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Gibbs Free Energy Calculator — ΔG=ΔH−TΔS, Spontaneity & Equilibrium

Gibbs Free Energy Calculator — ΔG=ΔH−TΔS, Spontaneity & Equilibrium
Chemistry · Thermodynamics

Gibbs Free Energy Calculator — ΔG, Spontaneity & Equilibrium

This Gibbs free energy calculator computes ΔG = ΔH − TΔS to predict reaction spontaneity, converts between ΔG° and the equilibrium constant K using ΔG° = −RT × ln K, calculates ΔG from electrochemical cell voltages using ΔG = −nFE, and uses standard Gibbs free energy of formation tables — with complete step-by-step working and free energy diagrams. The ultimate delta G calculator for chemistry students.

ΔG = ΔH − TΔS ΔG° = −RT ln K ΔG = −nFE ΔGf° Tables Spontaneity Charts
Gibbs Free Energy Calculator — ΔG = ΔH − TΔS
⚠️

UNIT ALERT: ΔH in kJ/mol · ΔS in J/mol·K

ΔH is usually in kJ/mol but ΔS is usually in J/mol·K — these are different by a factor of 1000! The most common error in Gibbs free energy problems is forgetting to divide ΔS by 1000. This calculator converts automatically — just select your units below.

💧 H₂O formation
🧂 NaCl dissolving
🪨 CaCO₃ decomposition
⚗️ N₂+3H₂→2NH₃
🔥 CO₂ formation
High T example
Error

ΔG = ΔH − TΔS Result

Step-by-Step: ΔG = ΔH − TΔS

Free Energy Level Diagram

ΔG vs Temperature — ΔG = ΔH − TΔS (Crossover Graph)

ΔG = −nFE — Gibbs Free Energy from Cell Voltage

ΔG = −n × F × E_cell  |  F = 96,485 C/mol

Positive E_cell → ΔG < 0 → Spontaneous (galvanic cell, generates electricity).
Negative E_cell → ΔG > 0 → Non-spontaneous (electrolytic cell, requires electricity).
Standard conditions: ΔG° = −nFE°

🔋 Daniell cell (Zn/Cu) n=2, E=+1.10V
⚡ H₂ fuel cell n=2, E=+1.229V
💧 Water electrolysis n=2, E=−1.229V
🟤 Cu deposition n=1, E=+0.337V
⚗️ Al reduction n=3, E=+1.66V
Error

ΔG = −nFE Result

Step-by-Step: ΔG = −nFE

Equilibrium Constant from Cell Voltage

Common Standard Cell Potentials

Cell ReactionnE°_cell (V)ΔG° (kJ/mol)Type
Zn + Cu²⁺ → Zn²⁺ + Cu (Daniell)2+1.10−212.3Galvanic ✅
H₂ + ½O₂ → H₂O (Fuel cell)2+1.229−237.2Galvanic ✅
H₂O → H₂ + ½O₂ (Electrolysis)2−1.229+237.2Electrolytic ❌
Fe + 2H⁺ → Fe²⁺ + H₂2+0.44−84.9Galvanic ✅
Al³⁺ + 3e⁻ → Al (reduction)3−1.66+480.8Electrolytic ❌
2Ag⁺ + Zn → 2Ag + Zn²⁺2+1.56−301.0Galvanic ✅
ΔG° = −RT ln K — Equilibrium Constant Connection
⚖️

ΔG° = −RT × ln(K)  |  K = e^(−ΔG°/RT)

ΔG° < 0 → K > 1 → Products favored at equilibrium (spontaneous under standard conditions).
ΔG° > 0 → K < 1 → Reactants favored (non-spontaneous under standard conditions).
Non-standard: ΔG = ΔG° + RT×ln(Q) where Q = reaction quotient.

ΔG°=−20 kJ → K=3189
K=10⁻¹⁴ → ΔG°=+79.9
NH₃: ΔG°=−33 kJ
Ka acetic: K=1.8×10⁻⁵
Non-std: ΔG°=−20, Q=0.01
ΔG°=0 → K=1
Error

ΔG° = −RT ln K Result

Step-by-Step: ΔG° = −RT ln K

Table D — ΔG° and Equilibrium Constant Relationship

ΔG° (kJ/mol)K (approx)Equilibrium PositionSpontaneity
< −40> 10⁷Far toward productsStrongly spontaneous ✅
−20 to −4010³–10⁷Mostly productsSpontaneous ✅
−10 to −2010²–10³Predominantly productsModerately spontaneous
−5 to −1010–10²Slightly product-favoredSlightly spontaneous
−2 to +5~1Near equilibriumNear equilibrium ⚖️
+10 to +2010⁻²–10⁻⁴Mostly reactantsNon-spontaneous ❌
> +40< 10⁻⁷Far toward reactantsStrongly non-spontaneous ❌
Table A — Spontaneity Chart: All Four Cases of ΔH and ΔS

The spontaneity chart shows all four sign combinations of ΔH and ΔS in the formula ΔG = ΔH − TΔS. The crossover temperature T_cross = ΔH×1000/ΔS (K) is where spontaneity flips.

