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Auxiallary functions and
Chemical reaction equilibria
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 𝑑𝑈 = 𝑑𝑄 − 𝑑𝑊
For reversible processes the second law states that
𝑑𝑆 =𝑑𝑄
𝑇or 𝑑𝑄 = 𝑇𝑑𝑆
And for mechanical work
𝑑𝑊 = 𝑃𝑑𝑉
so
𝑑𝑈 = 𝑇𝑑𝑆 − 𝑃𝑑𝑉
𝑑𝑈 = 𝑇𝑑𝑆 − 𝑃𝑑𝑉
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
• The system should be closed
• 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
𝑑𝐻 = 𝑑𝑈 + 𝑑 𝑃𝑉 = 𝑑𝑈 + 𝑃𝑑𝑉 + 𝑉𝑑𝑃
Replacing the relation for dU,
𝑑𝐻 = 𝑇𝑑𝑆 − 𝑃𝑑𝑉 + 𝑃𝑑𝑉 + 𝑉𝑑𝑃𝑑𝐻 = 𝑇𝑑𝑆 + 𝑉𝑑𝑃
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:
𝐺 = 𝐻 − 𝑇𝑆𝑑𝐺 = 𝑑𝐻 − 𝑇𝑑𝑆 − 𝑆𝑑𝑇
Replacing the relation for dH,
𝑑𝐺 = 𝑇𝑑𝑆 + 𝑉𝑑𝑃 − 𝑇𝑑𝑆 − 𝑆𝑑𝑇𝑑𝐺 = 𝑉𝑑𝑃 − 𝑆𝑑𝑇
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
𝐴 = 𝑈 − 𝑇𝑆𝑑𝐴 = 𝑑𝑈 − 𝑇𝑑𝑆 − 𝑆𝑑𝑇
Replacing the relationship for dU,
𝑑𝐴 = 𝑇𝑑𝑆 − 𝑃𝑑𝑉 − 𝑇𝑑𝑆 − 𝑆𝑑𝑇𝑑𝐴 = −𝑃𝑑𝑉 − 𝑆𝑑𝑇
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
𝑑𝐺 = 𝑉𝑑𝑃 − 𝑆𝑑𝑇
𝑑𝐺 =𝜕𝐺
𝜕𝑃𝑇
𝑑𝑃 +𝜕𝐺
𝜕𝑇𝑃
𝑑𝑇
𝑑𝐻 = 𝑇𝑑𝑆 + 𝑉𝑑𝑃
𝑑𝐻 =𝜕𝐻
𝜕𝑆𝑃
𝑑𝑆 +𝜕𝐻
𝜕𝑃𝑆
𝑑𝑃
𝑑𝑈 = 𝑇𝑑𝑆 − 𝑃𝑑𝑉
𝑑𝑈 =𝜕𝑈
𝜕𝑆𝑉
𝑑𝑆 +𝜕𝑈
𝜕𝑉𝑆
𝑑𝑉
𝑑𝐴 = −𝑃𝑑𝑉 − 𝑆𝑑𝑇
𝑑𝐴 =𝜕𝐴
𝜕𝑉𝑇
𝑑𝑉 +𝜕𝐴
𝜕𝑇𝑉
𝑑𝑇
𝑇 =𝜕𝐻
𝜕𝑆𝑃
=𝜕𝑈
𝜕𝑆𝑉
𝑃 = −𝜕𝑈
𝜕𝑉𝑆
= −𝜕𝐴
𝜕𝑉𝑇
𝑉 =𝜕𝐺
𝜕𝑃𝑇
=𝜕𝐻
𝜕𝑃𝑆
−𝑆 =𝜕𝐺
𝜕𝑇𝑃
=𝜕𝐴
𝜕𝑇𝑉
𝜕𝐺
𝜕𝑃𝑇
= 𝑉,𝜕𝐺
𝜕𝑇𝑃
= −𝑆
𝜕𝐻
𝜕𝑆𝑃
= 𝑇,𝜕𝐻
𝜕𝑃𝑆
= 𝑉
𝜕𝑈
𝜕𝑆𝑉
= 𝑇,𝜕𝑈
𝜕𝑉𝑆
= −𝑃
𝜕𝐴
𝜕𝑉𝑇
= −𝑃,𝜕𝐴
𝜕𝑇𝑉
= −𝑆
Other useful relationships called Maxwell Equations derive from the complete differentials
of the state functions by virtue of the exact differential function:
𝑑𝐺 = 𝑉𝑑𝑃 − 𝑆𝑑𝑇𝜕𝑉
𝜕𝑇𝑃
= −𝜕𝑆
𝜕𝑃𝑇
𝑑𝐻 = 𝑇𝑑𝑆 + 𝑉𝑑𝑃𝜕𝑇
𝜕𝑃𝑆
=𝜕𝑉
𝜕𝑆𝑃
𝑑𝑈 = 𝑇𝑑𝑆 − 𝑃𝑑𝑉𝜕𝑇
𝜕𝑉𝑆
= −𝜕𝑃
𝜕𝑆𝑇
𝑑𝐴 = −𝑃𝑑𝑉 − 𝑆𝑑𝑇𝜕𝑃
𝜕𝑇𝑉
=𝜕𝑆
𝜕𝑉𝑇
