The Van 't Hoff Equation

What the Van 't Hoff equation does

Le Chatelier's principle tells you the direction: heat an endothermic reaction and equilibrium shifts toward products; heat an exothermic reaction and it shifts back toward reactants. That's a useful arrow, but it's just an arrow. The Van 't Hoff equation puts a number on the arrow. It says exactly how much the equilibrium constant K moves per degree, and it names the single quantity that controls the movement: the standard reaction enthalpy, ΔH°.

In its compact differential form the equation is:

    d(ln K)          ΔH°
    ────────   =   ───────
      dT             R T²

Read it as a rate of change. The left side is how fast the logarithm of the equilibrium constant climbs as you warm the system. The right side is the reaction enthalpy divided by R T². Because R T² is always positive, the sign of ΔH° alone decides which way K goes:

• Endothermic reaction (ΔH° > 0): the right side is positive, so ln K rises with temperature. Heat shoves the equilibrium toward products. Heat behaves like a reactant.
• Exothermic reaction (ΔH° < 0): the right side is negative, so ln K falls with temperature. Heat shoves the equilibrium back toward reactants. Heat behaves like a product.

That is the whole physical story. Everything below is just how to use it — how to integrate it into a straight line, how to read a slope, and where it quietly breaks.

Where it comes from: Gibbs-Helmholtz in three lines

The Van 't Hoff equation isn't a new law — it falls straight out of thermodynamics you already have. Start from the bridge between free energy and the equilibrium constant:

    ΔG° = −R T ln K          ⟹      ln K = −ΔG° / (R T)

Now substitute ΔG° = ΔH° − TΔS° into that expression:

    ln K  =  −ΔH°/(R T)  +  ΔS°/R

Differentiate with respect to T (treating ΔH° and ΔS° as constant over the interval). The ΔS°/R term is a constant, so it dies; the first term gives:

    d(ln K)/dT  =  ΔH° / (R T²)          ← the Van 't Hoff equation

Equivalently, in the form that makes the straight line obvious, differentiate with respect to (1/T) instead of T:

    d(ln K) / d(1/T)  =  −ΔH° / R

The deep engine underneath is the Gibbs-Helmholtz equation, [∂(ΔG°/T)/∂T]_P = −ΔH°/T². Van 't Hoff simply fed ln K = −ΔG°/RT into it. The same machinery, applied to a phase equilibrium instead of a chemical one, gives the Clausius-Clapeyron equation for vapor pressure.

The linear form — why you plot ln K vs 1/T

Integrate ΔH° as a constant and the relationship becomes a line you can draw on graph paper:

                     ΔH°   1        ΔS°
    ln K   =   −  ─────  · ───   +   ─────
                      R      T          R

           y   =        m      · x    +    b        (a straight line!)

Set the y-axis to ln K and the x-axis to 1/T (in kelvin⁻¹). The data fall on a straight line whose:

• slope is m = −ΔH°/R, so ΔH° = −slope × R;
• intercept (at 1/T → 0) is b = ΔS°/R, so ΔS° = intercept × R.

This is the payoff that makes the equation a laboratory workhorse. You never have to run a calorimeter. Measure K at four or five temperatures, plot the points, fit a line, and both the enthalpy and the entropy of the reaction fall out of the geometry. An endothermic reaction gives a downward-sloping line (negative slope of ln K vs 1/T means positive ΔH° — mind the double negative); an exothermic reaction gives an upward-sloping line.

The two-point form — predicting K at a new temperature

Often you don't need a whole plot; you know K at one temperature and want it at another. Integrate the differential equation between two temperatures T₁ and T₂:

      K₂           ΔH°   ⎛  1     1  ⎞
   ln ───   =   −  ────  ⎜ ───  − ─── ⎟
      K₁            R    ⎝  T₂    T₁  ⎠

Everything on the right is known once you have ΔH° and one anchor point (K₁ at T₁), so you solve for K₂ at T₂ directly. This is how a chemical engineer takes an equilibrium constant measured in a benchtop flask at 298 K and extrapolates it to a reactor running at 700 K — provided ΔH° holds roughly constant across the gap.

Worked example: ammonia synthesis, N₂ + 3H₂ ⇌ 2NH₃

The Haber-Bosch reaction is strongly exothermic, ΔH° ≈ −92 kJ mol⁻¹ (about −46 kJ per mole of NH₃). The Van 't Hoff equation predicts a brutal consequence: K falls as you heat the reactor. Let's put numbers on it. Suppose K = 6.0 × 10⁵ at 298 K, and we want K at 700 K, a realistic industrial temperature.

    ln(K₂/K₁) = −(ΔH°/R)(1/T₂ − 1/T₁)

    ΔH° = −92 000 J/mol,   R = 8.314 J mol⁻¹ K⁻¹
    1/T₂ − 1/T₁ = 1/700 − 1/298 = 0.001429 − 0.003356 = −0.001927 K⁻¹

    ln(K₂/K₁) = −(−92000 / 8.314)(−0.001927)
              = −(−11066)(−0.001927)
              = −21.3

    K₂/K₁ = e^(−21.3) ≈ 5.5 × 10⁻¹⁰
    K₂ ≈ (6.0 × 10⁵)(5.5 × 10⁻¹⁰) ≈ 3 × 10⁻⁴

The equilibrium constant collapses by roughly nine orders of magnitude between room temperature and 700 K. Thermodynamically, cold is where the ammonia is. Yet every ammonia plant on Earth runs hot (~400-500 °C). Why? Because at 298 K the reaction, though favorable, is kinetically frozen — nothing happens in a human lifetime. Fritz Haber and Carl Bosch resolved the conflict with pressure (150-300 atm, which favors the side with fewer gas molecules, 4 → 2) and an iron catalyst to speed the kinetics, then accepted a temperature high enough to make the rate practical while still leaving a workable K. The Van 't Hoff equation is exactly the tool that quantifies that trade-off.

