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Chemical reaction equilibria
and Auxillary functions
Chemical reaction equilibria in metallurgical processes and the conditions that
maintain equilibrium are important to obtain maximum efficiency from production
processes
For example, 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
Iron oxide is reduced by CO to metallic iron while impurities in liquid iron are
subjected to reaction with gaseous oxygen in converting stage
Consider a general reaction in equilibrium:
ππ΄ + ππ΅ ππΆ + ππ·
Most metallurgical processes occur at constant T and P
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΅
The complete differential of G in terms of T and P is
ππΊ = πππ β πππ
ππΊ =ππΊ
ππππ +
ππΊ
ππππ
Consider the reaction in a mixture of ideal gases at constant temperature
The change in Gibbs free energy of each ideal gas component as a function
of its pressure is given asππΊπ
πππ= π
ππΊπ =π ππππ
ππ
ππΊπ = π ππππ
ππ= πΊπ = πΊπ
π + π π lnππ
πππ = πΊπ
π + π π ln ππ
The change in free energy of the system at constant temperature is the
sum of its componentsβ free energy change
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
β ππΊ = π πππ
πππππ + πΊπ πππ = πΊπ πππ
π ππΊ = ππ ππΊπ + 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(πππππ)
βπΊ = πΊπ πππ = 0
βπΊπ + π π lnππΆ
πππ·π
ππ΄πππ΅
π = 0
The equation can be written 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
Example - Consider the equilibria in which two salts dissolve in water to form aqueous
solutions of ions:
NaCl(s)Na+(aq) + Cl-(aq) ΞHΒ°soln(NaCl)= 3.6 kJ/mol, ΞSΒ°soln(NaCl)= 43.2 J/mol.K
AgCl(s)Ag+(aq) + Cl-(aq) ΞHΒ°soln(AgCl)= 65.7 kJ/mol, ΞSΒ°soln(NaCl)= 34.3 J/mol.K
a) Calculate the value of ΞGΒ° at 298 K for each of the reactions. How will ΞGΒ° for the solution process of NaCl
and AgCl change with increasing T? What effect should this change have on the solubility of the salts?
b) Is the difference between two free energies primarily due to the enthalpy term or the entropy term of the
standard free-energy change?
c) Use the values of ΞGΒ° to calculate the K values for the two salts at 298 K
d) Sodium chloride is considered a soluble salt, whereas silver chloride is considered insoluble. Are these
descriptions consistent with the answers to part c?
e)How will ΞGΒ° for the solution process of these salts change with increasing T? What effect should this change
have on the solubility of the salts?
Recall that βπΊ = βπ» β πβπ
πΊπ = π»π + πππΊπ
ππ πSince
ππΊπ
ππ π= βπ,
Multiplying both sides by dT and dividing by T2,
πΊπππ
π2 =π»πππ
π2 + πππΊπ
π2π
π»πππ
π2 =πΊπππ
π2 βπππΊπ
π2 = βππΊπ
π,
βπ»π
π2 =βπ
βπΊπ
π
ππ
Gibbs-
Helmholtz
Eqn
Example β Determine the heat exchange between system and surroundings for the
following reaction in order to keep the temperature of the system constant at 1300 K
π4 π 2π2 π
βπΊπ = β225000 + 18.2ππππ β 50.1π
Effect of temperature on equilibrium
At equilibrium, βπΊπ = βπ π lnπΎ
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
For the case ofβπ»π > 0, temperature increase shifts the reaction towards products
For the case ofβπ»π < 0, temperature increase shifts the reaction towards reactants
Vanβt Hoff equation
Slope>0
Slope<0
exothermic
endothermic
ln K
1/T
π lnπΎ
π 1 π
= ββπ»π
π
Effect of pressure on equilibrium
Although equilibrium constant is independent of pressure, Le Chetelierβs principle
states that an increase in total pressure at constant temperature will shift the
equilibrium in the direction which decreases the number of moles of gaseous
species in the system
πΎ =ππΆ
πππ·π
ππ΄πππ΅
π =(ππΆπ)π(ππ·π)π
(ππ΄π)π(ππ΅π)π
K is not affected by changes in pressure, but consists of two terms; KX and P:
πΎ = πΎππ(π+πβπβπ)
Change in pressure may have effect on KX, quotient of mole fractions depending on
the values of a, b, c, and d
If
c+d>a+b, increasing pressure decreases KX, reaction shifts towards reactants
c+d=a+b, pressure does not affect KX
c+d<a+b, KX is proportional to pressure, reaction shifts towards products with
increasing KX
Oxygen pressure dependence of spontaneity of oxidation reactions
The spontaneity of any process at constant T and P is dependent on the change in the Gibbs free
energy of the system:
βπΊ = βπΊπ + π π lnπ
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
Plotting the βπΊπ values of similar oxidation reactions as a function of T and comparing their
relative reactivities would be useful for engineering complex systems like furnace charge, if it
was possible to express βπΊπ of any reaction by a simple 2-term fit such as
βπΊπ = π΄ + π΅π
The following grouping lead to a condensed representation of βGo which can further
be simplified
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
2 β 2982
Tabulated thermochemical data such as βπ»π298, βππ
298, βπΆπ for a specific reaction are
replaced into the general equation for βπΊπ to obtain the variation of the spontaneity
with temperature
Alternatively experimental variation of βπΊπwith T can be calculated from the
measured oxygen partial pressure ππ2(πππ) that is in equilibrium with a metal and
metal oxide using equation:
βπΊπ = π π lnππ2(πππ)
T
298
T
298
Ellingham diagram
Example - Will the reaction
4Cu(l) + O2(g) = 2Cu2O(s)
go spontaneously to the right or to the left at 1500K when oxygen pressure is 0.01 atm?
