Calculate the lattice energy of sodium chloride (NaCl) using Born-Haber cycle and Hess's law. Given the following data:- Enthalpy of sublimation of sodium (Na): +108.6 kJ/mol- Enthalpy of dissociation of chlorine (Cl2): +243.0 kJ/mol- Bond energy of Na-Cl bond: -412.0 kJ/mol- Ionization energy of sodium (Na): +496.0 kJ/mol- Electron affinity of chlorine (Cl): -349.0 kJ/mol- Heat of formation of sodium chloride (NaCl): -411.0 kJ/mol.

Question

Calculate the lattice energy of sodium chloride (NaCl) using Born-Haber cycle and Hess's law. Given the following data:- Enthalpy of sublimation of sodium (Na): +108.6 kJ/mol- Enthalpy of dissociation of chlorine (Cl2): +243.0 kJ/mol- Bond energy of Na-Cl bond: -412.0 kJ/mol- Ionization energy of sodium (Na): +496.0 kJ/mol- Electron affinity of ch

Answer 1

To calculate the lattice energy of sodium chloride  NaCl  using the Born-Haber cycle and Hess's law, we need to follow these steps:1. Sublimation of sodium  Na : H_sublimation = +108.6 kJ/mol2. Dissociation of chlorine  Cl2 : H_dissociation = 1/2   +243.0 kJ/mol  = +121.5 kJ/mol3. Ionization of sodium  Na : H_ionization = +496.0 kJ/mol4. Electron affinity of chlorine  Cl : H_electron_affinity = -349.0 kJ/mol5. Formation of NaCl: H_formation = -411.0 kJ/molNow, we can use Hess's law to calculate the lattice energy. According to Hess's law, the sum of the enthalpy changes for each step in a reaction is equal to the overall enthalpy change for the reaction.H_lattice = H_sublimation + H_dissociation + H_ionization + H_electron_affinity + H_formationH_lattice =  +108.6 kJ/mol  +  +121.5 kJ/mol  +  +496.0 kJ/mol  +  -349.0 kJ/mol  +  -411.0 kJ/mol H_lattice = 108.6 + 121.5 + 496.0 - 349.0 - 411.0H_lattice = 725.1 - 760.0H_lattice = -34.9 kJ/molTherefore, the lattice energy of sodium chloride  NaCl  is -34.9 kJ/mol.