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# Answer to Question #65520 in Inorganic Chemistry for Abhishek mishra

Question #65520
Write Born-Haber cycle for BaO crystal
1
2017-02-24T14:46:05-0500
Step 1. Sublimation of solid Ba: Ba(s) â†’ Ba(g) , Î”H1 = 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) â†’ Ba2+ (g) + 2e- , Î”H2 = 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:
Â½ O2(g) â†’ O(g), Î”H3 = 249.2 kJ/mol The change in enthalpy is Â½ 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 - â†’ O 2- (g), Î”H4 = 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) â†’ BaO(s) , Î”H5 = -3048 kJ/mol
The value of Î”H5 is negative lattice energy of BaO crystals, because during their formation energy is released.
The equation Ba(s) + Â½ O2(g) â†’ BaO(s) , Î”Hf describes the overall process of BaO formation from solid Ba and gaseous oxygen molecules and is the sum of all above processes:
Î”Hf = Î”H1 + Î”H2 + Î”H3 + Î”H5 = -548 kJ/mol

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