06. Kesetimbangan Kimia dan Diagram Ellingham

06. Kesetimbangan Kimia dan Diagram Ellingham

Zulfiadi Zulhan

Teknik MetalurgiFakultas Teknik Pertambangan dan PerminyakanInstitut Teknologi BandungINDONESIA

Termodinamika Metalurgi (MG-2112)

201 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021

Jangan Mengunggah

Materi Kuliah ini di

INTERNET!


1. Pendahuluan, istilah-istilah dan notasi

2. Hukum I Termodinamika

3. Hukum II Termodinamika

4. Hubungan Besaran-Besaran Termodinamika

5. Kesetimbangan

6. Kesetimbangan Kimia dan Diagram Ellingham

7. Proses Elektrokimia dan Diagram Potensial - pH (Pourbaix)

8. Ujian Tengah Semester

9. Aktivitas Ion

10. Termodinamika Larutan

11. Penggunaan Persamaan Gibbs - Duhem

12. Penggunaan Metoda Elektrokimia untuk menentukan Sifat-Sifat / Besaran-Besaran

Termodinamika

13. Keadaan Standar Alternatif

14. Koefisien Aktivitas dalam Larutan Encer Multi-Komponen

15. Diagram Fasa

16. Ujian Akhir Semester

Materi Perkuliahan


dU = T dS - P dV

dH = T dS + V dP

dF = -S dT - P dV

dG = -S dT + V dP

dG = − S dT + V dP

At consntant temperature: dT = 0

dG = V dP

P ln d RT P

dP RT Gd ==

For ideal gas: V = RT/P

0 P as 1, f/P lim

f ln d RT Gd

→=

=At low pressure (pressure approaches zero), the

gas behaves ideally and its fugacity equal to its

pressure.

Thermodynamic Activity

For real gas, by analogy to pressure:

The change of molar Gibbs free energy of a single gas

with pressure at constant temperature


Fugasitas

https://id.wikipedia.org/

https://id.wikipedia.org/



For mixtures of gases, molar Gibbs free energy (G) is replaced by partial molar Gibbs free

energy, .

For component i:

if ln d RT Gd =

G

0 P as 1, f/P lim

f ln d RT Gd

→=

= At low pressure (pressure approaches zero), the gas

behaves ideally and its fugacity equal to its pressure.

For real gas, by analogy to pressure:

Fugacity is property of a gas ~ pressure of non ideal gas.



Fugacity of a condensed phase (solid or liquid) is equal to the fugacity of the vapor in equilibrium

with condensed phase.

At equilibrium, partial molar Gibbs free energies of vapor and condensed phase are equal.

For the difference in partial molar Gibbs free energy between two states at constant

temperature (in term of fugacity):

1

2

f

f

12

f

f ln RT f ln d RT G - G G

2

1

===

Sometimes, it is written as:

1

212

f

f ln RT - =

,...n,nP,T,a

a

cb

n

VV

=

i

nnS,V,innS,P,innV,T,innP,T,iijijijij

n

U

n

H

n

F

n

G=

=

=

=

if ln d RT Gd =



Thermodynamic activity of a material is defined as the ratio of fugacity

of material to the fugacity in its standard state:

i

ii

f

fa

1

2

f

f

12

f

f ln RT f ln d RT G - G G

2

1

=== 1

212

f

f ln RT - =

Thermodynamic Activity is evaluated relative to a standard state at the same temperature.

Activity is set equal to one at the standard state.

In terms of a quantity, Activity is called “fugacity”

Fugacity is property of a gas ~ pressure of non ideal gas.

Activity is the ratio of the fugacity of a material to its fugacity in the standard state.



Standard state of a material can be arbitrarily defined, certain “conventions” generally are followed:

▪ Standard state of a gas is “pure gas at one atmosphere pressure”

▪ Standard states for liquid and solids are “pure liquid or solid at one atmosphere pressure”

i

ii

f

fa

Standard state vs. reference state

Reference state defines the condition of material in term of temperature, pressure, and physical

form: gas, liquid and crystal structure (for solid). Example: reference state for enthalpy=0 for pure

element at 298K.

Standard state is defined only in terms of pressure and physical form. Thermodynamic activity is

an isothermal concept.

Thermodynamic Activity is evaluated relative to a standard state at the same temperature.

Activity is set equal to one at the standard state.