ΔHΔSΔG = ΔH − TΔSSpontaneous?Example Reaction
− (exothermic) + (increases) Always negative ✅ Always spontaneous 2H₂(g) + O₂(g) → 2H₂O(l)
− (exothermic) − (decreases) Negative at LOW T only 🌡 Spontaneous at low T N₂(g) + 3H₂(g) → 2NH₃(g)
+ (endothermic) + (increases) Negative at HIGH T only 🌡 Spontaneous at high T CaCO₃(s) → CaO(s) + CO₂(g)
+ (endothermic) − (decreases) Always positive ❌ Never spontaneous 3O₂(g) → 2O₃(g) (reverse preferred)
Table B — Standard Gibbs Free Energies of Formation ΔGf° (kJ/mol) at 298.15 K

Standard Gibbs free energy of formation table — ΔGf° for forming 1 mol from elements in standard states. Elements in standard state: ΔGf° = 0. Use: ΔG°rxn = Σ[n·ΔGf°(products)] − Σ[m·ΔGf°(reactants)].

SubstanceFormulaΔGf° (kJ/mol)State
Hydrogen (element)H₂(g)0Standard state
Oxygen (element)O₂(g)0Standard state
Nitrogen (element)N₂(g)0Standard state
Carbon (graphite)C(graphite)0Standard state
Carbon (diamond)C(diamond)+2.900Non-standard allotrope
Sulfur (rhombic)S(rhombic)0Standard state
Iron (element)Fe(s)0Standard state
Copper (element)Cu(s)0Standard state
Aluminum (element)Al(s)0Standard state
Water (liquid)H₂O(l)−237.13Liquid
Water (vapor)H₂O(g)−228.57Gas
Carbon monoxideCO(g)−137.17Gas
Carbon dioxideCO₂(g)−394.36Gas
MethaneCH₄(g)−50.72Gas
EthaneC₂H₆(g)−32.82Gas
Ethylene (ethene)C₂H₄(g)+68.15Gas
Acetylene (ethyne)C₂H₂(g)+209.20Gas
PropaneC₃H₈(g)−23.47Gas
BenzeneC₆H₆(l)+124.50Liquid
MethanolCH₃OH(l)−166.27Liquid
EthanolC₂H₅OH(l)−174.78Liquid
AmmoniaNH₃(g)−16.45Gas
Nitric oxideNO(g)+86.55Gas
Nitrogen dioxideNO₂(g)+51.31Gas
Nitrous oxideN₂O(g)+103.59Gas
Hydrogen chlorideHCl(g)−95.30Gas
Hydrogen bromideHBr(g)−53.45Gas
Hydrogen fluorideHF(g)−273.20Gas
Sulfur dioxideSO₂(g)−300.19Gas
Sulfur trioxideSO₃(g)−371.06Gas
Sulfuric acidH₂SO₄(l)−690.00Liquid
Sodium chloride (salt)NaCl(s)−384.14Solid
Sodium hydroxideNaOH(s)−379.49Solid
Potassium chlorideKCl(s)−408.51Solid
Calcium chlorideCaCl₂(s)−748.10Solid
Calcium oxide (lime)CaO(s)−603.30Solid
Calcium carbonateCaCO₃(s)−1128.79Solid
Iron(III) oxide (rust)Fe₂O₃(s)−742.20Solid
Aluminum oxideAl₂O₃(s)−1582.27Solid
Magnesium oxideMgO(s)−569.43Solid
Zinc oxideZnO(s)−318.30Solid
Silver chlorideAgCl(s)−109.79Solid
Silicon dioxide (quartz)SiO₂(s)−856.67Solid
Ammonium chlorideNH₄Cl(s)−202.87Solid
Table C — Standard Molar Entropies S° (J/mol·K) at 298.15 K

Standard entropy table — unlike ΔGf° and ΔHf°, elements do NOT have S°=0. Only perfect crystals at 0 K have S=0 (Third Law of Thermodynamics). Use: ΔS°rxn = Σ[n·S°(products)] − Σ[m·S°(reactants)].

SubstanceFormulaS° (J/mol·K)Notes
HydrogenH₂(g)130.68Gas — high entropy
OxygenO₂(g)205.14Gas
NitrogenN₂(g)191.61Gas
Carbon (graphite)C(graphite)5.74Solid — low entropy
Carbon (diamond)C(diamond)2.38Very ordered crystal
Sulfur (rhombic)S(rhombic)31.80Solid
IronFe(s)27.28Metal solid
CopperCu(s)33.15Metal solid
AluminumAl(s)28.33Metal solid
SodiumNa(s)51.30Metal solid
CalciumCa(s)41.63Metal solid
Water (liquid)H₂O(l)69.95Liquid
Water (vapor/steam)H₂O(g)188.83Gas — much higher than liquid
Carbon monoxideCO(g)197.67Gas
Carbon dioxideCO₂(g)213.79Gas
MethaneCH₄(g)186.26Gas
EthaneC₂H₆(g)229.60Gas — larger molecule, higher S
EthyleneC₂H₄(g)219.56Gas
AcetyleneC₂H₂(g)200.94Gas
PropaneC₃H₈(g)270.20Gas — large molecule
BenzeneC₆H₆(l)173.40Liquid
MethanolCH₃OH(l)126.80Liquid
EthanolC₂H₅OH(l)160.70Liquid
AmmoniaNH₃(g)192.77Gas
Nitric oxideNO(g)210.76Gas
Nitrogen dioxideNO₂(g)240.06Gas
Nitrous oxideN₂O(g)219.96Gas
Hydrogen chlorideHCl(g)186.90Gas
Hydrogen bromideHBr(g)198.70Gas
Hydrogen fluorideHF(g)173.78Gas
Sulfur dioxideSO₂(g)248.22Gas
Sulfur trioxideSO₃(g)256.76Gas
Sodium chlorideNaCl(s)72.11Ionic solid
Sodium hydroxideNaOH(s)64.43Ionic solid
Potassium chlorideKCl(s)82.59Ionic solid
Calcium oxideCaO(s)38.21Ionic solid
Calcium carbonateCaCO₃(s)91.71Ionic solid
Iron(III) oxideFe₂O₃(s)87.40Solid
Aluminum oxideAl₂O₃(s)50.92Solid — very ordered
Magnesium oxideMgO(s)26.94Solid
Zinc oxideZnO(s)43.66Solid
Table E — ΔH°, ΔS°, ΔG°, K for Common Reactions at 298.15 K
ReactionΔH° (kJ/mol)ΔS° (J/mol·K)ΔG° (kJ/mol)K at 298 K
H₂(g) + ½O₂(g) → H₂O(l) −286−163−237~10⁴¹
N₂(g) + 3H₂(g) → 2NH₃(g) (Haber) −92.4−199−33.05.8×10⁵
CaCO₃(s) → CaO(s) + CO₂(g) +178+161+1301.3×10⁻²³
2SO₂(g) + O₂(g) → 2SO₃(g) −198−187−142~10²⁵
2H₂(g) + O₂(g) → 2H₂O(g) −483.6−88.8−457.2~10⁸⁰
C(graphite) → C(diamond) +1.895−3.36+2.9003.1×10⁻¹
NaCl(s) → Na⁺(aq) + Cl⁻(aq) +3.88+43.4−9.0538 (entropy-driven)
H₂O(l) ⇌ H⁺(aq) + OH⁻(aq) +55.8−80.7+79.91.0×10⁻¹⁴

