REACTIONS INVOLVING PURE CONDENSED PHASES AND A GASEOUS
PHASE
It is desired here to determine equilibrium for the situation where one or more of the
reactants or products are in a condensed phase: Initially pure condensed phases. Many
practical systems are of this category e.g. how much O2 can be tolerated in a given system
without oxidizing a metal. To determine equilibrium, the following are necessary:
(1)
determine phase equilibrium between the individual condensed phase and the
gas phase
(2)
determine reaction equilibrium among the various species present in the gas
phase
Equilibrium between condensed and gaseous phases
Where M is metal, MO metal oxide, consider the equation
M(s) + 1/2O2(g) = MO
O2 insoluble in solid metal
Free energy in simplified form is written as
GoMO(s) - 1/2GoO2 - GoM(s) = -RTIn
1
1
P 2O2
It is assumed here that M and MO are pure condensed phases in their standard states and
their vapour pressures are negligible.
ΔGo = -RTlnk
K =
1
1
P 2O2
In the case of reaction equilibria involving pure condensed phases and a gas phase, the
equilibrium constant K can be written solely in terms of these species which occur in the
gas phase. Because both ΔGO and K are functions of T only, then at any fixed temperature
the establishment of reaction equilibrium occurs at a unique value of PO2 = PO2 (eq, T).
This equilibrium has one degree of freedom: If at any temperature T, the O2 partial
pressure in a closed ‘metal – metal oxide – oxygen’ system is greater that PO2(eq, T) then
spontaneous oxidation of the metal occurs, consuming oxygen decreasing the oxygen
partial pressure in the gas phase. When O2 pressure is lowered to PO2 (eq, T), as long as
both solid phases are still present, the oxidation reaction ceases and equilibrium is reestablished and the opposite is true.
For the reaction:
4Cu(s) + O2(g) = 2Cu2O(s)
ΔGoT = -339,000 – 14.2T1nT + 247T
298 – 1356
REACTIONS INVOLVING PURE CONDENSED PHASE AND A GASEOUS PHASE
-1nK = 1nPO2 =
logPO2(eq, T) =
2
G o
RT
339,000
- 14.2logT
2.303 x 8.3144T
8.3144
+
247 ____
2.3030 x 8.3144
1
is drawn. The line obtained is the continuous equilibrium of Cu(s)
T
and Cu2O(s) and O2(g) at that temperature T. Above the line, PO2 > PO2(eq, T), the metal
phase is unstable, the system exists as Cu2O(s) + O2(g); and below ab line PO2(eq, T) the
oxide is unstable and the system exists as Cu(s) + O2(g).
logPO2(eq, T) vs
Example 2
MO(s) + CO2(g) = MCO3(s)
Equilibrium occurs when
GoMO(s) + GoCO2 + RTlnPCO2 = GoMCO3(s)
i.e. when ΔGo = -RTlnK = RTlnPCO2(eq, T)
For the reaction
MgO(s) + CO2(g) = MgCO3(s) ; ΔGo = -117, 600 + 170T Joules
Log PCO2(eq, T) = -
117, 600
170
+
2.303  8.3144T
2.303  8.3144
REACTIONS INVOLVING PURE CONDENSED PHASE AND A GASEOUS PHASE
3
The variation is shown in the figure above. It shows the division of the diagram to
indicate the areas of stability for the gases and the corresponding solid phases.
Determination of ΔG at any temperature
Cp
dT
T
298
298
Can be determined at any temperature as follows: Let the applying heat capacity in
stated range of temperature:
T
o
ΔG
T
o
= ΔH
T
- TΔS
o
T
=
ΔHo298
+
 CpdT
T
-
TΔSo298
Cp = a + bT + CT-2
For the reaction we get
ΔCp = Δa + ΔbT + ΔcT-2
From Kirchoff,s equation,
 H o 
-2

 = ΔCp = Δa + ΔbT + ΔcT

T

p
dΔHo = (Δa + ΔbT + ΔcT-2)dT
ΔHo is the standard enthalpy change for the reaction. Integrating,
- T

REACTIONS INVOLVING PURE CONDENSED PHASE AND A GASEOUS PHASE
4
bT 2
- Δ cT-1 + ΔHo
10.8
2
ΔHo is an integration constant. ΔHo is normally evaluated by substituting a known value
of ΔHoT into equation 10.8. From the Gibbs Helmholtz equation,
ΔHoT = ΔaT +
 G 


o
 T  = - H
T2
T
H
T2
o
 G 


a b c Ho
T 

 3  2 = 
= T
2 T
T
T
 G o 
a b c Ho


 2 )dT
d
 = (T
2 T3
T
 T 
Integrating,
G o
= - a ln T
T
I an integration constant.

