Chemical reaction equilibria and Auxillary functions
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:
๐๐ด + ๐๐ต ๐๐ถ + ๐๐ท
Most metallurgical processes occur at constant T and P
The general criterion for equilibrium under constant T and P is โ๐บ = 0
โ๐บ = ๐บ๐๐๐๐๐ข๐๐ก๐ โ ๐บ๐๐๐๐๐ก๐๐๐ก๐
= ๐๐บ๐ถ + ๐๐บ๐ท โ ๐๐บ๐ด โ ๐๐บ๐ต
The complete differential of G in terms of T and P is
๐๐บ = ๐๐๐ โ ๐๐๐
๐๐บ =๐๐บ
๐๐๐๐ +
๐๐บ
๐๐๐๐
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 as๐๐บ๐
๐๐๐= ๐
๐๐บ๐ =๐ ๐๐๐๐
๐๐
๐๐บ๐ = ๐ ๐๐๐๐
๐๐= ๐บ๐ = ๐บ๐
๐ + ๐ ๐ ln๐๐
๐๐๐ = ๐บ๐
๐ + ๐ ๐ ln ๐๐
The change in free energy of the system at constant temperature is the
sum of its componentsโ free energy change
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
โ ๐๐บ = ๐ ๐๐๐
๐๐๐๐๐ + ๐บ๐ ๐๐๐ = ๐บ๐ ๐๐๐
๐ ๐๐บ = ๐๐ ๐๐บ๐ + 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(๐๐๐๐๐)
โ๐บ = ๐บ๐ ๐๐๐ = 0
โ๐บ๐ + ๐ ๐ ln๐๐ถ
๐๐๐ท๐
๐๐ด๐๐๐ต
๐ = 0
The equation can be written 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
Example - Consider the equilibria in which two salts dissolve in water to form aqueous
solutions of ions:
NaCl(s)Na+(aq) + Cl-(aq) ฮHยฐsoln(NaCl)= 3.6 kJ/mol, ฮSยฐsoln(NaCl)= 43.2 J/mol.K
AgCl(s)Ag+(aq) + Cl-(aq) ฮHยฐsoln(AgCl)= 65.7 kJ/mol, ฮSยฐsoln(NaCl)= 34.3 J/mol.K
a) Calculate the value of ฮGยฐ at 298 K for each of the reactions. How will ฮGยฐ 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 ฮGยฐ 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 ฮGยฐ 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 โ๐บ = โ๐ป โ ๐โ๐
๐บ๐ = ๐ป๐ + ๐๐๐บ๐
๐๐ ๐Since
๐๐บ๐
๐๐ ๐= โ๐,
Multiplying both sides by dT and dividing by T2,
๐บ๐๐๐
๐2 =๐ป๐๐๐
๐2 + ๐๐๐บ๐
๐2๐
๐ป๐๐๐
๐2 =๐บ๐๐๐
๐2 โ๐๐๐บ๐
๐2 = โ๐๐บ๐
๐,
โ๐ป๐
๐2 =โ๐
โ๐บ๐
๐
๐๐
Gibbs-
Helmholtz
Eqn
Example โ 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
๐4 ๐ 2๐2 ๐
โ๐บ๐ = โ225000 + 18.2๐๐๐๐ โ 50.1๐
Effect of temperature on equilibrium
At equilibrium, โ๐บ๐ = โ๐ ๐ ln๐พ
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
For the case ofโ๐ป๐ > 0, temperature increase shifts the reaction towards products
For the case ofโ๐ป๐ < 0, temperature increase shifts the reaction towards reactants
Vanโt Hoff equation
Slope>0
Slope<0
exothermic
endothermic
ln K
1/T
๐ ln๐พ
๐ 1 ๐
= โโ๐ป๐
๐
Effect of pressure on equilibrium
