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.
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.
ΔG = ΔH − TΔS Result
Step-by-Step: ΔG = ΔH − TΔS
Free Energy Level Diagram
ΔG vs Temperature — ΔG = ΔH − TΔS (Crossover Graph)
ΔG°rxn = Σ[n·ΔGf°(products)] − Σ[m·ΔGf°(reactants)]
Select substances from the standard Gibbs free energy of formation table. Elements in standard state have ΔGf° = 0.
⬆ PRODUCTS
⬇ REACTANTS
ΔG°rxn from Standard Gibbs Free Energy of Formation
Step-by-Step Working
Full Thermodynamic Calculation from ΔHf° and S° Tables
Select substances → auto-fills ΔHf° and S° → computes ΔH°rxn, ΔS°rxn, then ΔG° = ΔH° − TΔS°. Verifies against ΔGf° direct calculation.
⬆ PRODUCTS
⬇ REACTANTS
ΔG° from Full Thermodynamic Analysis
Step-by-Step Working
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°
ΔG = −nFE Result
Step-by-Step: ΔG = −nFE
Equilibrium Constant from Cell Voltage
Common Standard Cell Potentials
| Cell Reaction | n | E°_cell (V) | ΔG° (kJ/mol) | Type |
|---|---|---|---|---|
| Zn + Cu²⁺ → Zn²⁺ + Cu (Daniell) | 2 | +1.10 | −212.3 | Galvanic ✅ |
| H₂ + ½O₂ → H₂O (Fuel cell) | 2 | +1.229 | −237.2 | Galvanic ✅ |
| H₂O → H₂ + ½O₂ (Electrolysis) | 2 | −1.229 | +237.2 | Electrolytic ❌ |
| Fe + 2H⁺ → Fe²⁺ + H₂ | 2 | +0.44 | −84.9 | Galvanic ✅ |
| Al³⁺ + 3e⁻ → Al (reduction) | 3 | −1.66 | +480.8 | Electrolytic ❌ |
| 2Ag⁺ + Zn → 2Ag + Zn²⁺ | 2 | +1.56 | −301.0 | Galvanic ✅ |
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° = −RT ln K Result
Step-by-Step: ΔG° = −RT ln K
Non-Standard ΔG = ΔG° + RT·ln(Q)
Table D — ΔG° and Equilibrium Constant Relationship
| ΔG° (kJ/mol) | K (approx) | Equilibrium Position | Spontaneity |
|---|---|---|---|
| < −40 | > 10⁷ | Far toward products | Strongly spontaneous ✅ |
| −20 to −40 | 10³–10⁷ | Mostly products | Spontaneous ✅ |
| −10 to −20 | 10²–10³ | Predominantly products | Moderately spontaneous |
| −5 to −10 | 10–10² | Slightly product-favored | Slightly spontaneous |
| −2 to +5 | ~1 | Near equilibrium | Near equilibrium ⚖️ |
| +10 to +20 | 10⁻²–10⁻⁴ | Mostly reactants | Non-spontaneous ❌ |
| > +40 | < 10⁻⁷ | Far toward reactants | Strongly non-spontaneous ❌ |
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.
| ΔH | ΔS | ΔG = ΔH − TΔS | Spontaneous? | 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) |
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)].
