MME 2010 METALLURGICAL THERMODYNAMICS II
Transcript of MME 2010 METALLURGICAL THERMODYNAMICS II
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:
๐๐ด + ๐๐ต ๐๐ถ + ๐๐ท
The general criterion for equilibrium under constant T and P is โ๐บ = 0
โ๐บ = ๐บ๐๐๐๐๐ข๐๐ก๐ โ ๐บ๐๐๐๐๐ก๐๐๐ก๐
= ๐๐บ๐ถ + ๐๐บ๐ท โ ๐๐บ๐ด โ ๐๐บ๐ต
Recall that the complete differential of G is
Consider the reaction in a mixture of ideal gases at constant composition and
temperature (๐๐๐ = ๐๐ =0)
The change in Gibbs free energy of each component as a function of its pressure is given
as
๐๐บ๐
๐๐๐= ๐
๐๐บ๐ =๐ ๐๐๐๐
๐๐
๐ ๐๐บ = ๐๐ ๐๐ โ ๐๐ ๐๐ + ๐๐ ๐๐๐
๐๐บ๐ = ๐ ๐๐๐๐
๐๐
๐บ๐ = ๐บ๐๐ + ๐ ๐ ln
๐๐
๐๐๐ = ๐บ๐
๐ + ๐ ๐ ln๐๐ โ 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
๐๐บ = ๐๐๐บ๐
Since mole number and pressure of ideal gases are proportional, ni /Pi is constant
and since the total pressure of the system is constant, ๐๐๐ = 0
๐ ๐๐บ = ๐๐ ๐๐บ๐ + ๐บ๐ ๐๐๐
โ ๐๐บ = ๐ ๐๐๐
๐๐๐๐๐ + ๐บ๐ ๐๐๐
โ๐บ = ๐บ๐ ๐๐๐
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 ๐๐๐:
๐๐บ๐ถ๐ + ๐๐บ๐ท
๐ โ ๐๐บ๐ด๐ โ ๐๐บ๐ต
๐ + ๐ ๐ ln๐๐ถ๐ +๐ ๐ ln๐๐ท
๐ +๐ ๐ ln๐๐ดโ๐ +๐ ๐ ln๐๐ต
โ๐ = 0
where โ๐บ๐ = ๐๐บ๐ถ๐ + ๐๐บ๐ท
๐ โ ๐๐บ๐ด๐ โ ๐๐บ๐ต
๐
Absolute Gibbs free energy is computed for condensed phases as:
๐บ๐ = ๐บ๐๐ + ๐ ๐ ln ๐๐
where a can be taken as unity for pure condensed phases
โ๐บ = ๐บ๐๐ ๐๐๐ + ๐ ๐ ln(๐๐) ๐๐๐ = 0
โ๐บ = ๐บ๐ ๐๐๐ = 0
โ๐บ๐ + ๐ ๐ ln๐๐ถ
๐๐๐ท๐
๐๐ด๐๐๐ต
๐ = 0
The equation can be written for condensed phases as
โ๐บ = โ๐บ๐ + ๐ ๐ ln๐๐ถ
๐๐๐ท๐
๐๐ด๐๐๐ต
๐= โ๐บ๐ + ๐ ๐ ln๐
Q is called the reaction quotient
Q = K when โ๐บ = 0
โ๐บ = 0 = โ๐บ๐ + ๐ ๐ ln๐พ
โ๐บ๐ is readily given in literature for most compounds at STP
Since ๐๐บ๐
๐๐ ๐= โ๐,
โ๐บ๐ = โ๐ป๐ + ๐๐โ๐บ๐
๐๐๐
โ๐ ๐ ln๐พ = โ๐ป๐ โ ๐๐ ๐ ๐ ln๐พ
๐๐๐
๐ ln๐พ
๐๐=
โ๐ป๐
๐ ๐2
๐ ln๐พ
๐ 1 ๐
= โโ๐ป๐
๐ Vanโt Hoff equation
Equilibrium constant K of condensed phases
๐๐ด ๐ + ๐๐ต ๐ ๐๐ถ ๐ + ๐๐ท ๐ โ๐บ1๐
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
๐๐ด ๐ + ๐๐ต ๐ = ๐๐ถ ๐ + ๐๐ท ๐ โ๐บ1๐
๐๐ด ๐, ๐๐ด๐ = ๐๐ด ๐ โ๐บ2 = 0
๐๐ต ๐, ๐๐ต๐ = ๐๐ต ๐ โ๐บ3 = 0
