Gibbs Free Energy Calculator (ΔG = ΔH − TΔS)

Gibbs Free Energy Guide

What do I need to know about ΔG = ΔH − TΔS Explained?

Gibbs free energy combines enthalpy (ΔH) and entropy (ΔS) into a single spontaneity criterion. ΔG < 0: spontaneous (thermodynamically favourable). ΔG = 0: at equilibrium. ΔG > 0: non-spontaneous (reverse reaction is spontaneous). ΔH < 0 (exothermic) and ΔS > 0: always spontaneous (ΔG always negative). ΔH > 0 (endothermic) and ΔS < 0: never spontaneous (ΔG always positive). ΔH < 0 and ΔS < 0: spontaneous at low temperature (enthalpy drives). ΔH > 0 and ΔS > 0: spontaneous at high temperature (entropy drives). This is why some reactions that don't occur at room temperature — like the thermal decomposition of calcium carbonate to make lime — become spontaneous only above a specific temperature, the point at which the TΔS term grows large enough to overcome a positive ΔH.

What should I know about The Crossover Temperature?

When ΔH and ΔS have the same sign, there is a crossover temperature: T_crossover = ΔH / ΔS. Above T_crossover: TΔS > ΔH → entropy term dominates. Below T_crossover: ΔH > TΔS → enthalpy term dominates. Example: decomposition of limestone CaCO₃ → CaO + CO₂. ΔH = +178 kJ/mol (endothermic). ΔS = +165 J/mol·K (positive — produces gas from solid). T_crossover = 178,000/165 = 1,079 K = 806°C. Below 806°C: not spontaneous. Above 806°C: spontaneous. This is why lime kilns operate at 900°C.

What's the key thing to understand about ΔG and Equilibrium Constant?

Standard Gibbs energy and K are directly related: ΔG° = −RT ln(K). R = 8.314 J/mol·K. At 25°C (298K): ΔG° = −2.479 × ln(K) kJ/mol. ΔG° = −5.708 × log₁₀(K) kJ/mol. K > 1 (products favoured): ΔG° < 0. K < 1 (reactants favoured): ΔG° > 0. K = 1: ΔG° = 0. For water formation: H₂ + ½O₂ → H₂O. ΔG° = −237 kJ/mol. K = e^(237,000/8.314/298) ≈ 10^41 — essentially irreversible under standard conditions.

What do I need to know about Entropy Changes in Reactions?

Standard molar entropy values allow ΔS calculation: ΔS°reaction = Σ S°products − Σ S°reactants. Rules of thumb for entropy changes: gas produced: large positive ΔS. Gas consumed: large negative ΔS. Dissolution of solids in water: usually positive ΔS (more disorder). Precipitation: negative ΔS. Increase in moles of gas: positive ΔS per extra mole (~130-200 J/mol·K). Temperature always increases entropy. The second law of thermodynamics: the entropy of the universe always increases in any spontaneous process, even if entropy decreases locally within the reaction itself.

What should I know about a Worked Spontaneity Example?

Consider the melting of ice: H2O(s) to H2O(l), delta-H = +6.01 kJ/mol (endothermic), delta-S = +22.0 J/mol·K (positive — liquid is more disordered than solid). At 273K: delta-G = delta-H - T x delta-S = 6010 - (273 x 22.0) = 6010 - 6006 = +4 J/mol, essentially zero — ice and water are in equilibrium at 0°C, exactly as observed. Above 273K: T x delta-S exceeds delta-H, delta-G becomes negative, melting is spontaneous (ice melts above 0°C). Below 273K: delta-H exceeds T x delta-S, delta-G is positive, melting is non-spontaneous (water freezes below 0°C). This example shows why delta-G = 0 identifies a phase-transition temperature precisely, and shows that 'spontaneous' means thermodynamically favoured, not fast — melting ice and diamond slowly converting to graphite are both delta-G negative, but proceed at wildly different rates because of very different activation energies.

What do I need to know about Gibbs Free Energy in Biology?

