GROUP (17)
1
9.5 The Effect of Phase Transformation
 At the temperature of phase transformation (melting ,
boiling, sublimation , or solid phase changes) of a reactant
or product of the reaction:
𝟐𝑴(𝒔)+𝑶𝟐 (𝒈)=𝟐𝑴𝑶(𝒔)
(𝒊)
The Ellingham line will exhibit an elbow.
 Consider the reaction:
𝟐𝑴(𝒔)= 𝟐𝑴(𝒍)
(𝒊𝒊)
according to Hess’ law , Δ𝐻 ° (or Δ𝑆 ° ) of the reaction:
𝟐𝑴(𝒍)+ 𝑶𝟐 (𝒈) = 𝟐𝑴𝑶(𝒔)
(𝒊𝒊𝒊)
can be obtained by subtracting the standard enthalpy
change of reaction ( 𝑖𝑖 ) , Δ 𝑯° (𝒊𝒊) , form the standard
enthalpy change of reaction , Δ𝑯° (𝒊) , i.e.
Δ𝑯° (𝒊𝒊𝒊) = Δ𝑯° (𝒊) − Δ𝑯° (𝒊𝒊)
and,
𝜟𝑺° (𝒊𝒊𝒊) = Δ𝑺° (𝒊) − Δ𝑺° (𝒊𝒊)
Since Δ𝑯° (𝒊) < 0 , and Δ𝑯° (𝒊𝒊) > 0 , therefore Δ𝑯° (𝒊𝒊𝒊)
will be more negative than Δ𝑯° (𝒊).
Similarly , 𝜟𝑺° (𝒊𝒊𝒊) will be more negative than 𝜟𝑺° (𝒊).
This means that the Ellingham line for oxidation of liquid
metal to produce solid metal oxide has a larger slope than
that for the oxidation of solid metal that produces the solid
oxide ; hence at 𝑻𝒎 (𝑴) the line has an elbow upwards, as
is shown in figure (9.9a).
2
 On the other hand , Ellingham line for oxidation of the solid
metal to produce liquid oxide has a lesser slope than that for
oxidation of the solid metal to produce the solid oxide ;
hence at 𝑻𝒎 (𝑴𝑶) the line has an elbow downwards ; as is
shown in figure (9.9b).
 Thus if 𝑻𝒎 (𝑴) < 𝑻𝒎 (𝑴𝑶) , the Ellingham line will be as
shown in figure (9.10a).
and if 𝑻𝒎 (𝑴) > 𝑻𝒎 (𝑴𝑶) then the line will be as shown
in figure (9.10b). Examples are given in figures (9.11) and
(9.12).
3
Figure (9.9): (a) the effect of melting of the metal on the
Ellingham line for metal oxidation.
(b) the effect of melting of the metal oxide on the
Ellingham line for metal oxidation.
4
Figure (9.10) : the effects of phase changes of the reactants and
products of a reaction on the Ellingham line of an
oxidation reaction.
5
Figure (9.11): the Ellingham line for the oxidation of Copper.
6
Figure (9.12): the Ellingham line for the chlorination of iron.
7
9.6 The Oxides of Carbon in Equilibrium
with Graphite
 Carbon forms two gaseous oxides, 𝑪𝑶 and 𝑪𝑶𝟐 , according
to the following two reactions:
𝐶(𝑔𝑟)+ 𝑂2 (𝑔) = 𝐶𝑂2 (𝑔)
(𝑖)
for which , ΔG° (𝑖) (calories) = −94,200 − 0.20 T
2𝐶(𝑔𝑟)+ 𝑂2 (𝑔) = 2𝐶𝑂(𝑔)
(𝑖𝑖)
for which , Δ G° (𝑖𝑖) (calories) = −53,400 − 41.90 T
combination of reactions (i) and (ii) yields:
2𝐶𝑂(𝑔) + 𝑂2 (𝑔) = 2𝐶𝑂2 (𝑔)
(𝑖𝑖𝑖)
for which , Δ G° (𝑖𝑖𝑖) (calories) = −135000 + 41.50 T
 The Ellingham lines for reactions (𝑖 ) , (𝑖𝑖 ) 𝑎𝑛𝑑 (𝑖𝑖𝑖) are
included in figure (9.13).
