What does Gibbs free energy tell us?

If ΔG 0, the reaction is non-spontaneous (reverse direction is favored). If ΔG = 0, the system is at equilibrium. ΔG predicts equilibrium position, not reaction rate.

How is ΔG related to the equilibrium constant?

ΔG° = −RT ln(K). Solve for K: K = exp(−ΔG°/(RT)). At 298 K: ΔG° of −5.7 kJ/mol gives K ≈ 10 (favorable); −11.4 gives K ≈ 100; −17.1 gives K ≈ 1000. Each 5.7 kJ/mol decrease in ΔG° multiplies K by 10.

What's the difference between ΔG and ΔG°?

ΔG° is the standard Gibbs energy change (with all reactants and products at standard concentrations, 1 M or 1 atm). ΔG is the Gibbs energy change at actual conditions. They're related by ΔG = ΔG° + RT ln(Q), where Q is the reaction quotient.

Why is "spontaneous" not the same as "fast"?

ΔG tells you which direction the reaction wants to go (thermodynamics). Activation energy and catalysts determine how fast it actually gets there (kinetics). Diamond → graphite has ΔG < 0 but happens unmeasurably slowly at room T due to high activation energy.

How does temperature affect ΔG?

Directly through the −TΔS term. For reactions with ΔS > 0, increasing T makes ΔG more negative (more favorable). For ΔS < 0, increasing T makes ΔG less negative (less favorable). The crossover temperature is T = ΔH/ΔS — below this, ΔH dominates; above, TΔS dominates.

Can I have a reaction with positive ΔH and still spontaneous?

Yes, if ΔS is also positive AND T is high enough that TΔS > ΔH. Ice melting at room T is the classic example: ΔH = +6 kJ/mol (heat absorbed) but ΔS = +22 J/(mol·K) (disorder increase). At T > 273 K, the entropy term overwhelms the enthalpy and ΔG is negative.

Gibbs Free Energy Calculator

Q: Why divide ΔS by 1000?

ΔH is typically in kJ/mol while ΔS is in J/(mol·K). To compute ΔG = ΔH − TΔS, both terms must be in the same units. Dividing ΔS by 1000 converts from J to kJ. This calculator does the unit conversion automatically.

Q: How does temperature affect ΔG?

Directly through the −TΔS term. For reactions with ΔS > 0, increasing T makes ΔG more negative (more favorable). For ΔS < 0, increasing T makes ΔG less negative (less favorable). The crossover temperature is T = ΔH/ΔS — below this, ΔH dominates; above, TΔS dominates.

Q: Can I have a reaction with positive ΔH and still spontaneous?

Yes, if ΔS is also positive AND T is high enough that TΔS > ΔH. Ice melting at room T is the classic example: ΔH = +6 kJ/mol (heat absorbed) but ΔS = +22 J/(mol·K) (disorder increase). At T > 273 K, the entropy term overwhelms the enthalpy and ΔG is negative.

Determine whether a chemical reaction is spontaneous by calculating the Gibbs free energy change. Enter enthalpy change, temperature, and entropy change to find ΔG and predict reaction spontaneity.

Gibbs free energy is the thermodynamic quantity that tells you whether a chemical reaction will proceed spontaneously under a given set of conditions. The equation ΔG = ΔH − TΔS captures the trade-off between two fundamental driving forces: enthalpy (ΔH, the heat released or absorbed) and entropy (ΔS, the change in disorder). Reactions tend to proceed when they release heat (ΔH < 0) and/or increase disorder (ΔS > 0); both factors contribute to making ΔG negative. The temperature T determines how much weight is given to the entropy term.

This calculator computes ΔG given ΔH, T, and ΔS. A negative ΔG means the reaction is thermodynamically favorable (spontaneous in the forward direction); positive ΔG means non-spontaneous (the reverse reaction is favored); zero means the system is at equilibrium. The size of ΔG tells you how far from equilibrium the system is — ΔG = −50 kJ/mol is strongly product-favored; −5 kJ/mol means barely-favorable; +5 kJ/mol means barely-unfavorable.

A practical caution: "spontaneous" thermodynamically just means "thermodynamically favorable" — it doesn't say anything about how fast. The famous example: diamond is metastable; converting diamond to graphite has ΔG < 0 (graphite is the more stable form), but the activation energy is so high that the conversion is immeasurably slow at room temperature. Diamonds aren't forever, but they're close enough for human purposes. Kinetics and thermodynamics together determine real-world reaction behavior.

