The formula

Definition
G = H - TS
Change at constant temperature
ΔG = ΔH - TΔS
Link to the equilibrium constant
ΔG° = -RT·ln K
Temperature where the sign flips
T = ΔH / ΔS when ΔG = 0

What the symbols mean

SymbolMeaningUnit
ΔGFree energy change; negative is spontaneous, positive is not, zero is equilibriumkJ/mol (or J/mol)
ΔHEnthalpy change; negative releases heatkJ/mol
ΔSEntropy change; positive means more disorderJ/(mol·K)
TAbsolute temperatureK
RMolar gas constant8.314 J/(mol·K)
KEquilibrium constant for the reactiondimensionless
°Standard state marker: 1 bar and 1 M, at the stated temperaturedimensionless

When it applies

  • Temperature and pressure are constant, which is the situation almost every chemistry problem describes.
  • ΔH and ΔS have to share units before subtracting. Tables give ΔH in kilojoules and ΔS in joules, so divide ΔS by 1000 or multiply ΔH by 1000 first.
  • Spontaneous says nothing about speed. A reaction with a large negative ΔG can still be too slow to observe, which is what activation energy governs.
  • When ΔH and ΔS share a sign, spontaneity depends on temperature, and T = ΔH/ΔS locates the crossover.

Worked example

Problem. Limestone decomposes as CaCO₃(s) → CaO(s) + CO₂(g), with ΔH° = +178.3 kJ/mol and ΔS° = +160.6 J/(mol·K). Is it spontaneous at 298 K, and above what temperature does that change?

  1. Match the units by converting the entropy term: ΔS° = 160.6 J/(mol·K) = 0.1606 kJ/(mol·K).
  2. Evaluate the entropy term at 298 K: TΔS° = 298 × 0.1606 = 47.86 kJ/mol.
  3. Subtract: ΔG° = 178.3 - 47.86 = 130.44 kJ/mol, which is +130.4 kJ/mol. Positive, so the decomposition is not spontaneous at room temperature.
  4. Both terms are positive, so raising the temperature eventually wins. Set ΔG° = 0: T = ΔH° / ΔS° = 178.3 / 0.1606 = 1110 K.
  5. Above about 1110 K the TΔS° term exceeds ΔH°, ΔG° turns negative, and the decomposition becomes spontaneous, which is why lime kilns run hot.

Answer. ΔG° = +130.4 kJ/mol at 298 K, so not spontaneous, becoming spontaneous above about 1110 K.

Common mistakes

  • Subtracting joules from kilojoules. This single unit mismatch is the most common wrong answer on this equation, and it is off by a factor of a thousand.
  • Using Celsius for T. The equation needs kelvin, and at 25 °C the difference is a factor of about twelve in the entropy term.
  • Reading spontaneous as fast. Diamond turning into graphite has a negative ΔG and takes geological time.
  • Assuming a positive ΔH always blocks a reaction. A large positive ΔS can overcome it once the temperature is high enough.

Related formulas

Sources