So, the difference in Gibbs free energy between a material under an arbitrary set of conditions and a

standard state at the same temperature is:

Example: activity of an ideal gas at pressure of 2 atm is 2 (because the fugacity of the standard state

is one atmosphere).

Pure nitrogen at a pressure of 0.1 atm has an activity of 0.1, if pure nitrogen at one atmosphere is

taken as standard state.

Note: thermodynamic activity has no units.

a ln RT f

f ln RT G - G dG i

i

iii

G

G

i ===

Fugacity of liquid or solid is equal to fugacity of vapor in equilibrium with it. If equilibrium vapor

pressure of liquid is 0.01 atm at a specified temperature, then its fugacity is 0.01

if ln d RT Gd =

i

ii

f

fa

P ln d RT P

dP RT Gd ==

dG = -S dT + V dP

Gi = Gi° + RT ln ai



Value of thermodynamic activity changes not only with pressure but also with composition.

Fugacity of a material in an ideal solution varies linearly with mole fraction.

Pure A Pure B

Fugacity o

f B

, f B fB

p

Activity o

f B

, a

B

0

1

fB = XB fBp

Where fBp is the fugacity of pure B,then

If fBp = fB°, then aB = XB (ideal solution)

If in liquid A-B consisting of 80 mol % B, then aB

= 0.8, aA = 0.2 (ideal solution)

==

B

p

BB

B

BB

f

f X

f

fa

aB = B XB

aB = activity of B

B = activity coefficient of B

XB = mol fraction of B


Aktivitas

https://www.pikiran-rakyat.com/

https://lokadata.id/

Murni

Nelayan

Murni

Petani

https://dialeksis.com/

https://idxchannel.okezone.com/

Aktivitas N

ela

yan

Aktivitas P

eta

ni

Jumlah Petani

Pure A Pure B

Activity o

f A

, a

A

Activity o

f B

, a

B

0

1

XB

aB =fB

fB° =

fBpXB

fB°

aA =fA

fA° =

fApXA

fA°

ideal solution:

fAp = fa°, then aA = XA

ideal solution:

fBp = fB°, then aB = XB


Condition of Equilibrium

Partial molar Gibbs free energy of a material,

If G2 – G1 is negative, then reversible work (Wrev) is negative. It means

that material change spontaneously from state 1 to state 2, because no

work needs to be done to force the change.

δWrev = ഥG2 − ഥG2 dn = ΔഥG dn = ΔGAt equilibrim: δWrev = ΔG = 0

G

If G2 – G1 is positive, then reversible work (Wrev) is positive. Work would

have to be done on the system to force it to change from state 1 to state

2 (the change would not occur spontaneously).

https://www.tokopedia.com/

SISTEM

Q+ W+Lingkungan

G1

G2

i,1

i,2

i,1 > i,2

G1

G2

i,1

i,2i,1 < i,2



Chemical Equilibrium

Reaction: b B + c C = d D + e E

Gc -G b-G e G d GW CBEDrev +==

B

o

BB a ln RT G G +=

( ) ( ) ( ) ( )C

o

CB

o

BE

o

ED

o

D a ln RT Gc a ln RT G b-a ln RT G ea ln RT G d ΔG +−++++=

c

C

b

B

e

E

d

Do

C

o

B

o

E

o

Da a

a a ln RT G cG bG eG d ΔG +−−+=

K ln RT G a a

a a ln RT G ΔG

c

C

b

B

e

E

d

D +=+=

K = equilibrium constant for a reaction, expressed in terms of activities


δWrev = ΔG


Gaseous Equilibria

Case 1: a tank contains pure oxygen at total pressure of 1 atmosphere. Oxygen exists primarily in

diatomic form, O2. However, oxygen can also exist in monoatomic state. Calculate the equilibrium

concentration of monoatomic oxygen in the tank at 1000K.

Standard Gibbs free energy of formation of monoatomic oxygen is 187800 J per mole of

monoatomic oxygen at 1000K.