| Substance | Formula | ΔGf° (kJ/mol) | State |
|---|---|---|---|
| Hydrogen (element) | H₂(g) | 0 | Standard state |
| Oxygen (element) | O₂(g) | 0 | Standard state |
| Nitrogen (element) | N₂(g) | 0 | Standard state |
| Carbon (graphite) | C(graphite) | 0 | Standard state |
| Carbon (diamond) | C(diamond) | +2.900 | Non-standard allotrope |
| Sulfur (rhombic) | S(rhombic) | 0 | Standard state |
| Iron (element) | Fe(s) | 0 | Standard state |
| Copper (element) | Cu(s) | 0 | Standard state |
| Aluminum (element) | Al(s) | 0 | Standard state |
| Water (liquid) | H₂O(l) | −237.13 | Liquid |
| Water (vapor) | H₂O(g) | −228.57 | Gas |
| Carbon monoxide | CO(g) | −137.17 | Gas |
| Carbon dioxide | CO₂(g) | −394.36 | Gas |
| Methane | CH₄(g) | −50.72 | Gas |
| Ethane | C₂H₆(g) | −32.82 | Gas |
| Ethylene (ethene) | C₂H₄(g) | +68.15 | Gas |
| Acetylene (ethyne) | C₂H₂(g) | +209.20 | Gas |
| Propane | C₃H₈(g) | −23.47 | Gas |
| Benzene | C₆H₆(l) | +124.50 | Liquid |
| Methanol | CH₃OH(l) | −166.27 | Liquid |
| Ethanol | C₂H₅OH(l) | −174.78 | Liquid |
| Ammonia | NH₃(g) | −16.45 | Gas |
| Nitric oxide | NO(g) | +86.55 | Gas |
| Nitrogen dioxide | NO₂(g) | +51.31 | Gas |
| Nitrous oxide | N₂O(g) | +103.59 | Gas |
| Hydrogen chloride | HCl(g) | −95.30 | Gas |
| Hydrogen bromide | HBr(g) | −53.45 | Gas |
| Hydrogen fluoride | HF(g) | −273.20 | Gas |
| Sulfur dioxide | SO₂(g) | −300.19 | Gas |
| Sulfur trioxide | SO₃(g) | −371.06 | Gas |
| Sulfuric acid | H₂SO₄(l) | −690.00 | Liquid |
| Sodium chloride (salt) | NaCl(s) | −384.14 | Solid |
| Sodium hydroxide | NaOH(s) | −379.49 | Solid |
| Potassium chloride | KCl(s) | −408.51 | Solid |
| Calcium chloride | CaCl₂(s) | −748.10 | Solid |
| Calcium oxide (lime) | CaO(s) | −603.30 | Solid |
| Calcium carbonate | CaCO₃(s) | −1128.79 | Solid |
| Iron(III) oxide (rust) | Fe₂O₃(s) | −742.20 | Solid |
| Aluminum oxide | Al₂O₃(s) | −1582.27 | Solid |
| Magnesium oxide | MgO(s) | −569.43 | Solid |
| Zinc oxide | ZnO(s) | −318.30 | Solid |
| Silver chloride | AgCl(s) | −109.79 | Solid |
| Silicon dioxide (quartz) | SiO₂(s) | −856.67 | Solid |
| Ammonium chloride | NH₄Cl(s) | −202.87 | Solid |
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)].
| Substance | Formula | S° (J/mol·K) | Notes |
|---|---|---|---|
| Hydrogen | H₂(g) | 130.68 | Gas — high entropy |
| Oxygen | O₂(g) | 205.14 | Gas |
| Nitrogen | N₂(g) | 191.61 | Gas |
| Carbon (graphite) | C(graphite) | 5.74 | Solid — low entropy |
| Carbon (diamond) | C(diamond) | 2.38 | Very ordered crystal |
| Sulfur (rhombic) | S(rhombic) | 31.80 | Solid |
| Iron | Fe(s) | 27.28 | Metal solid |
| Copper | Cu(s) | 33.15 | Metal solid |
| Aluminum | Al(s) | 28.33 | Metal solid |
| Sodium | Na(s) | 51.30 | Metal solid |
| Calcium | Ca(s) | 41.63 | Metal solid |
| Water (liquid) | H₂O(l) | 69.95 | Liquid |
| Water (vapor/steam) | H₂O(g) | 188.83 | Gas — much higher than liquid |
| Carbon monoxide | CO(g) | 197.67 | Gas |
| Carbon dioxide | CO₂(g) | 213.79 | Gas |
| Methane | CH₄(g) | 186.26 | Gas |
