Formula used
- From dissociation constant: ΔG° = R·T·ln(Kd/C0)
- From association constant: ΔG° = −R·T·ln(Ka·C0)
- From thermodynamics: ΔG = ΔH − T·ΔS
How to use this calculator
- Select a calculation method: equilibrium constants or ΔH/ΔS.
- Enter the required values and choose appropriate units.
- Set temperature, and adjust C0 if needed.
- Press Compute to show results above the form.
- Use the download buttons to export CSV or PDF.
Example data table
| Case | Kd (nM) | T (K) | ΔG° (kJ/mol) | Comment |
|---|---|---|---|---|
| Strong binder | 1 | 298.15 | −51.3 | Typical high-affinity interaction |
| Moderate binder | 100 | 298.15 | −39.9 | Often observed in screening assays |
| Weak binder | 10000 | 298.15 | −28.6 | May require high ligand concentration |
Binding free energy guide
1) Why binding free energy matters
Binding free energy (ΔG) condenses affinity into one thermodynamic number that compares ligands across conditions. It links equilibrium constants to energetic driving forces, helping you rank candidates, rationalize selectivity, and report results in consistent units for analysis and documentation.
2) What the sign and magnitude mean
A negative ΔG indicates favorable binding, while a positive value indicates unfavorable binding under the stated conditions. At 298.15 K and C0 = 1 M, example values in the table show how Kd shifting from 10,000 nM to 1 nM produces a large ΔG change, reflecting much stronger affinity. It helps translate assay readouts into intuitive energetic differences directly.
3) Equilibrium route: Kd and Ka
For equilibrium inputs, the calculator converts your Kd or Ka to base units, forms a dimensionless logarithm, and applies the gas constant R = 8.314462618 J·mol⁻¹·K⁻¹. A one‑decade change in Kd (10×) shifts ΔG by RT·ln(10), which is about 5.71 kJ/mol at 298.15 K. This makes log‑scale affinity changes easy to interpret.
4) Temperature dependence and sensitivity
Temperature affects ΔG through the RT factor (and through ΔH/ΔS if you use the thermodynamic route). At 298.15 K, RT ≈ 2.48 kJ/mol; at 310 K, RT ≈ 2.58 kJ/mol. This shift can matter when tight binders are compared.
5) Standard state concentration (C0)
C0 ensures the logarithm argument is unitless. The common biochemical standard state is 1 M, but you may set mM or µM when you need a consistent reference for a specific workflow. Changing C0 rescales the reported ΔG° by RT·ln(1/C0 in M), so keep C0 fixed when comparing studies.
6) Thermodynamic route: ΔH and ΔS
When ΔH and ΔS are available, the calculator uses ΔG = ΔH − TΔS with consistent units. For example, ΔH = −40 kJ/mol and ΔS = −50 J/mol·K at 298.15 K gives ΔG ≈ −25.1 kJ/mol. This route helps separate enthalpic contributions from entropic penalties across temperatures.
7) Units, conversions, and numerical checks
Inputs accept common prefixes (pM to M for Kd and reciprocal units for Ka) and multiple energy units (J, kJ, cal, kcal). The calculator validates positivity where required, converts temperatures to kelvin, and avoids invalid logarithms by enforcing positive, dimensionless ratios for ln().
8) Reporting results for reuse
For reproducible reporting, record the method, temperature, and C0 alongside ΔG. Exporting CSV and PDF keeps inputs and outputs together for lab notebooks, slides, and review.
FAQs
1) What is the difference between ΔG and ΔG°?
ΔG° refers to a defined standard state (set by C0, commonly 1 M). ΔG can also be computed from ΔH and ΔS at a specific temperature. Always report temperature and C0 with the value.
2) Which temperature should I enter?
Use the temperature at which the affinity or thermodynamic parameters were measured. If comparing ligands, keep temperature consistent (for example, 298.15 K or 310 K). Even small temperature differences can shift ΔG.
3) Why does Ka use reciprocal units?
Ka is an association constant, so its units are the inverse of concentration (for example, 1/M). The calculator converts reciprocal prefixes consistently so Ka·C0 becomes dimensionless before taking the natural logarithm.
4) Why do we need C0 at all?
Logarithms require unitless arguments. C0 provides a reference concentration so Kd/C0 or Ka·C0 has no units. Using the same C0 across results makes comparisons consistent.
5) Can I report results in kcal/mol?
Yes. The result panel shows ΔG in J/mol, kJ/mol, and kcal/mol. Use kcal/mol when matching legacy biochemistry conventions, and kJ/mol for SI‑aligned reporting.
6) What does a negative ΔS imply here?
A negative ΔS often indicates reduced degrees of freedom upon binding. In ΔG = ΔH − TΔS, a negative ΔS makes −TΔS positive, which can oppose binding unless ΔH is sufficiently favorable.
7) Why does the calculator use ln instead of log10?
Thermodynamic relationships are defined with the natural logarithm. If you think in decades, remember ln(10) ≈ 2.303, so each 10× affinity change shifts ΔG by RT·ln(10) at your temperature.