J 800 187 0 - 800 187 G - G G o

2O,f21o

O,f ===Reaction:

½ O2 = O

1/2

O

O

1/2

O

O

22p

p ln RT

a

a ln RT G −=−=

22.58- 1000 8.314

187800

RT

G-

p

p ln

1/2

O

O

2

=−=

= 10 x 55.1 p

p 10-

1/2

O

O

2

=

-10

OOO 10 x 5.1p 1,pp2

==+ Activity of diatomic oxygen in the tank is very nearly one. O2 is

predominant at 1000 K


Gaseous Equilibria

205,1)*T - 0,5(0 - 161) * T - (249.000 G - G 2,21

, == o

Of

o

OfG

Reaction:

½ O2 = O

J/mol 205,1 * T - 0 2= o

OG

J/mol 161 * T - 249.200 = o

OG

) T * 58,45 - (249.000 G - G 2,21

, == o

Of

o

OfG

J/mol 190.550 G - G 2,21

,1000 == o

Of

o

Of

o

KG

J/mol 335.500- G - G 2,21

,000.10 == o

Of

o

Of

o

KG atm 99968,0p 1,pp OOO 2==+

http://hebasoffar.blogspot.co.id

Reaction:

b B + c C = d D + e E

ΔG = ΔG° + RT lnaDd aE

e

aBb aC

c


Gaseous Equilibria

Case 2: consider the equilibrium in water-hydrogen-oxygen system. At 2000K, a system will be

operated so that the partial pressure of oxygen is 10-10 atm. What should be the ratio of

hydrogen to water vapor to achieve this condition?

2/1

OH

OH

22

2

p p

p 3652 K

8.20 RT

G- K nl K, 2000 at

==

=

=

Reaction: H2 + ½ O2 = H2O G° = -246 000 + 54.84 T

atm 1ppp

10 x 3652p x 3652 p

p

222

2

2

2

OHOH

5-1/2

O

H

OH

=++

==

At one atmosphere, a hydrogen water mixture in which water content of 3.52% and hydrogen

content of 96.48% will have an oxygen partial pressure of 10-10 atm at 2000 K.

This illustrates that it is possible to obtain low oxygen pressures by chemical means.


Cu-O2 Equilibria

Consider the equilibrium among Cu2O, Cu and O2:

4 Cu (s) + O2 (g) = 2 Cu2O (s)

If pressure of oxygen exeeds equilibrium pressure, reaction proceeds to right (Cu2O will be formed).

If pressure of oxygen less than equilibrium pressure, reaction proceeds to left (Cu2O should

decompose into Cu and O2)

Cu2O

http://www.baofull.com

+ =

http://www.pometon.com


Source of Information on G°

For chemical reaction, Gibbs free energy change at temperature T, G°T, is the sum of Gibbs free

energies of formation of products less the sum of the Gibbs free energies of formation of the reactants

at the same temperature

=reactants

o

Tf,r

products

o

Tf,p

o

T G n-G n G

o

T

o

T

o

T S T-H G =

G°T can also be calculated from values of H°T and S°T as follows (assuming that there are no phase

transformation between 298 K and T):

Td T

CS S and

Td CH H :where

T

298

o

po

298

o

T

T

298

o

p

o

298

o

T

+=

+=

4 Cu (s) + O2 (g) = 2 Cu2O (s)

Δ𝐺° = − RT lnaCu2O2

aCu4 pO2

Δ𝐺° = −339000 − 14.24 T ln T + 247 T Joule



T ln T C T B A Go

T,f ++=

Above form of equation arises because Cp° is treated as indenpendent on

temperature for the purpose of calculating G°.

( )298TC H H o

p

o

298

o

T −+=

+=

298

Tln C S S o

p

o

298

o

T

( ) ( )298 ln - T ln C T S T298T C H G o

p

o

298

o

p

o

298

o

T −−−+=

( ) T ln C T 298 ln CCS TC 298 H G o

p

o

p

o

p

o

298

o

p

o

298

o

T −−−−−=

T ln T C T B A Go

T,f ++=

4 Cu (s) + O2 (g) = 2 Cu2O (s)


aCu4 pO2

Δ𝐺° = −339000 − 14.24 T ln T + 247 T J


T B T ln T C A Go

T,f ++=

4 Cu (s) + O2 (g) = 2 Cu2O (s)


aCu4 pO2

Δ𝐺° = −339000 − 14.24 T ln T + 247 T J


4 Cu (s) + O2 (g) = 2 Cu2O (s)

Calculate oxygen pressure where Cu2O dissociates into Cu and O2 at 1000K.

Joule T 247 T ln T 24.14339000G

p a

a ln RT G

2

2

O

4

Cu

2

OCu

+−−=

−=

Activity of Cu2O is 1 because during the reaction, Cu2O exists as pure solid and

standard state is pure Cu2O.