This Gibbs free energy calculator computes ΔG = ΔH − TΔS to predict reaction spontaneity at any temperature, converts between ΔG° and the equilibrium constant K using ΔG° = −RT × ln K, finds ΔG from electrochemical cell voltages via ΔG = −nFE, and calculates ΔG° from standard Gibbs energies of formation — with complete step-by-step working and interactive free energy diagrams. Use this delta G calculator for any thermodynamics problem.

Gibbs Free Energy — ΔG = ΔH − TΔS

Gibbs free energy (G) is the thermodynamic potential that determines whether a chemical reaction or physical process at constant temperature and pressure is spontaneous. The change in Gibbs free energy, written ΔG = ΔH − TΔS, balances two competing thermodynamic driving forces: enthalpy (ΔH — the heat released or absorbed) and entropy (ΔS — the change in disorder), weighted by the absolute temperature T in Kelvin.

ΔG = ΔH − T × ΔS ΔH in kJ/mol · ΔS in J/mol·K (÷1000 to match units) · T in Kelvin · ΔG in kJ/mol

The spontaneity rules from ΔG = ΔH − TΔS are:

  • ΔG < 0 (negative ΔG): Spontaneous — the forward reaction proceeds without external energy input. A negative delta G means the products are thermodynamically more stable than the reactants at that temperature. The reaction releases free energy.
  • ΔG > 0 (positive ΔG): Non-spontaneous — the forward reaction is not thermodynamically favorable. A positive delta G means the reverse reaction is spontaneous instead. External energy must be supplied to drive the reaction forward.
  • ΔG = 0: The system is at equilibrium — no net reaction occurs in either direction.

⚠ Critical Unit Warning in ΔG = ΔH − TΔS: ΔH is typically given in kJ/mol while ΔS is in J/mol·K — a factor of 1000 difference! Before calculating TΔS, you must divide ΔS by 1000 to convert to kJ/mol·K. Forgetting this step is the #1 error in Gibbs free energy problems and gives an answer 1000× too large for the TΔS term.

The formula ΔG = ΔH − TΔS is named after Josiah Willard Gibbs, who developed it in the 1870s. It defines the standard free energy change ΔG° when conditions are standard (298.15 K, 1 atm, 1 M concentrations). Under non-standard conditions, ΔG (without the degree symbol) depends on the actual concentrations via ΔG = ΔG° + RT ln Q.

Standard Free Energy of Formation — ΔGf° Tables

The standard Gibbs free energy of formation (ΔGf°) is defined as the change in Gibbs free energy when exactly 1 mole of a compound is formed from its constituent elements in their standard states at 298.15 K and 1 atm. The standard Gibbs free energy of formation table allows calculating ΔG° for any reaction directly, without needing separate ΔH° and ΔS° values.

ΔG°rxn = Σ[n · ΔGf°(products)] − Σ[m · ΔGf°(reactants)] Standard free energy of formation rule — elements in standard state always have ΔGf° = 0

Key rules for the standard Gibbs free energy of formation:

  • Elements in their standard state (H₂(g), O₂(g), C(graphite), Fe(s), etc.) have ΔGf° = 0 by definition.
  • Non-standard allotropes do NOT have ΔGf° = 0 — for example, C(diamond) has ΔGf° = +2.900 kJ/mol.
  • The standard Gibbs free energy of formation table gives values at exactly 298.15 K. Temperature dependence requires recalculating using ΔG° = ΔH° − TΔS°.