Inputs

Enthalpy Change (ΔH) kJ/mol

Temperature (K)

Entropy Change (ΔS) J/(mol·K)

Results

ΔG

-85.09 kJ/mol

TΔS

-14.91 kJ/mol

Result

Spontaneous (favorable)

Gibbs Free Energy Results

Parameter	Value
ΔH (Enthalpy Change)	-100.00 kJ/mol
Temperature (T)	298.15 K
ΔS (Entropy Change)	-50.00 J/(mol·K)
ΔS (converted)	-0.05000 kJ/(mol·K)
TΔS	-14.9075 kJ/mol
ΔG = ΔH - TΔS	-85.0925 kJ/mol
Spontaneity	Spontaneous (favorable)
Equilibrium Temperature	2000.00 K
ΔG (cal/mol)	-20337.62 cal/mol

Last updated: May 28, 2026

Formula

**Gibbs free energy change:** ΔG = ΔH − TΔS Where: - **ΔG**: Gibbs free energy change (kJ/mol or J/mol) - **ΔH**: enthalpy change (kJ/mol, often) - **T**: absolute temperature (Kelvin — never Celsius) - **ΔS**: entropy change (J/(mol·K), often) **Unit check:** ΔH and TΔS must be in the same units. If ΔH is in kJ/mol and ΔS in J/(mol·K), then TΔS will be in J/mol — divide by 1000 first or convert ΔH to J/mol. The calculator handles this automatically. **Spontaneity rules:** | ΔH | ΔS | ΔG | Spontaneity | |---|---|---|---| | − | + | − at all T | Always spontaneous (exothermic + disorder gain) | | + | − | + at all T | Never spontaneous (endothermic + order gain) | | − | − | − only at low T | Spontaneous only when |TΔS| < |ΔH| | | + | + | − only at high T | Spontaneous when TΔS > ΔH | **Crossover temperature** (where ΔG changes sign): T_crossover = ΔH / ΔS Below this T, the sign of ΔG depends on which of ΔH and ΔS dominates. **Relationship to equilibrium constant K:** ΔG° = −RT ln(K) Where: - **ΔG°**: standard Gibbs energy change (at standard conditions, 1 M for all species) - **R**: 8.314 J/(mol·K) = 0.008314 kJ/(mol·K) - **K**: equilibrium constant (dimensionless in thermodynamic form) **Predicting K from ΔG°:** | ΔG° (kJ/mol) at 298 K | K | |---|---| | −50 | 5.8 × 10⁸ | | −20 | 3,200 | | −10 | 57 | | 0 | 1 | | +10 | 0.018 | | +20 | 3.1 × 10⁻⁴ | | +50 | 1.7 × 10⁻⁹ | So a 10 kJ/mol change in ΔG° changes K by about a factor of 60 at room temperature. **Worked example: ammonia synthesis at 25 °C** N₂(g) + 3H₂(g) ⇌ 2NH₃(g): - ΔH° = −92.4 kJ/mol (exothermic) - ΔS° = −198.7 J/(mol·K) (negative — 4 mol gas → 2 mol gas) - ΔG° = −92.4 − (298.15 × −0.1987) = −92.4 + 59.2 = **−33.2 kJ/mol** (favorable at room T) - At 700 K (high T): ΔG° = −92.4 − (700 × −0.1987) = −92.4 + 139 = +46.7 kJ/mol (unfavorable) This is why Haber synthesis is paradoxical — favored at low T but kinetically slow; favored at high P (which K_p prefers) but unfavored at high T. The compromise: ~450 °C and 200 atm, with iron catalyst. **Non-standard conditions (any reaction quotient Q):** ΔG = ΔG° + RT ln(Q) At equilibrium: Q = K, so ΔG = 0 = ΔG° + RT ln(K), which gives ΔG° = −RT ln(K).

How to use this calculator

Enter the enthalpy change ΔH (kJ/mol) — positive for endothermic, negative for exothermic.
Enter the temperature in Kelvin (not Celsius). Room temp = 298.15 K.
Enter the entropy change ΔS (J/(mol·K)) — positive if disorder increases (e.g., solid → gas).
The calculator returns ΔG. Negative = spontaneous; positive = non-spontaneous; near zero = equilibrium-poised.
For predicting K: |ΔG°| > 30 kJ/mol → reaction effectively complete (K > 10⁵) or essentially blocked (K < 10⁻⁵).
Watch the sign of T: a reaction can switch from spontaneous to non-spontaneous as T changes (Haber example).

Worked examples

Combustion of methane

**Scenario:** CH₄ + 2O₂ → CO₂ + 2H₂O at 298 K. ΔH = −890 kJ/mol (exothermic). ΔS = −243 J/(mol·K) (decrease in entropy as 3 gas moles → 1 gas + 2 H₂O liquid). **Calculation:** ΔG = ΔH − TΔS = −890 − (298.15 × −0.243) = −890 + 72.4 = −817.6 kJ/mol. **Result:** Strongly negative ΔG → spontaneous and essentially irreversible. K = exp(817600/(8.314 × 298)) ≈ 10¹⁴³ — overwhelmingly product-favored. This is why methane burns completely once ignited; you can't "un-burn" it under normal conditions.