8
Figure (9.13): the Ellingham diagram for selected oxides
including the oxides of carbon.
9
 The line for reaction (𝑖) has virtually zero slope since:
Δ𝑆 ° (𝑖) = 0.2 𝑒. 𝑢. ;
for reaction (𝑖𝑖 ) , the line has a negative slope since:
Δ𝑆 ° (𝑖𝑖) = 41.9 𝑒. 𝑢. ;
and for reaction (iii) the line has a positive slope since:
Δ𝑆 ° (𝑖𝑖𝑖) = - 41.5 𝑒. 𝑢.
 For the equilibrium :
𝐶(𝑔𝑟) + 𝐶𝑂2 (𝑔) = 2𝐶𝑂(𝑔)
(𝑖𝑣)
the standard free energy change is given by :
Δ𝐺 ° (𝑖𝑣) (𝑐𝑎𝑙) = 40,800 − 41.70𝑇
 The temperature at which the Ellingham lines for reactions
(𝑖) 𝑎𝑛𝑑 (𝑖𝑖) intersect one another occurs when
Δ𝑮° (𝒊𝒗) = 0 ; i.e. at 705 ˚C.
 At this temperature , 𝑪𝑶 and 𝑪𝑶𝟐 , in their standard states
, i.e. each at 1 𝑎𝑡𝑚 , 𝑇 , are in equilibrium with 𝑪 ;
thus: 𝑪, 𝑪𝑶(1𝑎𝑡𝑚, 705 ℃) and 𝑪𝑶𝟐 (1𝑎𝑡𝑚, 705 ℃)
are in equilibrium with each other , so the total pressure of
gases = 2 𝑎𝑡𝑚.
10
 If the pressure is decreased to a total pressure of one
atmospheric pressure at 705 ℃ , and since the equilibrium
constant remains having the same value , i.e.
𝑃(𝐶𝑂)2
𝑃(𝐶𝑂2 )
=1
then 𝑷(𝑪𝑶) should be greater than 𝑷(𝐂𝐎𝟐 ); thus:
𝑃(𝐶𝑂)𝑒𝑞,705 ℃ + 𝑃(𝐶𝑂2 )𝑒𝑞,705 ℃ = 1
and:
𝑃(𝐶𝑂)2 𝑒𝑞,705 ℃ / 𝑃(𝐶𝑂2 )𝑒𝑞,705 ℃ = 1
𝑃(𝐶𝑂)𝑒𝑞,705 ℃ > 𝑃(𝐶𝑂2 )𝑒𝑞,705 ℃
 For the case where 𝑃(𝐶𝑂) = 𝑃(𝐶𝑂2 ) = 0.5 𝑎𝑡𝑚
the temperature at which
C , 𝑃(𝐶𝑂)𝑒𝑞,705 ℃ ,
𝑃(𝐶𝑂2 )𝑒𝑞,705 ℃ are in equilibrium can be obtained by
solving the equation :
2
𝑃(𝐶𝑂)
ΔG = Δ𝐺°(𝑖𝑣) + 𝑅𝑇 𝑙𝑛 [
]=0
𝑃(𝐶𝑂2 )
i.e.
Δ𝐺°(𝑖𝑣) + 𝑅𝑇 𝑙𝑛(0 .5) = 0
11
 This temperature can be obtained by rotating clockwise the
(Δ𝑮° − 𝑻) line of reaction (𝑖) by 𝑹𝑻𝒍𝒏 (𝟎. 𝟓) and rotating
clockwise the (Δ 𝑮° − 𝑻 ) line of reaction ( 𝑖𝑖 ) by
𝟐 𝑹𝑻𝒍𝒏 (𝟎. 𝟓) around the point of intersection with 𝑻 = 𝟎
axis , and the intersection of the new two lines gives the
value of 𝑻𝒆𝒒 for the system when
P(CO) = P(CO2 ) = 0.5
This is shown in figure (9.14) for the discussed case and
many other cases.
12
Figure (9.14): the effect of varying the pressures of the product
gases of the reactions:
C(gr) + O2 (g) = 𝐶𝑂2 (𝑔) , and
2C(gr) + O2 (g) = 2CO(g)
on the variations of ∆G with T for the two
reactions.
