Gibbs-Helmholtz Calculator

Calculate Gibbs free energy at any temperature from ΔH° and ΔS° using the Gibbs-Helmholtz equation.

Standard enthalpy ΔH° (kJ/mol) (kJ/mol)

Standard entropy ΔS° (J/mol·K) (J/mol·K)

Temperature T (K) (K)

Second temperature T₂ (K, for crossover) (K)

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The Four Thermodynamic Cases

Spontaneity is decided by ΔG, and the signs of ΔH and ΔS determine how it behaves with temperature. There are exactly four combinations, and knowing which one applies tells you immediately whether temperature can be used to make a reaction go.

ΔG = ΔH − TΔS

ΔH	ΔS	Spontaneous when	Example
Negative	Positive	Always — at all temperatures	Combustion of hydrocarbons
Negative	Negative	Low temperature only	Ammonia synthesis, freezing
Positive	Positive	High temperature only	Melting, evaporation, thermal decomposition
Positive	Negative	Never — at any temperature	Photosynthesis without light input

The two middle cases have a crossover temperature where ΔG passes through zero. Setting ΔG = 0 gives T = ΔH/ΔS, the temperature at which the reaction switches between spontaneous and non-spontaneous.

T_crossover = ΔH / ΔS

For a phase transition this crossover is the transition temperature. Ice melting has ΔH = +6.01 kJ/mol and ΔS = +22.0 J/(mol·K), giving 273 K — exactly 0°C. That is not a coincidence: at the melting point solid and liquid are in equilibrium, which is precisely the condition ΔG = 0.

Watch the Units

ΔH is almost always tabulated in kJ/mol while ΔS is in J/(mol·K). Multiplying TΔS without converting produces an answer 1,000 times too large, and it is the single most common error in this calculation. Convert ΔS to kJ, or ΔH to J, before combining.

Worked Examples

Example 1: N2+3H2→2NH3: ΔH=-92.4kJ, ΔS=-198 J/K

ΔG=−92.4−298×(−0.198)

Result: ΔG=−92.4+59.0=−33.4 kJ/mol at 25°C

Spontaneous at 25°C; crossover at 467K

Example 2: Ice melting: ΔH=+6.01kJ, ΔS=+22 J/K

Crossover T=6010/22

Result: T=273K=0°C — exactly the melting point!

ΔG=0 at phase equilibrium

Example 3: Ammonia synthesis crossover

ΔH = −92.4 kJ/mol, ΔS = −198 J/(mol·K)

Result: T_crossover = 467 K (194°C)

Above 467 K the reaction becomes non-spontaneous. Yet the Haber process runs near 450°C — far above it — because rate demands it, and the loss of equilibrium yield is offset by high pressure.

Example 4: Thermal decomposition

CaCO₃ → CaO + CO₂: ΔH = +178 kJ/mol, ΔS = +161 J/(mol·K)

Result: T_crossover = 1,106 K (833°C)

Non-spontaneous at room temperature but spontaneous above about 833°C, which is why lime kilns operate around 900°C.

Example 5: Always spontaneous

Combustion: ΔH strongly negative, ΔS positive

Result: No crossover exists

Both terms favour products at every temperature, so ΔG is negative throughout. Such reactions still need activation energy to start, which is why fuel does not ignite spontaneously.

Common Mistakes

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Mixing kJ and J

ΔH comes in kJ/mol and ΔS in J/(mol·K). Always convert one before computing TΔS — forgetting gives an answer off by a factor of 1,000.

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Using Celsius for temperature

TΔS requires absolute temperature. Using Celsius gives a meaningless product, and at temperatures below 0°C it even flips the sign.

⚠️

Assuming exothermic means spontaneous

Enthalpy is only half the picture. An exothermic reaction with a large negative entropy change becomes non-spontaneous above its crossover temperature.

⚠️

Treating ΔH and ΔS as temperature independent

They vary somewhat with temperature. The approximation is reasonable over modest ranges but degrades over hundreds of kelvin, making distant crossover predictions unreliable.

Frequently Asked Questions

Four thermodynamic cases?

ΔH<0, ΔS>0: always spontaneous. ΔH>0, ΔS<0: never spontaneous. ΔH<0, ΔS<0: spontaneous at low T. ΔH>0, ΔS>0: spontaneous at high T. Crossover T = ΔH/ΔS.

Why is ΔS negative for NH3 synthesis?

4 moles of gas (N2+3H2) become 2 moles (2NH3). Fewer gas molecules = lower entropy. High pressure in Haber process uses Le Chatelier to push equilibrium toward NH3 despite unfavorable entropy.

What are the four thermodynamic cases?

Negative ΔH with positive ΔS is always spontaneous; positive ΔH with negative ΔS never is. The two mixed cases are spontaneous only below or only above a crossover temperature.

How do I find the crossover temperature?

Set ΔG = 0 and solve, giving T = ΔH/ΔS. Remember to convert ΔS from J to kJ, or the answer will be wrong by a factor of 1,000.

Why is the crossover for ice melting exactly 0°C?

Because at the melting point solid and liquid coexist in equilibrium, which is the definition of ΔG = 0. The crossover temperature of a phase change is the transition temperature.

Why is ΔS negative for ammonia synthesis?

Four moles of gas become two, so the system becomes more ordered. Fewer gas molecules means substantially lower entropy.

Does a negative ΔG mean the reaction will be fast?

No. ΔG determines whether a reaction can proceed, not how quickly. Many spontaneous reactions are kinetically blocked by a high activation barrier.

Formula Explorer connections

Interpretation: This relationship tracks energy transfer, state-function change or the balance between enthalpy and entropy in a chemical process. Assumption: Keep energy units compatible, use kelvin for absolute temperature, and match standard states and reaction stoichiometry. Thermodynamic favorability does not determine reaction speed.

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