bT
2

c
2T 2

Ho
T

I
bT 2
c

 Ho  IT
…10.9
2
2T
The value of I can be determined if ΔGo at any temperature, usually 298 K is
known.
ΔGo = - ΔaTlnT -
From ΔGo = -RTlnK,
lnK =
a ln T
8.3144

bT
2  8.3144

c
2  8.3144  T 2

Ho
8.3144T

I
8.3144
the value of I can be determined if K at any temperature is known.
Example:
4Cu(s) + O2 = 2Cu2O(s)
ΔHo298 = -335,000 J/deg
ΔSo298 = -152.2 J/deg
ΔGo298 = ΔHo298 - T ΔSo298 = -335, 000 + 298 x 152.2 = -298,600
Cp
Cu(s)
= 22.6 + 6.3 x 10-3T
Cu2O(s) = 62.34 + 24 x 10-3T
O2(g) = 30 + 4.2 x 10-3T - 1.7 x 105T-2
298 – 1356K
298 – 1200K
298 – 3000K
REACTIONS INVOLVING PURE CONDENSED PHASE AND A GASEOUS PHASE
5
ΔGo can be obtained for the temperature range 298 - 1200K
ΔCp = 2CpCu2O(s) - 4CpCu(s) - CpO2(g)
= 4.28 + 18.6 x 10-3T + 1.7 x 105T-2 J/deg
d(ΔHoT) = (4.28 + 18.6 x 10-3T + 1.7 x 105T-2)dT
Integrating:
18.6 103 T 2
- 1.7 x 105T-1 + ΔHo
2
ΔHo an integration constant,
But ΔHo298 = -335,000 J, substituting,
ΔHoT = 4.28T +
18.6  103 (298) 2
ΔHo = -4.28 x 298 + 1.7 x 105(298)-1 -335, 000 = -336, 500
2
ΔHoT = 4.28T + 9.3 x 10-3T2 - 1.7 x 105T-1 - 336,500
 G 


