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Gibbs Energy Calculator (ΔG = ΔH − TΔS)

ΔG, ΔH, ΔS or T from the other three; the crossover temperature; ΔG° from tables; K; Q.

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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.

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  • 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.
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Result

ΔG —

When is it spontaneous?

AlwaysΔH < 0, ΔS > 0: spontaneous at every temperature
When coldΔH < 0, ΔS < 0: spontaneous below T*
When hotΔH > 0, ΔS > 0: spontaneous above T*
NeverΔH > 0, ΔS < 0: never spontaneous; the reverse always is

Values used

ΔG at other temperatures

ΔG against temperature

The line is ΔH − TΔS; it crosses ΔG = 0 at T*.

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    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

    1. Choose ΔH, ΔS and T, From a reaction, ΔG° and K or With Q.
    2. 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.
    3. 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.
    4. 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.
    5. Look at T* and the table to see the temperature where the answer flips.
    6. Copy the result or download it as CSV — with a Pro pass, or after unlocking this result.

    Examples

    Ammonia at 25 °C
    Input
    ΔH = −91.88 kJ/mol, ΔS = −198.11 J/(mol·K), T = 298.15 K
    Result
    ΔG = −32.81 kJ/mol: spontaneous; below T* = 463.8 K only

    Both terms negative: enthalpy-driven, so it stops being spontaneous when hot.

    When does a lime kiln work?
    Input
    ΔH = 178.3 kJ/mol, ΔS = 160.5 J/(mol·K), solve T for ΔG = 0
    Result
    T = 1110.9 K (837.8 °C): above it calcium carbonate decomposes
    From the tables
    Input
    CaCO3(s) -> CaO(s) + CO2(g)
    Result
    ΔrG° = +131.12 kJ/mol, ΔrH° = +178.49 kJ/mol, ΔrS° = +158.99 J/(mol·K); T* = 1122.7 K
    Silver chloride
    Input
    Ag^+(aq) + Cl^-(aq) -> AgCl(s)
    Result
    ΔrG° = −55.64 kJ/mol, K = 5.59 × 10⁹, so Ksp = 1/K = 1.79 × 10⁻¹⁰
    K to ΔG°
    Input
    K = 1.8 × 10⁻⁵ (acetic acid) at 298.15 K
    Result
    ΔG° = +27.08 kJ/mol: reactants favoured
    Which way will it go?
    Input
    ΔG° = −32.82 kJ/mol, Q = 10⁶ at 298.15 K
    Result
    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

    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.

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    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.

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