LSD25
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thermodynamic question: S-I cycle
OK,
Which product is likely to distill off first here?:
I2 + 2H2O + SO2 ==> 2HI + H2SO4
If there is insufficient water to take the reaction to this point - is oleum produced?
Alternatively, is the HI able to strip the water from the sulfuric acid?
Which numbers do I use to work this out? The energy of formation, disassociation, molar energy, entropy or enthalpy, or a combination of them all?
PS This is probably one of the more frustrating parts of the process of learning - because when learning takes place outside of a structured
environment people tend to ignore that which is too hard for them to take in - as I have done here (and elsewhere).  I am sick of the 'not-knowing' and 'excuses', help!
Whhhoooppps, that sure didn't work
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microcosmicus
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| Quote: |
Alternatively, is the HI able to strip the water from the sulfuric acid?
Which numbers do I use to work this out?
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To work out equilibria for chemical reactions, you use the free energy.
When a reaction is in equilibrium, there is no difference of free energy
between the reactants and the products.
| Quote: |
Which product is likely to distill off first here?:
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You also use free energy. When two phases, such as a liquid and
a vapor are in equilibrium, they have the same free energy.
[Edited on 18-3-2008 by microcosmicus]
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LSD25
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So which set of numbers do I use to work out the free energy? Trying to learn this, cos I really think I should know it...
If nothing else, it should save me a lot of time and effort trying things which were never going to work
Whhhoooppps, that sure didn't work
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microcosmicus
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By definition, G = H - TS.
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not_important
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| Quote: | Originally posted by LSD25
OK,
Which product is likely to distill off first here?:
I2 + 2H2O + SO2 ==> 2HI + H2SO4
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This is well know to proceed to the left when heated in an open system - that is a non-equilibrium system where you are removing one or more of
the reactants/products. It spontaneously goes to the right below 370 to 380 K.
| Quote: | If there is insufficient water to take the reaction to this point - is oleum produced?
Alternatively, is the HI able to strip the water from the sulfuric acid?
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SFAIK you will not get SO3 from the reaction.
HI will not strip water from H2SO4
The reaction is run with an excess of water to force it to the right; depending on the intent an excess of I2 (SI cycle for making H2) or SO2 (making
HI for lab use) are also used to drive it to the right.
the attached touches on the thermodynamics of the Bunsen reaction, the part of the SI cycle you are talking about, as well as the separation issues.
[Edited on 18-3-2008 by not_important]
Attachment: USFD - Final SI Cycle paper.pdf (260kB)
This file has been downloaded 734 times
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Nicodem
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| Quote: | Originally posted by LSD25
OK,
Which product is likely to distill off first here?:
I2 + 2H2O + SO2 ==> 2HI + H2SO4
If there is insufficient water to take the reaction to this point - is oleum produced? |
It is an equilibrium reaction (the <==> sign is preferred) meaning that the components are expected to behave in accordance to the equilbrium
equation:
K = ([HI]<sup>2</sup>[H2SO4]) / ([I2][H2O]<sup>2</sup>[SO2])
Where K is the equilibrium constant (variable on T!), the square brackets indicate the activity of those species (mol/L or partial pressure in case of
gases). But you can read more about equilibriums elsewhere. The point is, that when the reaction is out of equilibrium it has a free energy G not equal to zero and it will tend to reach that zero
value (for which the free energy has to be removed/added from/to the system, either by heat transfer or however else). So, if there is no more H2O in
the closed system you expect the equilibrium to compensate, so some H2SO4 will have to oxidize some HI in order to form some I2, SO2 and H2O (the
reaction to the left). (If the system is not closed, then a compensation can alternatively also be reached by heat transfer instead of chemical
transformation.)
For example, in the Karl-Fisher method for measurement of moisture traces exactly the discussed reaction is used. In order to make it unidirectional
(toward the right), pyridine is used so that it deprotonates both acids at the right (HI and H2SO4) efficiently lowering the [HI] and [H2SO4] values.