13
Table (𝟏): list of variation of
𝐏(𝐂𝐎)
in the gas with temperature
𝐏(𝐂𝐎𝟐 )
for the case 𝑷(𝑪𝑶) + 𝑷(𝑪𝑶𝟐 ) = 1 at different
temperatures.
14
 Figure (9.15) includes the points a, b, c, d and e drawn in
figure (9.14).
 In figure (9.15) :
Before point (a) :
𝐶+𝑂2 =𝐶𝑂2
Beyond point (d) :
2𝐶+ 𝑂2 = 2𝐶𝑂
 Figure (9.15) indicates that below 400 ℃ the equilibrium gas
is virtually 𝑪𝑶𝟐 at 1 𝑎𝑡𝑚 pressure ; and above 1000 ℃ the
equilibrium gas is virtually 𝑪𝑶 at 1 𝑎𝑡𝑚 pressure.
 At any temperature T, the 𝑪𝑶/𝑪𝑶𝟐 mixture in equilibrium
with 𝑪 exerts an equilibrium oxygen pressure which can be
calculated via the equilibrium:
2𝐶𝑂 +𝑂2 =2𝐶𝑂2
for which:
Δ 𝐺 (𝑅) =−135000 + 41.5𝑇= -𝑅𝑇 𝑙𝑛[
°
P(CO2 )2
𝑃(𝐶𝑂)2 .𝑃(𝑂2)
]
thus:
log 𝑃(𝑂2 )𝑒𝑞,𝑇 =
−135,000
4.575𝑇
+
41.50
4.575
+ 2log(
P(CO2 )2
)
𝑃(𝐶𝑂) eq with C
 Figure (9.15) is generated by reading of the equilibrium
𝐶𝑂/𝐶𝑂2 ratio for the reaction 𝐶+𝑂2 =𝐶𝑂2 up to the point
(a) in figure (9.14) , then the 𝐶𝑂/𝐶𝑂2 ratios along the line
𝑎𝑏𝑐𝑑𝑒 , and finally the 𝐶𝑂/𝐶𝑂2 ratios for the reaction
2𝐶+ 𝑂2 = 2𝐶𝑂 beyond point (e).
15
Figure (9.15): the variation of %𝐶𝑂 by value in the 𝐶𝑂/ 𝐶𝑂2 gas
with temperature.
16
9.7 THE Equilibrium 𝟐𝑪𝑶 +𝑶𝟐 =𝟐𝐂𝐎𝟐
 The Ellingham line for the reaction :
2𝐶𝑂(𝑔) (1𝑎𝑡𝑚, 𝑇𝑒𝑞 ) + 𝑂2(𝑔) (1𝑎𝑡𝑚, 𝑇𝑒𝑞 ) =
2𝐶𝑂2(𝑔) (1𝑎𝑡𝑚 , 𝑇𝑒𝑞 )
(𝑖)
is shown in figure (9.16) as the line 𝐶𝑆.
 The effect of producing 𝐶𝑂2 at any pressure P , other than
1 𝑎𝑡𝑚 is given by the relation:
Δ𝐺(𝑖) (𝑐𝑎𝑙) = Δ𝐺 ° (𝑖) + 𝑅𝑇 𝑙𝑛(𝑃(𝐶𝑂2 ))
which means the rotation of the line 𝐶𝑆 around the C point
clockwise if 𝑃(𝐶𝑂2 )< 1 and anticlockwise if 𝑃(𝐶𝑂2 ) > 1 as
shown in figure (9.16).
 Referring to figure (9.16) , this means that for the reaction:
𝑀𝑂 + 𝐶𝑂(1𝑎𝑡𝑚, 𝑇𝑒𝑞 ) = 𝑀 + 𝐶𝑂2 (1𝑎𝑡𝑚, 𝑇𝑒𝑞 )
the equilibrium 𝑀, 𝑀𝑂, 𝐶𝑂(1𝑎𝑡𝑚, 𝑇𝑒𝑞 ), 𝐶𝑂2 (1𝑎𝑡𝑚, 𝑇𝑒𝑞 )
will occur at 𝑇𝑠 ; i.e. at Δ 𝐺 ° (𝑅) of the previous reaction = 0
if 𝑇 > 𝑇𝑠 the system will move towards 𝑀, 𝐶𝑂 , 𝐶𝑂2 and if
𝑇 < 𝑇𝑠 the system will move towards 𝑀𝑂, 𝐶𝑂 , 𝐶𝑂2 .