Cu2O

http://www.baofull.com

+ =

http://www.pometon.com

Cu-O2 Equilibria

Reaction:

b B + c C = d D + e E


e

aBb aC

c


Copper

Cu2O

http://www.baofull.comhttp://www.pometon.com

https://dir.indiamart.com

CuO Cu(OH)2

http://www.instructables.com

http://www.nyctransitforums.com

http://www.reade.com

http://2il.orghttps://www.healingcrystals.com

CuFeS2


Solid Vapor Equilibria

atm 10 x 14.1p 22.90,- p ln

p ln 1000 * 8.314 360 190 G K, 1000 at

10-

OO

O

22

2

==

=−=

At higher pressure of oxygen, activitiy of copper will be less than 1, copper will not exist.

At oxygen pressure of 1.14 x 10-10 atm, metallic copper and cuprous oxide can exist in

equilibrium.

At lower oxygen pressure, metallic copper will exist

Activity of Cu will become 1, if the pressure of oxygen is in equilibrium with Cu2O and

Cu.

Cu2O+→


Gaseous Equilibria

Case 2: consider the equilibrium in water-hydrogen-oxygen system. At 2000K, a system will be

operated so that the partial pressure of oxygen is 10-10 atm. What should be the ratio of

hydrogen to water vapor to achieve this condition?

2/1

OH

OH

22

2

p p

p 3652 K

8.20 RT

G- K nl K, 2000 at

==

=

=

Reaction: H2 + ½ O2 = H2O G° = -246 000 + 54.84 T

atm 1ppp

10 x 3652p x 3652 p

p

222

2

2

2

OHOH

5-1/2

O

H

OH

=++

==

At one atmosphere, a hydrogen water mixture in which water content of 3.52% and hydrogen

content of 96.48% will have an oxygen partial pressure of 10-10 atm at 2000 K.

This illustrates that it is possible to obtain low oxygen pressures by chemical means.

Reaction:

b B + c C = d D + e E


e

aBb aC

c


Cu stable

Cu2O stable

CuO stable

Cu-O2 Equilibria0 = ΔG° + RT ln

aCuO4

aCu2O2 pO2

Reaction: b B + c C = d D + e E


e

aBb aC

c

ΔG = ΔG° + RT lnaCuO4

aCu2O2 pO2

2 Cu2O + O2 = 4 CuO

At equilibrium, ΔG = 0

ΔG° = − RT lnaCuO4

aCu2O2 pO2

For pure CuO, aCuO = 1

For pure Cu2O, aCu2O = 1

ΔG° = RT ln pO2

0 = ΔG° + RT lnaCu2O4

aCu4 pO2

Δ𝐺° = −339000 − 14.24 T ln T + 247 T Joule



T ln T C T B A Go

T,f ++=

Above form of equation arises because Cp° is treated as indenpendent on

temperature for the purpose of calculating G°.

( )298TC H H o

p

o

298

o

T −+=

+=

298

Tln C S S o

p

o

298

o

T

( ) ( )298 ln - T ln C T S T298T C H G o

p

o

298

o

p

o

298

o

T −−−+=

( ) T ln C T 298 ln CCS TC 298 H G o

p

o

p

o

p

o

298

o

p

o

298

o

T −−−−−=

T ln T C T B A Go

T,f ++=

4 Cu (s) + O2 (g) = 2 Cu2O (s)


aCu4 pO2

Δ𝐺° = −339000 − 14.24 T ln T + 247 T J


-350,000

-300,000

-250,000

-200,000

-150,000

-100,000

0 200 400 600 800 1000 1200 1400 1600

G

o=

RT

ln P

O2

Temperatur [K]

dG = A + C*T ln T + BT

dG = dH - T*dS

ΔGf,To = ΔHo - T ΔSo

ΔG = ΔG° + RT lnaCu2O2

aCu4 pO2

4 Cu + O2 = 2 Cu2O


ΔG° = − RT lnaCu2O4

aCu4 pO2

For pure Cu, aCu = 1

For pure Cu2O, aCu2O = 1

ΔG° = RT ln pO2


M + O2 = MO2

If the difference in heat capacities between

reactants and products is small, then G°T

equation assumes to form A - BT:

