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 15 broken link fixed edit approved May 5 at 21:42 Glorfindel 1,53544 gold badges1111 silver badges2020 bronze badges 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 edited Sep 7 '18 at 20:59 santimirandarp 1,27555 silver badges2525 bronze badges 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 edited Sep 7 '18 at 20:45 santimirandarp 1,27555 silver badges2525 bronze badges 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. 12 deleted 862 characters in body edited Sep 7 '18 at 20:31 santimirandarp 1,27555 silver badges2525 bronze badges 11 added 78 characters in body edited Sep 21 '17 at 1:51 santimirandarp 1,27555 silver badges2525 bronze badges 10 added 78 characters in body edited Sep 21 '17 at 1:35 santimirandarp 1,27555 silver badges2525 bronze badges 9 added 277 characters in body edited Sep 21 '17 at 0:45 santimirandarp 1,27555 silver badges2525 bronze badges 8 added 277 characters in body edited Sep 21 '17 at 0:38 santimirandarp 1,27555 silver badges2525 bronze badges 7 added 277 characters in body edited Sep 21 '17 at 0:31 santimirandarp 1,27555 silver badges2525 bronze badges 6 added 382 characters in body edited Sep 21 '17 at 0:12 santimirandarp 1,27555 silver badges2525 bronze badges 5 added 382 characters in body edited Sep 21 '17 at 0:04 santimirandarp 1,27555 silver badges2525 bronze badges 4 deleted 45 characters in body edited Sep 20 '17 at 23:55 santimirandarp 1,27555 silver badges2525 bronze badges 3 added 126 characters in body edited Sep 20 '17 at 23:46 santimirandarp 1,27555 silver badges2525 bronze badges 2 added 18 characters in body edited Sep 20 '17 at 23:26 santimirandarp 1,27555 silver badges2525 bronze badges 1 answered Sep 20 '17 at 23:19 santimirandarp 1,27555 silver badges2525 bronze badges