The enthalpies of formation of , , and are –70.97, –68.32, –22.1 and –188.84 (kcal mol). The enthalp — Thermodynamics and Thermochemistry Chemistry Question
Question
The enthalpies of formation of $\text{SO}_2\text{(g)}$, $\text{H}_2\text{O(l)}$, $\text{HCl(g)}$ and $\text{H}_2\text{SO}_4\text{(l)}$ are –70.97, –68.32, –22.1 and –188.84 (kcal mol$^{-1}$). The enthalpies of solution of $\text{SO}_2$, $\text{H}_2\text{SO}_4$ and HCl in dilute solution are –8.56, –22.05 and –17.63 (kcal mol$^{-1}$), respectively. What is the enthalpy change for the reaction?<br>$\text{SO}_2\text{(aq)} + \text{Cl}_2\text{(g)} + 2\text{H}_2\text{O(l)} \to \text{H}_2\text{SO}_4\text{(aq)} + 2\text{HCl(aq)}$
💡 Solution & Explanation
Evaluate $\Delta H_f$ for aqueous species: $\text{SO}_2\text{(aq)} = -70.97 - 8.56 = -79.53$; $\text{H}_2\text{SO}_4\text{(aq)} = -188.84 - 22.05 = -210.89$; $\text{HCl(aq)} = -22.1 - 17.63 = -39.73$. $\Delta H_{rxn} = [\Delta H_f(\text{H}_2\text{SO}_4\text{,aq}) + 2\Delta H_f(\text{HCl,aq})] - [\Delta H_f(\text{SO}_2\text{,aq}) + 2\Delta H_f(\text{H}_2\text{O,l})] = [-210.89 + 2(-39.73)] - [-79.53 + 2(-68.32)] = -290.35 - (-216.17) = -74.18\text{ kcal}$.