What free energy answers that enthalpy cannot
Enthalpy tells you whether a reaction gives out heat. It does not tell you whether the reaction happens. Ice melts at room temperature even though melting absorbs heat; ammonium nitrate dissolves and gets cold; a hot metal block and its surroundings equalise even though no bonds change. Something other than energy is driving these.
That something is entropy — the number of ways energy and matter can be arranged. The second law says the total entropy of the universe never decreases, which sounds like it requires accounting for the surroundings as well as the system. Gibbs free energy is the bookkeeping trick that removes the surroundings from the problem. At constant temperature and pressure, the entropy change of the surroundings is just −ΔH/T, so the total entropy change of the universe is ΔS − ΔH/T. Multiply by −T and you get ΔH − TΔS, which is ΔG. The sign flips, so ΔG negative corresponds to total entropy increasing, which is the second law's criterion for a spontaneous change.
The practical payoff is that you can decide the direction of a process using only quantities measured on the system itself. That is why ΔG, and not entropy, is the quantity chemists actually work with.
ΔG also has a second meaning worth carrying: it is the maximum non-expansion work the process can deliver. A reaction with ΔG = −237 kJ/mol — water formation — can in principle drive an electrical load with 237 kJ of work per mole, which is exactly what a hydrogen fuel cell attempts to extract. That link between free energy and electrical work is made explicit by the Nernst equation.
Reading the two terms and the temperature that decides between them
The equation is a competition between two terms with different temperature behaviour. ΔH does not depend on temperature to a first approximation. TΔS grows in proportion to T. So at low temperature the enthalpy term dominates and at high temperature the entropy term does. Every temperature-driven change of chemical behaviour — why calcium carbonate decomposes in a kiln but not on a shelf, why proteins denature when heated, why liquids boil — traces back to that asymmetry.
Four cases exhaust the possibilities, and the crossover temperature exists in only two of them:
ΔH < 0 and ΔS > 0. Both terms favour the reaction. ΔG is negative at every temperature and there is no crossover. Combustion of most fuels sits here.
ΔH > 0 and ΔS < 0. Both terms oppose it. ΔG is positive at every temperature. The reverse reaction is the spontaneous one.
ΔH < 0 and ΔS < 0. Enthalpy favours, entropy opposes. The reaction is spontaneous below the crossover temperature ΔH/ΔS and non-spontaneous above it. Ammonia synthesis and condensation of a gas both belong here — which is why the Haber process runs at high pressure to compensate for the high temperature the rate demands.
ΔH > 0 and ΔS > 0. Entropy favours, enthalpy opposes. The reaction is spontaneous above the crossover and non-spontaneous below. Melting, boiling, and the decomposition of carbonates all sit here, and the crossover temperature is the melting point, the boiling point, or the calcination temperature.
The crossover temperature itself is where ΔG = 0, so ΔH = TΔS and T = ΔH/ΔS. For a phase change this is not a coincidence: the melting point of a solid is the temperature at which the free energies of solid and liquid are equal. Water's enthalpy of vaporisation is 44.0 kJ/mol and its entropy of vaporisation 118.9 J/(mol·K), giving 44000 ÷ 118.9 = 370 K — within a few kelvin of the boiling point, and the small discrepancy comes from using 25 °C values at 100 °C.
Two unit traps. ΔH is tabulated in kilojoules per mole and ΔS in joules per mole per kelvin, a factor of a thousand apart; forgetting to reconcile them produces an answer that is wrong by roughly that factor. And T must be in kelvin, because it multiplies rather than merely shifts.
Worked example: is ammonia synthesis spontaneous at room temperature and at 500 °C?
The Haber process is N₂(g) + 3 H₂(g) → 2 NH₃(g), with ΔH° = −92.2 kJ and ΔS° = −198.7 J/K for the equation as written. Four moles of gas become two, so the entropy drop is no surprise.
- Convert the temperature. 25 °C = 298.15 K.
- Form the entropy term. TΔS = 298.15 K × (−198.7 J/(mol·K)) = −59 242 J/mol = −59.24 kJ/mol.
- Subtract. ΔG = −92.2 − (−59.24) = −32.96 kJ/mol. Negative, so the reaction is thermodynamically favourable at room temperature. The published value is −32.8 kJ.
- Find the crossover. T = ΔH ÷ ΔS = −92 200 ÷ −198.7 = 464.0 K, or 190.9 °C. Above this temperature ΔG turns positive.
- Check 500 °C. T = 773.15 K, so TΔS = −153.63 kJ/mol and ΔG = −92.2 + 153.63 = +61.4 kJ/mol. Positive: at 500 °C the equilibrium strongly favours the reactants.
- Convert ΔG° to K at 298.15 K. −ΔG°/RT = 32 957.6 ÷ (8.3145 × 298.15) = 13.295, so K = e^13.295 ≈ 5.9 × 10⁵ — the reaction goes essentially to completion at equilibrium.
This is the central paradox of industrial ammonia. Thermodynamics wants a low temperature; kinetics demands a high one, because nitrogen's triple bond makes the reaction unmeasurably slow below a few hundred degrees even with an iron catalyst. Haber's resolution was to accept a compromise near 400–500 °C and to push the equilibrium back with pressures of 150–300 bar, which favours the side with fewer moles of gas. The Arrhenius equation quantifies the kinetic half of that trade-off.
