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)