Although equilibrium constant is independent of pressure, Le Chetelierโs 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
๐พ =๐๐ถ
๐๐๐ท๐
๐๐ด๐๐๐ต
๐ =(๐๐ถ๐)๐(๐๐ท๐)๐
(๐๐ด๐)๐(๐๐ต๐)๐
K is not affected by changes in pressure, but consists of two terms; KX and P:
๐พ = ๐พ๐๐(๐+๐โ๐โ๐)
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:
โ๐บ = โ๐บ๐ + ๐ ๐ ln๐
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
Plotting the โ๐บ๐ 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 โ๐บ๐ of any reaction by a simple 2-term fit such as
โ๐บ๐ = ๐ด + ๐ต๐
The following grouping lead to a condensed representation of โGo which can further
be simplified
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
2 โ 2982
Tabulated thermochemical data such as โ๐ป๐298, โ๐๐
298, โ๐ถ๐ for a specific reaction are
replaced into the general equation for โ๐บ๐ to obtain the variation of the spontaneity
with temperature
Alternatively experimental variation of โ๐บ๐with T can be calculated from the
measured oxygen partial pressure ๐๐2(๐๐๐) that is in equilibrium with a metal and
metal oxide using equation:
โ๐บ๐ = ๐ ๐ ln๐๐2(๐๐๐)
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 ฮHm= 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
๐พ =๐๐ถ
๐๐๐ท๐
๐๐ด๐๐๐ต
๐
The partial pressures of the components are expressed as a function of the total P:
๐๐ด =๐๐ด. ๐
๐๐ด + ๐๐ต + ๐๐ถ + ๐๐ท
where ๐๐ด is the mole number of A under equilibrium
Equilibrium constant can be represented as
๐พ =๐๐ถ
๐๐๐ท๐
๐๐ด๐๐๐ต
๐ โ๐
๐๐ด + ๐๐ต + ๐๐ถ + ๐๐ท
๐+๐ โ(๐+๐)
๐๐ด(๐) + ๐๐ต(๐) ๐๐ถ(๐) + ๐๐ท(๐)
Suppose the reaction reaches equilibrium after a while and x moles of A is
converted to products
Then
๐๐ด =Moles of unreacted A = 1 โ ๐ฅ ๐๐๐ต = Moles of unreacted B = 1 โ ๐ฅ ๐๐๐ถ = Moles of formed C = ๐ฅ. ๐๐๐ท = Moles of formed D = ๐ฅ. ๐
and
๐พ =(๐ฅ. ๐)๐(๐ฅ. ๐)๐
๐ โ ๐๐ฅ ๐(๐ โ ๐๐ฅ)๐โ
๐
1 โ ๐ฅ ๐ + ๐ + ๐ฅ ๐ + ๐
๐+๐ โ(๐+๐)
If equilibrium temperature and the standard free energy change at that
temperature are given, the fraction x can be conveniently determined since
โ๐บ = โ๐บ๐ + ๐ ๐๐๐๐ ln๐พ = 0
โ๐บ๐ = โ๐ ๐๐๐๐ ln(๐ฅ. ๐)๐(๐ฅ. ๐)๐
๐ โ ๐๐ฅ ๐(๐ โ ๐๐ฅ)๐โ
๐
1 โ ๐ฅ ๐ + ๐ + ๐ฅ ๐ + ๐
๐+๐ โ(๐+๐)
Example โ Determine the equilibrium composition of the system when 1 mole of P4
reacts to form P2 at 1300 K
๐4 ๐ 2๐2 ๐
โ๐บ๐ = โ225000 + 18.2๐๐๐๐ โ 50.1๐
โ๐บ๐ = โ๐ ๐๐๐๐ ln(๐ฅ. ๐)๐(๐ฅ. ๐)๐
