Calculate the lattice energy of MgO using Born-Haber Cycle. Given the following information: - The enthalpy change of sublimation of Mg = +1472 kJ/mol- The first ionization energy of Mg = +738 kJ/mol- The second ionization energy of Mg = +1450 kJ/mol- The enthalpy change of electron affinity of O = -141 kJ/mol- The enthalpy change of formation of MgO = -601 kJ/mol- Assume that all other enthalpy changes are negligible.

Question

Calculate the lattice energy of MgO using Born-Haber Cycle. Given the following information: - The enthalpy change of sublimation of Mg = +1472 kJ/mol- The first ionization energy of Mg = +738 kJ/mol- The second ionization energy of Mg = +1450 kJ/mol- The enthalpy change of electron affinity of O = -141 kJ/mol- The enthalpy change of formation of MgO = -601 kJ/mol- Assu

Answer 1

To calculate the lattice energy of MgO using the Born-Haber cycle, we need to consider the following steps:1. Sublimation of Mg  s  to Mg  g : H_sub = +1472 kJ/mol2. Ionization of Mg  g  to Mg^+  g : H_IE1 = +738 kJ/mol3. Ionization of Mg^+  g  to Mg^2+  g : H_IE2 = +1450 kJ/mol4. Electron affinity of O  g  to O^-  g : H_EA1 = -141 kJ/mol5. Formation of MgO  s  from Mg^2+  g  and O^-  g : H_f = -601 kJ/molThe Born-Haber cycle states that the sum of all these enthalpy changes should be equal to the lattice energy  U  of MgO:U = H_sub + H_IE1 + H_IE2 + H_EA1 + H_fSubstitute the given values:U =  +1472 kJ/mol  +  +738 kJ/mol  +  +1450 kJ/mol  +  -141 kJ/mol  +  -601 kJ/mol U = 3919 kJ/mol - 742 kJ/molU = 3177 kJ/molTherefore, the lattice energy of MgO is 3177 kJ/mol.