Cu(s) S298=33.36 J/molK, Cp=22.65+0.00628T J/molK ΞHm= 13000 J/mole at 1356K
Cu(l) Cp= 31.40 J/molK
Cu2O(s) H298=-167440 J/mol S298=93.14 J/molK, Cp=83.6 J/molK
O2(g) S298=205.11 J/molK, Cp=33.44 J/molK
Determining the composition of reaction system under equilibrium
Consider the reacting A, B to produce C and D
πΎ =ππΆ
πππ·π
ππ΄πππ΅
π
The partial pressures of the components are expressed as a function of the total P:
ππ΄ =ππ΄. π
ππ΄ + ππ΅ + ππΆ + ππ·
where ππ΄ is the mole number of A under equilibrium
Equilibrium constant can be represented as
πΎ =ππΆ
πππ·π
ππ΄πππ΅
π βπ
ππ΄ + ππ΅ + ππΆ + ππ·
π+π β(π+π)
ππ΄(π) + ππ΅(π) ππΆ(π) + ππ·(π)
Suppose the reaction reaches equilibrium after a while and x moles of A is
converted to products
Then
ππ΄ =Moles of unreacted A = 1 β π₯ πππ΅ = Moles of unreacted B = 1 β π₯ πππΆ = Moles of formed C = π₯. πππ· = Moles of formed D = π₯. π
and
πΎ =(π₯. π)π(π₯. π)π
π β ππ₯ π(π β ππ₯)πβ
π
1 β π₯ π + π + π₯ π + π
π+π β(π+π)
If equilibrium temperature and the standard free energy change at that
temperature are given, the fraction x can be conveniently determined since
βπΊ = βπΊπ + π ππππ lnπΎ = 0
βπΊπ = βπ ππππ ln(π₯. π)π(π₯. π)π
π β ππ₯ π(π β ππ₯)πβ
π
1 β π₯ π + π + π₯ π + π
π+π β(π+π)
Example β Determine the equilibrium composition of the system when 1 mole of P4
reacts to form P2 at 1300 K
π4 π 2π2 π
βπΊπ = β225000 + 18.2ππππ β 50.1π
βπΊπ = βπ ππππ ln(π₯. π)π(π₯. π)π
π β ππ₯ π(π β ππ₯)πβ
π
1 β π₯ π + π + π₯ π + π
π+π β(π+π)
The power of thermodynamics lies in the provision of the criteria for spontaneity
within a system
The practical usefulness of this power to predict the outcome of processes is
determined by the practicality of the equations of state of the system, or the
relationships among the state functions
The relationships among thermodynamic functions P, V, T, S, U, H, A and G are well
determined which makes it possible to predict the spontaneity of any process at
certain conditions
Recall that ππ = ππ β ππ
For reversible processes the second law states that
ππ =ππ
πor ππ = πππ
And for mechanical work
ππ = πππ
so
ππ = πππ β πππ
ππ = πππ β πππ
This equation relates the dependent variable U to independent variables S and V as
result of the combined statement of the first and second laws
Restrictions on the applicability of this realation are
β’ The system should be closed
β’ The work due to volume change is the only form of work
Hence the criterion for equilibrium for constant entropy and constant volume is dU= 0
Recall that at constant pressure H= U+PV
ππ» = ππ + π ππ = ππ + πππ + πππ
Replacing the relation for dU,
ππ» = πππ β πππ + πππ + πππππ» = πππ + πππ
Hence the criterion for equilibrium for constant entropy and constant pressure is
dH= 0
The same restrictions apply to the system as the relation for dU