G°T =H° - T S°

ΔG = ΔG° + RT lnaMO2

aM pO2


ΔG° = − RT lnaMO24

aM pO2

For pure MO2, aMO2 = 1

For pure M, a𝑀 = 1

ΔG° = RT ln pO2

G

° T=

H°

-T

S

°=

RT

ln P

O2

0 200 400 600 800 1000

H°

G°T =H° - T S°


Ellingham

Diagram

twitter.com/kachra_peti

Ellingham diagram is a graph showing the

temperature dependence of the stability of

compounds. This analysis is usually used

to evaluate the ease of reduction of

metal oxides and sulfides. These diagrams

were first constructed by Harold

Ellingham in 1944.en.wikipedia.org

Harold Johann Thomas Ellingham, (1897–1975), British physical chemist, best known for Ellingham diagrams, which summarize a large amount of information concerning extractive metallurgy www.wikiwand.com

Increasing

Stability

https://en.wikipedia.org/wiki/Oxide

https://en.wikipedia.org/wiki/Sulfide

https://en.wikipedia.org/wiki/Harold_Ellingham

https://www.wikiwand.com/en/Ellingham_diagram

https://www.wikiwand.com/en/Metallurgy


Ellingham DiagramΔG°=

RTlnpO2

TTm,A

ΔG°=

RTlnpO2

TTm,AO2


ΔG°=

RTlnpO2

(kJ)

RT ln PO2 vs. T

O


ΔG°=

RTlnpO2

(kJ)

RT ln PO2 vs. T

O

At 1050°K, the

oxygen pressure

in equilibrium with

Cu2O and Cu is

10-9 atm


Ellingham Diagram

-273

ΔG° = RT ln pO2ΔG° = 2.303 RT log pO2

ΔG° = 2.303 RT log (10𝟎) Oxygen Potential (PO2)

ΔG° = 2.303 RT log (10−𝟑)

ΔG° = 2.303 RT log (10−𝟓)

ΔG° = 2.303 RT log (10−𝟏𝟒)

O


Ellingham Diagram

ΔG° = RT ln pO2ΔG° = 2.303 RT log pO2

ΔG° = RT ln (𝟏)

Oxygen Potential (PO2)

p O2 = 1 atm

ΔG° = RT ln (10−𝟏𝟎)

K

ΔG° = RT ln (10−𝟐𝟎)

O


Ellingham Diagram

-273

ΔG°

2.303= RT l𝑜𝑔 pO2 + RT log

1

1

2

Equilibrium pressure of

oxygen in a system by

controlling the ration of

H2 and H2O in vapor:

2H2 + O2 = 2H2O


ΔG° = − RT lnpH2O2

pH22 pO2

ΔG° = RT ln pO2 + RT lnpH2pH2O

2

ΔG°


103

1

2

ΔG°


107

1

2

ΔG°


1012

1

2

H2

H

OH

O p

p

K

1p

2

2

2

=


Ellingham Diagram

K





2H2 + O2 = 2H2O


ΔG° = − RT lnpH2O2

pH22 pO2

ΔG° = RT ln pO2 + RT lnpH2pH2O

2

ΔG° = RT ln pO2 + RT ln1

1

2

ΔG° = RT ln pO2 + RT ln10−3

1

2


1

2


1

2

H2

H

OH

O p

p

K

1p

2

2

2

=


Ellingham Diagram

-273

ΔG°


1

1

2

ΔG°


103

1

2

ΔG°


107

1

2

ΔG°


1012

1

2

C





2CO + O2 = 2CO2


ΔG° = − RT lnpCO22

pCO2 pO2

ΔG° = RT ln pO2 + RT lnpCOpCO2

2

2

CO

CO

O p

p

K

1p 2

2

=


Ellingham Diagram

K





2CO + O2 = 2CO2


ΔG° = − RT lnpCO22

pCO2 pO2

ΔG° = RT ln pO2 + RT lnpCOpCO2

2

ΔG° = RT l𝑜𝑔 pO2 + RT log1

1

2

ΔG° = RT l𝑜𝑔 pO2 + RT log10−3

1

2


1

2

ΔG° = RT l𝑜𝑔 pO2 + RT log1012

1

2

C

2

CO

CO

O p

p

K

1p 2

2

=


Mengapa kemiringan

kebanyakan garis (slope)

pada diagram tersebut

positif?

Apa maksudnya jika slopenya

negatif?

