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.
Ag(g) Ag+(g) + e- 7.576 eV
Br(g) + e- Br-(g) 324.5 kJ/mol
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%
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