Elements of
Chemical Reaction Engineering
6th Edition



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Chapter 6: Isothermal Reactor Design: Molar Flow Rates

Example 6.5: Acid Catalyzed Semibatch Reaction

A → B Acid Catalyzed

The irreversible liquid phase acid catalyzed isomerization reaction

is carried out isothermally in a semibatch reactor.

A 2 molar solution of H2SO4 is fed at a constant rate of 5 dm3/min to a reactor that initially contains no sulfuric acid. The initial volume of pure A solution in the reactor is 100 dm3. The concentration of pure A is 10 mol/dm3. The reaction is first order in A and first order in catalyst concentration and the specific reaction rate is 0.05 dm3/mol min–1. The catalyst, of course, is not consumed during the reaction.

 

a)  Write a complete POLYMATH program (line by line in POLYMATH notation) to determine both the number of moles of A and of H2SO4 in the reactor and the concentration of A and of H2SO4 as a function of time. Use the mole balance expressed in terms of NA.

b)  Obtain an analytical solution for the number of moles of A, NA and the concentration of A, CA, as a function of time. What is the concentration of A and of H2SO4 after 30 minutes?

 

     Additional Information

     

 


      Solution:

(a)    POLYMATH Solution

Mole Balance

        

        

Rate Law and Stoichiometry

        

Parameters

        

Polymath will combine

        

 

(b) Analytical Solution

AB

Mole Balance

                     (SS-1)

     (SS-2)

Rate Law

                  (SS-3)

Stoichiometry

                     (SS-4)

           (SS-5)

         (SS-6)

Combine

                               (SS-7)

                                                  (SS-8)

Dividing by v0 and recalling t0 = V0/v0

                                                  (SS-9)

Using the integral

                               (SS-10)

with

t = 0  NA = NA0

we obtain

                                        (SS-11)

                          (SS-12)

Note! If one does not divide by v0 in Equation (SS-8), then

                                          (SS-13)

Here b = v0, a = V0

The integral becomes

                             (SS-14)

Factoring out a v0 from the term in brackets

                                 (SS-15)

we see we get the same result as Equation (SS-11).

Evaluate

At t = 30 We substitute into Equation (SS-12)



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