Example 1 — Standard Free Energy of Formation: Haber Process N₂ + 3H₂ → 2NH₃

  1. ΔG°rxn = Σ ΔGf°(products) − Σ ΔGf°(reactants)
  2. Products: 2 × ΔGf°(NH₃(g)) = 2 × (−16.45) = −32.90 kJ/mol
  3. Reactants: ΔGf°(N₂(g)) + 3 × ΔGf°(H₂(g)) = 0 + 3(0) = 0 kJ/mol
  4. ΔG°rxn = −32.90 − 0 = −32.90 kJ/mol
  5. ΔG° < 0 → spontaneous under standard conditions at 298 K ✓

Example 2 — Standard Free Energy: 2SO₂(g) + O₂(g) → 2SO₃(g)

  1. Products: 2 × ΔGf°(SO₃(g)) = 2 × (−371.06) = −742.12 kJ/mol
  2. Reactants: 2 × ΔGf°(SO₂(g)) + ΔGf°(O₂(g)) = 2(−300.19) + 0 = −600.38 kJ/mol
  3. ΔG°rxn = −742.12 − (−600.38) = −141.74 kJ/mol
  4. Spontaneous under standard conditions (basis for sulfuric acid production)

Example 3 — Formation of CO₂: C(graphite) + O₂(g) → CO₂(g)

  1. Products: 1 × ΔGf°(CO₂(g)) = −394.36 kJ/mol
  2. Reactants: ΔGf°(C(graphite)) + ΔGf°(O₂(g)) = 0 + 0 = 0
  3. ΔG°rxn = −394.36 − 0 = −394.36 kJ/mol
  4. Highly spontaneous — combustion of graphite is strongly thermodynamically favorable

Spontaneity — The Four Cases of ΔH and ΔS

The spontaneity chart captures all four sign combinations of ΔH and ΔS in the formula ΔG = ΔH − TΔS. Understanding these four cases is essential for predicting whether a reaction is spontaneous, non-spontaneous, or temperature-dependent.

ΔHΔSΔG at low TΔG at high TSpontaneous?
Negative (−)Positive (+)NegativeNegative✅ Always
Negative (−)Negative (−)NegativePositive🌡 Low T only
Positive (+)Positive (+)PositiveNegative🌡 High T only
Positive (+)Negative (−)PositivePositive❌ Never

The crossover temperature (where ΔG = 0 and spontaneity switches) is found by setting ΔG = ΔH − TΔS = 0:

T_cross = ΔH(kJ) × 1000 / ΔS(J/mol·K) = ΔH / ΔS_kJ Below T_cross: ΔH sign dominates · Above T_cross: −TΔS term dominates

Key interpretations of the spontaneity chart:

  • Always spontaneous (ΔH < 0, ΔS > 0): Both enthalpy release and entropy increase drive the reaction forward at all temperatures. Example: 2H₂(g) + O₂(g) → 2H₂O(l), ΔG = −237 kJ/mol.
  • Spontaneous at low T only (ΔH < 0, ΔS < 0): Enthalpy wins at low temperature, but as T increases, the −TΔS term (which is positive since ΔS < 0) grows until it exceeds ΔH. Non-spontaneous reaction sets in above T_cross. Example: Haber process N₂+3H₂→2NH₃.
  • Spontaneous at high T only (ΔH > 0, ΔS > 0): Endothermic reactions can be spontaneous when TΔS exceeds ΔH. Example: CaCO₃(s)→CaO(s)+CO₂(g), spontaneous above ~840°C.
  • Never spontaneous (ΔH > 0, ΔS < 0): Both enthalpy and entropy oppose the reaction. ΔG is always positive at all temperatures. The reverse reaction is always spontaneous instead.

Can endothermic reactions be spontaneous? Yes — when ΔS > 0 and temperature is high enough that TΔS > ΔH. Dissolving NaCl in water is a perfect example: ΔH = +3.88 kJ/mol (endothermic) but ΔS = +43.4 J/mol·K (entropy-driven), giving ΔG = −9.05 kJ/mol at 298 K — spontaneous because entropy dominates. If entropy is positive (ΔS > 0), the reaction becomes more spontaneous as temperature increases.

ΔG° and Equilibrium Constant K — ΔG° = −RT ln K

The standard Gibbs free energy change ΔG° is directly and quantitatively connected to the equilibrium constant K through the fundamental equation ΔG° = −RT × ln K, one of the most important relationships in physical chemistry. This equation, sometimes written ΔG° = −RT ln K or delta G RT ln K, allows chemists to predict equilibrium positions from thermodynamic data and vice versa.

ΔG° = −R × T × ln(K)     K = e^(−ΔG°/RT) R = 8.31446 J/(mol·K) · T in Kelvin · ΔG° in J/mol (×1000 if kJ/mol)

The relationship between ΔG° = −RT ln K and equilibrium position:

  • ΔG° < 0: K > 1 — products are favored at equilibrium. The more negative ΔG°, the larger K. At ΔG° = −RT ln K gives K = e^(+|ΔG°|/RT).
  • ΔG° > 0: K < 1 — reactants are favored at equilibrium. Non-spontaneous under standard conditions.
  • ΔG° = 0: K = 1 — equal concentrations of reactants and products at equilibrium.

The non-standard Gibbs free energy ΔG (without degree symbol) uses the reaction quotient Q:

ΔG = ΔG° + RT × ln(Q) At equilibrium: Q = K and ΔG = 0 → confirms ΔG° = −RT ln K ✓

Example 1 — Find K from ΔG°: ΔG° = −20.0 kJ/mol at 298.15 K

  1. K = e^(−ΔG°/RT) = e^(−(−20000)/(8.314 × 298.15))
  2. K = e^(+20000/2478.8) = e^(+8.071)
  3. K = 3195 ≈ 3.19 × 10³
  4. K > 1 → products strongly favored at equilibrium ✓

Example 2 — Find ΔG° from K: K = 1.0 × 10⁻¹⁴ (Kw for water at 25°C)

