Question

Using standard thermodynamic tables and the following data, compute the crystal enthalpy of AgBr(c) using the Born-Haber cycle. The CRC Handbook lists the value as 904 kJ/mol. Compute the % difference between this value and your result.

Data:

Ag(g) Ag⁺(g) + e^-                    7.576 eV

Br(g) + e^- Br^-(g)                      324.5 kJ/mol

Answer 1

Lattice energy= Heat of formation- Heat of atomization- Dissociation energy- Ionization energies- Electron Affinities

Heat of formation of AgBr = −99.5 KJ/mol

Heat of atomization of Ag = 285 KJ/mol

BOnd dissociation energy of Br2 = 192 KJ/mol / 2 = 96 KJ/mol

Electron affinity = - 324.5 KJ/mol (energy is released)

Ionisation energy = 7.576 eV = 7.576*1.602*10^-19*6*10^23 = 728.205 KJ/mol

1 eV = 1.602*10^-19 J ( for 1 mole we are calculating)

Substituting all values

Lattice energy = - 884.205 KJ/mol

% difference = [904 - 884.205] *100 / 884.205 = 2.46%