# Gibbs free energy and internal energy

So I read that the sum of the energies (kinetic and potential) of the particles of a system is its internal energy, but then I read that the energy that they can use to do useful work is said to be in terms of internal gibbs energy...so I have a couple of doubts about the relation between internal energy/the energy of the components/free energy.

In a static system, is all the internal energy, energy of the microscopic particles or there is a part of energy "stored" as entropy?

I really don't how to work out the relation between these things, I don't have strong thermodynamics background

Thank you!

The internal energy $U$ is all of the energy in the system. The second law of thermodynamics, however, limits the amount of energy that you can actually extract from the system. You can think of it this way: you would have to expend at least $TS$ lots of energy to extract all of the energy $U$. So the available energy is $U-TS$.

(I've neglected the $PV$ term here from the enthalpy, for simplicity - I assume you understand its role).

• Sorry to ask after 2 years, but can you elaborate WHY do you think that $TS$ equals to energy which will be used in extraction "available energy"? – coobit Dec 20 '18 at 9:32

What happens when one of the potential driving forces behind a chemical reaction is favorable and the other is not? We can answer this question by defining a new quantity known as the Gibbs free energy (G) of the system, which reflects the balance between these forces.

The Gibbs free energy of a system at any moment in time is defined as the enthalpy of the system minus the product of the temperature times the entropy of the system.

$$G = H - TS$$

The Gibbs free energy of the system is a state function because it is defined in terms of thermodynamic properties that are state functions. The change in the Gibbs free energy of the system that occurs during a reaction is therefore equal to the change in the enthalpy of the system minus the change in the product of the temperature times the entropy of the system.

$$\Delta G = \Delta H - \Delta(TS)$$

If the reaction is run at constant temperature, this equation can be written as follows.

$$\Delta G = \Delta H - T\Delta S$$

The change in the free energy of a system that occurs during a reaction can be measured under any set of conditions. If the data are collected under standard-state conditions, the result is the standard-state free energy of reaction ($$\Delta G_o$$).

$$\Delta G_o = \Delta H_o - T\Delta S_o$$