open system or closed sytem? Of either gas or solution?
The first rxn is the one to consider with an excess of gas, the second one with an excess of solution. You must specify what the conditions are. Or if
it is a fixed amountkatchum - 21-11-2006 at 03:47
Okay,
I think open system: it's a packed wet scrubber.
SO2: 234 kg/h or 65 g/s and 0,087 mass fraction in N2 gas.
So total gas stream = 555 l/s.
NaOH liquid: 422 l/s, pH = unknown, how much pH do I need to absorb all that gas.
Would pH 12 do it? pH can be maintained with a recycle stream and NaOH injection.
[Edited on 21-11-2006 by katchum]YT2095 - 21-11-2006 at 06:14
I`ve just run this experiment for you, it gave me a chance to use my new air pump anyway.
half a gram of NaOH in 200ml of water SRTP.
in the water was also a universal indicator.
SO2 was bubbled through this soln at a rate of 3 l/min.
I found that a dense gas/smoke was evolved as it bubbles through the NaOH scruber, there was no particular smell only a slight H2S odour, nothing
major though.
as the ph reached neutral all was still good, same dense gas/smoke with no overpowering smell or SO2.
as the soln went just a Little bit acidic, the efficiency dropped of sharply and SO2 gas was smelled very strongly.
there was a Very maked drop-off, so that would say to me that your ph of 12 will be More than adequate for use as a scrubber, ph 8 wouldn`t make any
noticable difference either if it was maintained.
those are my findings anyway, it may be helpful katchum - 21-11-2006 at 08:00
Very helpful indeed!
My liquid: 422000ml/s || your liquid: 200 ml/s
My gas: 23 l/s || your gas: 0,05 l/s
So my liquid to your liquid is a factor 2110.
My gas to your gas is a factor 460.
This means you use more SO2 gas than I did, which means it would be a fail-safe prediction. Of course it also depends on mass transfer in the column
and all...
So pH 12,7->8 wouldn't let any of the SO2 go into the atmosphere, which is good.
That pH range apparently is very broad meaning I can really assume that reaction will be irreversibel at pH 12.
Meaning 1 molecule of SO2 will neutralise with 2 molecules of OH- ions.
Somewhere around pH9-10 the second neutralisation reaction won't occur, and at pH 7-8 the first reaction will stop.
I can't be 100% sure if the second reaction will occur at pH 12, but at least... when I have a column of pH 12 everything will be fine. I only need 1
liter/s of NaOH 20% w/w solution to keep the column at pH 12... all the rest 99,8% is recycling of the liquid to the feed, so it's not that important
to have a column at pH 9 or something.
[Edited on 21-11-2006 by katchum]YT2095 - 21-11-2006 at 08:10
well it certainly works on a small scale without a problem, but a Chem Engineer friend of mine told me once that not ALL reactions scale up so
perfectly.
all I can safely say is that it Does work on the scale I tried it at, beyond that I cannot comment.katchum - 21-11-2006 at 08:24
Hey, if the second reaction doesn't occur, then it's even better! One mole of SO2 with one mole of OH- meaning less pH drop. Thus fail-safe.
I'll calculate it with the second reaction in it just to be sure...
[Edited on 21-11-2006 by katchum]YT2095 - 21-11-2006 at 09:37
Quote:
Originally posted by katchum
The reactions are:
SO2 + NaOH = NaHSO3
NaHSO3 + NaOH = Na2SO3 + H2O
I`ve been thinking about this and maybe it`s just me, but something doesn`t quite seem right there?
it`s not that it actualy Backwards looking, but wouldn`t Na2SO3 exist almost right away when you consider that there will already be an excess of NaOH
when the reaction first starts?
so more or less right from the get-go you`ll have Na2SO3 surely?
just a thought woelen - 21-11-2006 at 11:06
I think YT2095 is right. At high pH you definitely will have mainly Na2SO3 and only very small amounts of NaHSO3.Maya - 21-11-2006 at 11:14
well yes, thats ovbious,the rxns as written are incorrect
THE Correct series of reactions are:
SO2 + H20 = H2SO3
H2SO3 + 2 NaOH = Na2SO3 + 2 H20
H2SO3 + Na2SO3 = 2 NaHSO3YT2095 - 21-11-2006 at 11:23
he wasn`t Technicaly wrong though in what in he said, both statement read individualy are quite right.
the addition of further NaOH to NaHso3 will indeed make Na2SO3.
the bit that didn`t sit right with me was his further explainations about the "Other reaction" occuring, when it`s clear that the "Other" reaction
would be the 1`st thing to occur katchum - 21-11-2006 at 15:56
Here they say the reaction SO2 + 2NaOH = Na2SO3 + H2O
Has a low probability of occurance. Which means I think, Maya's three reactions will occur but the further reaction of NaHSO3 with NaOH will not
occur. (And that is what I said earlier)
So, Maya:
The overall reaction from those 3 reactions is: SO2 + NaOH = NaHSO3. So your three equations are simplified to my first equation. But I think there is
another reaction occuring here:
Will this NaHSO3 react further with NaOH to form Na2SO3 or not? If they do, then I'm searching for the pH at which it would occur.
