Question

Calculate the enthalpy of formation of NaCl(s) using the following information:1. The enthalpy change of formation of Na(g) is +108.5 kJ/mol.2. The enthalpy change of formation of Cl2(g) is 0 kJ/mol.3. The enthalpy change of vaporization of Na is +97.42 kJ/mol.4. The enthalpy change of dissociation of Cl2 is +242.76 kJ/mol.

Answer 1

To calculate the enthalpy of formation of NaCl s , we will use the Born-Haber cycle, which relates the lattice energy, ionization energy, electron affinity, and other energy changes involved in the formation of an ionic solid.The formation of NaCl s  can be represented by the following equation:Na s  + 1/2 Cl2 g   NaCl s We can break this process down into several steps:1. Vaporization of Na s  to Na g : Na s   Na g  ; H = +97.42 kJ/mol2. Ionization of Na g  to Na+ g : Na g   Na+ g  + e- ; H = +108.5 kJ/mol3. Dissociation of Cl2 g  to 2 Cl g : 1/2 Cl2 g   Cl g  ; H = +242.76 kJ/mol / 2  since we only need 1 Cl atom 4. Formation of Cl- g  from Cl g : Cl g  + e-  Cl- g  ; H = Electron affinity of Cl = -349 kJ/mol  not given in the problem, but a known value 5. Formation of NaCl s  from Na+ g  and Cl- g : Na+ g  + Cl- g   NaCl s  ; H = Lattice energy of NaCl  unknown Now, we can write the overall enthalpy change for the formation of NaCl s  as:H_f NaCl  = H_vaporization Na  + H_ionization Na  + H_dissociation Cl2 /2 + H_electron_affinity Cl  + Lattice_energy NaCl Plugging in the given values:H_f NaCl  = 97.42 kJ/mol + 108.5 kJ/mol +  242.76 kJ/mol / 2  - 349 kJ/mol + Lattice_energy NaCl H_f NaCl  = 205.92 kJ/mol - 349 kJ/mol + Lattice_energy NaCl Now, we know that the lattice energy of NaCl is -787 kJ/mol  a known value, not given in the problem . Plugging this value in:H_f NaCl  = 205.92 kJ/mol - 349 kJ/mol - 787 kJ/molH_f NaCl  = -930.08 kJ/molSo, the enthalpy of formation of NaCl s  is -930.08 kJ/mol.