Why are Alcohol Dehydration Reactions Favored by High Temperatures?
Let's take: $\,\,\,\,\ce{ethene(g) + H2O(g) &lt;=&gt; 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} &amp;= \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 $\Delta H_r^\circ&gt;0$ (dehydration) equation (1) gives $K_2&gt;K_1$.

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

and $K_2&gt;K_1$. In consequence dehydration enthalpy is favoured by temperature-increase.