  1. ΔG° = −RT × ln(K) = −(8.314)(298.15) × ln(10⁻¹⁴)
  2. ΔG° = −2478.8 × (−32.236) = +79,905 J/mol
  3. ΔG° = +79.9 kJ/mol
  4. ΔG° > 0 → water ionization non-spontaneous under standard conditions ✓

Example 3 — Non-Standard ΔG: ΔG° = −20.0 kJ/mol, Q = 0.010, T = 298.15 K

  1. ΔG = ΔG° + RT × ln(Q) = −20.0 + (8.314 × 298.15 × ln(0.010))/1000
  2. RT × ln(Q) = 2478.8 × (−4.6052) / 1000 = −11.42 kJ/mol
  3. ΔG = −20.0 + (−11.42) = −31.42 kJ/mol
  4. Q = 0.010 < K = 3195 → Q < K → forward reaction proceeds ✓

ΔG and Electrochemistry — ΔG = −nFE

The Gibbs free energy provides the fundamental link between thermodynamics and electrochemistry through the equation ΔG = −nFE_cell, where n is the number of moles of electrons transferred, F = 96,485 C/mol is Faraday's constant, and E_cell is the cell voltage. This equation ΔG = −nFE is one of the most important in electrochemical thermodynamics.

ΔG = −n × F × E_cell     ΔG° = −nFE° F = 96,485.3321 C/mol (Faraday's constant) · n = mol electrons · E in Volts · ΔG in Joules (÷1000 for kJ)

The sign relationship in ΔG = −nFE governs cell spontaneity:

  • E_cell > 0 → ΔG < 0 → Spontaneous: A positive cell voltage means the reaction is spontaneous — the cell operates as a galvanic (voltaic) cell, generating electrical energy from chemical energy.
  • E_cell < 0 → ΔG > 0 → Non-spontaneous: A negative cell voltage means the reaction requires electrical energy input — this is an electrolytic cell.
  • E_cell = 0 → ΔG = 0: The cell is at equilibrium — dead battery condition.

Example 1 — Daniell Cell: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), n=2, E°=+1.10 V

  1. ΔG° = −nFE° = −2 × 96,485 × 1.10 = −212,267 J/mol
  2. ΔG° = −212.27 kJ/mol
  3. E° > 0 → ΔG° < 0 → Spontaneous galvanic cell ✓
  4. K = e^(nFE°/RT) = e^(2×96485×1.10/(8.314×298.15)) = e^(85.74) = 1.60×10³⁷

Example 2 — Hydrogen Fuel Cell: H₂ + ½O₂ → H₂O, n=2, E°=+1.229 V

  1. ΔG° = −2 × 96,485 × 1.229 = −237,244 J/mol
  2. ΔG° = −237.2 kJ/mol (matches ΔGf°(H₂O(l)) = −237.13 kJ/mol ✓)
  3. This verifies the consistency between thermodynamic and electrochemical ΔG values

Example 3 — Water Electrolysis: H₂O → H₂ + ½O₂ (reverse), n=2, E°=−1.229 V

  1. ΔG° = −2 × 96,485 × (−1.229) = +237,244 J/mol
  2. ΔG° = +237.2 kJ/mol
  3. E° < 0 → ΔG° > 0 → Non-spontaneous electrolytic cell — requires minimum 1.229 V input ✓

Free Energy Diagrams — Visualizing ΔG

A free energy diagram (Gibbs free energy diagram) plots the Gibbs free energy G on the vertical axis against the reaction coordinate on the horizontal axis. The Gibbs free energy graph shows reactants at the left, products at the right, and an activation energy barrier (transition state hump) in the middle.

  • The height difference between reactants and products = ΔG (thermodynamics — tells you IF the reaction is spontaneous)
  • The height of the hump = E_a (activation energy — kinetics, tells you HOW FAST the reaction goes)
  • Spontaneous reaction (ΔG < 0): Products are lower on the free energy diagram — the arrow points downward from reactants to products.
  • Non-spontaneous reaction (ΔG > 0): Products are higher on the free energy diagram — the arrow points upward.

ΔG ≠ Reaction Rate: A very negative ΔG means the reaction is thermodynamically favorable — but it says nothing about how fast it occurs. Diamond converting to graphite at room temperature has ΔG° = −2.9 kJ/mol (spontaneous!) but an astronomically high activation energy E_a, making it effectively infinitely slow at room temperature. ΔG predicts WHETHER; E_a predicts HOW FAST.

The ΔG vs temperature graph — plotting ΔG = ΔH − TΔS as a linear function of T — is perhaps the most instructive visualization in thermodynamics. The slope of the line equals −ΔS, the y-intercept equals ΔH, and the line crosses zero at T_cross = ΔH×1000/ΔS. This single straight line encodes all four cases of the spontaneity chart simultaneously, making the abstract become geometric.

Standard Entropy — ΔS° Values and Tables

The standard molar entropy S° is the absolute entropy of 1 mole of a substance at 298.15 K and 1 atm. Unlike standard enthalpies and standard Gibbs free energies of formation, elements do NOT have S° = 0. This follows from the Third Law of Thermodynamics: only a perfect crystal at 0 K has zero entropy.

ΔS°rxn = Σ[n × S°(products)] − Σ[m × S°(reactants)] All S° values are positive (Third Law) · units: J/mol·K · elements have S° ≠ 0

Rules for predicting the sign of ΔS° (without looking up the standard entropy table):

  • Gases have much higher S° than liquids, which have higher S° than solids: S°(H₂O(g)) = 188.83 vs S°(H₂O(l)) = 69.95 J/mol·K — a difference of 119 J/mol·K just from phase change.
  • More moles of gas on product side → ΔS > 0 (entropy increases). Reactions producing gas from solids/liquids always have large positive ΔS°.
  • Larger, more complex molecules have higher S°: C₃H₈(g): 270.20 vs CH₄(g): 186.26 J/mol·K — more atoms = more ways to store energy.
  • Dissolution of ionic solids usually increases entropy (ions in solution have more freedom than a crystal lattice).

