molecalculator

Reactions & Stoichiometry

Free Energy Equation: Gibbs, Helmholtz, and Examples

Calculator No. 05-0267 · Updated · Formula and assumptions shown below

Gibbs free-energy change ΔG -24.9075 kJ/mol
Temperature
298.15 K
= Temperature + 273.15
Enthalpy change ΔH
-10 kJ/mol
Set in the form below
Temperature
25 °C
Set in the form below
You enter
Enthalpy change ΔH (kJ/mol), Temperature (°C), Entropy change ΔS (J/(mol·K))
You get
Gibbs free-energy change ΔG (kJ/mol), Temperature (K)
Rounding
4 dp · 2 dp

For study and lab planning. The result follows from the numbers you enter and the formula shown. Check it against your method before you weigh, mix or dilute anything, follow the safety data sheet and your supervisor, and do not use it for clinical, dosing or medication work.

The free energy equation estimates how much energy in a system is available to do useful work under specified conditions. This calculator finds the Gibbs free-energy change, ΔG, from enthalpy change, ΔH, temperature, and entropy change, ΔS; enter ΔH in kJ/mol, temperature in °C, and ΔS in J/(mol·K), and it returns ΔG in kJ/mol. A negative ΔG indicates thermodynamic favorability for the stated process and conditions, not necessarily a fast reaction.

Formula

The usual equation meant by “the free energy equation” in introductory chemistry is the Gibbs free-energy relation:

ΔG = ΔH − TΔS

  • ΔG is the change in Gibbs free energy, in kJ/mol here.
  • ΔH is the enthalpy change, in kJ/mol.
  • T is absolute temperature, in kelvins (K). Convert Celsius with T = t + 273.15.
  • ΔS is the entropy change, in J/(mol·K). Divide the product TΔS by 1000 to convert joules to kilojoules.

Thus the calculator expression is ΔG = ΔH − [(t + 273.15) × ΔS / 1000]. The temperature conversion and the factor 1000 are essential: without them, the temperature may be on the wrong scale or the energy units will not match.

Worked Example

Using the calculator’s default values, ΔH = −10 kJ/mol, t = 25 °C, and ΔS = 50 J/(mol·K):

  1. Convert temperature: T = 25 + 273.15 = 298.15 K.
  2. Calculate the entropy term: TΔS = 298.15 × 50 = 14907.5 J/mol = 14.9075 kJ/mol.
  3. Substitute: ΔG = −10 − 14.9075 = −24.9075 kJ/mol.

The displayed result is −24.9075 kJ/mol (four decimal places). The negative sign means the forward process is thermodynamically favorable at 25 °C under the conditions represented by these thermodynamic changes.

Assumptions

  • The inputs are molar changes for the same process and use compatible sign conventions.
  • Temperature is converted to kelvins; enthalpy and entropy are treated as constant over the temperature range in this calculation.
  • The result is a thermodynamic criterion, not a prediction of reaction rate or a guarantee that a reaction will proceed measurably.

What “Free Energy” Means

Free energy is a thermodynamic state function used to assess the energy available for useful work under specified constraints. It is not a separate kind of energy that can be extracted without limits. The two common forms are Gibbs free energy, G = H − TS, and Helmholtz free energy, A = U − TS. Here H is enthalpy, U is internal energy, T is absolute temperature, and S is entropy.

Gibbs free energy is the relevant form for many chemical reactions and phase changes considered at constant temperature and pressure. Helmholtz free energy is especially useful for systems held at constant temperature and volume. In either case, a change in free energy—not simply the absolute value—is normally what is used to judge a process. Because the question gives no particular physical constraints, the calculator uses the Gibbs equation, the standard choice in general chemistry.

The equation originates in classical thermodynamics and is named for Josiah Willard Gibbs, who developed the thermodynamic potential now called Gibbs free energy. “Free energy equation” is an informal broad label, not the unique formal name of one equation: it can refer to the Gibbs relation, the Helmholtz relation, or equations connecting free energy with equilibrium. The page’s calculator specifies which one it uses.

