Temperature Dependence of K: The van't Hoff Equation

Ammonia synthesis runs at 400–500°C not because chemists like heat, but because the reaction is exothermic and equilibrium actually favors lower temperature. Industrial engineers accept a smaller equilibrium constant in exchange for faster kinetics, and they need to know precisely how much $K$ shrinks as $T$ climbs. That quantitative link between temperature and equilibrium position is the van't Hoff equation. This relation matters because it converts a purely thermodynamic quantity, $\Delta_rH^\circ$ , into something you can extract from ordinary equilibrium measurements at several temperatures. No calorimeter is required. You just measure $K$ at two or more temperatures and plot. Here you will see where the equation comes from. Then you will learn how to read a van't Hoff plot. Finally you will work two numerical examples, one foundational and one applied, and see where the underlying assumptions break down.

Key takeaways

The van't Hoff equation is $d\ln K/dT = \Delta_rH^\circ/RT^2$ , derived directly from the Gibbs–Helmholtz equation.
A plot of $\ln K$ against $1/T$ gives a straight line with slope $-\Delta_rH^\circ/R$ , assuming $\Delta_rH^\circ$ is constant.
Exothermic reactions ( $\Delta_rH^\circ < 0$ ) show $K$ decreasing as $T$ rises; endothermic reactions show the opposite.
The integrated two-point form, $\ln(K_2/K_1) = -(\Delta_rH^\circ/R)(1/T_2 - 1/T_1)$ , lets you find $\Delta_rH^\circ$ from just two data points.
Curvature in a real van't Hoff plot signals a temperature-dependent $\Delta_rH^\circ$ , governed by $\Delta_rC_p^\circ$ .
Key idea: the temperature slope of $\ln K$ is fixed entirely by the reaction enthalpy, because entropy's temperature dependence cancels out of $\Delta G^\circ/T$ .

The core insight

Because $\Delta_rG^\circ = -RT\ln K$ and the Gibbs–Helmholtz equation shows that $\partial(\Delta G^\circ/T)/\partial T$ depends only on $\Delta H^\circ$ , the temperature sensitivity of $\ln K$ is governed entirely by the reaction enthalpy — not directly by entropy — making a van't Hoff plot a calorimeter you never have to build.

The physical picture

Every equilibrium constant is a compromise between energetics and disorder, encoded in $\Delta_rG^\circ = \Delta_rH^\circ - T\Delta_rS^\circ$ . As temperature changes, both $\Delta_rH^\circ$ and $\Delta_rS^\circ$ can shift slightly, but the combination that determines $K$ through $\Delta_rG^\circ = -RT\ln K$ responds to temperature in a strikingly simple way once you divide by $T$ first. The trick, borrowed straight from the Gibbs–Helmholtz equation, is to look at $\Delta_rG^\circ/T$ rather than $\Delta_rG^\circ$ itself. Differentiating this ratio with respect to $T$ at constant pressure eliminates the entropy term entirely, leaving only enthalpy. Since $\ln K = -\Delta_rG^\circ/RT$ , this means the slope of $\ln K$ versus $1/T$ is a direct, uncontaminated readout of $\Delta_rH^\circ$ . Physically: raising $T$ pumps thermal energy into the system. For an exothermic reaction, extra thermal energy makes the reverse (energy-absorbing) direction more competitive, so equilibrium shifts back toward reactants and $K$ falls. For an endothermic reaction, the forward direction benefits from the extra thermal energy, and $K$ rises. This is exactly the qualitative content of Le Chatelier's principle, but the van't Hoff equation makes it quantitative: it tells you how much $K$ shifts per kelvin, in terms of a number you can look up or measure calorimetrically.

Figure

Van't Hoff plot: ln K versus 1/T for an exothermic reaction

The core relations

\frac{d\ln K}{dT} = \frac{\Delta_rH^\circ}{RT^2}

where: $K$ is the equilibrium constant, $\Delta_rH^\circ$ the standard reaction enthalpy, $R$ the gas constant, $T$ absolute temperature.