o
 T  = - H = - 4.28
T
T2
T

 G o 
4.28
H o dT
d
=
= (
2
T
T
 T 
9.3 103


9.3 103
1.7 105
T3


1.7 105
T3
336,500
T2

336,500
)dT
T2
336,500
1.7  105
G o
-3
= - 4.28lnT - 9.3 x 10 T + I
2
T
2T
T
1.7  105
ΔGoT = - 4.28TlnT - 9.3 x 10-3T2 - 336,500 + IT
2T
Substituting the value of ΔGo298 in this equation gives value of I i.e.
-298,600 = - 4.28(298)ln(298) – 9.3 x 10-3(298)2 - 0.85 x 105(298)-1 - 336,500 + 298I
I = 185.5
ΔGoT = -4.28TlnT - 9.3 x 10-3T2 - 0.85 x 105T-1 - 336,500 + 185.5T
Ellingham Diagrams
Ellingham considered the preceding ΔGo expressions for a number of sulphides and oxides
and found that when plotted, fairly straight lines resulted over the ranges where there was
no physical change in the state of the reactant and products. The relationships could thus
be expressed by means of a simple equation:
REACTIONS INVOLVING PURE CONDENSED PHASE AND A GASEOUS PHASE
6
ΔGoT = A + BT
Where A = ΔHoT , B = -ΔSo Both ΔH and ΔS Temp independent
e.g.
for 4Ag(s) + O2(g) = 2Ag2O(s)
When plotted gives the figure below.
ΔHo is the intercept of the line with T = 0 K axis, and ΔSo is the negative of the slope of the
line.
At 462 K, ΔGo for the reaction is 0. At this temperature pure solid silver and oxygen gas at
1 atm pressure are in equilibrium with pure solid silver oxide.
ΔGoT = -RTlnK = RTlnPO2(eq, T) = 0
At T1, ΔGoT = -ve : Ag2O more stable than Ag in O2 at 1 atm. And therefore Ag
spontaneously oxidizes.
ΔGo1 = RTlnPO2(eq, T), gives value of PO2(eq, T)
Because ΔGo is –ve, PO2(eq, T) is less than unity
REACTIONS INVOLVING PURE CONDENSED PHASE AND A GASEOUS PHASE
7
At T>462 K, i.e. T2, ΔGo is +ve. Ag2O less stable than Ag in O2 at 1 atm therefore Ag2O
decomposes to Ag and O2. ΔGoT is a +ve quantity and PO2(eq, T) is greater than unity.
ΔGo for an oxidation reaction is a measure of the chemical affinity of a metal for oxygen.
The more –ve the value of ΔGo at any temperature the more stable the oxide.
P272 Parker
Consider the following reaction:
A(s) + O2(g) = AO2(s)
A being oxidized
ΔSo = SAO2(s) - SO2(g) - SA(s)
Because when AO2(s) and A(s) are solid, SO2(g) is considerably greater than SoA(s) and SoAO(s)
therefore,
ΔSo ≈ -SoO2
Hence the standard entropy changes of oxidation reactions involving solid phases are
almost the same corresponding essentially to the entropy resulting from the disappearance
of 1 mole of O2 gas initially at 1 atm pressure. Because the slopes of the lines in an
Ellingham diagram are equal to - ΔSo, then the lines are more or less parallel to one
another.
2Co(s) + O2 = 2CoO(s) :
and
2Mn(s) + O2(g) = 2MnO(s) :
ΔGoT = -467, 800 + 143T
298 – 1763
ΔGoT = -769, 400 + 145.6T
298 - 1500
REACTIONS INVOLVING PURE CONDENSED PHASE AND A GASEOUS PHASE
8
The values of ΔSo for these reactions are virtually equal to one another. The more –ve the
value of ΔHo the more negative the value of ΔGo and hence the more stable the oxide.
Richardson’s Nomograph scale for two different oxides
Consider two oxidation reactions of intersecting Ellingham diagrams:
2X + O2 = 2XO
Y + O2 = YO2
…i
…ii
Subtracting reaction i from reaction ii,
Y + 2XO = 2X + YO2
….
Below TE, X and YO2 are stable with respect to Y and XO and above TE the reverse applies
REACTIONS INVOLVING PURE CONDENSED PHASE AND A GASEOUS PHASE
9
At TE, X, Y, XO, and YO2 occurring in their standard states are in equilibrium. The
Ellingham diagram shows that if pure X were to be used as a reducing agent to reduce pure
YO2 to form pure Y and pure XO, then the required reduction could only be carried out at
temperatures in excess of Te. Normally for the sake of comparison, the Ellingham diagram
must be drawn for oxidation reactions involving the reaction or consumption of one mole
of O2. Richardson’s nomograph scale does away with the need to calculate the value of
PO2(eq, T) for any oxidation reactions. The PO2 at any temperature are read off the scale.
Effect of phase transformation
Within the temperature ranges where no phase changes occur, the Ellingham diagram can
be approximated by a straight line. Because in changing from solid to liquid or liquid to
vapour etc. the entropy of the higher phase, i.e. liquid exceeds that of the solid lower phase
by the latent value of the phase change, the Ellingham diagram exhibits an elbow.
ΔGoT = A + BT
For a reaction
X(s) + O2(g) = XO2(s)
Enthalpy = ΔHo, and entropy is ΔSo at Tm mp of X the reaction:
X(s) = X(l)
Occurs whose enthalpy and entropy changes are ΔHoM and ΔSoM
ΔSoM, X
=
H o M , X
TM , x
For the reaction:
X(l) + O2(g) = XO2(s)
If the enthalpy and entropy changes are ΔHo(l), and ΔSo(l) then
X(s) + O2(g) = XO2(s)
(-1) x (X(l) + O2(g) = XO2(s))
X(s) = X(l)
And
-ΔHo(l) + ΔHo = ΔHom,x
ΔSo(l) + ΔSo = ΔSom,x
ΔH(l) = ΔHo - ΔHom,x
ΔS(l) = ΔSo - ΔSom,x
REACTIONS INVOLVING PURE CONDENSED PHASE AND A GASEOUS PHASE
10
Because melting is always an endothermic process, ΔHm,x and ΔSom,x are always +ve, then
ΔHo - ΔHom,x is a larger –ve quantity than ΔHo
ΔSo - ΔSom,x is a larger –ve quantity than ΔSo
Therefore the Ellingham diagram line for the oxidation of liquid X to form solid XO2 has a
greater slope than the line for the solid X. The line shows an elbow “ upward” at Tm,x. If
the melting point of the oxide , Tm, of XO2 is lower than the mp of the metal, then at Tm,
XO2, the reaction below occurs:
XO2(s) = XO2(l)
For which the entropy and enthalpy changes are ΔSom,x and ΔHom,x. For the reaction
X(s) + O2(g) = XO2(l)
The enthalpy change is ΔHo + ΔHom,x and the entropy change is ΔSo + ΔSom,x. Both are
less –ve than the corresponding ΔSo and ΔHo. In this case the Ellingham diagram line for
the oxidation of X to produce liquid XO2 has a lesser slope than for the solid oxide line. At
TM, XO2 the line has an elbow downwards.
REACTIONS INVOLVING PURE CONDENSED PHASE AND A GASEOUS PHASE
11