# 26.7: The Van't Hoff Equation

We can use Gibbs-Helmholtz to get the temperature dependence of $$K$$

$\left( \dfrac{∂[Δ_rG^o/T]}{∂T} \right)_P = \dfrac{-Δ_rH^o}{T^2}$

At equilibrium, we can equate $$Δ_rG^o$$ to $$-RT\ln K$$ so we get:

$\left( \dfrac{∂[lnK]}{∂T} \right)_P = \dfrac{-Δ_rH^o}{T^2}$

We see that whether $$K$$ increases or decreases with temperature is linked to whether the reaction enthalpy is positive or negative. If temperature is changed little enough that $$Δ_rH^o$$ can be considered constant, we can translate a $$K$$ value at one temperature into another by integrating the above expression, we get a similar derivation as with melting point depression:

$\ln \dfrac{K(T_2}{K(T_1)} = \dfrac{-Δ_rH^o}{R} \left( \dfrac{1}{T_2} - \dfrac{1}{T_1} \right)$

If more precision is required we could correct for the temperature changes of ΔrHo by using heat capacity data.

How $$K$$ increases or decreases with temperature is linked to whether the reaction enthalpy is positive or negative.

The expression for $$K$$ is a rather sensitive function of temperature given its exponential dependence on the difference of stoichiometric coefficients One way to see the sensitive temperature dependence of equilibrium constants is to recall that

$K=e^{−\Delta_r{G^o}/RT}\label{18}$

However, since under constant pressure and temperature

$\Delta_r{G^o}= \Delta_r{H^o}−T\Delta_r{S^o}$

Equation $$\ref{18}$$ becomes

$K=e^{-\Delta_r{H^o}/RT} e^{\Delta_r{S^o}/R}\label{19}$

Taking the natural log of both sides, we obtain a linear relation between $$\ln K$$and the standard enthalpies and entropies:

$\ln K = - \dfrac{\Delta_r{H^o}}{R} \dfrac{1}{T} + \dfrac{\Delta_r{S^o}}{R}\label{20}$

which is known as the van’t Hoff equation. It shows that a plot of $$\ln K$$ vs. $$1/T$$ should be a line with slope $$-\Delta_r{H^o}/R$$ and intercept $$\Delta_r{S^o}/R$$.

Figure: Endothermic Reaction (left) and Exothermic Reaction Van't Hoff Plots (right). Figures used with permission of Wikipedia

Hence, these quantities can be determined from the $$\ln K$$ vs. $$1/T$$ data without doing calorimetry. Of course, the main assumption here is that $$\Delta_r{H^o}$$ and $$\Delta_r{S^o}$$ are only very weakly dependent on $$T$$, which is usually valid.