Gibbs Free Energy Calculator

Enter your chemistry values to calculate Gibbs free-energy change and see whether the reaction is thermodynamically favorable under those conditions.

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How to use our Gibbs Free Energy Calculator

  1. Choose the quantity your problem asks for in "What do you want to calculate?" Use "Delta G from Delta H, Delta S, and temperature" for the usual Gibbs free-energy equation.
  2. Enter only the fields shown for that choice. A bare Delta H or Delta G value uses kJ/mol, a bare Delta S value uses J/(mol K), and a bare temperature uses K.
  3. Add a supported suffix when needed, such as "25 C", "-100000 J/mol", or "-0.2 kJ/(mol K)". For K and Q, enter a finite number greater than 0.
  4. Click "Calculate." Check that the displayed Kelvin temperature, signs, and units match your problem before using the answer.
Example inputs for Gibbs Free Energy Calculator
Example inputs for Gibbs Free Energy Calculator

Definitions

Gibbs free-energy change, Delta G: Energy change per mole for a reaction under stated conditions. Its sign describes thermodynamic favorability for the reaction direction as written.

Delta H: Enthalpy change, a heat-related energy change at constant pressure, usually reported in kJ/mol.

Delta S: Entropy change per mole per kelvin, usually reported in J/(mol K).

Absolute temperature: Temperature on the Kelvin scale. Gibbs free-energy equations require kelvins, not Celsius.

Standard Delta G: Gibbs free-energy change under the standard-state conditions used for the problem.

Equilibrium constant, K: A positive dimensionless quantity for a reaction at equilibrium.

Reaction quotient, Q: A positive dimensionless quantity that describes the current mixture using the same reaction definition as K.


Common mistakes and quick fixes

Mistake: Typing 25 for a temperature that was given as 25 C.
Fix: Type "25 C" so it converts to 298.15 K, or enter 298.15 directly.

Mistake: Mixing kJ/mol with J/(mol K).
Fix: Follow the units in each label. You can type a suffix such as J/mol or kJ/(mol K) instead of converting it yourself.

Mistake: Using K where the problem gives Q, or using Q where it gives K.
Fix: Choose "Standard Delta G from K" for an equilibrium constant and "Delta G from standard Delta G and Q" for current conditions.

Mistake: Entering 0 or a negative value for K or Q.
Fix: Use a finite value greater than 0 because the calculation takes the natural logarithm of K or Q.

Mistake: Treating a negative Delta G as proof that a reaction is fast.
Fix: A negative value means the forward direction is thermodynamically favorable under the stated conditions; reaction rate depends on kinetics.


Limitations & Key Assumptions / Boundary Conditions

  • The result applies to the reaction direction, temperature, composition, and values entered.
  • A negative Delta G means the forward reaction is thermodynamically favorable under the stated conditions; it does not predict reaction speed, mechanism, or completion time.
  • The direct equation uses the entered Delta H and Delta S at the entered temperature. In real systems, these quantities can change with temperature or phase.
  • K and Q must be positive, dimensionless values for the reaction as written. Replacing activities with uncorrected concentrations or pressures can change the result.
  • The temperature-solving path is undefined when Delta S is zero and accepts only a calculated temperature above 0 K.
  • Rounded worksheet or table values can make a result near zero appear slightly positive or negative.

Methodology

Equations used

The calculator converts Delta H and Delta G to kJ/mol, Delta S to kJ/(mol K), and temperature to K before using the Gibbs free-energy relationship. [1]

ΔG = ΔH - TΔS

Here, T is absolute temperature in K. The calculator also shows the signed T Delta S contribution so you can check the subtraction.

ΔH = ΔG + TΔS

ΔS = 1000 x (ΔH - ΔG) / T

T = (ΔH - ΔG) / ΔS

The factor 1000 in the Delta S equation changes kJ to J. Solving for temperature requires Delta S to be nonzero, and the result must be above 0 K.

K and Q calculations

For K and Q, the calculator uses the molar gas constant R = 8.31446261815324 J/(mol K). [2] ln means natural logarithm.

ΔG° = -RT ln(K) / 1000

ΔG = ΔG° + RT ln(Q) / 1000

Dividing by 1000 changes J/mol to kJ/mol. K and Q must both be greater than 0.

Worked example

For Delta H = -100 kJ/mol, Delta S = -200 J/(mol K), and 25 C, the calculator changes the temperature to 298.15 K and Delta S to -0.2 kJ/(mol K).

TΔS = 298.15 x -0.2 = -59.63 kJ/mol

ΔG = -100 - (-59.63) = -40.37 kJ/mol

The negative Delta G means the forward reaction is thermodynamically favorable for those entered conditions.


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