mme 2010 metallurgical thermodynamics ii
TRANSCRIPT
Chemical reactions take place in production of steel from liquid iron
Impurities in liquid iron are subjected to a reaction with gaseous oxygen
Steel production takes place in a blast furnace that is aimed to collect liquid iron, slag and
flue gases formed as a result of reaction with C and CO
The liquid phases iron and slag in the blast furnace consist of solutions of Fe, C, Si, Mn, P
and SiO2, Al2O3, CaO, FeO respectively
Flue gases typically contain CO, CO2 and N2 as main components
Mathematical relationships of extensive thermodynamic properties, particularly G, H and
S need to be considered for each solution in order to estimate the equilibrium condition
parameters for maximum efficiency
Consider a general equilibrium:
ππ΄ + ππ΅ ππΆ + ππ·
The general criterion for equilibrium under constant T and P is βπΊ = 0
βπΊ = πΊπππππ’ππ‘π β πΊπππππ‘πππ‘π
= ππΊπΆ + ππΊπ· β ππΊπ΄ β ππΊπ΅
Recall that the complete differential of G is
Consider the reaction in a mixture of ideal gases at constant composition and
temperature (πππ = ππ =0)
The change in Gibbs free energy of each component as a function of its pressure is given
as
ππΊπ
πππ= π
ππΊπ =π ππππ
ππ
π ππΊ = ππ ππ β ππ ππ + ππ πππ
ππΊπ = π ππππ
ππ
πΊπ = πΊππ + π π ln
ππ
πππ = πΊπ
π + π π lnππ β 0
Since pressure of gases at standard conditions is assigned a value of 1 atm
The change in free energy of the system at constant temperature is
ππΊ = πππΊπ
Since mole number and pressure of ideal gases are proportional, ni /Pi is constant
and since the total pressure of the system is constant, πππ = 0
π ππΊ = ππ ππΊπ + πΊπ πππ
β ππΊ = π πππ
πππππ + πΊπ πππ
βπΊ = πΊπ πππ
In the case of system equilibrium
If the number of molecules of each component in the ideal gas mixture is relatively high,
their coefficients a, b, c, d can be used to represent πππ:
ππΊπΆπ + ππΊπ·
π β ππΊπ΄π β ππΊπ΅
π + π π lnππΆπ +π π lnππ·
π +π π lnππ΄βπ +π π lnππ΅
βπ = 0
where βπΊπ = ππΊπΆπ + ππΊπ·
π β ππΊπ΄π β ππΊπ΅
π
Absolute Gibbs free energy is computed for condensed phases as:
πΊπ = πΊππ + π π ln ππ
where a can be taken as unity for pure condensed phases
βπΊ = πΊππ πππ + π π ln(ππ) πππ = 0
βπΊ = πΊπ πππ = 0
βπΊπ + π π lnππΆ
πππ·π
ππ΄πππ΅
π = 0
The equation can be written for condensed phases as
βπΊ = βπΊπ + π π lnππΆ
πππ·π
ππ΄πππ΅
π= βπΊπ + π π lnπ
Q is called the reaction quotient
Q = K when βπΊ = 0
βπΊ = 0 = βπΊπ + π π lnπΎ
βπΊπ is readily given in literature for most compounds at STP
Since ππΊπ
ππ π= βπ,
βπΊπ = βπ»π + ππβπΊπ
πππ
βπ π lnπΎ = βπ»π β ππ π π lnπΎ
πππ
π lnπΎ
ππ=
βπ»π
π π2
π lnπΎ
π 1 π
= ββπ»π
π Vanβt Hoff equation
Equilibrium constant K of condensed phases
ππ΄ π + ππ΅ π ππΆ π + ππ· π βπΊ1π
The standard state equilibrium free energy of the above reaction can be expressed in terms of
the partial pressures of the gas phase of the components
At equilibrium
For all components the equilibrium equations are
ππ΄ π + ππ΅ π = ππΆ π + ππ· π βπΊ1π
ππ΄ π, ππ΄π = ππ΄ π βπΊ2 = 0
ππ΅ π, ππ΅π = ππ΅ π βπΊ3 = 0
ππΆ π, ππΆπ = ππΆ π βπΊ4 = 0
ππ· π, ππ·π = ππ· π βπΊ5 = 0
ππ΄ π, ππ΄π + ππ΅ π, ππ΅
