ELLINGHAM DIAGRAMS
Ellingham diags.(1944)for M-oxides & M-sulphides
are 70 yrs.old.
Later on, others had given diags. for
M-carbonates,-sulphates,-chlorides,-fluorides.
Now diags.areavailable for
M-carbides,-nitrides,-hydroxides,-sillicates,-phoisphatesetc.
THERMODYNAMICFEASIBILITYOFVARIOUSREACTIONSCANBEFOUNDOUTFROMFREEENERGYCALCULATIONS,
i.e.
1)which metal compound can be reduced,
2)by which reducing agent,
3)at what temperature,
4)Comparatively stability of various M-compounds of similar nature(e.g. various M-oxides, M-sulfides etc.)
5)Ratio of equilibrium partial pressure of gases involved for a given reactions,
e.g. for oxides: Pco/Pco2,PH2/PH2O,PO2etc.
ELLINGHAM DIAG. FOR OXIDES
Consider M-oxide system, 2Ms+ O2g= 2MOs
Equili.const. -…….. assuming M & MO in Std.state
(pure state),aM= 1,aMO=1,aO2=Po2
K=1/PO2
Now, by std.relation, ΔG°= -RTlnK
= -RTln1/PO2
= -
ΔGf°= RTlnPO2………….eq.1
Where,…………………………………………………………………………..
ΔGf°= Std. free energy change of the reaction involving only 1 mole
of O2gas reacting with M to form MO, hence called,
Std. free energy change for formation of MO per mole of O2.
PO2= Equili.oxygenpressure for M-oxide system
OR
Oxygen dissociation pressure for M-oxide.
Since, ΔGf°= RTlnPO2--------eq.1 above,
for a metal oxide system, ΔGf°varies with Po2and T.
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For the reaction, 2M +O2 = 2MO
(1)If MO has very low Po2compare to Po2of external atmos.,
then MO is more stable or difficult to reduce and ΔGf°= -ve, for above reaction.
(2)If MO has high Po2compare to Po2of external atmos., then MO is less stable or easy to reduce and ΔGf°= +ve, for above reaction.
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Let us consider St.lineeq. y = mx+c……….…..eq.2
Also,asper std.relation, ΔGf°= ΔHf°–TΔSf°…...eq.3
Comparing eq.2 & 3 above,
slope=m= -ΔSf°….also called, std.entropyof formation of oxide,
Intercept=c = ΔHf°also called, std.heatof formation of oxide.
I
MP.FEATURESOFDIAG.FORM-MO SYSTEM
(1)..(a)In all cases: 2Ms+O2g=2MOs
(gas vol. in mole)...1O
Since gas phase has much higher entropy than condensed phase,
in above case, entropy decreases & hence entropy change,ΔSf°=-ve.
(when, ΔSf°=-ve,thenΔGf°= +ve& slope of the line will be positive,Fig1).
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(b)In case of: 2Cs+O2g=2COg
(gas vol.inmole)….1 2
i.e. entropy increases & entropy change, ΔSf°= +ve.
(when, ΔSf°=+ve,thenΔGf°=-ve& slope of the line will be negetive,
----------------------------------------------------------------------------------Fig 2).
(c) In case of: Cs+O2g = CO2g
(gas vol. in mole).. 1 1
(entropy does not change, ΔSf°= 0,ΔGf°= ΔHf°, slope = 0,&
line is horizontal ,Fig 2).
ASSEENINELLINGHAMDIAG.(FIG.3)
(2)Kinks:
The lines are straight lines with kinks, because of phasechangesin either metal or it’s oxides due melting, boiling, sublimation or transition point.
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Due to phase change at kink, ΔSf°will change and hence slope of the line changes.
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Change of slope is much sharper for boiling & sublimation temps.thanfor melting temp. because, ΔSf°at the melting pt. is much less than at boiling or sublimation pt.
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(3)Reduction of ‘MO’ by C:
As shown in fig.2,in case of,
2Cs+ O2g=2COg…….as T increases, ΔSf°& ΔGf°
(This is not true in almost all other cases).
And this makes Carbon, a useful reducing agent.
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The temp. at which M-MO & C-CO line intersects gives
equili.temp. of reduction of MO by C at 1 atm. Pressure.
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Thus from Ellingham diag. temp.ofreduction of MO by C is,
for MnO--->1400 °C & for Al2O3--->2000 °C
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(4)STABILITYOFCO & CO2
As seen in Fig.2,the lines C-CO & C-CO2cross each other at 723 °C,
Below 723 °C, CO2gis more stable than COg
and
above 723 °C, COgis more stable than CO2g
Hence below 723 °C, MO+C M+CO2
and above 723 °C, MO+C M+CO
e.g.
2PbOs+Cs=2Pbs+CO2g…………..at 100◦C
ZnSs+Cs=Zng+COg………………at1200◦C
(5)Reduction of ‘MO’ by reducing agent,otherthan carbon:
Itis not necessary that two curves of M-MO should always intersect each-other for reduction, as discussed for reduction by carbone.g. reduction of Cr2O3by Al or Pbat 1200 °C
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Each curve is for 1mole of O2,
hence distance between two curves at any temp. gives ΔGf
for the reduction of oxide.