The real calculation of Turn Over Number (TON) differs from one catalytic system to another. For metallic system, you need to define which is/are the active site(s) and if bimetallic, you will need to count the number of synergistic-linked metallic species. However, further information could be provided if you mention the reaction equation and the catalyst components.
@Sudeesh N: The two equations you give for TON have different units. I am that certain you mean the first one, which is unitless, not the second one that has units of mols/number or mols/area.
I would express TOF as (# molecules reacted)/(# active sites)(unit time). For heterogeneous catalytic reactions carried out in bulk reactors, you will work with volumetric reaction rate r (mols/m^3 reactor s), particle loading fV (m^3 particles/m^3 reactor), particle radius Rp (m), and number of active sites per catalyst particle Ns (# sites / particle) as
--> TOF = (r No / fV) (C Rp^3 / Ns)
where No is the Avagadro number and C is a conversion factor for particle radius to particle volume (nominally 4 Pi / 3 for spherical particles). You will have to define Ns either experimentally or theoretically.
For general information, these references may be useful (found by a search engine on "turnover number calculation" or "turnover frequency calculation").
Yes, TON is a unitless quantity. TOF = Moles of desired product formed/moles of catalyst/ time = TON/time. it just give the number of molecules reacted per active sites of catalyst. When you take TON at different time the quantity will be different. You may not be able to compare two sets of catalysts based on this. Turn over rate or TOF is the best one for catalytic studies. Also take care when you use moles of consumption of reactant for these calculations instead of desired product. It may mislead because the side reactions may also counted during these calculations.
The TOF is dened as the moles of reduced 4-NP molecules per mole of the surface noble metal atoms per hour. WhenM¼ Au, Pd and Au–Pd, the
TOF is calculated based on the dispersity of noble metal nanoparticles
Turnover Number (TON) -- the absolute number of passes through the catalytic cycle before the catalyst becomes deactivated.
Academic chemists sometimes report only the turnover number when the catalyst is very slow (they don’t want to be embarassed by reporting a very low TOF), or decomposes quite rapidly.
Industrial chemists are interested in both TON and TOF. A large TON (e.g., 10-6 - 10-10) indicates a stable, very long-lived catalyst. TON is defined as the amount of reactant (moles) divided by the amount of catalyst (moles) times the % yield of product.
I am post my question here as it is related to the same concept. 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
Hai, as all said, TOF can be calculated using TON divided by time. Also, you can calculate it from the first order rate equation for a catalytic reaction if you don't have the parameters to calculate TON. But it is always easy to calculate it using TON.
Please find the following link of pdf for better understanding and how to calculate the TOF and TON, and how those two are related, If we know TOF , then how to calculate TON
For homogeneous catalysis, TON is the "% yield divided by the catalyst loading (which is the mole% of the catalyst based on the moles of the limiting reactant)."
Turnover number 1 taking the of moles of product produced, dividing that by the of moles of catalyst used in the reaction, then dividing that by the time to produce the given amount of product
Turnover number 2 amount of reactant (moles) divided by the amount of Catalyst (moles) time the% yield of product
I want to explain why it was written in the first mall of the reactant and other written mole of the product
1. It depends on the nature of the reaction. if we have two or more reactant, we can consider the limiting reactant moles for calcuation or all the reactants moles are similar we can consider the moles of the desired product formation.
For TOF = TON/time, time of the reaction taking place is the main criteria, since some reaction can take long time and some ends short. hence better to choose an optimum shorter time between 30 to 50 time intervals will give better results for TOF
Moreover, active sites are also more important in aspect of catalyst, If we are choosing particular type of active sites such B4 or B5 sites then the edge atoms in the catalyst are more prominent than surface atoms.
hence we have to thing all these aspects for calculating TON and TOF based on my opinion.