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
𝛽𝑉 =𝜕𝑉
𝜕𝑇𝑃
= −𝜕𝑆
𝜕𝑃𝑇
𝜅𝑉 =𝜕𝑉
𝜕𝑃𝑇
=𝜕𝑉
𝜕𝑇𝑃
𝜕𝑇
𝜕𝑃𝑉
=𝜕𝑉
𝜕𝑇𝑃
𝜕𝑉
𝜕𝑆𝑇
𝐶𝑃 =𝜕𝐻
𝜕𝑇𝑃
Example – Develop a relationship for the variation of enthalpy with pressure for isothermal
processes as function of β, T and V and show that the enthalpy change with P for ideal
gases is 0
𝜕𝐻
𝜕𝑃𝑇
=
Since 𝑑𝐻 = 𝑇𝑑𝑆 + 𝑉𝑑𝑃𝜕𝐻
𝜕𝑃𝑇
= 𝑇𝜕𝑆
𝜕𝑃𝑇
+ 𝑉𝜕𝑃
𝜕𝑃𝑇
= 𝑇𝜕𝑆
𝜕𝑃𝑇
+ 𝑉
𝜕𝐻
𝜕𝑃𝑇
= −𝑇𝛽𝑉 + 𝑉
since
For ideal gases
𝜕𝐻
𝜕𝑃𝑇
= −𝑇𝜕𝑉
𝜕𝑇𝑃
+ 𝑉 = −𝑇𝑅
𝑃+ 𝑉 = −𝑉 + 𝑉 = 0
−𝜕𝑆
𝜕𝑃𝑇
=𝜕𝑉
𝜕𝑇𝑃
= 𝛽𝑉
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 β=
2.095*10-3 /K
−𝜕𝑆
𝜕𝑃𝑇
=𝜕𝑉
𝜕𝑇𝑃
= 𝛽𝑉
Example – 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
ΔHV= 131859 J/mole
𝑀𝑔(𝑙, 1400 𝐾, 20 𝑎𝑡𝑚) 𝑀𝑔(𝑔, 1400 𝐾, 20 𝑎𝑡𝑚)
𝛽𝑉 =𝜕𝑉
𝜕𝑇𝑃
= −𝜕𝑆
𝜕𝑃𝑇
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:
𝑎𝐴 + 𝑏𝐵 𝑐𝐶 + 𝑑𝐷
The general criterion for equilibrium under constant T and P is ∆𝐺 = 0
∆𝐺 = 𝐺𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 − 𝐺𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠
= 𝑐𝐺𝐶 + 𝑑𝐺𝐷 − 𝑎𝐺𝐴 − 𝑏𝐺𝐵
The complete differential of G in terms of T and P is
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 as𝜕𝐺𝑖
𝜕𝑃𝑖= 𝑉𝑖
𝑑𝐺𝑖 =𝑅𝑇𝑑𝑃𝑖
𝑃𝑖
𝑑𝐺 = 𝑉𝑑𝑃 − 𝑆𝑑𝑇
𝑑𝐺 =𝜕𝐺
𝜕𝑃𝑇
𝑑𝑃 +𝜕𝐺
𝜕𝑇𝑃
𝑑𝑇
𝑑𝐺𝑖 = 𝑅𝑇𝑑𝑃𝑖
𝑃𝑖
𝐺𝑖 = 𝐺𝑖𝑜 + 𝑅𝑇 ln
𝑃𝑖
𝑃𝑖𝑜
The change in free energy of the system at constant temperature is the
sum of the free energy change of its components
𝑛𝐺 = 𝑛𝑖𝐺𝑖
Since mole number and pressure of ideal gases are proportional, ni /Pi is
constant and since the total pressure of the system is constant, 𝑑𝑃𝑖 = 0
𝑑 𝑛𝐺 = 𝑛𝑖 𝑑𝐺𝑖 + 𝐺𝑖 𝑑𝑛𝑖
∆ 𝑛𝐺 = 𝑅𝑇𝑛𝑖
𝑃𝑖𝑑𝑃𝑖 + 𝐺𝑖 𝑑𝑛𝑖
∆𝐺 = 𝐺𝑖 𝑑𝑛𝑖
In the case of system equilibrium