 It is also clear from the figure that as P(CO2 ) decreases, 𝑇𝑒𝑞
for the reaction will decrease and vice versa.
17
Figure (9.16) : illustration of the effect of the ratio
𝑃(𝐶𝑂)/𝑃(𝐶𝑂2 ) in a 𝐶𝑂-𝐶𝑂2 gas mixture
on the temperature at which the reaction
equilibrium 𝑀 + 𝐶𝑂2 = 𝑀𝑂 + 𝐶𝑂 is
established.
18
 Thus a 𝐶𝑂/𝐶𝑂2 nomographic scale can be added to the
Ellingham diagram , and for any reaction of the type :
𝑀𝑂2 + 2𝐶𝑂 = 𝑀 + 2𝐶𝑂2
the equilibrium 𝐶𝑂/𝐶𝑂2 ratio at any temperature 𝑻 can be
read off the scale since this value is collinear with point 𝐶 and
the point ∆𝐺 ° (𝑇) , 𝑇 = 𝑇 for the mentioned reaction. This
scale is drawn in figure (9.13).
 In a similar manner an 𝐻2 /𝐻2 𝑂 nomographic scale is
added to figure (9.13) by considering the effect of variation of
the 𝐻2 𝑂 pressure on the equilibrium of the reaction:
2𝐻2 + 𝑂2 = 2𝐻2 𝑂
 The equilibrium 𝐻2 /𝐻2 𝑂 ratio at temperature 𝑇 for the
reaction:
𝑀𝑂2 + 2𝐻2 = 𝑀 + 2𝐻2 𝑂
can be read off the 𝐻2 /𝐻2 𝑂 scale since this value is
collinear with point 𝐻 and ∆𝐺 ° (𝑇) , 𝑇 = 𝑇 for the reaction:
𝑀 + 𝑂2 = 𝑀𝑂2
19
9.8 Graphical Representation of Equilibrium in
the System Metal – Oxygen – Carbon
 Consider the chemical reaction:
2𝐶𝑂(𝑔) + 𝑂2 (𝑔) = 2𝐶𝑂2 (𝑔)
for this reaction , we can write:
log(𝑃(𝐶𝑂2 )/𝑃(𝐶𝑂))𝑒𝑞,𝑇 =
1
2
log 𝑃(𝑂2 )𝑒𝑞,𝑇 +
135,000
2×4.575𝑇
−
41.5
2×4.575
 Thus , for any given value of log 𝑃(𝑂2 )𝑒𝑞,𝑇 ,
log(𝑃(𝐶𝑂2 )/𝑃(𝐶𝑂))𝑒𝑞,𝑇 can be calculated, so this
relation expresses the variation of the given oxygen isobar
with log(𝑃(𝐶𝑂2 )/𝑃(𝐶𝑂))𝑒𝑞,𝑇 and 𝑇.
 The Oxygen isobars in the range 10−29 to 10−4 atm are
drawn as a function of 𝑙𝑜𝑔( 𝑃(𝐶𝑂2 )/𝑃(𝐶𝑂)) and T in
figure (9.17) and is shown in figure(9.18) as a function of
1
𝑙𝑜𝑔( 𝑃(𝐶𝑂2 )/𝑃(𝐶𝑂)) and ( ).
𝑇
20
Figure (9.17) : the relationship between the partial pressure of
oxygen in 𝐶𝑂−𝐶𝑂2 gas mixture and
temperature. The broken line is the variation
with temperature of the composition of the gas
which is in equilibrium with graphite at 1atm
pressure. The lines 𝐴𝐵 and 𝐶𝐷 represent,
respectively, the equilibria 𝐹𝑒+𝐶𝑂2 =𝐹𝑒𝑂+𝐶𝑂
and 𝐶𝑜+𝐶𝑂2 =𝐶𝑜𝑂+𝐶𝑂.
