Van’t Hoff Equilibrium Temperature

The Van’t Hoff equation describes how the equilibrium constant of a chemical reaction changes with temperature. Assuming the standard reaction enthalpy is constant over the given temperature range, this equation connects thermodynamics to the observed equilibrium states without invoking kinetic rates or activation energy.

Worked Example Calculation

Let us verify the equilibrium characteristics for a reaction with a positive enthalpy change, given a chosen reference standard. We use the integrated Van’t Hoff equation: ln K(T) = ln Kref - (DeltaH / R) * (1/T - 1/298.15) with R = 8.31446261815324 J mol^-1 K^-1.

Reference K = 10, Enthalpy = 40 kJ/mol, and Temperature = 350 K.
First, calculate ln Kref = ln(10) = 2.302585092994046.
The term for (1/T - 1/298.15) = (1/350 - 1/298.15) = -0.000496874287860431.
Multiply by (40000 / 8.31446261815324) = 4810.894178652037 to get -2.39040621315842.
Subtract this from ln Kref: 2.302585092994046 - (-2.39040621315842) = 4.692991306152466 (Log equilibrium K).
Take the exponential to find the Equilibrium K: exp(4.692991306152466) = 109.17928048420983.
Calculate the slope dlnK/dT = DeltaH / (R * T^2) = 40000 / (8.31446261815324 * 350^2) = 0.03927260572823708.
Calculate Standard Gibbs energy = -R * T * lnK = -8.31446261815324 * 350 * 4.692991306152466 = -13656.89527381299 J/mol.

Experiment Simulation

Interact with the Van’t Hoff Equilibrium Temperature controls to see how enthalpy and temperature determine the equilibrium constant.

Set Reference K to 10, Enthalpy kJ per mol to 40, and Temperature kelvin to 350. Observe that Log equilibrium K is 4.692991306152466 and Equilibrium K increases to 109.17928048420983.
Set Enthalpy kJ per mol to -40. Notice the Equilibrium K drops to 0.9159247025305566, and the Log-K temperature slope becomes negative (-0.03927260572823708).
Set Reference K to 1. The Equilibrium K becomes 10.917928048420976 and Log equilibrium K shifts to 2.39040621315842. Standard Gibbs energy changes to -6956.230085527424.
Set Temperature kelvin to 298.15. See that Equilibrium K matches the reference exactly (10.000000000000002 due to precision), and Log equilibrium K is 2.302585092994046.
Set Enthalpy kJ per mol to 0. The Log-K temperature slope becomes 0.0, and Equilibrium K remains fixed at the reference value regardless of the temperature.

Limitations and Constraints

This derivation assumes the standard reaction enthalpy (DeltaH) remains completely constant over the chosen temperature range. In real chemical systems, heat capacities of reactants and products generally differ, causing enthalpy to be a function of temperature. The calculation describes purely a thermodynamic equilibrium state, not Arrhenius kinetics or an activation energy. Thus, it cannot predict how quickly the new equilibrium is reached, only the final state. Additionally, this calculation models dimensionless activities, so units are omitted from the K values.

External References

For more information on Van’t Hoff's original works on chemical dynamics and equilibrium, see the following reference:

Jacobus H. van 't Hoff - Nobel Lecture (1901)

Van’t Hoff Equilibrium Temperature