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:

If more precision is required we could correct for the temperature changes of Δ_{r}H^{o} 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

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 \(\Delta_r{S^o}/R\).

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.