The stoichiometric coefficients a, b, c, d of each component in the ideal gas
mixture can be used to represent 𝑑𝑛𝑖:
𝑐𝐺𝐶𝑜 + 𝑑𝐺𝐷
𝑜 − 𝑎𝐺𝐴𝑜 − 𝑏𝐺𝐵
𝑜 + 𝑅𝑇 ln𝑃𝐶𝑐 +𝑅𝑇 ln𝑃𝐷
𝑑 +𝑅𝑇 ln𝑃𝐴−𝑎 +𝑅𝑇 ln𝑃𝐵
−𝑏 = 0
where ∆𝐺𝑜 = 𝑐𝐺𝐶𝑜 + 𝑑𝐺𝐷
𝑜 − 𝑎𝐺𝐴𝑜 − 𝑏𝐺𝐵
𝑜
Absolute Gibbs free energy is computed for gas phases as:
𝐺𝑖 = 𝐺𝑖𝑜 + 𝑅𝑇 ln𝑃𝑖
∆𝐺 = 𝐺𝑖𝑜 𝑑𝑛𝑖 + 𝑅𝑇 ln(𝑃𝑖)𝑑𝑛𝑖
∆𝐺 = 𝐺𝑖 𝑑𝑛𝑖 = 0
∆𝐺𝑜 + 𝑅𝑇 ln𝑃𝐶
𝑐𝑃𝐷𝑑
𝑃𝐴𝑎𝑃𝐵
𝑏 = 0
The equation for gas phases can be written as
∆𝐺 = ∆𝐺𝑜 + 𝑅𝑇 ln𝑃𝐶
𝑐𝑃𝐷𝑑
𝑃𝐴𝑎𝑃𝐵
𝑏 = ∆𝐺𝑜 + 𝑅𝑇 ln𝑄𝑅
QR is called the reaction quotient
QR = K when ∆𝐺 = 0
∆𝐺 = 0 = ∆𝐺𝑜 + 𝑅𝑇 ln𝐾
∆𝐺𝑜 is readily given in literature for most compounds at STP
The relationship between DGo and K at 298 K
Example - Estimate DGo for the decomposition of NO2 at 25oC
At 25oC and 1.00 atmosphere pressure, K =4.3x10-13
FO
RW
AR
D R
EA
CT
ION
RE
VE
RS
E R
EA
CT
ION
∆𝐺 = 𝑅𝑇 ln𝑄𝑅 − 𝑅𝑇 ln𝐾 = 𝑅𝑇 ln𝑄𝑅
𝐾
∆𝐺 can be calculated for any temperature, since ∆𝐺𝑜 = ∆𝐻𝑜 − 𝑇∆𝑆𝑜
∆𝐺 = ∆𝐻𝑜298 +
298
𝑇
∆𝐶𝑃𝑑𝑇 − 𝑇 ∆𝑆𝑜298 +
298
𝑇 ∆𝐶𝑃𝑑𝑇
𝑇
where 𝐶𝑃 = 𝑎 + 𝑏𝑇 +𝑐
𝑇2
and ∆𝐶𝑃= ∆𝑎 + ∆𝑏𝑇 +𝑐
𝑇2 where ∆𝑎, 𝑏, 𝑐 = ∆𝑎, 𝑏, 𝑐𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 − ∆𝑎, 𝑏, 𝑐𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠
∆𝐺𝑜 is the free energy change that would accompany the conversion of reactants,
initially present in their standard states, to products in their standard states
DG is the free energy change for other temperatures and pressures
∆𝐺 = 𝑅𝑇 ln𝑄𝑅 − 𝑅𝑇 ln𝐾 = 𝑅𝑇 ln𝑄𝑅
𝐾
DG has a very large positive or negative value if QR and K are very different
The reaction releases or absorbs a large amount of free energy
