Can anyone tell me the reason of lower and higher turnover frequency values? In my case, I found that Standard P25 have higher photocatalytic activity than my sample (also TiO2 in the form of nanofibers). but my fibers have higher turnover frequency than P25 found by the formula TOF(h-1)= MMO.X/MTiO2.D.t,
Where MMO and MTiO2 is the amount of methyl orange dye and TiO2 nanofiber catalyst used (in moles), χ is the conversion of MO, t is the reaction time and D is the dispersion of TiO2 as a fraction of surface-active TiO2 catalyst, determined by the following equation,
D= Surface area/ [1/MTiO2.(NA.4pirTi2)],
Where SA and MTiO2 is the surface area and molecular weight of TiO2 (79.86 g/mol.) respectively. rTi is the atomic radius of Ti (0.176 nm) and NA is Avogadro number (6.02 × 1023).
What could be the possible reason of higher TOF in my case? can anyone please explain. Thanks
Where does you formula of TOF come from. Standard definition is TOF=1/Ntotal*dY/dt with Y being the Yield (or conversion for Selectivity =1), Ntotal beeing the total number of active sights in your catalyst. In case you know that a catalyst grain has for example d=50 nm (diameter) and contains Nactive=10 active sights Ntotal=mused/(Density*pi/6*d^3)*Nactive. Usually Nactive needs to be determined either by calculations or experimentally. Actually in photocatalysis it is very hard to define the amount of active sights Nactive as you'll find in literature surface defects like oxygen vacancies (bulk and surface defects) as well as the heterojunction between anatase and rutile are known active sights for excitone creation and separation. So you may want to rethink your way of calculating the TOF. I'd recommend determining the amount of oxygen vacancies by reactive frontal chromatography using N2O (if available) or for qualitative comparison of both products the measurement of the photoluminescense spectra (and integrate the PL signal + compare it).
However in case the formula is right some other issues may be the cause for your observation:
What was the yield or consversion you accieved during your reaction time?
To compare two catalysts according thoir TOF the conversion should be small as the TOF includes knowing dY/dt (or dX/dt for selectivity=1). So your conversion should be less then 10% - I guess 30-50% is still okay but the error will increase as dY/dt is a function of conversion. Ideally you want to know the slop dY/dt for t-->0.
Speaking in figures: In case you let your reaction run till for say 1h and with pure P25 being more active - your P25 may already be at a conversion of 93% after 10 min reaching e.g. 97% after 1h. On the other hand your prpared less active catalyst may have only 20% after 10 min and 95% after 1h as it is less active. however after 1h both catalysts have nearly the same conversion as you chose the reaction time to long. If your prepared catalyst e.g. has a smaller surface area / less active sights it will look as if the less active sights are more active as both P25 and your catalyst reached about 95% conversion during your reaction time. Actually this is only a misinterpretation due to a reaction time chosen to long. So in case you want to compare to catalysts according TOF you not only need to know their Nactive but also make sure to choose the reaction time such that the conversion remains less then 10%.
Hopefully this helps - Good luck.
In photocatalytic reactions the rate is commonly controlled by light intensity making measurements of TOF very complex. First of all you should confirm that the rate is independent on light intensity, stirring rate, etc.
Thanks for your kind response and help Prof. Sven Reichenberger and Prof. Yurii V Geletii
@Sven Reichenberger , I followed the following publication for TOF formula. http://www.rsc.org/suppdata/cc/c1/c1cc12264a/c1cc12264a.pdf
Ah I see, well first of all its very kind of you to call me Prof. but I am only a PhD just about to finish his thesis ;)
Secondly from the publication you send - a little comment:
be aware that they used the EAS in this publication which is the electrochemically active surfacearea. So this is NOT the surface area one determines from N2 adsorption experiments (e.g. BET). Not sure which one you used in your calculation as you only state"surface area" in your first comment which I would guess was determined by BET?
Additionally be aware that the calculation of the dispersion D usind the EAS should best only be used in case of electrochemical reactions e.g. electrocatalytic water splitting or in case of e.g. fuel cell research. I guess its OK to use the EAS for photocatalysis but although I am not an expert on electrochemistry I'd recommend to review your results carefully as the active sights for photocatalysis and cyclovotametric may differ depending on the material and reaction.
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