The formula is:
QE= 2 (number of evolved hydrogen molecules) / number of incident photons.
"Considering that I didn't use a sacrificial agent what mistakes Can I have made?"
Water splitting forms both H2 and O2 with the ratio 2:1. You must see both gases
Thank you for your useful response.
To calculate the H2 rate I oblviously considered that the Vtot= VH2+ VO2 and that VO2= 1/2 VH2. So VH2= VTOT* 2/3 considering the density of H2= 0.0899 gr/l I have the moles of H2....but obtaneid mmol
Maybe there are other parametres that I haven't consider.....
First, you must to prove that you have both O2 and H2. Second, the volumetric methods is extremely sensitive to the temperature. Third, the amount of the released gas is = {deltaV(ml)/22.7} = mmole. Fourth, QE= 2 (number of evolved hydrogen molecules) / number of ABSORBED photons
the number of incident photons can be measured from the power of light hitting your solution
the hydrogen production yield would be better to give in mole/s with specifying the reaction volume. The number calculated as ({H2 in mole/s)/(moles of catalyst)} is defined as an apparent TurnOver Frequency,TOF.
Thank yoy prof. Geletii for your clear and useful responses, maybe the heating of lamp allowed to have an higher rate of H2 products.
I will try to perform the experiments with a thermostated system.
which instrument via it get data from quantum efficiency for photocatalytic water splitting
about ammonia synthesis, in the literarures QE= 6 (number of evolved ammonia molecules) / number of ABSORBED photons, why 6 ? not 3