Why are Alcohol Dehydration Reactions Favored by High Temperatures?

Question

If I go by the Gibbs' free energy formula, it makes sense that high temperatures make the dehydration of alcohols more feasible, as elimination reactions usually have a positive value for the change in entropy ($\Delta S$) and the hydration of the sulfuric acid used is highly exothermic ($\Delta H$). This decreases the value for the change in free energy ($\Delta G$).

However, if I go by the modified Vant Hoff's formula (relationship between an initial equilibrium constant at an initial temperature and a final equilibrium constant at a final temperature), the $K_2/K_1$ ratio would be higher at lower temperatures, owing to the product $\Delta H \Delta T$ being positive (rather than negative if the reaction mixture is heated).

These are the formulas I'm referring to:

\begin{align} \frac{K_2}{K_1} &= \exp{\left(\frac{\Delta T \Delta H}{RT_1T_2}\right)} \tag{1}\\ \Delta G &= \Delta H -T \Delta S \tag{2} \end{align}

Answer 1

Let's take: $\,\,\,\,\ce{ethene(g) + H2O(g) <=> ethanol(g)} $

The standard enthalpy for hydration of ethene is $\Delta H^\circ_r=-45\frac{KJ}{mol}$. So we have $\Delta H^\circ_r=45\frac{KJ}{mol}$ for dehydration, which means that it is endothermic.

The expression:

\begin{align} \frac{K_2}{K_1} &= \exp{\left(\frac{\Delta T \Delta H}{RT_1T_2}\right)} \tag{1}\\ \end{align}

(This equation is valid when Ts are similar, because $\Delta H$ varies with $T$ and you are considering it constant.)

with $\Delta T= T_2-T_1$ and $\Delta H_r^\circ>0$ (dehydration) equation (1) gives $K_2>K_1$.

As $$K = \frac{[\ce{H2O}]\,[\ce{CH2CH2}]}{[\ce{CH3CH2OH}]}$$

and $K_2>K_1$. In consequence dehydration enthalpy is favoured by temperature-increase.