Why is the activation energy determined using a flow reactor different from that estimated in a batch-type reactor? If reactants used (i.e., substrate, catalyst, and solvent) are the same, properties of chemical reaction itself must be the same.
Mixing is different in two systems; thus, the diffusion of reactant in the catalyst surface and pores not equal. And hence, the interaction will be different.
Your questionis confusing. The entire CSTR is at or near the outlet concentration at all (most?) times because it is well mixed. A plug flow reactor has higher concentration of reactants at the inlet and higher concentration of products at the oultet, Pleae share some actlual data so we can understand and explain it better.
Your questionis confusing. The entire CSTR is at or near the outlet concentration at all (most?) times because it is well mixed. A plug flow reactor has higher concentration of reactants at the inlet and higher concentration of products at the oultet, Pleae share some actlual data so we can understand and explain it better.
Rate constant depends only on temperature: k=koexp(-Ea/RT). Therefore, the value of Ea (activation energy) should be the same and independently of the type of reactor if the determinations of the reaction rates are carried out in the kinetic regime.
Geomar Arteaga yes I agree with you but i found some articles reported different Ea for the same reaction under similar reaction conditions but using different reactor configuration and I am looking for explanation for that.
the pre exponential of Arrhenius eqn depends on two factors
rate=A exp(-E/RT)
A=PZ, Where Z is a collision number and P a ‘steric factor’
in a reactor the molecules must collide in the proper orientation for a reaction to occur. in different reactors there is different mixing patterns and for example in CSTR molecules will move faster than PFR; as molecules move faster, they are more likely to collide and therefore affect the collision frequency. thus A are different in each type of reactor.
is that possible??
Yes, that i's possible. The pre-exponential factor (A or ko) can be different even using the same reactor but changing any of the reaction conditions (reactants concentrations, temperature range, total pressure or partial pressures for reactions in gas phase,...). But, the activation energy (Ea) should be the same if the reaction is carried out in the kinetic regime.
As you know, the linearization of the Arrhenius' equation give: ln k = ln ko - Ea/RT, and plotting ln k vs. 1/T you get a straight line with negative slope (= - Ea/R) and intercept = ln ko. Under kinetic regime, but using different conditions, you can get a family of straight lines with the same slope but different intercepts.
You can investigate in the great work (books, papers, etc.) about this topic made by Geoffrey Bond.
Geomar Arteaga many thanks for your help and nice recommendation.
Prof. Geoffrey my favorite author
Yes I read some of his work ( Metal-Catalysed Reactions of Hydrocarbons) and some of Prof. Professor Zoltan Paal.
Highly appreciated
It is necessary to distinguish between activation energy and apparent activation energy. Real activation energy is independent on the type of reactor. It can be evaluated if the experiment is performed at kinetic regime, without an influence of diffusion. In the regime controlled by diffusion, the apparent activation energy is usually evauated as having around half value of real activation energy.
Besides the already mentioned apparent activation energy which can be different from the real one by diffusion control, also the isothermal behaviour of the reactor has to be considered. For strongly exothermic reactions a hotspot formation in the batch reactor can corrupt the results. Flow reactors with small diameters and thus high surface-to-volume ratios can often be accurately controlled.
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