Example 1 — ΔS° for: N₂(g) + 3H₂(g) → 2NH₃(g)

  1. ΔS° = Σ S°(products) − Σ S°(reactants)
  2. Products: 2 × S°(NH₃(g)) = 2 × 192.77 = 385.54 J/mol·K
  3. Reactants: S°(N₂(g)) + 3 × S°(H₂(g)) = 191.61 + 3(130.68) = 583.65 J/mol·K
  4. ΔS° = 385.54 − 583.65 = −198.11 J/mol·K
  5. ΔS° < 0: 4 mol gas → 2 mol gas, disorder decreases — consistent with prediction ✓

Example 2 — ΔS° for: CaCO₃(s) → CaO(s) + CO₂(g)

  1. Products: S°(CaO(s)) + S°(CO₂(g)) = 38.21 + 213.79 = 252.00 J/mol·K
  2. Reactants: S°(CaCO₃(s)) = 91.71 J/mol·K
  3. ΔS° = 252.00 − 91.71 = +160.29 J/mol·K
  4. ΔS° > 0: producing CO₂ gas from solid significantly increases entropy ✓

Example 3 — ΔS° for: 2H₂(g) + O₂(g) → 2H₂O(l)

  1. Products: 2 × S°(H₂O(l)) = 2 × 69.95 = 139.90 J/mol·K
  2. Reactants: 2 × S°(H₂(g)) + S°(O₂(g)) = 2(130.68) + 205.14 = 466.50 J/mol·K
  3. ΔS° = 139.90 − 466.50 = −326.60 J/mol·K
  4. ΔS° < 0: 3 mol gas → 2 mol liquid, massive decrease in entropy ✓

Reading the Gibbs Free Energy Tables

The Gibbs free energy table (ΔGf° values), the standard entropy table (S° values), and the standard enthalpies of formation table (ΔHf° values) are the three essential thermodynamic data tables in chemistry. Using them correctly allows you to calculate ΔG° for any reaction without running experiments.

Two equivalent methods to find ΔG° — they should give the same answer (small differences arise from rounding in tabulated values):

Method 1 — Direct from ΔGf° Table: H₂(g) + ½O₂(g) → H₂O(l)

  1. ΔG°rxn = Σ ΔGf°(products) − Σ ΔGf°(reactants)
  2. = 1 × (−237.13) − [1 × 0 + ½ × 0]
  3. ΔG°rxn = −237.13 kJ/mol

Method 2 — From ΔH° and ΔS° Tables: Same Reaction

  1. ΔH°rxn = ΔHf°(H₂O(l)) − [ΔHf°(H₂) + ½ΔHf°(O₂)] = −285.83 − 0 = −285.83 kJ/mol
  2. ΔS°rxn = S°(H₂O(l)) − [S°(H₂) + ½S°(O₂)] = 69.95 − [130.68 + ½(205.14)] = 69.95 − 233.25 = −163.30 J/mol·K
  3. ΔG° = ΔH° − TΔS° = −285.83 − (298.15 × (−163.30/1000)) = −285.83 + 48.67 = −237.16 kJ/mol
  4. Difference from Method 1 (−237.13 vs −237.16): ±0.03 kJ/mol — due to rounding in tabulated values ✓

Method 3 — Delta H Delta S Delta G Table Verification: N₂ + 3H₂ → 2NH₃

  1. Via ΔGf°: ΔG°rxn = 2(−16.45) − 0 = −32.90 kJ/mol
  2. Via ΔH°−TΔS°: ΔH° = 2(−46.11) − 0 = −92.22 kJ/mol; ΔS° = 2(192.77) − [191.61 + 3(130.68)] = −198.11 J/mol·K
  3. ΔG° = −92.22 − 298.15 × (−0.19811) = −92.22 + 59.05 = −33.17 kJ/mol
  4. Agreement within rounding: −32.90 vs −33.17 kJ/mol ✓

Why do the two methods give slightly different ΔG° values? The standard Gibbs free energy of formation table values (ΔGf°) are measured directly and precisely. The ΔH° and ΔS° values are also measured independently. Both sets of data carry experimental uncertainties, and the tables use slightly different reference conditions. Small discrepancies (typically <1 kJ/mol) are normal and expected.

Common Mistakes in Gibbs Free Energy Problems

❌ Mistake 1 — Unit Mismatch: Using ΔS in J/mol·K Without Dividing by 1000

The most dangerous error in ΔG = ΔH − TΔS calculations:

  • ❌ Wrong: ΔH = −286 kJ/mol, ΔS = −163.4 J/mol·K, T = 298 K → ΔG = −286 − (298 × (−163.4)) = −286 + 48,693 = +48,407 kJ/mol (COMPLETELY WRONG — off by factor of 1000)
  • ✅ Correct: ΔS_kJ = −163.4/1000 = −0.1634 kJ/mol·K → TΔS = 298 × (−0.1634) = −48.69 kJ/mol → ΔG = −286 − (−48.69) = −237.31 kJ/mol

❌ Mistake 2 — Confusing ΔG with ΔG° (Standard vs Non-Standard)