๐๐ถ ๐, ๐๐ถ๐ = ๐๐ถ ๐ โ๐บ4 = 0
๐๐ท ๐, ๐๐ท๐ = ๐๐ท ๐ โ๐บ5 = 0
๐๐ด ๐, ๐๐ด๐ + ๐๐ต ๐, ๐๐ต
๐ = ๐๐ถ ๐, ๐๐ถ๐ + ๐๐ท ๐, ๐๐ท
๐ โ๐บ6 = โ๐บ1๐
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 โ๐บ6 โ 0
The equilibirum reaction between gas phases should include actual partial pressures rather
than pure pressures of the components:
๐๐ด ๐, ๐๐ด + ๐๐ต ๐, ๐๐ต = ๐๐ถ ๐, ๐๐ถ + ๐๐ท ๐, ๐๐ท โ๐บ7 = 0
๐๐ด ๐, ๐๐ด๐ = ๐๐ด ๐ โ๐บ = 0
The change in reaction free energy due to the altered partial pressures can be calculated as
๐๐บ๐ = ๐ ๐๐ ln๐๐ ๐๐ โ๐บ = ๐ ๐ ln๐๐
๐๐
๐๐ด ๐, ๐๐ด๐ + ๐๐ต ๐, ๐๐ต
๐ = ๐๐ถ ๐, ๐๐ถ๐ + ๐๐ท ๐, ๐๐ท
๐ โ๐บ6๐ = โ๐บ1
๐
๐๐ด ๐, ๐๐ด = ๐๐ด ๐, ๐๐ด๐ โ๐บ8
๐ = ๐ ๐ ln๐๐
๐๐ด
๐
๐๐ต ๐, ๐๐ต = ๐๐ต ๐, ๐๐ต๐ โ๐บ9
๐ = ๐ ๐ ln๐๐
๐๐ต
๐
๐๐ถ ๐, ๐๐ถ = ๐๐ถ ๐, ๐๐ถ๐ โ๐บ10
๐ = ๐ ๐ ln๐๐
๐๐ถ
๐
๐๐ท ๐, ๐๐ท๐ = ๐๐ท ๐, ๐๐ท
๐ โ๐บ11๐ = ๐ ๐ ln
๐๐
๐๐ท
๐
๐๐ด ๐, ๐๐ด + ๐๐ต ๐, ๐๐ต๐ = ๐๐ถ ๐, ๐๐ถ
๐ + ๐๐ท ๐, ๐๐ท๐ โ๐บ7 = 0
โ๐บ7 = โ๐บ1๐ + โ๐บ๐ = 0
0 = โ๐บ1๐ + ๐ ๐ ln
๐๐ถ๐๐
๐ ๐๐ท๐๐
๐
๐๐ด๐๐
๐ ๐๐ต๐๐
๐ = โ๐บ1๐ + ๐ ๐ ln๐พ
โ๐บ ๐ = ๐ ๐ ln K
Fugacity
The equations used to calculate the free energy change of mixtures comprised of ideal gases
like ๐๐บ๐ = ๐ ๐๐ ln๐๐ can be modified by the fugacity concept for use with non-ideal gases:
๐๐บ๐ = ๐ ๐๐ ln ๐๐
For an ideal gas ๐๐ = ๐๐
If the free energy of a substance corresponding to the standard state fugacity is denoted as
โ๐บ๐๐,
โ๐บ1 โ โ๐บ1๐ = ๐ ๐ ln
๐๐
๐๐๐
As pressure approaches 0, ๐๐ approaches ๐๐
Activity rather than fugacity is used to relate ideal mixtures to condensed phase mixtures
๐๐ =๐๐
๐๐๐
โ๐บ๐ can be calculated for any temperature since โ๐บ๐ = โ๐ป๐ โ ๐โ๐๐,
โ๐บ๐ = โ๐ป๐298 +
298
๐
โ๐ถ๐๐๐ โ ๐ โ๐๐298 +
298
๐ โ๐ถ๐๐๐
๐
where ๐ถ๐ = ๐ + ๐๐ +๐
๐2
and โ๐ถ๐= โ๐ + โ๐๐ +๐
๐2 where โ๐, ๐, ๐ = โ๐, ๐, ๐๐๐๐๐๐ข๐๐ก๐ โ โ๐, ๐, ๐๐๐๐๐๐ก๐๐๐ก๐
โ๐บ๐ = โ๐ป๐298 +
298
๐
โ๐ + โ๐๐ + โ๐๐2 ๐๐ โ ๐ โ๐๐
298 + 298
๐ โ๐ + โ๐๐ + โ๐๐2 ๐๐
๐
โ๐บ๐ = โ๐ป๐298 + โ๐๐ +
โ๐๐2
2โ โ๐
๐ โ ๐ โ๐๐298 + โ๐ ln๐ + โ๐๐ โ โ๐
2๐2
Replacement of the upper and the lower limits yields
โ๐บ๐ = ๐ผ๐ + ๐ผ1๐ โ โ๐๐ ln๐ โโ๐
2๐2 โ
โ๐
2๐
where ๐ผ๐ = โ๐ป๐298 โ โ๐298 +
โ๐2982
2โ โ๐
298