Recall the general equation for Gibbs free energy:
πΊ = π» β ππππΊ = ππ» β πππ β πππ
Replacing the relation for dH,
ππΊ = πππ + πππ β πππ β πππππΊ = πππ β πππ
Hence the criterion for equilibrium for constant pressure and constant temperature is
dG= 0
This property is very important in metallurgical applications because most processes
occur under constant temperature and pressure
A less useful relation is used for the Helmholtz energy A
π΄ = π β ππππ΄ = ππ β πππ β πππ
Replacing the relationship for dU,
ππ΄ = πππ β πππ β πππ β πππππ΄ = βπππ β πππ
Hence the criterion for spontaneity for constant volume and temperature is dA= 0
Useful relationships between the partial derivatives of U, H, G, and A result in
valuable simplifications in thermodynamic equations
ππΊ = πππ β πππ
ππΊ =ππΊ
πππ
ππ +ππΊ
πππ
ππ
ππ» = πππ + πππ
ππ» =ππ»
πππ
ππ +ππ»
πππ
ππ
ππ = πππ β πππ
ππ =ππ
πππ
ππ +ππ
πππ
ππ
ππ΄ = βπππ β πππ
ππ΄ =ππ΄
πππ
ππ +ππ΄
πππ
ππ
π =ππ»
πππ
=ππ
πππ
π = βππ
πππ
= βππ΄
πππ
π =ππΊ
πππ
=ππ»
πππ
βπ =ππΊ
πππ
=ππ΄
πππ
ππΊ
πππ
= π,ππΊ
πππ
= βπ
ππ»
πππ
= π,ππ»
πππ
= π
ππ
πππ
= π,ππ
πππ
= βπ
ππ΄
πππ
= βπ,ππ΄
πππ
= βπ
Other useful relationships called Maxwell Equations derive from the complete
differentials of the state functions by virtue of the exact differential function:
ππΊ = πππ β πππππ
πππ
= βππ
πππ
ππ» = πππ + πππππ
πππ
=ππ
πππ
ππ = πππ β πππππ
πππ
= βππ
πππ
ππ΄ = βπππ β πππππ
πππ
=ππ
πππ
The value of Maxwell equations lies in the fact that they contain many experimentally
measurable quantities
Other equations may be developed for the changes in thermodynamic quantities that
are difficult to measure experimentally by the use of Maxwell equations
π½π =ππ
πππ
= βππ
πππ
π π =ππ
πππ
=ππ
πππ
ππ
πππ
=ππ
πππ
ππ
πππ
πΆπ =ππ»
πππ
Example β Develop a relationship for the variation of enthalpy with pressure for
isothermal processes as function of Ξ², T and V and show that the enthalpy change with
P for ideal gases is 0
ππ»
πππ
=
Since ππ» = πππ + πππππ»
πππ
= πππ
πππ
+ πππ
πππ
= πππ
πππ
+ π
ππ»
πππ
= βππ½π + π
since
For ideal gases
ππ»
πππ
= βπππ
πππ
+ π = βππ
π+ π = βπ + π = 0
βππ
πππ
=ππ
πππ
= π½π
Example - Estimate the change in enthalpy and entropy when liquid ammonia at 273 K
is compressed from its saturation pressure of 381 kPa to 1200 kPa. For saturated liquid
ammonia at 273 K, take volume and expansivity coefficient as V= 1.551*10-3 m3/kg,
and Ξ²= 2.095*10-3 /K
βππ
πππ
=ππ
πππ
= π½π
Example β Normal boiling point for Mg is 1393 K. By using entropy concept calculate
whether the evaporation is spontaneous or not at 1400 K under 20 atm pressure
Hint: Separate the process at 1400 K and 20 atm into reversible steps to bring to 1 atm
CP(Mg(l))= 31.0 J/mole.K
CP(Mg(g))= 20.8 J/mole.K
ΞHV= 131859 J/mole
ππ(π, 1400 πΎ, 20 ππ‘π) ππ(π, 1400 πΎ, 20 ππ‘π)
π½π =ππ
πππ
= βππ
πππ