How to read ΔG, and what “spontaneous” does not mean
The sign gives the direction; the magnitude gives the position of equilibrium. Because ΔG° = −RT ln K, the two are the same information in different units. At 298 K, every 5.71 kJ/mol of ΔG° corresponds to a factor of ten in K. So ΔG° = −5.7 kJ/mol means K = 10, ΔG° = −28.5 kJ/mol means K = 10⁵, and ΔG° = −57 kJ/mol means K = 10¹⁰ and the reaction is effectively complete. Beyond about ±60 kJ/mol the equilibrium is so lopsided that the reaction is best described as irreversible.
Three cautions about the word spontaneous.
Spontaneous does not mean fast. Diamond converting to graphite has ΔG° = −2.9 kJ/mol and the conversion is not observable in a human lifetime, because the activation barrier is enormous. Free energy tells you where the system is going, not how long it will take to arrive.
Spontaneous does not mean complete. A ΔG° of −2 kJ/mol gives K ≈ 2.2 at 298 K: at equilibrium a substantial fraction of reactant remains. Only large negative values imply near-complete conversion.
ΔG° is the standard value; your mixture is probably not standard. The standard state means every species at 1 M or 1 bar. Real reactions run away from that, and the working criterion is the non-standard ΔG = ΔG° + RT ln Q, which the optional field above computes. When Q reaches K, ΔG reaches zero and the reaction stops. That relation is also what the reaction quotient calculator uses to predict which way a mixture will shift.
A final caution on the temperature sweep in the table and chart above. It holds ΔH and ΔS constant with temperature, which is the standard approximation. It is good over a couple of hundred kelvin and degrades over a wider range, because heat capacities differ between products and reactants. Treat the crossover temperature as an estimate, accurate to perhaps a few tens of kelvin for a range far from 298 K.
ΔH, ΔS and ΔG at 298.15 K for reactions in each quadrant
| Reaction | ΔH° (kJ) | ΔS° (J/K) | ΔG° at 298 K (kJ) | Crossover (K) | Behaviour |
|---|---|---|---|---|---|
| CH₄ + 2O₂ → CO₂ + 2H₂O(l) | −890.5 | −242.9 | −818.1 | 3666 | Spontaneous below 3666 K |
| 2H₂ + O₂ → 2H₂O(l) | −571.6 | −326.7 | −474.2 | 1749 | Spontaneous below 1749 K |
| N₂ + 3H₂ → 2NH₃ | −92.2 | −198.7 | −33.0 | 464 | Spontaneous below 464 K |
| 2NO₂ → N₂O₄ | −57.2 | −175.8 | −4.8 | 325 | Spontaneous below 325 K |
| H₂O(l) → H₂O(g) | +44.0 | +118.9 | +8.6 | 370 | Spontaneous above 370 K |
| CaCO₃ → CaO + CO₂ | +178.3 | +160.5 | +130.4 | 1111 | Spontaneous above 1111 K |
Every crossover here is ΔH/ΔS with ΔS converted to kJ. The water row lands at 370 K against a true boiling point of 373 K; the 3 K gap is the error from using 298 K enthalpy and entropy values at 373 K.
Mistakes that flip the answer
- Mixing kilojoules and joules. ΔH is tabulated in kJ/mol, ΔS in J/(mol·K). Subtracting them without converting gives an answer wrong by about a factor of a thousand — and it usually still looks plausible.
- Using Celsius for T. T multiplies ΔS, so it must be absolute. At room temperature, using 25 instead of 298.15 shrinks the entropy term twelvefold.
- Reading a negative ΔG as a prediction of speed. Thermodynamics gives direction only. Diamond is thermodynamically unstable relative to graphite and does not visibly convert.
- Treating a small negative ΔG as complete conversion. ΔG° = −2 kJ/mol is K ≈ 2 at room temperature, which leaves plenty of reactant at equilibrium.
- Comparing ΔG° with a non-standard mixture. ΔG° assumes unit activity for every species. Use ΔG = ΔG° + RT ln Q for a real mixture, and remember ΔG reaches zero when Q reaches K.
- Extrapolating the crossover temperature too far. Holding ΔH and ΔS constant is an approximation that weakens as you move away from 298 K, so a crossover computed at 3000 K carries real uncertainty.
- Forgetting that ΔG is for the equation as written. Doubling every coefficient doubles ΔG and squares K.
Where the inputs come from and what to reach for next
ΔH and ΔS for a reaction are built the same way, as products minus reactants over tabulated standard values. Use the enthalpy of reaction calculator for the enthalpy half; standard molar entropies are tabulated alongside formation enthalpies and are combined identically. Note one asymmetry: standard entropies are absolute values measured from the third-law zero at 0 K, so an element in its standard state has a non-zero S° even though its ΔHf° is zero by definition.
If you measured your own ΔH in a calorimeter, the calorimetry calculator takes the temperature change to a molar enthalpy that you can bring straight here.
Downstream, ΔG° is the gateway to equilibrium. It gives K directly through ΔG° = −RT ln K, and from K you can compute equilibrium concentrations with the equilibrium constant calculator or judge the direction of an arbitrary mixture with the reaction quotient calculator. For an electrochemical reaction the same free energy appears as a cell voltage through ΔG = −nFE, which is where the Nernst equation calculator takes over. For solubility, the same ΔG governs Ksp and hence the solubility product.
What free energy will never tell you is rate. A reaction with ΔG° = −200 kJ/mol may sit inert for years behind a large activation barrier. Direction and speed are independent questions, answered by thermodynamics and kinetics respectively, and a complete picture of a process needs both.