8 1.00.0
0.2
0.4
0.6
0.8
1.0
T in C =
1450
800
450
silica speciesdisilicate species
molarfractionofspecies