Question

Calculate the lattice energy of MgO using the Born-Haber cycle given the following information: the standard enthalpies of formation of MgO(s), Mg(g), and O2(g) are -601.6 kJ/mol, 147.3 kJ/mol and 0 kJ/mol respectively, the ionisation energy of Mg(g) is 737.7 kJ/mol, the electron affinity of O2(g) is -141 kJ/mol, and the bond energy of the O2 molecule is 498 kJ/mol.

Answer 1

To calculate the lattice energy of MgO using the Born-Haber cycle, we need to consider the following steps:1. Sublimation of solid magnesium to gaseous magnesium: Mg s   Mg g 2. Ionization of gaseous magnesium: Mg g   Mg^+  g  + e^- 3. Dissociation of O2 molecule: 1/2 O2 g   O g 4. Formation of O^-  ion: O g  + e^-   O^-  g 5. Formation of MgO lattice: Mg^+  g  + O^-  g   MgO s Now, we can use the given information to calculate the enthalpy changes for each step:1. Sublimation of solid magnesium to gaseous magnesium: H_sublimation = 147.3 kJ/mol2. Ionization of gaseous magnesium: H_ionization = 737.7 kJ/mol3. Dissociation of O2 molecule: H_dissociation = 1/2 * 498 kJ/mol = 249 kJ/mol4. Formation of O^-  ion: H_electron_affinity = -141 kJ/mol5. Formation of MgO lattice: H_lattice = ?The overall enthalpy change for the formation of MgO s  is given as -601.6 kJ/mol. According to the Born-Haber cycle, the sum of the enthalpy changes for all the steps should be equal to the overall enthalpy change:H_sublimation + H_ionization + H_dissociation + H_electron_affinity + H_lattice = H_formationNow, we can plug in the values and solve for H_lattice:147.3 kJ/mol + 737.7 kJ/mol + 249 kJ/mol - 141 kJ/mol + H_lattice = -601.6 kJ/molH_lattice = -601.6 kJ/mol -  147.3 kJ/mol + 737.7 kJ/mol + 249 kJ/mol - 141 kJ/mol H_lattice = -601.6 kJ/mol - 993 kJ/molH_lattice = -1594.6 kJ/molTherefore, the lattice energy of MgO is -1594.6 kJ/mol.