Question

write Born-Haber cycle for BaO crystal.

Answer 1

The system below represents Born-Haber cycle for BaO crystal. Step 1. Sublimation of solid Ba: Ba(s) → Ba(g) , ΔH1 = 180.0 kJ/mol During this process, the change in enthalpy is positive, energy is required to change the state of barium from solid to gaseous (to break metal lattice). Step 2. The next step is ionization of atomized Ba gas: Ba(g) → Ba2+ (g) + 2e- , ΔH2 = 1468.1 kJ/mol Again change in enthalpy is positive, energy is required to take away two electrons from barium atom. Step 3. Dissociation of O2 molecule into atoms is presented with the following equation: ½ O2(g) → O(g), ΔH3 = 249.2 kJ/mol The change in enthalpy is ½ from dissociation energy. Step 4. Electron affinities of oxygen is the energy, required to maintain the process of electron accepting by oxygen atom: O(g) + 2e - → O 2- (g), ΔH4 = 603 kJ/mol Step 5. Now when Ba2+ ions and O2- ions are available, crystals of BaO are formed according to the scheme: Ba2+ (g) + O2- (g) → BaO(s) , ΔH5 = -3048 kJ/mol The value of ΔH5 is negative lattice energy of BaO crystals, because during their formation energy is released. The equation Ba(s) + ½ O2(g) → BaO(s) , ΔHf describes the overall process of BaO formation from solid Ba and gaseous oxygen molecules and is the sum of all above processes: ΔHf = ΔH1 + ΔH2 + ΔH3 + ΔH5 = -548 kJ/mol

Answer 2

This can occur either by direct combination or alternative process. Below is the cycle for this process.

Ba (s) +1/2O2(g)====>BaO(s) ∆H1=Q 
now,
alternatively
Ba(s)---------->Ba(g) ∆H=S
Ba(g)--------->Ba+ +e^- ∆H=I
1/2O2(g)-------->O(g) ∆H=D/2
O(g) +e^- ------>O-(g) ∆H=-E
Ba+ +O- ----->BaO(s) ∆H=-U