DG has a very small positive or negative value if QR and K are close
The reaction releases or absorbs a small amount of free energy
Example -The equilibrium constant at different temperatures for the following
reaction is given:
SO3(g) = SO2(g) + ½ O2(g)
K= 0.146 @ 900K
K= 0.516 @ 1000K
K= 1.45 @ 1100K
Estimate the enthalpy change of the reaction at 1000K and the equilibrium
composition at the same temperature
Example - Consider the equilibria in which two salts dissolve in water to form aqueous
solutions of ions:
NaCl(s)Na+(aq) + Cl-(aq) ΔH°soln(NaCl)= 3.6 kJ/mol, ΔS°soln(NaCl)= 43.2 J/mol.K
AgCl(s)Ag+(aq) + Cl-(aq) ΔH°soln(AgCl)= 65.7 kJ/mol, ΔS°soln(NaCl)= 34.3 J/mol.K
a) Calculate the value of ΔG° at 298 K for each of the reactions. How will ΔG° 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 ΔG° 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 ΔG° for the solution process of these salts change with increasing T? What effect should this change
have on the solubility of the salts?
Effect of pressure on equilibrium
Although equilibrium constant is independent of pressure, Le Chetelier’s 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
𝐾 =𝑃𝐶
𝑐𝑃𝐷𝑑
𝑃𝐴𝑎𝑃𝐵
𝑏 =(𝑋𝐶𝑃)𝑐(𝑋𝐷𝑃)𝑑
(𝑋𝐴𝑃)𝑎(𝑋𝐵𝑃)𝑏
K is not affected by changes in pressure, but consists of two terms; KX and P:
𝐾 = 𝐾𝑋𝑃(𝑐+𝑑−𝑎−𝑏)
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
Effect of temperature on equilibrium
At equilibrium, ∆𝐺𝑜 = −𝑅𝑇 ln𝐾
∆𝐺𝑜 = ∆𝐻𝑜 + 𝑇𝜕∆𝐺𝑜
𝜕𝑇𝑃
−𝑅𝑇 ln𝐾 = ∆𝐻𝑜 − 𝑇𝜕 𝑅𝑇 ln𝐾
𝜕𝑇𝑃
𝜕 ln𝐾
𝜕𝑇=
∆𝐻𝑜
𝑅𝑇2
𝜕 ln𝐾
𝜕 1 𝑇
= −∆𝐻𝑜
𝑅