๐ผ1 = โ๐ โ โ๐๐298 + โ๐ ln 298 + โ๐298 โ โ๐
2โ2982
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 ๐ + O2 ๐ = 2Cu2O(๐ )
โ๐ป๐298 = โ334400 J
โ๐๐298 = โ152.07 J/K
โ๐ถ๐ = 4.18 + 0.01839๐ +1.67 โ 105
๐2 J/K
๐ผ๐ = โ334400 โ 1500๐ผ1 = 4.18 + 152.07 + 28.35
Using the above thermochemical data
โ๐บ๐ = โ335900 + 184.59๐ โ 4.18๐ ln๐ โ 0.0092๐2 โ0.84 โ 105
๐J
Alternatively the standard free energy for formation of pure Cu2O from pure Cu and oxygen
is
โ๐บ๐ = ๐ ๐ ln๐๐2(๐๐๐)
Experimental variation of โ๐บ๐with T can be calculated from the measured oxygen partial
pressure ๐๐2(๐๐๐) that is in equilibrium with Cu and Cu2O
When experimental โ๐บ๐ vs T is fit to
โ๐บ๐ = ๐ด + ๐ต๐ log๐ + ๐ถ๐โ๐บ๐ = โ338580 โ 32.77๐ log๐ + 246.62๐ J
If a 2-term fit is used
โ๐บ๐ = ๐ด + ๐ต๐โ๐บ๐ = โ328580 + 137.94๐ J
The comparison of โ๐บ๐ 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 โ๐บ๐ 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 โ๐บ๐ 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 โ๐บ๐measurements
Therefore the variation of โ๐บ๐with 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 โ๐บ๐ 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
โ๐บ๐ = โ๐ป๐ โ ๐โ๐๐
โ๐บ๐ = ๐ด + ๐ต๐
A, which is the intercept at 0 K, is the temperature independent โ๐ป๐
B, the slope, โโ๐๐
These approximations imply that ฮCPโ 0 for oxidation reactions
Another important characteristic of Ellingham diagram is that; all the lines represent
reactions involving one mole of oxygen:
2๐ฅ
๐ฆM + O2
2
๐ฆMxOy
Therefore, the ordinate โ๐บ๐ of all the oxidation reactions become ๐ ๐ ln๐๐2(๐๐๐)
Almost all the โ๐บ๐ lines have positive slopes since โ๐๐ < 0:
M(s) + O2(๐) MO2(๐ )
โ๐๐ = โ๐๐๐๐2 โ โ๐๐
๐ โ โ๐๐๐2
๐๐๐2 is generally dominant in the temperature range where M and MO2 are solid
So โ๐๐ โ โ๐๐๐2
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 ๐ + O2 ๐ 2NiO ๐
๐๐๐2 = 205.11
๐ฝ
๐๐๐. ๐พ
๐๐๐๐(๐ ) = 29.8
๐ฝ
๐๐๐. ๐พ
๐๐๐๐๐(๐ ) = 38.09
๐ฝ
๐๐๐. ๐พ
โ๐๐ = โ188.53 J/K
Sn s + O2 ๐ SnO2 ๐
๐๐๐๐(๐ ) = 51.49
๐ฝ
๐๐๐. ๐พ
๐๐๐๐๐2(๐ ) = 48.59
๐ฝ
๐๐๐. ๐พ
โ๐๐ = โ208.04 J/K
