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.

 

[LudusRex]

I can provide you with some information about the amount of energy required to split zinc-oxide and how you can achieve this separation.

Firstly, the amount of energy required to split zinc-oxide into zinc and oxygen will depend on the specific reaction and conditions in which it takes place. However, generally, a significant amount of energy is required to break the strong bonds between zinc and oxygen atoms in zinc-oxide. This energy is known as the bond energy and varies depending on the type of bond. In the case of zinc-oxide, it has a high bond energy, making it challenging to separate into its individual components.

To split zinc-oxide into zinc and oxygen, you will need to provide enough energy to overcome the bond energy and break the bonds between the atoms. This can be achieved through various methods, such as chemical reactions, electrolysis, or thermal decomposition. In the case of thermal decomposition, the zinc-oxide would need to be heated to a temperature of 2500 degrees Celsius, as you mentioned. At this high temperature, the molecules in the zinc-oxide will vibrate more rapidly, and the bonds between the atoms will weaken, making it easier to break them apart.

However, it is essential to note that although a higher temperature may facilitate the process of splitting zinc-oxide, it does not necessarily reduce the amount of energy required. In fact, the higher temperature may require more energy input to maintain it and overcome the energy loss due to heat transfer and other factors. Therefore, the amount of energy required to split zinc-oxide will depend on the specific conditions and methods used.

In conclusion, splitting zinc-oxide into zinc and oxygen requires a significant amount of energy, and the specific amount will depend on the conditions and methods used. Heating zinc-oxide to 2500 degrees Celsius can facilitate the process, but it does not necessarily reduce the energy required. I hope this information helps answer your question.