For the case of∆𝐻𝑜 > 0, temperature increase shifts the reaction towards products
For the case of∆𝐻𝑜 < 0, temperature increase shifts the reaction towards reactants
Van’t Hoff equation
Slope>0
Slope<0
exothermic
endothermic
ln K
1/T
𝜕 ln𝐾
𝜕 1 𝑇
= −∆𝐻𝑜
𝑅
Recall that ∆𝐺 = ∆𝐻 − 𝑇∆𝑆
𝐺𝑜 = 𝐻𝑜 + 𝑇𝜕𝐺𝑜
𝜕𝑇 𝑃Since
𝜕𝐺𝑜
𝜕𝑇 𝑃= −𝑆,
Multiplying both sides by dT and dividing by T2,
𝐺𝑜𝑑𝑇
𝑇2 =𝐻𝑜𝑑𝑇
𝑇2 + 𝑇𝜕𝐺𝑜
𝑇2𝑃
𝐻𝑜𝑑𝑇
𝑇2 =𝐺𝑜𝑑𝑇
𝑇2 −𝑇𝑑𝐺𝑜
𝑇2 = −𝑑𝐺𝑜
𝑇,
∆𝐻𝑜
𝑇2 =−𝑑
∆𝐺𝑜
𝑇
𝑑𝑇
Gibbs-
Helmholtz
Eqn
Example – 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
𝑃4 𝑔 2𝑃2 𝑔
∆𝐺𝑜 = −225000 + 18.2𝑇𝑙𝑛𝑇 − 50.1𝑇
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:
∆𝐺 = ∆𝐺𝑜 + 𝑅𝑇 ln𝑄
∆𝐺 can be calculated for any temperature since
∆𝐺𝑜 = ∆𝐻𝑜 − 𝑇∆𝑆𝑜
∆𝐺 = ∆𝐻𝑜298 +
298
𝑇
∆𝐶𝑃𝑑𝑇 − 𝑇 ∆𝑆𝑜298 +
298
𝑇 ∆𝐶𝑃𝑑𝑇
𝑇
where 𝐶𝑃 = 𝑎 + 𝑏𝑇 +𝑐
𝑇2
and ∆𝐶𝑃= ∆𝑎 + ∆𝑏𝑇 +𝑐
𝑇2 where ∆𝑎, 𝑏, 𝑐 = ∆𝑎, 𝑏, 𝑐𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 − ∆𝑎, 𝑏, 𝑐𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠
∆𝐺 = ∆𝐻𝑜298 +
298
𝑇
∆𝑎 + ∆𝑏𝑇 + ∆𝑐𝑇2 𝑑𝑇 − 𝑇 ∆𝑆𝑜
298 + 298
𝑇 ∆𝑎 + ∆𝑏𝑇 + ∆𝑐𝑇2 𝑑𝑇
𝑇
Plotting the ∆𝐺𝑜 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 ∆𝐺𝑜 of any reaction by a simple 2-term fit such as
∆𝐺𝑜 = 𝐴 + 𝐵𝑇
The following grouping lead to a condensed representation of ∆Go which can further
be simplified
∆𝐺 = ∆𝐻𝑜298 + ∆𝑎𝑇 +
∆𝑏𝑇2
2− ∆𝑐
𝑇 − 𝑇 ∆𝑆𝑜298 + ∆𝑎 ln𝑇 + ∆𝑏𝑇 − ∆𝑐
2𝑇2
Replacement of the upper and the lower limits yields
∆𝐺 = 0 = ∆𝐺𝑜 − 𝐼𝑜 + 𝐼1𝑇 − ∆𝑎𝑇 ln𝑇 −∆𝑏
2𝑇2 −
∆𝑐
2𝑇
where 𝐼𝑜 = ∆𝐻𝑜298 − ∆𝑎298 +
∆𝑏2982
2− ∆𝑐
298
𝐼1 = ∆𝑎 − ∆𝑆𝑜298 + ∆𝑎 ln 298 + ∆𝑏298 − ∆𝑐
2 ∗ 2982
Tabulated thermochemical data such as ∆𝐻𝑜298, ∆𝑆𝑜
