# What is free energy?

##### 1 Answer
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, $T$.

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

Helmholtz free energy, $\Delta A$, is the same, but uses the change in internal energy of the reaction system, $\Delta U$, instead of $\Delta H$.

$\Delta A = \Delta U - T \Delta {S}_{s y s}$

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 the surrounding entropy change.

Therefore, the free energy equations were invented, because if $\Delta G < 0$ or $\Delta A < 0$, 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
$\Delta G = - T \Delta {S}_{o v e r a l l} = \Delta H - T \Delta {S}_{s y s} < 0$