From: lamunyj@jacobs.CS.ORST.EDU (James Lamunyon)
Newsgroups: sci.chem,rec.scuba
Subject: Re: Question on partial pressure of O_2 and combustion
Date: 28 Aug 92 17:31:02 GMT
Article-I.D.: leela.1992Aug28.173102.4440
Organization: Computer Science Outreach Services - Oregon State University

In article <1992Aug26.204553.14684@mailer.cc.fsu.edu> mayne@cs.fsu.edu writes:
>[Cross posted to rec.scuba.]
>
>Perhaps some chemists out there can help us software types (and
>others) in rec.scuba understand a problem regarding making
>oxygen enriched air mixtures for scuba diving. The problem centers
>around the role of partial pressure and fraction of oxygen in a gas
>leading to spontaneous combustion of hydrocarbons. To what extent
>does combustion depend only on partial pressure? Is roughly 60%
>nitrogen enough to inhibit combustion which would have resulted
>from pure oxygen at 0.4 times the total pressure coming into contact
>with hydrocarbons? Someone has given the example of hydrogen-oxygen
>mixtures with a low fraction of O_2 which do not burn even under
>high enough pressure that the partial pressure of O_2 is 1 atmosphere
>to argue that the fraction of O_2 is critical. I believe we are talking
>less than 4% O_2 and hence pressures above 25 atmospheres, equivalent
>to depths over 800 feet, not that many of us recreational divers use
>hydrogen-oxygen or go anywhere near that deep.
>
>Here is a specific application: A tank (constant volume) is filled to a
>given pressure X with pure oxygen. Obviously at this point the partial
>pressure of O_2 is X. Air is then added, reducing the fraction of oxygen
>to 40% or less, but obviously raising the partial pressure well above X.
>I estimate it is close to 1.6*X partial pressure of O_2 and 4*X total
>pressure for a 40% mix. Certainly the partial pressure of O_2 must go
>up as air is added. It is amazing that not everyone realizes that.
>Some who know better once their error is pointed out just think they are
>diluting the oxygen by adding air. Realistically the initial pressure of
>pure oxygen will be in the range of about 600 to 800 psi. Standards are
>that for mixtures above 40% oxygen-safe equipment should be used regardless
>of how the mix is made.
>
>If the final mixture is safe as far as spontaneous combustion from
>hydrocarbon residues go is it safe at the point that you have pure
>O_2 at a lower partial pressure, i.e. does the nitrogen somehow
>inhibit combustion? Or is there some other mechanism at work here?
>
>I'd appreciate an explanation of the chemistry understandable to
>non-chemist diver.
>
>Bill Mayne

     Spontaneity refers to the possibility that a process may take place.
We can't be definite that a spontaneous will proceed in a specific time,
but only that it will proceed given an infinite amount of time.  In other
words we can predict whether a reaction will proceed or not, but we can
not easily predict the rate.  If we can show that a process is not 
spontaneous, than we can guarantee that it will never occur.
      Think of a spontaneous reaction like a boulder perched on the top
of a hill.  If we were to give it a shove it would roll to the bottom of the
hill.  An adult would say that this is because of gravitational attraction, 
etc..  This is essentially correct, but we could also express the answer
in terms of potential energy which arise from the gravitational force.
      There is an equation of state called the Gibbs free energy which is

	G=H-TS   Where H is the enthalpy (energy liberated or consumed), T
		 is the temperature, and S is called the entropy.

If we take the differential we get:

	dG=dH-TdS-SdT  (We are looking at changes in these quantities)

If we hold the process at constant T, dT (change in temp)=0 thus

	dG=dH-TdS   

      If dG<0 at a given temperature and pressure, the process is said
to be spontaneous.  In the case of combustion, the process is obviously
spontaneous since we can burn gas in our autos.  However, if we increase
the pressure, we can increase the rate at which the combustion takes place.
Once again this is what happens in our cars when the piston compresses the
gas/air mixture.  
      Without getting too technical if we assume the gasses are "ideal"
gases we can actually say something about the rate.  
      There is another equation of state that says:

	    log K = dG/ -2.303 RT   R is the ideal gas constant, and K
				    is the equilibrium constant for the
				    process.  At equilibrium (no potential
				    energy left) K=0.  But if the process
				    is spontaneous, K is not 0.  Larger
				    K indicates faster process.

      The ideal gas law states pV= nRT.  If n is refers to the number of
molecules of gas you have, p is pressure and V is volume.  We can divide
both sides of the equation by n to give RT=pV/n  we can then substitute
into the equation for the equilibrium constant to give us a rate expression
dependant on the pressure.  Admittedly this has some major assumptions,
but at high enough of a pressure it won't really matter much since you will
have a major explosion on your hands if dealing with hydrocarbons.
      Back to the analogy of a boulder on a cliff.  We had to apply a force
(activation) energy to get it rolling.  Once on its way it proceeded just
fine without any help.  Like the boulder, to combust a hydrocarbon we have to
provide an activation energy (like from your spark plug in your car).  When
we increase the pressure we lower the activation energy thus at high pressures
any input of energy into a system of hydrocarbon/oxygen, look out.

      I hope this helps.  In the real world of diving you also have other
problems such as oxygen toxicity to worry about.  One side line to all of
this is that if you are ever placed in a hyperbaric chamber, often they
dress you in static resistant clothing before putting you in a high pressure
oxygen invironment.  A spark of static (activation energy) and the diver
himself will become an interesting sponataneous process.  Unfortunatelly,
you wouldn't be around long to comment on it.  ;-)

	James LaMunyon
	OSU Chemistry student

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