6.7 Gibbs Free Energy

Gibbs free energy ($G$) is a fundamental thermodynamic quantity that measures the amount of usable energy in a system at constant temperature and pressure. It is also known as available energy. This concept was introduced by the American scientist Josiah Willard Gibbs in 1876.

Definition and Equation

Gibbs free energy is defined as the enthalpy of a system minus the product of its absolute temperature and entropy:

$G=H-TS$

Where:

• $G$ = Gibbs free energy (J or kJ)
• $H$ = Enthalpy of the system (J or kJ)
• $T$ = Absolute temperature (Kelvin)
• $S$ = Entropy of the system (${\text{J K}}^{-1}$ or ${\text{kJ K}}^{-1}$)

Gibbs Free Energy as a State Function

Gibbs free energy is a state function — its value depends only on the initial and final states of the system, not on the path taken. Therefore, the change in Gibbs free energy ($\Delta G$) for a process is particularly significant.

For a reaction carried out at constant temperature ($\Delta T=0$), the change in Gibbs free energy is given by the Gibbs-Helmholtz equation:

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

This equation is central to predicting the spontaneity and equilibrium state of a chemical reaction.

Spontaneity and Equilibrium from $\Delta G$

The sign of $\Delta G$ directly indicates whether a process is spontaneous:

Effect of $\Delta H$ and $\Delta S$ on Spontaneity

The Gibbs-Helmholtz equation $\Delta G=\Delta H-T\Delta S$ shows that spontaneity depends on both enthalpy and entropy, and on temperature:

Application: Phase Transitions

Gibbs free energy is also valuable for understanding phase transitions such as melting, freezing, boiling, and condensation. Consider the melting of ice:

${\mathrm{H}}_2{\mathrm{O}}_{(\mathrm{s})}\to {\mathrm{H}}_2{\mathrm{O}}_{(\mathrm{l})}$

• At $0^{\circ }\text{C}$ (273.15 K): Solid ice and liquid water are in equilibrium, so $\Delta G=0$.
• Below $0^{\circ }\text{C}$: $\Delta G>0$ for melting — the solid phase is more stable (melting is non-spontaneous).
• Above $0^{\circ }\text{C}$: $\Delta G<0$ for melting — the liquid phase is more stable (melting is spontaneous).

The same logic applies to boiling: at the normal boiling point, $\Delta G=0$ for the liquid $\to$ vapour transition.

Worked Example

Problem: For a reaction at 298 K, $\Delta H=-92{\text{kJ mol}}^{-1}$ and $\Delta S=-198{\text{J K}}^{-1}{\text{mol}}^{-1}$. Calculate $\Delta G$ and determine if the reaction is spontaneous.

Solution:

Convert $\Delta S$ to kJ: $\Delta S=-0.198{\text{kJ K}}^{-1}{\text{mol}}^{-1}$

$\Delta G=\Delta H-T\Delta S=-92-(298)(-0.198)$ $\Delta G=-92+59.0=-33{\text{kJ mol}}^{-1}$

Since $\Delta G<0$, the reaction is spontaneous at 298 K.