Gibbs Energy Calculator (ΔG = ΔH − TΔS)
ΔG, ΔH, ΔS or T from the other three; the crossover temperature; ΔG° from tables; K; Q.
Try before you buy.
- Free preview: the Gibbs energy ΔG, or the value you solve for in full, with the first rows of each table, the first step and a watermarked chart; other figures are hidden.
- Locked until you unlock it: download and copy.
- Unlock: Pro pass, ₹179 for 30 days or ₹79 for 1 week, a one-time payment that never renews. Unlock: Pro pass, $5.99 for 30 days or $2.99 for 1 week, a one-time payment that never renews.
Ways to unlock shows how to get the full result.
Printing this result is locked in the free preview.
Result
When is it spontaneous?
Values used
ΔG at other temperatures
ΔG against temperature
The line is ΔH − TΔS; it crosses ΔG = 0 at T*.
How it was worked out
Look up a substance in the data table
| Substance | ΔfH° kJ/mol | ΔfG° kJ/mol | S° J/(mol·K) | Source |
|---|
Locked in the free preview. Opens the ways to unlock this result.
Locked in the free preview. Batch runs unlock with a pass.
Locked in the free preview. Query results unlock with a pass.
About the Gibbs Energy Calculator (ΔG = ΔH − TΔS)
Four calculations around the Gibbs energy, G = H − TS. ΔG = ΔH − TΔS: type any three of ΔG, ΔH, ΔS and T and get the fourth, with the units sorted out for you (kJ against J is the classic slip), whether the reaction is spontaneous, the crossover temperature T* = ΔH/ΔS where it changes, a table and a chart of ΔG at other temperatures, and the four-case map of the signs of ΔH and ΔS. From a reaction: type the equation and ΔrH°, ΔrS° and ΔrG° come from a table of 414 standard formation values (elements, compounds, ions in water, organic compounds), each value editable, with K and ΔrG at another temperature. ΔG° and K: ΔG° = −RT ln K both ways, with K far beyond the range of a calculator (10³⁰⁰ and more). With Q: ΔG = ΔG° + RT ln Q tells which way a mixture that is not at equilibrium will go.
Without a pass, the free preview shows the Gibbs energy ΔG, or the value you solve for in full, with the first rows of each table, the first step and a watermarked chart; other figures are hidden; with a Pro pass, or after unlocking this result, you can copy the full answer or download it as CSV.
How to use it
- Choose ΔH, ΔS and T, From a reaction, ΔG° and K or With Q.
- For ΔG = ΔH − TΔS, pick what to solve for and type the other three, each in its own unit: kJ/mol or J/mol, J/(mol·K) or kJ/(mol·K), K, °C or °F.
- For a reaction, type the equation, for example
CaCO3(s) -> CaO(s) + CO2(g); it is balanced for you. Check each substance’s state and values, or type your own. - Read the answer in the dark box and the verdict under it: ΔG < 0 is spontaneous (exergonic), ΔG > 0 is not, and the reverse reaction is.
- Look at T* and the table to see the temperature where the answer flips.
- Copy the result or download it as CSV — with a Pro pass, or after unlocking this result.
Examples
ΔH = −91.88 kJ/mol, ΔS = −198.11 J/(mol·K), T = 298.15 K
ΔG = −32.81 kJ/mol: spontaneous; below T* = 463.8 K only
Both terms negative: enthalpy-driven, so it stops being spontaneous when hot.
ΔH = 178.3 kJ/mol, ΔS = 160.5 J/(mol·K), solve T for ΔG = 0
T = 1110.9 K (837.8 °C): above it calcium carbonate decomposes
CaCO3(s) -> CaO(s) + CO2(g)
ΔrG° = +131.12 kJ/mol, ΔrH° = +178.49 kJ/mol, ΔrS° = +158.99 J/(mol·K); T* = 1122.7 K
Ag^+(aq) + Cl^-(aq) -> AgCl(s)
ΔrG° = −55.64 kJ/mol, K = 5.59 × 10⁹, so Ksp = 1/K = 1.79 × 10⁻¹⁰
K = 1.8 × 10⁻⁵ (acetic acid) at 298.15 K
ΔG° = +27.08 kJ/mol: reactants favoured
ΔG° = −32.82 kJ/mol, Q = 10⁶ at 298.15 K
ΔG = +1.43 kJ/mol: Q > K, so the reaction runs in reverse
Common uses
- Thermodynamics homework: ΔG, ΔH, ΔS or T from the other three, with the units converted and every step shown.
- Finding the temperature where a reaction becomes spontaneous, such as the decomposition of limestone.
- Turning ΔG° into an equilibrium constant (or a Ksp or Ka into ΔG°) and back.
- Checking which way a reaction mixture will move from its reaction quotient.
The four cases
- ΔH < 0 and ΔS > 0: ΔG is negative at every temperature, so the reaction is always spontaneous.
- ΔH < 0 and ΔS < 0: ΔG is negative when cold: spontaneous below T* = ΔH/ΔS (enthalpy-driven).
- ΔH > 0 and ΔS > 0: ΔG is negative when hot: spontaneous above T* = ΔH/ΔS (entropy-driven).
