Study the following thermodynamic data given by E. H. P. Cordfunke, A. S. Booji and M. Y. Furkalionk — Thermodynamics and Thermochemistry Chemistry Question
Question
Study the following thermodynamic data given by E. H. P. Cordfunke, A. S. Booji and M. Y. Furkalionk.<br>(i) $\text{DyCl}_3\text{(s)} \to \text{DyCl}_3\text{(aq, in 4.0 M-HCl)}; \Delta H^\circ = -180.06\text{ kJmol}^{-1}$<br>(ii) $\text{Dy(s)} + 3\text{HCl(aq, 4.0 M)} \to \text{DyCl}_3\text{(aq, in 4.0 M-HCl)} + 3/2\text{ H}_2\text{(g)}; \Delta H^\circ = -699.43\text{ kJ mol}^{-1}$<br>(iii) $1/2\text{ H}_2\text{(g)} + 1/2\text{ Cl}_2\text{(g)} \to \text{HCl (aq, 4.0 M)}; \Delta H^\circ = -158.31\text{ kJ mol}^{-1}$<br>What is $\Delta H_f^\circ$ of $\text{DyCl}_3\text{(s)}$ from these data?
💡 Solution & Explanation
Target: $\text{Dy(s)} + 3/2\text{ Cl}_2\text{(g)} \to \text{DyCl}_3\text{(s)}$. Combine via Hess's law: Reverse (i) [+180.06], add (ii) [–699.43], and add $3 \times$ (iii) [$3 \times -158.31 = -474.93$]. Sum = $180.06 - 699.43 - 474.93 = -994.30\text{ kJ/mol}$.