Re: Continuations

On Fri, 19 Jul 2024 11:28:17 -0000 (UTC)
Thomas Koenig <tkoenig@netcologne.de> wrote:

Michael S <already5chosen@yahoo.com> schrieb:

On Thu, 18 Jul 2024 18:48:45 -0000 (UTC)
Thomas Koenig <tkoenig@netcologne.de> wrote:
 

Michael S <already5chosen@yahoo.com> schrieb: 

On Thu, 18 Jul 2024 17:06:52 -0000 (UTC)
Thomas Koenig <tkoenig@netcologne.de> wrote:   

It does work far away from the equilibrium (which is the
point).   

>
But how far?
Does it work for EA/RT outside of, say [-10:+12] ?   

 
Ea is almost always positive; if if was negative, it would mean
that a reaction is is slowed down by increasing temperature (and
by this don't mean the reverse reaction).  

>
In the branch of physical chemistry that I was taught before
switching to completely different profession, that's not that rare.
For example, when temperature grows from 1500 degrees toward 1550
the transfer of phosphorus from molten steel to slag decelerates
greatly. 

 
Do you happen to recall if that process was limited by thermodynamic
equilibrium, by mass transfer or indeed by a chemical reaction?

I don't know, but would think that what causes it is an approuch to
equilibrium. But at 1550 degrees equilibrium is still pretty far away
(50-70 degrees away? I don't remember) and despite that a slowdown is
already very significant.

To me, that would sound like maybe one of the first two, but
knowing nothing about this, I am now firmly speculating :-)
 

And to make sense, -Ea/(RT) is always less then zero.  Reasonable
example: Ea = 30 kJ/mol, R=8.314, T=600 K gives you -6.01 as the
argument of the exponent, so the term is then ~ 0.002444 . 

>
So, something like [-10:0] ? 

 
Activation energies can also be 180 kJ/mol, but I don't really
know the range that people might want to calculate these values.

I would think that when reaction is slowed down by factor of more than
10K  then an exact rate is of lesser interest and the aithmetic with
precision of one octave should be o.k. since precision of the model is
worse than that anyway.