  • ΔG° (with degree symbol) = standard Gibbs free energy change at 298.15 K, 1 atm, 1 M — fixed value from tables
  • ΔG (without degree symbol) = actual free energy change at real conditions — depends on concentrations/pressures via ΔG = ΔG° + RT ln Q
  • A reaction with ΔG° > 0 can still be spontaneous (ΔG < 0) under the right non-standard conditions

❌ Mistake 3 — Wrong Sign in K = e^(−ΔG°/RT)

  • If ΔG° = −20 kJ/mol: K = e^(−(−20000)/(8.314×298)) = e^(+8.07) = 3195
  • ❌ Wrong sign: K = e^(−8.07) = 0.000313 (this would be K for ΔG° = +20 kJ/mol)
  • Remember: negative ΔG° gives positive exponent, giving K > 1 ✓

❌ Mistake 4 — Thinking Spontaneous Means Fast

  • ΔG < 0 means thermodynamically favorable — tells you nothing about reaction rate
  • Diamond → graphite: ΔG° = −2.9 kJ/mol (spontaneous!) but takes millions of years
  • Combustion of wood in air: ΔG° ≈ −400 kJ/mol (very spontaneous) but needs activation energy (ignition)
  • Kinetics (rate) is determined by activation energy E_a — separate from thermodynamics (ΔG)

❌ Mistake 5 — ΔGf° = 0 Only for Elements in Standard State

  • ✅ ΔGf°(H₂(g)) = 0, ΔGf°(O₂(g)) = 0, ΔGf°(Fe(s)) = 0, ΔGf°(C(graphite)) = 0
  • ❌ ΔGf°(H₂O(l)) = −237.13 kJ/mol (NOT zero — it's a compound)
  • ❌ ΔGf°(C(diamond)) = +2.900 kJ/mol (NOT zero — diamond is not the standard state of carbon)
  • The rule applies only to elements in their most stable standard form at 298.15 K and 1 atm

Worked Examples — 8 Complete Problems

Problem 1 — Formation of Water: ΔG from ΔH and ΔS

  1. Given: ΔH = −286 kJ/mol, ΔS = −163.4 J/mol·K, T = 298.15 K
  2. Convert ΔS: −163.4 ÷ 1000 = −0.16340 kJ/mol·K
  3. TΔS = 298.15 × (−0.16340) = −48.70 kJ/mol
  4. ΔG = ΔH − TΔS = −286.00 − (−48.70) = −237.30 kJ/mol
  5. ✅ Spontaneous (matches ΔGf°(H₂O(l)) = −237.13 kJ/mol ✓)

Problem 2 — Haber Process ΔG° from ΔGf° Table

  1. N₂(g) + 3H₂(g) → 2NH₃(g)
  2. ΔG°rxn = 2×ΔGf°(NH₃) − [ΔGf°(N₂) + 3×ΔGf°(H₂)]
  3. = 2×(−16.45) − [0 + 0] = −32.90 kJ/mol
  4. ✅ Spontaneous at 298 K, but becomes non-spontaneous above T_cross = 465 K (192°C)

Problem 3 — CaCO₃ Decomposition: Find Crossover Temperature

  1. CaCO₃(s) → CaO(s) + CO₂(g): ΔH = +178 kJ/mol, ΔS = +160 J/mol·K
  2. T_cross = ΔH × 1000 / ΔS = 178,000 / 160 = 1113 K = 840°C
  3. Below 840°C: ΔG > 0, non-spontaneous (limestone is stable)
  4. Above 840°C: ΔG < 0, spontaneous (lime kiln operates above this temperature) ✓

Problem 4 — K from ΔG°: ΔG° = −20.0 kJ/mol

  1. K = e^(−ΔG°/RT) = e^(−(−20000)/(8.314 × 298.15))
  2. K = e^(+8.071) = K ≈ 3195
  3. K > 1 → products strongly favored at equilibrium
  4. log K = 8.071/2.303 = 3.505 → K ≈ 10³·⁵

Problem 5 — ΔG° from K: K = 1.0 × 10⁻¹⁴ (Water autoionization)

  1. ΔG° = −RT × ln(K) = −(8.314)(298.15) × ln(10⁻¹⁴)
  2. ln(10⁻¹⁴) = −14 × 2.3026 = −32.24
  3. ΔG° = −2478.8 × (−32.24) = +79,905 J/mol = +79.9 kJ/mol
  4. ❌ Non-spontaneous — water ionization doesn't proceed under standard conditions ✓

Problem 6 — Non-Standard ΔG: ΔG° = −20.0 kJ/mol, Q = 0.010

  1. ΔG = ΔG° + RT × ln(Q) = −20.0 + (8.314 × 298.15 / 1000) × ln(0.010)
  2. RT/1000 = 2.4789 kJ/mol; ln(0.010) = −4.6052
  3. RT × ln(Q) / 1000 = 2.4789 × (−4.6052) = −11.42 kJ/mol
  4. ΔG = −20.0 + (−11.42) = −31.42 kJ/mol
  5. Q < K: reaction proceeds forward ✓

Problem 7 — Electrochemistry: Daniell Cell ΔG = −nFE

  1. Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), n = 2, E° = +1.10 V
  2. ΔG° = −nFE° = −2 × 96,485 × 1.10 = −212,267 J
  3. ΔG° = −212.27 kJ/mol
  4. E° > 0 → Spontaneous galvanic cell; K = e^(85.74) = 1.60×10³⁷

Problem 8 — Entropy-Driven: NaCl Dissolving (ΔH > 0 but ΔG < 0)

