Sciencemadness Discussion Board

Combustion and O2 content of air

axehandle - 28-10-2004 at 13:35

I'm trying to locate information about combustion processes in air at different O<SUB>2</SUB> levels.

Specifically, I want to be able to predict the behaviour of combustion of a sample of sulfur burning in air inside a hermetically sealed chamber. I want to know at what O<SUB>2</SUB> percentage the sulfur combustion becomes unsustainable. The sulfur will be either solid or molten when it gets alit and the chamber is then sealed to the outside. Temperature and pressure will be NTP.

I realize this is probably an extremely complex problem involving burning shape, fluid mechanics, temperature, energy transfer ratios, etcetera etcetera ad infinitum.

However, a simple figure like 5% will do. :)

Does anyone have any ideas?

chemoleo - 28-10-2004 at 16:12

Thats not an easily answered question.
As you know, atmospheric oxygen is at about 20 something percent. (probably about 21.7, reading our oxygen monitor).

If you have something burning in a closed environment, it will be eventually used up. It means, the reaction rate will decrease with decreasing O2 concentration. Eventually the heat of combustion won't be sufficient to maintain the reaction, and this is when the flame will go out.
I know the problem you are speaking of (Bleikammer?), and I guess the solution to your problem is to use a molar EXCESS of sulphur versus oxygen, in your batch.
Even if you have some oxygen and unburned sulphur left, it won't inhibit the reaction in any way. YOu can always relit the sulphur, if necessary, with fresh air.
Anyway, to the very question itself - i doubt very much it is easily posssible to say at what percentage the sulphur flame will stop burning - after all even the temperature of the reaction vessel will determine diffusion rates, and therefore combustion rates will be highly temp dependent. It's a tough question indeed. Just heat the sulphur in an electric coil (i.e. melted), and I'd guarantee you you will use up close to 99% of oxygen.

Dont forget it's an equilibrium. Basically you need to know the equilibrium constant, if the sulphur was as perfectly distributed as the oxygen. So in other words, at some point you can't go beyond the old 98% or whatever the equilibrium value is.

I know it's not a definite value - but I doubt anyone could come up with it. Other than trends, that's all I can give :)

unionised - 30-10-2004 at 04:52

Combustion doesn't happen at NTP.
The equlibrium constant for the reaction will be very large. There will be practically no oxygen left at eqm.

[Edited on 30-10-2004 by unionised]

mick - 30-10-2004 at 09:26

If you burn O2 with S in a sealed container it can not be at NTP. If you get the amounts right, in a sealed container when you open it up you might end up with liquid SO2 or solid SO3 or a mixture. Just a guess.

tokat - 7-11-2004 at 00:37

Use sulpher with KNO3 in knowing amounts, slow lower the KNO3 until the sulpher doesn't burn. ????

unionised - 7-11-2004 at 02:58

In the limit, you will lower the amount of KNO3 to zero and the sulpur will still burn. What will this experiment have shown?