Carefully observe the given diagrams which indicate standard enthalpy of formation of different stat — Thermodynamics and Thermochemistry Chemistry Question
Question
Carefully observe the given diagrams which indicate standard enthalpy of formation of different states of one mole of Mg and two moles of Cl atoms and match the entries in Column I and II provided. [Diagram provides energy levels (in kJ): $+2600$: $\text{Mg}^{2+}\text{(g)} + 2\text{Cl(g)}$; $+2360$: $\text{Mg}^{2+}\text{(g)} + \text{Cl}_2\text{(g)}$; $+1870$: $\text{Mg}^{2+}\text{(g)} + 2\text{Cl}^-\text{(g)}$; $+1110$: $\text{Mg}^{2+}\text{(g)} + 2\text{Cl}^-\text{(aq)}$; $+170$: $\text{Mg(g)} + \text{Cl}_2\text{(g)}$; $0$: $\text{Mg(s)} + \text{Cl}_2\text{(g)}$; $-640$: $\text{MgCl}_2\text{(s)}$; $-790$: $\text{Mg}^{2+}\text{(aq)} + 2\text{Cl}^-\text{(aq)}$]
💡 Solution & Explanation
Extracted from diagram differences: Lattice Enthalpy (D) $= 1870 - (-640) = +2510\text{ kJ}$. Hydration of $2\text{Cl}^- = 1110 - 1870 = -760\text{ kJ}$. Hydration of $\text{Mg}^{2+}$ (C) $= -790 - 1110 = -1900\text{ kJ}$. $\Delta H_f[\text{Cl}^-\text{(g)}] = (1870 - 2360)/2 = -245\text{ kJ}$. $\Delta H_f[\text{Cl}^-\text{(aq)}]$ (B) $= -245 + (-760/2) = -625\text{ kJ}$. $\Delta H_f[\text{Mg}^{2+}\text{(aq)}]$ (A) $= -790 - 2(-625) = +460\text{ kJ}$.