van 't Hoff Equation Calculator — K versus Temperature

Background.

The van 't Hoff equation describes how an equilibrium constant changes with temperature: ln(K₂/K₁) = −(ΔrH°/R) × (1/T₂ − 1/T₁). This calculator solves it four ways — project a known constant to a new temperature, run that backwards, fit a reaction enthalpy from two measured constants, or find the temperature at which the constant would reach a target value.

It is the quantitative form of Le Chatelier's principle for temperature. Heat is a reactant in an endothermic reaction and a product in an exothermic one, so raising the temperature pushes an endothermic equilibrium towards products and an exothermic one back towards reactants. **The sign convention that encodes this is the thing to get right: ΔrH° is positive for endothermic.** A positive enthalpy with T₂ above T₁ gives a positive ln(K₂/K₁), so K grows. A negative enthalpy under the same warming gives a negative logarithm, so K shrinks. Both directions are asserted by the calculator's tests rather than assumed, and both are reversed by cooling.

The same equation runs far beyond gas-phase reactions. It governs solubility products, acid dissociation constants, the ion product of water, binding constants in biochemistry, partition coefficients, and Henry's law constants — Sander's Henry's-law compilation publishes a van 't Hoff parameter alongside every one of its 46 434 constants, in the form d ln H/d(1/T), and the worked check on this page reproduces one of them. Anywhere a proportionality constant is really an equilibrium constant in disguise, the van 't Hoff equation moves it between temperatures.

**Two limits belong beside the answer rather than in an accordion.** First, the integrated form assumes ΔrH° is constant across the interval [T₁, T₂]. It is not: enthalpy drifts with temperature through the heat-capacity difference ΔrCp, so a van 't Hoff plot of ln K against 1/T is only approximately a straight line. Over a few tens of kelvin the error is usually small; over hundreds it is not, and a proper treatment needs the Kirchhoff correction or a full thermodynamic model. Second, the entropy output is not always an entropy. ΔrS° = R ln K₁ + ΔrH°/T₁ is only a reaction entropy when K is the **dimensionless standard constant K°** referenced to a defined standard state. Feed it a Kp in bar, a Kc in mol/L or a Henry's law constant and the arithmetic still produces a number, but that number is not an entropy. The first four outputs are unaffected, because they depend only on the ratio K₂/K₁.

Two conventions round it out. Temperatures are absolute, in kelvin, and strictly positive — a Celsius value entered here is not slightly wrong, it is a different quantity. Equilibrium constants must be strictly positive, because the equation takes their logarithm; a negative or zero K is rejected with the field named. Enthalpies, by contrast, are legitimately negative and are accepted. The solver carries every step at 30-digit precision, uses arbitrary-precision exponential and logarithm functions rather than the floating-point ones, and rounds once at the end — to twelve **significant figures** rather than a fixed number of decimal places, because equilibrium constants routinely sit far below 1 and a fixed decimal-place rounding would turn a perfectly good ion product of 1.008 × 10⁻¹⁴ into a zero.

What is van 't hoff equation calculator?

The van 't Hoff equation, named after Jacobus Henricus van 't Hoff, relates the temperature derivative of an equilibrium constant to the standard reaction enthalpy: d ln K / d(1/T) = −ΔrH°/R, where R is the molar gas constant, 8.314462618 J mol⁻¹ K⁻¹ (CODATA 2022, exact). Integrating between two temperatures, treating ΔrH° as constant over the interval, gives the working form ln(K₂/K₁) = −(ΔrH°/R)(1/T₂ − 1/T₁), and integrating indefinitely gives ln K = −ΔrH°/(RT) + ΔrS°/R — the equation of a straight line in the variables ln K and 1/T, whose slope is −ΔrH°/R and whose intercept is ΔrS°/R. That linear form is the van 't Hoff plot, the standard experimental method for extracting a reaction enthalpy from equilibrium measurements at several temperatures, and it explains why this calculator's two-point enthalpy mode is a crude version of the real procedure: two points define a line, but with no estimate of how well the line fits. The equation is not restricted to chemical reactions. IUPAC's 2021 recommendations on Henry's law constants state it in exactly the same form for a phase equilibrium, (1/Hs) dHs/d(1/T) = −ΔsolH/R, noting that a Henry's law constant is 'a special case of an equilibrium constant, describing an equilibrium between two phases'. Sander's compilation integrates it to ln Hs = A + B/T with B = −ΔsolH/R, and publishes B for thousands of species. The same structure appears in the Clausius-Clapeyron equation for vapour pressure, in the Arrhenius equation for rate constants with an activation energy in place of a reaction enthalpy, and in the temperature dependence of binding affinities in biochemistry.

