It is a kinetics homework from chemical engineering: We can obtain a 50.% conver

Question

It is a kinetics homework from chemical engineering:

We can obtain a 50.% conversion for a specific irreversible reaction which obeys first-order kinetics in a batch reactor at 50.C with a reaction time of 20. minutes. However, for a commerical process we will need 95% conversion in a CSTR. 1. What must the residence time for a CSTR be? 2. We decide that the residence time in the CSTR must be no more than 30. minutes to develop an economical process. If the activation energy for this reaction is 15 kcal/mole, what reactor temperature is necessary? 3. It turns out the reaction isn’t completely irreversible. The equilibrium conversion at 50C is found to be 99.5%. Will reversibility be a problem at the operating temperature found in the above problem if the rxn = 20. kcal/mol? What if rxn = 20. kcal/mol?

Explanation / Answer

Assume the elementary first-order irreversible reaction A+B=>C

Rate law: -rA=kCACB

Mole Balance: VCSTR=FA0X/-rA

Stoichiometry: Liquid phase (v=v0)

CA=FA/v=FA/v0=FA(1-X)/v0=CA0(1-X)

CA=CA0(1-X)

CB=CA0(1-X)

Combine: VCSTR=FA0X/kCA02(1-X)2

Relevant equation: k=k0exp[(E/R)(1/300-1/T)]

Now unless we know the value of FA0 & k we cannot procced further.

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