The more I think about all those reactions, the more I get confused.
The other site says:
My second reaction has a pKa of 6,91 = Ka = 0,000000123
So the second reaction is minimal.
If I assume that my first reaction is 100 % then the amount of HSO3- = 0,0024 mol/l.
Assume pH = 12= 0,01 mol/l OH-.
Ka = [SO3,2-][H3O+]/[HSO3-]
0,000000123 = (1.10^-12 + x)(x)/(0,0024-x)
x = 1,7 10^-5
This is the amount of SO3,2- formed. Right? It is negligible.
[Edited on 22-11-2006 by katchum]katchum - 22-11-2006 at 07:59
How much of those HSO3- ions will make SO4--? How can I calculate that?
I ask this because the water has to be dumped and there is an emission limit of 250 mg/l SO4--.YT2095 - 22-11-2006 at 08:17
you`ll not get SO4 from it unless you bubble it through H2O2 or something.
beleive me, a good many wish it was that easy
you`ll need some really good oxidiser involvement for that to happen.katchum - 22-11-2006 at 09:31
Oh, but then it's just perfect, I can concentrate more into the liquid phase. But now I have to research if there is a limit for HSO3-...and SO3--. I
just can't believe I can just dump all that sulfur in the water.
[Edited on 22-11-2006 by katchum]YT2095 - 22-11-2006 at 09:43
you do realise that it will release the SO2 gas over time in the enviroment don`t you?
sulphites do break down!
dumping this into the water is Not a good idea ecologicaly.
sulphate would be much safer ironicaly katchum - 22-11-2006 at 10:01
Are you sure that there isn't a little, tiny bit of SO4 formed? like almost... 250 mg/l?
And about that SO2 in the environment, if there isn't a law... we can't break it.katchum - 22-11-2006 at 10:14
I stumbled on this graph.
Why is it that at pH 12 everything is SO3--? (I thought my second reaction doesn't occur)
Is it because this chart only applies on H2O-SO2 and not H2O-SO2-NaOH?
[Edited on 22-11-2006 by katchum]
Attachment: naam.pdf (28kB) This file has been downloaded 741 times
YT2095 - 22-11-2006 at 10:36
SO2 in H2O alone will make mainly sulphurous acid, there will be a tiny amount of sulphuric present.
SO2 in the environment will indeed form further SO4, there are Many catalysts and Oxidisers available to do the job (think acid rain).
personaly if it were left to me, I`de seek a way to lock it up as a sulphate, Calcium Sulphate would ideal.
I`m certain the other guys here will know exactly how to do that (or better) and be cost effective.katchum - 23-11-2006 at 07:22
Um, is it physically possible to have:
HCl: 3,18 mol/l en SO2: 1,8 mol/l at pH 12?
[Edited on 23-11-2006 by katchum]YT2095 - 23-11-2006 at 08:43
no.katchum - 23-11-2006 at 09:13
And if it were Cl- and HSO3- molecules with Na+ molecules?YT2095 - 23-11-2006 at 09:22
well Cl is an Anion, Na is a Cation they will react, HSO3 is already "Reacted" as it contain both a Cation and Anion, the Na is stronger and so will
displace the H.
you could have the Ions present in a soln though, the same as if you dissolve house salt, you`ll have Na and Cl ions free to move about.
so your question lacks definition somewhat.
I can throw table salt into water and some NaHSO3, dissolve the lot and have exactly that, but at ph 12?katchum - 23-11-2006 at 10:06
NaCl: 36g/100g H2O.
Cl-: 3,18 mol/l or 112 g/l
Na+: 3,18 mol/l or 73 g/l
total: 185 g/l or 185 g/1000g or 18,5 g/100g H2O.
So HCl in H2O is possible up to 3,18 mol/l.
Now for SO2 I don't know how it will interact with pH and NaCl....
