What youre saying makes sense if in fact hydration is exothermic and dehydration is endothermic, but other sources like my book (Organic Chemistry Wade) say the latter are exothermic.More
Now Im confused because my book says alcohol dehydration is exothermic. They included an energy chart with the products (alkene) having lower energy than the reagents (alcohol). I will edit my original post to include it.More
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.
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.
More
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