Van 't Hoff vs the equations it's related to

Relation	Describes	Enthalpy term	Linear plot	Slope
Van 't Hoff	Equilibrium constant vs T	ΔH° (reaction)	ln K vs 1/T	−ΔH°/R
Clausius-Clapeyron	Vapor pressure vs T	ΔHvap	ln P vs 1/T	−ΔHvap/R
Arrhenius	Rate constant vs T	Ea (activation)	ln k vs 1/T	−Ea/R
Eyring	Rate constant vs T (TST)	ΔH‡ (activation)	ln(k/T) vs 1/T	−ΔH‡/R
Gibbs-Helmholtz	ΔG/T vs T (the parent)	ΔH	—	—

The family resemblance is not a coincidence: all four practical equations are the same "ln(something) is linear in 1/T with slope −energy/R" pattern. What differs is which equilibrium or barrier the energy describes. Van 't Hoff is about the position of a chemical equilibrium (a state function ΔH°); Arrhenius is about the height of a kinetic barrier (E_a). They are cousins, not the same equation — a common exam trap.

When the line curves: ΔCp and Kirchhoff's law

The straight-line form rests on one assumption: that ΔH° doesn't change with temperature. That is only ever approximately true. Kirchhoff's law says the reaction enthalpy drifts at a rate set by the heat-capacity change of the reaction:

    (∂ΔH°/∂T)_P  =  ΔCp°  =  Cp°(products) − Cp°(reactants)

When ΔCp is small — most gas-phase reactions over a modest range — the Van 't Hoff plot stays admirably straight and one slope gives one enthalpy. But when ΔCp is large, the plot curves, and the local slope changes with temperature. Two famous cases:

• Protein folding. Burying hydrophobic side chains releases ordered water, producing a large negative ΔCp. The Van 't Hoff plot of a folding equilibrium is visibly curved, and the folding enthalpy you read off depends on which temperature you read it at. This is also why proteins can cold-denature — the sign of ΔH flips at low temperature.
• Micelle formation. Surfactant self-assembly likewise has a big ΔCp from dehydration of the hydrocarbon tails, giving a curved plot.

There is a subtler warning too. A slope fitted to ln K vs 1/T gives the Van 't Hoff enthalpy — the enthalpy the equilibrium "feels." It only equals the true calorimetric enthalpy if the reaction is a clean two-state process. When intermediates or coupled equilibria hide in the data, the two enthalpies disagree, and the mismatch is itself a diagnostic that the mechanism is more complicated than two states.

Who was Van 't Hoff, and when

Jacobus Henricus van 't Hoff (1852-1911), a Dutch physical chemist, published this temperature relation in his 1884 book Études de dynamique chimique ("Studies in Chemical Dynamics"), the same landmark work in which he laid out the modern kinetic and thermodynamic treatment of reaction rates and equilibria. He had already, at age 22, co-founded stereochemistry by proposing the tetrahedral carbon atom (1874). In 1901 he received the very first Nobel Prize in Chemistry — cited for his work on chemical dynamics and osmotic pressure in solutions. His equilibrium-temperature equation and his osmotic-pressure law (Π = cRT, formally identical to the ideal gas law) both flow from the same conviction that the laws of thermodynamics govern solutions and reactions exactly, not merely qualitatively.

Where it earns its keep

• Industrial reactor design. Every equilibrium-limited process — ammonia synthesis, the contact process for sulfuric acid (2SO₂ + O₂ ⇌ 2SO₃, exothermic), methanol synthesis, the water-gas shift — uses the Van 't Hoff relation to choose an operating temperature that balances a favorable K against a fast enough rate.
• Solubility and Ksp. Because a solubility product is an equilibrium constant, plotting ln Ksp vs 1/T gives the enthalpy of dissolution. A salt whose solubility rises with temperature dissolves endothermically; one that falls (like cerium sulfate) dissolves exothermically — the plot tells you which.
• Binding thermodynamics. In drug discovery, measuring a binding constant K_a at several temperatures and building a Van 't Hoff plot yields the binding enthalpy and entropy — the same numbers isothermal titration calorimetry gives, and a cross-check on it.
• Analytical chemistry. Distribution and partition coefficients, acid dissociation constants (pKa vs T), and complex-formation constants are all routinely temperature-corrected through Van 't Hoff.

Common pitfalls

• Sign confusion on the slope. The slope of ln K vs 1/T is −ΔH°/R. A negative slope therefore means a positive ΔH° (endothermic). The double negative trips up nearly everyone once. Sanity-check against Le Chatelier: if heating raised your K, the reaction must be endothermic.
• Assuming a straight line over too wide a range. If your temperatures span hundreds of kelvin, ΔCp will bend the plot. Fit locally or include the ΔCp term.
• Using °C instead of K. The 1/T axis is strictly in kelvin. Slipping a Celsius value in silently ruins the enthalpy.
• Forgetting K is dimensionless. ln K only makes sense for a properly defined equilibrium constant relative to standard states. Using a concentration quotient with units gives a spurious intercept (a wrong ΔS°).
• Reading a Van 't Hoff enthalpy as calorimetric. They match only for a two-state process. Curvature or a slope/calorimetry mismatch is telling you the mechanism has hidden states.