Calculate the enthalpy of formation (in kcal/mol) of anhydrous from the following data:<br><br><br>< — Thermodynamics and Thermochemistry Chemistry Question
Question
Calculate the enthalpy of formation (in kcal/mol) of anhydrous $\text{Al}_2\text{Cl}_6$ from the following data:<br>$2\text{Al(s)} + 6\text{HCl(aq)} \to \text{Al}_2\text{Cl}_6\text{(aq)} + 3\text{H}_2\text{(g)}; \Delta H = -239.760\text{ kcal}$<br>$\text{H}_2\text{(g)} + \text{Cl}_2\text{(g)} \to 2\text{HCl(g)}; \Delta H = -44\text{ kcal}$<br>$\text{HCl(g)} + \text{aq} \to \text{HCl(aq)}; \Delta H = -17.315\text{ kcal}$<br>$\text{Al}_2\text{Cl}_6\text{(s)} + \text{aq} \to \text{Al}_2\text{Cl}_6\text{(aq)}; \Delta H = -153.690\text{ kcal}$
💡 Solution & Explanation
Target: $2\text{Al} + 3\text{Cl}_2 \to \text{Al}_2\text{Cl}_6\text{(s)}$. Combine equations: Eq1 + 3(Eq2) + 6(Eq3) - Eq4. $\Delta H = -239.760 + 3(-44) + 6(-17.315) - (-153.690) = -239.760 - 132 - 103.89 + 153.690 = -321.960\text{ kcal/mol}$.