Question

# Using standard thermodynamic tables and the following data, compute the crystal enthalpy of AgBr(c) using the...

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

#### Homework Answers

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%

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