# What is "Gibbs Free Energy"?

##### 2 Answers
Sep 26, 2014

Gibbs free energy, $G$, is just the enthalpy change of a reaction, $\Delta H$, minus the entropy change of the reaction system, $\Delta {S}_{s y s}$, multiplied by the temperature of the reaction, $T$.

$\Delta G = \Delta H - T \Delta {S}_{s y s}$

For a reaction to occur, it needs to cause the total entropy of the reaction system, ${S}_{s y s}$, and its surroundings, ${S}_{s u r}$, to increase.

$\Delta {S}_{o v e r a l l} = \Delta {S}_{s y s} + \Delta {S}_{s u r} > 0$

But, because "the surroundings" is effectively the whole rest of the universe, it's quite difficult to accurately measure $\Delta {S}_{s u r}$.

Therefore, the Gibbs free energy equation was invented, because if $\Delta G < 0$, then that means that the total entropy change of a reaction is greater than zero ($\Delta {S}_{o v e r a l l} > 0$), and so the reaction will happen.

Derivation of the Gibbs free energy equation:

$\Delta {S}_{o v e r a l l} = \Delta {S}_{s u r} + \Delta {S}_{s y s} > 0$

$\Delta {S}_{s u r} = - \frac{\Delta H}{T}$

$\Delta {S}_{o v e r a l l} = \frac{- \Delta H}{T} + \Delta {S}_{s y s} > 0$

Multiplying through by $- T$ gives
$\Delta G = - T \Delta {S}_{o v e r a l l} = \Delta H - T \Delta {S}_{s y s}$

NOTE: when doing actual calculations enthalpy $\Delta H$ values use kJ but entropy $\Delta S$ values are often expressed in J. Watch for this and convert $\Delta S$ to kJ (divide by 1000) in these types of calculations.

Sep 24, 2015

The Gibbs free energy is classically the energy associated with a chemical reaction that can be used to do work. It includes an enthalpy term, $\Delta H$, and an entropy term, $\Delta S$. It remains the unequivocal criterion for the spontaneity of chemical change.

#### Explanation:

By definition, the Gibbs free energy is defined by the relationship:

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

We can introduce specified standard state conditions, but if $\Delta G$ is negative, then the reaction as written is spontaneous. If $\Delta G$ is positive, then the reaction as written is non-spontaneous. It may be related to the equilibrium constant, ${K}_{e q}$, by the relationship:

$\Delta G = - R T \ln {K}_{e q}$.

Given this relationship, only negative $\Delta G$ values will give rise to an equilibrium constant that is $> 1$, and hence spontaneous.

Often you see values of $\Delta G$ tabulated for standard state conditions, $298$ $K$, and near atmospheric pressure, in which case $\Delta {G}^{\circ}$ is specified.

I recommend this video that explains in details Gibbs free energy.