Gibbs Free Energy

Combines enthalpy and entropy into one criterion for spontaneity: ΔG = ΔH − TΔS (and ΔG° = −RT ln K).

\Delta G = \Delta H - T\Delta S \quad ; \quad \Delta G^{\circ} = -RT \ln K

ΔG = ΔH − TΔS ; ΔG° = −RT ln K

What it means

Gibbs free energy settles the question enthalpy and entropy leave open: will this process actually happen? A reaction may release heat (favorable ΔH) yet create order (unfavorable ΔS), or vice versa — ΔG weighs the two against each other at temperature T, and its sign decides: negative means spontaneous, positive means it needs driving, zero means equilibrium. The second form ties standard free energy directly to the equilibrium constant, so ΔG° tells you how far the reaction goes. When to use it: deciding spontaneity, finding the temperature where a reaction turns favorable, or converting between free energy and equilibrium constants.

Variables

Symbol	Name	Unit	Meaning
$ΔG$	Gibbs free energy change	kJ/mol	Free-energy change; negative = spontaneous, positive = nonspontaneous, zero = equilibrium.
$ΔH$	Enthalpy change	kJ/mol	Enthalpy change of the process.
$T$	Temperature	K	Absolute temperature — always kelvin.
$ΔS$	Entropy change	kJ/(mol·K)	Entropy change of the process (often given in J — convert to kJ before subtracting).
$K$	Equilibrium constant	unitless	Equilibrium constant of the reaction.

Assumptions

The process occurs at constant temperature and pressure.
Units are consistent: ΔH and ΔS must both be in kJ or both in J (convert one).
ΔG° refers to standard conditions; ΔG = 0 marks equilibrium, not a rate statement.

Worked example

Problem

For a process with ΔH = −120 kJ/mol and ΔS = −0.300 kJ/(mol·K) at 298 K, find ΔG and state whether it is spontaneous.

Given:

ΔH = −120 kJ/mol
ΔS = −0.300 kJ/(mol·K)
T = 298 K

Write ΔG = ΔH − TΔS.
Convert-check: ΔH is in kJ/mol and ΔS in kJ/(mol·K) — consistent, so no conversion needed.
Compute TΔS: 298 K × (−0.300 kJ/(mol·K)) = −89.4 kJ/mol.
Substitute: ΔG = −120 − (−89.4) = −120 + 89.4 = −30.6 kJ/mol.
Since ΔG < 0, the process is spontaneous at 298 K.

ΔG = −30.6 kJ/mol → spontaneous under standard conditions

The negative ΔG means the process releases free energy and proceeds spontaneously at 298 K, despite the unfavorable entropy decrease.