van 't Hoff Equation (K at Two Temperatures)

Equilibrium constants move with temperature, and this says how. Take ΔG° = −RT ln K, substitute ΔG° = ΔH° − TΔS°, and the entropy term drops out of the difference between two temperatures, leaving ln(K₂/K₁) = (ΔH°/R)(1/T₁ − 1/T₂). Two constants at two temperatures give you the reaction enthalpy; an enthalpy and one constant give you the other constant.

Notice the shape. It is the Arrhenius two-temperature equation with ΔH° standing where E_a stands, and the resemblance is not a coincidence — both come from differentiating a logarithm against 1/T. The distinction worth holding on to is that Arrhenius describes how fast a reaction goes and this describes how far. A catalyst changes the first and cannot touch the second.

The sign is the payload. An endothermic reaction, ΔH° positive, has a larger K at higher temperature; an exothermic one has a smaller K. That is Le Chatelier's principle written as arithmetic, and it is why the great exothermic industrial syntheses are run cooler than their kinetics would like. Ammonia synthesis is exothermic at −92 kJ/mol, so heating the converter to get a usable rate costs equilibrium yield, and the entire design of a Haber plant — high pressure, a catalyst, recycle — is an argument about that trade.

Worked: N₂O₄ ⇌ 2NO₂ has ΔH° = +57.2 kJ/mol. With K = 0.113 at 298 K, moving to 318 K gives ln(K₂/K₁) = (57 200/8.3145)(1/298 − 1/318) = 6879.6 × 2.1105 × 10⁻⁴ = 1.452, so K₂ = 0.113 × e^1.452 = 0.483. A 20 K rise has more than quadrupled the constant — visible to the eye as a sealed tube of the colourless dimer turning brown when it is warmed.

The assumption is that ΔH° is constant across the interval. Over 20 or 30 K that is comfortable; over hundreds of kelvin it is not, and the ΔH° you get back is then an average over the range rather than the value at either end. Plotting ln K against 1/T and taking the slope, which is the graphical version of this equation, will show the curvature if there is any.