π = ππΆ π, ππΆπ + ππ· π, ππ·
π βπΊ6 = βπΊ1π
Reaction free energy is the same if the reaction between gas phases that are in equilibrium
with the condensed phases is taken into consideration
However the reaction between gas phases is not in equilibrium state since βπΊ6 β 0
The equilibirum reaction between gas phases should include actual partial pressures rather
than pure pressures of the components:
ππ΄ π, ππ΄ + ππ΅ π, ππ΅ = ππΆ π, ππΆ + ππ· π, ππ· βπΊ7 = 0
ππ΄ π, ππ΄π = ππ΄ π βπΊ = 0
The change in reaction free energy due to the altered partial pressures can be calculated as
ππΊπ = π ππ lnππ ππ βπΊ = π π lnππ
ππ
ππ΄ π, ππ΄π + ππ΅ π, ππ΅
π = ππΆ π, ππΆπ + ππ· π, ππ·
π βπΊ6π = βπΊ1
π
ππ΄ π, ππ΄ = ππ΄ π, ππ΄π βπΊ8
π = π π lnππ
ππ΄
π
ππ΅ π, ππ΅ = ππ΅ π, ππ΅π βπΊ9
π = π π lnππ
ππ΅
π
ππΆ π, ππΆ = ππΆ π, ππΆπ βπΊ10
π = π π lnππ
ππΆ
π
ππ· π, ππ·π = ππ· π, ππ·
π βπΊ11π = π π ln
ππ
ππ·
π
ππ΄ π, ππ΄ + ππ΅ π, ππ΅π = ππΆ π, ππΆ
π + ππ· π, ππ·π βπΊ7 = 0
βπΊ7 = βπΊ1π + βπΊπ = 0
0 = βπΊ1π + π π ln
ππΆππ
π ππ·ππ
π
ππ΄ππ
π ππ΅ππ
π = βπΊ1π + π π lnπΎ
βπΊ π = π π ln K
Fugacity
The equations used to calculate the free energy change of mixtures comprised of ideal gases
like ππΊπ = π ππ lnππ can be modified by the fugacity concept for use with non-ideal gases:
ππΊπ = π ππ ln ππ
For an ideal gas ππ = ππ
If the free energy of a substance corresponding to the standard state fugacity is denoted as
βπΊππ,
βπΊ1 β βπΊ1π = π π ln
ππ
πππ
As pressure approaches 0, ππ approaches ππ
Activity rather than fugacity is used to relate ideal mixtures to condensed phase mixtures
ππ =ππ
πππ
βπΊπ can be calculated for any temperature since βπΊπ = βπ»π β πβππ,
βπΊπ = βπ»π298 +
298
π
βπΆπππ β π βππ298 +
298
π βπΆπππ
π
where πΆπ = π + ππ +π
π2
and βπΆπ= βπ + βππ +π
π2 where βπ, π, π = βπ, π, ππππππ’ππ‘π β βπ, π, ππππππ‘πππ‘π
βπΊπ = βπ»π298 +
298
π
βπ + βππ + βππ2 ππ β π βππ
298 + 298
π βπ + βππ + βππ2 ππ
π
βπΊπ = βπ»π298 + βππ +
βππ2
2β βπ
π β π βππ298 + βπ lnπ + βππ β βπ
2π2
Replacement of the upper and the lower limits yields
βπΊπ = πΌπ + πΌ1π β βππ lnπ ββπ
2π2 β
βπ
2π
where πΌπ = βπ»π298 β βπ298 +
βπ2982
2β βπ
298
πΌ1 = βπ β βππ298 + βπ ln 298 + βπ298 β βπ
2β2982
T
298
T
298
Example - One important equilibrium between condensed phases such as metals and
oxides and a gaseous phase such as oxygen is oxidation of metals
Consider the oxidation of copper
4Cu π + O2 π = 2Cu2O(π )
βπ»π298 = β334400 J
βππ298 = β152.07 J/K
βπΆπ = 4.18 + 0.01839π +1.67 β 105
π2 J/K
πΌπ = β334400 β 1500πΌ1 = 4.18 + 152.07 + 28.35
Using the above thermochemical data
βπΊπ = β335900 + 184.59π β 4.18π lnπ β 0.0092π2 β0.84 β 105
πJ
Alternatively the standard free energy for formation of pure Cu2O from pure Cu and oxygen
is
βπΊπ = π π lnππ2(πππ)
Experimental variation of βπΊπwith T can be calculated from the measured oxygen partial
pressure ππ2(πππ) that is in equilibrium with Cu and Cu2O
When experimental βπΊπ vs T is fit to
βπΊπ = π΄ + π΅π logπ + πΆπβπΊπ = β338580 β 32.77π logπ + 246.62π J
If a 2-term fit is used
βπΊπ = π΄ + π΅πβπΊπ = β328580 + 137.94π J
The comparison of βπΊπ calculated from the thermochemical data and experimental data is