| Ethane | C₂H₆(g) | 229.60 | Gas — larger molecule, higher S |
| Ethylene | C₂H₄(g) | 219.56 | Gas |
| Acetylene | C₂H₂(g) | 200.94 | Gas |
| Propane | C₃H₈(g) | 270.20 | Gas — large molecule |
| Benzene | C₆H₆(l) | 173.40 | Liquid |
| Methanol | CH₃OH(l) | 126.80 | Liquid |
| Ethanol | C₂H₅OH(l) | 160.70 | Liquid |
| Ammonia | NH₃(g) | 192.77 | Gas |
| Nitric oxide | NO(g) | 210.76 | Gas |
| Nitrogen dioxide | NO₂(g) | 240.06 | Gas |
| Nitrous oxide | N₂O(g) | 219.96 | Gas |
| Hydrogen chloride | HCl(g) | 186.90 | Gas |
| Hydrogen bromide | HBr(g) | 198.70 | Gas |
| Hydrogen fluoride | HF(g) | 173.78 | Gas |
| Sulfur dioxide | SO₂(g) | 248.22 | Gas |
| Sulfur trioxide | SO₃(g) | 256.76 | Gas |
| Sodium chloride | NaCl(s) | 72.11 | Ionic solid |
| Sodium hydroxide | NaOH(s) | 64.43 | Ionic solid |
| Potassium chloride | KCl(s) | 82.59 | Ionic solid |
| Calcium oxide | CaO(s) | 38.21 | Ionic solid |
| Calcium carbonate | CaCO₃(s) | 91.71 | Ionic solid |
| Iron(III) oxide | Fe₂O₃(s) | 87.40 | Solid |
| Aluminum oxide | Al₂O₃(s) | 50.92 | Solid — very ordered |
| Magnesium oxide | MgO(s) | 26.94 | Solid |
| Zinc oxide | ZnO(s) | 43.66 | Solid |
| Reaction | ΔH° (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 | −33.0 | 5.8×10⁵ |
| CaCO₃(s) → CaO(s) + CO₂(g) | +178 | +161 | +130 | 1.3×10⁻²³ |
| 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 | +2.900 | 3.1×10⁻¹ |
| NaCl(s) → Na⁺(aq) + Cl⁻(aq) | +3.88 | +43.4 | −9.05 | 38 (entropy-driven) |
| H₂O(l) ⇌ H⁺(aq) + OH⁻(aq) | +55.8 | −80.7 | +79.9 | 1.0×10⁻¹⁴ |
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.
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.
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
- ΔG°rxn = Σ ΔGf°(products) − Σ ΔGf°(reactants)
- Products: 2 × ΔGf°(NH₃(g)) = 2 × (−16.45) = −32.90 kJ/mol
- Reactants: ΔGf°(N₂(g)) + 3 × ΔGf°(H₂(g)) = 0 + 3(0) = 0 kJ/mol
- ΔG°rxn = −32.90 − 0 = −32.90 kJ/mol
- ΔG° < 0 → spontaneous under standard conditions at 298 K ✓
Example 2 — Standard Free Energy: 2SO₂(g) + O₂(g) → 2SO₃(g)
- Products: 2 × ΔGf°(SO₃(g)) = 2 × (−371.06) = −742.12 kJ/mol
- Reactants: 2 × ΔGf°(SO₂(g)) + ΔGf°(O₂(g)) = 2(−300.19) + 0 = −600.38 kJ/mol
- ΔG°rxn = −742.12 − (−600.38) = −141.74 kJ/mol
- Spontaneous under standard conditions (basis for sulfuric acid production)
Example 3 — Formation of CO₂: C(graphite) + O₂(g) → CO₂(g)
- Products: 1 × ΔGf°(CO₂(g)) = −394.36 kJ/mol
- Reactants: ΔGf°(C(graphite)) + ΔGf°(O₂(g)) = 0 + 0 = 0
- ΔG°rxn = −394.36 − 0 = −394.36 kJ/mol
- 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.
| ΔH | ΔS | ΔG at low T | ΔG at high T | Spontaneous? |
|---|---|---|---|---|
| Negative (−) | Positive (+) | Negative | Negative | ✅ Always |
| Negative (−) | Negative (−) | Negative | Positive | 🌡 Low T only |
| Positive (+) | Positive (+) | Positive | Negative | 🌡 High T only |
| Positive (+) | Negative (−) | Positive | Positive | ❌ Never |
The crossover temperature (where ΔG = 0 and spontaneity switches) is found by setting ΔG = ΔH − TΔS = 0:
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.
