Question #44193

The stability of the $+4$ oxidation state in Lanthanides is primarily observed in their fluorides (e.g., \ce{CeF4}, \ce{TbF4}) due to the high lattice enthalpy compensating for the large fourth ionization energy ($IE_4$).

Consider a hypothetical Lanthanide, \ce{Ln}, with the following thermodynamic parameters at $300 \text{ K}$: \begin{itemize} \item Enthalpy of sublimation of \ce{Ln(s)}, $\Delta H_{sub} = 420 \text{ kJ mol}^{-1}$ \item Sum of first three ionization energies, $\Sigma IE_{1-3} = 3730 \text{ kJ mol}^{-1}$ \item Fourth ionization energy, $IE_4 = 3550 \text{ kJ mol}^{-1}$ \item Bond dissociation enthalpy of \ce{F2(g)}, $\Delta H_{diss} = 158 \text{ kJ mol}^{-1}$ \item Electron gain enthalpy of \ce{F(g)}, $\Delta H_{eg} = -328 \text{ kJ mol}^{-1}$ \item Lattice enthalpy of \ce{LnF3(s)}, $U_{L3} = -5900 \text{ kJ mol}^{-1}$ \item Lattice enthalpy of \ce{LnF4(s)}, $U_{L4} = -9616 \text{ kJ mol}^{-1}$ \item Standard molar entropy of \ce{F2(g)} = $200 \text{ J K}^{-1} \text{ mol}^{-1}$ \item Standard molar entropy of \ce{LnF3(s)} = $110 \text{ J K}^{-1} \text{ mol}^{-1}$ \item Standard molar entropy of \ce{LnF4(s)} = $160 \text{ J K}^{-1} \text{ mol}^{-1}$ \end{itemize}

Calculate the standard Gibbs free energy change ($\Delta G^\circ$) for the decomposition reaction at $300 \text{ K}$: $\ce{LnF4(s) -> LnF3(s) + 1/2 F2(g)}$