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mono-olefins to to N-diolefins and N-triolefins

mono-olefins to to N-diolefins and N-triolefins

mono-olefins to to N-diolefins and N-triolefins

The main reaction is the dehydrogenation of n-paraffins to form n-monoolefins which are the desired product (platinum catalyst).
it is written (To produce an environment favorable to olefin formation, high temperatures and low hydrogen partial pressures are required. Catalyst stability considerations limit reactor temperature levels, and safety consideration prevents designs below atmospheric pressure.As a result, the process must be designed for low conversion and, hence, high space velocity operation.)
Can anyone explain this passage and how process conditions (high temperature, low pressure) promote further dehydrogenation of n-mono-olefins to n-diolefins and n-diolefins?
Best regards

RE: mono-olefins to to N-diolefins and N-triolefins

Parafin + heat ---> mono-olefin + H2
mono-olefin + heat ---> bi-olefin + H2
bi-olefin + heat ---> tri-olefin + H2

Review Le Chatelier's principle:
Heat is needed to proceed - high temperature favors RXN.
More molecules are created as it proceeds - low pressure favors RXN.

Good Luck,

RE: mono-olefins to to N-diolefins and N-triolefins

Quote (BZLAM21)

safety consideration prevents designs below atmospheric pressure
Incorrect. Dehydrogenation reactions take place at vacuum conditions, e.g. butane to butadiene, propane to propylene, isobutane to isobutylene. Look for a description of CATADIENE or CATOFIN processes by Lummus/Clariant.

RE: mono-olefins to to N-diolefins and N-triolefins

This reaction for ethane conversion to ethylene must be endo thermic, so high temp would favour the forward reaction. Moreover, high temp very often increases reaction rate, even in exothermic reactions, so you get higher conversion per pass in either case.
But high temp can also cause thermal cracking and the risk of catalyst and reactor tube plugging, so its a good idea to keep hydrogen partial pressure high in the reactor to gasify any carbon that may form from this perspective. But agreed, high H2 pressure decreases the net equilibrium conversion, so you have to balance out these contradicting objectives.

RE: mono-olefins to to N-diolefins and N-triolefins

yes i know it is an endothermic reaction and increasing the temperature shifts the equilibrium and increases the conversion rate.and the reaction entropy is positive which requirespressure reduction.reducing the pressure of the system by reducing the partial pressure of hydrogen is the technique used.
how to reduce the hydrogen pressure?
for information paraffins and hydrogen are mixed in a packinox exchanger before sending to reactor and that reactor contains 2 compartments in series and a separator the first is for the production of n-olefin and the 2 is for the conversion of di-olefins into olefins.
Best regards.

RE: mono-olefins to to N-diolefins and N-triolefins

Reducing H2 partial pressure can be done by decreasing recycle hydrogen to the mixed feed or decreasing reactor pressure. In reality, decreasing reactor pressure would increase space velocity, reduce residence time in the reactor and reduce net conversion. This is because reactors are often built too small, and exit products are still some way from reaching theoretical equilibrium.

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