TOF, turnover frequency, and TON, turnover number, are two important characteristics of a catalyst. — Physical Chemistry Chemistry Question
TOF and TON
TOF, turnover frequency, and TON, turnover number, are two important characteristics of a catalyst. According to the definitions given by the International Union of Pure and Applied Chemistry (IUPAC), TOF is the maximum number of molecules of a reagent that a catalyst can convert to a product per catalytic site per unit of time. TON is the number of moles (or molecules) of a reagent that a mole of catalyst (or a catalytic site) can convert before becoming inactivated. TON characterizes the stability (life time) of a catalyst, while TOF is a measure of its best efficiency. Very important is the word “maximum” in the definition of TOF!
In Russian, TOF and TON sound like names of two clowns.
TON is a dimensionless value. What is the dimension of TOF? Derive a relation between TON and TOF.
Model Answer
The TOF unit is {time–1}. SI unit for TOF is {s–1}.
TOF relates to TON by the equation:
TOF × t = TON
where t is the time till the moment of inactivation of a catalyst. This formula gives the upper limit for TON. It assumes that the catalyst works with its best efficiency (TOF) all the time and becomes inactivated suddenly, in a moment. It is more realistic to assume that TOF goes down gradually. Then the following relation is valid:
TOF × t ≥ TON (or TON ≤ TOF × t).
Let a catalytic reaction А + Cat → B proceed in a closed system. А and В are gases, Cat is a solid catalyst.
а) The dependence of the amount of B produced at 1 cm2 of a catalytic surface upon time is given in Fig. 1a. There are 1015 catalytic sites in 1 cm2 of the surface. Estimate TOF.
[VISUAL]
Model Answer
TOF is the maximum value of ΔNB / (Δt × 10^15).
Maximum of ΔNB/Δt corresponds to the initial linear part of the curve in Fig. 1a and is equal to:
ΔNB / Δt = tg α = (7 × 10^-8 mol) / (2 s × cm^2) = 3.5 × 10^-8 mol / (cm^2 s) = 2.1 × 10^16 molecules / (cm^2 s) = 21 × 10^15 molecules / (cm^2 s)
Dividing by the number of catalytic sites per 1 cm^2 (10^15):
TOF = (21 × 10^15 molecules / (cm^2 s)) / 10^15 sites/cm^2 = 21 s^-1.
b) The dependences of the amount of B formed in 1 cm2 of the catalytic surface upon time are given in Fig. 1b. Different curves correspond to different initial pressures of the reagent A. These pressures (in arbitrary units) are shown by red numbers. There are 1015 catalytic sites in 1 cm2 of the surface. Calculate TOF for the catalyst. This catalyst worked during 40 minutes and then became inactivated. Estimate TON.
[VISUAL]
Model Answer
With the increase of the initial pressures of the reagent A, the initial slope ΔNB/Δt increases. However, for curves (10) and (11) the initial slopes are identical, indicating that the maximum efficiency of the catalyst is achieved (saturation/pressure independent).
Using the initial slopes of the curves (10) and (11):
ΔNB / (Δt × 10^15) = 210 s^-1
Thus, TOF = 210 s^-1.
To estimate the TON for t = 40 minutes (40 × 60 = 2400 seconds) of catalytic activity:
TON ≤ TOF × t = 210 × 40 × 60 = 5 × 10^5.
9.3 a) TOF is often used to describe the operation of deposited catalysts. To make a deposited catalyst one has to deposit atoms of metal on the inert surface. These atoms form catalytic sites. The dependence of the rate of the catalytic reaction upon the amount of metal atoms deposited on 1 cm2 of the surface (less than one monolayer) is shown in Fig. 2а. Calculate TOF.
[VISUAL]
Model Answer
The slope of the linear dependence in Fig. 2a is used to calculate TOF:
TOF = tg α = 6 s^-1
It is assumed that every single atom of the metal forms a catalytic site and works independently, making TOF independent of the amount of atoms deposited.
b) Russian scientist professor Nikolay I. Kobozev has shown that the dependence of NВ on NCat can be much more complicated. The corresponding curve in Fig. 2b has maximum! According to the Kobozev’s theory (a simplified version) a structure consisted of n deposited atoms rather than a single atom form a catalytic site. Maximum rate of catalytic reaction was observed when
(number of deposited atoms per surface unit) / n = (number of catalytic sites per surface unit)
From the data shown in Fig. 2b calculate n, the number of atoms forming a catalytic site. TOF for the point of maximum rate in Fig. 2b is given in SI units.
[VISUAL]
Model Answer
The number of catalytic sites at the maximum rate (NB = 18 × 10^-11 mol / (s cm^2)) with TOF = 35 s^-1 is:
N_sites = NB / TOF = (18 × 6.02 × 10^23 × 10^-12 molecules/s/cm^2) / 35 s^-1 = 3.1 × 10^12 sites/cm^2
The number of deposited atoms at this maximum is NCat = 7 × 10^12 atoms/cm^2.
Thus, the number of atoms n forming one catalytic site is:
n = NCat / N_sites = (7 × 10^12) / (3.1 × 10^12) = 2.25 ≈ 2.
Calculate the ratio of TOF for the atoms of Au in the upper layer in Fig. 3а (all red spherical particles), to TOF for the monolayer in Fig. 3b (all yellow spherical particles). In the former case, every single Au atom is a catalytic site. The rate of the catalytic reaction on each yellow site in Fig. 3a and Fig. 3b is the same if the site is accessible to reactants and is equal to zero if the access is blocked.
[VISUAL]
Model Answer
In monolayer structure (b), all yellow spheres participate. Let NAu be the number of yellow spheres. The rate of reaction is r2.
In bilayer structure (a), the top layer contains red spheres equal in number to 1/3 NAu. These block 2/3 of the yellow spheres beneath them, so only 1/3 NAu yellow spheres participate in case (a) along with the 1/3 NAu red spheres.
The rate of the reaction in bilayer (a) is r1 = 4 r2.
Summing the partial rates:
r1 = r2(red) + 1/3 r2 = 4 r2 ⇒ r2(red) = 11/3 r2
Thus, the TOF ratio of the atoms in the upper layer to the monolayer is:
TOF(upper red) : TOF(monolayer yellow) = (r2(red) / (1/3 NAu)) : (r2 / NAu) = (11/3 r2 / (1/3 NAu)) : (r2 / NAu) = 11 : 1.