Question

The following first-order gas-phase reaction occurs in an isothermal, isobaric PFR reactor: A->2B. There is 60%...

The following first-order gas-phase reaction occurs in an isothermal, isobaric PFR reactor: A->2B. There is 60% A and 40% inert in the initial feed. If the volumetric flow rate at the outlet of the reactor is 1.3 times of the one at entrance, calculate the conversion rate of the reactor. Assume ideal gas law (PQ = NRT)

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Answer #1

Given,

  • First order gas phase reaction
  • Isothermal, Isobaric reactor
  • A 2B
  • Initial Conditions: 60% A and 40% Inert
  • Volumetric flow rate at the exit (Vexit) of the reactor is 1.3 times of the one at the entrance (Vo)

Assumption:

Ideal gas law  PQ=NRT

Change in the volumetric flow rate is due to the difference between the number of moles of reactants and products. By ideal gas law Volumetric flow rate also increases with increase in number of moles. To account for it expansion factor is introduced as,

V100 = Volume at 100 % conversion.

V0 = Volume at 0% conversion.

Nf = Final number of moles

Ni = Initial number of moles

Volume(V) in batch reactor is replaced by Volumetric flow rate(Q) in the case of plug flow reactor.

Initially at 0 % conversion for one mole of feed there is 0.6 mole A and 0.4 mole Inert (I) Ni= 0.6+0.4=1

At 100 % conversion there will be 1.2 mole of B and 0.4 mole inert (I) Nf = 1.2 + 0.4 = 1.6.

Volumetric flow rate at a conversion of XA is given by,

Conversion is 50%.

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