Van 't Hoff Equation Calculator

Predict the Equilibrium Constant at a New Temperature

Van 't Hoff Equation Formula

Formula

\[ \ln\!\left(\frac{K_2}{K_1}\right) = -\frac{\Delta H^\circ}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]

Where:

\(K_1\) — equilibrium constant at the reference temperature \(T_1\) (dimensionless)
\(K_2\) — equilibrium constant at the target temperature \(T_2\)
\(\Delta H^\circ\) — standard reaction enthalpy (J/mol; positive for endothermic, negative for exothermic)
\(R\) — gas constant, \(8.314\ \text{J/(mol·K)}\)
\(T_1, T_2\) — absolute temperatures in kelvin

The equation tells you how strongly the equilibrium constant depends on temperature: the effect is driven by the reaction enthalpy and the difference between the reciprocals of the two temperatures.

Van 't Hoff Equation – Calculation Example

Example: a reaction has an equilibrium constant \(K_1 = 1\) at \(T_1 = 298.15\ \text{K}\) (25 °C) and a standard reaction enthalpy of \(\Delta H^\circ = +50\ \text{kJ/mol}\). What is \(K\) at \(T_2 = 373.15\ \text{K}\) (100 °C)?

\[ \ln\!\left(\frac{K_2}{K_1}\right) = -\frac{50{,}000}{8.314}\left(\frac{1}{373.15} - \frac{1}{298.15}\right) = -6013.9 \times (-0.000674) = 4.054 \]

\[ K_2 = K_1 \cdot e^{4.054} = 1 \times 57.6 \]

Result: \(K_2 \approx 57.6\) — the equilibrium constant grows by a factor of about 57.6. The calculator also gives \(\Delta G^\circ_2 \approx -12.57\ \text{kJ/mol}\), confirming the reaction becomes product-favoured at 100 °C.

How to Use the Calculator

Enter the Equilibrium Constant at Reference Temperature (K₁). It must be a positive, dimensionless number.
Enter the Standard Reaction Enthalpy (ΔH°) and choose its unit, kJ/mol or kcal/mol. Use a positive value for an endothermic reaction and a negative value for an exothermic one.
Enter the Reference Temperature (T₁), the temperature at which K₁ was measured, and select K, °C or °F.
Enter the Target Temperature (T₂), the temperature you want the new K for, with its unit.
Read the results: K₂ at the target temperature, the change factor K₂/K₁, ΔG° at the target temperature, plus an interpretation of the shift and a full calculation trace.

What the Van 't Hoff Equation Tells You

Chemical equilibria respond to temperature, and the direction of that response is fixed by the sign of the reaction enthalpy. Heat an endothermic reaction (ΔH° > 0) and the equilibrium shifts towards the products, so K increases; heat an exothermic reaction (ΔH° < 0) and K decreases. This is Le Chatelier's principle made quantitative: the Van 't Hoff equation converts a qualitative rule into a number you can act on.

The result is easiest to read through the change factor \(K_2/K_1\). A factor of 57.6 means the product-to-reactant ratio at equilibrium becomes 57.6 times larger — a dramatic shift. A factor near 1 means the temperature change barely matters, which typically happens when ΔH° is small or the two temperatures are close together. Because the equation works with \(1/T\), the same temperature jump has a much bigger effect at low temperatures than at high ones.

The calculator also reports \(\Delta G^\circ\) at the target temperature, derived from the relation \(\Delta G^\circ = -RT\ln K\) together with the assumption that \(\Delta S^\circ\) stays constant. A negative ΔG° means the reaction is product-favoured under standard conditions (K > 1); a positive value means reactants dominate. If you work the other way around — from ΔG° to K — the Gibbs Free Energy calculator and the Equilibrium Constant calculator are useful companions.