๐ โ ๐๐ฅ ๐(๐ โ ๐๐ฅ)๐โ
๐
1 โ ๐ฅ ๐ + ๐ + ๐ฅ ๐ + ๐
๐+๐ โ(๐+๐)
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 ๐๐ = ๐๐ โ ๐๐
For reversible processes the second law states that
๐๐ =๐๐
๐or ๐๐ = ๐๐๐
And for mechanical work
๐๐ = ๐๐๐
so
๐๐ = ๐๐๐ โ ๐๐๐
๐๐ = ๐๐๐ โ ๐๐๐
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
โข The system should be closed
โข 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
๐๐ป = ๐๐ + ๐ ๐๐ = ๐๐ + ๐๐๐ + ๐๐๐
Replacing the relation for dU,
๐๐ป = ๐๐๐ โ ๐๐๐ + ๐๐๐ + ๐๐๐๐๐ป = ๐๐๐ + ๐๐๐
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:
๐บ = ๐ป โ ๐๐๐๐บ = ๐๐ป โ ๐๐๐ โ ๐๐๐
Replacing the relation for dH,
๐๐บ = ๐๐๐ + ๐๐๐ โ ๐๐๐ โ ๐๐๐๐๐บ = ๐๐๐ โ ๐๐๐
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
๐ด = ๐ โ ๐๐๐๐ด = ๐๐ โ ๐๐๐ โ ๐๐๐
Replacing the relationship for dU,
๐๐ด = ๐๐๐ โ ๐๐๐ โ ๐๐๐ โ ๐๐๐๐๐ด = โ๐๐๐ โ ๐๐๐
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
๐๐บ = ๐๐๐ โ ๐๐๐
๐๐บ =๐๐บ
๐๐๐
๐๐ +๐๐บ
๐๐๐
๐๐
๐๐ป = ๐๐๐ + ๐๐๐
๐๐ป =๐๐ป
๐๐๐
๐๐ +๐๐ป
๐๐๐
๐๐
๐๐ = ๐๐๐ โ ๐๐๐
๐๐ =๐๐
๐๐๐
๐๐ +๐๐
๐๐๐
๐๐
๐๐ด = โ๐๐๐ โ ๐๐๐
๐๐ด =๐๐ด
๐๐๐
๐๐ +๐๐ด
๐๐๐
๐๐
๐ =๐๐ป
๐๐๐
=๐๐
๐๐๐
๐ = โ๐๐
๐๐๐
= โ๐๐ด
๐๐๐
๐ =๐๐บ
๐๐๐
=๐๐ป
๐๐๐
โ๐ =๐๐บ
๐๐๐
=๐๐ด
๐๐๐
๐๐บ
๐๐๐
= ๐,๐๐บ
๐๐๐
= โ๐
๐๐ป
๐๐๐
= ๐,๐๐ป
๐๐๐
= ๐
๐๐
๐๐๐
= ๐,๐๐
๐๐๐
= โ๐
๐๐ด
๐๐๐
= โ๐,๐๐ด
๐๐๐
= โ๐
Other useful relationships called Maxwell Equations derive from the complete
differentials of the state functions by virtue of the exact differential function:
๐๐บ = ๐๐๐ โ ๐๐๐๐๐
๐๐๐
= โ๐๐
๐๐๐
๐๐ป = ๐๐๐ + ๐๐๐๐๐
๐๐๐
=๐๐
๐๐๐
๐๐ = ๐๐๐ โ ๐๐๐๐๐
๐๐๐
= โ๐๐
๐๐๐
๐๐ด = โ๐๐๐ โ ๐๐๐๐๐
๐๐๐
=๐๐
๐๐๐
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
๐ฝ๐ =๐๐
๐๐๐
= โ๐๐
๐๐๐
๐ ๐ =๐๐
๐๐๐
=๐๐
๐๐๐
๐๐
๐๐๐
=๐๐
๐๐๐
๐๐
๐๐๐
๐ถ๐ =๐๐ป
๐๐๐
Example โ Develop a relationship for the variation of enthalpy with pressure for
isothermal processes as function of ฮฒ, T and V and show that the enthalpy change with
P for ideal gases is 0
๐๐ป
๐๐๐
=
Since ๐๐ป = ๐๐๐ + ๐๐๐๐๐ป
๐๐๐
= ๐๐๐
๐๐๐
+ ๐๐๐
๐๐๐
= ๐๐๐
๐๐๐
+ ๐
๐๐ป
๐๐๐
= โ๐๐ฝ๐ + ๐
since
For ideal gases
๐๐ป
๐๐๐
= โ๐๐๐
๐๐๐
+ ๐ = โ๐๐
๐+ ๐ = โ๐ + ๐ = 0
โ๐๐
๐๐๐
=๐๐
๐๐๐
= ๐ฝ๐
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 ฮฒ= 2.095*10-3 /K
โ๐๐
๐๐๐
=๐๐
๐๐๐
= ๐ฝ๐
Example โ 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
ฮHV= 131859 J/mole
๐๐(๐, 1400 ๐พ, 20 ๐๐ก๐) ๐๐(๐, 1400 ๐พ, 20 ๐๐ก๐)
๐ฝ๐ =๐๐
๐๐๐
= โ๐๐
๐๐๐