Why are Alcohol Dehydration Reactions Favored by High Temperatures?

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}