\frac{d\ln K}{d(1/T)} = -\frac{\Delta_rH^\circ}{R}

where: this is the same relation rewritten so that plotting $\ln K$ against $1/T$ gives a straight line of slope $-\Delta_rH^\circ/R$ .

\ln\frac{K_2}{K_1} = -\frac{\Delta_rH^\circ}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)

where: the two-point integrated form, valid when $\Delta_rH^\circ$ is essentially constant between $T_1$ and $T_2$ .

\Delta_rG^\circ(T) = \Delta_rH^\circ - T\Delta_rS^\circ, \qquad \ln K = -\frac{\Delta_rG^\circ}{RT}

where: the starting relations linking thermodynamics to the equilibrium constant, from the standard reaction Gibbs energy.

Derivation AFrom Gibbs–Helmholtz to van't Hoff

Goal. Show that $d\ln K/dT = \Delta_rH^\circ/RT^2$ follows directly from the Gibbs–Helmholtz equation, with no new physical assumptions beyond those already used to define $K$ .

Step 1. Start from the Gibbs–Helmholtz equation. For the standard reaction Gibbs energy, the Gibbs–Helmholtz equation states

\left(\frac{\partial (\Delta_rG^\circ/T)}{\partial T}\right)_p = -\frac{\Delta_rH^\circ}{T^2}

This form is special because differentiating $\Delta G^\circ/T$ , rather than $\Delta G^\circ$ itself, makes the entropy term vanish identically — a short algebraic fact worth re-deriving: writing $\Delta G^\circ/T = \Delta H^\circ/T - \Delta S^\circ$ and differentiating with $\Delta H^\circ$ , $\Delta S^\circ$ treated as slowly varying gives exactly $-\Delta H^\circ/T^2$ .

Step 2. Bring in the equilibrium constant. The standard reaction Gibbs energy is related to $K$ by

\Delta_rG^\circ = -RT\ln K

so dividing by $T$ ,

\frac{\Delta_rG^\circ}{T} = -R\ln K

Step 3. Differentiate both sides with respect to $T$ at constant pressure.

\frac{\partial}{\partial T}\left(\frac{\Delta_rG^\circ}{T}\right) = -R\frac{d\ln K}{dT}

Step 4. Equate the two expressions for the left-hand side. Combining Step 1 and Step 3,

-R\frac{d\ln K}{dT} = -\frac{\Delta_rH^\circ}{T^2}

Step 5. Solve for the derivative. Dividing through by $-R$ ,

\boxed{\frac{d\ln K}{dT} = \frac{\Delta_rH^\circ}{RT^2}}

Step 6. Recast in $1/T$ form for plotting. Using $d(1/T) = -dT/T^2$ , the chain rule gives

\frac{d\ln K}{d(1/T)} = \frac{d\ln K}{dT}\cdot\frac{dT}{d(1/T)} = \frac{\Delta_rH^\circ}{RT^2}\cdot(-T^2) = -\frac{\Delta_rH^\circ}{R}

Step 7. Integrate between two temperatures. If $\Delta_rH^\circ$ is treated as constant over the interval $[T_1, T_2]$ ,

\int_{\ln K_1}^{\ln K_2} d\ln K = -\frac{\Delta_rH^\circ}{R}\int_{1/T_1}^{1/T_2} d\left(\frac{1}{T}\right)

\boxed{\ln\frac{K_2}{K_1} = -\frac{\Delta_rH^\circ}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)}

This integrated form is the workhorse for two-temperature determinations of $\Delta_rH^\circ$ ; the full plot form from Step 6 is preferred whenever three or more temperatures are available, since it also reveals curvature from $\Delta_rC_p^\circ \neq 0$ .

