Hello, With my colleagues we are working on a project. We are currently designing a reactor, for the reaction isopropanol -> acetone + H2
For that reaction a catalyst is needed, and we do not know how to calculate the amount of catalyst necessary for the reaction. One of our ideas was to use the entering flow of isopropanol (7,02 mol/s) or the amount od isopropanol (256,2 mol) we can fit in the reactor (at 3 bas and 350 degrees) and substitute it in the mathematical equation of the chemical kinetics (view image).
However we are not sure if this is how we should do it.
Yes of course. Assuming an ideal mix continuous reactor your reactant balance is,
VR(dCA1/dt)= Q0CA0 – (viso)eVR –Q1CA1 (1)
Where VR is reactor volume CA is reactant (isopropanol) concentration Q0 is feed flow CA0 is reactant concentration in feed (Q0CA0=7.02 mol/s) Q1 is output flow and CA1 is reactant concentration at output (conversion is 1-CA1/CA0) viso is your reaction rate and “e” is the catalysis volume fraction (m3 catalysis/m3 reactor). With Eq. (1) you can calculate “e” in order that the reaction in steady state (dCA1/dt=0) in a given reactor volume reach the desired conversion. You can adapt Eq. (1) for a continuous tubular reactor.
First, type of reactor - process conditions indicate fixed-bed catalytic reactor. Second, stoichiometry - the change in number of moles should be included. Third, mass balance - at the beginning, one can assume the plug flow. Regards,
You have reason Prof. Grzesik. Eq.(1) suggested in my response can describe this situation with Q0 and Q1 equal to zero. I assume that Zwolan included stoichiometry in his kinetics, otherwise he must to include in v_iso.
Fixed bed reactor catalyst volume is not set by kinetic calculations it is set by catalyst deactivation (aka activity maintenance) and run length requirements.
The catalyst volume for the given initial molar flow rate of the substrate and the known form of the kinetic equation with parameters can only be determined from a correctly formulated mass balance. It is worth remembering that the reaction is to take place in the gas phase (To = 350 ?) and at increased pressure. Regards,
From:
https://www.icis.com/explore/resources/news/2007/11/01/9074860/acetone-production-and-manufacturing-process/
Early processes for the manufacture of acetone were based on the thermal decomposition of calcium acetate or the carbohydrate fermentation of corn starch or molasses. The ready availability of propylene in the 1960s led to routes based on the dehydrogenation of isopropyl alcohol or cumene peroxidation.
The cumene route, in which the acetone is coproduced with phenol, is the preferred technology because of its lower costs and nearly 90% of acetone is produced this way. The main process for manufacturing cumene involves the reaction of propylene and benzene in the presence of phosphoric acid-based catalysts or, more recently, zeolite catalysts. The cumene is oxidised in the liquid phase to cumene hydroperoxide which is then cleaved in the presence of sulphuric acid to phenol and acetone. About 0.62 tonnes of acetone is produced with each tonne of phenol.
The isopropyl alcohol route, where the alcohol is dehydrogenated to acetone over a metal, metal oxide or salt catalyst, has been declining. A newer process that has been gaining importance in Japan is the direct oxidation of propylene but it suffers from high capital costs and corrosion problems. Small quantities of acetone are also produced as a byproduct in the manufacture of propylene oxide (Lyondell), acetic acid and hydroquinone (Eastman Chemical and Goodyear).
Shell Chemical has developed a phenol process which coproduces both acetone and methyl ethyl ketone (MEK). According to consultants Nexant ChemSystems, it involves the co-oxidation of cumene and sec-butylbenzene, which is made via alkylation of benzene with n-butenes. The process has the potential to change the acetone/MEK ratio within reasonable limits to meet varying market demands. Although the cost of making the phenol is higher than the conventional cumene route due to higher raw material costs, the large byproduct credit received for the MEK more than makes up for these costs, says Nexant.
Rick, you are fully right. But the questioner wants to learn to solve this type of student task. One can give him a solution. But it's better that he thinks himself based on some suggestions. Regards
Miroslaw, Thank you for pointing out that I lost track of the question.
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