How to use this calculator.

1. 01Pick what you want to work out. The first two modes move a constant between temperatures; the third fits an enthalpy from two constants; the fourth finds the temperature that reaches a target constant.
2. 02Enter the equilibrium constant or constants. They must be positive, and both must be on the same basis — the same standard state, the same units, the same reaction written the same way round.
3. 03Enter the temperatures in kelvin: K = °C + 273.15. Both must be absolute and positive.
4. 04Enter the reaction enthalpy in kJ/mol, positive for endothermic and negative for exothermic. If you do not have one, use the third mode to fit it from two measured constants instead.
5. 05Read the answer, then sanity-check the direction against the sign: an endothermic reaction being heated must give K₂ > K₁ and a positive ln(K₂/K₁). If it does not, the sign of your enthalpy is wrong.
6. 06Ignore the ΔrS° output unless your K is the dimensionless standard constant K°. For a Kp in bar, a Kc in mol/L or a Henry's law constant it is arithmetic without meaning.
7. 07Keep the temperature interval modest. Across a few tens of kelvin the constant-enthalpy assumption is usually fine; across hundreds it degrades, and a two-point fit over a wide interval returns an average enthalpy rather than a value at either end.

The formula.

Start from the Gibbs relation ΔrG° = −RT ln K and substitute ΔrG° = ΔrH° − TΔrS°:

ln K = −ΔrH° / (RT) + ΔrS° / R

Differentiating with respect to 1/T, with ΔrH° and ΔrS° treated as constants over the interval, gives the differential van 't Hoff equation d ln K / d(1/T) = −ΔrH°/R. Evaluating the indefinite form at two temperatures and subtracting gives the working equation this calculator uses:

ln(K₂/K₁) = −(ΔrH° / R) × (1/T₂ − 1/T₁)

The four rearrangements are:

K₂ = K₁ · exp( −(ΔrH°/R)(1/T₂ − 1/T₁) ) K₁ = K₂ · exp( +(ΔrH°/R)(1/T₂ − 1/T₁) ) ΔrH° = −R · ln(K₂/K₁) / (1/T₂ − 1/T₁) 1/T₂ = 1/T₁ − (R/ΔrH°) · ln(K₂/K₁)

Enthalpies are entered and reported in kJ/mol and converted to J/mol internally so they match R's units; the entropy output comes back in J/(mol·K).

Rounding stage: FINAL ONLY, with one deliberate deviation from this site's usual convention. Every step is carried at 30-digit Decimal.js precision using arbitrary-precision exponential and natural-logarithm functions rather than floating-point ones, and a single rounding is applied at the return boundary — to twelve SIGNIFICANT FIGURES rather than to ten decimal places. The reason is that equilibrium constants have no characteristic magnitude: the ion product of water is about 1.0 × 10⁻¹⁴, silver chloride's solubility product is about 1.8 × 10⁻¹⁰, Henry's law solubilities are around 10⁻⁵, and metal-complex formation constants reach 10³⁰. Rounding any of the first three to ten decimal places would return exactly zero, which is not a rounding artefact but a wrong answer. A unit test pins the ion-product case.

Directional behaviour, read off the equation rather than from memory. Take the worked example below: ΔrH° = +57.12 kJ/mol (endothermic) and heating from 298.15 K to 350 K. Then 1/T₂ − 1/T₁ is negative, so −(ΔrH°/R) × (negative) is positive, the exponent is +3.4135, and K grows by a factor of e³·⁴¹³⁵ = 30.37 — from 0.1481 to 4.4970. Heat pushes an endothermic equilibrium towards products. Flip the sign of ΔrH° to −57.12 and the same warming shrinks K by the same factor. Cool instead of heat and both reverse. All four combinations are unit-tested.

Singularities and invalid domain. In the enthalpy mode, T₂ must differ from T₁: two constants measured at the same temperature carry no information about how K varies with temperature, and the expression is 0/0 there — the calculator refuses rather than returning an infinity, and near the singularity a tiny difference in K implies an enormous enthalpy, which is why a two-point fit over a narrow interval is so imprecise. In the temperature mode, ΔrH° must be non-zero, because a temperature-independent constant never reaches a different value; and some targets are simply unreachable — for an endothermic reaction, K rises towards a finite ceiling as T goes to infinity, and asking for a K beyond that ceiling would need a negative absolute temperature, so the calculator rejects it with an explanation rather than returning one. Constants and temperatures must be strictly positive; enthalpies may be negative.

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.