Free Energy Functions

Vijay Ratna

Let us consider that in a system, PV work is the only useful work or external energy. Let us disregard the electric and other forms of energy. Then the heat content or total energy of a system is divided into internal energy and external energy.

1. H = E + PV

Total energy Internal energy External energy

We can also divide the same total energy (free energy) H into isothermally available energy (TS).

2. H = F + TS

Total energy Isothermally available Isothermally unavailable

Energy energy

3. The internal energy E can be divided into isothermally available internal energy or work function (A) and isothermally unavailable energy TS. Thus, for an isothermal process,

E = A + TS

(Internal energy) Isothermally available Isothermally unavailable

Internalenergy energy

A number of relationships may be obtained by rearranging these quantities and placing various restrictions on the processes described. Thus, equation (2) is rearranged to

F = H – TS

And substituting H = E + PV for H

We have,

F = E + PV –TS

Since A = E – TS

F = A + PV

∆S _Total = ∆S _System – ∆S _Surroundings

It is this ∆S _Total that decides the spontaneity of the process. From this we can derive,

∆H _System – T ∆ S _System = - T ∆ S _Total

From this Gibbs energy or Gibbs function G has been defined as

G = H – TS

Therefore, the change in Gibbs energy for the system, ∆G _System is

∆G _System = ∆H _System – T ∆ S _System – S ∆ T _System

At constant temperature ∆T = O

∆G _System = ∆H _System – T ∆ S _System ------ 3

Usually the subscript ‘system’ is dropped and we simply write this equation as

∆G = ∆H – T ∆S

Thus Gibbs energy change = enthalpy change – temp (entropy change)

-this is Gibbs energy equation.

By combining equations (2) and (3) we get

∆G _system = - T ∆ S _total ------ 4

The second law states that ∆S_total increases (∆S_total is positive) in a spontaneous process, and so it follows that G_system should decrease, i.e. ∆G_system should have a negative value in a spontaneous process.

The criteria of spontaneity based at constant temperature and pressure are

(i) If ∆G_system is negative (<O), the process is spontaneous.

(ii) If ∆G_system is positive (>O), the process is not spontaneous

(iii) If ∆G_system = 0, the system has attained equilibrium with respect to the transformation.

Standard Gibbs energy change is related to equilibrium constant by

= - RT In k

When is large and negative the equilibrium lies strongly in favour of products; when it is large and positive, the equilibrium lies strongly in favour of reactants. The maximum quantity of non-expansion work per mole of reaction under standard conditions is equal to . Several reactions with positive Gibbs energy change can be made to take place on coupling with reactions having large negative Gibbs energy change. Such reactions are important in industry, metallurgy and in biosystems. Temperature is the important factor in the equation

∆ G = ∆H - T∆S. Many reactions which are nonspontaneous at low temperatures are made spontaneous by raising the temperature of the system having positive entropy of reactions.

Let us take the example of manufacture of iron. Iron can be obtained from is oxide are, Fe₂O₃. Is it possible to get iron by simply heating Fe₂O₃ and decomposing the oxide into iron and oxygen?

Fe₂O₃ (s) → 2F3 (s) + 3/2 O₂ (g)

Feasibility of the reaction can be predicted on the basis of the value of .
At room temperature (+
824 kJmol^-1) is unfavourable and (+ 0.276 kJ K^-1 mol^-1)
is favourable. Unfortunately, has
large positive value and it cannot be outweighed by -T(=
- 82kJ at 298K) at any reasonable temperature. The value of at
298K is +742 kJ mol^-1.

The usual method of reducing Fe₂O₃ to iron is to use carbon, the cheapest reducing agent. The decomposition of iron oxide which is nonspontaneous is compensated by highly spontaneous process of combustion of carbon, C_(graphite) + O₂ (g) → CO₂ (g)

The overall process can be represented by 2Fe₂O₃ (s) + 3 C (s) → 4 Fe (s) + 3 CO₂ (g)

Although entropy change is positive, the -T(= -187 kJ) is not large enough to counter the positive enthalpy change at 198K). However, at higher temperature the contribution of -Tbecomes more negative and finally overtakes the positive enthalpy change. Let us calculate temperature T at which is zero from the equation.

, substituting the values of

0= +467.9 kJ mol^-1 – T (0.5603 kJ mol^-1 K^-1)

T= 835.1 K (or 561.9^OC = 562^OC)

The free energy change for the reaction becomes zero at 562^OC, and at higher temperature it will be negative. Therefore, for achieving the spontaneity, temperature of the furnace in iron metallurgy is kept sufficiently high (much higher than 562^OC).