[ooiuniscience]

Dear all and respective expert,

Good days to everyone! Hope everyone are doing great!

I'm conducting watersplitting reaction to generate hydrogen and oxygen. The H2 and O2 gas generated after 1 hour reaction will be cummulated at headspace (volume headspace = 10ml; initial pressure = 1atm) and the H2 and O2 gas will be transferred into online-GC by using peristaltic pump for quantification.

From GC measurement, the area of H2 to O2 gas is 10:1. The pressure after 1 hour is measured to be 1.5 atm.

Can I determine the “TOTAL AMOUNT OF H2 GAS GENERATED IN 10ml HEADSPACE” in such a way:

TOTAL volume OF H2 and O2 GAS GENERATED IN 10ml HEADSPACE, V2 = P1(V1) / P2
                                                                                               = 1 (10) / 1.5
                                                                                               = 6.67 ml

Since area measured from GC is:
total H2: Total O2 = 10: 1, hence, O2 gas is 0.606 ml
                                               H2 gas is 6.67 ml - 0.606ml = 6.06 ml

P/s corrects me if I’m no correct.

Hopefully anyone of you can shed a light on this base on your expertise.

Jeff

[Arkcon]

I've seen this post all over, and I'm sorry, but I don't understand what you're asking. I was hoping someone would have an idea, or maybe you could ask this question a different way, but all we're getting is reposts, so I'll try to break it down.

Dear all and respective expert,

Good days to everyone! Hope everyone are doing great!

Uh ... hi, I guess?

I'm conducting watersplitting reaction to generate hydrogen and oxygen.

This is perfectly clear and easy to understand.

The H2 and O2 gas generated after 1 hour reaction will be cummulated at headspace (volume headspace = 10ml;

This is not an area of expertise for me, I haven't done GC.  But I think I can follow this much, but ...

initial pressure = 1atm)

OK, this seems logical for a sample of gas, in an flexible container, at the surface of the Earth.  But ... umm ... do you really have that sort of container?  And do you have a barometer handy, so you can record the exact pressure, at your location, at the time?  Still out of my expertise area, but this is general info you need whenever you actually measure pressure of gas and expect to make a calculation (As distinct from all the textbook questions we usually get.)

and the H2 and O2 gas will be transferred into online-GC by using peristaltic pump for quantification.

Wow.  That works?  The peristaltic pump moves a gas?  I suppose you're sure the tubing is non-permeable but ... how do you prime the peristaltic pump?  How do the metal rollers "push" a bubble of gas?  I mean, does that really work?  I only use peristaltic pumps for liquids.  I'm still out of my depth here.  Maybe someone else will help with just this part, even just for my information.  Shouldn't you use a diaphragm pump for moving gases, with gas valves, and the like?  Or an airtight syringe, or a sampling bag?  Somebody either agree or correct me, here.  I'm also worried about pulsations, it seems like, how much is actually injected, depend on what the pump is doing at that moment.

From GC measurement, the area of H2 to O2 gas is 10:1. The pressure after 1 hour is measured to be 1.5 atm.

OK, this is better.  A measurement we can use, buried in stuff we couldn't.  I usually force the kids to figure this next question out for themselves, but to help you since you're more advanced -- How did you compensate for the partial pressure of water vapor, which is part of your collected gas, at the temperature and pressure of the headspace?  You can't use the ambient temp, and not the temp of the electrolysis chamber, because you've pumped it.  Did you get a GC peak for water vapor?  That would make compensation for the water vapor easier.  But the bottom line is, the 1.5 atm is not just O₂ and H₂.  And ... um ... shouldn't you be a bit more accurate in the measurement of pressure, since you seem to need to use pressure to determine volume?

Can I determine the “TOTAL AMOUNT OF H2 GAS GENERATED IN 10ml HEADSPACE” in such a way:

TOTAL volume OF H2 and O2 GAS GENERATED IN 10ml HEADSPACE, V2 = P1(V1) / P2
                                                                                               = 1 (10) / 1.5
                                                                                               = 6.67 ml

Well, by significant figures ...no, that's not the correct answer.  Assuming 1.0 atm, and 1.5 atm and 10.0 ml, your answer should be 6.7 ml.  But I suspect its actually 1 atm, so the answer is more like 7 ml.  That is as accurate precise as you can be, given the accuracy precision of your measurements.  Or maybe not even, given you're using a pump and a flexible vessel.

Since area measured from GC is:
total H2: Total O2 = 10: 1, hence, O2 gas is 0.606 ml
                                               H2 gas is 6.67 ml - 0.606ml = 6.06 ml

P/s corrects me if I’m no correct.

Hopefully anyone of you can shed a light on this base on your expertise.

I think we can work on this, after we've worked on the above.

*[EDIT]*  Dang.  Broke my own rules about precision and accuracy.  Embarrassing.

[Stepan]

Ratio of the H2/O2 peak area  is not equal to ratio of their concentrations. You need to calibrate your detector first using mixtures of known concentration.

What is the initial gas in the Headspace? I hope it is not air.

Why do you need GC testing to determine H2/O2 ratio? shouldn't it be 2/1?

If your yield is about 90% or better, just measure extra gas volume accumulated in headspace. 

[ooiuniscience]

Ratio of the H2/O2 peak area  is not equal to ratio of their concentrations. You need to calibrate your detector first using mixtures of known concentration.

What is the initial gas in the Headspace? I hope it is not air.

Why do you need GC testing to determine H2/O2 ratio? shouldn't it be 2/1?

If your yield is about 90% or better, just measure extra gas volume accumulated in headspace.

Hi Stepan,

Thanks for your feedback.

But photosplitting of water hardly produce ratio H2/O2 = 2/1 as told by other researchers bc O2 that tends dissolve in reactant.

Initial headspace gas is nitrogen only.

I need GC to know the ratio of H2 to O2 base on the area under the graph, then only i can quantify the total amount of H2 generated.

Does the way i used Boyle's Law correct or not bc i'm doubtful Boyle's Law only valid for a single gas and also varying the container volume?

Jeff

[Borek]

But photosplitting of water hardly produce ratio H2/O2 = 2/1 as told by other researchers bc O2 that tends dissolve in reactant.

Just because it dissolves doesn't mean it wasn't produced. Perhaps you need to specify what the real question is, as when you split water can't produce hydrogen and oxygen in different molar ratio than 2:1. That's what we know since around 1800.

[Stepan]

I would calculate it differently.
I would prepare H2, O2 N2 mixtures with known H2 and O2 concentrations by volume. Then I would inject a standard volume of the calibrated gas into GC TCD (let say 0.05-0.1 mL. and find the calibration coefficients for H2 and O2 which allows you to convert areas  into concentrations. Then I would measured initial and final head-space volumes. Finally I would analyse head-space using the same injection volume  I used for GC calibration. After I would calculate H2 and O2 concentrations using instrument calibration,and finally you can calculate H2 and O2 volumes multiplying the final headspace volute times concentrations (exprecced as fractions or relative partial pressure). Volume of H2 + O2 should match increase in headspace air volume. If they do not - you have either losses, or secondary reactions