Calculator Inputs
Use one calculation mode at a time. Matching fields appear automatically.
Plotly Graph
The graph updates after each calculation and highlights how the selected variable changes free energy.
Example Data Table
| Scenario | Sample Inputs | Example Output | Quick Reading |
|---|---|---|---|
| ΔH° and ΔS° method | ΔH° = -95 kJ/mol, ΔS° = -120 J/mol·K, T = 298.15 K | ΔG° ≈ -59.22 kJ/mol | Spontaneous at standard conditions. |
| Equilibrium constant method | K = 4.20 × 105, T = 298.15 K | ΔG° ≈ -32.09 kJ/mol | Products are strongly favored. |
| Cell potential method | n = 2, E° = 1.10 V, T = 298.15 K | ΔG° ≈ -212.27 kJ/mol | Forward redox reaction is favorable. |
| Nonstandard method | ΔG° = -20 kJ/mol, Q = 50, T = 298.15 K | ΔG ≈ -10.30 kJ/mol | Still spontaneous forward. |
| Formation data method | Products: 1×(-350), 2×(-120); Reactants: 1×(-80), 1×(-60) | ΔG°rxn = -450 kJ/mol | Products dominate the standard balance. |
Formula Used
From enthalpy and entropy
ΔG° = ΔH° − TΔS°. Convert entropy to kJ when needed before combining terms.
From equilibrium constant
ΔG° = −RT ln K. Use the natural logarithm and absolute temperature in Kelvin.
From cell potential
ΔG° = −nFE°. Multiply electrons transferred by Faraday constant and standard cell potential.
From nonstandard conditions
ΔG = ΔG° + RT ln Q. This corrects the standard value using the reaction quotient.
From formation free energies
ΔG°rxn = ΣνΔGf°(products) − ΣνΔGf°(reactants). Multiply each formation value by its stoichiometric coefficient before summing.
How to Use This Calculator
- Choose the calculation mode that matches your known chemistry values.
- Enter temperature in Kelvin and confirm all signs and units.
- Fill only the fields shown for your selected mode.
- Click Calculate to place the result above the form.
- Review the table, interpretation line, and graph trend.
- Use the CSV or PDF buttons to save the output.
- For formation data, keep coefficient and energy lists aligned carefully.
Frequently Asked Questions
1. What does a negative standard free energy change mean?
A negative value means the forward reaction is thermodynamically favorable under standard conditions. It suggests the products are lower in free energy than the reactants.
2. Why must temperature be entered in Kelvin?
Thermodynamic equations require absolute temperature. Kelvin keeps the energy relationships consistent and avoids sign or scaling errors found with Celsius values.
3. When should I use the equilibrium constant mode?
Use it when K is known from equilibrium data. This method is useful for linking thermodynamics with reaction extent and product favorability.
4. What is the difference between ΔG and ΔG°?
ΔG° applies to standard-state conditions. ΔG adjusts that value for current concentrations or pressures through the reaction quotient Q.
5. Why does the calculator estimate K from ΔG°?
The relationship between ΔG° and K lets you convert energy into equilibrium behavior. It helps explain whether a reaction strongly favors products or reactants.
6. Can this calculator handle formation free energies?
Yes. Enter matching coefficient and formation-energy lists for products and reactants. The calculator multiplies each pair and subtracts reactant totals from product totals.
7. Why is entropy entered in joules while enthalpy uses kilojoules?
Many tables report entropy in J/mol·K and enthalpy in kJ/mol. The calculator converts the entropy term internally before combining both values.
8. What does the graph add to the calculation?
The graph shows how free energy responds to a changing variable, such as temperature, K, Q, or cell potential. It makes sensitivity easier to inspect.