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How do I calculate TON and TOF?

Thank you sir Yurii V Geletii for response.

Infact I am new to this question and at the movement working on chemical catalysis and you are 100 percent correct that this question must have been discussed numerous times on RG. Moreover, I request you sir if you can provide me any reference or link regarding to my query.

Regards.

" TOF is number moles of catalyst required to reduce the specie under investigation per unit time. " WRONG.

The oversimplified definition: TOF = TON/(reaction time). TON is the amount of formed product(s)/amount of the catalyst. "Amount" might be in mole, or M.

"I have catalyst amount in moles = 3.93593E-05 and catalyzed specie moles are = 1.44231E-06 and rate constant 0.054 per minute." You have more catalyst than a reactant. For a catalytic process TON must be >> 1.

Thank you very much sir for clarification. I better go for your first suggestion then may be I will be able to solve my problem.

Best regards.

the TOF is the total number of moles transformed into the desired product by one mole of active site per time (hour, sec...). The larger the TOF, the more active the catalyst.

TOF = TON/time

For TON:

Turn over number = Moles of desired product formed/moles of catalyst (for a homogeneous catalyst)

TON = Moles of desired product formed/number of active centers or surface area (heterogeneous catalyst)

For a single reaction which has more than one product, the TOF may vary with respect to each product.

If you are using a heterogeneous catalyst the amount of active catalyst is not certainly equal to the amount of catalyst you added into the system, which means not 100% catalyst becomes true catalytic species during the reaction. You have to report how you measure or make a sensible assumption of the percentage of the active catalytic sites.

Thank you dear Muhammad Ibrahim for Knowledge sharing...

Thanks Dr Peng Li for clarification. your answer has solved my problem but How may I calculate active sites for my catalyst? as my catalyst is in composite form containing hard to remove silica.

Best Regards

Usually, we make an ICP test to measure the catalytic metal amount. You can assume a percentage base on that result.

TOF is: number of mole of formed product (np) over number of mole of the Catalysts (ncat) by the time

TOF= np/ncat.time ; the unite will be 1/time unite

number of mole of formed product= numbre of mole of transformed limiting reagent

TOF was calculated as [mmol of product] / [mmol of catalyst] ×(Reaction time in hours

Maybe I am wrong or things have changed but TOF is the amount of reactant converted/amount of catalyst/time into products (it does no matter which one if you have plenty of products) TOF is a measure of activity not of the selectivity that is different. furthermore the amount of catalysts should be less than 10% (in my opinion even less) the amount of reactant to be considered as catalytic. Depending if the catalysis is homogeneous, heterogeneous the TOF calculation may vary a bit since in homogeneous you have every single site involved in the catalytic reaction and in heterogeneous you have some sites that are not accessible by the reactants.

I have calculated milli moles of product that comes out to be 1.34615E-09 and mili moles of catalyst and found 1.07344E-14 so TON as per Dr Yurii V Geletii answer, evaluated as 125405.98 is this result usually comes or it signifies better performance of catalyst?

Please answer.

First, a quick comment. The number of significant digits in your data should correspond to experimental error.

If your experimental data are correct, then you have TON=1.3e+5, which is an extremely high value. I doubt that your numbers are correct. You claim that you have 1.3e-9 mmole of products. If your product has a molecular weight in the range 100-1000, then you obtained in your reaction less than 1e-6 mg of the product. How did you quantify so miniscule amount?

If your product has a molecular weight in the range 100-1000 ( I do not have specie having Mwt in the range 100-1000 instead I am dealing with inorganic specie which has mas around 52.)

How did you quantify so miniscule amount? ( I am studying chemical catalysis of Inorganic specie dealing with milligram quantities of composite catalyst from where I back calculate the actual product keeping in view all quantities of reactant and product formed.

Moreover, sir please find attached the research article which deals with such type of calculations like TOF etc, please go through if you have time and interest then might be you may help me in better way.

Best Regards.

Something is wrong in your calculations.

You wrote " dealing with milligram quantities of composite catalyst ". At the same time " I have calculated milli moles of product that comes out to be 1.34615E-09," which corresponds to 52*(1.3e-9)~ (8e-8) mg of products. Plz, provide raw experimental data.

Zulfiqar download the article below

Thanks dear Ismail your work is highly useful and I got correct answer of my question by your accepted manuscript and congratulations on acceptance of this new research article. Really this is very interesting article in the field of reduction of hazardous pollutant to environmental benign species.

Best Regards.

You are welcome Zulfiqar Ali

Wish you gudluck

Muhammad,

I read your paper. What is the TurnOver Number (TON) in your system?

TON = (mmol of NP reduced)/(mmol of catalyst). TON must be >> 1 for a catalytic process and TOF = TON/time.

Yurii V Geletii Sir I have calculated direct TOF

TOF = (mmol of NP reduced)/(mmol of catalyst) * time

In future I will mentioned TON also

Thanks

You write

TOF = (mmol of NP reduced)/(mmol of catalyst) * time

which is the same as

TOF = TON/time

You did not describe how you determined (mmol of catalyst).

My estimations:

You have 3 mL of 1 mM 4 NP, which is 3 micromole. Your Cu-Ag/GP catalyst has about 20% of Cu. Then, 10 mg of the catalyst has 2 mg of Cu. Atomic weight of Cu is 68. Thus, you have in your system 2mg/68 = 30 micromole of Cu (which is assumed to be a part of catalyst). TON < 3/30 = 0.1. Your system is NOT catalytic while the concept of TOF is applicable exclusively for catalytic systems.

The basic problems on RG is to give too exact numbers (three significant figures are usually enough) and no ability to convert units. The easiest is sometimes the most difficult. Regards,

figured out my problem!

simple TOF is: moles of catalyzing species per mole of catalyst per minute.

so TOF in my case is: moles of Cr(VI) ions reduced per mole of Ni-MWCNTs per minute.

TOF is : (mmol of sub. * conversion)/ (mmol of cat. * time of reaction)

Dr abolfazal please comment on my understood definition of TOF (mentioned above) whether it is correct or not?

Why not just use: conversion, yied as well as in case of complex reactions selectivity. Regards,