298, ∆𝐶𝑃 for a specific reaction are
replaced into the general equation for ∆𝐺𝑜 to obtain the variation of the spontaneity
with temperature
Alternatively experimental variation of ∆𝐺𝑜with T can be calculated from the
measured oxygen partial pressure 𝑃𝑂2(𝑒𝑞𝑚) that is in equilibrium with a metal and
metal oxide using equation:
∆𝐺𝑜 = 𝑅𝑇 ln𝑃𝑂2(𝑒𝑞𝑚)
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 ΔHm= 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
𝐾 =𝑃𝐶
𝑐𝑃𝐷𝑑
𝑃𝐴𝑎𝑃𝐵
𝑏
The partial pressures of the components are expressed as a function of the total P:
𝑃𝐴 =𝑛𝐴. 𝑃
𝑛𝐴 + 𝑛𝐵 + 𝑛𝐶 + 𝑛𝐷
where 𝑛𝐴 is the mole number of A under equilibrium
Equilibrium constant can be represented as
𝐾 =𝑛𝐶
𝑐𝑛𝐷𝑑
𝑛𝐴𝑎𝑛𝐵
𝑏 ∗𝑃
𝑛𝐴 + 𝑛𝐵 + 𝑛𝐶 + 𝑛𝐷
𝑐+𝑑 −(𝑎+𝑏)
𝑎𝐴(𝑔) + 𝑏𝐵(𝑔) 𝑐𝐶(𝑔) + 𝑑𝐷(𝑔)
Suppose the reaction reaches equilibrium after a while and x moles of A is
converted to products
Then
𝑛𝐴 =Moles of unreacted A = 1 − 𝑥 𝑎𝑛𝐵 = Moles of unreacted B = 1 − 𝑥 𝑏𝑛𝐶 = Moles of formed C = 𝑥. 𝑐𝑛𝐷 = Moles of formed D = 𝑥. 𝑑
and
𝐾 =(𝑥. 𝑐)𝑐(𝑥. 𝑑)𝑑
𝑎 − 𝑎𝑥 𝑎(𝑏 − 𝑏𝑥)𝑏∗
𝑃
1 − 𝑥 𝑎 + 𝑏 + 𝑥 𝑐 + 𝑑
𝑐+𝑑 −(𝑎+𝑏)
If equilibrium temperature and the standard free energy change at that
temperature are given, the fraction x can be conveniently determined since
∆𝐺 = ∆𝐺𝑜 + 𝑅𝑇𝑒𝑞𝑚 ln𝐾 = 0
∆𝐺𝑜 = −𝑅𝑇𝑒𝑞𝑚 ln(𝑥. 𝑐)𝑐(𝑥. 𝑑)𝑑
𝑎 − 𝑎𝑥 𝑎(𝑏 − 𝑏𝑥)𝑏∗
𝑃
1 − 𝑥 𝑎 + 𝑏 + 𝑥 𝑐 + 𝑑
𝑐+𝑑 −(𝑎+𝑏)
Example – Determine the equilibrium composition of the system when 1 mole of P4
reacts to form P2 at 1300 K
∆𝐺𝑜 = −225000 + 18.2𝑇𝑙𝑛𝑇 − 50.1𝑇 𝑃4 𝑔 2𝑃2 𝑔
∆𝐺𝑜 = −𝑅𝑇𝑒𝑞𝑚 ln(𝑥. 𝑐)𝑐(𝑥. 𝑑)𝑑
𝑎 − 𝑎𝑥 𝑎(𝑏 − 𝑏𝑥)𝑏∗
𝑃
1 − 𝑥 𝑎 + 𝑏 + 𝑥 𝑐 + 𝑑
𝑐+𝑑 −(𝑎+𝑏)