Diagram Ellingham

Decreasing

Oxygen

Potential

Increasing

Stability

AlAl2O3

FeO

Fe


Diagram Baur – Glaessner / Diagram Chaudron

0

10

20

30

40

50

60

70

80

90

100

0 200 400 600 800 1000 1200 1400

Temperature [°C]

CO

/ (

CO

+ C

O2)

[%]

Fe3O4

FeO

Fe

Ptot = 0.2 atm

Ptot = 1.0 atm

Ptot = 5.0 atm

Reaction:

b B + c C = d D + e E


e

aBb aC

c

Fe3O4 + 4CO = 3 Fe + 4CO2

Fe3O4 + CO = 3 FeO + CO2

FeO + CO = Fe + CO2

3 Fe2O3 + CO = 2 Fe3O4 + CO2

CO2+C = 2CO

AISE, 1999Fe2O3


Diagram KelloggReaction:

b B + c C = d D + e EΔG = ΔG° + RT ln

aDd aE

e

aBb aC

c = ΔG° + RT ln K

Log pO2

Log p

SO

2

MeS

Me

MeO

MeSO4

4. 2MeO + 2SO2 + O2 = 2MeSO42 log pSO2

+ log pO2 = −log K4

2. 2Me + O2 = 2 MeOlog pO2 = − log K2

3. 2MeS + 3O2 = 2MeO + 2SO22 log pSO2

− 3 log pO2 = log K3

1. Me + SO2 = MeS + O2log pO2 − log pSO2

= log K1

5. MeS + 2O2 = 2 MeSO42 log pO2 = − log K5

6. S2 + 2O2 = 2 SO22 log pSO2

− 2 log pO2 = log K6 + log pS2

7. 2SO2 + O2 = 2SO32 log pSO2

+ log pSO2= −log K7 + 2 log pSO3

At equilibrium, ΔG = 0 ΔG° = -RT ln K

1 2

3

4

5


http://www.aimehq.org/

http://www.aimehq.org/


Diagram Kellogg


Diagram Kellog


Gases Dissolved in Metals (Sievert‘s Law)

Gases, such as hydrogen, oxygen and nitrogen, dissolve in liquid and solid metals.

In gaseous state, these gases exist in diatomic form, H2, O2 and N2.

If these gases dissolve in metals, they are found in monoatomic form, H, O and N.

For hydrogen dissolving in copper:

H2 (g) = 2H ( in copper solution)

Equilibrium constant:

2H

2

H

p

a K =

Activity of diatomic gases dissolved as dillute

solution is linear to their concentration in metal.

aH = [H]

2H

2

p

H K = 2/1

H

2/1

2p K H =

Sievert‘s Law: the amount of diatomic gases dissolved in metals at a given temperature

is proportional to sequare root of diatomic gas partial pressure.

en.wikipedia.org


Property Relations Derived from U, H, F, G

dU = T dS - P dV

dH = T dS + V dP

dF = -S dT - P dV

dG = -S dT + V dP

( ) dT S Gd −=

−= S T H G

( ) dT S Gd −=

−= S T H G

T

G H S

−=

( ) dTT

H dT

T

G-Gd

−=

Multiply by 1/T

( ) dT

T

H dT

T

G-

T

Gd22

−=

T

1dH

T

Gd

=

( ) T

1d

R

H K lnd

−=

( ) T

1d

R

H K lnd

2

1

2

1

T

T

K

K

−=

T

1

T

1

R

H

K

K nl

121

2

−

−=

Van‘t Hoff relationship

Gibbs free energy change for a reaction as a function of temperature:

en.wikipedia.org

At equilibrium, ΔG = 0 ΔG° = -RT ln K


Variation of Equilibrium Constant with Temperature

Example: decomposition of calcium carbonate (lime stone) into calcium oxide

and carbon dioxide

CaCO3 = CaO + CO2

Following data are available for pressure of carbon dioxide in equilibrium with

CaO and CaCO3:

Temperature (K) Pressure (atm)

1030 0.10

921 0.01

a. Calculate change in enthalpy

T

1

T

1

R

H

K

K nl

121

2

−

−= H° = 166000 J per

mol CaCO3

b. Estimate temperature, at which the equilibrium

partial pressure of CO2 will be one atmosphere T3 = 1168 K

T

1

T

1

R

H

K

K nl

121

2

−

−=

K =aCaO PCO2aCaCO3


Terima kasih!

Zulfiadi Zulhan

Program Studi Teknik Metalurgi

Fakultas Teknik Pertambangan dan Perminyakan

Institut Teknologi Bandung

Jl. Ganesa No. 10

Bandung 40132

INDONESIA

www.metallurgy.itb.ac.id

[email protected]

06. Kesetimbangan Kimia dan Diagram Ellingham

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