Worked examples

Example 1 — Foundational: extracting $\Delta_rH^\circ$ from two $K_p$ values. For the dissociation $N_2O_4(g) \rightleftharpoons 2NO_2(g)$ , experiments give $K_p = 0.113$ at $298$ K and $K_p = 0.663$ at $320$ K (both dimensionless, referenced to $p^\circ = 1$ bar). Step 1: identify the two-point van't Hoff form.

\ln\frac{K_2}{K_1} = -\frac{\Delta_rH^\circ}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)

Step 2: compute the left-hand side.

\ln\frac{K_2}{K_1} = \ln\frac{0.663}{0.113} = \ln(5.867) = 1.769

Step 3: compute the bracket on the right.

\frac{1}{T_2} - \frac{1}{T_1} = \frac{1}{320\ \text{K}} - \frac{1}{298\ \text{K}} = 3.1250\times10^{-3} - 3.3557\times10^{-3} = -2.307\times10^{-4}\ \text{K}^{-1}

Step 4: solve for $\Delta_rH^\circ$ .

\Delta_rH^\circ = -R\cdot\frac{\ln(K_2/K_1)}{(1/T_2 - 1/T_1)} = -(8.314\ \text{J K}^{-1}\text{mol}^{-1})\cdot\frac{1.769}{-2.307\times10^{-4}\ \text{K}^{-1}}

\Delta_rH^\circ = -(8.314)\times(-7669\ \text{K}) = 6.376\times10^4\ \text{J mol}^{-1} \approx +63.8\ \text{kJ mol}^{-1}

Sanity check: this matches the accepted literature value for N₂O₄ dissociation (≈ +57 to +58 kJ mol⁻¹ from tabulated $\Delta_fH^\circ$ values; our two-point estimate of +63.8 kJ mol⁻¹ is consistent within the scatter expected from using only two temperatures). The sign is positive, correctly reflecting that dissociation into two moles of gas is endothermic and entropy-driven, and that $K$ rising with $T$ is the expected signature of $\Delta_rH^\circ > 0$ .

Example 2 — Applied: predicting $K_p$ for ammonia synthesis at a new temperature, and checking the assumption. For $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ , $K_p = 6.8\times10^{-2}$ at $T_1 = 673$ K (400°C), and the standard reaction enthalpy is $\Delta_rH^\circ = -92.4$ kJ mol⁻¹ (essentially constant over a modest range near this temperature). Estimate $K_p$ at $T_2 = 773$ K (500°C), the temperature actually used industrially for faster kinetics. Step 1: compute the bracket.

\frac{1}{T_2} - \frac{1}{T_1} = \frac{1}{773} - \frac{1}{673} = 1.2937\times10^{-3} - 1.4859\times10^{-3} = -1.922\times10^{-4}\ \text{K}^{-1}

Step 2: apply the integrated van't Hoff equation.

\ln\frac{K_2}{K_1} = -\frac{\Delta_rH^\circ}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right) = -\frac{(-92400\ \text{J mol}^{-1})}{8.314\ \text{J K}^{-1}\text{mol}^{-1}}\times(-1.922\times10^{-4}\ \text{K}^{-1})

\ln\frac{K_2}{K_1} = (11114\ \text{K})\times(-1.922\times10^{-4}\ \text{K}^{-1}) = -2.136

Step 3: exponentiate.

\frac{K_2}{K_1} = e^{-2.136} = 0.1182, \qquad K_2 = 0.1182\times(6.8\times10^{-2}) = 8.0\times10^{-3}

Sanity check: raising the temperature by 100 K drops $K_p$ by roughly a factor of 8.5, consistent with the strongly exothermic nature of ammonia synthesis and with tabulated values showing $K_p$ falling by roughly one to two orders of magnitude between 400°C and 500°C. This is precisely why the Haber process is run hot for kinetic reasons despite an unfavorable equilibrium shift — high pressure and catalysis are used to compensate, exactly the trade-off Le Chatelier's principle anticipates.

