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What role does $\ce{CuSO4}$ has in the gas sampling of an $\ce{SO2/SO3}$ gas mix in a catalytic converter for sulphuric acid production?

To give you a little context: We take gas samples from the beds of the catalytic converter to determine the efficiency of the catalyst. We use "cartridges" that we make with glass tubes filled with $\ce{CuSO4}$. The $\ce{CuSO4}$ (lab grade) comes in crystal version as a pentahydrate. We heat it for around $\pu{30 minutes}$ at $\pu{105 ^\circ C}$ and then we use it to make the cartridges. The cartridges are placed between the sampling point and the sampling bag (made of Teflon). This is a process that has been made since many years but nobody here is able to tell me why. The answer they give me is that the cartridges are used to "remove" the $\ce{SO3}$ from the sample but I can't see how that could happen. My theory is that because anhydrous $\ce{CuSO4}$ is hygroscopic (I've read it's been used as a desiccant) it may be used to avoid the formation of $\ce{H2SO4}$ that could damage the gas analyzer (a SICK S700). I've really googled it but can´t find the answer. Hope someone could enlighten me in the subject.

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    $\begingroup$ Where do you install your cartridges? I suppose you are oxidizing a gaseous current of SO2 + O2 over a catalyst (either vanadium oxide or platinum). SO2 is oxidized in SO3, and this gaseous current is dissolved into somewhat diluted sulfuric acid. Where do you put your cartridges in this setup ? Does the gas go through your cartridges ? Before or after catalytic oxidation? $\endgroup$ – Maurice Nov 23 '19 at 21:34
  • $\begingroup$ That's exactly the process as you describe it. We use the cartridges on sampling points after and before each bed. The gas passes through the cartridges and ends in the sampling bag. The gas analyzer measures %SO2 and %O2 v/v and with this data we determine the conversion to SO3 un each bead and of the full reactor. $\endgroup$ – Ed Corso Nov 24 '19 at 21:32
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Heating hydrate copper sulfate at 105 degrees does nothing. The chemically bound water takes a higher temperature to remove all water molecules (water is lost in steps as a function of temperature). This low temperature removes only two water molecules out of five. I think the major role of copper sulfate, besides being a gentle dessicant, is to oxidize sulfur dioxide to trioxide. Cu(II) ion is a mild oxidizing agent, and mild enough that it can oxidize sulfur dioxide to the trioxide.

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  • $\begingroup$ I don't think that would be the case because the gas analyzer only measures SO2 and O2 in %volume. Or do you think this oxidation has another role? $\endgroup$ – Ed Corso Nov 24 '19 at 21:29
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    $\begingroup$ You would have to share the whole protocol with us. By sharing a small section of a sample preparation, nobody can predict what is going on and what is being analyzed. $\endgroup$ – M. Farooq Nov 24 '19 at 22:56
  • $\begingroup$ That's all the sample preparation really, and the copper sulphate it's just prepared that way: heated 30 min at 105 Celsius. The copper sulphate it's poured in the glass tubes (they are about 15 cm long) each and both sides of the tube closed with a small piece of ceramic fiber. The sample is taken before and after each catalyst bed on bags then it's taken to the analyzer and we get %SO2 and %O2 v/v values. With these we calculate %conversion on each bed. Hope it helps. Thanks $\endgroup$ – Ed Corso Nov 25 '19 at 1:43
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    $\begingroup$ Actually, I could not find any reference on the reaction of sulfur trioxide with semihydrated copper sulfate. One speculation is that sulfur trioxide may react with the remaining water in the crystals. Then the question is why just semi hydrated copper sulfate? SO3 is also a good dessicant. At ambient temperature CuSO4 may not be a strong oxidant to oxidize sulfur dioxide. $\endgroup$ – M. Farooq Nov 25 '19 at 2:15
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    $\begingroup$ Yes, it may do that. If you pour conc. on hydrated CuSO4 i.e. CuSO4.5H2O crystals, it becomes white because the acid dehydrates it. $\endgroup$ – M. Farooq Nov 28 '19 at 17:51

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