How to Interpret ΔG

For a process at constant temperature and pressure, ΔG < 0 means the forward direction is thermodynamically favorable; ΔG > 0 means the forward direction is unfavorable as written; and ΔG = 0 describes equilibrium. The reverse process has the opposite sign. These statements assume the relevant conditions and reaction direction are clearly defined.

  • Negative ΔG: the process can proceed spontaneously in the forward direction in the thermodynamic sense.
  • Positive ΔG: the forward process requires a coupled process or a change in conditions to become favorable.
  • Zero ΔG: there is no net thermodynamic driving force at equilibrium.

“Spontaneous” does not mean “instantaneous.” A favorable reaction may be slow because it faces a large activation-energy barrier. A catalyst can change the rate by changing the pathway, but it does not change the overall ΔG between the same initial and final states.

Standard and Actual Free-Energy Changes

A key distinction is between standard free-energy change, ΔG°, and actual free-energy change, ΔG. The calculator’s direct equation works when the entered ΔH and ΔS describe the process at the stated temperature and conditions. Standard-state thermodynamic data are often used to calculate ΔG°; they do not automatically give the actual value in a mixture whose concentrations or pressures differ from standard states.

For a reaction mixture, the relation is ΔG = ΔG° + RT ln Q, where R = 8.31446261815324 J/(mol·K), T is in kelvins, and Q is the dimensionless reaction quotient formed using activities (often approximated with properly normalized concentrations or pressures in dilute, near-ideal cases). At equilibrium, Q = K and ΔG = 0, so ΔG° = −RT ln K. This explains why the same reaction may be favorable in one mixture and unfavorable in another.

Common Input and Interpretation Mistakes

  • Mixing units: ΔH and TΔS must use the same energy units. This calculator accepts ΔH in kJ/mol and converts TΔS from J/mol to kJ/mol by dividing by 1000.
  • Using Celsius directly: thermodynamic equations require absolute temperature in kelvins. Enter °C here; the calculator adds 273.15.
  • Ignoring signs: use signed ΔH and ΔS values for the process as written. A negative entropy change does not become positive just because it appears after a minus sign in the equation.
  • Applying the equation at the wrong conditions: ΔH and ΔS can vary with temperature, and actual ΔG depends on mixture composition. Values from a table may be standard-state values or tied to a different temperature.
  • Equating favorability with speed: ΔG gives a direction of thermodynamic driving force, not a reaction time.

For a more accurate calculation across a broad temperature range, temperature-dependent heat capacities and appropriate thermodynamic data may be needed. This simple form treats the supplied ΔH and ΔS as applicable at the temperature entered.

Frequently Asked Questions

What is the free energy equation in chemistry?
The commonly intended equation is ΔG = ΔH − TΔS, for Gibbs free-energy change. It is used for processes considered at constant temperature and pressure.

What does a negative ΔG mean?
It means the forward process is thermodynamically favorable under the stated conditions. It does not tell you how quickly the process occurs.

Why must temperature be in kelvins?
Temperature in thermodynamic equations is absolute temperature, measured from absolute zero. Convert Celsius to kelvins by adding 273.15.

Is free energy the same as enthalpy?
No. Gibbs free energy combines enthalpy and entropy through G = H − TS. A process’s favorability depends on both contributions and temperature.

When should I use Helmholtz free energy instead?
Use the Helmholtz form, A = U − TS, when analyzing a system at constant temperature and volume. Gibbs free energy is commonly used for constant-temperature, constant-pressure chemistry.

Related

Formula used
in dh | Enthalpy change ΔH | kJ/mol | -10 | -
in t | Temperature | °C | 25 | -273.15
in ds | Entropy change ΔS | J/(mol·K) | 50 | -
out dg | Gibbs free-energy change ΔG | kJ/mol | dh - ((t + 273.15) * ds / 1000) | 4
out tk | Temperature | K | t + 273.15 | 2

Each line is one input (in, sel, list) or one result (out: label, unit, formula, decimals). How the calculators work

Method: the standard formula for this calculation with the constants and conversion factors written on the page, rounded as each result states. See method and constants, the disclaimer or report an error.

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