Question

What is the lattice energy of NaCl(s) at 25°C given that the enthalpy of formation of NaCl(s) is -411.2 kJ/mol, the enthalpy of sublimation of Na(s) is 107.29 kJ/mol, and the first ionization energy of Na(g) is 495.8 kJ/mol?

Answer 1

To calculate the lattice energy of NaCl s , we can use the Born-Haber cycle, which relates the lattice energy to other known thermodynamic quantities. The lattice energy  U  can be calculated using the following equation:U = Hf - Hsub - IE - EA - Hdisswhere:Hf = enthalpy of formation of NaCl s  = -411.2 kJ/molHsub = enthalpy of sublimation of Na s  = 107.29 kJ/molIE = first ionization energy of Na g  = 495.8 kJ/molEA = electron affinity of Cl g   we need to find this value Hdiss = enthalpy of dissociation of Cl2 g   we need to find this value First, let's find the electron affinity  EA  of Cl g . The electron affinity of Cl is -349 kJ/mol.Next, we need to find the enthalpy of dissociation  Hdiss  of Cl2 g . The bond dissociation energy of Cl2 is 242 kJ/mol. Since we are breaking one mole of Cl2 into two moles of Cl, the enthalpy of dissociation is half of the bond dissociation energy:Hdiss = 242 kJ/mol / 2 = 121 kJ/molNow we can plug all the values into the equation:U = -411.2 kJ/mol - 107.29 kJ/mol - 495.8 kJ/mol -  -349 kJ/mol  - 121 kJ/molU = -411.2 + 107.29 + 495.8 + 349 - 121U = 419.89 kJ/molThe lattice energy of NaCl s  at 25C is approximately 419.89 kJ/mol.