Ice melting at room temperature

**Scenario:** H₂O(s) → H₂O(l) at 298 K. ΔH = +6.0 kJ/mol (endothermic — ice has to absorb heat to melt). ΔS = +22 J/(mol·K) (liquid more disordered than crystal). **Calculation:** ΔG = 6.0 − (298.15 × 0.022) = 6.0 − 6.56 = −0.56 kJ/mol. **Result:** Slightly negative — melting is spontaneous at 298 K (room temperature). Crossover T = ΔH/ΔS = 6000/22 = 273 K = 0 °C — exactly the freezing point of water. Below 273 K, ΔG > 0 (freezing is spontaneous); above 273 K, ΔG < 0 (melting is spontaneous). The thermodynamics define the phase transition.

ATP hydrolysis in biochemistry

**Scenario:** ATP + H₂O → ADP + Pᵢ at 310 K (body temperature). Cellular conditions (not standard): ΔG ≈ −50 kJ/mol (vs ΔG° ≈ −30 kJ/mol at standard state). **Calculation:** Under cellular concentrations (ATP/ADP ratio ~10), the reaction is much more favorable than its standard value. The cell maintains high [ATP] / [ADP] ratio specifically to keep this reaction strongly product-favored. **Result:** ATP hydrolysis releases ~50 kJ/mol of usable energy in cells — the "energy currency" used to drive endergonic reactions (protein synthesis, muscle contraction, active transport). A typical adult uses about 50 kg of ATP per day, all of which is regenerated by oxidative phosphorylation.

When to use this calculator

**Use Gibbs free energy calculations to:**

- **Predict reaction spontaneity**: sign of ΔG tells whether a reaction proceeds in the forward direction. - **Compute equilibrium constants**: K = exp(−ΔG°/RT) at any temperature. - **Find the crossover temperature**: T at which a reaction switches between spontaneous and non-spontaneous. - **Design industrial processes**: Haber, methanol synthesis, ethanol production all use ΔG analysis to choose conditions. - **Biochemistry**: coupling reactions, metabolic pathway analysis, ATP-driven processes. - **Phase transitions**: predict whether a substance will melt, vaporize, sublime at given T. - **Electrochemistry**: ΔG = −nFE relates to cell voltage (used in Nernst equation). - **Geological / planetary processes**: mineral stability, magma equilibria, atmospheric chemistry.

**Practical interpretation guide:**

- **ΔG < −30 kJ/mol**: reaction essentially complete (>99.99% conversion) at equilibrium. - **−30 < ΔG < −10**: strongly product-favored but reversible (K = 100–10⁵). - **−10 < ΔG < +10**: significant amounts of both reactants and products at equilibrium. - **+10 < ΔG < +30**: strongly reactant-favored; minimal product forms. - **ΔG > +30**: essentially no reaction.

**Beware: ΔG describes equilibrium position, not rate.**

- A reaction with ΔG = −500 kJ/mol can still be vanishingly slow if the activation energy is high (diamond, gasoline at room T). - A reaction with ΔG = −5 kJ/mol can be very fast with a good catalyst. - Thermodynamics is necessary but not sufficient for a reaction to happen.

**Standard state conventions for ΔG°:**

- **Gases**: 1 atm partial pressure (or 1 bar in newer conventions). - **Solutions**: 1 M concentration. - **Pure solids/liquids**: their natural state. - **Temperature**: usually 298.15 K (25 °C) but ΔG° is reported at any T.

**Computing ΔG from formation energies:**

ΔG°_rxn = Σ(ν × ΔG°_f) products − Σ(ν × ΔG°_f) reactants

Where ν is stoichiometric coefficient and ΔG°_f is the standard Gibbs energy of formation (look up in tables for each species).

**Why divide ΔS by 1000:**

ΔH is typically in kJ/mol while ΔS is in J/(mol·K). The "TΔS" product needs to match ΔH's units; this calculator does the conversion automatically.

Common mistakes to avoid

Using Celsius instead of Kelvin. ΔG = ΔH − TΔS requires absolute temperature. Wrong T units flip the sign of TΔS contribution.
Mixing kJ/mol (ΔH) with J/(mol·K) (ΔS) without unit conversion. Always check that ΔH and TΔS have matching units before subtracting.
Confusing ΔG° (standard conditions) with ΔG (actual conditions). Only ΔG° relates simply to K; under non-standard concentrations use ΔG = ΔG° + RT ln(Q).
Predicting reaction speed from ΔG. ΔG only describes thermodynamics, not kinetics. A favorable reaction can still be infinitely slow without a catalyst.
Forgetting to update ΔG when T changes. ΔG changes both directly (the explicit T factor) and indirectly (ΔH and ΔS also depend on T, though usually weakly).
Using activities ≠ 1 without adjusting. Standard state values assume 1 M concentrations, 1 atm pressures. Real cellular or industrial conditions can be very different.
Misreading the sign convention. ΔG = G(products) − G(reactants). A negative value means products have lower G, so the reaction is favored.

Frequently Asked Questions

Sources & further reading

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