van 't Hoff Equation Calculator — K versus Temperature

Background.

Warm a reaction that absorbs about 50 kJ of heat per mole by ten degrees from room temperature and its equilibrium constant almost doubles. Cool an exothermic reaction and the same arithmetic runs the other way. The van 't Hoff equation turns that piece of Le Chatelier's principle into a number: ln(K₂ ⁄ K₁) = −(ΔrH° ⁄ R)(1 ⁄ T₂ − 1 ⁄ T₁).

The sign of the reaction enthalpy sets the direction. ΔrH° is positive for an endothermic reaction, one that takes in heat, and its K rises with temperature; a negative ΔrH° means K falls on warming. The size of the enthalpy sets how steeply K responds. Because the equation works in reciprocal temperature, a ten-degree step near 300 K moves K further than the same step near 600 K.

You can put four questions to it: carry a known constant to a new temperature, carry one back to the original temperature, fit an enthalpy from constants measured at two temperatures, or find the temperature at which K reaches a target value. Only the ratio K₂ ⁄ K₁ enters, so any consistently defined constant works, whether a gas-phase K°, an acid dissociation constant, the ion product of water, a solubility product, a binding constant or a Henry's-law constant.

The working form assumes ΔrH° stays fixed between the two temperatures. In reality it drifts with the difference in heat capacity between products and reactants, so a few tens of kelvin is usually safe and long extrapolations are not. The reaction-entropy output is a separate matter: it is meaningful only when K is the dimensionless standard constant K°.

What is van 't hoff equation calculator?

The van 't Hoff equation, published by Jacobus Henricus van 't Hoff, links the temperature dependence of an equilibrium constant to the standard reaction enthalpy. It follows from two standard relations, ΔrG° = −RT ln K and ΔrG° = ΔrH° − TΔrS°, which combine to ln K = −ΔrH° ⁄ (RT) + ΔrS° ⁄ R. Plotted as ln K against 1 ⁄ T this is a straight line with slope −ΔrH° ⁄ R and intercept ΔrS° ⁄ R, the van 't Hoff plot used to extract reaction enthalpies from equilibrium measurements. The Clausius–Clapeyron equation for vapour pressure has the same form.

How to use this calculator.

1. 01Choose what to solve for: K₂ at a new temperature, K₁ back at the original one, the enthalpy from two measured constants, or the temperature T₂ that gives a target K.
2. 02Enter the constants. Both must be positive and on the same basis: the same reaction written the same way round, the same standard state and the same units. A Kp in bar and a Kc in mol/L cannot be mixed.
3. 03Enter temperatures in kelvin (°C + 273.15). A Celsius figure typed here is a different quantity, not a slightly wrong one.
4. 04Enter ΔrH° in kJ/mol, positive if the reaction absorbs heat. Standard formation enthalpies from a table such as NIST-JANAF give it as products minus reactants.
5. 05Check the direction before trusting the number: heating an endothermic reaction must give K₂ ⁄ K₁ above 1 and a positive ln(K₂ ⁄ K₁). If it does not, the enthalpy's sign is wrong.

The formula.

Writing ln K = −ΔrH° ⁄ (RT) + ΔrS° ⁄ R at two temperatures and subtracting removes the entropy term, leaving ln(K₂ ⁄ K₁) = −(ΔrH° ⁄ R)(1 ⁄ T₂ − 1 ⁄ T₁). R is the molar gas constant, 8.314462618 J mol⁻¹ K⁻¹, exact since the 2019 SI redefinition. The enthalpy is entered in kJ/mol and multiplied by 1,000 so its units match R.

Each mode is a rearrangement. K₂ = K₁ · exp[−(ΔrH° ⁄ R)(1 ⁄ T₂ − 1 ⁄ T₁)], and K₁ is the same with the sign of the exponent flipped. The enthalpy comes from ΔrH° = −R ln(K₂ ⁄ K₁) ⁄ (1 ⁄ T₂ − 1 ⁄ T₁), and the temperature from 1 ⁄ T₂ = 1 ⁄ T₁ − (R ⁄ ΔrH°) ln(K₂ ⁄ K₁). Whatever you solve for, all nine outputs are filled in, including the entropy estimate ΔrS° = R ln K₁ + ΔrH° ⁄ T₁, returned in J/(mol·K).

