Q.Calculate the lattice energy of $CaCl_2$ from the given data $Ca(s) + Cl_2(g) \rightarrow CaCl_2(s)$, $\Delta H_f^0 = -795$ kJ mol$^{-1}$; Sublimation: $Ca(s) \rightarrow Ca(g)$, $\Delta H_1^0 = +121$ kJ mol$^{-1}$; Ionisation: $Ca(g) \rightarrow Ca^{2+}(g) + 2e^-$, $\Delta H_2^0 = +2422$ kJ mol$^{-1}$; Dissociation: $Cl_2(g) \rightarrow 2Cl(g)$, $\Delta H_3^0 = +242.8$ kJ mol$^{-1}$; Electron affinity: $Cl(g) + e^- \rightarrow Cl^-(g)$, $\Delta H_4^0 = -355$ kJ mol$^{-1}$

Lattice energy $\Delta H_{lattice}$ is the energy required to completely separate the ions of one mole of an ionic crystal to infinite distance -- for NaCl, $NaCl(s)\rightarrow Na^+(g)+Cl^-(g)$, $\Delta H_{lattice}=+788$ kJ mol$^{-1}$. Because this cannot be measured directly (no calorimeter watches a crystal fly apart into gaseous ions), it is calculated indirectly via the Born-Haber cycle (Max Born and Fritz Haber), which applies Hess's law to the formation of an ionic solid $MX$ by breaking it into five measurable steps: sublimation of the metal ($\Delta H_1$, $M(s)\rightarrow M(g)$), dissociation of the halogen ($\Delta H_2$, $\tfrac{1}{2}X_2(g)\rightarrow X(g)$), ionisation of the metal ($\Delta H_3$, $M(g)\rightarrow M^+(g)$), electron affinity of the halogen ($\Delta H_4$, $X(g)\rightarrow X^-(g)$), and the lattice enthalpy $U$ itself ($M^+(g)+X^-(g)\rightarrow MX(s)$), summed as $\Delta H_f=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4+U$. Since every …