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 \(ΔH^o\) 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 \(ΔH^o\) 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:
\[\ln K = - \dfrac{\Delta {H^o}}{R} \dfrac{1}{T} + \dfrac{\Delta{S^o}}{R}\label{20} \]
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 26.7.1
: 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.