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:
Example 1 — Find K from ΔG°: ΔG° = −20.0 kJ/mol at 298.15 K
- K = e^(−ΔG°/RT) = e^(−(−20000)/(8.314 × 298.15))
- K = e^(+20000/2478.8) = e^(+8.071)
- K = 3195 ≈ 3.19 × 10³
- K > 1 → products strongly favored at equilibrium ✓
Example 2 — Find ΔG° from K: K = 1.0 × 10⁻¹⁴ (Kw for water at 25°C)
- ΔG° = −RT × ln(K) = −(8.314)(298.15) × ln(10⁻¹⁴)
- ΔG° = −2478.8 × (−32.236) = +79,905 J/mol
- ΔG° = +79.9 kJ/mol
- 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
- ΔG = ΔG° + RT × ln(Q) = −20.0 + (8.314 × 298.15 × ln(0.010))/1000
- RT × ln(Q) = 2478.8 × (−4.6052) / 1000 = −11.42 kJ/mol
- ΔG = −20.0 + (−11.42) = −31.42 kJ/mol
- 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.
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
- ΔG° = −nFE° = −2 × 96,485 × 1.10 = −212,267 J/mol
- ΔG° = −212.27 kJ/mol
- E° > 0 → ΔG° < 0 → Spontaneous galvanic cell ✓
- 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
- ΔG° = −2 × 96,485 × 1.229 = −237,244 J/mol
- ΔG° = −237.2 kJ/mol (matches ΔGf°(H₂O(l)) = −237.13 kJ/mol ✓)
- This verifies the consistency between thermodynamic and electrochemical ΔG values
Example 3 — Water Electrolysis: H₂O → H₂ + ½O₂ (reverse), n=2, E°=−1.229 V
- ΔG° = −2 × 96,485 × (−1.229) = +237,244 J/mol
- ΔG° = +237.2 kJ/mol
- 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.
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)
- ΔS° = Σ S°(products) − Σ S°(reactants)
- Products: 2 × S°(NH₃(g)) = 2 × 192.77 = 385.54 J/mol·K
- Reactants: S°(N₂(g)) + 3 × S°(H₂(g)) = 191.61 + 3(130.68) = 583.65 J/mol·K
- ΔS° = 385.54 − 583.65 = −198.11 J/mol·K
- ΔS° < 0: 4 mol gas → 2 mol gas, disorder decreases — consistent with prediction ✓
Example 2 — ΔS° for: CaCO₃(s) → CaO(s) + CO₂(g)
- Products: S°(CaO(s)) + S°(CO₂(g)) = 38.21 + 213.79 = 252.00 J/mol·K
- Reactants: S°(CaCO₃(s)) = 91.71 J/mol·K
- ΔS° = 252.00 − 91.71 = +160.29 J/mol·K
- ΔS° > 0: producing CO₂ gas from solid significantly increases entropy ✓
Example 3 — ΔS° for: 2H₂(g) + O₂(g) → 2H₂O(l)
- Products: 2 × S°(H₂O(l)) = 2 × 69.95 = 139.90 J/mol·K
- Reactants: 2 × S°(H₂(g)) + S°(O₂(g)) = 2(130.68) + 205.14 = 466.50 J/mol·K
- ΔS° = 139.90 − 466.50 = −326.60 J/mol·K
- 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)
- ΔG°rxn = Σ ΔGf°(products) − Σ ΔGf°(reactants)
- = 1 × (−237.13) − [1 × 0 + ½ × 0]
- ΔG°rxn = −237.13 kJ/mol
Method 2 — From ΔH° and ΔS° Tables: Same Reaction
- ΔH°rxn = ΔHf°(H₂O(l)) − [ΔHf°(H₂) + ½ΔHf°(O₂)] = −285.83 − 0 = −285.83 kJ/mol
- 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
- ΔG° = ΔH° − TΔS° = −285.83 − (298.15 × (−163.30/1000)) = −285.83 + 48.67 = −237.16 kJ/mol
