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

The standard enthalpy for hydration of ethene is $\Delta H^\circ_r=-45\frac{KJ}{mol}$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.

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.

14 deleted 250 characters in body
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Let's suppose a reaction liketake: $\,\,\,\,\ce{ethene(g) + H2O(g) <=> ethanol(g)} $

\begin{align} \ce{ethene(g) + H2O(g) <=> ethanol(g)} \end{align} 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 endothermicendothermic.

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 T'sTs 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 we were expecting.  

Finally, you can think on the whole process as:As $$K = \frac{[\ce{H2O}]\,[\ce{CH2CH2}]}{[\ce{CH3CH2OH}]}$$

\begin{align} \ce{ethanol <=> ethene + H2O} \\ \Delta H^\circ_r>0 \end{align} Where I have assumed and $\Delta H>0$ for the aqueous phase reaction$K_2>K_1$. Remember ethanol will be more stabilized in waterIn consequence dehydration enthalpy is favoured by hydrogen bondingtemperature-increase.

Let's suppose a reaction like:

\begin{align} \ce{ethene(g) + H2O(g) <=> ethanol(g)} \end{align} 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 T's 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 we were expecting.  

Finally, you can think on the whole process as:

\begin{align} \ce{ethanol <=> ethene + H2O} \\ \Delta H^\circ_r>0 \end{align} Where I have assumed $\Delta H>0$ for the aqueous phase reaction. Remember ethanol will be more stabilized in water by hydrogen bonding.

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.

13 deleted 250 characters in body
source | link

Let's suppose a reaction like:

\begin{align} \ce{ethene(g) + H2O(g) <=> ethanol(g)} \end{align} 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 T's 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 we were expecting.

Finally, you can think on the whole process as:

\begin{align} \ce{ethanol <=> ethene + H2O} \\ \Delta H^\circ_r>0 \end{align} Where I have assumed $\Delta H>0$ for the aqueous phase reaction. Remember ethanol will be more stabilized in water by hydrogen bonding. \begin{align} \ce{H2SO4 <=>HSO4- + H+}\\ \Delta H^\circ_r<0 \end{align} and using Hess Law,the reaction becomes possible. (We are asuming net entropy=0). I must say I am not totally sure of this last point, that's the best I can think for now..

Let's suppose a reaction like:

\begin{align} \ce{ethene(g) + H2O(g) <=> ethanol(g)} \end{align} 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 T's 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 we were expecting.

Finally, you can think on the whole process as:

\begin{align} \ce{ethanol <=> ethene + H2O} \\ \Delta H^\circ_r>0 \end{align} Where I have assumed $\Delta H>0$ for the aqueous phase reaction. Remember ethanol will be more stabilized in water by hydrogen bonding. \begin{align} \ce{H2SO4 <=>HSO4- + H+}\\ \Delta H^\circ_r<0 \end{align} and using Hess Law,the reaction becomes possible. (We are asuming net entropy=0). I must say I am not totally sure of this last point, that's the best I can think for now..

Let's suppose a reaction like:

\begin{align} \ce{ethene(g) + H2O(g) <=> ethanol(g)} \end{align} 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 T's 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 we were expecting.

Finally, you can think on the whole process as:

\begin{align} \ce{ethanol <=> ethene + H2O} \\ \Delta H^\circ_r>0 \end{align} Where I have assumed $\Delta H>0$ for the aqueous phase reaction. Remember ethanol will be more stabilized in water by hydrogen bonding.

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