Gibbs Energy and Ellingham Diagrams

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Gibbs Energy and Ellingham Diagrams   Background
  ?G lt0        for a spontaneous change to be
possible      a CLOSED SYSTEM      at constant
temperature and pressure.
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Under certain conditions   Metal Oxide CO (g) ?
Metal CO2 (g)   Under different conditions
  i.e. Metal CO2 (g)?Metal oxide CO(g)
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       Equilibrium (i.e. state of maximum
stability) will be attained in a closed system at
constant temperature and pressure when G reaches
its minimum possible value        At any
temperature T, G is given by ?G?H-T.?S       
For chemical reactions involving gases the sign
and magnitude of ?S is governed mainly by
increase/decrease in number of gas moles, ie. If
gas moles increase as a result of reaction then
?Sgt0, whereas if gas moles decrease then ?Slt0.
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Since ?G depends on both ?H and ?S then we have
four possibilities  
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  (1)                 Oxidation of C(s) by O2(g)
to form CO(g)   2C(s) O2(g)?2CO(g)   ?Hlt0, ?Sgt0
(increase in gas numbers), therefore, ?G always
lt0 at any T.
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  2) Oxidation of a metal by O2(g)   e.g. 2Al(s)
3/2 O2(g) ?Al2O3 (s)   ?Hlt0 (exothermic), ?Slt0,
provided T is low enough, then ??H?gt?T?S? then
?Glt0, reaction will work. But there will be a
temperature at which ?G0 and beyond which ?G is
positive then metal oxide will decompose to metal
and oxygen.
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  (3)                 Reaction between C(s) and
CO2(g) to CO(g) i.e. C(s) CO2(g)?2CO(g)   The
reaction is endothermic (?Hgt0), but ?S is
positive (increase in gas moles). If T is high
enough reaction will occur. But if T is low
enough the reverse reaction will occur.
If ?H of reaction is positive and there is a
decrease in number of gas moles from L to R then
there can be no temperature where ?G become
negative and reaction cannot occur spontaneously.
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Variation of ?Go with temperature  
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This means that ?G versus T is a straight line
for many chemical reactions   i.e.
?G?H-T?S   Rearrange ?G-T?S?H, ymxc   Where
y?G, xT slope m-?S, intercept C?H
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Plot several ?Gs on same diagram ?Ellingham
Diagram (?rGo plotted in each case for 1 mole of
O2(g))
The relation could be thus be expressed by    
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The variation of with T for the
oxidation reaction   shown in Fig. 12.2 as an
Ellingham diagram.
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• For the oxidation reaction
•  
•  
• In the temperature range in which A and AO2 are
  solid, is considerably greater than
  both and . Then
•  

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Thus the standard entropy change for oxidation
reactions involving solid phase have almost the
same value, which corresponds with the
disappearance of 1 mole of oxygen gas initially
at 1 atm pressure. As the slopes of the lines in
an Ellingham diagram are equal to - , then
the lines are more or less parallel to one
another, as will be seen in Fig. 12.13.
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Consider two oxidation reactions, the Ellingham
lines of which intersect one another, e.g.  Which
are shown in Fig. 12.4.
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is more negative than and that
is more negative than . Subtraction of
reaction (i) from reaction (ii)
gives   B2AO2ABO2
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For which the variation of with T is
as shown in Fig. 12.5.