Standard entropies at 298 K
Standard entropies at 298 K
The exceptions to the general trends in oxidation lines are
C ๐ + O2 ๐ CO2 ๐ โ๐๐ โ 0
and
2C ๐ + O2 ๐ 2CO ๐ โ๐๐ > 0
Lines on Ellingham diagrams often have sharp breaks in them which are caused by phase
transformations
The straight line representation of โ๐บ๐ 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 ๐ + O2 ๐ MO2 ๐
M ๐ + O2 ๐ MO2 ๐ 1
2
T
โ๐ป๐
Tm
T
โ๐๐
Tm
โ๐ป๐1> โ๐ป๐
2
โ๐๐1> โ๐๐
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
โ๐โ๐๐
Tm
M(s)
M(l)
T
โ๐บ๐
Tm
M(s)
M(l)
Consider the transformation of product MO2(s) to MO2(l) at Tm
M ๐ + O2 ๐ MO2 ๐
M ๐ + O2 ๐ MO2 ๐
1
2
TTm
T
โ๐๐
โ๐ป๐1< โ๐ป๐
2
โ๐๐1< โ๐๐
2
โ๐ป๐
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
โ๐โ๐๐
Tm
MO2 (s)MO2 (l)
T
โ๐บ๐
Tm
M(s)
M(l)
Effect of evaporation on โ๐บ๐ vs T plots is similar to melting, but more pronounced since
โ๐ป๐๐ฃ is about 10 times greater than โ๐ป๐
๐
Slope changes for allotropic transformations will be less than that for melting
T
โ๐บ๐
Tm
M(s)
M(l)
M(g)
Tb
The Ellingham line for the M/MO2 equilibrium is shown in the figure
โ๐บ = 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
โ๐บ = โ๐บ๐ + ๐ ๐ ln1
๐๐2(๐๐๐ก๐ข๐๐), โ๐บ = ๐ ๐ ln๐๐2 ๐๐๐ โ๐ ๐ ln๐๐2 ๐๐๐ก๐ข๐๐
At any point above the line, ๐๐2 ๐๐๐ก๐ข๐๐ > ๐๐2 ๐๐๐ , then โ๐บ < 0This implies that MO2 formation is spontaneous and MO2 is stable above the line
At any point below the line, ๐๐2 ๐๐๐ก๐ข๐๐ < ๐๐2 ๐๐๐ , then โ๐บ > 0This implies that MO2 formation is impossible and M is stable below the line
T
โ๐บ๐
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 ๐ 2M ๐ + O2(๐)
2N ๐ + O2 ๐ 2NO(๐ )
2MO ๐ + 2N ๐ 2M ๐ + 2NO ๐ Net reaction
T
โ๐บ๐
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
โ๐บ๐
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 ๐๐2 values, โ๐บ๐ vs T is represented by straight lines with R ln๐๐2slope and
โ๐บ๐ = 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 ยฐC, 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
โ๐บ๐
XO2, O2
X, O2
700 ยฐC
0PO2
1 atm
10-11 atm