- 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
- Via ΔGf°: ΔG°rxn = 2(−16.45) − 0 = −32.90 kJ/mol
- 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
- ΔG° = −92.22 − 298.15 × (−0.19811) = −92.22 + 59.05 = −33.17 kJ/mol
- 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
- Given: ΔH = −286 kJ/mol, ΔS = −163.4 J/mol·K, T = 298.15 K
- Convert ΔS: −163.4 ÷ 1000 = −0.16340 kJ/mol·K
- TΔS = 298.15 × (−0.16340) = −48.70 kJ/mol
- ΔG = ΔH − TΔS = −286.00 − (−48.70) = −237.30 kJ/mol
- ✅ Spontaneous (matches ΔGf°(H₂O(l)) = −237.13 kJ/mol ✓)
Problem 2 — Haber Process ΔG° from ΔGf° Table
- N₂(g) + 3H₂(g) → 2NH₃(g)
- ΔG°rxn = 2×ΔGf°(NH₃) − [ΔGf°(N₂) + 3×ΔGf°(H₂)]
- = 2×(−16.45) − [0 + 0] = −32.90 kJ/mol
- ✅ Spontaneous at 298 K, but becomes non-spontaneous above T_cross = 465 K (192°C)
Problem 3 — CaCO₃ Decomposition: Find Crossover Temperature
- CaCO₃(s) → CaO(s) + CO₂(g): ΔH = +178 kJ/mol, ΔS = +160 J/mol·K
- T_cross = ΔH × 1000 / ΔS = 178,000 / 160 = 1113 K = 840°C
- Below 840°C: ΔG > 0, non-spontaneous (limestone is stable)
- Above 840°C: ΔG < 0, spontaneous (lime kiln operates above this temperature) ✓
Problem 4 — K from ΔG°: ΔG° = −20.0 kJ/mol
- K = e^(−ΔG°/RT) = e^(−(−20000)/(8.314 × 298.15))
- K = e^(+8.071) = K ≈ 3195
- K > 1 → products strongly favored at equilibrium
- log K = 8.071/2.303 = 3.505 → K ≈ 10³·⁵
Problem 5 — ΔG° from K: K = 1.0 × 10⁻¹⁴ (Water autoionization)
- ΔG° = −RT × ln(K) = −(8.314)(298.15) × ln(10⁻¹⁴)
- ln(10⁻¹⁴) = −14 × 2.3026 = −32.24
- ΔG° = −2478.8 × (−32.24) = +79,905 J/mol = +79.9 kJ/mol
- ❌ Non-spontaneous — water ionization doesn't proceed under standard conditions ✓
Problem 6 — Non-Standard ΔG: ΔG° = −20.0 kJ/mol, Q = 0.010
- ΔG = ΔG° + RT × ln(Q) = −20.0 + (8.314 × 298.15 / 1000) × ln(0.010)
- RT/1000 = 2.4789 kJ/mol; ln(0.010) = −4.6052
- RT × ln(Q) / 1000 = 2.4789 × (−4.6052) = −11.42 kJ/mol
- ΔG = −20.0 + (−11.42) = −31.42 kJ/mol
- Q < K: reaction proceeds forward ✓
Problem 7 — Electrochemistry: Daniell Cell ΔG = −nFE
- Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), n = 2, E° = +1.10 V
- ΔG° = −nFE° = −2 × 96,485 × 1.10 = −212,267 J
- ΔG° = −212.27 kJ/mol
- E° > 0 → Spontaneous galvanic cell; K = e^(85.74) = 1.60×10³⁷
Problem 8 — Entropy-Driven: NaCl Dissolving (ΔH > 0 but ΔG < 0)
- NaCl(s) → Na⁺(aq) + Cl⁻(aq): ΔH = +3.88 kJ/mol, ΔS = +43.4 J/mol·K, T = 298 K
- TΔS = 298 × (43.4/1000) = +12.93 kJ/mol
- ΔG = +3.88 − (+12.93) = −9.05 kJ/mol
- ✅ Spontaneous even though endothermic — entropy drives the dissolution!
- This is the clearest example of "can endothermic reactions be spontaneous? YES" ✓
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