Question #44179

Uranium hexafluoride (\ce{UF6}) exists as a volatile covalent molecule at $298\text{ K}$. To understand why the ionic state is thermodynamically forbidden despite a massive theoretical lattice energy, a student models it as a purely ionic solid lattice, \ce{U^{6+}(F^-)_6}.

Given the following standard thermodynamic data at $298\text{ K}$:

• Enthalpy of sublimation of \ce{U(s)}, $\Delta_{sub}H^\circ = 530 \text{ kJ/mol}$
• Cumulative ionization energy ($\sum_{i=1}^{6} IE_i$) of \ce{U(g)} = $14700 \text{ kJ/mol}$
• Bond dissociation enthalpy of \ce{F2(g)}, $\Delta_{diss}H^\circ = 158 \text{ kJ/mol}$
• Electron gain enthalpy of \ce{F(g)}, $\Delta_{eg}H^\circ = -328 \text{ kJ/mol}$
• Theoretical lattice formation enthalpy of \ce{U^{6+}(F^-)_6(s)}, $\Delta_{lat}H^\circ = -11800 \text{ kJ/mol}$
• Actual standard enthalpy of formation of covalent \ce{UF6(g)}, $\Delta_f H^\circ_{\text{actual}} = -2147 \text{ kJ/mol}$

Calculate the thermodynamic penalty for forcing the ionic model, defined as $\Delta E = \Delta_f H^\circ_{\text{ionic}} - \Delta_f H^\circ_{\text{actual}}$.