08 Exercise Oxide Activities

7/28/2019 08 Exercise Oxide Activities

1/16

(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:


2/16

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


3/16

(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)



4/16

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!):


5/16

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


6/16

)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.


7/16

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.


8/16

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.


9/16

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


10/16

+=

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 +

+=


11/16

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)



12/16

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)



13/16

][][][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


14/16

( )

( )

.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

++++

+++==

===

====


15/16

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


constitutional model &

associated species model


16/16

0.4 0.6 0.8 1.0

T in C =

400

800

1450


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

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08 Exercise Oxide Activities

Documents