21
Figure (9.18): Figure ( 9.17 ) reproduced as
1
𝑙𝑜𝑔( 𝑃(𝐶𝑂2 )/𝑃(𝐶𝑂)) vs ( ).
𝑇
22
 The lower limit of 𝑃(𝐶𝑂2 )/𝑃(𝐶𝑂) ratio can be obtained at
any temperature by considering the equilibrium:
𝐶(𝑠) + 𝐶𝑂2 (𝑔) = 2𝐶𝑂(𝑔)
for which:
Δ𝐺 ° = − 𝑅𝑇 ln[𝑃(𝐶𝑂)𝟐 /𝑃(𝐶𝑂2 )] = 40,800 − 41.7𝑇
 The minimum obtainable ratio of 𝑃(𝐶𝑂2 )/𝑃(𝐶𝑂) at the
temperature 𝑻 can be obtained from the relation:
𝑃(𝐶𝑂2 )/𝑃(𝐶𝑂)=
(2+𝑥)−√(2+𝑥)2 −4
√(2+𝑥)2 −4 −𝑥
where: 𝑥 =𝑒𝑥𝑝(- 40,800 / 1.986T).𝑒𝑥𝑝(41.7 / 1.986)
 On both diagrams , the solid line 𝐴𝐵 , which is representing
the (𝑃(𝐶𝑂2 )/𝑃(𝐶𝑂)) – 𝑇(or 1/𝑇) relation for the reaction:
𝐹𝑒𝑂(𝑠) + 𝐶𝑂(𝑔) = 𝐹𝑒(𝑠) + 𝐶𝑂2 (𝑔) ,
and the solid line 𝐶𝐷 , which is representing the
𝑃(𝐶𝑂2 )/𝑃(𝐶𝑂) – 𝑇(or 1/𝑇) relation for the reaction:
𝐶𝑜(𝑠) + 𝐶𝑂(𝑔) = 𝐶𝑜(𝑠) + 𝐶𝑂2 (𝑔) ,
are plotted.
 The intersection point of the 𝐴𝐵 line or 𝐶𝐷 line with any of
the oxygen isobar yields the temperature of the equilibrium
𝑀, 𝑀𝑂, 𝐶𝑂 , 𝐶𝑂2 under the 𝑃(𝑂2 ) of this isobar, i.e. yield
the equilibrium point, (𝑃(𝐶𝑂2 )/𝑃(𝐶𝑂))𝑒𝑞 , 𝑇𝑒𝑞 for the
system 𝑀, 𝑀𝑂, 𝐶𝑂 , 𝐶𝑂2 at the corresponding pressure of
the isobar.
23
 It should worth to point out that the states under the 𝐴𝐵 line
represent the system 𝑀, 𝑃(𝐶𝑂), 𝑃(𝐶𝑂2 )(i.e. the reduction
case) while the states above the 𝐴𝐵 line represent the
system 𝑀, 𝑃(𝐶𝑂), 𝑃(𝐶𝑂2 ) (i.e. the oxidation case).
 This means that 𝑻 of the intersection point represents the
minimum temperature at which 𝐹𝑒𝑂 can be reduced to solid
𝐹𝑒.
 The line 𝐴𝐵 stops at 560 ℃ since below which 𝐹𝑒𝑂 is not
stable as iron oxide anymore because it dissociates into:
𝐹𝑒 + 𝐹𝑒3 𝑂4 ; i.e. 4𝐹𝑒𝑂 → 𝐹𝑒 + 𝐹𝑒3 𝑂4 .
 According to Gibbs-Helmholtz relation:
𝜕(∆𝐺 ° /𝑇)
(
so:
𝜕(1/𝑇)
𝜕ln(𝐾)
1
𝜕( )
𝑇
)𝑃 = ∆𝐻°
=−
∆𝐻 °
𝑅
 Therefore, the slopes of lines AB and CD are linear , since
∆ 𝐶𝑃 = 0 , and each slope = −∆𝑯˚/𝑹 for the
corresponding reaction, i.e. for 𝐹𝑒𝑂 + 𝐶𝑂 = 𝐹𝑒 + 𝐶𝑂2 ,
and 𝐶𝑜𝑂 + 𝐶𝑂 = 𝐶𝑜 + 𝐶𝑂2 .
24