Revision - Criterion of spontaneity

The criterion of spontaneity has two ideas involved. Decrease in energy and increase in disorder.

Spontaneity means one the chemical reaction is started it will continue further without additional work being done on it.

Both endothermic and exothermic reactions were found to be spontaneous.

Gibbs free energy concept combines both ideas being spontaneity.


Gibbs Free Energy

The second law of thermodynamics provides a criterion for the determination of the direction of a thermodynamic process, depending on the value of ∆S(total).

∆S(total)[entropy change due to a process] requires calculation of the entropy change for the system and the surroundings. Hence it is desirable to have a function of state that predicts the directionality (feasibility) of a process in a system without the need for explicit calculations for the surroundings.

Such a function exists and is known as Gibbs free energy (or Gibbs function) (G), after its originator, John Willard Gibbs.

Derivation of Gibbs free energy

Consider a process being carried out at constant pressure and constant temperature.
Under these conditions,

qp = ∆Hsys for the system,
and for the surroundings, -qp = - ∆Hsys
Now, ∆Ssurr = - ∆Hsys/T

Since ∆S(total) = ∆Ssys + ∆Ssurr,
= (∆Ssys - ∆Hsys)/T

= -(∆Hsys - T∆Ssys)/T

= -∆(Hsys - TSsys)T

(since T is constant)

If we define the term (H - TS) as G, Gibbs free energy, then, ∆S(total) = - ∆G /T (or ∆G = -T∆S(total))

Therefore, ∆Gsys is a criterion of spontaneity or directionality of the system:

Value of ∆G and Feasibility of process
< 0 - Spontaneous
0 - Reversible
> 0 - Non-spontaneous

Gibbs Free Energy Calculations at Constant Temperature

Example 1: Phase transition of ice to water

This example uses ∆G = ∆H - T∆S: Consider the ice – water phase transition

∆Hfus = 6007 J mol-1 Tfus = 273.15 K

Find ∆G at –10°C (263.15 K), 0°C (273.15 K) and 10°C (283.15 K)
For all the calculations here, ∆S = ∆Hfus/Tfus = 6007/273.15

At 263.15 K, ∆G = +6007 - 263.15 x 6007/273.15 = +213 J
At 273.15 K, ∆G = +6007 - 273.15 x 6007/273.15 = 0
At 283.15 K, ∆G = +6007 - 283.15 x 6007/273.15 = -213 J

*Below 0° C, water freezes spontaneously (i.e. the right to left direction).
**At 0° C, ice and water co-exist (they are in equilibrium and the process is reversible).
***Above 0° C, ice melts spontaneously (i.e. the left to right direction).

### It is impossible for water to freeze on its own accord above 0oC, or ice to melt on its own accord below 0°C.