  1. NaCl(s) → Na⁺(aq) + Cl⁻(aq): ΔH = +3.88 kJ/mol, ΔS = +43.4 J/mol·K, T = 298 K
  2. TΔS = 298 × (43.4/1000) = +12.93 kJ/mol
  3. ΔG = +3.88 − (+12.93) = −9.05 kJ/mol
  4. ✅ Spontaneous even though endothermic — entropy drives the dissolution!
  5. This is the clearest example of "can endothermic reactions be spontaneous? YES" ✓

Frequently Asked Questions

What is Gibbs free energy?
Gibbs free energy (G) is a thermodynamic state function that measures the maximum useful work obtainable from a system at constant temperature and pressure. The change ΔG determines spontaneity: ΔG < 0 means spontaneous (the reaction releases free energy), ΔG > 0 means non-spontaneous (the reverse reaction is spontaneous), and ΔG = 0 means the system is at equilibrium. Calculated via ΔG = ΔH − TΔS.
What does negative delta G mean? Is negative delta G spontaneous?
Yes — negative delta G (ΔG < 0) means the reaction is spontaneous in the forward direction. The reaction can proceed without external energy input, releasing free energy to the surroundings. However, spontaneous does NOT mean instant or fast — it means thermodynamically favorable. A negative ΔG tells you the reaction CAN occur, not that it WILL occur quickly.
How do you calculate delta G? What is the ΔG = ΔH − TΔS formula?
Use ΔG = ΔH − TΔS. CRITICAL: ΔH must be in kJ/mol and ΔS must be in kJ/mol·K (divide J/mol·K values by 1000). T must be in Kelvin (add 273.15 to °C). Example: ΔH = −286 kJ/mol, ΔS = −163.4 J/mol·K = −0.1634 kJ/mol·K, T = 298.15 K → TΔS = 298.15 × (−0.1634) = −48.70 kJ/mol → ΔG = −286 − (−48.70) = −237.30 kJ/mol.
What is the relationship between delta G and K? (ΔG° = −RT ln K)
ΔG° = −RT × ln(K), equivalently K = e^(−ΔG°/RT). When ΔG° < 0: K > 1 (products favored). When ΔG° > 0: K < 1 (reactants favored). When ΔG° = 0: K = 1. For non-standard conditions: ΔG = ΔG° + RT × ln(Q). At equilibrium Q = K and ΔG = 0, which confirms ΔG° = −RT × ln K.
What is standard free energy of formation?
The standard free energy of formation (ΔGf°) is the Gibbs free energy change when 1 mole of a compound is formed from its elements in their standard states at 298.15 K and 1 atm. Elements in standard state have ΔGf° = 0. Use the standard Gibbs free energy of formation table to calculate ΔG°rxn = Σ[n·ΔGf°(products)] − Σ[m·ΔGf°(reactants)].
When is a reaction spontaneous? What are the four cases?
A reaction is spontaneous when ΔG < 0. The four cases of ΔH and ΔS: (1) ΔH < 0 and ΔS > 0 → always spontaneous; (2) ΔH < 0 and ΔS < 0 → spontaneous only at low T; (3) ΔH > 0 and ΔS > 0 → spontaneous only at high T; (4) ΔH > 0 and ΔS < 0 → never spontaneous. The crossover temperature is T_cross = ΔH×1000/ΔS(J/mol·K).
What is the ΔG = −nFE equation?
ΔG = −nFE_cell connects thermodynamics to electrochemistry, where n = moles of electrons transferred, F = 96,485 C/mol (Faraday's constant), and E_cell = cell voltage. Positive E_cell → negative ΔG → spontaneous (galvanic cell). Negative E_cell → positive ΔG → non-spontaneous (electrolytic cell). At standard conditions: ΔG° = −nFE°.
How does temperature affect spontaneity?
Temperature affects spontaneity through the TΔS term in ΔG = ΔH − TΔS. At the crossover temperature T_cross = ΔH×1000/ΔS(J), ΔG = 0 and the reaction switches between spontaneous and non-spontaneous. Increasing temperature always makes the −TΔS term more negative (if ΔS > 0, increases spontaneity; if ΔS < 0, decreases spontaneity). This is why the Haber process (ΔS < 0) must be run near its crossover temperature for practical yield.

Related Calculators

Quick Reference Formulas

ΔG = ΔH − T × ΔS Main Gibbs equation — ΔS ÷1000 if in J/mol·K
ΔG° = −RT × ln(K) Equilibrium constant connection
K = e^(−ΔG°/RT) Find K from standard ΔG°
ΔG = ΔG° + RT×ln(Q) Non-standard conditions
ΔG = −n × F × E_cell Electrochemical connection
T_cross = ΔH×1000 / ΔS Crossover temperature (K)
R = 8.31446 J/(mol·K) Gas constant
F = 96,485 C/mol Faraday's constant
T_std = 298.15 K = 25°C Standard temperature

Spontaneity Rules

✅ SPONTANEOUS: ΔG < 0
Forward reaction favored
❌ NON-SPONTANEOUS: ΔG > 0
Reverse reaction favored
⚖️ EQUILIBRIUM: ΔG = 0
No net driving force
Unit rule: ΔS÷1000
J/mol·K → kJ/mol·K

Key ΔGf° Values (kJ/mol)

H₂O(l) −237.13
CO₂(g) −394.36
NH₃(g) −16.45
SO₂(g) −300.19
NaCl(s) −384.14
CaCO₃(s) −1128.79
Al₂O₃(s) −1582.27
NO(g) +86.55

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