Question

Use the Born-Haber Cycle to determine the lattice energy of chromium (Il) oxide given the information below. For full credit, write out the chemical equation for each step. 

Answer 1

Cr₂O₃ has 2 chromium and 3 oxygen atoms.

Enthalpy of sublimation of chromium: 368 kJ/mol

2 Cr(s) 2 Cr(g)         kJ

Total ionization energy of chromium (IE₁ + IE₂ + IE₃) = 652.9 + 1590.6 + 2987 = 5230.5 kJ/mol

2 Cr(g) 2 Cr³⁺(g)      kJ

Dissociation energy of oxygen gas : 498 kJ/mol

3/2 O₂ (g) 3 O (g)    kJ

Electron affinity

EA₁ + EA₂ = - 142 + 844 = 702 kJ/mol

3 O(g) 3 O^2- (g) kJ

Enthalpy of formation of 1 mol solid chromium trioxide, kJ

Let us say that lattice energy of 1 mol Cr₂O₃ = U kJ

Now for the cycle:

or,

736 + 10461 + 747 + 2106 - 1128 + U = 0

or, U = - 12922 kJ/mol

Hence, the lattice energy of chromium(III) trioxide = - 12922 kJ/mol