Question #44183

The standard enthalpy of formation of solid magnesium bromide, $\ce{MgBr2(s)}$, is $-524 \text{ kJ/mol}$ at $298 \text{ K}$. Calculate the standard lattice enthalpy of $\ce{MgBr2(s)}$ given the following thermodynamic data at $298 \text{ K}$:

Enthalpy of sublimation of $\ce{Mg(s)} = 148 \text{ kJ/mol}$ First ionization enthalpy of $\ce{Mg(g)} = 738 \text{ kJ/mol}$ Second ionization enthalpy of $\ce{Mg(g)} = 1451 \text{ kJ/mol}$ Enthalpy of vaporization of $\ce{Br2(l)} = 31 \text{ kJ/mol}$ Bond dissociation enthalpy of $\ce{Br2(g)} = 193 \text{ kJ/mol}$ Electron gain enthalpy of $\ce{Br(g)} = -325 \text{ kJ/mol}$

(Note: Lattice enthalpy is defined as the standard enthalpy change for the conversion of one mole of an ionic solid into its constituent gaseous ions).