Assumptions, Limitations and Practical Tips

ΔH° is treated as constant. In reality it drifts with temperature because of heat capacity differences between reactants and products. Over a range of 50–100 K the error is usually small; over hundreds of kelvin it can become significant.
Temperatures must be absolute. The equation requires kelvin. The calculator converts °C and °F automatically, but if you use the formula by hand, convert first.
K must be dimensionless. Use the thermodynamic equilibrium constant (activities or the properly standardised Kc/Kp), not a ratio with units. If you need to convert between pressure and concentration constants, see the Kp to Kc calculator.
Watch the exponential behaviour. Because \(K_2 = K_1 e^{\ln(K_2/K_1)}\), large enthalpies or big temperature gaps can shift K by many orders of magnitude. A prediction of K = 10¹² is mathematically correct but practically meaningless if the reaction becomes diffusion-limited or a competing pathway takes over.
The equation says nothing about rate. A favourable K does not guarantee a fast reaction; kinetics and thermodynamics are separate questions.

A related temperature-dependence tool is the Clausius–Clapeyron calculator, which applies the same mathematics to vapour pressures instead of equilibrium constants.

Frequently Asked Questions

How do I calculate the Van 't Hoff equation by hand?

Convert both temperatures to kelvin, then compute \(\ln(K_2/K_1) = -\Delta H^\circ/R \times (1/T_2 - 1/T_1)\). Keep ΔH° in J/mol so it matches R = 8.314 J/(mol·K). Finally, multiply: \(K_2 = K_1 \cdot e^{\ln(K_2/K_1)}\). Watch the sign of the \(1/T_2 - 1/T_1\) term — if \(T_2 > T_1\), this difference is negative, which is what makes K increase for endothermic reactions.

Can I use Celsius temperatures in the formula?

No. The Van 't Hoff equation requires absolute temperatures in kelvin. If you plug in °C values, the reciprocals \(1/T\) are meaningless and the result will be badly wrong. Always add 273.15 first (for Fahrenheit, use \((T - 32) \times 5/9 + 273.15\)). This is one of the most common mistakes students make with this equation.

What is a 'normal' or 'good' value of the equilibrium constant K?

K is dimensionless, and its meaning is relative: \(K \gg 1\) means products dominate at equilibrium, \(K \ll 1\) means reactants dominate, and \(K \approx 1\) means a roughly balanced mixture. There is no universal 'good' value — it depends on the reaction and purpose. A useful benchmark: \(K = 10\) corresponds to \(\Delta G^\circ \approx -5.7\) kJ/mol at 25 °C, and \(K = 0.01\) to about +11.4 kJ/mol.

Does the Van 't Hoff equation use natural log or log base 10?

The standard form uses the natural logarithm (ln) together with R = 8.314 J/(mol·K). If you prefer base-10 logarithms, the equation becomes \(\log(K_2/K_1) = -\Delta H^\circ/(2.303 R)\,(1/T_2 - 1/T_1)\). Mixing log bases with the wrong constant is a frequent source of errors — the result will be off by a factor of 2.303 in the exponent, which changes K enormously.

Is the Van 't Hoff equation the same as the Arrhenius or Clausius–Clapeyron equation?

Mathematically they share the same structure — a straight line of ln(quantity) versus \(1/T\) — but they describe different things. Van 't Hoff applies to equilibrium constants, Arrhenius to rate constants (with activation energy \(E_a\)), and Clausius–Clapeyron to vapour pressures (with enthalpy of vaporisation). You can often extract \(\Delta H^\circ\) from the slope of a plot of ln K versus 1/T.

When does the Van 't Hoff equation give inaccurate results?

The main weakness is that it treats ΔH° as constant. Over a span of 50–100 K the error is usually negligible, but across hundreds of kelvin heat-capacity differences between reactants and products make ΔH° drift, so predictions can be off substantially. It also fails for reactions that change mechanism, become diffusion-limited, or involve phase changes in the interval. For high-precision work, use a heat-capacity correction (Kirchhoff's law).

How is the Van 't Hoff equation related to ΔG° and Le Chatelier's principle?

Combining it with \(\Delta G^\circ = -RT\ln K\) shows that \(\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ\) when ΔS° is constant. It quantifies Le Chatelier's principle: heating an endothermic reaction (ΔH° > 0) raises K, while heating an exothermic one (ΔH° < 0) lowers it. Cooling reverses both effects, and if ΔH° = 0, K does not depend on temperature at all.