GIBBS FREE ENERGY: UNITS, HOW TO CALCULATE IT, SOLVED EXERCISES - CHEMISTRY - 2025

The Gibbs free energy (commonly known as G) is a thermodynamic potential defined as the difference of the enthalpy H, minus the product of the temperature T, the entropy S of the system:

Gibbs free energy is measured in Joules (according to the International System), in ergs (for the Cegesimal System of Units), in calories or in electron volts (for electro Volts).

Figure 1. Diagram showing the definition of Gibbs energy and its relationship with the other thermodynamic potentials. Source: nuclear-power.net.

In processes that occur at constant pressure and temperature, the variation of Gibbs free energy is ΔG = ΔH - T ΔS. In such processes, (G) represents the energy available in the system that can be converted into work.

For example, in exothermic chemical reactions, enthalpy decreases while entropy increases. In the Gibbs function these two factors are counteracted, but only when the Gibbs energy decreases does the reaction occur spontaneously.

So if the variation in G is negative, the process is spontaneous. When the Gibbs function reaches its minimum, the system reaches a stable equilibrium state. In summary, in a process for which the pressure and temperature remain constant, we can affirm:

- If the process is spontaneous, then ΔG <0

- When the system is in equilibrium: ΔG = 0

- In a non-spontaneous process G increases: ΔG> 0.

How is it calculated?

Gibbs free energy (G) is calculated using the definition given at the beginning:

In turn, the enthalpy H is a thermodynamic potential defined as:

- Step by Step

Next, a step-by-step analysis will be made to know the independent variables of which the Gibbs energy is a function:

1- From the first law of thermodynamics we have that the internal energy U is related to the entropy S of the system and its volume V for reversible processes through the differential relationship:

From this equation it follows that the internal energy U is a function of the variables S and V:

2- Starting from the definition of H and taking the differential, we obtain:

3- Substituting the expression for dU obtained in (1) we have:

From this it is concluded that the enthalpy H depends on the entropy S and the pressure P, that is:

4- Now the total differential of the Gibbs free energy is calculated obtaining:

Where dH has been replaced by the expression found in (3).

5- Finally, when simplifying, we obtain: dG = VdP - SdT, making it clear that the free energy G depends on the pressure and the temperature T as:

- Maxwell's thermodynamic relations

From the analysis in the previous section it can be deduced that the internal energy of a system is a function of the entropy and the volume:

Then the differential of U will be:

From this partial derivative expression, the so-called Maxwell thermodynamic relations can be derived. Partial derivatives apply when a function depends on more than one variable and are easily calculated using the theorem in the next section.

Maxwell's first relationship

∂ _V T- _S = -∂ _S P- _V

To arrive at this relationship, the Clairaut - Schwarz theorem on partial derivatives has been used, which states the following:

Maxwell's second relationship

Based on what is shown in point 3 of the previous section:

It can be obtained:

We proceed in a similar way with the Gibbs free energy G = G (P, T) and with the Helmholtz free energy F = F (T, V) to obtain the other two Maxwell thermodynamic relations.

Figure 2. Josiah Gibbs (1839-1903) was an American physicist, chemist and mathematician who made great contributions to thermodynamics. Source: Wikimedia Commons.

Maxwell's four thermodynamic relationships

Exercise 1

Calculate the variation of Gibbs free energy for 2 moles of ideal gas at a temperature of 300K during an isothermal expansion that takes the system from an initial volume of 20 liters to a final volume of 40 liters.

Solution

Recalling the definition of Gibbs free energy we have:

Then a finite variation of F will be:

What applied to the case of this exercise remains:

Then we can get the change in Helmholtz energy:

Exercise 2

Taking into account that Gibbs free energy is a function of temperature and pressure G = G (T, P); determine the variation of G during a process in which the temperature does not change (isothermal) for n moles of a monatomic ideal gas.

Solution

As demonstrated above, the change in Gibbs energy only depends on the change in temperature T and volume V, so an infinitesimal variation of it is calculated according to:

But if it is a process in which the temperature is constant then dF = + VdP, so a finite pressure variation ΔP leads to a change in the Gibbs energy given by:

Using the ideal gas equation:

During an isothermal process it occurs that:

That is:

So the previous result can be written as a function of the variation of the volume ΔV:

Exercise 3

Considering the following chemical reaction:

N ₂ 0 (g) + (3/2) O ₂ (g) ↔️ 2NO ₂ (g) at temperature T = 298 K

Find the variation of the Gibbs free energy and, using the result obtained, indicate whether or not it is a spontaneous process.

Solution

Here are the steps:

- First step: reaction enthalpies

- Second step: the reaction entropy variation

- Third step: variation in the Gibbs function

This value will determine the balance between the decreasing energy and the increasing entropy to know if the reaction is finally spontaneous or not.

Since it is a negative variation of the Gibbs energy, it can be concluded that it is a spontaneous reaction at the temperature of 298 K = 25 ºC.

References

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2. Cengel, Y. 2012. Thermodynamics. 7th Edition. McGraw Hill.
3. Libretexts. Gibbs Free Energy. Recovered from: chem.libretexts.org
4. Libretexts. What are Free Energies. Recovered from: chem.libretexts.org
5. Wikipedia. Gibbs free energy. Recovered from: es.wikipedia.com
6. Wikipedia. Gibbs free energy. Recovered from: en.wikipedia.com