But I found solubility of NaHSO3 to be 4,95 mol/l
So I think my assumption of their solubility is correct, unless the interaction between NaCl and NaHSO3 does something to their solubility. (when you
did your experiment, did you see any crystallization?)
The problem is if I add them I get 5 mol/l and maybe that's too much...
I think I'll have to design for a higher liquid flow.
Question: how do you calculate the solubility when there are two kinds of salts involved?
[Edited on 23-11-2006 by katchum]YT2095 - 23-11-2006 at 11:09
it depends upon Loads of things, saturation, common ion effect etc...
salting out by pure displacement (not to be confused with Displacement reactions).
I`m going to step out of this thread for now, as you`re getting into areas I`m not trained in, and I`de rather give you No advice than Bad advice.
I`m more than sure the others here will be able to answer these things for you FAR better than I can.katchum - 23-11-2006 at 11:36
If someone replies, but thanks a lot for all your help!katchum - 26-11-2006 at 11:24
Just to make sure:
Can someone please verify to me that when I add NaOH to a level of pH 12 that all SO2 will convert to SO3--?
See this figure:
Attachment: so2.pdf (61kB) This file has been downloaded 673 times
woelen - 26-11-2006 at 12:30
Keep in mind that adding SO2 reduces the pH, so you have to add fresh NaOH all the time to keep pH at 12.unionised - 26-11-2006 at 12:38
If you add enough NaOH to get the pH up to 12 all the SO2 will be converted to sulphite. In fact, most of it would be converted at pH9katchum - 26-11-2006 at 16:37
Okay, I'll make sure the pH is kept at 12.
I could calculate that this way:
When you have 0,2 mol/s SO2 and you want to have 0,01 mol/l OH- (pH 12) in 2 l/s liquid in the end (or 0,02 mol/s), we can calculate the begin pH
necessary to keep pH at 12.
I know 1 mole of SO2 reacts with 2 moles of OH- so 0,2 mol/s SO2 reacts with 0,4 mol/s OH-. 0,4 mol/s + 0,02 mol/s = 0,42 mol/s at 2 l/s.
So the pH at the beginning is: 0,21 mol/l or 13,3 pH.
Summary: To keep a pH of 12 I need a pH of 13,3 in the beginning to absorb 0,2 mol/s SO2 with liquid flow of 2 l/s.
I could use a pH of 10, maybe that's better for cost. But I have to be sure that the reaction occurs super fast. So it's for some safety protocol or
something.
[Edited on 27-11-2006 by katchum]woelen - 27-11-2006 at 04:53
You should also keep in mind that the curves you posted are equilibrium curves. They do not tell how fast an SO2/NaOH mix reaches the equilibrium and
how well the SO2 is absorbed by the solution. With a pH, equal to 10 (and sufficient buffering to keep it equal to 10), most SO2 will be absorbed,
when left in contact for sufficiently long time, but I can imagine that simply bubbling SO2/air mix through such a liquid does not remove all of it.
Maybe you could use some opposite flow mechanism. Liquid is gpoing in one direction and SO2 is going in the other direction. Where highest
concentration of SO2 is encountered, the liquid has lowest pH, but at that point, this is not of real concern, because a lot of SO2 can be absorbed
yet. Higher up in the scrubber, more alkaline liquid is used and at the top, it is most alkaline. Over there, the last traces of SO2 can be absorbed.
In this way, you could save on costs of NaOH, using a pH equal to 12 or even higher near the end, where concentration of SO2 already has dropped
considerably, while using a lower pH of 10 or so, at the start of the SO2 stream, where concentration still is high.Maya - 27-11-2006 at 06:22
You can do away completely with almost all the NaOH by freezing the SO2 first below the B.P. , -15 or so?katchum - 27-11-2006 at 10:25
I'm doing it counterflow, yes.
How do I estimate the kinetics of the reaction? I really don't know, so keeping the pH high enough is the message. It's a matter of mass transfer,
people who sell scrubbers will give HTU data and we can rely on that. Also, I'm not just bubbling it through NaOH, I'm using a packed scrubber which
gives huge turbulence and good mass transfer. So I hope it will reach equilibrium at every HTU.
Freezing the gas. Hmm, maybe, but you need an installation for cooling which requires space and besides you still need to build that scrubber. Also
that liquid SO2 has to be neutralised so it won't get into the air again. How are we going to do that? (with NaOH? So work for nothing)
You could think, maybe make H2SO4 out of that SO2 but I know there will be a huge amount of HCl in the liquid making things difficult.