Living organisms couple thermodynamically unfavourable reactions to favourable ones so the overall process has a negative delta-G. ATP hydrolysis releases roughly -30.5 kJ/mol under cellular conditions, and this is coupled to otherwise non-spontaneous biological reactions, such as active transport against a concentration gradient or synthesis of large molecules, so the combined delta-G of the coupled pair is negative overall. Photosynthesis is the reverse: converting low-energy CO2 and H2O into high-energy glucose has delta-G° = +2,870 kJ/mol per mole of glucose (strongly non-spontaneous), only possible because light energy is captured and supplied to drive it. Metabolic pathways are sequences of individually small, often near-equilibrium steps rather than one large reaction, letting cells regulate flux precisely at specific control points — it is the sum of delta-G across the whole pathway, not any single step, that must be negative for the overall process to proceed.

What's the key thing to understand about ΔG, K, and the Reaction Quotient Q?

The full relationship between Gibbs free energy and a reaction's progress is ΔG = ΔG° + RT ln Q, where Q is the reaction quotient describing the current (not necessarily equilibrium) concentrations or pressures. At equilibrium, Q equals the equilibrium constant K and, by definition, ΔG = 0 — substituting this condition gives the well-known ΔG° = −RT ln K, linking a reaction's standard free energy change directly to how far its equilibrium lies toward products (large positive K, strongly negative ΔG°) or reactants (small K, positive ΔG°).

What should I know about Standard States and ΔGf°?

Standard Gibbs free energy of formation, ΔGf°, is defined for a compound as the free energy change when it forms from its elements in their standard states — the most stable physical form of each element at 1 bar pressure and a specified temperature, usually 298K, such as graphite rather than diamond for carbon, O2 gas for oxygen, and solid Na for sodium. By convention, every element in its standard state is assigned ΔGf° = 0, which gives a zero baseline that lets ΔG° for any reaction be calculated as the difference between the sum of ΔGf° of products and the sum of ΔGf° of reactants, without needing to know absolute free energy values, which don't exist in any measurable sense. This mirrors how enthalpy of formation is used in Hess's law calculations, and the two can be combined: tabulated ΔHf° and S° values for each species are used to calculate ΔG° for virtually any reaction, even ones never performed in a lab, as long as thermodynamic data for all the individual compounds exists. This is why large tables of standard formation data for thousands of compounds are maintained and widely used in chemistry.

Worked Example: Spontaneity Check at a Given Temperature

A reaction has ΔH = -120 kJ/mol and ΔS = -150 J/(mol·K) at T = 298 K. Using ΔG = ΔH - TΔS, with units carefully converted (ΔS in kJ: -0.150 kJ/(mol·K)): ΔG = -120 - (298 × -0.150) = -120 - (-44.7) = -120 + 44.7 = -75.3 kJ/mol Since ΔG is negative, the reaction is spontaneous at 298 K despite the entropy decrease, because the large negative enthalpy dominates at this temperature. Find the crossover temperature where ΔG becomes zero (the reaction stops being spontaneous): T = ΔH/ΔS = -120 / -0.150 = 800 K. Above 800 K, the entropy term (-TΔS, which becomes increasingly positive as T rises since ΔS is negative) would outweigh the favourable enthalpy, making ΔG positive and the reaction non-spontaneous in the forward direction.

Sign Combinations of ΔH and ΔS: Four Scenarios

This four-case table is worth memorising because it lets you predict spontaneity qualitatively without doing any arithmetic at all, just from the signs of ΔH and ΔS — ice melting (ΔH positive, absorbs heat; ΔS positive, more disorder as solid becomes liquid) is a textbook example of the third row, spontaneous only above a crossover temperature (0°C at standard pressure), which is exactly the melting point.

Worked Example: ΔG From the Equilibrium Constant

A reaction has an equilibrium constant K = 100 at T = 298 K. Using ΔG° = -RT·ln(K) with R = 8.314 J/(mol·K): ΔG° = -8.314 × 298 × ln(100) = -8.314 × 298 × 4.605 ≈ -11,410 J/mol ≈ -11.4 kJ/mol The negative ΔG° confirms that a reaction with K > 1 (products favoured at equilibrium) is indeed spontaneous under standard conditions, consistent with the direct relationship between the sign of ΔG° and whether K is greater than or less than 1. If K had been exactly 1 (equal amounts of products and reactants at equilibrium), ln(1) = 0 and ΔG° would be exactly zero — the system would already be at equilibrium with no net driving force in either direction.