made, the difference in temperature range 400 to 1200 K is seen between 293 to 794 J
The comparison of βπΊπ calculated from the experimental data and 2-term fit shows that
the difference in the 400 to 1200 K temperature range is even less, between 286 to 788 J
T (K) Thermochemical Experimental 2-term fit a-b b-c a-c
400 -273748 -274041 -273324 293 -717 -424
600 -244655 -245241 -245736 586 495 1081
800 -216566 -217360 -218148 794 7881582
1000 -189480 -190274 -190560 794 286 1080
1200 -163270 -163689 -162972 419 -717 -298
Table shows that overall difference between βπΊπ calculated from the thermochemical data
and 2-term fit in the temperature range is at most 1.6 kJ which is within the experimental
error for most of the βπΊπmeasurements
Therefore the variation of βπΊπwith temperature for oxidation and sulfidation reactions can
be approximated to linear forms
This observation lead to representation of the reaction free energies as Ellingham
diagram:
Ellingham plotted the experimentally determined βπΊπ values as a function of T for the
oxidation and sulfidation of a series of metals
The general form of the relationships approximated to straight lines over temperature
ranges in which no change in physical state occurred
The complex terms TlnT, T2 and 1/T does not affect the linearity of the curves for the
considered range
βπΊπ = βπ»π β πβππ
βπΊπ = π΄ + π΅π
A, which is the intercept at 0 K, is the temperature independent βπ»π
B, the slope, ββππ
These approximations imply that ΞCPβ 0 for oxidation reactions
Another important characteristic of Ellingham diagram is that; all the lines represent
reactions involving one mole of oxygen:
2π₯
π¦M + O2
2
π¦MxOy
Therefore, the ordinate βπΊπ of all the oxidation reactions become π π lnππ2(πππ)
Almost all the βπΊπ lines have positive slopes since βππ < 0:
M(s) + O2(π) MO2(π )
βππ = βππππ2 β βππ
π β βπππ2
πππ2 is generally dominant in the temperature range where M and MO2 are solid
So βππ β βπππ2
The slopes are approximately equal in the temperature range where metal and oxide are
solid
Therefore almost all the lines are parallel to each other in this temperature range
Examples
2Ni π + O2 π 2NiO π
πππ2 = 205.11
π½
πππ. πΎ
ππππ(π ) = 29.8
π½
πππ. πΎ
πππππ(π ) = 38.09
π½
πππ. πΎ
βππ = β188.53 J/K
Sn s + O2 π SnO2 π
ππππ(π ) = 51.49
π½
πππ. πΎ
πππππ2(π ) = 48.59
π½
πππ. πΎ
βππ = β208.04 J/K
Standard entropies at 298 K
Standard entropies at 298 K
The exceptions to the general trends in oxidation lines are
C π + O2 π CO2 π βππ β 0
and
2C π + O2 π 2CO π βππ > 0
Lines on Ellingham diagrams often have sharp breaks in them which are caused by phase
transformations
The straight line representation of βπΊπ vs T relationships are valid if there is no physical
change taking place for any one of the components taking part in the reaction equilibrium
Consider the transformation of M(s) to M(l) at Tm
M π + O2 π MO2 π
M π + O2 π MO2 π 1
2
T
βπ»π
Tm
T
βππ
Tm
βπ»π1> βπ»π
2
βππ1> βππ
2
The net effect of phase transformation of a reactant from a low temperature to a high
temperature phase is an increase in slope
