The Amount of energy, which has the capacity to to do work, which is acquired from a thermodynamic system is called Gibbs free energy. Gibbs free energy is calculated by using the formula ,

G = Gibbs free energy,

H = Enthalpy,

A = Helmoltz free energy,

U = Internal energy of the system.

G = H + A - U

Where,G = Gibbs free energy,

H = Enthalpy,

A = Helmoltz free energy,

U = Internal energy of the system.

**Step 1 :**From the question, note down the values given.

**Step 2 :**Using the formula, substitute the given values and get the appropriate values accordingly.

G = H + A - U. Based on Gibbs free energy, some of the problems are solved below.

### Solved Examples

**Question 1:**Find the Gibbs free energy of the reaction given,

2H$_{2}$(g) + O$_{2}$(g) $\rightarrow$ 2H$_{2}$O(g).

If -241.8 KJ is the Enthalpy, 100 J is the Helmoltz free energy and 10 J is the Internal energy.

**Solution:**

Given : Enthalpy, H = -241.8 KJ,

Helmoltz free energy, A = 100 J,

Internal energy, U = 10 J

Gibbs free energy of the given reaction is,

G = H + A - U

G = - 241.8 KJ + 100 J - 10 J

G = - 151.8 KJ

**Question 2:**Find the Enthalpy of the reaction given,

C$_{2}$H$_{4}$(g) + H$_{2}$O(l) $\rightarrow$ C$_{2}$H$_{5}$OH(l).

If -6 KJ is the Gibbs free energy, -175 KJ is the Helmoltz free energy and 68 KJ is the Internal energy.

**Solution:**

Given : Gibbs free energy, G = -6 KJ

Helmoltz free energy, A = -175 KJ,

Internal energy, U = 68 KJ

Gibbs free energy of the given reaction is,

G = H + A - U

H = G - A + U

H = -6 KJ - (-175 KJ) + 68 KJ

H = 237 KJ