Thus introducing another equilibrium which forces the first to go to practical completion.
…there is a human touch of the cultist “believer” in every theorist that he must struggle against as being
unworthy of the scientist. Some of the greatest men of science have publicly repudiated a theory which earlier they hotly defended. In this lies their
scientific temper, not in the scientific defense of the theory. - Weston La Barre (Ghost Dance, 1972)
Read the The ScienceMadness Guidelines!
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LSD25
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not_important,
I chose this one because I could look up the answer to some of the questions raised... I don't really want the bare answer, what I am looking for is
how to arrive at that answer...
I found this
http://www.saskschools.ca/curr_content/chem30_05/appendix/ta...
Which is useful at present, then I found the MIT courses again...
http://ocw.mit.edu/OcwWeb/Chemistry/5-111Fall-2005/VideoLect...
Give me a couple of days to work through the lectures and find my way - I'll need time for this
Nicodem,
Yeah, pretty much why I chose this reaction - it is well known thus there is plenty of information available on it - this allows me to work out why I
am wrong when I am wrong. For once I ain't looking for the shortcut to the answer, I want to know how it works.
Microsmicus,
Thanks for the initial help - that helped me find all of the information that is currently causing me to pull my hair out
Whhhoooppps, that sure didn't work
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LSD25
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Double-post, but hell...
I probably picked a hell of a reaction to try and comprehend 'cos the more I look the harder it gets to understand...
Here's what might be an easier one, if 2NaOH is heated, does it convert to Na2O? If so, this might offer a better (simpler) way by which for me to
begin to understand this stuff (why is it that the first year sub's are so bloody difficult to get your head around?)
Oh while looking into getting access to the data for sodium oxide/hydroxide I found this:
http://en.wikipedia.org/wiki/Wikipedia:WikiProject_Chemicals
I dunno, but it looks like they are seeking information which some here might be able (or willing?) to provide... Seems like a better wikipedia would
be a good thing
Whhhoooppps, that sure didn't work
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microcosmicus
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Since you are looking for an easier reaction, may I suggest
starting with a reaction that takes place completely in
the gaseous phase, such as one of the following two
reactions which have been discussed elsewhere on
this site:
Br2 + Cl2 <---> 2 BrCl
SO2Cl2 <----> SO2 + Cl2
Because they occur completely in the gaseous phase,
if we restrict attention to dilute gases we can ignore
condensation of the gases to liquid and the only
thermodynamics required is the ideal gas equation
of state:
PV = nRT
The only other input needed is the difference in
energy between the reactant molecules and the
product molecules or some measurement from
which that could be deduced.
With this little bit of data, we can deduce everything
we would want to know about the thermodynamics
of these systems, such as
* Composition at any temperature and pressure
* Density at any temperature and pressure
* Heat capacity of fixed volume such as Woelen's bottle
* Work done in compressing such a gas adiabatically
* Work done in compressing such a gas isothermally
* Heat released during isothermal compression
* Temperature change during compression
Therefore, I suggest, that before returning to the S-I
cycle or the NaOH, you first work out the thermodynamics
of these two systems or some similar system involving a
single reaction in the gaseous phase. When you
understand how the thermodynamics of this system
works, can derive the equilibrium constant from first
principles, and and compute the quantities mentioned
above, you will have a good grasp on thermodynamics
and will be in a better position to move on to other systems
By starting simple and adding extra complexity like
multiple phases or multiple reactions later, you should
find things easier, as opposed to harder, to understand
the more you look at them.
The system you proposed with NaOH is not much harder ---
it involves two phases but the composition of the gaseous
phase is fixed. That also be a good example to consider.
If you find yourself pulling your hair out again, please don't
go bald over this  Just let me know either via the forum
or private and I will be glad to help as I can.
[Edited on 23-3-2008 by microcosmicus]
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