Calculate the enthalpy of formation (in kcal/mol) of HI(g) from the following data:<br><br><br><br>< — Thermodynamics and Thermochemistry Chemistry Question
Question
Calculate the enthalpy of formation (in kcal/mol) of HI(g) from the following data:<br>$\text{H}_2\text{(g)} + \text{Cl}_2\text{(g)} \to 2\text{HCl(g)}; \Delta H = -44.20\text{ kcal}$<br>$\text{HCl(g)} + \text{aq} \to \text{HCl(aq)}; \Delta H = -17.31\text{ kcal}$<br>$\text{HI(g)} + \text{aq} \to \text{HI(aq)}; \Delta H = -19.21\text{ kcal}$<br>$\text{KOH(aq)} + \text{HCl(aq)} \to \text{KCl(aq)}; \Delta H = -13.74\text{ kcal}$<br>$\text{KOH(aq)} + \text{HI(aq)} \to \text{KI(aq)}; \Delta H = -13.67\text{ kcal}$<br>$\text{Cl}_2\text{(g)} + 2\text{KI(aq)} \to 2\text{KCl(aq)} + \text{I}_2\text{(s)}; \Delta H = -52.42\text{ kcal}$
💡 Solution & Explanation
Target reaction: $1/2\text{H}_2\text{(g)} + 1/2\text{I}_2\text{(s)} \to \text{HI(g)}$. Summing appropriately to cancel intermediates evaluates to $\Delta H = \frac{1}{2} [ -44.20 + 2(-17.31) + 2(-13.74) - (-52.42) - 2(-13.67) - 2(-19.21) ]$. This resolves to $\frac{1}{2} [11.88] = +5.94\text{ kcal/mol}$.