Q.Construct a Born-Haber cycle for $\text{MgO}(s)$ and calculate its lattice enthalpy from the following data: enthalpy of sublimation of $\text{Mg}(s) = +146\ \text{kJ mol}^{-1}$; sum of first and second ionization enthalpies of $\text{Mg}(g) = +2189\ \text{kJ mol}^{-1}$; bond dissociation enthalpy of $\text{O}_2(g) = +498\ \text{kJ mol}^{-1}$; first electron gain enthalpy of $\text{O}(g) = -141\ \text{kJ mol}^{-1}$; second electron gain enthalpy of $\text{O}^-(g) = +780\ \text{kJ mol}^{-1}$; standard enthalpy of formation of $\text{MgO}(s) = -601.7\ \text{kJ mol}^{-1}$.

[!TLDR] Rearrange the Born-Haber cycle equation for MgO to solve for the one unknown, lattice enthalpy. [!ANSWER] Lattice enthal …

The Born-Haber cycle equation for $\text{MgO}(s)$, analogous to $\text{NaCl}$ but using BOTH ionization enthalpies of Mg (since $\text{Mg}^{2+}$ requires two electrons removed) and BOTH electron gain enthalpies of oxygen (since $\text{O}^{2-}$ requires two electrons added), is: $\Delta_fH^\circ = \Delta_{sub}H(\text{Mg}) + \left[IE_1+IE_2\right](\text{Mg}) + \tfrac{1}{2}\Delta_{diss}H(\text{O}_2) + \left[\Delta_{eg1}H + \Delta_{eg2}H\right](\text{O}) + \left[-\Delta_{lattice}H\right]$. Substituting: $-601.7 = 146 + 2189 + \tfrac{1}{2}(498) + (-141) + (780) + \left[-\Delta_{lattice}H\right]$. Computing $\tfrac12(498)=249$, then summing the known terms in order: $146+2189=2335$; $2335+249=2584$; $2584+(-141)=2443$; $2443+780=3223$. So the equation becomes $-601.7 = 3223 - \Delta_{lattice}H$, giving $\Delta_{lattice}H = 3223-(-601.7) = 3223+601.7 = 3824.7\ \text{kJ mol}^{-1}$. This is roughly five times larger than the …

Write the Born-Haber cycle equation for MgO including BOTH ionization steps and BOTH electron-gain steps, substitute …

Using only the first ionization enthalpy of Mg (forgetting Mg2+ requires removing 2 electrons); using only the first, exothermic electron gain enthalpy of O and …