Three inputs are refused because no answer exists. Fitting an enthalpy with T₂ = T₁ divides zero by zero: two constants at one temperature say nothing about temperature dependence. Asking for a temperature with ΔrH° = 0 fails because such a constant never changes. And for an endothermic reaction K approaches a finite ceiling, K₁ · exp(ΔrH° ⁄ RT₁), as temperature grows without limit; a target above it would need a negative absolute temperature, so it is rejected. An exothermic reaction has the mirror-image floor.

The constant-enthalpy assumption is the main source of error. Kirchhoff's law, dΔrH° ⁄ dT = ΔrCp, says the enthalpy drifts by the heat-capacity difference for every kelvin, so a two-point fit over a wide interval returns an average enthalpy rather than the value at either end.

A worked example.

Frequently asked questions.

References& sources.

1. [1]Sander, R., Acree, W. E., De Visscher, A., Schwartz, S. E. & Wallington, T. J. (2022). Henry's law constants (IUPAC Recommendations 2021). Pure and Applied Chemistry 94(1), 71–85, §4.2 'Temperature dependence': 'The Henry's law constant is a special case of an equilibrium constant, describing an equilibrium between two phases. To describe the temperature dependence of Hs, the van 't Hoff equation can be used: (1/Hs)(dHs/d(1/T)) = −Δsol H / R, where Δsol H is the enthalpy of solvation, R is the gas constant and T is the thermodynamic temperature. For a small temperature range, Δsol H can be considered constant.' The sign convention and the constant-enthalpy caveat on this page both come from this clause. Text verified from the open-access author manuscript (BNL-220956-2021-JAAM) 2026-07-29.
2. [2]Sander, R. (2023). Compilation of Henry's law constants (version 5.0.0) for water as solvent. Atmospheric Chemistry and Physics 23, 10901–12440, §2.4 Eqs. (1)–(4): the differential form d ln Hs/d(1/T) = −ΔsolH/R, its integration to ln Hs = A + B/T, and the definition B = −ΔsolH/R with reference temperature T° = 298.15 K. Source of the independent worked check on this page (oxygen in water, B = 1500 K ⇒ ΔsolH = −12.47 kJ/mol). Open access; PDF text extracted and read 2026-07-29.
3. [3]Chase, M. W. (1998). NIST-JANAF Thermochemical Tables, 4th edition, J. Phys. Chem. Ref. Data Monograph 9, as served by the NIST Chemistry WebBook SRD 69. Nitrogen dioxide (NO₂, CAS 10102-44-0), gas phase: ΔfH° = 33.10 kJ/mol and S° = 240.04 J/(mol·K) at 298.15 K and 1 bar. One of the two entries from which the worked example's ΔrH° = 57.12 kJ/mol and ΔrS° = 175.70 J/(mol·K) are derived. Open access; values retrieved and confirmed 2026-07-29.
4. [4]Chase, M. W. (1998). NIST-JANAF Thermochemical Tables, 4th edition, as served by the NIST Chemistry WebBook SRD 69. Dinitrogen tetroxide (N₂O₄, CAS 10544-72-6), gas phase: ΔfH° = 9.08 kJ/mol and S° = 304.38 J/(mol·K) at 298.15 K and 1 bar. The second entry behind the worked example. Open access; values retrieved and confirmed 2026-07-29.
5. [5]Tiesinga, E., Mohr, P. J., Newell, D. B. & Taylor, B. N. CODATA Recommended Values of the Fundamental Physical Constants: 2022. NIST Standard Reference Database 121, 'molar gas constant' R = 8.314 462 618… J mol⁻¹ K⁻¹, listed with standard uncertainty '(exact)' following the 2019 SI redefinition. The only physical constant this calculator uses. Independent national metrology institute; open access; retrieved 2026-07-29.
6. [6]International Union of Pure and Applied Chemistry. Compendium of Chemical Terminology (the 'Gold Book'), 5th edition, online version 5.0.0 (2025), entry 'standard pressure' (S05921): 'In 1982 IUPAC recommended the value 10⁵ Pa, but prior to 1982 the value 101 325 Pa (= 1 atm) was usually used.' Cited because the entropy output is only meaningful for a standard equilibrium constant K°, and K° depends on which standard pressure the tabulated data assumed. Open access; retrieved 2026-07-29.