T
βπβππ
Tm
M(s)
M(l)
T
βπΊπ
Tm
M(s)
M(l)
Consider the transformation of product MO2(s) to MO2(l) at Tm
M π + O2 π MO2 π
M π + O2 π MO2 π
1
2
TTm
T
βππ
βπ»π1< βπ»π
2
βππ1< βππ
2
βπ»π
Tm
The net effect of phase transformation of a product from a low temperature to a high
temperature phase is a decrease in slope
T
βπβππ
Tm
MO2 (s)MO2 (l)
T
βπΊπ
Tm
M(s)
M(l)
Effect of evaporation on βπΊπ vs T plots is similar to melting, but more pronounced since
βπ»ππ£ is about 10 times greater than βπ»π
π
Slope changes for allotropic transformations will be less than that for melting
T
βπΊπ
Tm
M(s)
M(l)
M(g)
Tb
The Ellingham line for the M/MO2 equilibrium is shown in the figure
βπΊ = 0 all along the line and oxygen partial pressure is PO2(eqm)
Let the partial pressure of oxygen be PO2(actual) when a metal is exposed to an oxidizing
atmosphere
βπΊ = βπΊπ + π π ln1
ππ2(πππ‘π’ππ), βπΊ = π π lnππ2 πππ βπ π lnππ2 πππ‘π’ππ
At any point above the line, ππ2 πππ‘π’ππ > ππ2 πππ , then βπΊ < 0This implies that MO2 formation is spontaneous and MO2 is stable above the line
At any point below the line, ππ2 πππ‘π’ππ < ππ2 πππ , then βπΊ > 0This implies that MO2 formation is impossible and M is stable below the line
T
βπΊπ
M, O2
MO2, O2
Most of the lines on Ellingham diagram are almost parallel to each other
Consider the Ellingham lines for the M/MO2 and N/NO2 equilibria:
The oxide with the larger region of stability is more stable
It is evident from the figure that NO is relatively more stable than MO
The element with the more stable oxide is more reactive
The element of the less stable oxide is more stable in elemental form, M is more stable
than N
2MO π 2M π + O2(π)
2N π + O2 π 2NO(π )
2MO π + 2N π 2M π + 2NO π Net reaction
T
βπΊπ
M, NO2
MO2, NO2
M, N
Ellingham lines sometimes intercept each other
Consider the two lines for X/XO and Y/YO equilibria:
Relative stability of oxides changes with temperature in this case
Below the equilibrium temperature, YO2 is more stable, XO2 becomes more stable above TE
All the components X, Y, XO2, YO2 coexist at the equilibrium temperature
There is no stability region for Y and XO2 together below TE, and for X, YO2 together above
TE
If Y and XO2 are brought together at a temperature below TE, Y will be oxidized while XO2
will be reduced
T
βπΊπ
X, YO2
XO2, YO2
X, Y
XO2, Y
TE
Partial pressure grid lines
Ellingham diagram offers a simple and useful way to estimate equilibrium oxygen
pressures as a function of temperature
For constant ππ2 values, βπΊπ vs T is represented by straight lines with R lnππ2slope and
βπΊπ = 0 intercept
Constant oxygen partial pressures can be read from the oxygen partial pressure scale
when these lines are superimposed
The intersections of the constant oxygen partial pressure lines and the X-XO2equilibirum
line give the equilibrium oxygen partial pressures for this reaction at various temperatures
Oxygen partial pressure at 700 Β°C, in equilibrium with X and XO2 is 10-11atm
Oxide is stable if the oxide lines are under the partial pressure line at a temperature
T
βπΊπ
XO2, O2
X, O2
700 Β°C
0PO2
1 atm
10-11 atm