Amount of energy required to split zinc-oxide

[goleynik]

How could I split zinc-oxide into zinc and oxygen? The zinc-oxide would be at 2500 degrees Celsius. Because it is as such a high temperature, would that reduce the amount of energy that I would need to put into separate the two?

Thanks!

 

[Nguyen Quang]

Yes, definitely.

When the substance is heated up such high temperature, they already gain a substantial amount of energy from the the heating process. The energy is used to overcome the electrostatic force of attraction between the Zinc ions and Oxygen ions.

If you are comparing between the required energy to break the bond between Zn and O at room temperature and at 2500 C, you would definitely need to put in more energy in the case of room temperature. This is because at 2500 C, the substance already possesses more energy that is used to break the bond.

 

[goleynik]

What temperature would it have to get to split. Also would the amount of energy to electrolyze it decrease because of the temperature?

 

[Nguyen Quang]

The boiling point of Zinc Oxide is 2360 degree C. At this temperature, the ionic structure is completely destroyed, Zinc and Oxygen ions will escape. With this, the Zinc and Oxygen can be said to have been 'split'.

Well, through my research, there is no specific calculated values for heat of vaporisation for ZnO. As such, I cannot possibly calculate for you the value of energy required for ZnO to be 'split' due to vaporisation. However, there is calculation values for conversion to Zn2+ and O2- ions at 25 degree C (298K), which not likely to happen in reality. I will show it to you anyway.

ZnO(s) [2000 C] → ZnO(s) [25 C] → Zn2+(g) + O2-(g) [25 C]

The calculation of the the energy to split will be as follow:

1) The temperature of ZnO solid at 2000 C will be lowered to 25 C

Heat Capacity: 40.3 J/(mol K) (http://chemicals.etacude.com/z/zinc_oxide.php)

ΔHrxn = -40.3 x (2000-25) = -79600 J/mol

2) The ZnO solid is split into gasous ions at 25 C

Lattice energy of ZnO at 25 C: 4142 kJ/mol (http://www.webelements.com/zinc/lattice_energies.html )

3) Overall ΔH = -796000 + 4142 = -75450kJ/mol



Hope that will help to satiate your curiosity.