- ΔH > 0 and ΔS < 0: ΔG is positive at every temperature: never spontaneous, and the reverse reaction always is.
At T* itself ΔG = 0: the reaction is at equilibrium with every substance in its standard state. Use the same energy unit for both terms: ΔS in J/(mol·K) is divided by 1000 before TΔS is subtracted from ΔH in kJ/mol.
ΔG°, K and Q
- ΔG° = −RT ln K. The standard Gibbs energy of reaction fixes the equilibrium constant at that temperature: ΔG° < 0 gives K > 1 (products favoured), ΔG° > 0 gives K < 1. Every 5.7 kJ/mol at 25 °C is a factor of 10 in K.
- ΔG = ΔG° + RT ln Q = RT ln(Q/K). Q has the form of K but uses the present pressures (in bar) or concentrations (in mol/L). Q < K: ΔG < 0, the reaction runs forward; Q > K: it runs backwards; Q = K: equilibrium, ΔG = 0.
- R = 8.314 462 618 J/(mol·K), exact since the SI was redefined (it is the Avogadro constant times the Boltzmann constant).
- K here is the thermodynamic constant: dimensionless, with pressures relative to 1 bar and concentrations relative to 1 mol/L.
About the data
- 414 entries at 298.15 K and 1 bar: ΔfH° and ΔfG° in kJ/mol and S° in J/(mol·K), each with its source — the same table as the reaction enthalpy calculator.
- CODATA Key Values for Thermodynamics (ΔfG° worked out from ΔfH° and S°), the NIST-JANAF tables, the NBS tables (every entry with all three values checked against ΔfG° = ΔfH° − TΔfS°) and the NIST Chemistry WebBook.
- An element in its reference state has ΔfH° = ΔfG° = 0 by definition, but its entropy S° is not zero. Ions in water use the convention that H⁺(aq) has zero for all three.
- Any two of the reaction’s sums are enough: when an S° is missing, ΔrS° comes from (ΔrH° − ΔrG°)/T.
Sources
- IUPAC, Quantities, Units and Symbols in Physical Chemistry (the Green Book), 3rd edition: G = H − TS, ΔrG° = −RT ln K°
- NIST, molar gas constant R (CODATA recommended values)
- CODATA, Key Values for Thermodynamics
- NIST, NIST-JANAF Thermochemical Tables
- Wagman et al., The NBS Tables of Chemical Thermodynamic Properties, J. Phys. Chem. Ref. Data 11, Supplement 2
- NIST, Chemistry WebBook
Limitations
- ΔH and ΔS are taken as constant with temperature, so ΔG far from 298 K is an estimate; heat capacities would refine it.
- Spontaneous means thermodynamically favourable, not fast: the rate is a question of kinetics.
- Table values are for the standard state (1 bar, 1 mol/L, infinite dilution for ions); real solutions and high pressures need activities.
- Substances that are not in the table need their ΔfH°, ΔfG° or S° typed in.
Privacy
On this device: nothing is sent
Everything happens in your browser. What you enter or open here is not uploaded or stored by MySmartCoPilot.
Frequently asked questions
What do I get without a pass?
Without a pass, Gibbs Energy Calculator (ΔG = ΔH − TΔS) shows the Gibbs energy ΔG, or the value you solve for in full, with the first rows of each table, the first step and a watermarked chart; other figures are hidden. Until you unlock it, the result can’t be downloaded or copied. A Pro, Premium or Ultimate pass, a one-time payment that never renews, unlocks the full result. The pricing page lists the passes and their prices.
How do I know if a reaction is spontaneous?
Work out ΔG = ΔH − TΔS at the temperature you care about. A negative ΔG means spontaneous (exergonic); a positive ΔG means not spontaneous, and the reverse reaction is; ΔG = 0 is equilibrium.
Why do I divide ΔS by 1000?
ΔH is usually in kJ/mol and ΔS in J/(mol·K). TΔS must be in kJ/mol before you subtract it from ΔH, so ΔS is divided by 1000 (or ΔH multiplied by 1000). Mixing them is the most common error, and the calculator warns when a value looks like the wrong unit.
What is the crossover temperature?
When ΔH and ΔS have the same sign, ΔG changes sign at T* = ΔH/ΔS (with ΔS in kJ/(mol·K)). For CaCO₃ → CaO + CO₂, T* is about 1110 K: the decomposition is spontaneous above it.
How do I get K from ΔG°?
K = e^(−ΔG°/RT), with ΔG° in J/mol, R = 8.314 J/(mol·K) and T in kelvins. At 25 °C, ΔG° = −32.82 kJ/mol gives K = e^13.24 = 5.6 × 10⁵.
What is the difference between ΔG and ΔG°?
ΔG° is for every substance in its standard state (1 bar for gases, 1 mol/L in solution). ΔG is for the mixture you actually have: ΔG = ΔG° + RT ln Q. At equilibrium ΔG = 0 while ΔG° keeps its value.
Is ΔG° = ΔH° − TΔS° valid at any temperature?
Exactly, at the temperature where ΔH° and ΔS° apply. Using 298.15 K values at another temperature assumes they do not change, which is a fair estimate near room temperature and rougher far from it.