In order to account for the change in both entropy and enthalpy, J.Willard Gibbs defined a new function, which we now call Gibbs energy change or Gibbs potential G. This function is independent of variations in pressure and temperature.

**Gibbs energy change depends only on the state of a system, not on how this state has been attained.**The Gibbs energy change associated with the formation of a compound from its constituent elements under standard conditions is termed as Standard Gibbs free energy of formation.Â The term free energy of a process such as a rate or equilibrium and the free energy of standard process often an equilibrium, which could be the process under investigation or some other standard reaction.

## Standard Gibbs Free Energy

*The reversible work in thermodynamics means a special method in which work is carried out such that the system is in perfect equilibrium with all its surroundings.*

**The maximum (or reversible) work that can be done by a thermodynamic system at constant temperature and pressure is known as Gibbs energy.**In terms of chemical reactions, the word reversible means that the reaction can be carried out in either direction simultaneously and a dynamic equilibrium is always maintained. This further means that reactions in both directions should proceed with a decrease in free energy, which seems impossible. This is only possible if at equilibrium the Gibbs energy of the system reaches its minimum value. Otherwise, the system will spontaneously change to the configuration of lower free energy.

## Gibbs Free Energy Equation

“A thermodynamic system is said to be in equilibrium if itâ€™s

**intensive properties**(temperature, pressure) and**extensive properties**(U, G, A) are constant. Or the total change in any of the property is zero”. Looking at the following equation we can say if the reaction is reversible and the**Gibbs free energy is zero then the system is said to be in equilibrium.**\( A + B â‡Œ C + D\); \(\triangle_r G\) = \(0\)

The Gibbs energy for a reaction which is in the standard state, \(\triangle_r Gá¶¿\) is related to the equilibrium constant as follows:

\(0\) = \(\triangle_r Gá¶¿ + RT~ ln~ K\)

or \(\triangle_r Gá¶¿\) = \(- RT ~ln ~K\)

or \(\triangle_r Gá¶¿\) = \(-2.303~ RT~ log~ K\)

It is also known that:

\(\triangle_r Gá¶¿\) =\( \triangle_rHá¶¿ – T\triangle_r Sá¶¿\) =\(- RT ~ln ~K\)

For endothermic reactions the value of \(\triangle_rHá¶¿\) maybe large and positive; if the value of K is less than 1 then it is unlikely to form much of the product. In the case of an exothermic reaction, the value of \(\triangle_rHá¶¿\) is large and negative, even the value of \(\triangle_rGá¶¿\) is likely to be large and negative.

In these cases, the value of K will be much larger than 1. It can be said that strong exothermic reactions will have a larger value of K. The

**unit for Gibbs free energy is Kilo Joule**and generally represented by KJ/K.## Gibbs Free Energy Problem

**Question:**

Determine the standard free energy change for the following reaction at 25^{o}C.

**N _{2} + 3H_{2}Â â†’ 2NH_{3}**

Given Î”H and S are -81.5KJ and -189.0J/K

**Solution:**

We have an equation

**Î”G = Î”H â€“ TÎ”S**

Substitute the above values in this equation

Î”G = -81.5KJ â€“ (298 K) (-0.1890KJ/K)

Î”G = -24.7KJ

## Summary ofÂ Gibbs Free Energy

- The reaction is favourable means spontaneous if the value of
.*Î”G is negative* - The reaction is not favourable means non-spontaneous if the value of
.*Î”G is positive* - The reaction is said to be in equilibrium if the value of
.*Î”G is zero*

So far we have read Gibbs free energy and the conditions that affect the equilibrium of the system. For any further query on these topics kindly install Byjuâ€™s the learning app.