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12.5: The Van't Hoff Equation

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    202949
  • We can use Gibbs-Helmholtz to get the temperature dependence of \(K\)

    \[ \left( \dfrac{∂[ΔG^o/T]}{∂T} \right)_P = \dfrac{-ΔH^o}{T^2}\]

    At equilibrium, we can equate \(ΔG^o\) to \(-RT\ln K\) so we get:

    \[ \left( \dfrac{∂[\ln K]}{∂T} \right)_P = \dfrac{ΔH^o}{RT^2} \]

    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 \(ΔH^o\) 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:

    \[\ln \dfrac{K(T_2)}{K(T_1)} = \dfrac{-ΔH^o}{R} \left( \dfrac{1}{T_2} - \dfrac{1}{T_1} \right)\]

    If more precision is required we could correct for the temperature changes of \(ΔH^o\) 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

    \[\Delta{G^o}= \Delta{H^o}−T\Delta{S^o}\]

    Equation \(\ref{18}\) becomes

    \[ K=e^{-\Delta{H^o}/RT} e^{\Delta {S^o}/R}\label{19}\]

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

    \[\ln K = - \dfrac{\Delta {H^o}}{R} \dfrac{1}{T} + \dfrac{\Delta{S^o}}{R}\label{20}\]

    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 \(\PageIndex{1}\): 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.

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