The enthalpy of formation of KCl(s) from the following data is<br>(i) <br>(ii) <br>(iii) <br>(iv) <b — Thermodynamics and Thermochemistry Chemistry Question
Question
The enthalpy of formation of KCl(s) from the following data is<br>(i) $\text{KOH(aq)} + \text{HCl(aq)} \to \text{KCl(aq)} + \text{H}_2\text{O(l)}; \Delta H = -13.7\text{ kcal}$<br>(ii) $\text{H}_2\text{(g)} + 1/2 \text{O}_2\text{(g)} \to \text{H}_2\text{O(l)}; \Delta H = -68.4\text{ kcal}$<br>(iii) $1/2 \text{H}_2\text{(g)} + 1/2 \text{Cl}_2\text{(g)} + \text{aq} \to \text{HCl(aq)}; \Delta H = -39.3\text{ kcal}$<br>(iv) $\text{K(s)} + 1/2 \text{O}_2\text{(g)} + 1/2 \text{H}_2\text{(g)} + \text{aq} \to \text{KOH(aq)}; \Delta H = -116.5\text{ kcal}$<br>(v) $\text{KCl(s)} + \text{aq} \to \text{KCl(aq)}; \Delta H = +4.4\text{ kcal}$
💡 Solution & Explanation
Target: $\text{K(s)} + 1/2\text{Cl}_2\text{(g)} \to \text{KCl(s)}$. Perform Eq(iv) + Eq(iii) + Eq(i) - Eq(ii) - Eq(v): $(-116.5) + (-39.3) + (-13.7) - (-68.4) - (+4.4) = -169.5 + 68.4 - 4.4 = -105.5\text{ kcal/mol}$.