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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}

enter image description here

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  • $\begingroup$ I agree with your first point (Vant Hoff only working when ΔH is constant); however, ΔT is T2−T1. In the link you sent me, the negative sign was assigned to ΔH, but it will cancel out with the other negative sign once we convert the reciprocal temperature difference to T1-T2/T1T2 which is equal to -ΔT/T1T2. $\endgroup$ – Sam202 Sep 20 '17 at 22:39
  • $\begingroup$ Heating implies starting with a lower temperature (T1) and ending with a higher temperature (T2); in other words, ΔT = T2-T1 > 0. If we say that ΔT = T1-T2, then we would get ΔT < 0, which would correspond to cooling. $\endgroup$ – Sam202 Sep 20 '17 at 22:53
  • $\begingroup$ I agree. I think it would make sense anyway since the sulfuric acid is acting as a catalyst and therefore does not participate in the reaction itself. $\endgroup$ – Sam202 Sep 21 '17 at 0:05
  • $\begingroup$ You're calculations are correct, but now I'm doubting the situation with the acid. Even if the acid itself does not suffer any chemical alterations in the reaction, it does give off heat once it dehydrates the alcohol. Wouldn't this heat have to be considered in the enthalpy analysis as well? $\endgroup$ – Sam202 Sep 21 '17 at 0:30
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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.

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  • $\begingroup$ Yes, but if the reaction is exothermic, that means ΔH<0 which would make the argument negative as ΔT is positive. $\endgroup$ – Sam202 Sep 20 '17 at 23:22
  • $\begingroup$ Now I'm 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. $\endgroup$ – Sam202 Sep 20 '17 at 23:40
  • $\begingroup$ What you're 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. $\endgroup$ – Sam202 Sep 20 '17 at 23:48

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