Thermodynamics and ThermochemistryhardMCQ SINGLE

Study the following thermodynamic data given by E. H. P. Cordfunke, A. S. Booji and M. Y. FurkalionkThermodynamics 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?

Answer: B

💡 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 = $180.06 - 699.43 - 474.93 = -994.30\text{ kJ/mol}$.

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