Limiting cases & common traps

Situation	Behavior	Trap to avoid
$\Delta_rH^\circ > 0$ (endothermic)	$K$ increases with $T$	Don't assume all reactions behave like combustion; check the sign
$\Delta_rH^\circ < 0$ (exothermic)	$K$ decreases with $T$	Industrial high- $T$ operation is a kinetic choice, not an equilibrium one
$\Delta_rH^\circ \approx 0$	$\ln K$ nearly flat vs. $1/T$	Small $K$ changes may be dominated by experimental error, not real curvature
$\Delta_rC_p^\circ \neq 0$ over wide $T$ range	van't Hoff plot curves	Fitting a straight line anyway gives a poorly defined "average" $\Delta_rH^\circ$
Two-point formula	Only valid if $\Delta_rH^\circ$ roughly constant between $T_1, T_2$	Using widely separated temperatures inflates error silently

The chemist behind it

Jacobus Henricus van't Hoff built this relation in the 1880s while developing the broader framework of chemical thermodynamics and the theory of solutions, work recognized when he received the first Nobel Prize in Chemistry in 1901. He was among the first to apply thermodynamic reasoning systematically to chemical equilibria, connecting the abstract Gibbs energy formalism to quantities a bench chemist could actually measure: concentrations, pressures, and temperatures. His equation remains one of the most direct experimental routes to reaction enthalpies, requiring nothing more sophisticated than an equilibrium measurement repeated at a few temperatures.

Origins · 1 minJacobus Henricus van't Hoff and how this was discovered

Jacobus Henricus van't Hoff

1852–1911 · Dutch physical chemist who founded stereochemistry and chemical thermodynamics; first Nobel laureate in Chemistry, 1901

Ask the tutor: How does the van't Hoff equation connect quantitatively to Le Chatelier's principle when pressure, not temperature, is the variable being changed?

Why is a van't Hoff plot linear at all, if enthalpy and entropy can both vary with temperature

Over a modest temperature range, $\Delta_rH^\circ$ and $\Delta_rC_p^\circ$ are often small enough that $\Delta_rH^\circ$ can be treated as approximately constant. Under that approximation, integrating $d\ln K/d(1/T) = -\Delta_rH^\circ/R$ gives an exact straight line. The apparent linearity is really an assumption of constant $\Delta_rH^\circ$ , valid piecewise, not a universal law. Over very wide temperature ranges the plot curves, revealing that $\Delta_rC_p^\circ \neq 0$ .

Can I use the van't Hoff equation with concentration-based Kc instead of Kp

Yes, provided you are consistent: the derivation only uses $\Delta_rG^\circ = -RT\ln K$ and the Gibbs–Helmholtz equation, both of which hold for any properly defined standard-state equilibrium constant. The extracted $\Delta_rH^\circ$ is the same reaction enthalpy either way, since $K_p$ and $K_c$ differ only by a factor involving $\Delta n_{gas}$ and $RT$ , which does not itself carry an enthalpy of reaction.

How many temperature points do I actually need for a reliable ΔrH°

Two points give a value but no way to check linearity or estimate uncertainty. Three or more temperatures, spread over a reasonably wide but not excessive range, let you fit a line, assess scatter, and detect curvature from $\Delta_rC_p^\circ$ . In careful thermochemical work, five to ten temperatures spanning 50 to 100 K are typical for a trustworthy slope.

Is the van't Hoff enthalpy the same as a calorimetric enthalpy

Usually close but not always identical. The van't Hoff $\Delta_rH^\circ$ assumes the reaction mechanism and species distribution are the same across the temperature range and that $K$ is measured under genuinely standard-state, equilibrated conditions. Discrepancies between van't Hoff and calorimetric values often signal a hidden temperature-dependent equilibrium, such as a coupled protonation or aggregation step, especially in biochemical binding studies.