Born-Haber cycle exercise

Question

Calculate the lattice energy of CuBr(s) using a Born-Haber cycle. Enthalpy of sublimation of Cu(s) = 337.7 kJ/mol 1^st ionization energy of Cu(g) = 745 kJ/mol Enthalpy of vaporization of Br_2 (l) = 29.96 kJ/mol Bond dissociation energy of Br_2(g) = 193.87 kJ/mol 1^st electron affinity of Br(g) = -324.54 kJ/mol Standard enthalpy of formation of CuBr(s) = -104.6 kJ/mol



Answer 1

Standard heat of formation Cu(m) + 1/2 Br₂ (l) -> CuBr₂(s) is -104.6 kJ and, using the BH cycle we will use a bunch of processes to get from the reactants to the products and, from Hess's Law, the sum of the enthalpies will give us the heat of formation. (The lattice energy will be one of those energies: So

Cu -> Cu(g) 337.7 Sublimation

1/2 Br₂(l) ->1/2 Br₂ (g) 1/2 (29.96 ) Vaporization energy of 1/2 mole of Br₂

1/2Br₂ -> Br(g) 1/2(193.87) Dissociation Energy of1/2 mole of Br₂

Br(g) + e^- -> Br^- 324.54 EA of Br

Cu -> Cu + e^- 745 IE₁ of Cu

Cu⁺ + Br^- --> CuBr ΔH_Lat Lattice Energy

The sum of the column = heat of Formation of CuBr (Solve for Lattice E after you make sure you understand each step and check the numbers)

Answer 2

Cu + 1/2 Br₂ ==> CuBr

Cu(s) ==> Cu(g) .. 337.7 ∆H_sub

Cu(g) ==> Cu⁺ .. 745 IE

Br₂(l) ==> Br₂(g) .. 29.96 Vaporization

Br₂(g) ==> 2Br(g) .. 1/2 x 193.87 = 96.93 Dissociation E

Br(g) ==> Br^-(g) .. -324.54 EA

Cu⁺(g) + Br^-(g) ==> CuBr(s) LE = ?

∆H_formation = ∆H_sub + IE + Dissociation E + EA + LE

-104.6 = 333.7 + 745 + 29.96 + 96.93 - 324.54 + LE

-104.6 = 881,05 + LE

LE = -985.45 kJ/mol