Van't Hoff Equation Calculator

Calculate equilibrium constant at any temperature using the Van't Hoff equation.

K1 at temperature T1

Temperature T1 (K) (K)

Standard enthalpy delta H (kJ/mol) (kJ/mol)

Target temperature T2 (K) (K)

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How Temperature Changes K

Le Chatelier tells you which direction an equilibrium shifts when heated. The van’t Hoff equation tells you by how much. It is the quantitative version of the same physics, and the link between them is the sign of ΔH°.

ln(K₂/K₁) = −(ΔH°/R) × (1/T₂ − 1/T₁)

Term	Meaning	Note
K₁, K₂	Equilibrium constants at T₁ and T₂	Dimensionless
ΔH°	Standard enthalpy of reaction	J/mol — convert from kJ/mol
R	Gas constant	8.314 J/(mol·K)
T	Absolute temperature	Kelvin, never Celsius

The sign of ΔH° decides everything. For an endothermic reaction (ΔH° positive) heating increases K — more product at equilibrium. For an exothermic reaction (ΔH° negative) heating decreases K. This is Le Chatelier’s heat-as-a-reagent argument expressed numerically.

Notice that the equation contains no entropy term. ΔS° affects the absolute value of K but not how K changes with temperature, because the entropy contribution to ΔG° scales with T and cancels in the ratio. Only enthalpy drives the temperature dependence.

The Linear Plot

Rearranged, the equation becomes ln K = −ΔH°/(RT) + ΔS°/R. Plotting ln K against 1/T therefore gives a straight line whose slope is −ΔH°/R and whose intercept is ΔS°/R. This van’t Hoff plot is the standard experimental route to both thermodynamic quantities from equilibrium measurements alone, without calorimetry.

Plot feature	Yields	Sign meaning
Slope	−ΔH°/R	Negative slope means endothermic
Intercept	ΔS°/R	Positive intercept means entropy-favoured
Curvature	ΔH° varies with T	Assumption of constant ΔH° is breaking down

Worked Examples

Example 1: Water Ka at body temp: K1=1e-14, dH=+55.8kJ, T1=298, T2=310

K2=1e-14*exp(-55800/8.314*(1/310-1/298))

Result: K2=2.4e-14 - neutral pH=6.8 at 37C

pH of neutrality shifts with temperature

Example 2: Exothermic: K1=100, dH=-40kJ, 298K to 400K

K2=100*exp(+40000/8.314*(1/298-1/400))

Result: K2=1.63 - K decreases

Le Chatelier: heat disfavors exothermic

Example 3: Endothermic dissolution

K₁ = 0.010 at 298 K, ΔH° = +25 kJ/mol, T₂ = 323 K

Result: K₂ ≈ 0.023

Heating by 25 K more than doubles K. This is why solubility of most salts increases with temperature — dissolution is usually endothermic.

Example 4: Reading a van’t Hoff plot

Slope of ln K vs 1/T measured as −4,200 K

Result: ΔH° = −slope × R = +34.9 kJ/mol

A negative slope gives a positive ΔH°, confirming an endothermic reaction. No calorimeter required.

Example 5: Small ΔH°, small effect

K₁ = 5.0 at 298 K, ΔH° = +2 kJ/mol, T₂ = 348 K

Result: K₂ ≈ 5.6

A reaction with near-zero enthalpy change is almost temperature independent, because the exponent stays close to zero regardless of the temperature span.

Common Mistakes

⚠️

Using Celsius instead of Kelvin

The equation uses 1/T with absolute temperature. Entering Celsius produces a completely wrong answer, and near 0°C it produces a division that is meaningless.

⚠️

Mixing kJ and J

ΔH° is usually tabulated in kJ/mol but R is in J/(mol·K). Failing to multiply by 1000 makes the exponent 1000 times too small and the predicted change essentially zero.

⚠️

Assuming ΔH° is constant over any range

The equation assumes enthalpy does not vary with temperature. Over tens of kelvin that is reasonable; over hundreds it is not, and a curved van’t Hoff plot is the diagnostic.

⚠️

Confusing a change in K with a change in rate

Van’t Hoff describes the equilibrium position; the Arrhenius equation describes rate. Heating an exothermic reaction lowers K while still speeding both directions up.

Frequently Asked Questions

Le Chatelier vs Van't Hoff?

Both give same direction. Van't Hoff is quantitative. Le Chatelier gives direction only.

Applications?

Industrial reactor T optimization, protein denaturation T, enzyme kinetics Q10 rule, atmospheric chemistry.

What is the van’t Hoff equation used for?

Calculating an equilibrium constant at one temperature from its value at another, given the reaction enthalpy — or extracting ΔH° and ΔS° from equilibrium data measured at several temperatures.

How does it relate to Le Chatelier’s principle?

Le Chatelier predicts the direction of shift qualitatively. Van’t Hoff gives the magnitude. Both come from the same thermodynamics, with the sign of ΔH° determining the direction.

Why does entropy not appear in the equation?

Because the entropy contribution to ΔG° scales with temperature and cancels when taking the ratio of two K values. ΔS° sets the absolute value of K but not its temperature dependence.

What does a van’t Hoff plot show?

ln K plotted against 1/T gives a straight line with slope −ΔH°/R and intercept ΔS°/R, allowing both quantities to be determined from equilibrium measurements alone.

Why must temperature be in kelvin?

The equation uses reciprocal absolute temperature. Celsius has an arbitrary zero, so 1/T would be meaningless and near 0°C would approach infinity.

Formula Explorer connections

Interpretation: This relationship converts chemical amount, mass, composition or balanced-equation ratios into a reaction quantity. Assumption: Use a balanced reaction, consistent units and the correct molar mass. Purity, side reactions and limiting reagents can change experimental results.

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